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Q #1
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#### Wipro Quants Question

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

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logx2(81-24x) = 1

81 – 24x = (<sub></sub>)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

Q #2
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#### Wipro Quants Question

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

Q #3
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#### Wipro Quants Question

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?
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16 men can complete one – fourth of the work in 12 days i.e. M1 = 16, D1 = 12 and W1 = 14

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 14

M2 = 32 men

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

Q #4
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#### Wipro Quants Question

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?
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Given: A = 41, B = 59, A?B?C = 98, A∩B = 15, B∩C = 20, C∩A = 25, A∩B∩C = 14

We know that;

A?B?C = A + B + C - A∩B - B∩C - C∩A + A∩B∩C

98 = 41 + 59 + C – 15 – 20 – 25 + 14

C = 44

Q #5
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#### Wipro Quants Question

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

2 = log2(22)

Thus, log2(p – 1) + log2(22) = log2(3p + 1)

log2((p – 1)*(22)) = log2(3p + 1)

(p - 1)(22) = 3p + 1

4p – 4 = 3p + 1

4p – 3p = 1 + 4

p = 5