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Number System

Let’s say you have some apples and if somebody asks how many apples you have, what will you answer? One got to put a "label" on them. And here comes "numbers" to help you out. We need numbers to define the quantity and to perform mathematical operations like addition and subtraction. Numbers can classify in various depending on what they represent and where they are used. Here they are:

Natural Number

And how will counting start if you are counting your apples? You have to start with 1 and then go on. (One doesn't have to count, if they have no apple to begin with).

Natural Numbers are those numbers that starts from 1, followed by 2,3 and so on.

Natural Numbers= {1, 2,3,4,5...}

Whole Number

Whole numbers are natural numbers with the inclusion of zero. Whole numbers starts from Zero, followed by 1, 2 and so on.

Whole Numbers = {0, 1,2,3,4...}

or Whole Numbers = 0 + Natural Numbers.

Integers

Integers are whole numbers plus the inclusion of negatives numbers from -1, -2, -3, and so on. Integers = Whole Numbers + { -1, -2, -3, -4, -5,...}


Real Number

Number line is a line where we can represent the numbers and their fractional or decimal parts. A number lines start from 0 to either sides, both positive and negative and goes on till positive and negative infinity.

Real Numbers are those numbers that can be represented on a number line.

Natural Numbers, Whole Numbers, Integers all belongs to Real Numbers.

Decimal Number

Decimal Numbers have an integer plus a decimal part or the fractional part.

For example, the number 15.02 is a decimal number having 15 as the integral part and the .02 as the fractional part.

Irrational Numbers

Irrational Numbers are the numbers that cannot be expressed on the number line. These numbers can't be expressed as a terminating decimal.

For example, √2, ∛2, √3 are some of the irrational numbers.

Negative Numbers

Negative numbers have the same absolute value as a positive numbers, but with a different sign. For Example the absolute value of -5 and 5 is 5 but their sign is opposite.

One can't see negative numbers in nature. One can observe that there are 5 cows in the field but one cannot say that there are -5 cows in the field. Negative numbers are used to represent opposing quantities. For example credit and debit, positive charge and negative charge.

For example, a credit $5 and debit of $5 adds to 0 because their absolute value is same but of opposite nature.

5 unit positive charge and 5 units of negative charge add to zero.


Prime Numbers

Prime number are those numbers that have only two divisors - the number 1 and the number itself. Example: 1,2,3,5,7,11... are some prime numbers. Try to divide these numbers using various divisors and you will know that they are only divided by the number 1 and the dividend itself.

Special case of number 1: The number 1 is not a prime number. The smallest prime number is the number 2 not the number 1.

Composite Numbers

Any number that is not a prime number is a composite number.Example of composite numbers are 1,4,6,8,9,10,12,14,15,16,18,20,...These numbers have more than two divisors.

Place Value

Place value is calculated by multiplying the digit whose place value is to be calculated and the place (hundredth, thousandth etc...). For example: if we have to calculate the place value of 5 in the number 858412. Then first we have to decide the place at which 5 is located. Here we can see that 5 have 4 digits on the right side. Therefore we have to multiply 5 by 10000 to get the place value. ( 5 x 10000 = 50000).

An easier method is to discard all the digits before the digit whose place value we are calculating. For the above number we will discard 8 and we will left with 58412. Now replace all digits after the digit 5 to get the place value for the required digit. i.e. 58412 => 50000.

Predecessor and successor

Predecessor of a number X is a number that comes 1 count before the number X. To get predecessor of number X we have to subtract the number X by1. In the same way successor of a number X is number that immediately follows the number X


Example: Predecessor of number 45 is (45 - 1)= 44 and the successor of 45 is (45 +1) =46.

The second number will be read as five crore three lakh twenty one thousand six hundred fifty four.

Addition (+)

Commutative Property

We take a number 2 and added it to another number 8, the result is 10. Now, We take a number 8 and added it to another number 2, the result is again 10.

This property of addition, where addition of two numbers is always equal, irrespective of the order of addition is known as commutative property of addition.

a + b = b + c, where a and b are any two numbers.

Associative Property

We take a number 5 and added it to another number 9, the result is 14 and add the result of previous addition to 8 to arrive at 22. Now, We take a number 9 and add it to number 8 and the result of this addition is added to 5, the result is again 22.

This property of addition, where addition of a group of number is always the same irrespective of order of the addition is known as associative property of Addition.

(a + b) + c = a + (b + c) , where a,b and c are numbers.

Subtraction (-)

We have,

8 - 3 = 5

and

3 - 8 = - 5

Therefore, Subtraction is not a commutative.


Multiplication ( x )

We have,

8 X 3 = 24

and

3 X 8 = 24

Therefore, Multiplication is commutative.

And

(8 X 3) X 5 = 120

and

3 X (8 X 5) = 120

Therefore, Multiplication is also associative.

Division (÷)

We have,

8 ÷ 4 = 2

and

4 ÷ 8 = 0.5

Therefore, Division is not a commutative and obviously, not associative.


Solved Problems On Number System

Q1: In the following, which is the greatest number ?

(1). (4)²

(2). (2×2×2)²

(3). [(2 + 2)²]²

(4). (2 + 2 + 2)²

Answer: (3). [(2 + 2)²]²

Explanation:

1. (4)² = 4 x 4 = 16

2. (2×2×2)² = (8)² = 8 X 8 = 64

3. [(2 + 2)²]² = [(4)²]² = [16]²=16 x16 = 256 ( and hence this the answer)

4. (2 + 2 + 2)² = (6)² = 36

Q2. 407928 is read as

(1) Four lakh seventy nine thousand twenty eight

(2) Forty seven thousand nine hundred twenty eight

(3) Forty thousand nine hundred twenty eight

(4) Four lakh seven thousand nine hundred twenty eight

Answer: (4) Four lakh seven thousand nine hundred twenty eight


Q3.If an operator θ is defined as

4 θ 3 = 4 + 5 + 6

5 θ 4 = 5 + 6 + 7 + 8

6 θ 4 = 6 + 7 + 8 + 9

What will n θ 8 be equal to ?

(1) n + 28

(2) 8n + 28

(3) 8n + 36

(4) n + 36

Answer :(2) 8n + 28

Explanation: From the examples given in the question, the first number is the starting number of the series and the second number represents the number of numbers in the series inclusive the starting number of the series.

Using the above rule, we can expand n θ 8 to n +(n+1)+(n+2)+(n+3)+(n+4)+(n+5)+(n+6)+(n+7)

= 8n + (1 + 2 + 3 +4 +5 +6 +7)

= 8n +28

Q4. How many 4-digit numbers are there in the Hindu-Arabic Numeration System ?

(1) 99

(2) 8999

(3) 9999

(4) 9000

Answer :( 4) 9000


Explanation:

Don't get scared by "Hindu-Arabic Numeration" argon. Now to get at the answer just subtracted the lowest 4 digit number from the highest 4 digit number ( and add + 1 since both the ranges are included ) to arrive at the answer.

i.e. 9999 - 1000 + 1 = 8999 + 1 = 9000

Q5. The number 49532 rounded off to the nearest thousand is

(1) 49000

(2) 49500

(3) 41000

(4) 50000

Answer: (4) 50000

Explanation: The thousand part in the given number is 49 thousand. Now we will look at the rest of the number i.e. 532 which is greater than 500 and hence can be rounded off to 1000.

Therefore, the answer is 49,000+1000 = 50000

Q6. When faced with word problems, Rajan usually asks ‘‘should I add or subtract?’’ ‘‘Should I multiply or divide?’’. Such questions suggest

(1) Rajan seeks opportunities to disturb the class

(2) Rajan has problems in comprehending language

(3) Rajan lacks understanding of number operations

(4) Rajan cannot add and multiply

Answer :(2) Rajan has problems in comprehending language


Explanation: From his words it is clear that he is confused about using the mathematical operators.

Q7. Sum of place value of 6 in 63606 is

(1) 6606

(2) 6066

(3) 18

(4) 60606

Answer: (4) 60606

Explanation:

The number is 63606. Place value of first six in the number, from left to right, is 60000, place value of second six is 600 and last six is 6.

Adding place values of all the sixes we have,

60000 + 600 + 6 = 60606

Q8. The difference of 5671 and the number obtained on reversing its digits is

(1) 3916

(2) 7436

(3) 3906

(4) 4906

Answer :( 3) 3906

Explanation: The number is 5671 and the number obtained by reversing the digits is 1765.


i.e. 5671 - 1765 = 3906

Q9.Study the following pattern:

1 X 1 = 1

11 x 11 = 121

111 x 111 = 12321

.... .... What is 11111 X 11111?

(1) 12345421

(2) 123453421

(3) 1234321

(4) 123454321

Answer :( 4) 123454321

Explanation: From the pattern given in examples, we can observe that:

1. The numbers of digit in the answer = Sum of number of digits in each multiplier - 1.

2. The resulting number have digits starting from 1 to digit having value equal to the number of digits in either number and then decreases from there to 1.

i.e. 111 X 111 = 12321. The Digits in the answer start from 1 and goes to 3 ( Number of digits in 111) and decreases from there till digit 1 .

Applying the same rules to 11111 X 11111, we know the right answer is 123454321.

Q10 which of the following is correct?

(1) Predecessor of predecessor of 1000 is 999


(2) Successor of predecessor of 1000 is 1001

(3) Successor of predecessor of 1000 is 1002

(4) Predecessor of successor of 1000 is 1000

Answer: (4) predecessor of sucessor of 1000 is 1000

Explanation:

1. Predecessor of predecessor of 1000 is 999 => predecessor of (1000 - 1 )=> 1000- 1 - 1 = 998 ≠ 999.Rejected

2. Successor of predecessor of 1000 => successor of (1000 - 1) => 1000 - 1 + 1 => 1000 ≠ 1001 Rejected!

3. Successor of predecessor of 1000 => successor of (1000 - 1 ) => 1000 - 1 + 1 => 1000 ≠ 1002 Rejected!

4. Predecessor of successor of 1000 => predecessor of (1000 + 1) => 1000 + 1 -1 => 1000 = 1000 Success!

Q11.A shop has 239 toys. Seventy more toys were brought in.Then 152 of them were sold. The number of toys left was

(1) 239 - 70 + 152

(2) 239 + 70 -152

(3) 239 -70 -152

(4) 239 +70 +152

Answer: 239 + 70 -152

Explanations:


Let put a plus sign on the incoming toys and negative sign on all outgoing toys (negative signs because outgoing decreases the inventory). Then, we have 239 + 70 = 152

Q12.In the product 3759 x 9573, the sum of tens' digit and units' digit is

(1) 9

(2) 16

(3) 0

(4) 7

Answer: (4) 7

Explanation: Just multiply the last two digits from each number and you will get the tens' digit and the ones digit of the resultant. Add the digit to get the answer.

Q13. 19thousands + 19 hundreds + 19 ones is equal to

(1) 20919

(2) 19919

(3) 191919

(4) 21090

Answer :(1) 20919

Explanation: We can take one thousand out 19 hundreds as 19 hundreds makes 1 thousand and 900 hundred. Add 1 to 19,000 to get 20,000. Then add remaining 900 and 19 to get the answer, 20919.easy!!

Q14. If 567567567 is divided by 567, the quotient is


(1) 10101

(2) 1001001

(3) 3

(4) 111

Answer: 1001001

Explanation: The first impulse is to tick 111. nope. Just do it in your head or on piece of paper. your will get 1 quotient for first 567 and then you have to add 0 and 0 to quotient then 1 to clear 567 and then again 0 and 0 and then 1.

Q15.How many 1/8 are there in 1/2?

(1) 16

(2)8

(3)4

(4)2

Answer: (3) 4

Solution: Divide 1/2 by 1/8 to get the answer.

Therefore, No. of 1/8 in 1/2 = (1/2)/(1/8) = (1/2)x(8) = 4

16. A pencil costs two and half rupees. Amit buys one and half dozen pencils and gives hundred rupee note to the shop keeper.The money he will get back is

(1)55

(2)45

(3)65

(4)30


Answer: (1)55

Solution:

Cost of 1 pencil = 2.5

Cost of one and half dozen ( 1 dozen = 12) or 18 pencil = 2.5 x 18 = 45

Since he gave 100 rupee note to the shop keeper, he will get ( 100 - 45 ) = 55 rupees from the

him.

Q17. When 90707 is divided by 9, the remainder is

(1)3

(2)5

(3)6

(4)7

Answer: (2) 5

Solution:

Divide the 90707 by 9 and you will get 5 as remainder and 10078 as quotient.

Quicker Method: Try Rule of divisibility for 9. According to this rule if the sum of the all the digits of the dividend is divisible by 9 then the number itself is divisible by 9.

Here,

Sum of digits of 90707 = 9 + 7 + 7 = 23

and if subtract 5 from this number we will get 23 - 5 = 18 therefore it should be the remainder of the division.


Q18. The sum of the place value and the face value of the number 3 in 12345 is

(1) 0

(2) 295

(3) 297

(4) 305

Answer: (3) 297

Solution:

Place value of 3 in the number 12345 is 3 X 100 = 300

And the face value of 3 is 3 itself.

Then the difference = 300 - 3 =297

Q19. Look at the following pattern:

(9 - 1) / 8 = 1

(98 - 2) / 8 = 12

(987 - 3) / 8 = 123

(9876 - 4) / 8 = 1234

..

(987654 - 6) / 8 =

(1) 12345

(2)123456

(3)123465

(4) 123467


Answer :(2)123456

Solution:

Looking at the pattern, it is evident that the answer will have :

(a) 6 digits because there are six digit in the number beginning with 98.. and digit 6 is subtract from it

(b) Starting from 1 and ending on 6

Therefore the number is 123456

Q 20 The numbers of integers less than -3 and greater than -8 are :

(1) 2

(2) 3

(3) 4

(4) 6

Answer: (3) 4

Solution:

Numbers of integers greater than -8 and less than -3 are -7,-6,-5,-4. And hence 4 in numbers.

Q21. The value of 0.001 + 1.01 + 0.11 is

(1) 1.111 (2) 1.101

(3) 1.013 (4) 1.121

Answer: (4) 1.121

Solution:


Aligning the decimal point of all the number given in the problem , we have

0.001

1.010

0.110

1.121

Q22. In 1999, the population of a country was 30.3 million. The number which is the same as 30.3 million is

(1) 303000000 (2) 30300000

(3) 3030000 (4) 3030000000

Answer: (2) 30300000

Solution:

1 Million = 1000000

Therefore, 30.3 million = 30.3 x 1000000 = 30300000

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Q23. If a^3 = 1 + 7 , 3^3 = 1 + 7+ b and 4^3 =1 + 7 + c, where a, b and c are different positive integers, then the value of a + b + c is

(1) 58 (2) 68

(3) 77 (4) 79

Answer:

Solution:


Taking first equation, a^3 = 1 + 7 = 8

a^3 = 2^3

=> a = 2

Taking second equation, 3^3 = 1 + 7+ b

27 = 8 + b

19 = b

Taking third equation, 4^3 =1 + 7 + c

64 = 8 + c

56 = c

Therefore value of a,b and c came out to be 2,19 and56 respectively.

And a + b +c = 56 + 2 + 19 = 77

Q 24: We call a number perfect if it is the sum of all its positive divisors , except itself. For example 28 is a perfect number because 28 = 1 + 2 + 4 + 7 +14. Which of the following is a perfect number?

(1) 13 (2) 10

(3) 9 (4) 6

Answer: (4) 6

Solution:

(1) Factors of 13 other than itself are 1. Not this number.

(2) Factors of 10 other than itself are 1, 2, 5 and their sum = 1 + 2 + 5 = 8. Not this number.

(3) Factors of 9 other than itself are 1, 3. Not this number.


(4) Factors of 6 other than itself are 1,2,3 and their sum = 1 + 2 + 3 = 6.This is number.

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Q25. The product of two whole numbers is 24. The smallest possible sum of these numbers is

(1) 8 (2) 9

(3) 10 (4)12

Answer: (3) 10

Solution:

First let find all pairs of number that can give 24 upon multiplication. They are ( 8, 3) ,(12,2),(6,4) and upon addition they give out 11, 14 and 10 respectively. 10 is the smallest.

____________________________________________________________

Q26. The value of

( 3^502 - 3^500 + 16 ) /( 3^500 + 2) is

(1) 2 (2) 4

(3) 8 (4) 16

Answer: (3) 8

Solution:

Dividing both numerator and denominator by 3^500, we get

{ ( 3^2 - 1) + 16/3^500} / (1 + 2/3^500)


Both 16^/3^500 and 2/3^500 are negligible values and therefore can be ignored.

What's left is = ( 3^2 - 1) / 1 = 9 - 1 = 8

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Q27. If 800880 = 8 x 10^x + 8x10^y + 8x10^z where x, y and z are whole numbers, then the value of x + y + z is

(1) 11 (2)8

(3) 6 (4) 15

Answer:

Solution:

800880 can be broken down into

800880 = 800000 + 800 + 80

= 8x10^5 + 8x10^2 + 8x10^1

Comparing it with the equation given in the question, we have

x =5, y =2 and z=1

And x + y + z = 5 + 2 + 1 = 8

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Q28. The number n is doubled and then y is added to it.The result is then divided by 2 and the original number n is subtracted from it. The final result is

(1) y (2) y/2

(3) n + y (4) (n + y )/2

Answer: (2) y/2


Solution:

Going step by step

1. Taking a number => n

2. Double the number => 2n

3. Add y => 2n + y

4. Result is divided by 2 => n + y/2

5. Subtract n => y/2

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Q 29: If

1957

- a9

18b8

Then sum of a and b is

(1) 15 (2)14

(3)13 (4)12

Answer: (2)14

Solution:

Starting from the right hand side of the subtraction. Since 7 is less than 9, 1 has to be borrowed from 5. Again 9 is reduced to 8 in the answer, therefore 1 has to borrowed from 9.This means 5 becomes 14 after borrowing from 9 and giving 1 to 7.


And since we got b after subtracting a from 14 , this means a + b =14

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Q30. Which of the following is not a perfect square:

(1)548543213 (2)548543251

(3)548543215 (4)548543241

Answer: (4)548543241

Solution:

None of the perfect square will have 3 at the one's digit. This rules out option (1).

Again only a perfect square of a number that has digit 5 at ones place will have 5 at the unit digit but never alone. It will have 2 in the tenth place. This rules out option (3).

Try the division method for any one of the option (2) and (4) and you will get that only option (4) is a perfect square.

____________________________________________________________

Q31. When an integer K is divided by 3, the remainder is 1, and when K + 1 is divided by 5, the remainder is 0. Of the following, a possible value of K is (a) 62 (b) 63 (c) 64 (d) 65

Answer:

Solution:


Using Divisibility Rules, We know that any number that is divided by 5 has to have either 0 or 5 in the unit place. Therefore, (K+1) has 5 in the unit place and K should have either 4 or 9 in the unit place. While checking for options only option (c) has a number 4 in the unit place and hence the CORRECT answer.

Q32. The least prime number is

(a) 1 (b) 3

(c) 2 (d) 0

Answer: (c) 2

Solution:

2 is the lowest prime number and also 2 is the only even number.

Q33. If 999x + 888y =1332 and 888x +999y =555, then x² - y² is equal to

(a) 7 (b) 8

(c) 9 (d) 5

Answer :( a) 7

Solution:

999x + 888y =1332 ----------(i)

888x + 999y = 555 ----------(ii)

Subtracting (ii) from (i)

=> 111x - 111y = 777

=> (x - y) = 7 ----------(iii)

And adding (i) and (ii)

=> 1887x + 1887y = 1887


=> x + y = 1 --------(iv)

Now, (x² - y²) = (x-y)(x+y) = (7)(1) = 7

Q34. If {1/2(a-b)}² + ab = p(a + b)², then the value of p is (assume a≠b)

(a) 1/2 (b) 1/4

(c) 1/8 (d) 1

Answer:

Solution:

Taking LHS

{1/2(a-b)}² + ab = 1/4(a-b)² + ab

= 1/4(a² + b²- 2ab) + ab

= 1/4((a² + b²) -ab/2 + ab

= 1/4(a² + b²) + ab/2

=1/4(a² + b² + 2ab)

= 1/4(a+b)²

Thus the equation in the question becomes 1/4(a+b)² = p(a + b)²

Therefore. p = 1/4


HCF and LCM

Factors and multiple : If a number a divides number b exactly, we say that a is a factor of b in this case , b in called a multiple of a .

Highest common factor (H.C.F) or greatest common measure (G. C.M.)

or greatest common Divisor (G.C.D.). The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.

There are two methods of finding the HCF of a given set of numbers :

1. Factorizing method : Express each one of the given number as the product of prime factors. The product of least of common prime factors gives

HCF.

Example: Lets say we have to find the HCF of 24 and 54.

Lets prime factorized the number 24 and 54.

24 = 2 * 2 * 2 * 3

54 = 2 * 3 * 3 * 3

only 2 *3 common in both the numbers.

Therefore, 6 is the HCF of 24 and 54.

2.Long Division Method: Suppose we have to find the HCF of two given numbers. Divide the larger number by the smaller opne. Now divide the

larger number by the smaller number one.; Now, Divide the divisor by the remainder. Repeat the process of dividing the preceding number by the

remainder last obtained till zero is obtained as remainder. the last divisor is required HCF.

Taking above numbers 24 and 54.


Dividing 54 by 24 we will get 2 as quotient and 6 as remainder.

Now divide the divisor 24 by the remainder 6. We will get 0 remainder. Since the number 6 was the last divisor, we have 6 as the H.C.F.

Finding the HCF of more than two number: Suppose we have to find the HCF of three numbers. Then, HCF, of { HCF of any two and the third

number} gives the HCF of three given numbers.

Similarly, the HCF of more than three number is obtained.

Least Common Multiple( LCM): The least number which is exactly

divisible each one of the given numbers is called their LCM.

1 Factorization method of Finding L.C.M : Resolve each one of the given

numbers in to a P roduct of Prime factors then , L .C. M is the Product of highest Product of all the factors

2.Common Division Method (Short-cut Method ) of finding L.C.M :

Arrange the given numbers in a row in any order : Divide by a number which divides exactly at divisible . Repeat the above Process til no of the numbers

are divisible by the same number except 1. the Product of the divisors and the undivided numbers is the required L.c.M of the given numbers

4. Product of two numbers = Product of their H.C.F.. and L.C.M.

5. Co-primes : Two numbers are said to be co-primes if their H.C.F. is 1.

6. H.C.F and L.C .M of Fractions :

HCF= ( HCF . of Numbers / LCM. of Denominators ) 2. LCM.= (of

Numbers | HCF of Denominators


7. HCF and LCM of Decimal Fractions : in given numbers , make the same number of decimal places by annexing zeros in some numbers, if

necessary. Considering these numbers without decimal Point , find HCF or LCM as the case may be Now , in the result, mark off many decimal Places

as are there in each of the given numbers,

8,Comparison of Fractions : Find the LCM of the denominators of the

given fractions. Convert each of the fractions into an equivalent fraction with LCM as the denominator , by multiplying both the numerator and

denominator by the same number the resultant fraction with the greatest

numerator is the greatest

Solved Questions:

Q1. Find the Least Number which when divided by 12,18,36 and 45 leaves

the remainder 8,14,32 and 41 respectively.

(a) 180 (b) 178

(c) 186 (d) 176

Answer: (d) 176

Solution:

First we have to find the relationship between the Divisors and the

remainder.

We see that 12-8=4 , 18-14 = 4, 36-32=4 and 45-41 = 4

The rule here is to find the LCM of 12,18,36 and 45. That is 180.

Now, the required number is obtained by subtracting the common difference (4) obtained earlier from the LCM of the number.

i.e. 180 - 4 = 176


2. The ratio of the two numbers is 3:4 and their LCM is 180. The second number is

(a) 45 (b) 90

(c) 30 (d) 60

Answer: (d) 60

Solution:

The point to understand here is that,

When we find ratio of two numbers then the HCF of the numbers is eliminated while reducing the ratio to the lowest term. Thus, the two numbers are 3 x HCF and 4 x HCF

Now, Let the HCF be x. Then using the formula

LCM x HCF = First Number and Second Number

180 x HCF = (3 x HCF)(4 x HCF)

=> 180/(3x4) = HCF

=> 15 = HCF

Therefore, the second number is 4 x15 = 60

3. A farmer has 945 cows and 2475 sheep. He farms then into flocks,

keeping cows and sheep separate and having the same number of animals in each flock. If these flocks are as large as possible, then the maximum

number of animals in each flock and total number of flocks required for the purpose are respectively

(a) 15 and 228 (b) 9 and 380

(c) 45 and 76 (d) 46 and 75

Answer:(c) 45 and 76

Solution:


Here we just have to find the Highest Common Factor of these two numbers

(945,2475). It will come out to be 45. Therefore, looking at the options the CORRECT answer is option (c).

4. If x:y be the ratio of two whole numbers and z be their h.c.f. then the l.c.m. of those two numbers is

(a) yz (b) xz/y

(c) xy/z (d) xyz

Answer:(d) xyz

Solution:

The main point to note here is that when we find ratio of two numbers then

the what is reduced from them is the h.c.f. of both the numbers.

Thus the two numbers are (x.z ) and (y.z)

Again, using the formula,

Multiplication of two numbers = (l.c.m.)(h.c.f)

=> (x.z)(y.z) = (z)(l.c.m.)

=> l.c.m. = xyz


PROFIT AND LOSS

Cost price: Cost price is the price at which the seller has acquired the article in question.

Selling Price: Selling price is the price at which the seller is willing to sell

the product.

Profit: Profit is the amount the seller earns by selling an article at a price

greater than the cost price.

Profit = Selling Price - Cost price

Loss: Loss is the amount the seller loses by selling an article at a price less

than the cost price.

Loss = Cost price - Selling Price

Profit Percentage: Profit percentage is the profit percent that the seller earns on the sell the article. It is calculated on the cost Price.

Overheads: overheads are the expenses that the seller has to incur to sell

the products/articles. Overhead includes the rent of the shop, salaries of the employees, utilities bills, advertisement bill or any kind of expenses the

seller has to pay to keep his business running.

Fixed Costs: Fixed cost is the cost that the seller has to bear in spite of quantity he is selling. Fixed cost include the rent , utility bills, employee

salary and so


Variable Costs: Variable cost or direct cost is the cost the seller has to bear

to sell the article. The cost price of the articles, cost price of the raw material and others.

Formula for profit and loss:

1.Profit Percentage = {(S.P. - C.P)/C.P.}x100

2.Loss Percentage = {(C.P. - S.P.)/C.P.}x100

3.To find the Cost Price when Selling Price (S.P.) and Percentage profit (x) is

given

Cost Price = {S.P/(100 + x)} x 100

4.To find the Selling Price when the cost price (C.P.) and percentage profit

(x) is

given by

Selling Price = {(100 + x)(C.P.)/100}

5.To find the Cost Price when Selling Price (S.P.) and Percentage loss (x) is given

Cost Price = {S.P/(100 - x)} x 100

6.To find the Selling Price when the cost price (C.P.) and percentage loss

(x) is

given by

Selling Price = {(100 - x)(C.P.)/100}


7. If an article is sold at a profit of x% which is further sold at a profit of y% then

the final Selling price is

Final Selling Price = Initial Cost

Price{(100+x)/100}{(100+y)/100)

8. If two successive profits x and y are made on the same article then the

net

profit is

(x + y + xy/100)

Note: above formula could also be used for loss. Just make sure to use +ve

sign for profit and -ve sign for the loss.

9. If a false weight is used instead of the true weight then, the profit

percentage is

{(True Weight - false weight)/False Weight} x 100

Solved Examples:

1. An article is sold for 75 at a loss of 25%. Then find the cost price of the article.

Solution:


Using the formula 5 mentioned above we have ,

Cost Price = {S.P/(100 - x)} x 100

= {75/(100 - 25)} x 100

= {75/75}x 100

= (1/1)x100

= 100

2. An article with a cost price of Rs. 840 is sold for a profit of 10% and then

again sold for a profit of 20%. Find the final selling price.

Solution:

Using the formula 7 mentioned above, we have

Final Selling price = 840{(100+10)/100}{(100+20)/100}

= 840(110/100)(120/100)

= 840(11/10)(12/10)

= 840(1.1)(1.2)

= 1108.80

3. A sells a mobile to B worth Rs. 10,000. He makes a profit of 15%. Again B

sold it back to A at a loss of 15%. Find the A's profit or loss.

Solution:

Using the formula 8, taking Y as negative, We have


Net profit/loss = ( 15 - 15 -225/100)

= (-2.25)

First, A made a profit of 15% and again total loss of (-2.25)% in the entire

deal was profit to A.

Therefore, total Profit A made = 15+2.25 = 17.25 %

4. I gain 0.70 paisa on Rs. 70. My gain percent is :

(a)0.1% (b)1%

(c)7% (d) 10%

Solution:

Total profit = 0.70 paisa

And C.P. = Rs 70

Therefore, Gain percent = (0.70/70)x100 = 1%

5. In terms of percentage profit , which is the best transaction ?

C.P (in Rs.) profit (in Rs.)

(a)36 17

(b)50 24

(c)40 19

(d) 60 29

Solution:


We have to calculate the percentage profit in each case, the table becomes

C.P (in Rs.) profit (in Rs.) profit %

(a)36 17 47.22

(b)50 24 48

(c)40 19 47.5

(d) 60 29 48.33

Therefore, the best transaction is (d)

6. A shopkeeper purchased 70 kg of potatoes for Rs.420 and sold the whole

lot at the rate of Rs. 6.50 per kg. What will be his gain percent ?

(a) 4(1/6) % (b)6(1/4)

(c)8(1/3)% (d)20%

Solution:

Rate per kg at which the potatoes are bought = 420/70 = Rs 6 per kg

And they are sold at 6.5 kg per kg.

Therefore total profit = (0.5/6)x100 = 8(1/3) %

7. The price of the a radio is raised by 40% above the cost price and sold at

a discount of 10%. What will be percentage of profit?

(a) 14% (b) 30%

(c) 25% (d) 26%

Answer: (d) 26%


Solution:

Let the Cost price was 100 then the selling price at 40% profit will be 140.

And at this price a discount of 10% is offered then the final selling price will

be

140 - (0.10)(140) = 140 - 14 = 126

And net profit is 126 - 100 = 26 and in percentage it will be 26%

8. Profit after selling a commodity for 425 is the same as the loss after

selling it for 355.The cost of the commodity is

(a) 390 (b) 395

(c) 400 (d) 385

Answer:(a) 390

Solution:

Since the profit after selling the commodity at 425 and loss after selling it at

355 is the same.Then the Cost Price for the commodity lies exactly in the middle of the two numbers. It will be found by finding the averages of the

these two numbers.

There the Cost Price = (425 + 355)/2 = 390

9. A watch is purchased for 400 and sold for 460.The profit percentage is

(a) 15.5% (b) 12%

(c) 13% (d) 15%

Answer:

Solution:

Total profit = 60 and cost price is 400. Therefore the percentage profit is


(60/400)x100 = 15 %

10. By selling an article for 450 , a man loses 10%. The gain or loss percent if he sells it for 540 is

(a) gain 9% (b) loss 9%

(c) gain 8% (d) loss 8%

Answer: (a) gain 9%

Solution:

Here, Cost Price = 450(110/100) = 495

If he sold it for 540, the profit would have been (540 -495) = 45

And therefore the profit = {(45)/(495)} x100 = 9%

11. The cost price : selling price of an article is a : b. If b is 200% of a then the percentage of profit on cost price is

(a) 75% (b) 125%

(c) 100% (d) 200%

Answer:(c) 100%

Solution:

Given that b is 200% of a. This implies that if a = 1 then b = 2.

This Implies cost price : Selling price = 1:2

And profit percentage = {(2-1)/1}x 100 = 100%

12. The successive discounts of 10% and 20% are equivalent to a single discount of

(a) 30% (b) 28% (c) 25% (d) 27%


Answer: (b) 28%

Solution:

Let the Selling price be 100. Then the discount of 10% is 10. The selling price

will become 90. Again a 20% discount is offered then the selling price becomes 90 - (0.2)(90) = 72.

Hence the net discount becomes 100 - 72 = 28%

13. A dealer marks his goods at 40% above the cost price and allows a

discount of 20% on the marked price. The dealer has a (a) loss of 20% (b) gain of 25%

(c) loss of 12% (d) gain of 12%

Answer:(d) gain of 12%

Solution:

Let the Cost price be 100. Then after it was marked up 40% , the selling price would be 100 + (0.40)(100) = 140.

Again a discount of 20% was offered on the marked price then the final selling price become

140 - (0.20)(140) = 140 - 28 = 112

And the profit is 112 - 100 = 12% or gain of 12%

14. A person sells 400 mangoes at the cost price of 320 mangoes. His

percentage of loss is

(a) 10 (b) 15

(c) 20 (d) 25

Answer: (c) 20 %

Solution:


Out of 400 mangoes he sold,he was able to recover the cost price of 320

mangoes. And his lost is the cost price of the 80 mangoes.

Thus, total loss = {(80)/400} x 100 = 20 %

15. A shoe company sold 50 pairs of shoes on a day costing Rs. 189.50 each

for Rs. 10,000. Then the profit obtained in rupees is

(a) 522 (b) 525

(c) 573 (d) 612

Answer:

Solution:

Cost of 50 pairs of shoes = (189.5)(50) = 9475

And he sold them for 10000

Therefore , total profit = 10000 - 9475

= 525


Compound Interest

Compound interest: At the end of the interest cycle, if the interest is getting added to the sum, initially borrowed, to create a new principal, then this kind of arrangement is called as compound interest.

In compound interest for the new cycle , interest is calculated on the sum borrowed plus on the accumulated interest.

Formula for calculating compound interest

Amount = P { 1 + (r/100)}t

Where,

P = Principal Amount or the amount that was initially lend

r = rate of interest per annum at the principal was borrowed

t = time period ( in years ) for which the amount was

borrowed

Some Important Formulas:

1. If interest is compounded Quarterly

If the interest is compounded quarterly,

then,


the time 't' in years becomes (t/3) as there are 4 quarters in a year of 3 month each

and the rate of interest r becomes (r/3).

And the formula for compound interest becomes

Amount = P [ 1 + {(r/3)/100}]t/3

2. If interest is compounded Monthly

If the interest is compounded monthly,

then,

the time 't' in years becomes (t/12) as there are 12 months in a year

and the rate of interest r becomes (r/12).

And the formula for compound interest becomes

Amount = P [ 1 + {(r/12)/100}]t/12

Q1. The compound Interest on Rs. 1,800 at 10% per annum for a certain

period of time is 378. Find the time in years

(a) 3 years (b) 2.5 years

(c) 2 years (d) 2.8 years

Answer:(c) 2 years


Solution:

Using the formula used above, we have

2178 = 1800(1 + 10/100)t

=> 1.21 = (1.1)t

=> t = 2 Because 1.12= 1.21

Q2. The difference between the simple and compound interest on a certain sum of money for 2 years at 4% per annum is Rs.1. Find the sum.

(a) Rs .630 (b) Rs. 620

(c) Rs. 625 (d) Rs. 635

Answer:(c) Rs. 625

Solution:

Let the number be X then

Simple Interest at the end of 2nd year = (X x 4 x 2)/100 = ( X x 8)/100 = (2X)/25

Compound Interest at the end 2nd year = A - P = X {(1 + 4/100)2 - 1 }= X(0.0816)

Again it is given that the difference between the two types of interest is Re 1

Therefore,

0.0.0816X - (2X)/25 = 1

=> X(0.0816 -0.08 ) =1

=> X(0.0016) =1

=> X = 1/0..0016 = 625


Simple Interest:

When a borrower takes money from a lender then the borrower has to pay some interest to the lender for using that money. If the interest does not get added to principal amount at the end of the payment term, then this kind of arrangement is called as simple Interest.

Terms associated with Simple Interest:

Principal : The total Money borrowed.

Term : The Time period of calculation of interest. May be quarterly,

half -yearly or yearly.

Rate of interest: Rate of interest is the rate at the the money is lent.

Amount : Amount is the sum of principal and interest to be paid at the

end of the term.

Formula for calculating Simple Interest:

Simple Interest = (P x R x T) /100


Formula for calculating Amount:

Amount = Simple Interest + Principal

Few Solved Problems in Simple Interest:

Problem 1:

A sum of Money will double it self in 10 years at simple rate of interest. Find the rate of Interest?

Solution:

It is given to us that a certain sum of money will double itself in 10 years. We have to find the rate at which it was lent.

Let the principal be P

The the Amount after 10 years would be 2P

Using the formula for amount ,

we have,


Amount = SI + Principal

=> 2P = (PxRx10)/100 + P ( dividing both LHS and RHS by P)

=> 2 = R/10 + 1

=> 1 = R/10

=> R = 10 %

Thus, any kind of money lent at simple rate of interest of 10% will double itself in 10 years.

Problem 2:

The Simple interest earned on a certain sum of money in 5 years at 11 % per annum is Rs 330. Find The sum.

Solution:

The equation for simple interest is :


S.I. = (P x R x T)/100

Data given in the question,

R = 14% per annum

T = 5 years

S.I. = 330

P =?

Put all the data in the above equation, we get

330 = (P x 11 x 5)/100

=> 330 = (Px11)/20

=> P = (330 x 20)/11 = 30x20 = 600

Hence , a sum of Rs. 600 will earn a simple interest of 330 in 5 years at 11% rate of interest.

Problem 3:


If 600 amounts to 816 in 3 years at simple interest. If the rate of interest is increased by 3%, to how much would it amount to in the same time period?

Solution:

The equation for simple interest is :

S.I. = (P x R x T)/100

Data given in the question,

R = ?

T = 3 years

S.I. = 816 - 600 = 216

P = 600

Put all the data in the above equation, we get

216 = (600 x R x3)/100

=> 216 = 6 x R x 3

=> R = 216/18 =12 %

Now, the question was if it is to be increased by 3% then what will be the amount in 3


years. Now again using the new value of rate of interest ( 12 + 3 = 15) to get simple interest:

S.I. = (600x15x3)/100 = ( 6 x 15 x 3) = 270

And amount would have been = 600 + 270 = 870.


RATIO AND PROPORTION

1. If A:B = 3:4 B:C =8: 10 and C:D =15:17 then find A:B.C.D

(a) 9:12:15:17 (b) 6:12:20:32

(c) 9:22:32:34 (d) 7:19:27:42

Answer: (a) 9:12:15:17

Solution:

If there are three ratio a:b and c:d and e:f

Then the ratio among these 4 quantities is given by ace: bce : bde : bdf

Using this the required ratio is:

3 x 8 x 15:4 x 8 x 15: 4 x 10 x 15 : 4 x 10 x 17

= 9 : 12 : 15 : 17

2. Find the ratio compounded of the four ration:

4:3, 9:13, 26:5, and 2:15

Solution:

The ratio compounded = (4 x 9 x 26 x 2)/(3 x 13 x 5 x 15) = 16/25

3. An amount of Rs 750 is distributed among A,B and C in the ratio of 4:5:6 what is the share of B ?

(a) 500 (b) 250


(c) 1050 (d) 370

Answer: (b) 250

Solution:

In such question, the first we add all the part. we get 4 + 5 + 6 = 15. Out these, B got 5 parts.

Thus, B got (5/15) x 750 = 250

4. Two number are in the ratio of 9: 14 if the larger number is 55 more than the smaller number.Find the numbers.

(a) 99,154 (b) 48:88

(c) 17:96 (d) 59:74

Answer: (a) 99,154

Solution:

The smaller number be x, then the larger number be (x+ 55) . Then according to the question

x / (x+55) = 9/14

Thus,

14x = 9(x + 55) ( multiplying the LHS and RHS of the above equation by (x+55)/14 }

=> 14x = 9x + 495

=> 5x = 495

=> x = 99

Thus the number are 55 and 154.


5.The sum of the three is 98 if the ratio between the first and the second be 2:3 and that between the second and the third be 5:8, then find the second number .

(a) 90 (b) 45

(c) 77 (d)30

Answer: (d)30

Solution:

Let the three numbers be A , B and C.

Then the ratio of A to B is = 2:3 or 10:15

And the ratio of B to C is = 5 : 8 15:24

Thus, the ratio of A:B:C is 10:15:24.

Again, Given that A + B + C = 98.

Therefore, A = (10/49) x 98 = 20

B = (15/49) x 98 = 30

6. A sum of money is divided between two people in the ratio of 3:5. If the share of one person is Rs 20 less than that of other,find the sum .

(a) 88 Rs (b) 76 Rs

(c) 85 Rs (d) 80 Rs

Answer: (d) 80 Rs

Solution:

Let the share of one person be x then the share of other person is x + 20.

It is given that

(x / x +20) = 3:5


=> 5x = 3(x + 20)

=> 5x = 3x + 60

=> 2x = 60

=> x = 30

And ( x +20) = 30 + 20 = 50

And therefore the sum is 50 + 30 = Rs . 80

7. The ratio between two number is 3:4 If each number be increased by 6 the ratio becomes 4:5 find the number.

(a) 43,16 (b) 18,24

(c) 22,12 (d) 32, 22

Answer: (b) 18,24

Solution:

Let the number be 3x and 4x. Then (3x/4x) = 3/4

Again, (3x+6/4x+6) = 4/5 ( given that when both numbers are increased by 6 the ratio becomes 4/5)

=> 5(3x + 6 ) = 4(4x + 6)

=> 15x + 30 = 16x + 24

=> 15x - 16x = 24 - 30

=> -x = -6

=> x = 6

Thus, the first number is 6 x 3 = 18

and the seconder number is 4 x 6 = 24


8. The ratio between two number is 3: 4. If each number be increased by 2, the ratio becomes 7:9. Find the number.

(a) 43, 16 (b) 22,10

(c) 12, 16 (d) 18,9

Answer: (c) 12, 16

Solution:

Let the number be 3x and 4x. Then (3x/4x) = 3/4

Again, (3x + 2/4x+2)=7/9 (given that when both numbers are increased by 2 the ratio becomes 7/9)

=> 9(3x + 2 ) = 7(4x + 2)

=> 27x + 18 = 28x + 14

=> 27x - 28x = 14 -18

=> - x = -4

=> x = 4

Thus, the first number is 4 x 3 =12

and the seconder number is 4 x 4 =16

9. The students in three classes are in the ratio 2:3:5. If 20 students are increased in each class the ratio changes to 4:5:7. what was the total number of students in the three classes before the increases?

(a) 120 (b) 85

(c) 72 (d) 100

Answer: (d) 100


Solution:

Let the number be 2x,3x and 5x.

It is given that if 20 students in each class are increased then the ratio becomes 4:5:7

It can written as (2x +20) : (3x +20) : (5x + 20) = 4 : 5 : 7

Taking the first two ratio

we have,

(2x +20) / (3x +20) = 4/5

=> 5(2x +20) ) = 4(3x +20)

=> 10x + 100 = 12x + 80

=> 10x - 12x = 80 - 100

=> -2x = -20

=> x =10

Therefore, the student in each class before the increase are 2x10, 3x10, 5x10 = 20,30,50

And total student = 100

10. The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number

(a) 135 (b) 105

(c) 155 (d) 165

Answer: (b) 105

Solution: Let the two numbers be 3x and 4x.

When they are increased by 9 they become 3x + 9 and 4x + 9.


It is given that the ratio is 18:23

Thus, 3x + 9/4x + 9 = 18:23

23(3x + 9) = 18(4x + 9)

69x + 207 = 72x + 162

69x – 72x = 162- 207

-3x = -45

X = 15

Thus two numbers are 3x15 = 45 and 4 x 15 = 60

And the sum is 60+45 = 105 11. Find the number which, when added to the terms of the ratio 11:23 makes it equal to the ratio 4:7

(a) 15 (b) 5

(c) 25 (d) 20

Answer: (b) 5

Solution:

Let the number be x which when added to each terms of the ratio 11:23 makes it equal to 4:7.

Thus, (11 + x)/(23 + x ) = 4/7

7(11 + x ) = 4(23 + x)

77 + 7x = 92 + 4x

7x-4x = 92 – 77

3x = 15


X = 5

Thus, 5 should be added.

12. Find the number which, when subtracted from the terms of ratio 11:23 makes it equal to the ratio 3:7

(a) 2 (b) 8

(c) 16 (d) 6

Answer: (a) 2

Solution:

Let the number be x which when subtracted from each terms of the ratio 11:23 makes it equal to 3:7.

Thus, (11 - x)/(23 - x ) = 3/7

7(11 - x ) = 3(23 - x)

77 - 7x = 69 - 3x

-7x+3x = 69 – 77

-4x = -8

X = 2

Thus, 2 should be subtracted

13. In 40 liters mixture of milk and water the ratio of milk and water is 3: 1 how much water should be added in the mixture so that the ratio of milk to water becomes 2:1 ?

(a) 500ml (b) 20 liters

(c) 5 liters (d) 10 liters


Answer:

Solution:

Let x liters of water be added to the mixture to make the ratio of milk to water to become 2:1.

Thus,

30/(10 + x) = 2/1

30(1) = 2(10+x )

30 = 20 + 2x

10 = 2x

X = 5 liters

14. A mixture contains milk and water in the ratio of 3:2 liter of water is added to the mixture, milk, milk and water in the mixture become equal find the quantities of milk and water in the mixture .

(a) 12, 8 liter (b) 4,3 liter

(c) 12, 6 liters (d) 10,8 liters

Answer: (a) 12, 8 liter

Solution:

Let quantities of milk and water in the mixture be 3x and 2x. Then if 4 liters of water is added to the mixture the ratio of milk and water become 1:1.

It can be written as (3x): (2x + 4) = 1/1

Thus, 3x = 2x +4


x = 4

Therefore, the milk in the mixture is 4x3 = 12 litres and quantity of water = 4x2 = 8 liters

15. A bag contain equal number of one rupee, 50 paisa and 25 paisa coins respectively if the total value of rs 35, how many coins of each type are three?

(a) 40 (b)80

(c) 160 (d) 20

Answer: (d) 20

Solution:

50 paisa = .5 rupee. .25 paisa = .25 rupees.

Now, given that X (1 + 0.50 + 0.25) = 35

=> x = 35/(1.75) = 20

Thus, there are 20 coins of each type.

16. The speed of these cars is in the ratio of 2:3:4. what is the ratio among the time taken by these cars to travel the same distance ?

(a) 7:8:12 (b) 3:6:9

(c) 9:3:6 (d) 6:4:3

Answer: (d) 6:4:3

Solution:


Let the distance be 2x3x4 = 24 km.

Then, ratio of time taken by each car is 24/2: 24/3: 24/4

12:8:6

or 6:4:3

17. The incomes of A of B are in the ratio 3:2 and their expenditure are in the ratio 5:3 if each saves rs 2000, what is their income ?

(a) 32000 (b) 20000

(c) 1190 (d) 8000

Answer: (d) 8000

Solution:

Let the income be 3x and 2x. It is given that the saving of each is Rs. 2000.

Then, their expenditures are 3x – 2000 and 2x – 2000

Again, (3x – 2000)/(2x – 2000) = 5/3

=> 3(3x – 2000) = 5(2x – 2000)

=> 9x – 6000 = 10x – 10000

=> 9x -10x = -10000+ 6000

=> -x = -4000

=> x = 4000

Therefore, their salaries are 3 x 4000 = 12000 and 2 x 4000 = 8000

18. The contents of two vessels containing water and milk are in the ratio 1:2 and 2:5 are mixed in the ratio 1:4, the resulting mixture will have water and milk in the ratio.


(a) 11:22 (b) 71:58

(c) 31:74 (d) 91:49

Answer:(c) 31:74

Solution:

In vessel 1

The fraction of water = 1/3

The fraction of milk = 2/3

In vessel 2

The fraction of water = 2/7

The fraction of milk = 5/7

Now, From vessel 1, 1/5 of the mixture is taken out. Thus water taken out = (1/5)(1/3) = 1/15 and milk taken out = (1/5)(2/3) = 2/15

And from vessel 2, 4/5 of the mixture is taken out. Thus water taken out = (4/5)(2/7) = 8/35 and milk taken out = (4/5)(5/7) = 20/35

Thus Ratio of water and milk in resulting solution is (1/15 + 8/35) : (2/15+20/35) = 31/105 : 74/105

= 31:74

19. Find the mean proportion of 9, 16

(a)12 (b) 144

(c) 25 (d) 4/3


Answer:(a)12

Solution:

Mean proportion of 9 and 16 = √ (9 x 16) = 3 x 4 = 12

20. Find the third proportional to 15 and 20

(a)80/3 (b) 55/6

(c) 37/2 (d) 71/2

Answer:(a)80/3

Solution:

Here, given one ratio

15:20 and 20:x . We have to find the value of x.

Now 15/20 = 20/x

=> x = (20x20)/ 15 = 400/15 = 80/3

21. Find the fourth proportional to the number 6,8 and 15

(a) 48 (b) 48

(c) 41 (d) 20

Solution:

Let the fourth proportional be F.

Then,

6/8 = 15/F

=> F = (15 x 8 )/6 = 120/6 = 20


22. The ratio of three number is 3:4:5 and the sum of their squares is 1250. The sum of their number is :

(a) 30 (b) 50

(c) 60 (d) 90

Answer: (c) 60

Solution:Let the numbers be 3x, 4x and 5x.

Then,

Given that sum of their squares is 1250

=> (3x)² + (4x ) ² + (5x)² = 1250

=> 9x² + 16 x² + 25x² =1250

=> 50x² = 1250

=> x² = 25

=> x = 5

Thus, the number are 3 x 5 = 15, 4 x 5 = 20 and 5 x 5 = 25

Ad therefore their sum = 60

23. The ratio of three number is 3:4:7 and product 18144 the number are :

(a) 9,12,21 (b) 15,20,25

(c) 18,24,42 (d) None of the

Answer:(c) 18,24,42

Solution: Let numbers are 3x, 4x and 7x


Given that the product of the numbers is 18144

Therefore, (3x)(4x)(7x) = 18144

=> 84x³ = 18144

=> x ³ = 216

=> x = 6

Therefore the numbers are : 3 x 6 = 18

4 x 6 = 24

and 7 x 6 = 42

24. The value of K that must be added to 7,16,43,79 so that they are in proportion is.

(a) 7 (b) 5

(c) 9 (d) None of the

Answer: (b) 5

Solution:

k is added to 7,16,43,79 to make them in proportion then

(7 + k)/(16 + k) = (43 + k)/(79 + k)

=> (7 + k)(79 + k) = (43 + k)(16 + k)

=> 553 + 7k +79k + k² = 688 + 43k + 16k + k²

=> 553 + 86k = 688 + 59k

=> 553 - 688 = -86k + 59k

=> -135 = -27k

=> k = 5


25. What should be subtracted from 15,28,20 and 38 so that the remaining number may by proportional?

(a) 2 (b) 4

(c) 6 (d) None of the

Answer: (a) 2

solution:

Let k be subtracted from 15,28,20,38 such that yhey may be proportional. then

(15 - k)/(28-k) = (20 - k)/(38 - k)

=> (15 - k)( 38-k) = (20-k)(28-k)

=> 570 - 53k + k = 560 - 48k +k

=> 570 - 53k = 560 - 48k

=> 570- 560 = 53k - 48k

=> 10 = 5k

=> k = 2

26. Two number are respectively 20% and 50% more than a third number. The ratio of two numbers is :

(a) 2:5 (b) 3:5

(c) 4:5 (d) 6:7

Answer:(c) 4:5

Solution:

Let the third number be x.


Then the first number = (1.2) x

And the second number = (1.5)x

Therefore their ration = (1.2x)/(1.5x) = 12/15=4/5 or 4 : 5

27. A and B together have Rs 1210, if 4/15 of A's amount is equal to (2/5) of B's amount. How much amount does B have ?

(a) Rs.460 (b) Rs.484

(c) Rs.550 (d) Rs.664

Answer: (b) Rs.484

Solution:

Let A's amount to be a.

And B's amount be b.

Given that (4/15)a = (2/5)b => a = (2/5)(15/4)b = (3/2)b Thus, a + b = 1210

=> (3/2)b + b = 1210

=> (5/2)b = 1210

=> b = (1210)(2/5) = (242)x 2 = 484

28. Rs. 900 is to be distributed among A,B,C and in the proportion 2:3:4 how much C get?

(a) Rs.400 (b) Rs.450

(c) Rs.540 (d) Rs.None of the

Answer:(a) Rs.400


Solution:

Share of C = (4/9)x 900 = 4 x 100 = 400

29. If x:y is 9:7 then (x+y):(x-y) is ......

(a)8:1 (b) 1:8

(c)4:1 (d) 1:4

Answer:(a)8:1

Solution:

(x+y) :(x-y) = {(x/y) + 1 } /{(x/y) -1} = {9/7 +1}/{9/7 - 1} = 16/2 = 8/1 or 8:1

30. The ratio of the present ages of Sunita and Vinita is 4:5. Six years hence, the ratio of their ages will be 14:17. What will be their ages 12 years hence.

(a) 13:19 (b)16:19

(c) 17:19 (d) 15:19

Answers: (b)16:19

Solution:

The word to understand here is 'hence'. 'Hence' in such questions means 'after' the years mentioned.

Let the present ages of Sunita and Vinita be 4x and 5x. Now, according to the question,

(4x + 6)/(5x + 6 ) = 14/17

=> 68x + 102 = 70x + 84

=> 102-84 = 70x - 68x

=> 18 = 2x


=> x = 9

Therefore, Sunita's age = 4x9 = 36 years

Vinita's age = 5x9 = 45

After, 12 years Sunita's age = 36+ 12 = 48 years

12 years Vinita's age = 45+ 12 = 57 years

And ratio of their ages = 48/57 = 16/19 or 16:19

31. The ratio of the two numbers is 3:4 and their LCM is 180. The second number is

(a) 45 (b) 90

(c) 30 (d) 60

Answer: (d) 60

Solution:

The point to understand here is that,

when we find ratio of two numbers then the HCF of the numbers is eliminated while reducing the ratio to the lowest term. Thus, the two numbers are 3 x HCF and 4 x HCF

Now, Let the HCF be x. Then using the formula

LCM x HCF = First Number and Second Number

180 x HCF = (3xHCF)(4xHCF)

=> 180/(3x4) = HCF

=> 15 = HCF

Therefore, the second number is 4 x15 = 60


32. From each of the two given unequal numbers, half the smaller number is subtracted. Then, of the resulting numbers, the larger one is five times than the smaller one. Then the ratio of the larger to smaller one is (a) 2 : 1 (b) 3 : 2 (c) 3 : 1 (d) 1 : 4

Answer:(c) 3 : 1

Solution:

Let the numbers be X and Y and X >Y.

Then according to the conditions given in the problem, we have

(X-Y/2)/(Y/2) = 5/1

=> (2X-Y/(Y)) = 5/1

=> 2X-Y = 5Y

=> 2X = 6Y

=> X/Y = 6/2 = 3/1

or X:Y = 3:1

Click here for more problems on

Time and Work

Time Speed and distance


UNITARY METHOD

Unitary method is about calculating for a one item and then using it to get the solution for the given number of quantity.

For example:

If the cost of 30 m of cloth is 345, then find the cost of 16 m of such cloth?

Solution:

Given that the cost of 30 m of cloth is: Rs 345

Then the cost of 1 m of cloth is (345/30) = 11.5

Therefore, the cost of 16 m of cloth is (11.5) x 16 = 184

SOLVED PROBLEMS

1. If the cost of 30 meters of cloth is Rs 345 . Find the cost of 16 meters of this cloth.

(a) 184 (b) 162

(c) 875 (d) 321

Answer:(a)

Solution:

Cost of 30 m of cloth = Rs 345

Cost of 1 m of cloth = Rs 11.5

Cost of 16 meter of cloth = 16 x 11.5 = 184


2. If 4 men or 7 women do a work in 60 days then in how many days will 8 men and 7 women finish the same work

(a) 18 (b) 20

(c) 48 (d) 36

Answer:(b)

Solution :

In this problem, we to calculate the work done by a single men or a single women in one day and then we can easily find the answer to the above problem. We assume that the total work done is 1 unit.

Work done by 4 men in one day = 1/60

Work done by a single men = 1/(60 x 4) = 1/240

Work Done by 7 women in 1 day = 1/60

Work Done by 7 women in 1 day = 1/(60 x 7) = 1/(420)

Now, from the condition given in the problem,

Work done by 8 men and 7 women in a single day is 8 x (1/240) + 7 x(1/420) = 1/30 + 1/60 = 3/60=1/20

Time taken to do the work = 1/1/20 = 20 days.


3. A contractor undertook to do a work in 60 days employed 50 workers to carry the job but after 40 days he found that only half had work had been done now how many more worker should he employ to finish the work in time ?

(a) 81 (b) 50

(c) 20 (d) 26

Answer: (b)

Solution:

The hey here is only half the work is done in 40 days by 50 workers.

Now we will find the work done by a 50 workers in 40 days = (1/2)

And lets say x workers can do the remaining half of the work in 20 days.

the equation thus becomes

X x 20 = 50 x 40

X = 100

And since 50 workers are already employed, The contractor has to rope in 100 - 50 = 50 more workers to do the job.

4. If 6 people spend rs 2400 in 4 months then 2 less people will spend Rs 1600 in how many month?

(a) 4 (b) 6

(c) 63 (d) 24

Answer:(a)


Solution:

6 people spent 2400 in 4 month

6 people will spent in 1 month = 2400/4 = 600

1 person spent in 1 month = 600/6 = 100

Now to find how many month it will take to 4 person to spent 1800.

Since spent of 1 person is 100 and 4 person spent is 100 x 4 = 400

And the month it will take to spent 1800 is = 1600/400 = 4 months

5. If 20 men can complete a work in 3 days while 12 ladies can complete the same work in 12 days .in how many days will 20 men and ladies complete the same work ?

(a) 2/17 (b) 2/5

(c) 36/17 (d) 17/3

Answer:(c)

Solution:

First find the work done by 1 men and 1 lady in 1 day

Lets assume work done is 1 unit.

Now,

Work done by 1 men in 1 day = 1/(20 x 3) = 1/60

and work done by 1 lady in 1 day = 1/(12 x 12 ) = 1/144

Work done by 1 men and 1 lady in 1 day = 1/60 + 1/144 = 17/720

And Work done by 20 men and 20 lady in 1 days = (17/720)x 20 = 17/36


And days it will take to complete the work = 1/(17/36) = 36/17

6. If each boy takes two chocolates a day it will suffice to 7 boys for 12 days if each boy takes three chocolates a day then for how many days will suffice to 8 boys ?

(a) 10 (b) 15

(c) 16 (d) 7

Answer: (d)

Solution:

First calculate the total chocolate availabe.

There are 7 boys and they take 2 chocolate each day for 12 days = 7 x 12 x 2 = 168

Now there are 8 boys now and they each take 3 chocolate each day there total chocolate eaten in day = 24

and they will last 168/24 = 7 days

7. If the rate of exchange is 7.50 dollars for Rs 100 how many dollars be equal to Rs 550 ?

(a) 41.75 (b) 42.75

(c) 41.25 (d) 42.25

Answer: (c)


Solution:

Rate of exchange for 1 dollar = 7.5/100

And for 550 we will get = (7.5/100) x 550 = 41.25 dollars

8. If the weight of 13 meters long rod is 23.4 kg what is the weight of 6 meter long rod ?

(a) 7.1 kg (b) 10.8 kg

(c) 12.4 kg (d)18.0 kg

Answer: (b)

Solution:

First Calculate the weight of 1 m rod = 23.4/13 = 1.8 kg/m

And the weight of 6 m rod is = 1.8 x 6 = 10.8 Kg

10. If a scale of map represents 0.8 cm for 8.8 km if two points in a map are 80.5 cm apart what the actual distance between these points ?

(a) 885.5 km (b) 905 km

(c) 700 km (d) 990 km

Answer:(a)

Solution::

First calculate how much 1 cm represents = 8.8/0.8 Km

Therefore 80.5 cm represents = (8.8/0.8) x 80.5 = 885.5 Km


11. If 45 men can complete work in 30 days working 12 hours a day in how many day work 60 men complete the work working 10 hours days ?

(a) 31 (b) 27

(c) 25 (d) 33

Answer: (b)

Solution:

Work done by 1 men per day per hour = 1(/ 30 x 12 x 45)

Lets say it takes X days to complete the work if they work 10 hours a day and there are 60 men

then, work done by a men per hour comes out to be 1/ ( 10 x 60 x X)

Equating both the equations to get

30 x 12 x 45 = 10 x 60 x X

= > X = 27 days

12. A fort provision for 150 men for 45 days after 10 days 25 men left fort how long will the provision last from that day?

(a) 45 (b) 54

(c) 29 /15 (d) 42

Answer: (d)

Solution:

Total provision in the fort = 150 x 45

After 10 days provision left = 150 x 35 = 5250

Now, there are 125 men left and it will last = 5250/125 = 42 days


There fore the provision will last 42 days more after 25 men left.

13. If 6 men weave 168 shawls in 7 days how many shawls will be woven by 8 men is 5 days ?

(a) 126 (b) 90

(c) 160 (d) 104

Answer: (c)

Solution:

Shawls woven in 1 day by the 6 men = 168/7 = 24

And Shawls woven in 1 day by 1 men = 24/6 = 4 shawls

Therefore ,

8 men will weave 4 x 8 = 32 shawls in 1 day

and in 5 days they will weave 5 x 32 = 160 shawls.

14. If 51 people can eat some food in 50 days how many days will the same food be eaten 50 people ?

(a) 52 (b) 51

(c) 50 (d) 55

Answer: (b)

Solution:


Since Total food remains the same which is 51 people X 50 men.

Now there are 50 people there days = 51x 50 / 50 = 51 days

16. If 15 oxen or 20 cows can eat the grass of a field in 80 days then in how many days will 6 oxen and 2 cows eat the same grass ?

(a) 40 (b) 60

(c) 100 (d) 160

Answer:

Solution:

grass eaten by 1 oxen in 1 day = 1/(15 x 80)

grass eaten by 1 cow in 1 day = 1/(20 x 80)

Let it takes x days to eat the grass, then

x { 6/1200 + 2/1600) = x ( 30/4800) = x (1/160) = 1

And therefore x = 160 days

17. A contractor undertook at do a work in 80 days he employed 30 men to carry out the job but after 50 days he found that only half work had been done now how many more men should he employ to finish the work in time ?

(a) 30 (b) 50

(c) 20 (d) 10

Answer: (c)

Solution:

Half work was done in 50 days employing 30 men


And lets says he employs a total of M men to complete the rest of the half work in 30days

Then

M x 30 = 50 X 30

M = 50 men

And men that were added to complete the work in time = 50 30 = 20

18. A family has provision for 15 people for 1 week for if after 2 days 5 people are increased and the ration is reduced by 25 % per person how long will the provision last ?

(a) 5 (b) 6

(c) 7 (d) 4

Answer:

Solution:

Lets ration per day is R then accodring to the conditions given in the above problem, we have

15 x 7 x R = 15 x R x 2 + 20 x 0.75R x D

105 = 30 + 15 D

D = 5

Therefore Ration will last 5 more days after intial 2 days are over. Therefore total days = 7 days


19. A regular working day is 8 hours and regular week is 5 working days .A man is paid rs 2.40 per regular hour and rs 3.20 per hour overtime if he earns rs 432 in 4 week what is the total number of hours he works ?

(a) 180 (b) 195

(c) 160 (d) 175

Answer: (d)

Solution:

Total hours in 4 weeks for a regular week = 8 x 5 x 4 = 160 hours

Amount he should get without overtime = 160 x 2.4 = 384

He got 432, an extra 48 have come due to overtime. Rate of overtime is 3.20

Therefore overtime is done for 48/3.2 = 15 hours

Therefore total hours = 160 + 15 = 175

20. If one fourth kg of a substance costs rs 0.80 what will be the cost of 200 gram of the same substance?

(a) 40 paisa (b) 100 paisa

(c) 64 paisa (d) 60 paisa

Answer: (c)

Solution:

Cost of 1/4 kg of substance = 0.80

Cost of 1 kg of substance = 0.80/.25 = 3.2

And therefore cost of 200 g = 3.2 X .2 = 0.64


21. If 3/5 of a work is done by 12 men in 10 days how many men will complete the whole work in 20 days?

(a) 9 (b) 8

(c) 10 (d) 7

Answer:

Solution:

Total work done would have been done in (5/3) x 12 x 10 days

Let M men could complete the work in 20 days

Then,

Equating both the terms

We have,

(5/3) x 12 x 10 = M x 20

=> M= 10 men

22. 400 persons working 9 hours a day complete 1/4 the of a piece of work in 10 days the number of additional persons working 8 hours a day required to complete the remaining work in 20 days is ....(a) 675 (b) 275

(c) 250 (d) 225

Answer: (b)

Solution:

Work done by 400 people working 9 hours a day for 10 days = 400 x 9 x 10 = 1/4 of total work


Let total work be T

Then T= 4 x 400 x 9 x 10

And work remaining = 3/4 T= (3/4) x 4 x 400 x 9x 10

Again Let M is the number of total men after the addition of men to complete the work in 20 days, then

(3/4) x 4 x 400 x 9 x 10 = M X 20 x 8

108000 = M x 160

M = 675

Therefore Men increased by = 675 - 400 = 275


Arithmetic Average

The formula for calculating average is :

Average = (x1 + x2 + x3 +x4 +x5 + x6 + ... + xn)/n

where,

n is the number of quantities

x1 is the value of first quantity, x2 is the value of second

quantity and so on.

Example: There are seven boxes having weight of 7, 6 , 8, 5, 4 , 6 ,8.

Calculate

the average of all the boxes.

Solution: Using the above formula, we have seven boxes, therefore n =7

and x1 = 7, x2 = 6 and so on

put the values in the above formula to get

average = ( 7+ 6 + 8 + 5 +4 +6 +8)/7 = 44/7 = 6.28 approx.

Weighted Average


In case of weighted average we assign different weights to each quantity

and then calculate their average.

Formula for weighted average is

Average (weighted) = (w1x1 + w2x2 + w3x3 + .... + wnxn)/(w1 + w2 + w3 +.. + wn)

Where,

w1,w2,w3....wn are the weights of x1,x2,x3,...xn quantities.

Example:

A student scored 60, 80 and 70 in Physics, chemistry and math. If the

college assigns a weight of 3 to physics, 1 to chemistry and 2 to math for admission to the physics course, then calculate the weighted average of the

student.

Solution:

Here,

Weights of marks in physics, chemistry and math are 30, 40 and 80 respectively.

Therefore using the formula for the weighted average, we have,

weighted Average = (3x60+ 1x80 + 2x70)/(3 + 1 + 2 )

= (180 + 80 + 140)/(6)


= 66.67

and an arithmetic average would have been

= (210)/3= 71

Comparison between Weighted average and arithmetic average:

The weighted average calculated in this case is less than the arithmetic

average because of the weight attached to the various subjects. Since Maximum weight is attached to physics, and the student has scored less

mark in physics, this will pull down the average.

Take another scenario where admission is sought in chemistry course and

the weight attached are as:

Math = 1

Physics = 2

chemistry = 3

Then the weighted average would be = ( 120 + 240 + 70)/6 = 71.67

Now weighted average has come out to be greater than arithmetic average

because of the weight assigned to chemistry and the higher marks in chemistry has pull UP the weighted average.

Solved Problems on Averages Question 1: The average age of 20 students in a class is 12 years. If the age of the teachers is also added then the average becomes 13 years. Find out the age of the teacher?


Solution:

Method 1: solving by using the formula.

Let the age of the teacher be X years.

Then the formula for averages becomes

13 = (20 x 12 + X)/21 = (240 + X)/21

Multiplying both the sides by 21 to get

=> 273 = 240 + X

on solving for X

X = 33 years

So the age of the teacher is 33 years

Method 2 :

Adding the teachers age to the group increases the age of the students by 1 full year.

Since there are 20 students the total years added by the inclusion of the teacher = 20 years

And since the teacher is new addition to the group having an average of 13, 13 years more years are added.

Therefore total years added to the group due to the inclusion of the teacher = 13 + 20 = 33 years.

Question 2: Rakesh bought 10 shirts for 250 each, 15 pants for 350 each and 20 handkerchiefs for 50 each. Find the average cost of each item.


Solution:

Here simply use the average formula :

Average = (10 x 250 + 15 x 350 + 20x50)/(45) = (2500 + 5250 + 1000)/45

= 194.44

Question 3: The average contribution of an office towards the health fund is 120 month. If the officer level employees are contributing Rs. 460 and non-officers staff is paying Rs 110. Then, if there are 15 officers then find the number of non-officer staff.

Solution:

Let the non-officer staff be X.

Then total staff would be (15 + X)

Then total contribution by all the staff = 120 (15 + X )

Using the formula for the average calculation

120 (15 + X ) = 6900 + 110 X

Solving for X

1800 + 120X = 6900 + 110X

120X - 110X = 6900 - 1800 = 5100

10X= 5100

X = 510

There fore non-Officer staff is 510

Question 4: A certain college while taking admission in the math course is assigning following weight age to the subjects.


Math - 4

Science - 3

English - 2

Hindi - 1

If a student has secured following marks in these subjects then find out arithmetic average and the weighted average of the marks.

Math - 70

Science - 60

English - 70

Hindi - 80

Solution:

Using the Formula for arithmetic averages we have,

Arithmetic average = (70 + 60 + 70 + 80) / 4 = 70

Using the formula for the weighted average we have

weighted average = ( 4 x 70 + 3x60 + 2x70 + 1x80)/(4 + 3 + 2 + 1)

= (280 + 180 + 140 + 80)/10

= 68

Difference between the arithmetic and the weighted average = 2

and it is due to the different weight age to various subjects in calculating the weighted average.


Question 5: One-third of the journey is covered at the rate of 25 km/hr , one-fourth at the rate of 30 km/hr and the rest at 50 km/hr. Find the average speed for the whole journey.

Solution:

For calculating speed we have to first find the total distance covered and total time taken.

Let the total distance be X km.

Then according to the question

1. (X/3) distance is covered at 25 Km/hr and time taken = X/75 hr

2. (1/4)X distance is covered at 30 Km.hr and time taken = X/120

3. Rest of the journey ( X - ( X/3 + X/4)) = 5X/12 at the rate 50 Km/hr and time taken = X/120

Therefore calculating the average time taken

Average Speed = Total Distance Covered/Total Time Taken

= X/ ( X/75 + X/120 + x/120)

= 1/(3/100) = 100/3

= 33(1/3) Km/Hr

Question 6: The average weight of 8 persons is increased by 2 Kg when one of them who weighs 56 kg is replaced by another person. The weight of the new person is :

Solution:


The new person added 2 kg to the average weight of 8 people. Then total new weight added to the group is 8 X 2 = 16 kg. In addition to this weight , The new person also replaced a member with 56 kg of weight. Then, then total weight of the new person = 56 + 16 = 72 Kg.

Question 7: In an examination, one section of class X, having 60 students scored an average marks of 55 and the second section, having 40 students scored an average of 45 marks.What is the overall average.

Solution:

Average = Total Marks scored by all the students / Total numbers of students

1. Total Marks scored by 60 students of first section = 55x60 = 3300

2. Total marks scored by 40 students of second section = 40 X 45 = 1800

Therefore total marks scored by all the students = 5100

and total students = 60 + 40 = 100

Using the formula for average mentioned above,

we have ,

average marks = 5100/100 = 51

Question 8: The average marks of a student in four subjects is 65. If the student scored 75 marks in the fifth subject then the average of the total marks is ?

Solution:

Marks in the four subjects = 65 x 4 = 260

Marks in the fifth subject = 75


Total marks in five subjects = 335

Therefore average marks in five subjects = (335)/5 = 67

Question 9: The average of 13 results is 50. If the average of the first 6 results is 49 and that of the last 6 is 53, then find the 7th results.

Solution:

Given that the averages of the first 6 results and the last six result are 49 and 53 respectively.

Then the total value of all the 12 result ( except the 7 th one) is = 49x6 + 53x6 = 612

And the total value of all the 13 result = 650

Therefore the seventh value would be = 650 -612 = 38

Question 10: A driver drives to the office at 50 km/her and returns at 30km/hr. Find the average speed.

Solution:

The key here is that the driver drives the same distance to and fro. Let the distance be D in one direction.

Then the total distance travelled wound be 2D.

Now, time taken while driving at 50 km/hr = D/ 50 hr

And time taken to while driving at 30 km/hr = D/30 hr

Now,


Average Speed = Total Distance Traveled/ Total Time Taken = 2D/(D/ 50 + D/30)

= 2/8/150 = 150/4 = 37.50 km/hr

Q 11. Out of 30 teachers of a school, a teacher of age 60 years retired. In his place another teacher of 30 years was appointed. As a result , the mean age of the teachers will

(a) decrease by 1 year (b) remain same

(c) decrease by 2 years (d) decrease by 6 month.

Answer:

Solution:

First Method:

Net result of the retirement and appointment of the teachers result in a net decrease of 30 years. These 30 years are distributed across 30 teachers. Thus, the mean decreases by 1 years.

Second method:

Let the mean of the 29 teachers that remain the same be x years, then

Final Mean - Initial mean = (29x +30 )/30 - (29x + 60)/30 = (29x + 30 -29x - 60)/30

= - 30/30 = - 1

Thus, the final mean is less than the initial mean by 1 year.

Q12. Average age of A, B and C is 84 years.When D joins them the average age becomes 80 years. A new person E, whose age is 4 years more than D replaces A and the average of B, C , D and E becomes 78 years. What is the age of A?

(a) 70 years (b) 80 years

(c) 50 years (d) 60 years


Answer: (b) 80 years

Solution:

Since Addition of D to the group reduces the average age of the group to 80 years. This means addition of D, reduces 4 x 4 = 16 years from the total age. Thus, the age of D is 84 - 16 = 68 years.

Again, given that the age of E is 4 years less than D then the age of E is 68 + 4 = 72 years.

Addition of E and exit of A reduces the group average to 78 years. This means A total of 4x2 = 8 years have been taken from the group total. Thus, age of A is 72 + 8 = 80 years.

Q13. The average of six numbers is 32. If each of the first three numbers is increased by 2 and each of the remaining three numbers is decreased by 4, then the new average is (a) 35 (b) 34 (c) 31 (d) 30

Answer:(c) 31

Solution:

Each of the first three numbers is increased by 2 then the increases in total

= 3 x 2 = +6

And each of the 3 remaining three numbers are decreased by 4, then total decreased is 4 x 3 = - 12

Therefore, net increase/decrease = + 6 - 12 = -6

And the decrease in the mean = -6/6 = -1. And new mean = 32- 1 = 31

14. Five years ago, the average age of P and Q was 25. The average age of P, Q and R today is 25. Age of R after 5 years will be


(a) 15 (b) 20 (c) 40 (d) 35

Answer: (b) 20

Solution:

Sum of ages of P and Q, five years ago, is {(P - 5) + (Q - 5 ) = 50 years} And their present ages =P + Q = 60 years

Again, When R is added to the group then sum = (P + Q + R) = 60 + R

and average = (60 + R)/3 = 25 (given)

=> 60 + R = 75

=> R =15 years

And age of R after 5 years = 15+ 5 = 20 years

15. The average marks scored by 36 students was 52. But it was discovered that an item 64 misread as 46.What is the correct mean of marks?

(a) 54 (b) 53.5

(c) 53 (d) 52.5

Answer:(d) 52.5

Solution:

First method:

Total marks of the student with the misread item = 36 x 52 = 1872

Total marks of the student with the correct item = 1872 - 46 + 64 = 1890

Therefore, average = 1840/36 = 52.5


Faster Method:

Due to the error in reading, the total reduced by 64-46= 18. And the loss of 18 was distributed among 36 members. That is average reduced by 19/36 = 0.5. Therefore correct average is 52+ 0.5 = 52.5


SPEED TIME AMD DISTANCE

1. A train is running with the speed of 45 km per hour? what is its speed in meter per second ?

(a) 12.5 m/s (b) 13 m/s

(c) 16 m/s (d) 81 m/s (e)None of the

Answer: (a) 12.5 m/s

Solution:

Speed of train = 45 km per hour

Convert km to meters= 45 x 1000 = 45000 m ( because 1 km = 1000 m)

Convert Hour to seconds=

60 x 60 seconds = 3600 seconds (because 1 hour = 60 min, 1 min =

60 seconds)

Therefore 45 km/hr = (45000/3600) meter per second = 12.5 meter per second

Short Cut:

1 Km/hr = (5/18) m/s

Simply Multiply the speed in km/hr by (5/18) to get the answer in m/s.

here, 45 km/hr = (45)(5/18) = 12.5 m/s

2. A train 100 m long is running at the speed of 21 km /hr and another train

150 m long is running at the speed of 36 km /hr in the same direction.How long will the faster train take to pass the first train ?


(a) 2 min (b)9 min

(c) 1 min (d) 7 min (e) None of the

Answer: (c) 1 min

Solution:

The faster train is the one that is running at 36 km/hr.

The difference in speed of the two trains is (36 - 21 ) km/hr = 15 km/hr

And the combined length of the train is (100 + 150 ) m = 250 m.

Therefore, the time taken is = (250m )/(15 km/hr) = (250 m)/ (15 x 5/18

m/s)

= 60 seconds

= 1 minutes.

3. A plat form is 131 meter long if the speed of the train, 67 meter long,

is 45 km per hour, how long will it take to pass the platform ?

(a) 15.84 sec (b) 59 sec

(c) 439 sec (d) 36.7 sec (e) None of the

Solution:

Here, the platform is stationary. There its speed = 0

The difference in speed between the two objects = 45 - 0=45 km/hr=45

(5/18) m/s= 12.5 m/s

And the combined length of the platform and the train = (131 + 67) =

198 m


Thus the time taken by the train to pass the platform = 198m/12.5m/s = 15.84 seconds

4. Find the distance covered by a man walking for 10 minutes at a speed of 6 km/hr.

(a) 3 km (b) 4 km

(c) 8 km (d) 1 km

Answer: (d) 1 km

Solution:

Total distance covered = speed x time = (6 km/hr)x(10minutes) =

(6km/hr)(10/60)hr

= 1 Km

5. A motor car does a journey in 10 hrs for the first half at 21 km/hr and the second half at 24 km/ hr. Find the distance?

(a) 244 km (b) 238 km

(c) 268 km (d)224 km

Answer: (d)224 km

Solution:

Total distance covered = speed x time

Let the Total distance be X.

Then, for the first half (X/2) speed was 21 Km/hr.

Therefore the time taken for the first half = (X/2)/(21 Km/hr) = X/42 hrs --

------- (i)


And the time taken for the second half is = (X/2)/(24) = X/48 hrs ---------

(ii)

But given that total time taken is 10 hrs. Equating (i) and (ii) to get

X/42 + X/48 = 10

=> X(8 + 7)/336 = 10

=> X(15) = 3360

=> X = (3360)/15

=> X = 224 km/hr

6. Walking at (3/4)th of his usual speed a person is 10 min late to office.

Find his usual time to cover the distance.

(a) 35 min (b) 30 min

(c) 38 min (d) 25 min

Answer: (b) 30 min

Solution:

Let the distance between his house and office is X km and his usual speed

be S.

Then, his usual time to reach office = X/S

And if he is walking at (3/4)th of his usual speed then the time taken is=

(X)/ (3S/4) = 4X/3S

Again, It is given that he was 10 min or 1/6 hours late.

Therefore the equation thus forms is

4X/3S - X/S = 1/6


=> X/S( 4/3 - 1) = 1/6

=> X/S (1/3)= 1/6

=> X/S = (1/6)(3/1) = 3/6 = (1/2) km/hr

Therefore, the time taken walking at 1/2 km/hr = 2X hours

And time taken walking at (3/4)(1/2) or 3/8 km/hr = 8X/3 hours.

Again it is given walking at (3/4) speed he is late by 10 min or 1/6 hours

=> (8X/3) - 2X = 1/6

=> X(2/3) =1/6

=> X = (1/6)(3/2) = 1/4 km

And usual time = (1/4)/(1/2) = 1/2 hour or 30 minutes

Short cut:

Given that he is walking at 3/4 of his usual speed.

And Since time taken is inversely proportional to the speed then his time taken would be (4/3) of his usual time.

Then, Let the time be T,

then (4T/3) - T = 1/6

(1/3)T = 1/6

T = 3/6 = 1/2 hours or 30 minutes

7. Two men A and B start from a place P walking at 3 km and 3.5 km an hour respectively. How many km will they be apart at the end of 3 hours if

they walk in same direction ?

(a) 6.5 km (b) 1.5 km

(c) 7.7 km (d) 3.2 km


Answer: (b) 1.5 km

Solution:

Distance Covered by A in 3 Hours = (3)(3.5) = 10.5 km

Distance Covered by B in 3 Hours = (3)(3) = 9 km

Thus they will be 10.5 - 9 = 1.5 km apart walking in the same direction.

8. Two men A and B walk from P to Q , a distance of 21 km, at 3 and 4 km an hour respectively B reaches Q, returns immediately and meets A at R.

Find the distance from P to R .

(a) 18 km (b) 34 km

(c) 20 km (d) 22 km

Answer: (a) 18 km

Solution:

Time Taken by B to reach Q is 21/4 hours.

In this time, A covered distance of = (21/4)(3) = 63/4 km.

He is still short of 21 - 63/4 = (84-63)/4 = 21/4 km.

Now, after reaching Point Q, B turns back to go to point P.

The relative speed of A and B is = 3 + 4 = 7 km

And distance Between them = 21/4 km.

Therefore, time taken to cover this distance = (21/4)/7 = 3/4 hours.


And distance traveled by A in this time to reach point R is = (3/4)3 = 9/4

km.

Thus, the distance between Point P and R = (63/4) + (9/4) = 72/4 = 18 km.

9. A man covers a certain distance by car driving at 70 km/hr and he

returns back to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.

(a) 31.5 km/hr (b) 71.5 km/hr

(c) 11.5 km/hr (d) 61.5 km/hr

Answer: (d) 61.5 km/hr

Solution:

For such questions, where the distance covered is same but at different speeds the formula for the average speed is

average speed = (2xy)/(x + y), where x and y

are the different speeds.

=> average speed = (2 x 70 x 55)/(70 + 55)

= 7700/125

= 61.5 km/hr

10. Peter can cover a certain distance in 1 hour 24 minutes by covering two third of the distance at 4 km/hr and rest at 5 km/hr. Find the total distance.

(a) 6 km (b) 19 km

(c) 22 km (d) 32 km

Answer: (a) 6 km

Solution:


Let the total distance be X km.

Then Peter covered (2/3)X of the distance at 4 km/hr in time = (2X/3)/4

= X/6 hours

And Peter covered (1/3)X of the distance at 3 km/hr in time = (X/3)/5 = ( X/15) hours

Given that total time taken was 1 hour 24 minutes = 1 + 24/60 = 1 + 2/5 = 7/5 hours

There the equation thus formed is X/6 + (X/15) = 7/5

X(5 + 2 )/30=7/5

X(7/30) = 7/5

X = (7/5)(30/7)

X = 6 km

11. An hero plane flies along the four sides of a square at the speed of 200,

400, 600 and 800 km/hr . Find the average speed of the plane around the field .

(a) 384 km (b) 333 km

(c) 324 km (d) 312 km

Answer:(a) 384 km

Solution:

For such questions , where the distance covered is the same for all the speeds, the formula for 4 different speeds is

4abcd/(abc + bcd + cda + dab)

where, a,b,c,d are the different speeds to cover the

same distance


Therefore, average speed =

{4(200)(400)(600)(800)}/{(200)(400)(600) + (400)(600)(800) + (600)(800)(200) + (800)(200)(400)}

= (400)(2)(4)(6)(8)/{(2)(4)(6) + (4)(6)(8) + (6)(8)(2)+(8)(2)(4)}

= (153600)/( 48 + 192 + 96 + 64)

= (153600)/(240 + 160) = 384 km

12. A man is walking at a speed of 12 km per hour. After every km he takes rest for 12 minutes. How much time will he take to cover a distance of 36

km ?

(a) 30 hours (b) 10 hours

(c) 25 hours (d) 40 hours

Answer: (b) 10 hours

Solution:

Total time taken to cover the distance of 36 km (without the rest time) =

36/12 = 3 hours

Now, number of rest he will take to complete the journey = 36 - 1 = 35. 35 because he would have covered the entire distance by the time he takes

36th rest break.

Therefore total rest taken = 35 x 12 = 420 min = 7 hours.

Thus Total time taken = 3 + 7 = 10 hours.

13. I walk a certain distance, ride back taking a total time of 37 minutes . I could walk both ways in 55 minutes. How long would it take me to ride both

ways .


(a) 22 min (b) 28 min

(c) 26 min (d) 19 min

Answer: (d) 19 min

Solution:

Given that

riding time + walking time = 37 minutes ----------(i)

and walking time + walking time = 55 minutes

=> 2 x walking time = 55

=> walking time = 55/2 = 27.5 minutes

Put this in the equation (i)

Therefore, 27.5 minutes + riding time = 37 minutes

riding time = 37 - 27.5 = 9.5 minutes

Therefore, total time taken to ride back = 2 x 9.5 = 19 minutes

14. A man travels a distance of 18 km from his house to an exhibition by tonga at 15 km/hr and returns back on cycle at 10 km/hr . Then the average

speed for the whole journey is

(a) 12 km/hr (b) 10 km/hr

(c) 15 km/hr (d) 18 km/hr

Answer:

Solution:

Since the distance traveled is the same for both the journey, therefore using the formula mentioned in the Short Cut formula for time speed and distance

. we have


Average Speed = 2(15)(10)/(15+10) = 2(15)(10)/(25) = 12 km/hr

15. A gun is fired at a distance of 1.34 km from Geeta. She hears the sound after 4 secs. The speed at which the sound travels is

(a) 330 m/s (b) 300 m/s

(c) 325 m/s (d) 335 m/s

Answer: (d) 335 m/s

Solution:

Speed of sound = (1.34 x1000 m)/ (4 sec)

= (1340 m)/ 4 sec

= 335 m/s

16. If I walk at 5 km/hr , I miss a train by 7 minutes. However, if I walk at

6 km/hr I reach

the station 5 minutes before the departure of the train. The distance between my house and the station is

(a) 7 km (b) 6 km

(c) 5 km (d) 6.5 km

Answer:

Solution:

Using the formula mentioned in the Short Cut formula for time speed and

distance, the required distance is

= {(5x6)/(6-5)} x {(7+5)/60}


= (30) x {(12)/60}

= 6 km

17. Two trains starts from the stations A and B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their

meeting, the second train has traveled 120 km more than the first. The distance between A and B is

(a) 1200 km (b) 1440 km

(c) 1320 km (f) 990 km

Answer:(c) 1320 km

Solution:

The trains were traveling towards each other then the relative speed is (50+60)=110 kmph

Again it is given that the second train traveled 120 km more than first.

The difference in their speed = 60 - 50 = 10 kmph

And time taken to cover 120 kmph = 120/10 = 12 hours.

Thus, it took 12 hours for the trains to cover as distance of 110x12 = 1320

km

5. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times

while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps is

equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is

(a) 48:35 (b) 28:15

(c) 24:20 (d) 20:21


Answer: (b) 28:15

Solution:

Distance covered by Tom in 7 jump = Distance covered by Jerry in 5 jumps.

Distance covered by Tom in 1 jump = (5/7) jumps of Jerry.

Therefore, ratio of speed of Tom to Jerry = (8x1)/{6x(5/7)}

= 56/30 = 28/15


TRIANGLES

Properties of Triangles:

1. A triangle has three sides and three interior angles

2. Sum of any two sides of a triangle is always greater than the third side.

i.e. for a triangle ABC, we have

AB + BC > AC

AB + AC > BC

AC + BC > AB

3. Sum of all angles of a triangle is 180.

i.e For a triangle ABC

∠A + ∠B + ∠C = 180.


4. The exterior angle of a triangle is equal to the sum of the vertically

opposite interior angles.

5. If an angle of a triangle is greater than 90 then the other two angles of

triangle must be acute angles.

Types of triangle

1. Scalene : Scalene triangle is a triangle that has length of its three sides

different from one other.

(i). All angles are different. None of angles are equal.

2. Isosceles: Isosceles triangle is a triangle that has two of it's sides equal to

each other.


(i) The angles opposite to equal sides are also equal.

3. Equilateral: Equilateral triangle has all its sides equal to each other

(i) All angles are equal and are of 60 each.

4. Right Angled: One of the angle of the triangle

is equal to 90.

(i). The side opposite to the right angle is greatest side in the

triangle.


Terms related to Triangles:

1. Median and centroid : Median is a line that joins a vertex of a triangle to

the mid-point of the opposite side. Median divides the

triangle into two equal halves.If we draw all the medians of a

triangle then they will intersect at a point. That point is called

a centroid.

2. Height and orthocentre: A straight line from any vertex of the triangle to


the opposite side and perpendicular to it , is called as a height

or altitude. The point where all the altitudes meet is called as

circumcentre.

3. Perpendicular bisector and circumcentre : If we draw perpendicular

bisector of each side, then they will meet at a point known as

circumcentre of the triangle. Circumcentre because if we put

compass on the circumcentre and open the mouth to match the

distance between the circumcentre and any vertex, then we can

be able to draw a circle that contains all the vertices of the

triangle.


Formula for calculating area of triangles:

1. Using base and the height:

Area of triangle = ½ x b x h

where, b is the length of base or the side of the triangle where the

altitude from the opposite vertex was drawn

h is the height of the altitude from the vertex to the base.

2. Using Hero's formula


_______________

Area of triangle = √s(s-a)(s-b)(s-c)

where, s = (a +b + c)/2

and a , b, c are the length of the sides of the triangle.

3. For equilateral Triangle

______

Area = (√3 x a² )

4

where, a is the side of the equilateral triangle

4. For isosceles Triangle

______

Area = b √4a² - b²

4

where, a is the length of the equal sides

and b is the length of the third side.


Solved Questions:

1. The perimeter of an equilateral triangle is 72√3 cm. Find its height.

(1) 63 m

(2) 24 m

(3) 18 m

(4) 36 m

Solution: Let Perimeter of equilateral triangle = 3 a = 72 √3

=> a = 72√3/ 3 => a = 24√3

we also know,

Height of equilateral triangle = a (√3 / 2)

= 24 √3(√3/2)

= (24 x 3) /2

= 12 x 3

= 36


2. In the triangle below angle ∠PQR and angle ∠QRP have the measure of 4m° , ∠QSP

have the measure of m° and ∠RPS has a measure of 45°, then what is the measure

of ∠QPS?

(1) 45°

(2) 60°

(3) 90°

(4) 105°

(5) 120°


Solution: ∠QRP is an exterior angle to triangle PRS.

Therefore, ∠QRP = ∠RPS + ∠RSP

=>∠QRP = 45° + m

But ∠QRP = 4m (given)

=> 4m = 45° + m

=> 3m = 45°

=> m = 15°

Thus,

∠RSP = 15°


and ∠PQS = 4 x 15 = 60°

Now in triangle PQS, We know that sum of all angles of a triangle =180

Therefore, ∠QPS + ∠PQS + PSR = 180°

=> ∠QPS = 180 - 15 - 60°

=> ∠QPS = 105°

3 : In the triangular shown below, if x = √6, then what is the area of the triangle?

_

(1) √ 3/8 (2) 3/8

_


(3) 3√ 3/8 (4) 3/4

_

(5) 3√ 3/4

Solution:

Given,

x = √6

To find,

Area of the given triangle

Since the triangle is a right angled triangle, then using the Pythagoras theorem, we have

i.e. (x)² = (x √3/2)² + y²

where 'y' is the base of the triangle in the figure.


=> (√6)² = {(√6 x √3)/2}² + y²

=> (√6)² = {(√18)/2}² + y²

=> 6 = {(3√2)/2}² + y²

=> 6 = (3/√2)² + y²

=> 6 = 9/2 + y²

=> y² = 6 - 4.5

=> y² = 1.5 = 3/2

=> y =√ (3/2)

Area of triangle = ½ x b x h

Putting in the values

Area of triangle = ½ x √ (3/2) x (3/√ 2)

= ½ x (3√ 3/2)

_

= 3√ 3/4


Volume Volume is the total space that is enclosed by the 3D figures like cuboid, cube, cone and so on. It is an indication of the space it will occupy or liquid or gases it can store.

cube

Cube

A Cube has 6 equal square faces. A good example is a box with equal height, length and breadth.

Volume of cube = l x l x l ,

where l is the length of any side.

Total surface area of cube = 6 X l


Cuboid

A Cuboid has 6 faces just like cube but they are not all equal. For a cuboid at least one of the length, breadth or height is different .

Cuboid with length l , height h, and breadth b

Volume of cuboid = l x h x b

Where, l is the length,

h is the height,

and b is the breadth of the cuboid

Total surface area = 2 ( l x h + l x b + b x h)


Where, l is the length,

h is the height,

and b is the breadth of the cuboid

Cone

Cone

A cone has circular base and then it tapers with increasing height to a point.

A good example is ice cream cone, party hats.

Volume of cone = (1/3) x π x r² x h

Where,π is approximately equal to (22/7) or 3.14


h is the height of cone,

r is radius of circular base

Lateral Surface Area = π x r x l

Total surface area = π x r x l + π x r²

Where, π is approximately equal to (22/7) or 3.14


l is the slant height of cone

Cylinder: A cylinder has circular base and then it extends to a certain

height "h". Good example is soda can, cookie jars, sanitation pipes.


Volume of cylinder: π x r² x h Where,

π = 3.14 or 22/7

r is the radius of the

circular base

h is the height of the cylinder

Lateral Surface area is the surface area of the side( in figure it is the area in yellow color). It does not include the area of the base and the top

(area in the blue color).

Lateral Surface area = 2 x π x r x h

Total surface area = Lateral surface area + area of the circular base and top

Total surface area = 2 x π x r x h + 2 x π x r²

= 2 x π x r ( h + r)

Sphere

A sphere has round 3D figure.


Sphere

A good example is cricket ball or tennis ball

Volume of sphere = (4/3)π x r³

Where,

π is approximately equal to

(22/7) or 3.14

r is radius of sphere

Total surface area = π

where,

π is approximately equal to (22/7) or 3.14



Hemisphere

An hemisphere, as the name suggests is half sphere. A sphere cut in half.

Hemisphere

Volume of hemisphere = (2/3)π x r³

Where,

π is approximately equal to

(22/7) or 3.14

r is radius of sphere

Total surface area = 3 x π x r²

Solved geometry problems:

Q1 Which is true for a hexagonal pyramid ?


(1) It has six faces and each face is a hexagon

(2) It has a hexagonal base with six triangular faces meeting at a point

(3) It has two hexagonal faces and six rectangular faces

(4) It has six hexagonal faces joined by six rectangular faces

Answer: (2) It has a hexagonal base with six triangular faces meeting at a point

Explanation : The keyword here is "pyramid". A pyramid has a polygon base at one end and an apex at another ( ex cone). An hexagonal pyramid has hexagonal base and then tapers to a point or an apex.

Q2


Answer: (1) 60

Explanation: Let consider the front face. It has 5 cubes in one rows and there are three such rows. Therefore in front face there are ( 5 x 3) 15 cubes. And there are 4 such rows of (5 x 3) cubes, That's makes a total of 4 x 15 = 60 cubes.

Q3

Answer: (3) 36

Solution: Volume of a cylinder = (π x r² x h)

Since, diameter and hence radius is increases by 25 % but the volume remained the same, the height should have been decreased. Let the new height

be h2. And the volume = (π x (1.25 r)² x h2). Equating both the equation, we have,

(π x r² x h) = (π x (1.25 r)² x h2)

=> h = (25/16)h2

=> (h2/h)= 16/25


=> (h - h2/h) x 100 = (25-16/25) x 100 = 9 x 4 = 36

=> % height decrement = 36

Q4. The internal length, breadth and height of a rectangular box A is 20cm, 18 cm and 15 cm respectively and that of the box B are 18cm, 12cm and 5 cm respectively. The volume of Box A is how many times that of B?

(1) 4

(2) 5

(2) 6

(5) 3

Answer:

(1) 5

Solution:

Volume of a rectangular box = height x length x width

Therefor volume of Box A = 20 x 18 x 15 = 5400

And the volume of the box B = 18 x 12 x5 = 1080

Divide the volume of A by the volume of B to get the answer = 5400/1080 = 5

Q5. The Whole surface of a cube is 150 sq. cm. Then the volume of the cube is

(a) 125 cm³ (b) 216 cm³

(c) 343 cm³ (d) 512 cm³


Answer:(a) 125 cm³

Solution:

A cube has 6 equal square faces that makes the whole surface area. Let the area of the square be X sq. cm.

Then, 6 x X = 150

=> X = 150/6 = 25 sq. cm.

Since, area of the square = l x l =25

=> l = 5 cm

And the volume = 5 x 5 x 5 = 125 cm³

Q6. If each edge of a square be doubled, then the increase percentage of its area

is

(a) 200% (b) 250%

(c) 280% (d) 300%

Answer: (d) 300%

Solution:

Let the edge of the square be X cm then the area = X x X = x²

Then if the edge is double then = (2x)(2x) = 4x²

Therefore. the percentage increase in area is = {(4x²-x²)/x²} x100 = 300%

Q7.A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with a diameter of the base as 12 cm.The height of the cone is


(a) 2cm (b) 3cm

(c) 4cm (d) 6 cm

Answer:3 cm

Solution:

Volume of a sphere = (4/3)π r³ = (4/3)(22/7)(6/2) ³ = (4/3)(22/7)(3) ³

Again, Volume of the cone = (1/3)(22/7)(12/2)²(h) = (1/3)(22/7)(6)²(h)

Since same metal is used in recast the cone from the metal ball

then, the

(4/3)(22/7)(3) ³ = (1/3)(22/7)(6)²(h)

=> h = (4)(3)³/(6)²

= (4)(27)/(36)

= 3 cm

Geometry: Area and Perimeter

1.Triangle :


Area of triangle = (1/2) x b x p

Where,

b = the length of the side where perpendicular is drawn from opposite

point

p = the length of the perpendicular

Perimeter of triangle :

Perimeter of a triangle = (a + b + c),

where,

a, b, c are the length of the sides AB, BC and AC.

2. Rectangle : A rectangle is a parallelogram, whose opposite sides are equal and the angle at which the lines intersect is 90. The diagonals of the rectangle are equal in length.


Area of rectangle = l x b

where,

l = length of rectangle

b = breadth of rectangle

Perimeter of rectangle = 2 ( l + b )

where,

l = length of rectangle

b = breadth of rectangle

Diagonal of a rectangle is a line segment joining two opposites vertices of the rectangle.

Relationship of the diagonal to the sides of rectangle:

______


diagonal =√l² + b²

3. Square : A square is a rectangle, whose all sides are equal. Diagonal of a square are perpendicular bisector of each other.

Area of square = l x l

where l = length of one side

Perimeter of square = 2 ( l + l ) = 4 l

Diagonal of a square is a line segment joining two opposites vertices of the square.

______ __


diagonal =√l² + l² = l√2

4. Rhombus: A rhombus is a parallelogram, whose all sides are equal.

Rhombus with side "a" and diagonals d1 and d3

Area of rhombus = (1/2) x d1 x d2

where d1 = length of shorter diagonal

d2 = length of larger diagonal

Perimeter of rhombus = 2 ( a + a ) = 4 a


Solved Problems

Q1. The length of a rectangle is ‘l’ and its width is half of its length. What will be the perimeter of the rectangle if the length is doubled keeping the width same ?

(1) 4l (2) 5l (3) 6l (4) 3l

Answer: (2) 5l

Solution: Here,for first rectangle length = l and the width = (l/2). A

new rectangle is formed by keeping the width same i.e (l/2) and doubling the length i.e. 2l .

therefore, perimeter becomes = 2( 2l + l/2)

= 2(5l/2)

= 5l

Q2 To introduce the concept of area, a teacher can start with

(1) comparing area of any figure with the help of different objects like palm, leaf, pencil, notebook, etc.


(2) calculating area of a rectangle by finding length and breadth of a

rectangle and using the formula for area of a rectangle (i.e. length x breadth)

(3) calculating area of figures with the help of counting unit square

(4) explaining of formulae for finding area of figures of different shapes

Answer:(1) comparing area of any figure with the help of different objects like palm, leaf, pencil, notebook, etc.

Q3. A rhombus has diagonals of length 8 cm and 6 cm. Find its perimeter.

(1) 18 cm

(2) 20 cm

(3) 24 cm

(4) 28 cm

Answer:(3) 24 cm

Explanation: Area of rhombus = (1/2 )x d1 x d2 ,

where d1 and d2 are the two the diagonal of a rhombus.

here, d1 = 8, d2=6, Therefor, area = (1/2) X 8 X 6 =48/2 = 24

Q4 .The concept of areas of plane figures can be introduced to the students of Class V by

(1). calculating the area of a rectangle by finding length and breadth of a rectangle and using the formula for area of a rectangle


(2) stating the formula for area of rectangle of rectangle and square

(3). calculating the area of figures with the help of counting unit squares.

(4). measuring the area of any figure with the help of different objects like palm, leaf, pencil etc.

Answer:(3). calculating the area of figures with the help of counting unit squares.

Q5: The figure consists of five squares of the same size. The area of the figure is 180 square centimeters. The perimeter (in cm) of the figure will be

(1) 48

(2) 72

(3) 36

(4) 45

Answer: (2) 72


Explanation: The area of the total figure = 180 sq. cm. Since it is made up of 5 identical square, the area of one square will come out be = 180/5 = 36 sq.cm. And the length of

the square will work out to be = √36 = 6 cm.

Now, going around the figure we will see, that there are 12 sides in total of al the square that makes the perimeter of this figure.Hence, the perimeter would come out to be 12 x 6 = 72

Q6

Answer: (4) 4

Q7:


Answer: (2) 120

Solution: Since it is given that <d = (1/2) <E. And both angle D and angle E are two angle of right angled triangle CED. We have,

∠D + ∠E + 90° = 180°

=> ∠D + 2∠D = 180° -90°


=> 3∠D = 90°

=> ∠D = 90° x (3)

=> ∠D = 30°

Also, it is given that (1/2)∠A = ∠D

=> ∠A = 60° = ∠B ( since CA and CB are equal and so

are angle)


And therefore, ∠ BCA = 60°

But ∠ BCA + ∠ACD = 180° => ∠ACD = 180° - 60° = 120°

Q8.

Answer: (1) ∆ PQR ≡ TRQ


Solution: In each option, compare the triangles on the left and the right of the similarity symbol, for similar angles and sides. For example taking option (1), the angle QRP and similarly taking angle from the other triangle it will be angle RQT and they are equal. Similarly comparing each sides and angles, we will confirm that option (1) is correct.

Q9. Perimeter of a square is 24 cm and the length of the rectangle is 8 cm.If the perimeters of the square and the rectangle are equal , then the area of the rectangle is

(1) 16

(2) 24

(3) 32

(4) 64

Answer: (3) 32

Solution:

Since perimeter of the square and rectangle are equal, then

4S = 2( L + B)

where, 4S = 24, L = 8cm.

Put the values to get

24 = 2(8 + B)

=>12 = 8 + B

=>4 = B


Therefore area of the square is = 4 x 8 = 32 square cm

Q10 : The perimeter of the square is 20 cm. A rectangle has the same width as the square. The length of the rectangle is twice the width. The area in ,square cm, of the rectangle is :

(1) 30

(2) 50 (3) 100

(4) 25

Answer: (2) 50

Solution:

Perimeter of the square = 20 cm

Therefore length of the side of the square = 20/4 = 5cm

A rectangle has the same width as the side of the square. So, width of the rectangle is 5 cm and since it length is double the width, the length comes out to be 10 cm.

Therefore area of rectangle = l x b = 5 x 10 =50 square cm

Percentage

Percentage literally means per Cent or Per 100. In percentage, we convert the fraction, such that the denominator becomes 100. It is used to compare fraction as we have to convert all the fractions to a single denominator i.e. 100.


Let say we have to compare two fractions 7/10 and 6/9. To do this comparison either we take LCM and then make the denominator of each equal to LCM. Or we can simply convert them to denominator 100 by use of percentage and then we can compare them.

Now taking 7/10 and converting it percentage, we have (7/10) x 100 = 70 %

and converting 6/9 to percentage , we have (6/9) x 100 = 66.67 %

It is clear now that 7/10 > (6/9) .It is easier than use of LCM to compare them.

Percentage have a wide area of use from schools to managements decisions, scientific applications, data interpretation and so on.

Percentage Formula

Percentage = fraction x 100

the fraction could be any ratio, marked scored against total marks, run scored against ball faced, energy used against total energy and so on.

Converting a decimal to percentage

Percentage = Decimal x 100

Expressing a quantity as a percentage on another quantity.

Lets say we have to express x as percentage of another quantity y


then the formula is (x/y) x 100

Expressing the increase and decrease of a quantity in percentage

If |x| is the absolute increase or decrease in the value of a quantity say 'y' then the percentage increase or decrease is expressed as

absolute percentage change = { |x|/y } x 100

where the sign of x will determine the increase or the decrease in value.

Expressing percentage as a fraction

Let us say the percentage change is x %, then the corresponding fraction is

Fraction = x/100 i.e. just divide the x by 100 to get the corresponding fraction.

1. The ratio 5:4 expressed as percent equals:

(a) 12.5% (b) 40%

(c) 80% (d) 125%

Answer:(d)

Solution:


Applying the formula for percentage,

required percentage = 5/4 x 100 = 125 %

3, Half of 1 percent written as a decimal is :

(a)0.005 (b) 0.05

(c) 0.02 (d) 0.2

Answer: (a)

Solution:

Half of 1 percent is 0.5 % = 0.5/100 = .005

(to convert % to decimal you divide it by 100)

4. what is 15 percent of Rs. 34?

(a) Rs. 3.40 (b) Rs. 3.75

(c) Rs. 4.50 (d) Rs. 5.10

Answer: (d)


Solution:

Applying the formula for conversion of percentage to actual

Actual = (x %/100) x N where x is the percentage given and N is the total

Amount.

Here, x = 15 %, N= 34.

Now, put the values to get (15/100) x 34 = 0.15 x 34 = 5.1

5. 63% of 3 x (4/7) is :

(a) 2.25 (b) 2.40

(c) 2.50 (d) 2.75

Answer: (a)

Solution:

First convert the mixed fraction 3 (4/7) to fraction which is = (21 + 4)/7 =

25/7.

Now, to calculate 63% of 25/7.

we have,

( 63/100) x (25/7) = 9/4 = 2.25

6. 88% of 370 + 24% of 210 - ? = 118

(a) 256 (b) 258

(c) 268 (d) 358

Answer:(b)


Solution :

88 % of 370 = 0.88 x 370 = 325.6

24 % of 210 = 0.24 x 210 = 50.4

put these terms in the equation and let the unknown be x then we have,

325.6 + 50.4 - x = 118

376 - x = 118

x = 258

7. 860% of 50 +50% of860 =?

(a) 430 (b) 516

(c) 860 (d) 960

Answer: (c)

Solution:

860 % of 50 = 8.6 x 50 = 430

50% of 860 = 0.5 x 860 = 430

put the values in the above equation to get 430 + 430 = 860

8. 45% of 750 - 25% of 480 = ?

(a) 216 (b) 217.50

(c) 236.50 (d) 245


Answer: (b)

Solution: 45 % of 750 = 337.5

25 % of 480 = 120

Put the values in the above equations to get 337.5 - 120 = 217.5

9. 40% of 1640 + ? = 35%of 980 +150%of 850

(a) 372 (b) 842

(c) 962 (d) 1052

Answer: (c)

Solution:

40% of 1640 = 656

35% of 980 = 343

150% of 850 = 1275

Put these values and x as the unknown value in the equation to get

656 + x = 343 + 1275

x = 1618 - 656

x = 962

10. 218% of 1674 = ? x 1800

(a) 0.5 (b) 4


(c) 6 (d) none of these

Answer:(d)

Solution:

218% of 1674 = 2.18 x 1674 = 3649.32

Put this value in the above equation to get

3649.72 = x x 1800

x = 2.0274

11. 60% of 246 is the same as :

(a) 10% of 44 (b) 15% of 1056

(c) 30% of 132 (d) none of these

Answer:(d)

Solution:

60% 0f 246 = 0.6 x 246 = 147.6

or

Checking the options

(a) => 4.4

(b) => 158.4

(c)=> 39.6

12. 270 candidates appeared for an examination, of which 252 passed . the pass percentage


(a) 80% (b) 83(1/3) %

(c) 90(1/3) % (d) 93(1/3)

Answer:(d)

Solution:

Using the formula for calculating percentage we have,

(Total passed/ Total appeared ) x 100 = (252/270 ) x 100 = 93(1/3)

13 . 0.01 is what percent of 0.1

(a) (1/100) (b) (1/10)

(c) 10 (d) 100

Answer:(c)

Solution:

(0.01/0.1) x 100 = (1/10) x 100 = 10 %

14 . what percent of Rs. 2650 is Rs 1987.50

(a) 60% (b) 75%

(c) 80 % (d) 90%

Answer: (b)

Solution:

(1987.50/2650) x 100 = 75 %

15. What percent of a day is 3 hours?

(a) 12(1/2)% (b) 16(2/3)%


(c) 18(2/3)% (d) 22(1/2)%

Answer:(a)

Solution:

(3/24) x 100 = 12 (1/2) %

16. It costs Re 1 to photocopy a sheet of paper. However, 2% discount is allowed on all photocopies done after first 1000 sheets. How much will it

cost to copy 5000 sheets of paper?

(a)Rs. 3920 (b) Rs. 3980

(c)Rs. 4900 (d) Rs. 4920

Answer: (d)

Solution:

Cost of photocopying first 1000 sheets = Rs 1000

Cost of photocopying next 4000 sheets is = 4000 - 0.02 x 4000 = 3920

Therefore total cost = 4920

17. A housewife saved Rs. 2.50 in buying an item on sale.If she spents Rs.

25 for the item, approximately how much percent she saved in the transaction?

(a) 8% (b) 9%

(c)10% (d) 11%

Answer: (b)

Solution:


Total saved 2.5

Total Spent after the discount = 25

Therefore original cost of item = 27.5

And percentage saved = (2.5/27.5)x 100 = 9.09

18.The number which exceeds 16% of it by 42 is :

(a)50 (b)52

(c) 58 (d)60

Answer: (d)

Solution:

Let the number be x then

{(x - 42) /x} x 100 = 16

(x - 42) x 100 = 16x

100x - 4200 = 16x

84x = 4200

x = 50

19. If 15% of 40 is greater than 25% of number by 2. then the number is :

(a)12 (b)16

(c)24 (d)32


Answer : (b)

Solution: Making it an equation and converting the % into numbers and unknown into x

0.15 x 40 - 0.25 x X = 2

6 - 0.25X = 2

4 = 0.25 X

X = 16

20 . The sum of two numbers is 2490. if 6.5% of one number is equal to

8.5% of the other

(a)989,1501 (b)1011,1479

(c)1401, 1089 (d)1411,1079

Answer: (d)

Solution:

We will have two equations in two variables

i.e. x + y = 2490

65 x =85 y

Solving these equation , we will have y = 1079 and x = 1411

21. The difference of two numbers is 20% of the larger number . if the

smaller number is 20, then the larger number is :

(a)25 (b)45

(c)50 (d)80


Answer: (a)

Solution:

Writing the above condition into equations,

( x - y ) = 0.2 x x

( x - 20 ) = 0.2 x x

0.8 x = 20

x = 25

22. If one number is 80% of the other and 4 times the sum of their squares

is 656, then the numbers are :

(a)4,5 (b)8,10

(c)16,20 (d) None of these

Answer:(b)

Solution:

Here we can check for options or form two equations in two variable to get the answer.

We will follow the first approach

We can observe that the first condition is satisfied by the first three options.

Now check for second condition

(a) 4 ( 16 +25) = 4 (41) = 164 Not satisfied

(b) 4 ( 64 + 100) 4 (164) = 656 satisfied. Hence this is the answer.

23. A person who spends 66(2/3)% of his income is able to save Rs. 1200 per month. His monthly expenses ( in Rs. ) are :


(a) Rs. 1200 (b) Rs. 2400

(c) Rs. 3000 (d) Rs. 3200

Answer: (b)

Solution:

His saving in Percentage is 33 (1/3) % and it is equal to 1200

Therefore, x x (100/3%) = 1200

x / 3 = 1200

x = 3600

And expense = 3600 - 1200 = 2400

24. 10% of of the voters did not cast their in an election between two

candidates. 10% of the votes polled were found invalid;id.The successful candidate got 54% of the valid votes and won by a majority of 1620 votes.

The number of voters enrolled on the voters' list was:

(a) 25000 (b) 33000

(c) 35000 (d) 40000

Answer:(a)

Solution:

First we calculate the valid votes, They are divided into two candidates and the winning candidate got 54 % and losing got 46 %. the difference is 8 %

and it is equal to 1620.

i.e. 0.08 x x = 1620

x = 20250 Votes that were valid

and since 10 % votes were invalid.


0.9 x vote polled = 20250

Votes polled = 22500

And Since 10 % did not cast their votes

i.e.e0.9 x Total Voters = 22500

Total Voters = 25000

25. Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid

per week?

(a) Rs. 200 (b) Rs. 250

(c) Rs. 300 (d) None of them

Answer: (b)

Solution:

Total Paid to tailor in 1 week = 550

Let Y is Paid at the rate of P per week

Then X is paid at the rate of 1.2 P per week

And this total comes to be 550 = 1P + 1.2P

2.2P = 550

P= 250

26 . 1100 boys and 700 girls are examined in a test , 42 % of the boys and 30 % of the girls pass . the percentage of the total who failed is :


(a)58% (b) 62(2/3) %

(c)64% (d) 78%

Answer: (b)

Solution:

42% boys passed that means 58 boys failed which is equal to = 638

30% girls passed that means 70 % girls failed which is equal to = 490

Total Appeared = 1800

Total Failed % = (1128/1800) x 100 = 62 (2/3)

27 . If 20% of a= b, then b% of 20 is the same as :

(a) 40%o of a (b)5% of a

(c)20% of a (d)None of these

Answer: (a)

Solution:

Given ,

0.2 x a = b

Then ,

b% X 20 = (b/100) x 20 = {(0.2 x a)/100} x 20 = 0.4 x a or 40 percent of a

29 If p% of p is 36, then p is equal to :

(a) 15 (b)60

(c) 600 (d)3600


Answer: (b)

Solution:

Given

(p/100) x p = 36

P X p = 3600

p = 60

30. If x % of y is 100 and y% of z is 200 then find a relation between x and

z

(a)z= (x/ 2) (b) z = 2 x

(c) z= (x/4) (d)z=4 x

Answer: (a)

Solution:

First

(x/100) y = 100

(xy) = 10000 --------- (i)

And (y/100)z = 200

(yz)=20000------- (ii)

Dividing Equation (i) by equation (ii) we get

(x/z) = (1/2)

2x = z

z = (x/2)


31. A man loses 20(1/2) % of his money and after spending 80% of the

remainder, he is left with 159. How much money did he have first?

(a) 800 (b) 1000

(c) 1,200 (d) 500

Answer: (b) 1000

Solution:

Let the money was Rs x .Then after losing 20.5 % he is left with (0.795)x

Now, he spend 80% of the remainder and left with 159.

This means 20% of 0.795x is 159.

=> (0.795x)(0.20) = 159

=> 0.159x = 159

=> x = 159/0.159 = 1000

32. If 120% of a is equal to 80% of b, then (b+a)/(b-a) is equal to (a) 5 (b) 6

(c) 7 (d) 8

Answer:(a) 5

Solution:

Given that 120% of a = 80% of b,

=> 1.20 x a = (0.80)b


=> a = (8/12)b

=> a = (2/3)b

Thus, (b+a)/(b-a) = {b+(2/3)b }/{b - (2/3)b}

= (5/3)/(1/3) = 5

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