UPSC CSAT Quiz – 2021: IASbaba’s Daily CSAT Practice Test – 18th March 2021
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 Date March 18, 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.
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Question 1 of 5
1. Question
Let k be a positive integer such that k+4 is divisible by 7. Then the smallest positive integer n, greater than 2 such that k+2n is divisible by 7 equals
Correct
Solution (a)
Let’s say k+4 = 7m
k = 7m4
Now, for k+2n or 7m+ (2n4) is also multiple of 7.
Or, 2n4 should be a multiple of 7
So, 2n4 = 7p
2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer >2)
Incorrect
Solution (a)
Let’s say k+4 = 7m
k = 7m4
Now, for k+2n or 7m+ (2n4) is also multiple of 7.
Or, 2n4 should be a multiple of 7
So, 2n4 = 7p
2n = 7p+4
For p=2; n=9 (p cannot be 1 as n is an integer >2)

Question 2 of 5
2. Question
Every day Ramya’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
Correct
Solution (d)
We know that Ramya’s husband drives at a uniform speed to and from his residence
If he saved 10 min overall travel time, he should have driven 5 min less towards the railway station and 5 min less while driving towards the residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Ramya, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 pm.
So, Ramya must have walked for 55 minutes from 5 pm.
Incorrect
Solution (d)
We know that Ramya’s husband drives at a uniform speed to and from his residence
If he saved 10 min overall travel time, he should have driven 5 min less towards the railway station and 5 min less while driving towards the residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Ramya, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 pm.
So, Ramya must have walked for 55 minutes from 5 pm.

Question 3 of 5
3. Question
A garment shop which sold the same marked price shirts was running an offer which said if one buys 3 shirts they would get every fourth shirt at a discounted price of 100 Rs only. Dev availed the offer and left shop with 20 shirts after paying 20,000 Rs. What was the marked price of the shirt?
Correct
Solution (a)
Dev bought 20 shirts and he paid 20,000 Rs, which means he must have bought 15 shirts at the marked price and 5 shirts at the discounted price of 100 Rs
Let the marked price of 1 shirt be X
According to the question,
15*X+100*5 = 20,000
15X = 19,500
X = 1,300 Rs
Incorrect
Solution (a)
Dev bought 20 shirts and he paid 20,000 Rs, which means he must have bought 15 shirts at the marked price and 5 shirts at the discounted price of 100 Rs
Let the marked price of 1 shirt be X
According to the question,
15*X+100*5 = 20,000
15X = 19,500
X = 1,300 Rs

Question 4 of 5
4. Question
Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?
Correct
Solution (c)
Out of 8 consonants and 5 vowels selecting 4 consonants and 3 vowels can be done in 8C4*5C3 = [8*7*6*5/4*3*2]*[5*2*3] = 700
Internal arrangement of 7 letters = 7!
Thus, the total number of ways = 7!*700 = 352800
Hence, option c is the correct answer.
Incorrect
Solution (c)
Out of 8 consonants and 5 vowels selecting 4 consonants and 3 vowels can be done in 8C4*5C3 = [8*7*6*5/4*3*2]*[5*2*3] = 700
Internal arrangement of 7 letters = 7!
Thus, the total number of ways = 7!*700 = 352800
Hence, option c is the correct answer.

Question 5 of 5
5. Question
The average of 7 consecutive numbers is a. If the next three numbers are also added, the average shall
Correct
Solution (c)
Let the 7 consecutive numbers be a3, a2, a1, a, a+1, a+2 and a+3.
Sum of the numbers = 7a and the average of these numbers = a
If next 3 numbers a+4, a+5 and a+6 are also added then the average of these 10 numbers = (10a + 15)/10 = a + 1.5
Thus, the average increases by 1.5
Hence, option c is correct.
Incorrect
Solution (c)
Let the 7 consecutive numbers be a3, a2, a1, a, a+1, a+2 and a+3.
Sum of the numbers = 7a and the average of these numbers = a
If next 3 numbers a+4, a+5 and a+6 are also added then the average of these 10 numbers = (10a + 15)/10 = a + 1.5
Thus, the average increases by 1.5
Hence, option c is correct.