Breakup G Strategy | Percentages + Equations | EASY | CAT 2025 Slot 1
Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was _____
Answer
(Total investment 100000; B = 25% of G; returns: stocks 10%, bonds 6%, gold 8%; total gain 8200) — find gain from bonds.
Let bonds = B, gold = G, stocks = S. Given B = G/4 ⇒ G = 4B.
Total money: S + B + G = 100000 ⇒ S + B + 4B = 100000 ⇒ S = 100000 − 5B.
Total gain: 0.10·S + 0.06·B + 0.08·G = 8200.
Substitute S and G: 0.10(100000 − 5B) + 0.06B + 0.08(4B) = 8200.
Compute: 10000 − 0.5B + 0.06B + 0.32B = 8200 ⇒ 10000 − 0.12B = 8200.
Solve: 0.12B = 1800 ⇒ B = 1800/0.12 = 15000.
Gain from bonds = 6% of B = 0.06 × 15000 = 900.
Answer (gain from bonds) = Rs 900.
Breakup 1.2/0.8 | Percentages | MEDIUM | CAT 2025 Slot 1
A shopkeeper offers a discount of 22% on the marked price of each chair, and gives 13 chairs to the customer for the discounted price of 12 chairs to earn a profit of 26% on the transaction. If the cost price of each chair is Rs 100, then the marked price, in rupees, of each chair is
Answer
Percentages 1.2 / 0.8
Let marked price be M. Discount 22% ⇒ selling price per chair = 0.78M.
Customer pays for 12 discounted chairs but receives 13 chairs, so total revenue = 12 × 0.78M = 9.36M.
Total cost = 13 × 100 = 1300. For 26% profit, revenue = 1.26 × 1300 = 1638.
So 9.36M = 1638 ⇒ M = 1638 / 9.36.
Compute M = 163800 / 936 = 2275 / 13 = 175.
Answer: Rs 175 per chair.
Breakup G Strategy | Percentages + Maxima Minima | MEDIUM | CAT 2025 Slot 1
In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was _____.
Answer
Let B = number of boys, G = number of girls (both integers), and B > 10.
After departures: remaining girls = 60% of G = 3G/5, remaining boys = 40% of B = 2B/5.
Given 3G/5 = 2B/5 + 8. Multiply by 5: 3G = 2B + 40. (Equation ★)
For 3G to be integer, G must be integer. For the remaining counts 3G/5 and 2B/5 to be integers (no fractional people), G and B should both be multiples of 5.
Let B = 5m (m integer) with B > 10 ⇒ m ≥ 3. Try smallest multiples of 5 for B:
• B = 15 ⇒ 2B + 40 = 70 ⇒ G = 70/3 (not integer)
• B = 20 ⇒ 2B + 40 = 80 ⇒ G = 80/3 (not integer)
• B = 25 ⇒ 2B + 40 = 90 ⇒ G = 90/3 = 30 (integer, works)
So the smallest B (multiple of 5) that satisfies (★) is B = 25, with G = 30.
Total initial students = B + G = 25 + 30 = 55.
Therefore the minimum possible number of students initially in the class (with realistic integer people and integer remaining counts) is 55.
Breakup G Strategy | Interest | MEDIUM | CAT 2025 Slot 1
At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to
1. 3221
2. 3180
3. 3096
4. 3156
Answer
Let P be the principal and r the annual simple rate (in decimal).
Amount after 3 years = P + 3·(P·r) = P(1 + 3r) = 13,920.
Amount after 6.5 years = P(1 + 6.5r) = 18,960.
Difference = interest for 3.5 years = 18,960 − 13,920 = 5,040.
So interest for 1 year = 5,040 ÷ 3.5 = 1,440.
Thus P·r = 1,440. Use A3: P + 3·1,440 = 13,920 ⇒ P = 13,920 − 4,320 = 9,600.
So annual simple rate r = 1,440 ÷ 9,600 = 0.15 = 15% p.a.
For 2 years compounded half-yearly: number of half-year periods n = 4; half-year rate = 15%/2 = 7.5% = 0.075.
Amount = P(1 + 0.075)^4 = 9,600 × (1.075)^4 ≈ 9,600 × 1.335469 ≈ 12,820.50.
Compound interest = 12,820.50 − 9,600 ≈ 3,220.50 → nearest rupee = 3,221. (Option 1)
Breakup G Strategy | Ratio + Table | MEDIUM | CAT 2025 Slot 1
The ratio of the number of students in the morning shift and afternoon shift of a school was 13 : 9. After 21 students moved from the morning shift to the afternoon shift, this ratio became 19 : 14. Next, some new students joined the morning and afternoon shifts in the ratio 3 : 8 and then the ratio of the number of students in the morning shift and the afternoon shift became 5 : 4. The number of new students who joined is
- 88
- 121
- 99
- 110
Answer
Solve using Matrix table | Answer 3. 99
| Stage | Morning Shift | Afternoon Shift | Ratio | Explanation |
| Initial | 13k | 9k | 13 : 9 | Given initial ratio |
| After 21 move M → A | 13k − 21 | 9k + 21 | 19 : 14 | 13k – 21/9k + 21 = 19/14 k=63 |
| After solving k = 63 | 798 | 588 | 19 : 14 | Substitute k = 63 into expressions |
| New students join (3x, 8x) | 798 + 3x | 588 + 8x | 5 : 4 | 798 + 3x/588 + 8x = 5/4 x = 9 |
| Total new students | 3x | 8x | Total: 11x | 11x = 11 × 9 = 99 |
