Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
Quiz-summary
0 of 5 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
Information
Best of Luck! 🙂
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 5 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- 5
- Answered
- Review
-
Question 1 of 5
1. Question
In a college fest, every student participates in at least one of the events—Debate, Poster-making, or Sports—and no one participates in all three. If 80 students participate in Debate, 60 in Poster-making, 75 in Sports, and 25 students participate in exactly two events, how many students are there in total?
Correct
Answer: (c)
Explanation:
Given:- |D|=80, |P|=60, |S|=75
- Everyone in at least one event
- No one in all three
- Exactly two events = 25
Let total = N.
Using inclusion-exclusion:
N = 80+60+75 − (sum of pairwise intersections) + (triple intersection)Triple intersection = 0.
Now, “exactly two” = 25 and “none in all three” ⇒
sum of pairwise intersections = exactly two (because no triple to add multiple counts) = 25.So:
N = 215 − 25 = 190Incorrect
Answer: (c)
Explanation:
Given:- |D|=80, |P|=60, |S|=75
- Everyone in at least one event
- No one in all three
- Exactly two events = 25
Let total = N.
Using inclusion-exclusion:
N = 80+60+75 − (sum of pairwise intersections) + (triple intersection)Triple intersection = 0.
Now, “exactly two” = 25 and “none in all three” ⇒
sum of pairwise intersections = exactly two (because no triple to add multiple counts) = 25.So:
N = 215 − 25 = 190 -
Question 2 of 5
2. Question
In a contest, 25 balls are placed in a straight line at an interval of 8 meters each, with the first ball being 32 meters away from the starting point. A contestant is required to bring all the balls back to the starting place one at a time. What is the total distance the contestant has to cover?
Correct
Answer: B
Explanation
Here, 25 balls are placed in a straight line at an interval of 8 meters each, with the first ball being 32
meters away from the starting point
Distance travelled to pick up and drop the first ball = 32 + 32 = 64
Distance travelled to pick up and drop the second ball = 2(32 + 8) = 80
Distance travelled to pick up and drop the third ball = 2(32 + 8 + 8) = 96
So, Distances travelled to pick up and drop the balls are 64, 80, 96, …………
Here, a = 64, d = 80 – 64 = 16
Sum of the terms, Sn = (n/2) [2a + (n – 1)d]
So, Total distance travelled by the contestant, S25 = [(25/2) {2×64 + (25 – 1)16] = [(25/2) {128 + 384}] =
(25/2) × 512 = 6400 meters = 6.4 km
Incorrect
Answer: B
Explanation
Here, 25 balls are placed in a straight line at an interval of 8 meters each, with the first ball being 32
meters away from the starting point
Distance travelled to pick up and drop the first ball = 32 + 32 = 64
Distance travelled to pick up and drop the second ball = 2(32 + 8) = 80
Distance travelled to pick up and drop the third ball = 2(32 + 8 + 8) = 96
So, Distances travelled to pick up and drop the balls are 64, 80, 96, …………
Here, a = 64, d = 80 – 64 = 16
Sum of the terms, Sn = (n/2) [2a + (n – 1)d]
So, Total distance travelled by the contestant, S25 = [(25/2) {2×64 + (25 – 1)16] = [(25/2) {128 + 384}] =
(25/2) × 512 = 6400 meters = 6.4 km
-
Question 3 of 5
3. Question
Consider the following statements:
- A coin is tossed 7 times. The probability of getting exactly 3 heads is 35/128.
- A coin is tossed 4 times. The probability of getting at most 1 head is 5/16.
Which of the above statements is/are correct?
Correct
Answer: (c)
Explanation:
- 7 tosses, exactly 3 heads:
Total outcomes = 2⁷ = 128
Favourable = 7C3 = 35
Probability = 35/128. Correct. - 4 tosses, at most 1 head = 0 head or 1 head.
Total outcomes = 16
Ways for 0 head = 4C0 = 1
Ways for 1 head = 4C1 = 4
Probability = (1 + 4)/16 = 5/16. Correct.
Incorrect
Answer: (c)
Explanation:
- 7 tosses, exactly 3 heads:
Total outcomes = 2⁷ = 128
Favourable = 7C3 = 35
Probability = 35/128. Correct. - 4 tosses, at most 1 head = 0 head or 1 head.
Total outcomes = 16
Ways for 0 head = 4C0 = 1
Ways for 1 head = 4C1 = 4
Probability = (1 + 4)/16 = 5/16. Correct.
-
Question 4 of 5
4. Question
A loan of Rs. 2,40,000 is repaid in monthly instalments starting at Rs. 5,000 and increasing by Rs. 500 each month. Then consider the following statements:
- After the 20th instalment, Rs. 15,000 remains unpaid.
- It will take more than 22 instalments to fully repay the loan.
Which one of the above statements is/are correct?
Correct
Answer: (b)
Explanation:
a = 5000, d = 500S₂₀ = 20/2 × [2×5000 + 19×500]
= 10 × [10000 + 9500]
= 10 × 19500
= 1,95,000Remaining = 2,40,000 − 1,95,000 = 45,000
So statement 1 is false.S₂₂ = 22/2 × [10000 + 21×500]
= 11 × [10000 + 10500]
= 11 × 20500
= 2,25,500Still remaining 14,500, so it takes more than 22 instalments.
Statement 2 is true.Incorrect
Answer: (b)
Explanation:
a = 5000, d = 500S₂₀ = 20/2 × [2×5000 + 19×500]
= 10 × [10000 + 9500]
= 10 × 19500
= 1,95,000Remaining = 2,40,000 − 1,95,000 = 45,000
So statement 1 is false.S₂₂ = 22/2 × [10000 + 21×500]
= 11 × [10000 + 10500]
= 11 × 20500
= 2,25,500Still remaining 14,500, so it takes more than 22 instalments.
Statement 2 is true. -
Question 5 of 5
5. Question
A mason has 120 bricks. He stacks them such that the first layer has 15 bricks, the second layer has 14 bricks, and so on. Which of the following is correct?
Correct
Answer: (c)
Explanation:
First term = 15
Common difference = −1Let layers = n
Last layer = 15 − (n−1)Total = n/2 × (15 + last)
= n/2 × (15 + 15 − (n−1))
= n/2 × (31 − n)Given total = 120:
n(31 − n) = 240
n² − 31n + 240 = 0
(n − 15)(n − 16) = 0Valid solution: n = 15
Last layer = 15 − 14 = 1
So both statements are correct.
Incorrect
Answer: (c)
Explanation:
First term = 15
Common difference = −1Let layers = n
Last layer = 15 − (n−1)Total = n/2 × (15 + last)
= n/2 × (15 + 15 − (n−1))
= n/2 × (31 − n)Given total = 120:
n(31 − n) = 240
n² − 31n + 240 = 0
(n − 15)(n − 16) = 0Valid solution: n = 15
Last layer = 15 − 14 = 1
So both statements are correct.
