Arithmetic (UPSC) Part- I

Total Questions: 50

11. The number of times in a day the Hour-hand and the Minute- hand of a clock are at right angles, is [1997]

(a) 44
(b) 48
(c) 24
(d) 12

Correct Answer: (b) 48

Solution:No. of right angles in one hour = 2
∴ No. of right angles in 24 hours = 24 × 2 = 48

12. There are 50 students admitted to a nursery class. Some students can speak only English and some can speak only Hindi. 10 students can speak both English and Hindi. If the number of students who can speak English is 21, then how many can speak Hindi, how many can speak only Hindi and how many can speak only English? [1998]

(a) 21, 11 and 29 respectively
(b) 28, 18 and 22 respectively
(c) 37, 27 and 13 respectively
(d) 39, 29 and 11 respectively

Correct Answer: (d) 39, 29 and 11 respectively

Solution:Number of students who speak only English = (Number of students who speak English – Number of students who speak both Hindi and English) = 21 – 10 = 11

Number of students who speak Hindi = (Total no of students – No of students who speak only English) = 50 – 11 = 39

∴ Number of students who speak only Hindi = (Number of Hindi speaking students – no of students who speak both languages) = 39 – 10= 29

13. An accurate clock shows the time as 3.00. After hour hand has moved 135°, the time would be [1998]

(a) 7.30
(b) 6. 30
(c) 8.00
(d) 9.30

Correct Answer: (a) 7.30

Solution:Hour hand covers an angle of 360° in 12 hours.

∴ Time taken to cover an angle of 135° =12/360 × 135 = 4.5 h

∴ Required time = 3 + 4.5 = 7.5 = 7:30

14. A man purchases two clocks A and B at a total cost of ₹ 650. He sells A with 20% profit and B at a loss of 25% and gets the same selling price for both the clocks. What are the purchasing prices of A and B respectively? [1998]

(a) ₹ 225; ₹ 425
(b) ₹ 250; ₹ 400
(c) ₹ 275; ₹ 375
(d) ₹ 300; ₹ 350

Correct Answer: (b) ₹ 250; ₹ 400

Solution:Let the cost price of clocks A and B be ‘a’ and (650-a) respectively.

Selling price for A = Selling price for B

a (1 +20/ 100 ) = (650 – a) (1 -25/100)

(120a/100) = (650 – a) (75 /100)

a = 250

Cost price for B = 650 – 250 = 400

15. If 15 pumps of equal capacity can fill a tank in 7 days, then how many extra pumps will be required to fill the tank in 5 days? [1998]

(a) 6
(b) 7
(c) 14
(d) 21

Correct Answer: (a) 6

Solution: No. of pumps required to fill a tank in 7 days = 15

∴ No. of pumps required to fill a tank in 1 day

= 15 × 7 = 105 .......(i)

Let the extra pumps required be n.

Now, no. of pumps required to fill the tank in 1 day = (n + 15) 5 .......(ii)

From (i) and (ii),

(n + 15) 5 = 105⇒ n + 15 = 21 ⇒ n = 6

16. Out of the three annual examination, each with a total of 500 marks, a student secured average marks of 45% and 55% in the first and second annual examinations. To have an overall average of 60%, how many marks does the student need to secure in the third annual examination? [1998]

(a) 450
(b) 400
(c) 350
(d) 300

Correct Answer: (b) 400

Solution:Let the average marks in the third Annual examination be x.

Total marks = (Marks in first + second + third) Annual
examination

3(60)(500/100) = (45/100)(500) + (55/100)(500) + (x/100)(500)

3(60) = 45 + 55 + x ⇒ x =80

∴ Average marks in the third annual examination

= (80/100)(500) = 400

17. In a family, a couple has a son and daughter. The age of the father is three times that of his daughter and the age of the son is half of his mother. The wife is nine years younger to her husband and the brother is seven years older than his sister. What is the age of the mother? [1998]

(a) 40 years
(b) 45 years
(c) 50 years
(d) 60 years

Correct Answer: (d) 60 years

Solution:Let the mother's age be y years.
∴ The age of father = (y + 9) years

The age of son = y/2 years

The age of daughter = (y/2 - 7) years

Now according to the given condition,

(y + 9) = 3(y/2 - 7)

⇒ y + 9 = 3y - 42

⇒ 2y + 18 = 3y - 42 ⇒ y = 60 years

18. The missing fraction in the series given below is: [1998]

4/9, 9/20, ...., 39/86

(a) 17/40
(b) 19/42
(c) 20/45
(d) 29/53

Correct Answer: (b) 19/42

Solution:Given pattern :

4 × 2 + 1/9 × 2 + 2 = 9 × 2 +1/20 × 2 + 2 = 19 × 2 + 1/42 × 2 + 2 = 39/86

19. If x + 2y = 2x + y, then x²/y² is equal to [1999]

(a) 0
(b) 1
(c) 2
(d) 4

Correct Answer: (b) 1

Solution:2x + y = x + 2y ⇒ x = y

Now, x²/y² = x²/x² = 1

20. In an office, the distribution of work hours is as shown in the following table [1999]

Members	Work hours
5	0-19
1	20-24
25	25-29
40	30-34
15	35-39
8	40-45

Consider the following inferences drawn from the table:

The average number of hours worked by a staff member is about 30
The percentage of those who worked 3.5 or more hours is less than 25
At least 5 staff members worked more than 44 hours

Which of these inferences is/are valid?

(a) 1 alone
(b) 2 alone
(c) 1 & 2
(d) 1, 2 & 3

Correct Answer: (c) 1 & 2

Solution:Average number of hours 5(17) + 1(22) + 25(27) + 40(32) + 15(37) + 8(42.5)/5 + 1 + 25 + 40 + 15 + 8

= 2957/94 = 30

Number of persons who worked 35 or more hours
= 18 + 8 = 23

∴ % of such persons = 23/94× 100 = 24.468 < 25

So, Inference 1 and 2 are valid

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