🎉 Ace XAT 2025 with 2IIM's Crash Course - Enrol now


🎉Effortlessly calculate your score with the 2IIM Score Calculator!🎉

CAT 2023 Question Paper | Quant Slot 1

CAT Previous Year Paper | CAT Quant Questions | Question 10

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 10 : In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is

  1. 19
  2. 21
  3. 20
  4. 22

🎉 Ace XAT 2025 with 2IIM's Crash Course - Enrol now


Best CAT Online Coaching
Try upto 40 hours for free
Learn from the best!


2IIM : Best Online CAT Coaching.

🎉Effortlessly calculate your score with the 2IIM Score Calculator!🎉


2IIM : Best Online CAT Coaching.


Video Explanation


Best CAT Coaching in Chennai


CAT Coaching in Chennai - CAT 2022
Limited Seats Available - Register Now!


Explanatory Answer

Let the marks of each girl in the class be g and the marks of each boy be b.
Since, the average marks of 4 girls and 6 boys is 24…
4g + 6b = 24 × 10 = 240
A girl scores more than or equal to the score of a boy but never more than double the score.
Therefore, g = kb, where 1 ≤ k ≤ 2
4(kb) + 6b = 240
b(4k + 6) = 240
b(2k + 3) = 120
b = \( \frac { 120 } { 2 k + 3 } \)
Finally we need 2g + 6b = b(2k + 6) to be an integer.
\( 120 \times \left( \frac { 2 k + 6 } { 2 k + 3 } \right) \) needs to be an integer.
\( 120 \times \left( 1 + \frac { 3 } { 2 k + 3 } \right) \) needs to be an integer.
∴ \( 120 \times \left( \frac { 3 } { 2 k + 3 } \right) \) needs to be an integer.
\( \left( \frac { 360 } { 2 k + 3 } \right) \) needs to be an integer.
Let \( \frac { 360 } { 2 k + 3 } = n \), where n is an integer.
\( \frac { 360 } { n } - 3 = 2 k \)
Since, 1 ≤ k ≤ 2
2 ≤ 2k ≤ 4
2 ≤ \( \frac { 360 } { n } - 3 \) ≤ 4
5 ≤ \( \frac { 360 } { n } \) ≤ 7
\( \frac { 5 } { 360 } \) ≤ \( \frac { 1 } { n } \) ≤ \( \frac { 7 } { 360 } \)
\( \frac { 360 } { 7 } \) ≤ n ≤ \( \frac { 360 } { 5 } \)
51.42 ≤ n ≤ 72
52 ≤ n ≤ 72
Therefore, n can take 21 values, 52 to 72 both inclusive.
For all these values k takes a distinct value from 1 to 2 and 2g + 6b takes a distinct integral value.


The answer is '21'

Choice B is the correct answer.

CAT Questions | CAT Quantitative Aptitude

CAT Questions | Verbal Ability for CAT


Where is 2IIM located?

2IIM Online CAT Coaching
A Fermat Education Initiative,
58/16, Indira Gandhi Street,
Kaveri Rangan Nagar, Saligramam, Chennai 600 093

How to reach 2IIM?

Mobile: (91) 99626 48484 / 94459 38484
WhatsApp: WhatsApp Now
Email: info@2iim.com