Saturday 16 September 2017

Quantitative Aptitude



1.A sum of money is divided among A, B, C and D in the ratio 3 : 5 : 7 : 11. If the share of C is Rs 1,668 more than the share of A, then what is the total amount of money of B and D together? 
(1) Rs 6,762
(2) Rs 6,672
(3) Rs 7,506
(4) Rs 6,255
(5) None of these

2.Jay spent Rs 55,475 on his birthday party, Rs 28,525 on buying home appliances and the remaining 25% of the total amount he had as cash with him. What was the total amount?
(1) Rs 1, 05,000
(2) Rs 1, 00,000
(3) Rs 1, 12,000
(4) Rs 1, 24,000
(5) None of these

3.Vinay started a business investing Rs 42,000. After 4 month Nitin joined him with a capital of Rs 57,000. At the end of the year the total profit was Rs 26,000. What is the difference between the share of profits of Vinay and Nitin?
(1) Rs 1,200
(2) Rs 1,400
(3) Rs 1,600
(4) Rs 1,800
(5) None of these

4.The profit earned after selling an article for Rs 522 is the same as the loss incurred after selling the article for Rs 378. What is the cost price (in Rs) of the article?
(1) Rs 4602
(2) Rs 4903
(3) Rs 520
(4) Rs 5505
(5) None of these

5.The average age of a class of 65 boys was 14 years, the average age of 20 of them was 14 years, and that of another 15 was 12 years. Find the average age of the remaining boys.
(1) 16 years
(2) 13 years
(3) 17 years
(4) 15 years
(5) None of these

6.The Simple interest accrued on an amount of Rs 15,500 at the end of three years is Rs 5,580. What would be the compound interest accrued on the same amount at the same rate in the same period?
(1) Rs 6726.348
(2) Rs 6276.384
(3) Rs 6267.834
(4) Rs 6627.438
(5) None of these

7.The ages of Sachin and Jatin are in the ratio 8 : 11. After 10 years the ratio of their ages will be 13 : 16. What is the difference in their ages?
(1) 16 years
(2) 3 years
(3) 8 years
(4) 6 years
(5) None of these


Directions (8-12): Study the following table carefully and answer the questions given below it. 

8.If 20% of the students playing football from school ‘A’ also play Badminton, what would be the total number of students playing Badminton from school ‘A’?
(1) 110
(2) 120
(3) 95
(4) 100
(5) None of these

9.The number of students playing Basketball from school ‘C’ is approximately what percentage of total number of students playing Basketball from school ‘E’?
(1) 75
(2) 87
(3) 94
(4) 70
(5) 81

10.What is the difference between the average number of students playing cricket from all the school and the average number of student splaying tennis from all the schools?
(1) 31
(2) 26
(3) 29
(4) 33
(5) None of these

11.The number of students playing football from school ‘D’ is what percent of the total number of students playing all the given games from that school? (Rounded off two digits after decimal)
(1) 20.61
(2) 21.59
(3) 22.60
(4) 20.59
(5) None of these

12.What is the difference between the average number of students playing all the given games from school ‘B’ and the number of students playing Badminton from that school?
(1) 72
(2) 65
(3) 78
(4) 69
(5) None of these


Answers:

1.(2): Total share of B and D together = Rs (5 + 11) x (1668/ (7 – 3))
= Rs 6672

2. (3): 75% of total amount = 55,475 + 28525
= 84000
Hence, total amount = (4/3 x 8400)
= Rs 1, 12,000

3. (5): Ratio of their shares = 42 x 12 : 57 x 8
= 21: 19
Hence, required difference in shares = (21 – 19) x (26000/(26 + 19))
= 2/45 x 26000
= Rs 1300

4. (5): Cost Price = (522 x 378)/2
= Rs 450

5. (4)

6. (2)

7. (4) : Let the Sachin and jatin age be 8x and 11x. Now
= (8x + 10)/ (11x + 10) = 13/16
= 128x + 160 = 143x + 130
=143x-128x = 160-130
= 15x = 30
= x = 30/15
= 2

8. (4) :Total no. of students playing badminton from school ‘A’ = 75 + [ (125×20)/100] = 100

9. (2) : required % =[ (100×195)/225] = 86.67%

10. (3) :  required difference = 1/5 [(250-240)+(200-210)+(225-200)+(215-130)+(200-165)]
= (10-10+25+85+35)/5
= 145/5
=29

11. (2) : required % = [175/(175+245+215+130+45)]×100
= (175/810)×100
= 21.60 approx.

12. (1) : Average  no. of students playing all the given games from schools = [1/5 (250+200+200+210+125)]
= 985/5
=197
Hence, required difference = 197- 125
= 72

No comments:

Post a Comment