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Placement Materials & Answers-Free Download


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1. The average of all odd numbers up to 100 is:

A. 50 B. 60 C. 40 D. 55

Ans. A

Solution:

1, 2, 3… n If n is odd, the formula is (n+1)/2 th term The average of all odd numbers up to 100 is 1+3+5+7+9+…+99 Therefore here n = 99 (99 + 1) / 2 = 100/2 = 50 2. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?

A. 16C7 * 7!

B. 12C4 * 4C3 * 7!

C. 12C3 * 4C4

D. 12C4 * 4C3

Ans. B

Solution: 4 consonants out of 12 can be selected in 12C4 ways. 3 vowels can be selected in 4C3 ways. Therefore, total number of groups each containing 4 consonants and 3 vowels = 12C4 * 4C3 each group contains 7 letters, which can be arranging in 7! ways. Therefore required number of words = 124 * 4C3 * 7!

3. The ratio of boys to girls in a class is 5: 3. The class has 16 more boys than girls. How many girls are there in the class?

A. 16 B. 6 C. 24 D. 64

Ans. C

Solution:

Let the number of boys in the class be 5k. And the number of girls in the class is 3k. The class has 16 more boys. i.e., the difference between the number of boys and girls is 16 i.e., 5k – 3k = 16 2k = 16 so, k = 8. Number of girls = 3k = 2 * 8 = 24.

4. The average of 7 consecutive numbers is 33. The largest of these numbers is:

A. 30 B. 80 C. 57 D. 36

Ans. D

Solution: Basic Formula: Let the numbers x, x+1, x+2, x+3, x+4, x+5, x+6 Answer with Explanation: Given ( x +x+1+ x+2+ x+3+ x+4+ x+5+ x+6) /7 = 33 7x +21/7 = 33 7x +21 = 33×7 7x + 21 = 231 7x = 231 – 21 = 210 X = 210/7 = 30. There for the largest number is x +6 = 30+6 = 36.

5. Kiran had 85 currency notes in all, some of which were of Rs.100 denomination and the remaining of Rs.50 denomination the total amount of all these currency note was Rs.5000. How much amount did she have in the denomination of Rs.50?

A. Rs. 3000 B. Rs. 3500 C. Rs.3200 D. Rs. 3100

Ans. B

Solution:

Let the number of fifty rupee notes be x Then, number of 100 rupee notes = (85-x) 50x + 100(85-x) = 5000 x + 2(85-x) = 100 x=70 So, required amount=Rs. (50*70) = Rs. 3500

6. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:

A. 14 years B. 18 years C. 20 years D. 22 years

Ans. D

Solution: Let the son’s present age be x years. Then, man’s present age = (x + 24) years. (x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 x = 22.

7. A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?

A. 40 B. 30 C. 50 D. 60

Ans. C

Solution: Let the number of horses = x Then the number of pigeons = 80 – x. Each pigeon has 2 legs and each horse has 4 legs. Therefore, total number of legs = 4x + 2(80-x) = 260 4x + 160 – 2x = 260 2x = 100 x = 50

8. How many kgs of Basmati rice costing Rs.42/kg should a shopkeeper mix with 25 kgs of ordinary rice costing Rs.24 per kg so that he makes a profit of 25% on selling the mixture at Rs.40/kg?

A. 20 kgs B. 12.5 kgs C. 16 kgs D. 200 kgs

Ans. A

Solution: Let the amount of Basmati rice being mixed be x kgs. As the trader makes 25% profit by selling the mixture at Rs.40/kg, his cost /kg of the mixture = Rs.32/kg. i.e. (x * 42) + (25 * 24) = 32 (x + 25) 42x + 600 = 32x + 800 10x = 200 or x = 20 kgs

9. Which is greater 2^300 or 3^200?

A. 2^300 B. 3^200 C. Both are equal D. Both are equal

Ans. B

Solution: 2300 = (23)100 = 8100 3200 = (32)100 = 9100 and since 9 > 8, 3200 is greater than 2300 3^200 > 2^300.

10. The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?

A. 13.5 B. 14 C. 15 D. 12.5

Ans. D

Solution: At present the total age of the family = 5 * 20 = 100 the total age of the family at the time of the birth of the youngest member = [100-10-(10*4)] = 50 Therefore, average age of the family at the time of birth of the youngest member = 50/4 = 12.5

11. Study the following information and answer the questions based on it: A, B, C, D, E, F and G are sitting on a wall and all of them are facing east. C is on the immediate right of D. B is at an extreme end and has E as his neighbor. G is between E and F. D is sitting third from the south end. Who is sitting to the right of E?

A. A B. C C. D D. F E. None of these

Ans. E

Solution:

C is to the right of D. D is third from south. So, B will be at the extreme end from north because it should have E as its neighbour. G is between E and F. SO, the sequence is B-> E-> G-> F-> D-> C-> A-> G is sitting to the right of E.

12. Name the person who should change places with C such that he gets the third place from the north end?

A. E B. F C. C D. G

Ans. D

Solution: G should change place with C to make it third from north.

13. Which of the following pairs of people are sitting at the extreme ends?

A. AB B. AE C. CB D. FB

Ans. A

Solution: A and B are sitting at the extreme ends.

14. Who is sitting to the right of A?

A. B B. C C. D D. F E. None of these

Ans. E

Solution: We know that sitting arrangement as follows: B-> E-> G-> F-> D-> C-> A-> G And G is sitting right of A. So, answer is “None of these”

15. Immediately between which of the following pairs of people is D sitting?

A. A C B. A F C. C E D. C F

Ans. D

Solution: D is sitting between C and F.

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