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Arithmetic (UPSC) Part- III

📆 April 17, 2025

Total Questions: 50

1. W can do 25% of a work in 30 days, X can do 1/4 of the work in 10 days, Y can do 40% of the work in 40 days and Z can do 1/3 of the work in 13 days. Who will complete the work first? [2016-II]

Correct Answer: (d) Z
Solution:W can do 25% of a work in 30 days.
⇒ W can do 1/4 of a work in 30 days
∴ W can do the complete work in 120 days.
X can do 1/4 of the work in 10 days
∴ X can do the complete work in 40 days.
Y can do 40% of the work in 40 days
⇒ Y can do 100% of the work in 100 days
∴ Y can do the complete work in 100 days.
Z can do 1/3 of the work in 13 days.
∴ Z can do the complete work in 39 days.
Hence, it is clear from above results that Z will complete
the work first.

2. The average monthly income of a person in a certain family of 5 is ₹10,000. What will be the average monthly income of a person in the same family if the income of one person increased by ₹1,20,000 per year? [2016-II]

Correct Answer: (a) ₹12,000
Solution:Average monthly income of a person in family of 5
= ₹ 10,000
∴ Monthly income of family = ₹ 10,000 × 5 = ₹ 50,000

After Increament
Increase in monthly income of one person

= 1,20,000 / 12 = ₹ 10,000
Now, the average monthly income of a person in family
= (50,000 + 10,000) / 5 = ₹ 12,000

3. In a race, a competitor has to collect 6 apples which are kept in a straight line on a track and a bucket is placed at the beginning of the track which is a starting point. The condition is that the competitor can pick only one apple at a time, run back with it and drop it in the bucket. If he has to drop all the apples in the bucket, how much total distance he has to run if the bucket is 5 meters from the first apple and all other apples are placed 3 meters apart? [2016-II]

Correct Answer: (c) 150 m
Solution:

Distances covered by the competitor to collect the apples in
the bucket are as follows:
1st apple : 2(5) = 10 m
2nd apple : 2(5 + 3) = 16 m
3rd apple : 2(5 + 2 × 3) = 22 m
4th apple : 2(5 + 3 × 3) = 28 m
5th apple : 2(5 + 4 × 3) = 34 m
6th apple : 2(5 + 5 × 3) = 40 m
Therefore, total distance covered
= 10 + 16 + 22 + 28 + 34 + 40 = 150 m

4. A round archery target of diameter 1 m is marked with four scoring regions from the centre outwards as red, blue, yellow and white. The radius of the red band is 0.20 m. The width of all the remaining bands is equal. If archers throw arrows towards the target, what is the probability, that the arrows fall in the red region of the archery target? [2016-II]

Correct Answer: (c) 0.16
Solution:

Area of red circle = π(20)²
Radius of Archry board = 1/2 m = .50 m
Area of Archry Board = π(.50)²
P(ε) = π(.20)² / π(.50)² = 0.400 / 0.2500 = 0.16

5. A person allows 10% discount for cash payment from the marked price of a toy and still he makes a 10% gain. What is the cost price of the toy which is marked ₹770? [2016-II]

Correct Answer: (c) ₹630
Solution:Marked price (M.P.) of toy = ₹ 770
Discount = 10% of MP = 10/100 × 770 = ₹ 77

Price after discount = ₹ (770 – 77) = ₹ 693
Let cost price (C.P.) of toy be ₹ x.
According to Question,
693 – x = 10/100 of x ⇒ 693 – x = 0.1 x

⇒ 1.1 x = 693 ⇒ x = 630
∴ Cost price of the toy = ₹ 630; MP = 770

SP after discount = 90/100 × 770

CP getting 10% gain = 100/110 × 90/100 × 770 = 630

6. A class starts at 11:00 am and lasts till 2:27 pm. Four periods of equal duration are held during this interval. After every period, a rest of 5 minutes is given to the students. The exact duration of each period is: [2016-II]

Correct Answer: (a) 48 minutes
Solution:Class start at 11:00 am
Class ends at 02:27 pm
So, duration of class = 3h 27min = 207 min
There will be 4 periods.

After every period, there will be a 5 min break. Here, only
3 breaks will be taken as there are only 4 periods and end
of 4th period means end of the class. 4th break is not needed
to be considered.
So, duration of the periods = 207 – 3×5 = 192 min.
∴ Time assigned to each period = 192/4 = 48 min.

7. 30 g of sugar was mixed in 180 ml water in a vessel A, 40 g of sugar was mixed in 280 ml of water in vessel B and 20 g of sugar was mixed in 100 ml of water in vessel C. The solution in vessel B is: [2016-II]

Correct Answer: (d) less sweeter than that in C
Solution:Vessel A : 30g sugar mixed in 180ml water.
So, the concentration of sugar in vessel A is 30/180 g/ml i.e.,
1/6 g/ml.

Vessel B : 40g sugar mixed in 280ml water.
So, the concentration of sugar in vessel B is 40/280 g/ml i.e.,
1/7 g/ml.

Vessel C : 20g sugar mixed in 100ml water.
So, the concentration of sugar in vessel C is 20/100 g/ml i.e.
1/5 g/ml.

More the concentration of sugar, more will be the sweetness.
Therefore, in terms of sweetness : C > A > B

8. In aid of charity, every student in a class contributes as many rupees as the number of students in that class. With the additional contribution of ₹2 by one student only, the total collection is ₹443. Then how many students are there in the class? [2016-II]

Correct Answer: (b) 21
Solution:Let the number of students in the class be x.
Total collection without considering the additional contribution by one student = 443 – 2 = ₹ 441

So, x × x = 441 ⇒ x² = 441 ⇒ x = √441 = 21

9. Anita’s mathematics test had 70 problems carrying equal marks i.e., 10 arithmetic, 30 algebra and 30 geometry. Although she answered 70% of the arithmetic, 40% of the algebra and 60% of the geometry problems correctly, she did not pass the test because she got less than 60% marks. The number of more questions she would have to answer correctly to earn a 60% passing marks is: [2016-II]

Correct Answer: (b) 5
Solution:Questions correctly answered by Anita are as follows:
Arithmetic – 70% of 10 = 7
Algebra – 40% of 30 = 12
Geometry – 60% of 30 = 18
∴ Total questions correctly answered = 7 + 12 + 18 = 37
Anita had to attempt 60% of 70 questions i.e., 42 questions
correctly to pass the test.
∴ Number of more questions required to be answered
= 42 – 37 = 5

10. In a class, there are 18 very tall boys. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class? [2016-II]

Correct Answer: (b) 12
Solution:Here, 3/4 × x = 18 ⇒ x = 24
Boys = 24 and 2/3 × y = 24 ⇒ y = 36

Total students = 36
The number of girls in the class = 36 – 24 = 12.

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