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 - logx2(81-24x) = 1
81 – 24x = (x2)1
x2 + 24x -81 = 0
x2 + 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?
1 1 1 1
A] B] C] D]
4 2 5 20
Ans - C
Solution- 20 packs contain 4 B’s
4 1
The probability that 1 pack will have a B is 20 or 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.
4 5 2 1
A] B] C] D]
9 9 7 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
10 5
Probability = 18 = 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 = a2
196⁄ 2
10000 = a
√196⁄10000 = a
14
a = 100 = 0.14
,9. Convert 4.3333…….. into p/q form
39 39
A] B] C] 39 D] None of the above
9 10
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
7
a particular day exactly one out of A and B attends the class is 10; exactly one out of B and C
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 100
, then find the probability that at least one attends the class.
46 63 74 99
A] 100
B] 100 C] 100 D] 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.
7
The probability of exactly one of A or B attending = a(1 - b) + b(1 - a) = 10
7
= a + b – 2ab = 10 (1)
4
Similarly, for B or C = b + c – 2bc = (2)
10
7
And, for C or A = c + a – 2ca = 10
(3)
99
Probability of all three attending (i.e. abc) = 100
(4)
7 4 7 18 9
From (1), (2) and (3): 2(a + b + c – ab – bc - ca) = 10 +10 + 10
= 10
=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
1 99
Probability of at least one attending = 1 – 100 = 100
11. If a = 2 and b2 – ab = - 1 then what is the value of log(a+b)(a3 + b3)?
A] 1 B] – 1 C] 2 D] – 2
Ans- C
Solution- a = 2; b2 – ab = - 1
b2 – 2b + 1 = 0 (given a = 2)
(b - 1)2 = 0
b=1
, log(a+b)(a3+b3) = ?
Put values: log(1+2)(13+23) = log3(1+8) = log3(9) = log3(32) = 2log3(3) = 2
Alternatively;
a3+b3 = (a+b)(a2+b2-ab)
log(a+b)(a+b)(a2+b2-ab) = log(a+b)(a+b) + log(a+b)(a2+b2-ab)
= 1 + log(3)(4 - 1) = 1 + 1 = 2
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 22
100 100 100 100 100 100
Largest power of 2 in 100! = [ 2 ] + [ 4 ] + [ 8 ] + [ 16 ] + [ 32 ] + [ 64 ]
= 50 + 25 + 12 + 6 + 3 + 1 = 97
[] = only integer part is to be considered.
Highest power of 4 in 100! = 297 = (22 )48*2 48 times
100 100
Highest power of 5 in 100! = [ 5 ] + [ 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
1
W1 =
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
1
16 x 12 x 1 = M2 x 24 x 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
1
Solution- 40 part cool by first AC in 1 minute
1
45
part cool by second AC in 1 minute
17 1 1
= + = part cool by both AC in 1 minute.
360 40 45
360
Total time = = 21.2 ≈ About 22 minutes
17
