math tricks

Fast Arithmetic Tips

Mental Calculations - Getting the result fast

1. Multiplication by 5It's often more convenient instead of multiplying by 5 to multiply first by 10 and then divide by 2.For example,

137×5 = 1370/2 = 685.

2. More examples and explanation 3. Division by 5

Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.For example,

1375/5 = 2750/10 = 275.

4.5. More examples and explanation 6. Division/multiplication by 4

Replace either with a repeated operation by 2.For example,

124/4 = 62/2 = 31. Also,124×4 = 248×2 = 496.

7.8. Division/multiplication by 25

Use operations with 4 instead.For example,

37×25 = 3700/4 = 1850/2 = 925.

9.10. Division/multiplication by 8

Replace either with a repeated operation by 2.For example,

124×8 = 248×4 = 496×2 = 992.

11.

12. Division/multiplication by 125Use operations with 8 instead.For example,

37×125 = 37000/8 = 18500/4 = 9250/2 = 4625.

13.14. Squaring two digit numbers.

i. You should memorize the first 25 squares:

1 2 3 4 5 6 7 8 9 10 11 12 13 14

1 4 9 16 25 36 49 64 81 100 121 144 169 196

15 16 17 18 19 20 21 22 23 24 25

225 256 289 324 361 400 441 484 529 576 625

ii. If you forgot an entry.Say, you want a square of 13. Do this: add 3 (the last digit) to 13 (the number to be squared) to get 16 = 13 + 3. Square the last digit: 3² = 9. Append the result to the sum: 169.

As another example, find 14². First, as before, add the last digit (4) to the number itself (14) to get 18 = 14 + 4. Next, again as before, square the last digit: 4² = 16. You'd like to append the result (16) to the sum (18) getting 1816 which is clearly too large, for, say, 14 < 20 so that 14² < 20² = 400. What you have to do is append 6 and carry 1 to the previous digit (8) making 14² = 196.

More examples and explanation

iii. Squares of numbers from 26 through 50.Let A be such a number. Subtract 25 from A to get x. Subtract x from 25 to get, say, a. Then A² = a² + 100x. For example, if A = 26, then x = 1 and a = 24. Hence

26² = 24² + 100 = 676.

iv.v. More examples and explanation

vi. Squares of numbers from 51 through 99.

If A is between 50 and 100, then A = 50 + x. Compute a = 50 - x. Then A² = a² + 200x. For example,

63² = 37² + 200×13 = 1369 + 2600 = 3969.


15. Squares Can Be Computed SquentiallyIn case A is a successor of a number with a known square, you find A⊃ by adding to the latter itself and then A. For example, A = 111 is a successor of a = 110 whose square is 12100. Added to this 110 and then 111 to get A²:

111² = 110² + 110 + 111 = 12100 + 221 = 12321.

16.17. More examples and explanation 18. Squares of numbers that end with 5.

A number that ends in 5 has the form A = 10a + 5, where a has one digit less than A. To find the square A² of A, append 25 to the product a×(a + 1) of a with its successor. For example, compute 115². 115 = 11×10 + 5, so that a = 11. First compute 11×(11 + 1) = 11×12 = 132 (since 3 = 1 + 2). Next, append 25 to the right of 132 to get 13225!


19. Product of 10a + b and 10a + c where b + c = 10.Similar to the squaring of numbers that end with 5:

For example, compute 113×117, where a = 11, b = 3, and c = 7. First compute 11×(11 + 1) = 11×12 = 132 (since 3 = 1 + 2). Next, append 21 (= 3×7) to the right of 132 to get 13221!


20. Product of two one-digit numbers greater than 5.This is a rule that helps remember a big part of the multiplication table. Assume you forgot the product 7×9. Do this. First find the excess of each of the multiples over 5: it's 2 for 7 (7 - 5 = 2) and 4 for 9 (9 - 5 = 4). Add them up to get 6 = 2 + 4. Now find the complements of these two numbers to 5: it's 3 for 2 (5 - 2 = 3) and 1 for 4 (5 - 4 = 1). Remember their product 3 = 3×1. Lastly, combine thus obtained two numbers (6 and 3) as 63 = 6×10 + 3.


21. Product of two 2-digit numbers.

The simplest case is when two numbers are not too far apart and their difference is even, for example, let one be 24 and the other 28. Find their average: (24 + 28)/2 = 26 and half the difference (28 - 24)/2 = 2. Subtract the squares:

28×24 = 26² - 2² = 676 - 4 = 672.

If the difference is odd add or subtract 1 from one of the numbers, compute the product as before and then subtract or add the other number. For example,

37×34 = 37×35 - 37 = (36² - 1²) - 37 = 1296 - 1 - 37 = 1258.

The same product could be found differently:

37×34 = 37×33 + 37 = (35² - 2²) + 37 = 1225 - 4 + 37 = 1258.

The ancient Babylonian used a similar approach. They calculated the sum and the difference of the two numbers, subtracted their squares and divided the result by four. For example,

33×32 = (65² - 1²)/4 = (4225 - 1)/4 = 4224/4 = 1056.


22. Product of numbers close to 100.Say, you have to multiply 94 and 98. Take their differences to 100: 100 - 94 = 6 and 100 - 98 = 2. Note that 94 - 2 = 98 - 6 so that for the next step it is not important which one you use, but you'll need the result: 92. These will be the first two digits of the product. The last two are just 2×6 = 12. Therefore, 94×98 = 9212.


23. Multiplying by 11.To multiply a 2-digit number by 11, take the sum of its digits. If it's a single digit number, just write it between the two digits. If the sum is 10 or more, do not forget to carry 1 over.For example, 34×11 = 374 since 3 + 4 = 7. 47×11 = 517 since 4 + 7 = 11.

24. Faster subtraction.Subtraction is often faster in two steps instead of one.For example,

427 - 38 = (427 - 27) - (38 - 27) = 400 - 11 = 389.

25. A generic advice might be given as "First remove what's easy, next whatever remains". Another example:

1049 - 187 = 1000 - (187 - 49) = 900 - 38 = 862.

26.27. Faster addition.

Addition is often faster in two steps instead of one.For example,

487 + 38 = (487 + 13) + (38 - 13) = 500 + 25 = 525.

28. A generic advice might be given as "First add what's easy, next whatever remains". Another example:

1049 + 187 = 1100 + (187 - 51) = 1200 + 36 = 1236.

29.30. Faster addition, #2.

It's often faster to add a digit at a time starting with higher digits. For example,

583 + 645 = 583 + 600 + 40 + 5 = 1183 + 40 + 5 = 1223 + 5 = 1228.

31.32. Multipliply, then subtract.

When multiplying by 9, multiply by 10 instead, and then subtract the other number. For example,

23×9 = 230 - 23 = 207.

33. More examples and explanation 34. The same applies to other numbers near those for which multiplication is simplified:

23×51 = 23×50 + 23= 2300/2 + 23

= 1150 + 23= 1173.

87×48 = 87×50 - 87×2

= 8700/2 - 160 - 14= 4350 - 160 - 14= 4190 - 14= 4176.

35.36. Multiplication by 9, 99, 999, etc.

There is another way to multiply fast by 9 that has an analogue for multiplication by 99, 999 and all such numbers. Let's start with the multiplication by 9.

To multiply a one digit number a by 9, first subtract 1 and form b = a - 1. Next, subtract b from 9: c = 9 - b. Then just write b and c next to each other:

9a = bc.

For example, find 6×9 (so that a = 6.) First subtract: 5 = 6 - 1. Subract the second time: 4 = 9 - 5. Lastly, form the product 6×9 = 54.

Similarly, for a 2-digit a:

bc = 100b + c = 100(a - 1) + (99 - (a - 1)) = 100a - 100 + 100 - a = 99a.

Do try the same derivation for a three digit number. As an example,

543×999 = 1000×542 + (999 - 542) = 542457.


Related materialRead more...

A Broken Calculator Has Its Uses Defensive Tips for Mental

Arithmetic Stunning Friends With Math Magic Divisibility Criteria Abacus and Its Relatives

To find out if a number is divisible by seven:Take the last digit, double it, and subtract it from the rest of the number. If the answer is more than a 2 digit number perform the above again.If the result is 0 or is divisible by 7 the original number is also divisible by 7.

Example 1 ) 2599*2= 18.25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 27933*2= 6279-6= 273

now 3*2=627-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 28412) 38733) 13934) 2877

TO FIND SQUARE OF A NUMBER BETWEEN 40 to 50Sq (44) .

1) Subtract the number from 50 getting result A.2) Square A getting result X.3) Subtract A from 25 getting result Y4) Answer is xy

EXAMPLE 1 : 4450-44=6 Sq of 6 =3625-6 = 19So answer 1936

EXAMPLE 2 : 4750-47=3Sq 0f 3 = 09

25-3= 22So answer = 2209

NOW TRY To Find Sq of 48 ,26 and 49

TO FIND SQUARE OF A 3 DIGIT NUMBER :

LET THE NUMBER BE XYZ

SQ (XYZ) is calculated like this

STEP 1. Last digit = last digit of SQ(Z)STEP 2. Second Last Digit = 2*Y*Z + any carryover from STEP 1.STEP 3. Third Last Digit 2*X*Z+ Sq(Y) + any carryover from STEP 2.STEP 4. Fourth last digit is 2*X*Y + any carryover from STEP 3.STEP 5 . In the beginning of result will be Sq(X) + any carryover from Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP 1.= 6STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP 2.= 2*4*1 +9= 17. so 7 and 1 carryoverSTEP 4. Fourth last digit is 2*4*3 + any carryover (which is 1) . = 24+1=25. So 5 and carry over 2. STEP 5 . In the beginning of result will be Sq(4) + any carryover from Step 4. So 16+2 =18.

So the result will be 185761.

If the option provided to you are such that the last two digits are different, then you need to carry out first two steps only , thus saving time. You may save up to 30 seconds on each calculations and if there are 4 such questions you save 2 minutes which may really affect UR Percentile score.

PYTHAGORAS THEROEM :

In any given exam there are about 2 to 3 questions based on pythagoras theorem. Wouldn’t it be nice that you remember some of the pythagoras triplets thus saving up to 30 seconds in each question. This saved time may be used to attempt other questions. Remember one more right question may make a lot of difference in UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one of the side less than 20 are as follows :

(3,4,5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (33, 56, 65), (16, 63, 65), (48, 55, 73), (36, 77, 85), (13, 84, 85), (39, 80, 89), and (65, 72, 97).(15,112,113), (17,144,145), (19,180,181), (20,99,101)

If you multiply the digits of the above mentioned sets by any constant you will again get a pythagoras triplet .

Example : Take the set (3,4,5).Multiply it by 2 you get (6,8,10) which is also a pythagoras triplet.Multiply it by 3 you get ( 9,12,15) which is also a pythagoras triplet.Multiply it by 4 you get (12,16,20) which is also a pythagoras triplet.You may multiply by any constant you will get a pythagoras triplet

Take another example (5,12,13)Multiply it by 5,6 and 7 and check if you get a pythagoras triplet.

TIPS FOR SMART GUESSING :

You will notice that in any case, whether it is a unique triplet or it is a derived triplet (derived by multiplying a constant to a unique triplet), all the three numbers cannot be odd.

In case of unique triplet , the hypotenuse is always odd and one of the remaining side is odd the other one is even.

Below are the first few unique triplets with first number as Odd.

3 4 55 12 137 24 259 40 4111 60 61

You will notice following trend for unique triplets with first side as odd.

Hypotenuse = (Sq(first side) +1) / 2Other side = Hypotenuse -1

Example : First side = 3 , so hypotenuse = (3*3+1)/2= 5 and other side = 5-1=4

Example 2: First side = 11so hypotenuse = (9*9+1)/2= 41 and other side = 41-1=40

Please note that the above is not true for a derived triplet for example 9,12 and 15, which has been obtained from multiplying 3 to the triplet of 3,4,5. You may check for other derived triplets.

Below are the first few unique triplets with first number as Even .

4 3 58 15 1712 35 3716 63 6520 99 101

You will notice following trend for unique triplets with first side as Even.

Hypotenuse = Sq( first side/ 2)+1Other side = Hypotenuse-2

Example 1. First side =8So hypotenuse = sq(8/2) +1= 17Other side = 17-2=15

Example 2. First side = 16So hypotenuse = Sq(16/2) +1 =65Other side = 65-2= 63

PROFIT AND LOSS : In every exam there are from one to three questions on profit and loss, stating that the cost was first increased by certain % and then decreased by certain %. How nice it would be if there was an easy way to calculate the final change in % of the cost with just one formula. It would really help you in saving time and improving UR Percentile. Here is the formula for the same :

Suppose the price is first increase by X% and then decreased by Y% , the final change % in the price is given by the following formula

Final Difference % = X- Y – XY/100.

EXAMPLE 1. : The price of T.V set is increased by 40 % of the cost price and then decreased by 25% of the new price . On selling, the profit for the dealer was Rs.1,000 . At what price was the T.V sold.

From the above mentioned formula you get : Final difference % = 40-25-(40*25/100)= 5 %.

So if 5 % = 1,000then 100 % = 20,000.C.P = 20,000S.P = 20,000+ 1000= 21,000.

EXAMPLE 2 : The price of T.V set is increased by 25 % of cost price and then decreased by 40% of the new price . On selling, the loss for the dealer was Rs.5,000 . At what price was the T.V sold.

From the above mentioned formula you get : Final difference % = 25-40-(25*45/100)= -25 %.

So if 25 % = 5,000then 100 % = 20,000.C.P = 20,000S.P = 20,000 – 5,000= 15,000.

Now find out the difference in % of a product which was :First increased by 20 % and then decreased by 10 %. First Increased by 25 % and then decrease by 20 %.First Increased by 20 % and then decrease by 25 %.First Increased by 10 % and then decrease by 10 %.First Increased by 20 % and then decrease by 15 %.

TIPS TO IMPROVE UR PERCENTILE :

HOW ABOUT SOLVING THE FOLLOWING QUESTION IN JUST 10 SECONDS

Ajay can finish work in 21 days and Blake in 42 days. If Ajay, Blake and Chandana work together they finish the work in 12 days. In how many days Blake and Chandana can finish the work together ?

(21*12 )/(24-12) = (21*12)/9= 7*4= 28 days.

NOW CAREFULLY READ THE FOLLOWING TO SOLVE THE TIME AND WORK PROBLEMS IN FEW SECONDS.

TIME AND WORK :

1. If A can finish work in X time and B can finish work in Y time then both together can finish work in (X*Y)/ (X+Y) time.

2. If A can finish work in X time and A and B together can finish work in S time then B can finish work in (XS)/(X-S) time.

3. If A can finish work in X time and B in Y time and C in Z time then they all working together will finish the work in(XYZ)/ (XY +YZ +XZ) time

4. If A can finish work in X time and B in Y time and A,B and C together in S time then :C can finish work alone in (XYS)/ (XY-SX-SY)B+C can finish in (SX)/(X-S)and A+ C can finish in (SY)/(Y-S)

Here is another shortcut

TYPE 1 : Price of a commodity is increased by 60 %. By how much % should the consumption be reduced so that the expense remain the same.

TYPE 2 : Price of a commodity is decreased by 60 %. By how much % can the consumption be increased so that the expense remain the same.

Solution : TYPE1 : (100* 60 ) / (100+60) = 37.5 %TYPE 2 : (100* 60 ) / (100-60) = 150 %

Numbers

1) 2^2n-1 is always divisible by 3

2^2n-1 = (3-1)^2n -1 = 3M +1 -1 = 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2?

ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6 Funda : if a number ‘n’ is represented as a^x * b^y * c^z …. where, {a,b,c,.. } are prime numbers then

Quote:(a) the total number of factors is (x+1)(y+1)(z+1) …. (b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)…. (c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)…. (d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * …../(x*y*…)

3) what is the highest power of 10 in 203!ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually 203/2 = 101 101/2 = 50

50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5 203/5 = 40 40/5 = 8 8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT 2002 has similar question )

ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5

5) In how many ways can 2310 be expressed as a product of 3 factors?

ANS: 2310 = 2*3*5*7*11 When a number can be expressed as a product of n distinct primes, then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways

6) In how many ways, 729 can be expressed as a difference of 2 squares?

ANS: 729 = a^2 – b^2 = (a-b)(a+b), since 729 = 3^5, total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27. So 4 ways Funda is that, all four ways of expressing can be used to findout distinct a,b values, for example take 9*81 now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.

7) How many times the digit 0 will appear from 1 to 10000

ANS: In 2 digit numbers : 9, In 3 digit numbers : 18 + 162 = 180, In 4 digit numbers : 2187 + 486 + 27 = 2700,

total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between 10 and 20 with denominator as 3?

ANS : sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66……. = 21 + 23 + …… = 300

9) if n = 1+x where x is the product of 4 consecutive number then n is, 1) an odd number, 2) is a perfect square

SOLN : (1) is clearly evident (2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number ‘n’ the reminder is also same, what is that number if the number is a odd prime number?

ANS : since both leave the same reminder, let the reminder be ‘r’, then, 987 = an + r and 643 = bn + r and thus 987 – 643 is divisible by ‘r’ and 987 – 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43 hence ‘r’ is 43

11) when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?

ANS : whenever such problem is given, we need to write the numbers in top row and rems in the bottom row like this

11 7 4 | \ \ 5 6 3

( coudnt express here properly ) now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5 that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2 75 mod 7 = 5 10 mod 11 = 10

12) a^n – b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc

EXAMPLE: 40^3-17^3-23^3 is divisble by since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.

14) There is a seller of cigerette and match boxes who sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found that there are no takers. So he reduced the price of cigarette and managed to sell all the cigerattes, realising Rs. 77.28 in all. What is the number of cigerattes?

a) 49 b) 81 c) 84 d) 92

ANS : (d) since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer

Quote:

i have given this question to make the funda clear

15) What does 100 stand for if 5 X 6 = 33 ANS : 81 SOLN : this is a number system question, 30 in decimal system is 33 in some base ‘n’, by solving we will have n as 9 and thus, 100 will be 9^2 = 81

16) In any number system 121 is a perfect square, SOLN: let the base be ‘n’ then 121 can be written as n^2 + 2*n + 1 = (n+1)^2 hence proved

17) Most of you ppl know these, anyways, just in case

Quote:(a) sum of first ‘n’ natural numbers – n*(n+1)/2 (b) sum of the squares of first ‘n’ natural numbers –

n*(n+1)*(2n+1)/6 (c) sum of the cubes of first ‘n’ natural numbers – n^2*(n+1)^2/4

(d) total number of primes between 1 and 100 – 25

18 ) See Attachment to know how to find LCM, GCF of Fractions

Quote:CAT 2002 has 2 questions on the above simple concept

19) Converting Recurring Decimals to Fractions

let the number x be 0.23434343434……..

thus 1000 x = 234.3434343434…… and 10 x = 2.3434343434……… thus, 990 x = 232 and hence, x = 232/990

20) Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n

EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with

9? ( never seen such question though ) the reminder would be (6 + 8 + 1) % 9 = 6

(b) (a*b*c) % n = (a%n * b%n * c%n) %n

EXAMPLE: What is the remainder left when 1073 * 1079 * 1087 is divided by 119 ? ( seen this

kinda questions alot ) 1073 % 119 = ? since 1190 is divisible by 119, 1073 mod 119 is 2 and thus, “the remainder left when 1073 * 1079 * 1087 is divided by 119 ” is 2*8*16 mod 119

and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that is 18

Glossary : % stands for reminder operation

find the number of zeroes in 1^1* 2^2* 3^3* 4^4………….. 98^98* 99^99* 100^100

the expresion can be rewritten as (100!)^100 / 0!* 1!* 2!* 3!….99!

Now the numerator has 2400 zeros

the formular for finding number of zeros in n! is

[n/5]+[n/5^2]…[n/5^r] where r is such that 5^r<=n<5^(r+1)

and [..] is the grestest integer function

for the numerator find the number of zeros using the above formulae..

for 0!…4! number of zeros ..0 5!…9!.number os zeros ..1 9!…14!… 2 15!..19!………………3 20!..24!………………4! now at 25! the series makes a jump to 6 25!…29!……………..6 30!…34!……………..7 this goes on and again makes a jump at 50! and then at 75!

so the number of zeros is…

5(1+2….19) + 25+ 50+ 75

the last 3 terms 25 50 and 75 are because of the jumps..

this gives numerator has 1100 zeros

now total number of zeros in expression is no of zeros in denominator – no of zeros in numerator

2400 – 1100

the Answer 1300

Quantitative Ability – POINTS TO REMEMBER

1. If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.

Eg: x3+3x2+2x+6=0 has no positive roots

2. For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.

3. For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)

4. Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.

5.

For a cubic equation ax3+bx2+cx+d=o

5. Sum of the roots = – b/a Sum of the product of the roots taken two at a time = c/a Product of the roots = -d/a

For a bi-quadratic equation ax4+bx3+cx2+dx+e = 0

5. Sum of the roots = – b/a Sum of the product of the roots taken three at a time = c/a Sum of the product of the roots taken two at a time = -d/a Product of the roots = e/a

6. If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)

7. Consider the two equations

a1x+b1y=c1

a2x+b2y=c2

Then,

If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.

If a1/a2 = b1/b2 <> c1/c2, then we have no solution.

If a1/a2 <> b1/b2, then we have a unique solution.

8. Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1

9. |a| + |b| = |a + b| if a*b>=0

else, |a| + |b| >= |a + b|

10. The equation ax2+bx+c=0 will have max. value when a<0 and min. value when a>0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a

11.

If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.

If for two numbers x*y=k (a constant), then their SUM is MINIMUM ifx=y (=root(k)). The minimum sum is then 2*root (k).

12. Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.

13. For any 2 numbers a, b where a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)

(GM)^2 = AM * HM

14. For three positive numbers a, b, c

(a + b + c) * (1/a + 1/b + 1/c)>=9

15. For any positive integer n

2<= (1 + 1/n)^n <=3

16. a2 + b2 + c2 >= ab + bc + ca

If a=b=c, then the case of equality holds good.

17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1)

18. (n!)2 > nn

19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s

20. If n is even, n(n+1)(n+2) is divisible by 24

21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + …….+ a^(n-1) ) ……Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 – 14^3)

22. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity

Note: 2 < e < 3

23. log(1+x) = x – (x^2)/2 + (x^3)/3 – (x^4)/4 ………to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]

24. (m + n)! is divisible by m! * n!

25. When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.

26. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)

27. To Find Square of a 3-Digit Number

Let the number be XYZ

Step No.

Operation to be Performed

1 Last digit = Last digit of Sq(Z)2 Second last digit = 2*Y*Z + any carryover from STEP 13 Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 24 Fourth last digit is 2*X*Y + any carryover from STEP 35 Beginning of result will be Sq(X) + any carryover from Step 4

Eg) Let us find the square of 431

Step No.

Operation to be Performed

1 Last digit = Last digit of Sq(1) = 12 Second last digit = 2*3*1 + any carryover from STEP 1=6+0=63 Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 =

17 i.e. 7 with carry over of 14 Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e.

5 with carry over of 25 Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 =

18THUS SQ(431) = 185761

If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only.

28.

The sum of first n natural numbers = n(n+1)/2

The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

The sum of cubes of first n natural numbers is (n(n+1)/2)2/4

The sum of first n even numbers= n (n+1)

The sum of first n odd numbers= n2

29. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then

the total number of factors is (x+1)(y+1)(z+1) ….

the total number of relatively prime numbers less than the number isN * (1-1/a) * (1-1/b) * (1-1/c)….

the sum of relatively prime numbers less than the number isN/2 * N * (1-1/a) * (1-1/b) * (1-1/c)….

the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * …../(x * y *…)

30.

Total no. of prime numbers between 1 and 50 is 15



31.

The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6

The number of rectangles in n*m board is given by n+1C2 * m+1C2

32. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.

33. Certain nos. to be remembered

210 = 45 = 322 = 1024

38 = 94 = 812 = 6561

7 * 11 * 13 = 1001

11 * 13 * 17 = 2431

13 * 17 * 19 = 4199

19 * 21 * 23 = 9177

19 * 23 * 29 = 12673

34. Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.

35. If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.

36. To find out the sum of 3-digit nos. formed with a set of given digits

This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)

Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.

Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)

= 25 * 24 * 11111

=6666600

37. Consider the equation x^n + y^n = z^n

As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.

38. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then,N^(p-1) – 1 is always divisible by p.

39. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.

145 = 1! + 4! + 5!

40.

Where a no. is of the form a^n – b^n, then,

The no. is always divisible by a – b

Further, the no. is divisible by a + b when n is even and not divisible bya + b when n is odd

Where a no. is of the form a^n + b^n, then,

The no. is usually not divisible by a – b

However, the no. is divisible by a + b when n is odd and not divisible bya + b when n is even

41. The relationship between base 10 and base ‘e’ in log is given bylog10N = 0.434 logeN

42. WINE and WATER formula

Let Q – volume of a vessel, q – qty of a mixture of water and wine be removed each time from a mixture, n – number of times this operation is done and A – final qty of wine in the mixture, then,

A/Q = (1-q / Q)^n

43. Pascal’s Triangle for computing Compound Interest (CI)

The traditional formula for computing CI is

CI = P*(1+R/100)^N – P

Using Pascal’s Triangle,

Number of Years (N)

——————-

1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

… 1 …. …. … … ..1

Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?

Step 1:

Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331

The coefficients – 1,3,3,1 are lifted from the Pascal’s triangle above.

Step 2:

CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)

If N =2, we would have had,

Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210

CI = 2 * 100 + 1* 10 = Rs.210

44. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:

Final Difference% = X – Y – XY/100

Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?

Applying the formula,

Final difference% = 40 – 25 – (40*25/100) = 5 %.

So if 5 % = 1,000

Then, 100 % = 20,000.

Hence, C.P = 20,000

& S.P = 20,000+ 1000= 21,000

45. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.

46.

Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)2

The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003

47.

If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.

If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.

If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time

If A can finish a work in X time and B in Y time and A, B & C together in S time then

C can finish that work alone in (XYS)/ (XY-SX-SY)

B+C can finish in (SX)/(X-S); and

A+C can finish in (SY)/(Y-S)

48. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5

49.

When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0

When an unbiased coin is tossed even no. (2n) of times, then,P (no. of heads=no. of tails) = 1-(2nCn/22n)

50. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!

Eg)1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?

Here n=10, m=3 (i.e. A, B, C)

Hence, P (A>B>C) = 1/3!

= 1/6

Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?

P (M>S) = 1/2!

= 1/2

51. CALENDAR

Calendar repeats after every 400 years.

Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.

Century has 5 odd days and leap century has 6 odd days.

In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.

January 1, 1901 was a Tuesday.

52.

For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)

For any regular polygon, the sum of interior angles =(n-2)*180 degrees

So measure of one angle is (n-2)/n *180

If any parallelogram can be inscribed in a circle, it must be a rectangle.

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).

53. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)

54.

For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is

0.5*d1*d2, where d1, d2 are the length of the diagonals.

For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), wheres=(a + b + c + d)/2

Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.

Area of a Rhombus = Product of Diagonals/2

55. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for

[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]

56. Area of a triangle

1/2*base*altitude

1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B

root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2

a*b*c/(4*R) where R is the circumradius of the triangle

r*s ,where r is the inradius of the triangle

57. In any triangle

a=b*cos C + c*cos B

b=c*cos A + a*cos C

c=a*cos B + b*cos A

a/sin A=b/sin B=c/sin C=2R, where R is the circumradius

cos C = (a^2 + b^2 – c^2)/2ab

sin 2A = 2 sin A * cos A

cos 2A = cos^2 (A) – sin^2 (A)

58. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1

59. Appollonius Theorem

In a triangle ABC, if AD is the median to side BC, then

AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2)

60.

In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.

61. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.

62. Let W be any point inside a rectangle ABCD, then,

WD2 + WB2 = WC2 + WA2

63. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))

64.

Distance between a point (x1, y1) and a line represented by the equationax + by + c=0 is given by |ax1+by1+c|/Sq(a2+b2)

Distance between 2 points (x1, y1) and (x2, y2) is given bySq((x1-x2)2+ (y1-y2)2)

65. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.

66. To find the number of factors of a given number, express the number as a product of powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (24 * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.

67. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

68. 69. The sum of first n natural numbers = n (n+1)/270. 71. The sum of squares of first n natural numbers is n (n+1)(2n+1)/672. 73. The sum of first n even numbers= n (n+1)74.

75. The sum of first n odd numbers= n^276. 77. ++++++++++++++++++++++++++++++++++++++++++

++++++++++++++++78. To find the squares of numbers near numbers of which squares are

known79. 80. To find 41^2 , Add 40+41 to 1600 =168181. 82. To find 59^2 , Subtract 60^2-(60+59) =348183. ++++++++++++++++++++++++++++++++++++++++++

++++++++++++++++84. If an equation (i:e f(x)=0 ) contains all positive co-efficient of any

powers of x , it has no positive roots then. eg: x^4+3x^2+2x+6=0 has no positive roots .

85. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

86. For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) . Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

87. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

88.For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = – b/a sum of the product of the roots taken two at a time = c/a product of the roots = -d/a ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

89. For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = – b/a sum of the product of the roots taken three at a time = c/a sum of the product of the roots taken two at a time = -d/a product of the roots = e/a +++++++++++++++++++++++++++++++++++++++++++++++++++++++++

90. If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if x=y(=k/2). The maximum product is then (k^2)/4

91. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++If for two numbers x*y=k(=constant), then their SUM is MINIMUM if x=y(=root(k)). The minimum sum is then 2*root(k) .

92. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++

93. 94. 95. 96. 97. 98.

|x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)

99. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Product of any two numbers = Product of their HCF and LCM . Hence product of two numbers = LCM of the numbers if they are prime to each other

100. 101. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++102. For any regular polygon , the sum of the exterior angles is equal

to 360 degrees hence measure of any external angle is equal to 360/n. ( where n is the number of sides)

103.

104. For any regular polygon , the sum of interior angles =(n-2)180 degrees

105.

106. So measure of one angle in107.

108. Square =90109. Pentagon =108110. Hexagon =120111. Heptagon =128.5112. Octagon =135113. Nonagon =140114. Decagon = 144115. 116.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++If any parallelogram can be inscribed in a circle , it must be a rectangle.

117. 118. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sides equal).

119. 120. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++

For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .

121. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a regular hexagon : root(3)*3/2*(side)*(side)122. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++123. For any 2 numbers a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)

(GM)^2 = AM * HM

124. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

125. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++For any positive integer n

2<= (1+1/n)^n <=3126. 127. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++

a^2+b^2+c^2 >= ab+bc+ca If a=b=c , then the equality holds in the above.

a^4+b^4+c^4+d^4 >=4abcd128. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++

(n!)^2 > n^n (! for factorial)129. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++130. 131. If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s

will be maximum if a/p = b/q = c/r = d/s .

132. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Consider the two equations

a1x+b1y=c1 a2x+b2y=c2

Then , If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations. If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to ) If a1/a2 <> b1/b2 , then we have a unique solutions for these equations.. ++++++++++++++++++++++++++++++++++++++++++

++++++++++++++++ For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is 0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.

133. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is , the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .

134. The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) .

135. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

136. If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4

137. 138. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for [(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

139. 140. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++

Area of a triangle 1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2 =a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .

In any triangle a=b*CosC + c*CosB b=c*CosA + a*CosC c=a*CosB + b*CosA

141. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If a1/b1 = a2/b2 = a3/b3 = ………….. , then each ratio is equal to (k1*a1+ k2*a2+k3*a3+…………..) / (k1*b1+ k2*b2+k3*b3+…………..) , which is also equal to (a1+a2+a3+…………./b1+b2+b3+……….)

142. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

(7)In any triangle a/SinA = b/SinB =c/SinC=2R , where R is the circumradius

143. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + …….+ a^(n-1) ) ……Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 – 14^3)

144. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity 2 < e < 3

145. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

log(1+x) = x – (x^2)/2 + (x^3)/3 – (x^4)/4 ………to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]

146. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

In a GP the product of any two terms equidistant from a term is always constant .

147. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2

148. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.

(m+n)! is divisible by m! * n! .149. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++++++++150. If a quadrilateral circumscribes a circle , the sum of a pair of

opposite sides is equal to the sum of the other pair .

151. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a

152. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f) is((a+c+e)/3 , (b+d+f)/3) .

153. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .

154. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a parallelogram = base * height155. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++

APPOLLONIUS THEOREM:

In a triangle , if AD be the median to the side BC , then AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .

156. 157. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++158. 159. 160. 161. 162. 163. 164. 165. 166. 167. 168. 169.

170. for similar cones , ratio of radii = ratio of their bases.

The HCF and LCM of two nos. are equal when they are equal .171. 172. +++++++++++++++++++++++++++++++++++++++

++++++++++++++

Volume of a pyramid = 1/3 * base area * height173. +++++++++++++++++++++++++++++++++++++++

++++++++++++++

In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.

174. 175. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++

In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.

176. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.

177. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1178. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++++++++

|a|+|b| = |a+b| if a*b>=0 else |a|+|b| >= |a+b|

179. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

2<= (1+1/n)^n <=3180. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++++++++WINE and WATER formula:

If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture

then , A/Q = (1-q/Q)^n

181. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a hexagon = root(3) * 3 * (side)^2182. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++++++++

(1+x)^n ~ (1+nx) if x<<<1183. +++++++++++++++++++++++++++++++++++++++

++++++++++++++++++++++++

Some pythagorean triplets:

3,4,5 (3^2=4+5) 5,12,13 (5^2=12+13) 7,24,25 (7^2=24+25) 8,15,17 (8^2 / 2 = 15+17 ) 9,40,41 (9^2=40+41) 11,60,61 (11^2=60+61) 12,35,37 (12^2 / 2 = 35+37) 16,63,65 (16^2 /2 = 63+65) 20,21,29(EXCEPTION) ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.

184. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height where median is the line joining the midpoints of the oblique sides.

185. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .

186. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

187. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Let W be any point inside a rectangle ABCD . Then WD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn inside this triangle touching each other then each’s radius = a/(2*(root(3)+1))

188. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Let ‘x’ be certain base in which the representation of a number is ‘abcd’ , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d

189. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

190. when you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.

191. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

192. 193. To find the squares of numbers from 50 to 59194. 195. For 5X^2 , use the formulae196. 197. (5X)^2 = 5^2 +X / X^2198. 199. Eg ; (55^2) = 25+5 /25200. =3025201. (56)^2 = 25+6/36202. =3136203. (59)^2 = 25+9/81204. =3481205. +++++++++++++++++++++++++++++++++++++++

+++++++++++++++++++206. 207. 208. many of u must b aware of this formula, but the ppl who don’t know it must b

useful for them. a+b+(ab/100)

this is used for succesive discounts types of sums. like 1999 population increses by 10% and then in 2000 by 5%

so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999

and if there is a decrease then it will be preceeded by a -ve sign and likeiwse

Question PaperUMERICAL ABILITY TEST1. Sum of smallest six digit no. and greatest five digit no. is:

a. 199999

b. 201110

c. 211110

d. 1099999

e. None of these

2. Value of 112 * 54. is :

a. 6700

b. 70000

c. 76500

d. 77200

e. None of these

3. 1399*1399

a. 1687401

b. 1901541

c. 1943211

d. 1957201

e. None of these

4. When a no. is multiplied by 13 product consist of all 5’s. The smallest such no. is

a. 41625

b. 42135

c. 42515

d. 42735

e. None of these

5. If n is –ve no. then which of the following is least.

a. 0

b. –n

c. 2n

d. n2

e. None of these

6. If -1<=x <=2 and 1<= y <= 3 then least value of 2y-3x is :

a. 0

b. -3

c. -4

d. -5

e. None of these

7. The least prime no. is :

a. 0

b. 1

c. 2

d. 3

e. None of these

8. The sum of prime no.’s b/w 60 and 75 is:

a. 199

b. 201

c. 211

d. 272

e. None of these

9. Total no. of even prime no.’s is :

a. 0

b. 1

c. 2

d. None

e. None of these

e. None of these

10. How many No’s B/w 400 and 600 begin with or end with digit 5.

a. 40

b. 100

c. 110

d. 120

11. The digit in unit’s place of product 81*82……..*89 is:

a. 0

b. 2

c. 6

d. 8

e. None of these

12. The sum of first 45 natural no’s is :

a. 1035

b. 1280

c. 2070

d. 2140

e. None of these

13. The unit’s digit in the product of 771 * 659 * 3 65 is :

a. 1

b. 2

c. 4

d. 6

e. None of these

14. Which no. is exactly divisible by 11

a. 235641

b. 245642

c. 315624

d. 415624

e. None of these

15. The largest natural no. which exactly divides the product of any 4 consecutive natural no’s is:

a. 6

b. 12

c. 24

d. 120

e. None of these

16. The diff. b/w squares of 2 consecutive odd integers is always divisible by:

a. 3

b. 6

c. 7

d. 8

e. None of these

17. The smallest no. to be added to 1000 so that 45 divides the sum exactly is :

a. 10

b. 20

c. 35

d. 80

e. None of these

18. The least no. which must be subtracted from 6709 to make it exactly divisible by 9 is :

a. 2

b. 3

c. 4

d. 5

e. None of these

19. Find the no. nearest to 99547 and exactly divisible by 687:

a. 98928

b. 99479

c. 99615

d. 100166

e. None of these

20. The least no. by which 72 must be multiplied in order to produce a multiple of 112 is

a. 6

b. 12

c. 14

d. 18

e. None of these

21. Which largest no. of 5 digits is divisible by 99:

a. 99909

b. 99981

c. 99990

d. 99999

e. None of these

22. A no. when divided by 114 leaves the remainder 21 if same no. is divided by 19 the remainder will be

a. 1

b. 2

c. 7

d. 21

e. None of these

23. The diff. b/w 2 no’s is 1365 when larger no. is divided by smaller the quotient is 6 and the remainder is 15. The smaller no. is :

a. 240

b. 270

c. 295

d. 360

e. None of these

24. The divisor is 10 times the quotient and 5 times the remainder if remainder is 46 then dividend is :

a. 4236

b. 4306

c. 4336

d. 5336

e. None of these

e. None of these

25. A four digit no. divisible by 7 becomes divisible by 3 when 10 is added to it the largest such no. is :

a. 9947

b. 9987

c. 9989

d. 9996

ANSWERS WITH EXPLANATION :

1) a

2) 1120000/16 = b

3) (1400-1)*(1400-1) = d

4) 555555/13 = d

5)c 6)c 7)c 8)d 9)b 10)c 11)a 12)a 13)c 14)d

15)c 16)d 17)a 18)c 19)c 20)c 21)c

22)d 23)b 24)d 25)b

1. A clock shows the time as 6 a.m. If the minute hand gains 2 minutes every hour, how many minutes will the clock gain by 9 p.m.?

(a) 30 minutes

(b) 25 minutes

(c) 28 minutes

(d) 34 minutes

2. Find the right number, from the given options, at the place marked by the question mark: 2, 4, 8, 32, 256, ?

(a) 4096

(b) 8192

(c) 512

(d) 1024

3. Find the number missing at question mark:

10, 11, 23, 39, 64, ?, 149

(a) 100

(b) 103

(c) 78

(d) 128

4. A super fast bus of KSRTC starting from ‘Trivandrum’ and reaches ‘Attingal’ in 45 minutes with an average speed of 40 km/hr. If the speed is increased by 10 km/hr how much time it will take to cover the same distance?

(a) 34 minutes (b) 36 minutes (c) 38 minutes (d) 40 minutes

5. The difference between 6 times and 8 times of a figure is 14. What is the figure?

(a) 12 (b) 9 (c) 7 (d) 6

6. If 92y = 36 what is 9y?

(a) 4 (b) 6 (c) 9 (d) 18

7. One fourth percent of 180 is:

(a) 4.5 (b) 0.45 (c) 0.045 (d) 45

8. A candidate appearing for an examination has to secure 40% marks to pass paper I. But he secured only 40 marks and failed by 20 marks. What is the maximum mark for paper I?

(a) 100 (b) 200 (c) 180 (d) 150

9. Find the missing number 32, 52, 74, 112, 135 ……………

(a) 16 (b) 15 (c) 17 (d) 14

10. If 250 is increased to 300, what is the percentage increase?

(a) 16.67 (b) 20 (c) 23 (d) 17

11. The ratio of 9 seconds to 10 hours is ………….

(a) 1:40 (b) 1:4000 (c) 9:10 (d) 1:400

12. A person lost 10% when he sold goods at Rs.153. For how much should he sell them to gain 20%?

(a) 204 (b) 250 (c) 240 (d) 210

13. What will be xy if 7862xy is to be divisible by 125?

(a) 25 (b) 00 (c) 75 (d) 50

14. A train of 100 meters long is running at the speed of 36 km per hour. In what time it passes a bridge of 80 meters long?

(a) 30 seconds (b) 36 seconds (c) 20 seconds (d) 18 seconds

15. If two-third of a bucket is filled in one minute then the time taken to fill the bucket completely will be …….

(a) 90 seconds (b) 70 seconds (c) 60 seconds (d) 100 seconds

16. If a quarter kilogram costs Rs. 60 then how much will cost for 150 grams?

(a) Rs. 30 (b) Rs. 24 (c) Rs. 36 (d) Rs. 40

17. If 3 men or 6 boys can do a piece of work in 20 days then how many days with 6 men and 8 boys take to do the same work?

(a) 5 (b) 8 (c) 10 (d) 6

18. Find the sum of first 100 natural numbers

(a) 5050 (b) 5005 (c) 9900 (d) 9050

19. Two poles of height 6 meters and 11 meters stand on a plane ground. If the distance between their feet is 12 meters then find the difference in the distance between their tops:

(a) 12m (b) 5m (c) 13m (d) 11m

20. How many balls of radius 4 cm can be made from a solid sphere of lead of radius 8 cm?

(a) 4 (b) 8 (c) 16 (d) 2

21. The solution to x2 +6x+9 = 0 is ……..

(a) x1 = + 3, x2 = -3 (b) x1 = 3, x2 = 3

(c) x1 = -3, x2 = -3 (d) No solution

22. What is the chance of getting a 2 or 4 in rolling a die?

(a) 2/3 (b) 1/6 (c) 1/3 (d) 1/2

23. At what rate of simple interest per annum an amount will be doubled in 10 years?

(a) 10% (b) 7.5% (c) 16% (d) 15%

24. Five times an unknown number is 5 less than 50. The unknown number

(a) 10 (b) 11 (c) 9 (d) 5

25. The acute angle between the hour hand and minute hand of a clock at 4 PM

(a) 900 (b) 1200 (c) 1500 (d) 2400

26. Water is filled in a cylindrical vessel in such a way that its volume doubles after every five minutes. If it takes 30 minutes for the vessel to be full, then the vessel will be one fourth full in

(a) 20 minute (b) 25 minutes

(c) 7 minutes 30 seconds (d) 10 minutes

27. If 10 cats can kill 10 rats in 10 minutes how long will it take 100 cats to kill 100 rats

(a) 1 minutes (b) 10 minute (c) 100 minutes (d) 10000 minutes

28. If 75 % of a number is added to 75, the result is the number itself, then the number is:

(a) 250 (b) 750 (c) 400 (d) 300

29. A school has enough food for 400 children for 12 days. How long will the food last if 80 more children join them?

(a) 6 days (b) 7 days (c) 10 days (d) 8 days

30. The sum of two consecutive numbers is 55, which is the largest number?

(a) 25 (b) 28 (c) 26 (d) 27

31. When a shop keeper sold 2/3 of an item, he got the cost price of the whole lot. What is the percentage of his profit?

(a) 33 1/8 % (b) 66 2/3 % (c) 25 % (d) 50 %

32. The perimeter of a rectangular field is 480 meters and the ratio between the length and breadth is 5:3. The area of the field is:

(a) 7,200m2 (b) 15,000m2 (c) 54,000m2 (d) 13,500m2

33. If you add 100 to a certain number, the result will be more than, if you multiply that number by 1000 what is that number?

(a) 1.5 (b) 1.0 (c) 2.5 (d) 2.0

34. A student has to secure 40 % marks to pass. He gets 150 marks and fails by 30 marks. What is the maximum marks?

(a) 400 (b) 500 (c) 475 (d) 450

35. The circumcentre of an obtuse triangle will always be in the

(a) Interior of the triangle

(b) Midpoint of the diameter

(c) Exterior of the triangle

(d) Midpoint of the side of the triangle

36. What is the degree measure of a semicircle?

(a) 360 (b) 90 (c) 270 (d) 180

37. Which among the following is the point of intersection of the medians of a triangle?

(a) Circumcentre (b) Centroid (c) Orthocenter (d) Incentre

38. The height of a cone and its base diameter are equal. If the base radius is ‘r’ what is its slant height?

(a) 3r (b) 4r (c) v5r (d) v3r

39. The radii of two spheres are in the ratio 2:3. What is the ratio of their surface areas?

(a) 4:9 (b) 2:3 (c) 8:27 (d) 4:6

40. What is the common ratio of the progression 3v2, 6, 6v2 ?

(a) 3 (b) 2 (c) v2 (d) v3

41. In class of 100 students 50 students passed in Mathematics and 70 passed in English, 5 students failed in both Mathematics and English. How many students passed in both the subjects?

(a) 25 (b) 30 (c) 50 (d) 45

42. Speed of a boat in still water is 9 km/hr. It goes 12 km down stream and comes back to the starting point in three hours.What is the speed of water in the stream?

(a) 3 km/hr (b) 4 km/hr (c) 4.5 km/hr (d) 5 km/hr

43. A student was asked to add 16 and subtract 10 from a number.He by mistake added 10 and subtracted 16. If his answer is 14 what is the correct answer?

(a) 20 (b) 26 (c) 30 (d) 32

44. Find the area of a right angled triangle whose hypotenuse is 10 cm and base 8 cm.

(a) 48 sq.cm (b) 34 sq.cm (c) 24 sq.cm (d) 42 sq.cm

45. Find the next term of the series: 3, 6, 9, 18, 27, 54, ……

(a) 81 (b) 69 (c) 63 (d) 57

46. A number consists of 20 plus 20% of its value. The number is:

(a) 20 (b) 22 (c) 25 (d) 30

47. 20% of 5 + 5% of 20 =

(a) 5 (b) 2 (c) 6 (d) 21

48. The angle between the minute hand and the hour hand of a clock, when the time is 8.30

(a) 800 (b) 600 (c) 1050 (d) 750

49. Rs. 1581 is divided among A, B and C in the ratio 10 : 15 : 6. What is the share of B?

(a) 306 (b) 765 (c) 700 (d) 510

50. The sum of four consecutive counting numbers is 154. Find the smallest number:

(a) 36 (b) 37 (c) 38 (d) 31

Answers

1 A 11 B 21 C 31 D 41 A

2 B 12 A 22 B 32 D 42 A

3 A 13 D 23 A 33 B 43 B

4 B 14 D 24 C 34 D 44 C

5 C 15 A 25 B 35 C 45 A

6 B 16 C 26 A 36 D 46 C

7 C 17 D 27 B 37 B 47 B

8 D 18 A 28 D 38 C 48 D

9 C 19 C 29 C 39 C 49 B

10 B 20 B 30 B 40 C 50 B

math tricks

Documents

explanation squares

product of numbers close

digit numbers greater

squaring of numbers

product aa

previous digit

known square

result fast1