Mock AMCAT Quant CHANDIGARH UNIVERSITY “DEPARTMENT OF CAREER DEVELOPMENT” Quantitative Ability

CHANDIGARH UNIVERSITY “DEPARTMENT OF CAREER DEVELOPMENT” Quantitative Ability 

1. logx2(81-24x) = 1; solve for x.

A] x = 3 or -7  

 

B] x = 3 or -27            C] x = 9 or -67  

             D] x = 67 or 3 

 

Ans - B

 

Solution - logx

2

(81-24x) = 1 

81 – 24x = (x

2

1

)

 

2

x

 + 24x -81 = 0 

2

x

 + 27x – 3x – 81 = 0 

x(x+27) – 3(x+27) = 0

(x+27)(x-3) = 0

x = 3 or – 27

 

2. A detergent powder company is having a contest. Each pack of 1kg contains one of the

letters B, A, M and O. In every 20 packs, there are four Bs, five As, ten Ms and one O. What is

the probability that a pack will have a B?

A] 

1

4

 

 

 

B] 

1

2

 

 

          C] 

1

5

1

 

             D] 

 

20

 

Ans - C

 

Solution- 20 packs contain 4 B’s

The probability that 1 pack will have a B is 

4

 or 

20

1

5

. 

 

3. A jar contains 5 white, 8 red, 2 blue and 3 black balls. Find the probability that a ball drawn

at random is red or blue. 

A] 

4

9

 

 

 

B] 

5

9

                                C] 

2

7

                                        D] 

1

5

 

 

Ans- B

 

Solution- Total number of balls in the bag = 5 + 8 + 2 + 3 = 18

Number of red or blue balls in the bag = 8 + 2 = 10 

Probability = 

10

18

 = 

5

9

 

 

4. Which smallest number should be multiplied by 45 so that it will have 3 distinct prime

factors?

A] 2 

 

 

B] 3 

 

          C] 5 

             D] 7 

 

Ans- A

 

Solution- Prime factors of 45 are 3 x 3 x 5 i.e. 3 and 5.

The smallest number to be multiplied by 45 so that it may have 3 distinct prime factors is 2.

Number become 90 (45x2) and factors are 2x3x3x5.

 

5. The LCM and HCF of two numbers are 2970 and 30 respectively. Prime factors of the product

of two numbers are: 

A] 2, 3, 5, 11 

 

B] 2, 3, 7, 11 

           C] 2, 4, 5, 11 

               D] 2, 3, 7, 13 

 

Ans-  A

 

Solution- LCM = 2970; HCF = 30

Factors of LCM = 2x3x3x3x5x11 and of HCF = 2x3x5

Product of two numbers = Product of LCM and HCF of the numbers

= 2x3x3x3x5x11x2x3x5

 

6. Let P be the greatest number that will divide 522, 762 and 1482, leaving the same remainder

in each case. What is the sum of the digits in P?

A] 4 

 

 

B] 6 

 

            C] 8 

               D] 10 

 

Ans- B

 

Solution- Numbers 522, 762 and 1482 leaves same remainder when divided by a common

number.

Number = HCF of (1482-762), (762-522) and (1482 - 522) 

                 = HCF of (720, 240 and 960) = 240

Sum of the digits = 2 + 4 + 0 = 6                                           

 

7. Which number should be subtracted from 876905 so that it can be divisible by 8?

A] 1 

 

 

B] 2 

 

            C] 3 

                D] 4 

 

Ans- A

 

Solution- Divisibility test for 8             if last 3 – digits of number is divisible by 8 then the

complete number is divisible by 8.

Number = 876905 is divisible by 8 if 905 is divisible by 8.

                           113

 

       8     905 

 

              - 8 

 

 

   10 

 

 

  -8 

 

 

     25 

 

 

    -24 

 

 

      1 

 

As remainder is 1; 1 should be subtracted from given number to make it divisible by 8.

 

8. The value of a in loga 0.0196 = 2 is

A] 0.14  

 

B] 1.4  

            C] 0.7 

                D] 0.07 

 

Ans- A

 

Solution- loga (0.0196) = 2

0.0196 = a

2 

196

10000

⁄

2

 = a

 

√

196

10000

⁄

 = a 

14

a = 

 = 0.14 

100

 

9. Convert 4.3333…….. into p/q form

A] 

39

9

                                   B] 

39

10

                                C] 39                                   D] None of the above

 

Ans- A

 

Solution- Let x = 4.333…… 

(1) 

Multiply both sides by 10

10x = 43.333…… 

(2) 

Subtract (1) from (2)

9x = 39

x =

39

9

⁄ 

 

10. A, B and C are three students who attend the same tutorial classes. If the probability that on

a particular day exactly one out of A and B attends the class is 

7

; exactly one out of B and C 

10

4

7

attends is 

; exactly one out of C and A attends is 

. If the probability that all the three 

10

10

9

attend the class is 

, then find the probability that at least one attends the class. 

100

46

63

74

A] 

                               B] 

99

                               C] 

                                    D] 

 

100

100

100

100

 

Ans- D

 

Solution- Probability of at least one attending the class = 1- probability of none attending

Let probability of A, B and C attending the class be a, b, c respectively.

Thus, probability of not attending the class by A, B and C is (1 - a), (1 - b) and (1 – c)

respectively.

The probability of exactly one of A or B attending = a(1 - b) + b(1 - a) = 

7

 

10

                                                                                        = a + b – 2ab = 

7

 

 

(1) 

10

4

Similarly, for B or C = b + c – 2bc = 

  

 

 

 

(2) 

10

7

And, for C or A = c + a – 2ca = 

 

 

 

 

 

(3) 

10

99

Probability of all three attending (i.e. abc) = 

  

 

(4)  

100

7

4

From (1), (2) and (3): 2(a + b + c – ab – bc - ca) = 

7

 +

 + 

 = 

10

10

10

18

10

 = 

9

5

 

9

a + b + c – ab – bc – ca = 

 

10

Probability of none attending = (1 - a)(1 - b)(1 – c) = 1 – a – b – c + ab + bc + ca – abc 

Probability of none attending = 1 – (a + b + c – ab – bc – ca + abc) 

9

99

1

Probability of none attending = 1 – (

 + 

) = 

 

10

100

100

Probability of at least one attending = 1 – probability of none attending

Probability of at least one attending = 1 – 

1

99

 = 

 

100

100

 

11. If a = 2 and b

2

 – ab = - 1 then what is the value of log(a+b)(a

3

3

 + b

)? 

A] 1 

 

             B] – 1 

 

           C] 2 

               D] – 2  

 

Ans- C

 

Solution- a = 2; b

2

 – ab = - 1  

2

b

 – 2b + 1 = 0      

(given a = 2) 

2 

(b - 1)

= 0 

b = 1 

3

log(a+b)(a

) = ?

Put values: log(1+2)(1

3

+b

) = 2log3(3) = 2

 Alternatively;

a

3

3

+2

) = log3(1+8) = log3(9) = log3(3

2

3

3

-ab)

log(a+b)(a+b)(a

+b

2

 = (a+b)(a

2

+b

-ab)

                                      = 1 + log(3)(4 - 1) = 1 + 1 = 2

 

2

2

+b

-ab) = log(a+b)(a+b) + log(a+b)(a

2

2

+b

12. What is the largest power of 20 contained in 100!?

A] 56 

 

             B] 1 

 

           C] 24 

                 D] 2 

 

Ans- C

 

Solution- 20 = 5 x 4 = 5 x 2

2

 

Largest power of 2 in 100! = [

100

2

] + [

100

4

] + [

100

8

] + [

100

16

] + [

100

32

] + [

100

64

]  

                                                = 50 + 25 + 12 + 6 + 3 + 1 = 97

[] = only integer part is to be considered.

Highest power of 4 in 100! = 2

2

48

97

 = (2

)

*2               48 times 

Highest power of 5 in 100! = [

100

5

] + [

100

25

] = 20 + 4 = 24 times 

So, power of 20 in 100! = least of the powers = 24.

 

13. 16 men complete one – fourth of a piece of work in 12 days. What is the additional number

of men required to complete the work in 12 more days?

A] 48 

 

             B] 36 

 

           C] 30  

                 D] 16 

 

Ans- D

 

Solution- 16 men can complete one – fourth of the work in 12 days i.e. M1 = 16, D1 = 12 and

W1 = 

1

4

 

M2 = ?, D2 = 12 more days i.e. 24 days and W2 = 1

Acc to chain rule: M1 x D1 x W2 = M2 x D2 x W1 

                                16 x 12 x 1 = M2 x 24 x 

1

4

 

                                 M2 = 32 men

So, in order to complete the work in 12 more days (32 - 16) 16 more men will be needed.

 

14. An air conditioner can cool the hall in 40 minutes while another takes 45 minutes to cool

under similar conditions. If both air conditioners are switched on at same instance, then how

long will it take to cool the room?

A] About 22 minutes  

 

 

 

B] About 20 minutes    

C] About 30 minutes 

 

 

 

D] About 25 minutes 

 

Ans- A

 

Solution- 

1

 part cool by first AC in 1 minute 

40 

1

 part cool by second AC in 1 minute 

45

17

1

1

 = 

 + 

 = part cool by both AC in 1 minute. 

360

40

45

Total time = 

360

17

 = 21.2 ≈ About 22 minutes 

 

15. A vendor purchases binder clips at 12 for Rs. 60. How many clips should he sell for Rs. 60 to

earn a profit of 20%?

A] 5 

 

 

B] 8 

 

          C] 6  

  

D] 10 

 

Ans- D

 

Solution- Cost of 12 clips = Rs. 60

Profit = 20% 

Selling Price of clips = CP x 

120

100

 = 60 x 

120

100

 = Rs. 72 

Selling Price of 12 clips = Rs. 72

Rs. 72 is the selling price of 12 clips.

Rs. 60 will be the selling price of 

12

72

 x 60 = 10 clips 

 

16. A bag is full of 20 bananas and no other fruit. Rajeev draws a fruit from the bag. What is the

probability that he will draw a banana?

A] 1 

 

 

B] 0 

 

          C] 

1

2

 

                D] None of these 

 

Ans- A

 

Solution- As any drawn fruit will be a banana so the probability of drawing a banana is 1

 

17. What is the value of (0.027)

1

3

? 

A] 0.3                                 B] 0.03                        C] 0.003                     

 D] None of these 

 

Ans- A

 

Solution- (0.027)

1

3

 

3

3

√0.027

 = √

27

1000

⁄

 = 

3

10

⁄

 = 0.3 

 

18. What is the probability of getting an even sum of score in a throw of 2 dice?

A] 

1

3

                                 B] 

17

36

                                  C] 

1

4

 

   D] 

1

2

 

 

Ans- D

 

Solution- The total outcome when two dice are rolled is 36.

We need the outcomes with the even sum i.e. 2, 4, 6, 8, 10, 12.

Cases when sum is 2 = (1,1) i.e. 1

Cases when sum is 4 = (1,3), (2,2), (3,1) i.e. 3

Cases when sum is 6 = (1,5), (2, 4), (3, 3), (4, 2), (5, 1) i.e. 5

Cases when sum is 8 = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) i.e. 5

Cases when sum is 10 = (4, 6), (5, 5), (6, 4) i.e. 3

Cases when sum is 12 = (6,6) i.e. 1

Favorable cases = 1 + 3 + 5 + 5 + 3 + 1 = 18

Probability = 

18

36

 = 

1

2

 

 

19. The possibility that a student passes a subject A, B or C is 98%. The probability that he or she

passes A is 41%, B is 59%. The probability that he or she passes A and B is 15%, A and C is 

25% and B and C is 20%. The probability that he or she passes all the three subjects is 14%.

What is the probability that he or she passes subject C?

A] 0.44%  

 

B] 50%   

             C] 44%  

  

 D] 38% 

 

Ans- C

 

Solution- Given: A = 41, B = 59, AᴜBᴜC = 98, A∩B = 15, B∩C = 20, C∩A = 25, A∩B∩C = 14

We know that; 

AᴜBᴜC = A + B + C - A∩B - B∩C - C∩A + A∩B∩C

98 = 41 + 59 + C – 15 – 20 – 25 + 14 

C = 44

 

20. The rate at which a sum will become 10 times itself in 20 years at simple interest is:

A] 45%   

 

B] 50%   

             C] 47.50%  

  

 D] 49% 

 

Ans- A

 

Solution- Let Principal be Rs. X

Amount = Rs. 10x

S.I. = (A - P) Rs. 9x

S. I. =

P ∗ R ∗ T

100

⁄

 

 9x = 

x ∗ R ∗ 20

100

⁄

 

R = 45%

 

21. The largest number that always divides the difference of a three – digit number and the

number formed by reversing its digits is:

A] 1 

 

 

B] 3 

 

      C] 9 

D] 11 

 

E] 99 

 

Ans- E

 

Solution- Let the digits at hundreds, tens and unit place be x, y and z respectively.

The number will be 100x+10y+z

On reversing the positions, number will be 100z+10y+x

On subtracting, we get ((100x+10y+z) – (100z+10y+x)) = 99x-99z = 99(x-z)

The largest number dividing the difference is 99.

 

22. A shopkeeper offers ‘Buy 1, Get 1 Free’ offer on a T – shirt marked at Rs. 2, 400. If after a 

sale, the shopkeeper earns a profit of 33.33% then what is the actual price of the T- shirt? 

A] Rs. 900  

 

B] Rs. 800 

C] Rs. 1200  

D] Rs. 1, 000 

E] Rs. 1, 500 

 

Ans- A

 

Solution- The shopkeeper sold two T-shirts for Rs. 2400

Selling Price of one T-shirt = Rs. 1200

Profit = 33.33%

Cost Price = S.P. * 

100

100

100

 = 1200 * 

 = 1200 * 

 = 1200 * 

100+P%

100+33.33

133.33

3

4

 ≈ Rs. 900 

 

23. Find the number to be multiplied by (-6)

-1

, so as to get (-8)

-1

 as the product? 

A] 

3

4

                                    B] - 

3

4

                                    C] 

4

3

 

  D] - 

4

3

 

 

Ans- A

 

Let the number multiplied be x

(-6)

-1

 

x = 

-1

 * x = (-8)

−1

(−8)

−1

 = (

(−6)

−8

−6

−1

)

 = (

4

3

−1

)

 = 

3

4

 

 

24. (17

3

*17

5

2

 *(17

3

)

3

2

)*(17

10

7

∗ (17

5

)

3

35

∗ (17

6

)

1

7

)

 = 

A] 17

-1

51

7

                             B] 17

43

7

                              C] 175                   D] 1712               E] 1 

 

Ans- A

 

Solution- (17

3

*17

5

2

 *(17

3

)

3

2

)*(17

10

7

∗ (17

5

)

3

35

∗ (17

6

)

1

7

)

 

17

-1

3

*17

5

2

 *(17

3∗

3

2

10

7

∗ 17

5∗

3

35

)*(17

6∗

1

7

 

(17

∗ 17

-1

)

3+

5

2

+

9

2

)*(17

10

7

+

3

7

+

6

7

)

 

17

-1

20

2

∗ 17

19

7

∗(−1)

−

10

 = 17

 * 17

19

7

 = 

10

17

10− 

 = 17

19

7

= 17

51

7

 

19

7

17

 

25. Find the value of p which satisfies the relation log2(p – 1) + 2 = log2(3p + 1).

A] 1 

 

 

B] 3 

 

               C] 5  

 

 

D] 7 

 

Ans- B

 

Solution- log2(p – 1) + 2 = log2(3p + 1)

2 = log2(2

2

) 

Thus, log2(p – 1) + log2(2

2

) = log2(3p + 1) 

log2((p – 1)*(2

2

)) = log2(3p + 1) 

2

(p - 1)(2

) = 3p + 1 

4p – 4 = 3p + 1

4p – 3p = 1 + 4

p = 5

 

26. Namita has 4.2 kg of flour. She has been asked to make 5 cup cakes out of every 

1

2

 kg of flour. 

How many cup cakes can she bake out of the flour she has?

A] 21   

 

B] 24 

 

 

C] 30   

 

D] 42 

 

Ans- D

 

Solution- 0.5 kg flour can make 5 cup cakes i.e. one cup cake can be made using 100 gm of

flour.

4.2 kg or 4200 gm flour can make (

4200

100

⁄

) = 42 cup cakes. 

 

27. There are four prime numbers written in ascending order of magnitude. The product of the

first three is 385 and that of the last three is 1001. Find the first number.

A] 5 

 

 

B] 7 

 

 

C] 11 

 

 

D] 17 

 

Ans- A

 

Solution- Factors of 385 = 5 * 7 * 11 

Factors of 1001 = 7 * 11 *13

First prime number = 5

 

28. What is the value of the expression 4

?

A] 

23

20

-2 

 * 5

2

 * 6

-5

* 3

-46

 * 5

-10

 * 2

 * 5 

0

-5

 * 11

 * 5

10

5

  B] 

4

1

4

 

  C] 4                                  D] 5

10 

 

Ans- B

 

23

20

2

0

2

23

4

∗5

20

∗3

2

∗11

46

(2

20

)

2

∗5

∗3

∗1

2

∗5

∗3

Solution- 

∗1

1

2

5

46

10

5

 = 

6

∗5

∗2

2

∗5

5+10+5

46

 = 

∗5

(3∗2)

∗5

2

2

20

46

 = 

2

 = 

∗2

2

∗3

∗5

∗2

2

1

4

 

 

29. Which number should be subtracted from 321 so that it becomes prime?

A] 2 

 

 

B] 4 

 

 

C] 6 

 

 

D] 9 

 

Ans- B

 

Solution- To make 321 a prime number by subtracting a number from it, we have a number

of possibilities (any number between 1 to 320). In this case, we solve the question with the

help of options.

By carefully analyzing the options, only when option b i.e. 4 is subtracted from 321, we get a

prime number i.e. 317.

 

30. A person buys a mobile phone for Rs. 7, 500 and sold it for Rs. 6, 000. What is the loss

percentage?

A] 0.05  

 

B] 0.1   

 

C] 0.15   

 

D] 0.2 

 

Ans- D

 

Solution- Cost Price = Rs. 7500

Selling Price = Rs. 6000

Loss = (CP - SP) = Rs. 1500 

Loss% = 

Loss

CP

 x 100 = 

1500

7500

 x 100 = 20% or 0.2 

 

31. What will be the value of x in the expression [72

2

2

 – 28

 = 50x]? 

A] 44 

 

 

B] 46 

 

 

C] 86 

 

 

D] 88 

 

Ans- D

 

Solution- (72

2

2

 - 28

) = (72 + 28)(72 - 28) = 50x 

                   = 100 * 44 = 50x

               x = 88

 

32. What is the value of log 2205?

Given that log 5 = a, log 7 = b and log 3 = c.

A] 2b – a – 2c          B] a + 2b + 2c  

    C] 2a – b + 2c        D] a – 2b + 2c  

 E] a + 2b – 

2c

 

Ans- B

 

Solution- log(2205) = ?

factors of 2205 = 3 x 3 x 5 x 7 x 7 = 3

2

2

x5x7

 

2

2

log (3

)

                       = 2*log(3) + log(5) + 2*log(7)

                       = 2c + a + 2b

 

x5x7

2

) = log (3

) + log (5) + log(7

2

33. In how many ways can the digits 2, 3, 5, 7 and 9 be placed to form a three – digit number so

that the higher order digit is always greater than the lower order digits? (Assume digits are

all different)

A] 8 

 

 

B] 9 

 

               C] 10   

  

D] 15 

 

Ans- C

 

Solution- Digit at hundred’s place is always greater than the digit at ten’s as well as unit’s

place.

Possible numbers for hundred’s place = 9, 7 and 5

If we select ‘9’ at hundred’s place, other places can have four possibilities (2, 3, 5 and 7) 

i. e. 

4

C2 = 6  

If we select ‘7’ at hundred’s place, other places can have three possibilities (2, 3 and 5) 

i. e. 

3

C2 = 3 

If we select ‘5’ at hundred’s place, other places can have two possibilities (2 and 3) 

i. e. 

2

C2 = 1 

Total possibilities = 6 + 3 + 1 =10 ways.

 

34. A goods carriage of length 2 km, headed to Srinagar from Punjab was running at a speed of

30 Km/hr. it crosses a tunnel which is 58 km long with that speed. Find the time taken by the

goods carriage to cross the tunnel.

A] 4 hours 

               B] 3 hours 

              C] 2 hours 

                D] 1 hour 

 

Ans – C

 

Solution- Total distance covered = length of carriage + length of tunnel = 2 + 58 = 60 km

Speed = 30 km/h 

Time taken = 

Distance

Speed 

 = 

60

30

 = 2 hours 

 

35. Wagonar car was in rage two years back and it costs Rs. 5, 60, 000 then. Now, however, with

many new hi-tech cars coming into the market, the price of the car has dipped to Rs. 4, 00,

000. Find the decrease in price of the cars as a percentage of the old price.

 

A] 28%   

 

B] 28.57%  

                C] 40%  

                D] 71.42% 

 

Ans- B

 

Solution- Depreciate in price (5, 60, 000 – 4, 00, 000) = Rs. 1, 60, 000 

Percentage decrease in price = 

1,60,000

5,60,000

 x 100 = 

200

7

 = 28.57% 

 

36. A home security system provides a security codes for a door, which consist of five buttons.

Code may be generated by pressing any one button, any two, any three, any four, or all five

buttons. How many such codes are possible?

A] 32 

 

 

B] 5 

 

 

C]31 

 

 

D] 10 

 

Ans- C 

 

Solution- Where the code consists of only 1 digit. We could have 5 possible codes 

Where the code consists of 2 digits. Since all the digits have to be keyed in at once there is

no possibility of repetition. Both the digits have to be unique. So for the first digit we have 5

options and for the second we have 4 options. 5*4=20. However, we have considered the

different orders as well which we need to convert back into the unordered combinations by

dividing by 2!. 20/2=10 possible codes 

Where the code consists of 3 digits. Using the principles mentioned in statement 2 we have

5*4*3/3!=10 possible codes 

Where the code consists of 4 digits. We have 5*4*3*2/4!=5possible codes 

 Where the code consists of 5 digits. We have 5*4*3*2*1/5!=1 possible codes 

Summing up we have 5+10+10+5+1=31

OR

5C1+5C2+5C3+5C4+5C5=5+10+10+5+1=31.

 

37. 8 friends A, B, C, D, E, F, G, H are to be seated around a round table. Find the probability that

A and B never sit next to each other.

A] 2/7  

 

B] 5/7  

 

C] 3/8  

 

D] 5/8 

 

Ans- B

 

Solution- P(never seat together)=1-P(seat together) 

CIRCULAR PERMUTATION=(n-1)! =7! 

A and B can interchange their positions in 2! ways. 

the no. of favourable cases is (n-2)! 2 ! =6 ! 2 ! 

P(seat together)= 6! 2 ! / 7! 

=6! 2!/ 7 .6 ! 

=2!/7 =2/7 

P(never seat together)=1- (2/7) 

=(7-2) / 7 

=5/7

 

38. In a match awards are given to each of 11 members of the team and a trophy to the team. In

all winning team gets 2.75 kg weight awards. If the weight of the match winning trophy is

1.275kg, what is the weight of the award given to each player?

A] 200grams 

 

B] 150 grams 

 

C] 124 grams 

 

D] 134 grams. 

 

Ans- D

 

Solution- Total weight of all the awards for winning team is 2.75kg = 2750gms 

trophy weight is 1.275 kg=1275gms 

now rest weight = 2750-1275= 1475gms 

rest weight divided in 11 players is 

= 1475/11 = 134.09 gms  

 134 gms 

 

39. A trend was observed in the growth of the population in Saya islands. The population tripled

every month. Initially the population of the Saya Island was 100. What would be its