The ratio of the number of learners enrolled in the forenoon batch and the afternoon batch of a coaching institute was 13 : 9. When 42 learners shifted from the forenoon batch to the afternoon batch, the ratio of the two batches became 19 : 14. Subsequently, a few fresh learners were admitted to the forenoon and afternoon batches in the ratio 3 : 8, after which the ratio of the number of learners in the forenoon batch to that in the afternoon batch became 5 : 4. What is the total number of fresh learners admitted?
Answer and solution
Correct answer: 198
Solution
Answer: 198
1. Let the original number of learners in the forenoon and afternoon batches be 13x and 9x.
2. After 42 learners move from forenoon to afternoon, the ratio becomes 19 : 14.
3. (13x − 42)/(9x + 42) = 19/14
4. Cross-multiplying:
5. 14(13x − 42) = 19(9x + 42)
182x − 588 = 171x + 798
11x = 1386
x = 126
6. So, after the shift:
7. Forenoon = 13 × 126 − 42 = 1596
Afternoon = 9 × 126 + 42 = 1176
8. Now let the fresh learners admitted be 3k in the forenoon and 8k in the afternoon.
9. The new ratio is 5 : 4:
10. (1596 + 3k)/(1176 + 8k) = 5/4
11. Cross-multiplying:
12. 4(1596 + 3k) = 5(1176 + 8k)
6384 + 12k = 5880 + 40k
504 = 28k
k = 18
13. Total fresh learners = 3k + 8k
= 11 × 18
= 198.
14. Answer: 198 (TITA).
Why this method works:
The transfer of 42 learners changes the two batch sizes in opposite directions but does not change their combined total. The later admission adds new learners in 3:8, so a new multiplier k must be used; it is not the original x.
Independent check and common trap:
After adding 54 and 144 students, the batches become 1650 and 1320. Dividing both by 330 gives 5:4. Their combined increase is 198, confirming that the requested total includes both batches.
Apply: (13x−42)/(9x+42)=19/14 →x 126. Then (1596 + 3k)/(1176+8k)=5/4 →k 18. Fresh total 11k=198.
Check: After adding 54 and 144 students, the batches become 1650 and 1320. Dividing both by 330 gives 5:4. Their combined increase is 198, confirming that the requested total includes both batches.
1. Let the original number of learners in the forenoon and afternoon batches be 13x and 9x.
2. After 42 learners move from forenoon to afternoon, the ratio becomes 19 : 14.
3. (13x − 42)/(9x + 42) = 19/14
4. Cross-multiplying:
5. 14(13x − 42) = 19(9x + 42)
182x − 588 = 171x + 798
11x = 1386
x = 126
6. So, after the shift:
7. Forenoon = 13 × 126 − 42 = 1596
Afternoon = 9 × 126 + 42 = 1176
8. Now let the fresh learners admitted be 3k in the forenoon and 8k in the afternoon.
9. The new ratio is 5 : 4:
10. (1596 + 3k)/(1176 + 8k) = 5/4
11. Cross-multiplying:
12. 4(1596 + 3k) = 5(1176 + 8k)
6384 + 12k = 5880 + 40k
504 = 28k
k = 18
13. Total fresh learners = 3k + 8k
= 11 × 18
= 198.
14. Answer: 198 (TITA).
Why this method works:
The transfer of 42 learners changes the two batch sizes in opposite directions but does not change their combined total. The later admission adds new learners in 3:8, so a new multiplier k must be used; it is not the original x.
Independent check and common trap:
After adding 54 and 144 students, the batches become 1650 and 1320. Dividing both by 330 gives 5:4. Their combined increase is 198, confirming that the requested total includes both batches.
G Strategy shortcut
Spot: Arithmetic – Ratios.Apply: (13x−42)/(9x+42)=19/14 →x 126. Then (1596 + 3k)/(1176+8k)=5/4 →k 18. Fresh total 11k=198.
Check: After adding 54 and 144 students, the batches become 1650 and 1320. Dividing both by 330 gives 5:4. Their combined increase is 198, confirming that the requested total includes both batches.
G Strategy: Spot: Arithmetic – Ratios. Apply: (13x−42)/(9x+42)=19/14 →x 126. Then (1596 + 3k)/(1176+8k)=5/4 →k 18. Fresh total 11k=198. Check: After adding 54 and 144 students, the batches become 1650 and 1320. Dividing both by 330 gives 5:4. Their combined increase is 198, confirming that the requested total includes both batches.