UPSC CSAT Quiz – 2021: IASbaba’s Daily CSAT Practice Test – 15th March 2021
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 Date March 15, 2021
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Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
Make the best use of the initiative. All the best!
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Question 1 of 5
1. Question
The sum of money is to be divided amongst A, B and C in the respective ratio of 3:4:5 and another sum of money is to be divided between E and F equally. If F got Rs 1,050 less than A, how much amount did B receive?
Correct
Solution (d)
Since the total sum of A, B and C is not given and also total sum of E and F is also not given, we cannot find the share of B.
Incorrect
Solution (d)
Since the total sum of A, B and C is not given and also total sum of E and F is also not given, we cannot find the share of B.

Question 2 of 5
2. Question
A watch is one minute slow at 1 p.m. on Tuesday and 2 minutes fast at 1 p.m. on Thursday. When does it show the correct time?
Correct
Solution (b)
A watch is a minute slow at 1 p.m. on Tuesday and 2 minutes fast at 1 p.m. on Thursday.
So, the watch gains 3 minutes in 48 hours.
It will show the exact time when it would have gained 1 minute from the initial state.
If it gains 3 min in 48 hours, to gain 1 min it requires
1/3 = x/48
x = 16 hours.
i.e. at 5:00 am Wednesday.
Incorrect
Solution (b)
A watch is a minute slow at 1 p.m. on Tuesday and 2 minutes fast at 1 p.m. on Thursday.
So, the watch gains 3 minutes in 48 hours.
It will show the exact time when it would have gained 1 minute from the initial state.
If it gains 3 min in 48 hours, to gain 1 min it requires
1/3 = x/48
x = 16 hours.
i.e. at 5:00 am Wednesday.

Question 3 of 5
3. Question
The average of 4 distinct prime numbers a, b, c, d is 35, where a < b < c < d. a and d are equidistant from 36 and b and c are equidistant from 34 and a, b are equidistant from 30 and c and d are equidistant from 40. The difference between a and d is
Correct
Solution (b)
Given,
The average of the four prime numbers = 35.
a + b + c + d = 35 * 4 = 140.
Since, a and d are equidistant from 36.
a + d = 72 à(1)
b + c = 68 à(2)
a + b = 60 à(3) and c + d = 80 à(4)
Using the equation (3), let us look for the prime values of a and b and the corresponding values of c and d using equation 2 and 1
Also given that a < b < c < d.
(a, b, c, d) = 29, 31, 37, 43
d – a = 43 – 29 = 14
Hence, option b is correct.
Incorrect
Solution (b)
Given,
The average of the four prime numbers = 35.
a + b + c + d = 35 * 4 = 140.
Since, a and d are equidistant from 36.
a + d = 72 à(1)
b + c = 68 à(2)
a + b = 60 à(3) and c + d = 80 à(4)
Using the equation (3), let us look for the prime values of a and b and the corresponding values of c and d using equation 2 and 1
Also given that a < b < c < d.
(a, b, c, d) = 29, 31, 37, 43
d – a = 43 – 29 = 14
Hence, option b is correct.

Question 4 of 5
4. Question
Shubha sells a book at a profit of 20%. Had she bought it at 20% less and sold it for Rs 18 less, she would have gained 25%. The cost price of the book is
Correct
Solution (a)
Let the CP of the book be Rs 100.
So, old SP = 120. New CP = 80.
Hence, new SP = 80 × 1.25 = 100.
So, if difference in SP = 120 – 100 = 20, CP = 100.
So, if difference = 18, then CP = 100/20 × 18 = Rs 90.
Hence, option a is correct.
Incorrect
Solution (a)
Let the CP of the book be Rs 100.
So, old SP = 120. New CP = 80.
Hence, new SP = 80 × 1.25 = 100.
So, if difference in SP = 120 – 100 = 20, CP = 100.
So, if difference = 18, then CP = 100/20 × 18 = Rs 90.
Hence, option a is correct.

Question 5 of 5
5. Question
A train is scheduled to cover the distance between two stations 46 km apart in one hour. If it travels 25 km at a speed of 40 km/h, find the speed for the remaining journey to complete it in the scheduled time
Correct
Solution (a)
The time taken to cover 25 km = (25/40)*60 = 37.5 m
Remaining time = 60 – 37.5 = 22.5 m and the remaining distance = 21 km
Therefore, speed = (21/22.5)*60 = 56 km/hr
Incorrect
Solution (a)
The time taken to cover 25 km = (25/40)*60 = 37.5 m
Remaining time = 60 – 37.5 = 22.5 m and the remaining distance = 21 km
Therefore, speed = (21/22.5)*60 = 56 km/hr