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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.
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Question 1 of 10
1. Question
If 25% of a number is subtracted from a second number, the second number reduces to its fivesixth. What is the ratio of the first number to the second number?
Correct
Solution: D
Let two numbers be x and y.
It is given that, y – x/4 = 5y/6
Therefore, y 5y/6 = x/4
Or y/6 = x/4
Therefore, x/y = 4/6 or 2:3
Incorrect
Solution: D
Let two numbers be x and y.
It is given that, y – x/4 = 5y/6
Therefore, y 5y/6 = x/4
Or y/6 = x/4
Therefore, x/y = 4/6 or 2:3

Question 2 of 10
2. Question
A number when divided by 136, leaves remainder 36. If the same number is divided by 17, then what will be the remainder?
Correct
Solution: C
If the first divisor be a multiple of the second divisor, then required number = Remainder obtained by dividing the first remainder by the second divisor.
Since, 17 is a factor of 136.
Therefore, 36 divided by 17 is 2.Incorrect
Solution: C
If the first divisor be a multiple of the second divisor, then required number = Remainder obtained by dividing the first remainder by the second divisor.
Since, 17 is a factor of 136.
Therefore, 36 divided by 17 is 2. 
Question 3 of 10
3. Question
In 11 innings a certain average has been made by a player. In the next innings he made 90 runs and his average was decreased by 5 runs. What will be the average of that player after 12 innings?
Correct
Solution: A
If x is the average of run in 11 innings. Then,
11x+90 =(x5)*12 ⇒ x= 150
Average run rate after 12th innings = (11x+90 )/12 =(11*150+90)/12= 145
Incorrect
Solution: A
If x is the average of run in 11 innings. Then,
11x+90 =(x5)*12 ⇒ x= 150
Average run rate after 12th innings = (11x+90 )/12 =(11*150+90)/12= 145

Question 4 of 10
4. Question
Rs.2500 is divided into two parts such that if one part to be charged at 5% per annum simple interest and the other at 6% per annum. If the annual income from interest alone is Rs.140. How much amount was lent at 5% per annum interest?
Correct
Solution: D
Simple interest = Principal * Time * Rate of Interest /100
Let Rs x was given at 5 % interest, then
Total income from interest = x*1*5/100 + (2500x) * 1 * 6/100 = 140
Or 5x/100 + (15000 – 6x)/100 = 140
Or 15000 – x = 14000
Or x= 1000
Incorrect
Solution: D
Simple interest = Principal * Time * Rate of Interest /100
Let Rs x was given at 5 % interest, then
Total income from interest = x*1*5/100 + (2500x) * 1 * 6/100 = 140
Or 5x/100 + (15000 – 6x)/100 = 140
Or 15000 – x = 14000
Or x= 1000

Question 5 of 10
5. Question
The average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 24 years. Find the average age of the women?
Correct
Solution: C
Let the average age of men be x
Therefore, sum of age of 8 men = 8x
If average increases by 2 when two women are included in place of two men of ages 20 and 24 years, then average age will be x+2.
Since, total number of people remain 8, sum of age new group = 8 * (x+2) = 8x +16
Therefore, total age of two women = 20 + 24 + 16 = 60
Therefore, average of women = 30 years.
Incorrect
Solution: C
Let the average age of men be x
Therefore, sum of age of 8 men = 8x
If average increases by 2 when two women are included in place of two men of ages 20 and 24 years, then average age will be x+2.
Since, total number of people remain 8, sum of age new group = 8 * (x+2) = 8x +16
Therefore, total age of two women = 20 + 24 + 16 = 60
Therefore, average of women = 30 years.

Question 6 of 10
6. Question
In a 100m race, A beats B by 10m and C by 13m. In a race of 180m, by how much distance B will beat C ?
Correct
Solution: B
A: B :C = 100: 90:87
Therefore, B/C = 90/87
Therefore, by the time B covers 180 m, C covers 87*2 m or 174 m
Thus, B beats C by 6 m
Incorrect
Solution: B
A: B :C = 100: 90:87
Therefore, B/C = 90/87
Therefore, by the time B covers 180 m, C covers 87*2 m or 174 m
Thus, B beats C by 6 m

Question 7 of 10
7. Question
The length, breadth and height of a room are 363m, 528m and 693m respectively. What is the longest tape that can measure the three dimensions of the room exactly?
Correct
Solution: D
Maximum length of tape
= HCF of 363m, 528m and 693m
363 can be simplified as 3 * 11 * 11
528 can be simplified as 2 * 2 * 2 * 2 * 3 * 11
693 can be simplified as 3 * 3 * 7 * 11
HCF of 363m, 528m and 693m = 3 * 11= 33 metreIncorrect
Solution: D
Maximum length of tape
= HCF of 363m, 528m and 693m
363 can be simplified as 3 * 11 * 11
528 can be simplified as 2 * 2 * 2 * 2 * 3 * 11
693 can be simplified as 3 * 3 * 7 * 11
HCF of 363m, 528m and 693m = 3 * 11= 33 metre 
Question 8 of 10
8. Question
How many numbers between 100 and 300 are divisible by 11?
Correct
Solution: C
Least number between 100 and 300 which is divisible by 11 is 110.
Highest number between 100 and 300 which is divisible by 11 is 297
Therefore, total number divisibly 11 including 110 and 297 are
(297 – 110)/11 + 1 = 187/11 + 1
= 17 + 1 = 18
Incorrect
Solution: C
Least number between 100 and 300 which is divisible by 11 is 110.
Highest number between 100 and 300 which is divisible by 11 is 297
Therefore, total number divisibly 11 including 110 and 297 are
(297 – 110)/11 + 1 = 187/11 + 1
= 17 + 1 = 18

Question 9 of 10
9. Question
What is the least number which when divided by 8, 12 and 16, leave the common remainder 3 ?
Correct
Solution: D
The least number which when divided by 8, 12 and 16, leave the common remainder 3 = LCM (8, 12 and 16) + 3
= 48 + 3 = 51.
Incorrect
Solution: D
The least number which when divided by 8, 12 and 16, leave the common remainder 3 = LCM (8, 12 and 16) + 3
= 48 + 3 = 51.

Question 10 of 10
10. Question
A man walked 20 m to cross a rectangular field diagonally. If the length of the field is 16 cm. Find the breadth of the field?
Correct
Solution: A
In the case of rectangle, Diagnol² = Length² + Breadth ²
Therefore 20² = 16 ² + Breadth ²
Therefore, breadth of rectangle = √(400256 = 12 m
Incorrect
Solution: A
In the case of rectangle, Diagnol² = Length² + Breadth ²
Therefore 20² = 16 ² + Breadth ²
Therefore, breadth of rectangle = √(400256 = 12 m