Arithmetic (UPSC) Part- II

Total Questions: 50

21. A contract on construction job specifies a penalty for delay in completion of the work beyond a certain date is as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day etc., the penalty for each succeeding day being ₹50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days? [2011-II]

(a) ₹4950
(b) ₹4250
(c) ₹3600
(d) ₹650

Correct Answer: (b) ₹4250

Solution:Given series is 200,250,300, .....n

Here, $a$ = first term = 200

$d$ = common difference = 50

$n$ = 10

Since, given series is A.P

$∴$ Sum of A.P = $n [2 a + (n - 1) d]$

$= 10/2 [2 \times 200 + 9 \times 50] = 5 \times 850 = 4250$

Required penalty = sum of the series = ₹ 4250.

22. Consider the following figure and answer the item that follows: [2011-II]

A square is divided into four rectangles as shown above. The lengths of the sides of rectangles are natural numbers. The areas of two rectangles are indicated in the figure. What is the length of each side of the square?

(a) 10
(b) 11
(c) 15
(d) Cannot be determined as the given data are insufficient

Correct Answer: (b) 11

Solution:

Area of ABIH = 15 $⟹$ AH = 3 and AB = 5

Now, area of EFID = 48 $⟹$ possibilities of length and breadth are (1,48), (2,24), (3,16), (4,12), (6,8)

Since, BC > AB $⟹$ BC > 5

Now, BC = ID $⟹$ BC = ID = 6.

Hence, length of square = 11

23. A person has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively [2011-II]

(a) 15 and 35
(b) 35 and 15
(c) 30 and 20
(d) 25 and 25

Correct Answer: (d) 25 and 25

Solution:

Let the no. of ₹ 1 coins = $x$

and the no. of ₹ 2 coins = $y$

According to the question:

$x + y = 50$ ...(i)

$x .1 + 2. y = 75$ ...(ii)

on solving (i) and (ii), we get

$2 y - y = 75 - 50$

$⟹ y = 25$

put value of $y$ in equation (i), we get

$x = 50 - y$

$= 50 - 25 = 25$

Hence, number of ₹ 1 and ₹ 2 coins are 25 and 25 respectively.

24. Three persons start walking together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? [2011-II]

(a) 25 m 20 cm
(b) 50 m 40 cm
(c) 75 m 60 cm
(d) 100 m 80 cm

Correct Answer: (a) 25 m 20 cm

Solution:Required minimum distance = L.C.M of 40, 42 and 45

L.C.M = 2 × 2 × 2 × 5 × 3 × 3 × 7 = 2520
Required distance = 2520 ÷ 100 = 25 m 20 cm

25. A student on her first 3 tests received an average score of N points. If she exceeds her previous average score by 20 points on her fourth test, then what is the average score for the first 4 tests? [2011-II]

(a) N + 20
(b) N + 10
(c) N + 4
(d) N + 5

Correct Answer: (d) N + 5

Solution:Average score for the first
4 tests = (3N + N + 20) / 4 = (4N + 20) / 4 = N + 5

26. In a group of persons, 70% of the persons are male and 30% of the persons are married. If two-sevenths of the males are married, what fraction of the females is single? [2011-II]

(a) 2/7
(b) 1/3
(c) 3/7
(d) 2/3

Correct Answer: (d) 2/3

Solution:Let the no. of total persons = 100
No. of male = 70% = 100 × (70 / 100) = 70
∴ No. of female = 100 – 70 = 30
Total married person = 30% = (30 / 100) × 100 = 30

∴ Total unmarried person = 100 – 30 = 70
Married male = 70 × (2 / 7) = 20

∴ Married female = 30 – 20 = 10
Now, unmarried female = 30 – 10 = 20

∴ Required fraction = 20 / 30 = 2 / 3

27. In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1 : 2. Based on the information, the total number of coins in the collection now becomes [2013-II]

(a) 90
(b) 80
(c) 60
(d) 50

Correct Answer: (a) 90

Solution:Let gold coin be represented by ‘G’.
Let non-gold coin be represented by ‘N’.

Initial ratio = G/N = 1/3 ...(i)

After adding 10 gold coins, the new ratio will be
(G + 10)/N = 1/2 ...(ii)

Putting the value of G = N/3 in equation (ii).

(N/3 + 10)/N = 1/2

On solving, we get N = 60 and G = 30
Presently, the total number of coins in the collection
= 10 + 20 + 60 = 90

28. A gardener has 1000 plants. He wants to plant them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of plants that he needs more for this purpose? [2013-II]

(a) 14
(b) 24
(c) 32
(d) 34

Correct Answer: (b) 24

Solution:If the number of rows and columns are to be equal, then the total number of trees would represent a perfect square. Since, 1000 is not a perfect square, we need to check for a perfect square above and nearest to 1000. It's 1024, which is square of 32. So he needs 24 more trees to get 1024.

29. A sum of ₹700 has to be used to give seven cash prizes to the students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, what is the least value of the prize? [2013-II]

(a) ₹30
(b) ₹40
(c) ₹60
(d) ₹80

Correct Answer: (b) ₹40

Solution:Let the least value be x. Then the next value is x + 20
and the next value is x + 40 and so on.
According to question

(x) + (x + 20) + (x + 40) + (x + 60) + (x + 80) + (x + 100)
+ (x + 120) = 700
⇒ 7x + (20 + 40 + 60 + 80 + 100 + 120) = 700
⇒ 7x + 20 (1 + 2 + 3 + 4 + 5 + 6) = 700
⇒ 7x + 20 × 21 = 700
⇒ 7x + 420 = 700
⇒ 7(x + 60) = 7(100) ⇒ x = 40

30. Out of 120 applications for a post, 70 are male and 80 have a driver's license. What is the ratio between the minimum to maximum number of males having driver’s license? [2013-II]

(a) 1 to 2
(b) 2 to 3
(c) 3 to 7
(d) 5 to 7

Correct Answer: (c) 3 to 7

Solution:Out of 120, there are 70 males.
∴ 120 – 70 = 50 females
For a maximum, all 70 male shall have driver’s license.
For a minimum, if all 50 females posses driver’s license, then the remaining 30 licenses, shall be possessed by males.
So minimum male driver’s licenses = 30.
Min. to max. = 30 to 70 = 3 to 7

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