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MCQ Practice Test & Solutions: Daily Passage Test for CLAT - Jan 17 (5 Questions)

You can prepare effectively for CLAT Daily Passage Practice for CLAT with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Daily Passage Test for CLAT - Jan 17". These 5 questions have been designed by the experts with the latest curriculum of CLAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 5

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Daily Passage Test for CLAT - Jan 17 - Question 1

Directions: Study the following information carefully and answer the questions given beside.

There were five centres (A, B, C, D and E) in a city for CLAT 2023 exam. The minimum distance between any two centres was 4 km. The combined number of students in all five centres was 2400. The number of students in Centre A was 25% more than the number of students in Centre B. The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D. There were 1020 students in Centre E.

Q. What was the difference between the number of students in Centre A and the number of students in Centre D?

Detailed Solution: Question 1

From the common explanation, we have

Reqd. difference = 480 – 300 = 180

Hence, Option B is correct.

Common explanation :

The combined number of students in all five centres was 2400. There were 1020 students in Centre E.

The combined number of students in A, B, C and D = 2400 – 1020 = 1380

The number of students in Centre A was 25% more than the number of students in Centre B.

Let the number of students in B = 100x

The number of students in A


The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D.

The number of students in Centre C = 100x + 120

The number of students in Centre D = 100x + 120 + 120 = 100x + 240

So,

125x + 100x + 100x + 120 + 100x + 240 = 1380

425x + 360 = 1380

425x = 1020

x = 2.40

The number of students in Centre A = 125x = 125 × 2.40 = 300

The number of students in Centre B = 100x = 100 × 2.40 = 240

The number of students in Centre C = 100x + 120 = 100 × 2.40 + 120 = 360

The number of students in Centre D = 100x + 240 = 100 × 2.40 + 240 = 480

Daily Passage Test for CLAT - Jan 17 - Question 2

Directions: Study the following information carefully and answer the questions given beside.

There were five centres (A, B, C, D and E) in a city for CLAT 2023 exam. The minimum distance between any two centres was 4 km. The combined number of students in all five centres was 2400. The number of students in Centre A was 25% more than the number of students in Centre B. The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D. There were 1020 students in Centre E.

Q. What was the combined number of students in centres A, B and C?

Detailed Solution: Question 2

From the common explanation, we have

Reqd. sum = 300 + 240 + 360 = 900

Hence, Option C is correct.

Common explanation :

The combined number of students in all five centres was 2400. There were 1020 students in Centre E.

The combined number of students in A, B, C and D = 2400 – 1020 = 1380

The number of students in Centre A was 25% more than the number of students in Centre B.

Let the number of students in B = 100x

The number of students in A


The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D.

The number of students in Centre C = 100x + 120

The number of students in Centre D = 100x + 120 + 120 = 100x + 240

So,

125x + 100x + 100x + 120 + 100x + 240 = 1380

425x + 360 = 1380

425x = 1020

x = 2.40

The number of students in Centre A = 125x = 125 × 2.40 = 300

The number of students in Centre B = 100x = 100 × 2.40 = 240

The number of students in Centre C = 100x + 120 = 100 × 2.40 + 120 = 360

The number of students in Centre D = 100x + 240 = 100 × 2.40 + 240 = 480

Daily Passage Test for CLAT - Jan 17 - Question 3

Directions: Study the following information carefully and answer the questions given beside.

There were five centres (A, B, C, D and E) in a city for CLAT 2023 exam. The minimum distance between any two centres was 4 km. The combined number of students in all five centres was 2400. The number of students in Centre A was 25% more than the number of students in Centre B. The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D. There were 1020 students in Centre E.

Q. The number of students in Centre E was what percentage of the number of students in Centre A?

Detailed Solution: Question 3

From the common explanation, we have


Hence, Option D is correct.

Common explanation :

The combined number of students in all five centres was 2400. There were 1020 students in Centre E.

The combined number of students in A, B, C and D = 2400 – 1020 = 1380

The number of students in Centre A was 25% more than the number of students in Centre B.

Let the number of students in B = 100x

The number of students in A


The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D.

The number of students in Centre C = 100x + 120

The number of students in Centre D = 100x + 120 + 120 = 100x + 240

So,

125x + 100x + 100x + 120 + 100x + 240 = 1380

425x + 360 = 1380

425x = 1020

x = 2.40

The number of students in Centre A = 125x = 125 × 2.40 = 300

The number of students in Centre B = 100x = 100 × 2.40 = 240

The number of students in Centre C = 100x + 120 = 100 × 2.40 + 120 = 360

The number of students in Centre D = 100x + 240 = 100 × 2.40 + 240 = 480

Daily Passage Test for CLAT - Jan 17 - Question 4

Directions: Study the following information carefully and answer the questions given beside.

There were five centres (A, B, C, D and E) in a city for CLAT 2023 exam. The minimum distance between any two centres was 4 km. The combined number of students in all five centres was 2400. The number of students in Centre A was 25% more than the number of students in Centre B. The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D. There were 1020 students in Centre E.

Q. The number of students in Centre C was what percentage of the number of students in Centre D?

Detailed Solution: Question 4

From the common explanation, we have


Hence, Option A is correct.

Common explanation :

The combined number of students in all five centres was 2400. There were 1020 students in Centre E.

The combined number of students in A, B, C and D = 2400 – 1020 = 1380

The number of students in Centre A was 25% more than the number of students in Centre B.

Let the number of students in B = 100x

The number of students in A


The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D.

The number of students in Centre C = 100x + 120

The number of students in Centre D = 100x + 120 + 120 = 100x + 240

So,

125x + 100x + 100x + 120 + 100x + 240 = 1380

425x + 360 = 1380

425x = 1020

x = 2.40

The number of students in Centre A = 125x = 125 × 2.40 = 300

The number of students in Centre B = 100x = 100 × 2.40 = 240

The number of students in Centre C = 100x + 120 = 100 × 2.40 + 120 = 360

The number of students in Centre D = 100x + 240 = 100 × 2.40 + 240 = 480

Daily Passage Test for CLAT - Jan 17 - Question 5

Directions: Study the following information carefully and answer the questions given beside.

There were five centres (A, B, C, D and E) in a city for CLAT 2023 exam. The minimum distance between any two centres was 4 km. The combined number of students in all five centres was 2400. The number of students in Centre A was 25% more than the number of students in Centre B. The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D. There were 1020 students in Centre E.

Q. Which centre had the second least number of students?

Detailed Solution: Question 5

From the common explanation, we have

Centre A had the second lowest number of students.

Hence, Option A is correct.

Common explanation :

The combined number of students in all five centres was 2400. There were 1020 students in Centre E.

The combined number of students in A, B, C and D = 2400 – 1020 = 1380

The number of students in Centre A was 25% more than the number of students in Centre B.

Let the number of students in B = 100x

The number of students in A


The number of students in Centre C was 120 more than the number of students in Centre B and 120 less than the number of students in Centre D.

The number of students in Centre C = 100x + 120

The number of students in Centre D = 100x + 120 + 120 = 100x + 240

So,

125x + 100x + 100x + 120 + 100x + 240 = 1380

425x + 360 = 1380

425x = 1020

x = 2.40

The number of students in Centre A = 125x = 125 × 2.40 = 300

The number of students in Centre B = 100x = 100 × 2.40 = 240

The number of students in Centre C = 100x + 120 = 100 × 2.40 + 120 = 360

The number of students in Centre D = 100x + 240 = 100 × 2.40 + 240 = 480

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