AMCAT TRAINING

C.S.VEERARAGAVAN

SIXPHRASE

AMCAT NAC

AMCAT

QUANTITATIVE ABILITY

LOGICAL

ABILITY

NAC

AMCAT QUANTITATIVE ABILITYBASIC NUMBERS

• Decimals & Fractions Basic Algebra• Factorization Divisibility• HCF & LCM

WORD

PROBLEMS

• Application of algebra to real world Common Applications• Direct and Inverse Proportion Speed–time–distance• Profit/Loss Percentage Age Relations Mixtures

Probabilit

y

• Permutation & Combination Factorial Notation• Principles of counting P&C with repetition, restriction & counting• Probability Addition, Multiplication Law, Events Dependent, Independent, exclusive events

Power

• Fractional and Negative powers Graphical, Intuitive, Understanding • Logarithm/antilog definition and properties Word Problems Mixed with Algebra

AMCAT LOGICAL ABILITY

DEDUCTIVE

• NUMERICAL• NON – NUMERICAL

INDUCTIVE

• N• NN• ANALOGY

ABDUCTIVE

• GIVEN STATEMENT FIND UNDERLYING ASSUMPTION• CRITICAL ANALYSIS ASSUMPTIONS TESTING

AMCAT LOGICAL ABILITY

Deduction. Rule: All the beans from this bag are white.

Case: These beans are from this bag.

Result: These beans are white.

Induction. Case: These beans are [randomly selected] from this bag.

Result: These beans are white.

Rule: All the beans from this bag are white.

AbductionRule: All the beans from this bag are white.

Result: These beans [oddly] are white.

Case: These beans are from this bag.

DECIMALS & FRACTIONS

• If 2505 / 0.5 = 5010 then 25.05 / 0.5 = ?a. 5.010 b. 50.10 c. 501.0 d. None of these

• 50.10

DECIMALS & FRACTIONS

• Which pair of rational numbers lie between 1/5 and 2/5 -a. 262/1000, 275/1000 b. 362/1000, 562/1000

c. 451/1000, 552/1000 d. 121/1000,131/1000• BETWEEN 0.2 AND 0.4 • ANS A.

BASIC ALGEBRAA, B and C are three typists, who working simultaneously can type 216 pages in four hours. In one hour, C can type as many pages more than B as B can type more than A. During a period of five hours, C can type as many pages as A can during seven hours. How many pages does each of them type per hour respectively?

A. 14,17,20B. 16,18,22C. 15,17,22D. 15,18,21

Let A's typing in 7 hour be x pages.Then his 1 hour typing = pages.C's typing in 5 hour = x pagesThen his typing in 1 hour = pages.B's typing in 1 hour ===216 x=105⇒Required ratio =:: = 15 : 18 : 21

BASIC ALGEBRA

• A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which are in A.P. When 30 of the instalments are paid he dies leaving one-third of the debt unpaid. The value of the 8th instalment is:

A. Rs 35B. Rs 50C. Rs 65D. Rs 70

Let the first instalment be 'a' and the common difference between any two consecutive instalments be 'd'Using the formula for the sum of an A.P.S=[2a+(n−1)d]We have, 3600=[2a+(40−1)d] 180=2a+39d -------- (i)⇒And 2400=[2a+(30−1)d] 160=2a+29d -------- (ii)⇒On solving both the equations we get: d=2 and a=51Value of 8th instalment=51+(8−1)2 = Rs 65

BASIC ALGEBRAA student was asked to divide a number by 6 and add 12 to the quotient. He, however first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been:

A. 122B. 118C. 114D. 124

Let the number be x, then operations undertook by the student:==112

⇒ x=660Correct answer =+12= 122

FACTORIZATION• A number has exactly 3 prime factors, 125 factors of this number are

perfect squares and 27 factors of this number are perfect cubes, totally how many factors does this number have?

• N = where p1,p2,p3 are primes and their powers are x, y and z.• The total number of factors are (x+1)(y+1)(z+1).• The total number of factors which are perfect squares are ([+1) ([+1)

([+1)=125.• [ = 4 x =8, y = 8 and z = 8• The total number of factors which are perfect squares are ([+1) ([+1)

([+1)=27.• [ = 2 x = 6,7 or 8. similarly for y,z• Hence x,y,z can take the only value 8 which satisfy both the conditions.• Hence total number of factors = (8+1)(8+1)(8+1) = 729.

DIVISIBILITY

• What least number must be subtracted from 9400 to get a number exactly divisible by 65?

• a. 40 b. 20 c. 80 d. none of these• 65 = 5 X 13• Last digit should be zero or 5.• 9400–40=9360 divisible by 13 & 5• 9400–20=9380, 9400–80=9320. none of these

divisible by 13.• Hence answer is A.

HCF & LCM

• If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:A. B. C. D.

• Let numbers are x and 55-x• x(55-x) = 5*120 = 600• solving it , we get numbers as 15 and 40.• sum of reciprocals = = ..... option C

HCF & LCM

• The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:A. 276 B. 299 C. 322 D. 345

• Numbers are • 23 x 13 = 299 and • 23 x 14 = 322.

So, Larger number = (23 x 14) = 322.

HCF & LCM

• The product of three natural number is 24000 and HCF is 10. How many such a triplets exists ?

• Let the numbers be 10x,10y and 10z.• Given xyz = 24 and x,y,z are co–prime.• The possible values of x,y,z are (1,2,12),(1,3,8),

(1,4,6),(2,3,4).• only four triplets are possible. they are

10,20,120;10,30,80;10,40,60 and 20,30,40

ODD, EVEN,PRIME, RATIONAL NUMBERS

• A school purchased x computers at y dollars each for a total of $640,000. Both x and y are whole numbers and neither is a multiple of 10. Find the absolute value of the difference between x and y.

• The prime factorization of 640,000 is 21054.• 210 = 1024 and 54 = 625.• Hence the absolute difference is 399.

APPLICATION OF ALGEBRA TO REAL WORLD

• Three maths classes: x,y and z take an algebra test.• The average score in class X is 83.• The average score in class Y is 76.• The average score in class Z is 85.• The average score of all students in classes X and Y together is 79.• The average score of all students in classes Y and Z together is 81.• What is the average score for all the three classes, taken together?• = 79 => 4X = 3Y• Solving x = 3y/4• Similarly = 81 => 4Z = 5Y• Solving z = 5y/4• Average score for all the three classes taken together

(83x+76y+85z)/(x+y+z) = 81.5

ALGEBRA

• Three consecutive whole numbers are such that the square of the middle number is greater than the product of the other two by 1. Find the middle number.

• a. 6 b. 18 c. 12 d. All of these• Let the numbers be x–1, x, x + 1. • Since x2 = (x–1) * (x+1) + 1 it is true for any

values of x.

A, B, C and D play a game of cards. A says to B "If I give you 8 cards, you will have as many as C has and I shall have 3 less than what C has. Also if I take 6 cards from C, I shall have twice as many as D has". If B and D together have 50 cards, how many cards have A got?

A. 40B. 37C. 23D. 27

B+8=CA−8=C–3A+6=2DB+D=50solving these we get A= 40

BASIC ALGEBRANaveen purchased two oranges, 3 apples and 5 bananas and paid Rs 40. Had Naveen purchased 3 oranges, 5 apples and 9 bananas, He would have to pay Rs 64. Gagan demanded only 1 orange, 1 apple and 1 banana. If Naveen purchased only what was demanded by Gagan, then how much would he have paid (in Rs)?

A. 16B. 28C. 36D. 24

Let the cost of each orange, apple and banana be o, a and b respectively.Given, 2(o)+3a+5b=40 -------- (i) 3(o)+5a+9b=64-------- (ii)2(i) - (ii): o+a+b= ⇒ 16

BASIC ALGEBRAA,B,C and D each had some money. D doubled the amounts with the others. C then doubled the amounts with the others. B then doubled the amounts with the others. A then doubled the amounts with the others. At this stage, each of them has Rs 80. Find the initial amount with C (in Rs).

A. 75B. 80C. 95D. 85

Before doubling, the amounts B,C and D, each of them must have had = Rs 40.A must have then had Rs 80+ Rs 120 = Rs 200. Similarly, we can work out the amounts with each of them before the other doubled the amounts. The results are summarized below:Finally --------> A(80), B(80), C(80), D(80)Before A doubles --------> A(200), B(40), C(40), D(40)Before B doubles --------> A(100), B(180), C(20), D(20)Before C doubles --------> A(50), B(90), C(170), D(10)Before D doubles --------> A(25), B(45), C(85), D(165)

A man purchased 40 fruits; apples and oranges for Rs 17. Had he purchased as many as oranges as apples and as many apples as oranges, he would have paid Rs 15. Find the cost of one pair of an apple and an orange.

A. 70 paiseB. 60 paiseC. 80 paiseD. 1 rupee

Man buys x apples at m price and y oranges at n price,then: x+y=40 -------- (i)mx+ny=17 -------- (ii)nx + my = 15 –-––(iii)On solving the equations we get:(m+n)(x+y)=17+15 m+n= = Rs 0.80 = 80 paise

SPEED–TIME–DISTANCE

• a train travels a distance of 300km at a constant speed.• if the speed of the train is increased by 5km/hr,the

journey would have taken two hours less.• find the initial speed of the train? • we know that, s = distance = 300 km

initial speed ( s) , new speed = s + 5initial time taken ( t ) = new time taken =

• s = 25 or -30answer: 25 km / hr..

TRAINS

• Two trains start from stations Chennai and Villupuram spaced 150 km apart at the same time and speed.

• As the trains start, a bird flies from one train towards the other and on reaching the second train; it flies back to the 1st train.

• This is repeated till the trains collide. If the speed of the train is 75kmph and that of the bird is 100kmph.

• How much did the bird travel till collision?• (a) 100km (b) 120km (c) 220km (d) 175km

• distance between trains is 150 km. • one train travels 75 km / hr. • other train travels 75 km / hr. • they will meet when the total

distance traveled by both of them is equal to 150 km.

• rate * time = distance. • We know that the time that each

train travels will be the same. • We let h = the time each train

travels. • train 1 travels 75 km / hr * h

hours.train 2 travels 75km / hr * h hours.

• the total distance they travel is 150 km.

• 75 * h + 75 * h = 150 • h = 1• They will meet in 1 hour. • the bird was flying at 100 km/h. • in the same 1 hour the bird flew a

total of 100 km. • bird flies 50 km in one direction,

then travels back 50 km when the trains meet.

PROFIT/ LOSS

• The price of a computer varies depending on the additional configurations. If a DVD drive is included, the price increases by 10% and comes to Rs. 10000. What is the price of the computer excluding the DVD drive?

• A) Rs. 9191 B) Rs. 8181 C) Rs. 9000 D) Rs. 9090.90

•

PROFIT / LOSS

• A man buys a watch for Rs.135. His overhead expenses were Rs.50. If he sells the watch for Rs.254, what is his approximate profit/loss in percent?

• A) 39% profit B) 37% profit C) 37% loss D) 39% loss• C.P = 135+50 = 185.• PROFIT % = 37%

PROFIT / LOSS

• A mango vendor bought 120 mangoes for Rs. 528 and sold them at a profit of 20%. What is the selling price of each mango?

• A) Rs.4.60 B) Rs.5.28 C) Rs.5.50 D) Rs.6• C.P OF ONE MANGO = =.• S.P OF ONE MANGO = ==Rs.5.28.

PROFIT / LOSS

• After giving a discount of Rs.45 the shopkeeper still gets a profit of 20%, if the cost price is Rs.180. Find the mark up percent.

• A) 0.4 B) 0.55 C) 0.45 D) 0.48• LET THE S.P = X• C.P = 180. S.P = 180 + 36 = 216.• M.P = S.P + 45 = 216+45 = 261.• M.P% = 0.48

PROFIT

COST PRICE

DISCOUNT

MARKED PRICESELLING PRICE

PROFIT / LOSS

• On selling an article for Rs.550, a man gains 10%. What will be the selling price, if desired profit is 20%?

(a) 500 (b) 600 (c) 660 (d) 715S.P = 550. PROFIT% = 10. C.P = =500.DESIRED PROFIT = 20%. S.P = =600.

PERCENTAGE

• If x% of a is the same as y% of b, then z% of b is :

• a. (xy/z)% of a b. (yz/x)% of a• c. (xz/y)% of a d. None of these

• Therefore b = • Z% of b = • Choice C is correct.

PERCENTAGE

• If 7% of A is 42, then find 27% of A.• A) 162 B) 62 C) 600 D) 126• Ans: 162• Another method• 7 is prime number and the last digit of given

number is 2. Hence 27% of A also should have last digit is 2 and so C and D are not possible.

• B is also not possible because it is not divisible by 9(ie 27).

PERCENTAGE

• A fruit seller had some apples. He sold 40% of them and was left with 420 apples. How many apples did he have initially?

• A) 280 B) 600 C) 700 D) None of the options

PERCENTAGE

• Tom is 25 % taller than Amy. By what percent is Amy shorter than Tom?

• A) 75% B) 80% C) 25% D) 20%• =

PERCENTAGE

• If 75% of a number is 90, what is the number?

• A) 100 B) 125 C) 120 D) 67.5

• Of =

PERCENTAGE

• 4% is equal to one out of every:• A) 12 B) 20 C) 40 D) 25• Answer• 25

PERCENTAGE

• Find the percentage of prime numbers present between 1 and 100 (including both).

• A) 20 B) 25 C) 24 D) 23• 2, 3, 5, 7, 11, 13, 17, 19 =8• 23, 29, 31, 37, 41, 43, 47, 53 =8• 61, 67, 71, 73, 79, 83, 89, 97 =8• 24%

AVERAGE

• The average age of a group of nine men today is the same as it was six years ago, with one of the men having been replaced by another much younger man. The difference in ages of the man being replaced and the man replacing him is how many years?

(a) 53 (b) 54 (c) 55 (d) 56• Six years ago the average is x and now also x. the

number is same. Hence• No of persons * time elapsed = 9 * 6 = 54.

MISCELLANEOUS

RACES & GAMES

• Q1. In a kilometre race, if A gives B a 40 m start, A wins by 19 s.

• But if A gives B a 30 s start, B wins by 40 m. • Find the time taken by B to run 5,000 m?• a. 150 s b. 450 s c. 750 s d. 825 s• If A takes x seconds and B takes y seconds to run 1km,

then:• x+19= and +30=y• ⇒ y=150 sec and x=125 sec• Answer =×5000 = 750 sec

PIPES & CISTERNS• Pipe A takes 16 min to fill a tank. Pipes B and C, whose cross-sectional

circumferences are in the ratio 2:3, fill another tank twice as big as the first. If A has a cross-sectional circumference that is one-third of C, how long will it take for B and C to fill the second tank? (Assume the rate at which water flows through a unit cross-sectional area is same for all the three pipes.)

• a. 66/13 b. 40/13 c. 16/13 d. 32/13• If A has a cross – sectional circumference x and area will be proportional to x2 • cross section of C is 3x and area will be proportional to 9x2 .• Hence cross sectional circumference of B will be (2/3)*3x = 2x • so area is proportional to 4x2. • Pipe A fills in 16 min. • Pipe B and C sum of area is proportional to 13x2. will fill the same tank in

16/13 min.• The second tank is twice big in size require 2 * 16 / 13 = 32/13 min

A.M & G.M

• The arithmetic mean of 2 numbers is 34 and their geometric mean is 16. One of the numbers will be

• a. 4 b. 16 c. 18 d. 12• Let the numbers be a,b,c.• Then and =16.that is ab = 256.• a-b= = 60. Hence a = 64 and b = 4.

PERMUTATION• The letters of the word WOMAN are

written in all possible orders and these words are written out as in a dictionary ,then the rank of the word 'WOMAN' is

• a. 117 b. 120 c. 118 d. 119• 5!=120

120/5=24 • Since W is the last in rank WOMAN is in

the last 24.• O is in the last rank of the remaining 4

letters. 24/4 = 6• Therefore WOMAN is in the last 6 in rank.• M is between A and N so WOMAN is in

rank 117 or 118.• A is before N so WOMAN's rank is 117.

There are 120 possible arrangements of the letters in WOMAN. Since WOMAN is near the end, I'll list out the ones after WOMAN:

WONMAWONAMWOMNA

There are only three, so these account for the 118th, 119th, 120th words, and it places WOMAN 117th.

PERMUTATION

• A father with 8 children takes 3 children at a time to the zoological garden, as often as he can without taking the same 3 children together more than once. Then:A. number of times he will go to zoological garden is 56.B. number of times each child will go to the zoological garden is 21.C. number of times a particular child will not go to the zoological garden is 35.D. All of the above.

The number of times the father would go to the zoological garden = Number of ways of selection of 3 children taken at a time = C3 = 56Number of times a child will go to the zoological garden = Number of times he is accompanied by two other =1×7C2 =21

⇒ Number of times a child will not go to the zoological garden: =56−21= 35

PROBABILITY

• What is the probability of getting the sum 5 in two throws of the dice?

• a. b. c. d. None of these• Favourable {(1,4),(2,3),(3,2),(4,1)}. Total = 36.• Answer c.

PROBABILITY

• The probability of A solving a problem is .• The probability of B solving a problem is. • If the same problem is given to both of them,

what is the probability of neither of them solving it?

(a) (b) x (c) + (d) • D.

Permutation with restriction

Probability

• what is the probability of 2 man will be born in a same month ?

• 1/12 is the probability of 2 man will be born in a same month as first person can be born in any month and

• second person has 1/12 probability of birth in same month.

Probability

• What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '5'?

• Total two digit numbers = 90• The numbers multiple of 3 but not 5 • 12,18,21,24,27,33,36,39,42,48,51,54,57,63,66,6

9,72,78,81,84,87,93,96,99 • = 24 numbers• Hence probability = 24/90 = 4/15

Probability

• The bacteria had a probability of splitting into three and a probability to die is one third of total bacteria. Let the probability be P. Some of them survived with probability 1/5 than which among the following relation is true?(a) P=1/3+1/5*3(b) P=1/5*(1/8-3)

• here prob for dying = 1/3• prob of survival = 1/5• if they survive, they will be splited into 3 more bacteria

hence, p = (1/3) + (3* 1/5)

POWERS

• If x increases linearly, how will a-x behave (a>1) ?

• a. Increase linearly• b. Decrease linearly• c. Increase exponentially• d. Decrease exponentially• Since x increases linearly, –x decreases linearly

and a–x also decreases linearly. Ans b.

LOGARITHM

• What is the value of the following expression: 2 log105 + log104 ?a. 2 b. 2.5 c. 3 d. None of these

• Log10 25*4 = log10100 = 2.

POWERS

• What is x if xx 33

19

81

13

27

1 100

Rewrite each term as a power of three, then combine exponents:

12932

4

100

3333

3

13

3

1 xxx

Since the bases are the same, the exponents must be equal:

1293 xx

which has the solution x = –94

Power – graphical

• what is maximum no. of point of intersection of graph of 2 different cubic polynomial function with leading coefficient equal to 1 a.)2 b.)3 c.)9 d.)6

• Finding the number of solutions to p(x) = q(x) will find the number of intersections of the two graphs.

• This is also equivalent to the number of roots of p(x) - q(x) = 0. • Since p(x) and q(x) are both third degree polynomials with a

leading term of x^3, the x^3 term will drop out, leaving at most a second degree polynomial (quadratic) on the left side. By the Fundamental Theorem of Algebra, a 2nd degree polynomial can have at most 2 real solutions, leading to an answer of 2.