CAT 2026 · Slot 2 · Expected paper

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CAT 2026 Slot 2 Quant: Arithmetic Questions with Solutions

6 questions: Q8, Q13, Q16, Q17, Q20, Q21. Solve each one here: tap the answer, press Check, then open the solution.

Question 8Mixtures & Alligation

MCQ · +3 / −1 · medium

A vessel contains a certain quantity of a solution of acid and water. When 3 litres of water is added to it, the new solution has 50% acid concentration. When 12 litres of acid is further added to this new solution, the final solution has 75% acid concentration. The ratio of water and acid in the original solution was:

Answer and solution

Correct answer: A

Solution

Answer: A — 1 : 2

1. Let the original amount of acid be A litres and water be W litres.
2. After adding 3 litres of water, the solution is 50% acid. This means the amount of acid and water are equal.
3. So,
4. A = W + 3. ...(1)
5. Next, 12 litres of acid is added. The final solution is 75% acid.
6. Therefore,
7. (A + 12)/(A + W + 15) = 3/4
8. Cross-multiplying:
9. 4(A + 12) = 3(A + W + 15)
4A + 48 = 3A + 3W + 45
A = 3W − 3. ...(2)
10. From equations (1) and (2):
11. W + 3 = 3W − 3
2W = 6
W = 3
12. So, A = 3 + 3 = 6.
13. The original ratio of water to acid is 3 : 6, or 1 : 2.
14. Answer: 1 : 2.

Why this method works:
A 50% acid mixture contains equal volumes of acid and water. After only acid is added, the water amount stays fixed, while 75% concentration requires acid to be three times that fixed water amount.

Independent check and common trap:
At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.

G Strategy shortcut

Spot: Arithmetic – Mixtures and Alligation.
Apply: After water addition acid=water. Final acid: water=3: 1, so adding 12 acid equals 2 parts: each part 6. Originally water=6−3=3, acid 6 →1: 2.
Check: At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.

G Strategy: Spot: Arithmetic – Mixtures and Alligation. Apply: After water addition acid=water. Final acid: water=3: 1, so adding 12 acid equals 2 parts: each part 6. Originally water=6−3=3, acid 6 →1: 2. Check: At the intermediate stage both volumes are 6 litres. Adding 12 litres of acid gives acid 18 and water 6, or 75% acid. Undoing the earlier water addition leaves original water 3 and acid 6; the requested order is water: acid=1:2.

Question 13Averages

MCQ · +3 / −1 · medium

A company has 40 employees whose names are listed in a certain order. In 2022, the average bonus of the first 30 employees was Rs. 36,000, of the last 30 employees was Rs. 56,000, and of the first 10 and last 10 employees together was Rs. 46,000. In the next year, the average bonus of the first 10 employees doubled, the average bonus of the last 10 employees became three times its previous value, and the bonuses of the remaining 20 employees remained unchanged. Then, the average bonus, in rupees, of all 40 employees together in 2023 was:

Answer and solution

Correct answer: C

Solution

Answer: C — 88,000

1. Divide the 40 employees into three groups:
2. First 10 employees = F
Middle 20 employees = M
Last 10 employees = L
3. Work in thousands of rupees.
4. The average of the first 30 employees is 36,000:
5. F + M = 30 × 36 = 1080 ...(1)
6. The average of the last 30 employees is 56,000:
7. M + L = 30 × 56 = 1680 ...(2)
8. The average of the first 10 and last 10 together is 46,000:
9. F + L = 20 × 46 = 920 ...(3)
10. Add equations (1) and (2), then subtract equation (3):
11. 2M = 1080 + 1680 − 920
2M = 1840
M = 920
12. Now:
13. F = 1080 − 920 = 160
L = 1680 − 920 = 760
14. In 2023:
The first 10 employees' bonus doubles, so their total becomes 2 × 160 = 320.
The middle 20 employees' total remains 920.
The last 10 employees' total becomes 3 × 760 = 2280.
15. New total bonus = 320 + 920 + 2280 = 3520 thousand.
16. Average = 3,520,000 ÷ 40 = Rs. 88,000.
17. Answer: Rs. 88,000.

Why this method works:
F, M, L denote total bonuses of groups of 10,20,10 employees, not their averages. The first-30 and last-30 groups overlap in the middle 20, which is why adding those totals counts M twice.

Independent check and common trap:
Original group averages are ₹16,000,₹46,000,₹76,000. Next-year totals are ₹320,000,₹920,000,₹2,280,000, summing to ₹3,520,000. Divide by all 40 employees, not by three groups of unequal size.

G Strategy shortcut

Spot: Arithmetic – Averages.
Apply: In thousands: F+M=1080, M+L=1680, F+L=920. M=920, F=160, L=760. New total 2F+M+3L=3520; divide 40 →88 thousand.
Check: Original group averages are ₹16,000,₹46,000,₹76,000. Next-year totals are ₹320,000,₹920,000,₹2,280,000, summing to ₹3,520,000. Divide by all 40 employees, not by three groups of unequal size.

G Strategy: Spot: Arithmetic – Averages. Apply: In thousands: F+M=1080, M+L=1680, F+L=920. M=920, F=160, L=760. New total 2F+M+3L=3520; divide 40 →88 thousand. Check: Original group averages are ₹16,000,₹46,000,₹76,000. Next-year totals are ₹320,000,₹920,000,₹2,280,000, summing to ₹3,520,000. Divide by all 40 employees, not by three groups of unequal size.

Question 16Profit, Loss & Discount

TITA (type the answer) · +3 / 0 · easy

Riya incurs a 20% loss when she sells a product at Rs. 4800 to Meera, who in turn sells the product to Kavya. If Riya had sold the product to Meera at the purchase price of Kavya, she would have obtained a 15% profit. Then, the profit, in rupees, made by Meera is:

Your answer

Answer and solution

Correct answer: 2100

Solution

Answer: 2100

1. Riya sells the product for Rs. 4,800 at a loss of 20%.
2. So, Rs. 4,800 is 80% of her cost price.
3. Cost price = 4800 × 100/80 = Rs. 6,000.
4. If Riya had sold the product at the price paid by Kavya, she would have made a profit of 15%.
5. Therefore, Kavya's purchase price was:
6. 6000 × 115/100 = Rs. 6,900.
7. Meera bought the product for Rs. 4,800 and sold it for Rs. 6,900.
8. Meera's profit = 6900 − 4800 = Rs. 2,100.
9. Answer: Rs. 2,100 (TITA).

Why this method works:
The loss and hypothetical profit are both measured against Riya’s original cost price. Meera’s profit, however, must be measured as her own selling price minus her purchase price of ₹4,800. Mixing these two cost bases gives the wrong result.

Independent check and common trap:
₹4,800 is 80% and ₹6,900 is 115% of the same ₹6,000 base. Their difference is 35% of ₹6,000=₹2,100. This directly verifies Meera’s rupee profit; 35% is not Meera’s own profit percentage.

G Strategy shortcut

Spot: Arithmetic – Profit and Loss.
Apply: Riya CP=4800/0.8=6000. Kavya price=1.15×6000=6900. Meera profit=6900−4800=2100.
Check: ₹4,800 is 80% and ₹6,900 is 115% of the same ₹6,000 base. Their difference is 35% of ₹6,000=₹2,100. This directly verifies Meera’s rupee profit; 35% is not Meera’s own profit percentage.

G Strategy: Spot: Arithmetic – Profit and Loss. Apply: Riya CP=4800/0.8=6000. Kavya price=1.15×6000=6900. Meera profit=6900−4800=2100. Check: ₹4,800 is 80% and ₹6,900 is 115% of the same ₹6,000 base. Their difference is 35% of ₹6,000=₹2,100. This directly verifies Meera’s rupee profit; 35% is not Meera’s own profit percentage.

Question 17Percentages

TITA (type the answer) · +3 / 0 · hard

A fruit seller has a stock of mangoes, bananas and apples, with at least one fruit of each type. At the beginning of a day, mangoes constitute 40% of his total stock. During the day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he finds that he has sold exactly 50% of the total fruits. The smallest possible total number of fruits in the stock at the beginning of the day is:

Your answer

Answer and solution

Correct answer: 340

Solution

Answer: 340

1. Let the total number of fruits at the beginning be N.
2. Mangoes make up 40% of the stock, so the seller has 0.4N mangoes. He sells half of them, which means he sells 0.2N mangoes.
3. He also sells 96 bananas and 40% of the apples. Let the number of apples be A.
4. The total number of fruits sold is 50% of the stock:
5. 0.2N + 96 + 0.4A = 0.5N
6. So,
7. 0.4A = 0.3N − 96
A = 0.75N − 240
8. For the number of mangoes and apples, the required percentages must give whole numbers. Since 40% of the stock is an integer, N must be a multiple of 5. Since 40% of the apples must also be a whole number, A must be a multiple of 5. This makes N a multiple of 20.
9. Also, there must be at least one apple, so:
10. 0.75N − 240 > 0
N > 320
11. The smallest multiple of 20 greater than 320 is 340.
12. At N = 340, the number of apples is 15, mangoes are 136 and bananas are 189. All conditions are satisfied.
13. Answer: 340 (TITA).

Why this method works:
Integer fruit counts constrain the solution beyond the sales equation. From 4A=3N−960, N must be divisible by 4; the mango fraction requires N divisible by 5. Together these force N divisible by 20.

Independent check and common trap:
At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.

G Strategy shortcut

Spot: Arithmetic – Percentages.
Apply: A=3N/4−240>0 gives N>320. Whole-fruit sales require N multiple 20. Smallest is 340; check M 136, A 15, B 189 and sales 68+6+96=170.
Check: At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.

G Strategy: Spot: Arithmetic – Percentages. Apply: A=3N/4−240>0 gives N>320. Whole-fruit sales require N multiple 20. Smallest is 340; check M 136, A 15, B 189 and sales 68+6+96=170. Check: At the boundary N=320 the apple count is zero, contradicting at least one of each fruit. The next allowed total is 340. Stocks 136,189,15 allow sales 68,96,6, totaling 170, exactly half the original stock.

Question 20Simple & Compound Interest

MCQ · +3 / −1 · medium

Arjun invests Rs. 25,000 for 4 years in a scheme offering 4% interest per annum, compounded half-yearly. Separately, Varun invests a certain amount in the same scheme for 3 years. At the end of 3 years, he reinvests the entire amount received for one more year at 8% simple interest. If the amounts received by both at the end of 4 years are equal, then the initial investment, in rupees rounded to two decimal places, made by Varun is:

Answer and solution

Correct answer: B

Solution

Answer: B — 24,083.33

1. Arjun invests Rs. 25,000 at 4% per annum, compounded half-yearly.
2. The rate for each six-month period is 2%. In 4 years, there are 8 such periods.
3. Arjun's final amount = 25,000 × (1.02)⁸.
4. Let Varun's initial investment be P.
5. Varun invests for 3 years at the same half-yearly compound rate. There are 6 half-year periods, so his amount becomes P × (1.02)⁶.
6. He then reinvests this amount for one year at 8% simple interest. The final amount becomes:
7. P × (1.02)⁶ × 1.08
8. Both final amounts are equal:
9. 25,000 × (1.02)⁸ = P × (1.02)⁶ × 1.08
10. Cancel (1.02)⁶ from both sides:
11. P = 25,000 × (1.02)² ÷ 1.08
12. P = Rs. 24,083.33 approximately.
13. Answer: Rs. 24,083.33.
14. Rounding check: 25000×1.0404/1.08 = 72250/3 = 24083.333… . Option B is the value rounded to two decimal places, not an exact equality.

Why this method works:
A 4% annual nominal rate compounded half-yearly means 2% per half-year. Both investments share the first six compound periods, so those factors cancel from the equality. Varun’s last year is one simple-interest factor 1.08.

Independent check and common trap:
The remaining growth factor for Arjun is 1.02²=1.0404, less than 1.08. Varun therefore needs less than₹25,000 initially, consistent with₹24,083.33. Keep the exact fraction 72250/3 until the final rounding.

G Strategy shortcut

Spot: Arithmetic – Compound Interest.
Apply: Cancel the common first 3 years. P=25000×1.02²/1.08=72250/3=24083.333…; nearest paisa 24083.33.
Check: The remaining growth factor for Arjun is 1.02²=1.0404, less than 1.08. Varun therefore needs less than₹25,000 initially, consistent with₹24,083.33. Keep the exact fraction 72250/3 until the final rounding.

G Strategy: Spot: Arithmetic – Compound Interest. Apply: Cancel the common first 3 years. P=25000×1.02²/1.08=72250/3=24083.333…; nearest paisa 24083.33. Check: The remaining growth factor for Arjun is 1.02²=1.0404, less than 1.08. Varun therefore needs less than₹25,000 initially, consistent with₹24,083.33. Keep the exact fraction 72250/3 until the final rounding.

Question 21Ratios & Proportion

TITA (type the answer) · +3 / 0 · medium

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?

Your answer

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.

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.

An expected paper by Cetking, not the actual CAT paper. CAT is conducted by the IIMs.