Arithmetic (UPSC) Part- II

Total Questions: 50

41. The monthly incomes of Peter and Paul are in the ratio of 4 : 3. Their expenses are in the ratio of 3 : 2. If each saves ₹6,000 at the end of the month, their monthly incomes respectively are (in ₹) [2015-II]

(a) 24,000 and 18,000
(b) 28,000 and 21,000
(c) 32,000 and 24,000
(d) 34,000 and 26,000

Correct Answer: (a) 24,000 and 18,000

Solution:Let the Income of Peter and Paul are 4x and 3x
Let the Expenses of Peter and Paul are 3y and 2y
So, According to question
4x – 3y = 6000 ... (i)
3x – 2y = 6000 ... (ii)
Now, From equation (i) and equation (ii)
4x – 3y = 3x – 2y
∴ x = y ⇒ y = x
Now, put the value of y in equation (i), we get
⇒ x = 6000
∴ 4x = 24000
3x = 18000
Monthly Incomes of Peter and Paul are ₹ 24000 and ₹ 18000 respectively.

42. A cow costs more than 4 goats but less than 5 goats. If a goat costs between ₹600 and ₹800, which of the following is a most valid conclusion? [2015-II]

(a) A cow costs more than ₹2,500.
(b) A cow costs less than ₹3,600.
(c) A cow costs between ₹2,600 and ₹3,800.
(d) A cow costs between ₹2,400 and ₹4,000.

Correct Answer: (d) A cow costs between ₹2,400 and ₹4,000.

Solution:Let ‘g’ be the cost of goats.
4g < Cow < 5g
Now, Given a goat price = ₹ 600
2400 < Cow < 3000 ... (i)
Given, a goat price = ₹ 800
3200 < Cow < 4000 ... (ii)
Hence, it can be seen that from above equations that cow cost is between ₹ 2400 and ₹ 4000.

43. Candidates in a competitive examination consisted of 60% men and 40% women. 70% men and 75% women cleared the qualifying test and entered the final test where 80% men and 70% women were successful. Which of the following statements is correct? [2015-II]

(a) Success rate is higher for women.
(b) Overall success rate is below 50%.
(c) More men cleared the examination than women.
(d) Both (a) and (b) above are correct.

Correct Answer: (c) More men cleared the examination than women.

Solution:Let initial population be 100

It is clear that more men cleared the examination than women.

44. The sum of the ages of 5 members comprising a family, 3 years ago, was 80 years. The average age of the family today is the same as it was 3 years ago, because of an addition of a baby during the intervening period. How old is the baby? [2016-II]

(a) 6 months
(b) 1 year
(c) 2 years
(d) 2 years and 6 months

Correct Answer: (b) 1 year

Solution:Let x₁, x₂, x₃, x₄ and x₅ be the present ages of the family
comprising of 5 members
3 years ago,
(x₁ + x₂ + x₃ + x₄ + x₅) – 3 × 5 = 80
⇒ (x₁ + x₂ + x₃ + x₄ + x₅) – 15 = 80 ...(i)
Now, let the age of new born baby be x₆.
Given,
Average age of family today = Average age of family 3 years ago

((x₁ + x₂ + x₃ + x₄ + x₅) – 15) / 5
⇒ (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) / 6

80 / 5 = (15 + 80 + x₆) / 6 [from (i)] ⇒ x₆ = 1
∴ Baby’s age = 1 year

45. The total emoluments of two persons are the same, but one gets allowances to the extent of 65% of his basic pay and the other gets allowances to the extent of 80% of his basic pay. The ratio of the basic pay of the former to the basic pay of the latter is: [2016-II]

(a) 16 : 13
(b) 5 : 4
(c) 7 : 5
(d) 12 : 11

Correct Answer: (d) 12 : 11

Solution:Let the basic pay of the first person be ₹ x and that of
second person be ₹ y.
According to the question,
Emolument of both the persons are same.
So, x + 0.65x = y + 0.8y

⇒ x(1 + 0.65) = y(1 + 0.8) ⇒ x/y = 1.8 / 1.65

⇒ x/y = 12/11 or 12 : 11

46. A person is standing on the first step from the bottom of a ladder. If he has to climb 4 more steps to reach exactly the middle step, how many steps does the ladder have? [2016-II]

(a) 8
(b) 9
(c) 10
(d) 11

Correct Answer: (b) 9

Solution:A person is standing on the first step from the bottom of a ladder.
Now, he has to climb 4 more steps to reach exactly the middle step,

So, it is clear that the ladder have 9 steps.

47. AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at a height of 12 meters, broken part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B touches the ground at D which is at a distance of 5 meters from A, then the original height of the trunk is: [2016-II]

(a) 20 m
(b) 25 m
(c) 30 m
(d) 35 m

Correct Answer: (b) 25 m

Solution:

According to pythagoras theorem,
CD² = AD² + AC²
⇒ CD² = 5² + 12² ⇒ CD = √25 + 144

⇒ CD = √169 = 13 m
Now, AB = AC + CD (∵ CD is the broken part of the trunk AB touching ground at D)
⇒ AB = (12 + 13) m = 25 m

48. Ram and Shyam work on a job together for four days and complete 60% of it. Ram takes leave then and Shyam works for eight more days to complete the job. How long would Ram take to complete the entire job alone? [2016-II]

(a) 6 days
(b) 8 days
(c) 10 days
(d) 11 days

Correct Answer: (c) 10 days

Solution:Let the work done by Ram and Shyam be R and S respectively.
Given, (R + S)’s 4 days work = 60/100 of work

⇒ (R + S)’s 1 day work = (60/100) × (1/4) of work = 15/100 of work

After Ram takes leave :
S’s 8 days work = (1 – 60/100) of work = 40/100 of work

⇒ S’s 1 day work = (40/100) × (1/8) of work = 5/100 of work

So, Ram’s 1 day work = (R + S)’s 1 day work – S’s 1 day work
= (15/100 – 5/100) of work = 1/10 of work

∴ Ram will take 10 days to complete the entire job alone.

49. If R and S are different integers both divisible by 5, then which of the following is not necessarily true? [2016-II]

(a) R – S is divisible by 5
(b) R + S is divisible by 10
(c) R × S is divisible by 25
(d) R² + S² is divisible by 5

Correct Answer: (b) R + S is divisible by 10

Solution:By looking at all the options, we observe that option (b) is not necessarily true.
We know that, a number is divisible by 10 if it has 0 at the unit’s place.
(R + S) may or may not have 0 at the unit’s place. Therefore, it may or may not be divisible by 10.
Thus, it is not necessarily true.

50. How many numbers are there between 100 and 300 which either begin with or end with 2? [2016-II]

(a) 110
(b) 111
(c) 112
(d) None of the above

Correct Answer: (a) 110

Solution:From 100 to 199, there are 10 numbers ending with 2.
They are 102, 112, 122, 132, 142, 152, 162, 172, 182, 192.
And from 200 to 300, there are 100 numbers beginning with 2. They are 200, 201, 202...299.
∴ There are 110 numbers between 100 and 300 which either begin with or end with 2.

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