Insta–DART (Daily Aptitude and Reasoning Test) 2020  21
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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 CSP2021. 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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
Average of a, b and c is 11; average of c, d and e is 17; average of e and f is 22 and average of e and c is 17. Find out the average of a, b, c, d, e and f.
Correct
Answer : a
According to the question,
a+b+c = 11×3= 33 ………….Eq No 1
c+d+e = 17×3 = 51 …………..Eq No 2
e+f = 22×2 = 44 ……………Eq No 3
e+c = 17 x 2 = 34 ……………..Eq No 4
From Eqs. ( 2 ) and ( 4 ), we get
34 + d = 51 or d = 17 ………..Eq No 5
Now by adding Eqs. ( 1 ) , ( 3 ) and ( 5 ), we get
a + b + c + d + e + f = 33 + 44 + 17 = 94
So Average of a, b, c, d, e and f =94/6
= 47/3 = 15 2/3
Incorrect
Answer : a
According to the question,
a+b+c = 11×3= 33 ………….Eq No 1
c+d+e = 17×3 = 51 …………..Eq No 2
e+f = 22×2 = 44 ……………Eq No 3
e+c = 17 x 2 = 34 ……………..Eq No 4
From Eqs. ( 2 ) and ( 4 ), we get
34 + d = 51 or d = 17 ………..Eq No 5
Now by adding Eqs. ( 1 ) , ( 3 ) and ( 5 ), we get
a + b + c + d + e + f = 33 + 44 + 17 = 94
So Average of a, b, c, d, e and f =94/6
= 47/3 = 15 2/3

Question 2 of 5
2. Question
The average age of students of a class is 15.8 yr. The average age of boys in the class is 16.4 yr. and that of the girls is 15.4 yr. Find out the ratio of the number of boys to the number of girls in the class.
Correct
Answer : c
Incorrect
Answer : c

Question 3 of 5
3. Question
A car reached Raipur from Somgrah in 35 min with an average speed of 69 km/h. If the average speed is increased by 36 km/h, how much time will it take to cover the same distance?
Correct
Answer : c
Distance between Raipur and Somgarh = Average speed ´ Time
= (69×35)/60 = 161/4km
New speed=(69+36)km/h
=105km/h
Time =distance/time=161/4*105hour
=161*60/4*105
=23min
Incorrect
Answer : c
Distance between Raipur and Somgarh = Average speed ´ Time
= (69×35)/60 = 161/4km
New speed=(69+36)km/h
=105km/h
Time =distance/time=161/4*105hour
=161*60/4*105
=23min

Question 4 of 5
4. Question
The average weight of 4 men A,B,C and D, is 67 kg. The 5^{th} man E is included and the average weight decreases by 2 kg. A is replaced by F.The weight of F is 4kg more than E. Average weight decreases because of the replacement of A and now the average weight is 64 kg. Find the weight of A .
Correct
Answer : b
Sum of the weights of A, B, C and D = 67×4 = 268kg
And average weight of A,B,C,D AND e = 67 – 2 = 65 KG
So Sum of the weights of A, B, C, D and E = 65×5 = 325 kg
So Weight of E = 325 – 268 = 57 kg
So Weight of F = 57 + 4 = 61 kg
Now, average weight of F, B, C, D and E = 64 kg
So Sum of the weights of F, B, C, D and E = 64x 5 = 320 KG
So Sum of the weights of B, C and D = 320 – 57 – 61 = 202 kg
So Weight of A = 268 – 202 = 66 kg
Incorrect
Answer : b
Sum of the weights of A, B, C and D = 67×4 = 268kg
And average weight of A,B,C,D AND e = 67 – 2 = 65 KG
So Sum of the weights of A, B, C, D and E = 65×5 = 325 kg
So Weight of E = 325 – 268 = 57 kg
So Weight of F = 57 + 4 = 61 kg
Now, average weight of F, B, C, D and E = 64 kg
So Sum of the weights of F, B, C, D and E = 64x 5 = 320 KG
So Sum of the weights of B, C and D = 320 – 57 – 61 = 202 kg
So Weight of A = 268 – 202 = 66 kg

Question 5 of 5
5. Question
The average marks obtained by 120 students in an examination is 30. If the average marks of passed students is 40 and that of the failed students is 10, what is the 25% of the number of students who passed the examination?
Correct
Answer: b
Let number of passed students = x
Total marks = 120*30
According to the question,
40x + (120 – x ) *10 = 120 *30
3600 = 40x + 1200 – 10x
30x = 2400
X = 2400/30= 80
So 25% of 80 = 80/4= 20
Incorrect
Answer: b
Let number of passed students = x
Total marks = 120*30
According to the question,
40x + (120 – x ) *10 = 120 *30
3600 = 40x + 1200 – 10x
30x = 2400
X = 2400/30= 80
So 25% of 80 = 80/4= 20