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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 5
1. Question
The percent profit made when an article is sold for Rs.56 is thrice as when it is sold for Rs.42. What is the profit percentage if the article sold at Rs.56?
Correct
Answer B
Let Rs.’x’ be the profit made when an article sold at Rs.42.
Therefore, 3x is the profit made when the article sold at
Rs.56.
CP = 42-x = 56-3x
=> 2x = 14
=> x = 7
Therefore CP = 42-x
= 56-3x
= 42-7
= 56- 21
= Rs.35
Profit percentage if sold @ Rs.56 = (56-35)/35 = 60%
Incorrect
Answer B
Let Rs.’x’ be the profit made when an article sold at Rs.42.
Therefore, 3x is the profit made when the article sold at
Rs.56.
CP = 42-x = 56-3x
=> 2x = 14
=> x = 7
Therefore CP = 42-x
= 56-3x
= 42-7
= 56- 21
= Rs.35
Profit percentage if sold @ Rs.56 = (56-35)/35 = 60%
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Question 2 of 5
2. Question
How much percentage more with respect to the previous bill a consumer must pay when Goods and Services Tax was raised from 5% to 18%?
Correct
Your Answer C
Let us assume the bill before tax as Rs.100.
Earlier tax was 5%. So, the earlier bill after tax is Rs.105. Now the tax is 18%.
So, the new bill after tax is Rs.118.
The extra tax paid = 118 – 105 = Rs.13
Extra bill paid percentage wise = (13/105) × 100 = 12.38%
Incorrect
Your Answer C
Let us assume the bill before tax as Rs.100.
Earlier tax was 5%. So, the earlier bill after tax is Rs.105. Now the tax is 18%.
So, the new bill after tax is Rs.118.
The extra tax paid = 118 – 105 = Rs.13
Extra bill paid percentage wise = (13/105) × 100 = 12.38%
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Question 3 of 5
3. Question
A man covers a certain distance by car. Had he moved 6 km/h faster, he would have taken 4 hours less. If he had moved 4 km/h slower, he would have taken 4 hours more. The distance (in km) is?
Correct
Answer : C
Let ‘D’ km be distance to be covered.
Let ‘S’ km/h be original speed of car.
Let ‘T’ h be original time taken.
By formula for D, we have
D= TS ———–> Eq.
1 Case 1: If speed increased by 6 km/h, the time taken is 4h less i.e,
D=(T-4)(S+6) D=TS + 6T – 4S – 24 ———-> Eq2
From Eq1 and Eq2,
TS = TS + 6T – 4S – 24
6T – 4S – 24=0 ————> Eq3
Case 2: If speed decreased by 4 km/h, the time taken is 4h more i.e,
D= (T+4) (S-4) D= TS – 4T + 4S – 16———–>Eq4
From Eq1 and Eq4,
TS = TS – 4T + 4S – 16
4T – 4S + 16 = 0 ———–> Eq5
Solving Eq3 and Eq5,
T = 20 h
S = 24 km/h
Therefore,
D = TS = 20 X 24 = 480 km.
Incorrect
Answer : C
Let ‘D’ km be distance to be covered.
Let ‘S’ km/h be original speed of car.
Let ‘T’ h be original time taken.
By formula for D, we have
D= TS ———–> Eq.
1 Case 1: If speed increased by 6 km/h, the time taken is 4h less i.e,
D=(T-4)(S+6) D=TS + 6T – 4S – 24 ———-> Eq2
From Eq1 and Eq2,
TS = TS + 6T – 4S – 24
6T – 4S – 24=0 ————> Eq3
Case 2: If speed decreased by 4 km/h, the time taken is 4h more i.e,
D= (T+4) (S-4) D= TS – 4T + 4S – 16———–>Eq4
From Eq1 and Eq4,
TS = TS – 4T + 4S – 16
4T – 4S + 16 = 0 ———–> Eq5
Solving Eq3 and Eq5,
T = 20 h
S = 24 km/h
Therefore,
D = TS = 20 X 24 = 480 km.
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Question 4 of 5
4. Question
If a/3 = b/4 =c/ 5 then (a+b+c)/b =?
Correct
Answer B
From the given equation, we can conclude that
a/b=3/4;
c/b=5/4;
Therefore, (a+b+c)/b
=(a/b)+(b/b)+(c/b)
=(3/4)+1+(5/4) =3
Incorrect
Answer B
From the given equation, we can conclude that
a/b=3/4;
c/b=5/4;
Therefore, (a+b+c)/b
=(a/b)+(b/b)+(c/b)
=(3/4)+1+(5/4) =3
-
Question 5 of 5
5. Question
How many two-digit numbers are there which on reversing digits will leave difference of 18 and multiples of 18 with original number?
Correct
Answer B .
All the digit numbers on reversing the digits, will leave a difference of 9 or multiples of 9 as difference with original number.
If the difference between two digits is 2 or multiple of 2, then the difference between the numbers and its reverse will be 18 or multiples of 18.
Case 1: If a ten’s place has odd numbers, then, in 5 ways are there, unit digit number should not be same number as the ten’s digit. There are 4 ways possible for unit’s digit.
Therefore, 5×4=20 ways are there.
Case 2: If ten’s place has even numbers, there are 4 ways. Unit digit can be filled with 3 other even numbers are ‘0’.
Therefore 4×4=16 ways are there.
Total = 20+16 = 36.
Incorrect
Answer B .
All the digit numbers on reversing the digits, will leave a difference of 9 or multiples of 9 as difference with original number.
If the difference between two digits is 2 or multiple of 2, then the difference between the numbers and its reverse will be 18 or multiples of 18.
Case 1: If a ten’s place has odd numbers, then, in 5 ways are there, unit digit number should not be same number as the ten’s digit. There are 4 ways possible for unit’s digit.
Therefore, 5×4=20 ways are there.
Case 2: If ten’s place has even numbers, there are 4 ways. Unit digit can be filled with 3 other even numbers are ‘0’.
Therefore 4×4=16 ways are there.
Total = 20+16 = 36.
