**Wipro Quantitative Ability** **Questions**

**Q1) There are 2 different blue toys, 3 different green toys and 4 different red toys. In how many ways one can choose 3 toys, such that there is at least one green toy and one blue toy the chosen three?**

A. 36 ways

B. 38 ways

C. 33 ways

D. 40 ways

**Answer -**Option C

**Explanation: **Here there are 2 blue toys, 3 green toys & 4 red toys

So total number of toys are 9.

Now we have to select 3 toys of which there must be at least 1 blue & 1 green toy.

Case 1.

There is 1 blue, 1 green & 1 red toy

No of selections = 2C1 × 3C1 × 4C1 = 3 × 2 × 4= 24

Case 2.

There are 2 blue & 1 green toys

No of selections = 2C2 × 3C1 × 4C0= 3 × 2 × 1 = 6

Case 3.

There are 1 blue toy & 2 green toys

No of selections = 2C1 ×3C2 × 4C0= 3 ×1 ×1 = 3

Hence the total number of ways to select three toys of which at least 1 is blue & 1 is green are 24 + 6 + 3 = 33

**Q2) There are 3 terms in an A.P. If the sum of terms is 30 and the product of extreme terms is 75, find the second term.**

A. 30

B. 25

C. 20

D. 10

**Answer -**Option D

**Explanation: **Let, three terms in A.P. are a-d, a and a+d.

Given sum of the above three terms is 30.

So, a -d + a + a + d = 30., 3a = 30. a = 10.

Since the second term in A.P. is ‘a’ in our series.

a = 10, is the second term. So, option D is the correct answer.

**Q3) The difference between the compound interest and the simple interest on a certain sum at 10% p.a. for two years is Rs. 90. What will be the value of the amount at the end of 3 years, assuming SI?**

A. Rs 11900

B. Rs 11800

C. Rs 20000

D. Rs 11700

**Answer -**Option D

**Explanation:**

**Q4) A merchant makes a profit of 20% even after allowing a discount of 20% on the marked price. What should be the marked price if the cost price of the article is Rs.300?**

A. Rs 380

B. Rs 450

C. Rs 590

D. Rs 250

**Answer -**Option B

**Explanation: **Given 20% discount, so 80% is what he is marketing his good for. And while marketing for 80%, still he got 20% profit on his Cost Price, which he bought for.

80% MP = 120%CP

80MP = 120*300

MP = 450.

**Q5) If log 27 = 1.431, then what is the value of log 9?**

A. 0.600

B. 0.369

C. 0.954

D. 0.856

**Answer -**Option C

**Explanation: **Log27 = 1.431

Log = 1.431

3 log3 = 1.431

Log3 = 0.477

Log9 = log(3)^{2}

Log9 = 2 log3

Log9 = 2*0.477

Log9 = 0.954

**Q6) Ram is five years elder to his youngest sibling Shreya. Shreya is two years younger than her brother Ritesh. Ritesh is 13 years old and is Ram’s brother. How old will Ram be in two years from now?**

A. 16

B. 17

C. 20

D. 15

E. 18

**Answer -**Option E

**Explanation: **Age of Ritesh = 13 years

Age of Shreya = Ritesh’s age – 2

Age of shreya = 13 – 2

Age of shreya = 11 years

Age of Ram = age of shreya + 5

Age of ram = 11 + 5

Age of ram = 16 years

After 2 years from now, he will be (his present age + 2)

(16 + 2) = 18 years

**Q7) How many litres of a 90% solution of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 L solution of 78% concentrated acid?**

A. 24 L

B. 22.5 L

C. 6 L

D. 17.5 L

**Answer -**Option C

**Explanation: **Let the amount of 90% solution of concentrated acid added be A.

Let the amount of 75% solution of concentrated acid added be B.

Since total volume must be 30 L,

A + B = 30 – equation 1

Amount of acid in A L of 90% concentrated acid = 0.9A

Amount of acid in B L of 75% concentrated acid = 0.75B

Amount of acid in 30 L of 90% concentrated acid = 30 x 78/100 = 23.4

Thus 0.9 A + 0.75 B = 23.4 - equation 2

By multiplying equation 1 by 0.9 we get

0.9 A + 0.9 B = 27 - equation 3

Subtracting equation 2 from 3

We get 0.15 B = 3.6

B = 24

From equation 1

A + B = 30, A = 30 – B, A = 30 – 24, A = 6

Thus 6 litres of a 90% solution of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30L solution of 78% concentrated acid.

**Q8) What is the unit digit of the following sum: 1+ 2**

^{2}+ 3^{3}+ 4^{4}+ 5^{5}+ 6^{6}?A. 0

B. 4

C. 7

D. 9

**Answer -**Option D

**Explanation: **

1^{1 }= 1

2^{2 }= 4

3^{3 }= 27, unit digit is 7.

4^{4 }= 6, when 4 to the even power, the unit digit is always = 6.

5^{5 }= 5, 5 to the any power, unit digit is 5.

6^{6 }= 6, 6 to the any power, unit digit is 6.

So, 1 + 2^{2 } + 3^{3 } + 4^{4 }+ 5^{5 }+ 6^{6 }= 1 + 4 + 7 + 6 + 5 + 6 = unit digit is = 9

**Q9) A pie has to be divided amongst few kids. Raj gets 2/7th portion of the pie and Sheela gets 5/8th portion of the pie. Who amongst the two gets lesser share?**

A. Raj

B. Sheela

C. Both get equal share

D. Cannot be determined

**Answer -**Option A

**Explanation: **Raj gets 2/7th portion ⇒ 28.56 %

Sheela gets 5/8th portion ⇒ 62.5 %

So clearly Raj gets a lesser share.

**Q10) 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

**Answer -**Option A

**Explanation: **We have been given that LCM and HCF of two numbers are 2970 and 30.

Now, we know that the product of LCM and HCF of two numbers is equal to the product of the numbers.

Therefore, in order to find the prime factors of product of the numbers, we can find the prime factors of the product of their LCM and HCF.

2970 * 30 = 89100

The Prime Factorization of 89100 is:

2 x 2 x 3 x 3 x 3 x 3 x 5 x 5 x 11

Therefore, the prime factors are 2,3,5 and 11.

**Q11) In an examination, a candidate is required to answer 5 questions in all, from 2 sections having 5 questions each. What are the total number of ways in which a candidate can select the questions, provided that at least two questions are to be attempted from each section?**

A. 200

B. 20

C. 100

D. 10

**Answer -**Option A

**Explanation: **The student has to attempt 2 questions from each section, so either he can do 2 questions from each section and one question from section 1 or do 2 questions from each section and one question from section 2.

The number of ways of choosing questions will be:

5C3*5C2 + 5C2*5C3

= 2(5C2*5C3)

= 2* 10* 10 = 200

The required number of possibilities are 200.

**Q12) If from a deck of 52 cards,4 cards are selected and one card of it should be spade and another should be heart, in how many ways can these cards be selected?**

A. 13C1 * 13C1 * 26C2

B. ^{52}C_{4}

C. 26 * ^{50}C_{2}

^{13}C4

**Answer -**Option A

**Explanation: **4 cards to be selected. One from spades (13 cards) and other from hearts (13 cards) and remaining 2 from 26 cards of clubs and diamonds.

So, the number of ways = 13C1 * 13C1 * 26C2. Option A.

**Q13) A quiz has one multiple choice question with answer choices A, B, and C, and two true/false questions. What is the probability of answering all three questions correctly by guessing?**

A. 1/5

B. 1/4

C. 1/3

D. 1/12

**Answer -**Option D

**Explanation: **probability of getting mcq answer correct = 1/3

probability of getting one true/false answer correct = 1/2

probability of getting all 3 answers correct = 1/3 * 1/2 *1/2

=1/12

**Q14) Which number should be multiplied by 43 so that it will have 3 prime factors?**

A. 2

B. 3

C. 6

D. 8

**Answer -**Option C

**Explanation: **Every natural number can be categorised into two parts of prime and composite numbers. We first discuss the characteristics of the prime and composite numbers. The numbers which have only two factors as 1 and that number itself are called prime numbers.

The prime numbers are only divisible by 1 and that number. For example, the numbers 5, 7, 11 are the prime numbers. The factorisation of prime numbers is the multiplication of 1 and that number. 43 can be written as 43=1×4343=1×43. The number of factorisations for prime numbers is always one but for composite numbers it is at least 2.

Now we multiply 2 and 3 with 43 to get 43×2×3=25843×2×3=258.

It has 3 unique prime numbers which are 2, 3, 43.

Therefore, the number is 6.

**Q15) In a poultry farm, 50 hens give 200 eggs in 2 days. In how many days will 20 hens give 400 eggs?**

A. 15

B. 10

C. 5

D. 8**Answer -**Option B

**Explanation: **Let the no. of hen be H1, no. of eggs E1, no. days D1 separately

Thus,

H1=50, E1=200, D1= 2

We are supposed to find what amount of time will 20 hens require to give 400 eggs

Allow x to be the no. of days

So,

H2= 20, E2= 400, D2=x

Therefore, the formula here will be

(H1 * D1)/E1 = (H2 * D2)/ E2

(50 * 2)/200 = (20 * x)/400

(50 * 2) * 400/(200 * 20) = x

Hence, x = 10

So, we can conclude that 20 hens will take 10 days to lay 400 eggs.

Consequently 20 hens require 10 days to give 400 eggs.