Quant 2 in CMAT 2025 Slot 1 Actual Paper
11. Alpha and Beta are two chemical fertilizers. Alpha consists of N, P, K and Beta consists of N and P. A mixture of Alpha and Beta is prepared in which the ratio of N, P and K is 26%, 68% and 6%. The ratio of N, P, K in Alpha is 20%, 70% and 10%. What is the ratio of N and P in Beta?
Correct Option: C
Step-by-Step Solution:
Let Alpha = ‘a’ and Beta = ‘b’.
In Alpha: N=20%, P=70%, K=10%. In Beta: N=x%, P=y%.
Final Mixture: N=26%, P=68%, K=6%.
Since K is only in Alpha: 10% of a = 6% of (a+b)
0.10a = 0.06a + 0.06b → 0.04a = 0.06b → a:b = 3:2.
Apply weighted average for Nitrogen: (3*20 + 2*x)/5 = 26
60 + 2x = 130 → 2x = 70 → x = 35%.
Phosphorus = 100 – 35 = 65%.
Let Alpha = ‘a’ and Beta = ‘b’.
In Alpha: N=20%, P=70%, K=10%. In Beta: N=x%, P=y%.
Final Mixture: N=26%, P=68%, K=6%.
Since K is only in Alpha: 10% of a = 6% of (a+b)
0.10a = 0.06a + 0.06b → 0.04a = 0.06b → a:b = 3:2.
Apply weighted average for Nitrogen: (3*20 + 2*x)/5 = 26
60 + 2x = 130 → 2x = 70 → x = 35%.
Phosphorus = 100 – 35 = 65%.
12. A man takes 4 hours 20 minutes in walking to a certain place and riding back. If he walks on both sides, he loses 1 hour. The time he would take by riding both ways is:
Correct Option: C
Step-by-Step Solution:
W + R = 260 min (4hr 20min).
Walking both ways (2W) = 260 + 60 (loses 1hr) = 320 min.
W = 160 min.
R = 260 – 160 = 100 min.
Riding both ways (2R) = 200 min = 3 hours 20 min.
W + R = 260 min (4hr 20min).
Walking both ways (2W) = 260 + 60 (loses 1hr) = 320 min.
W = 160 min.
R = 260 – 160 = 100 min.
Riding both ways (2R) = 200 min = 3 hours 20 min.
13. In a race A, B and C take part. A beats B by 30 meters, B beats C by 20 meters and A beats C by 48 meters.
Statement I: The length of the race is 300 meters.
Statement II: The speed of A, B and C are in the ratio 50:45:40.
Statement I: The length of the race is 300 meters.
Statement II: The speed of A, B and C are in the ratio 50:45:40.
Correct Option: C
Step-by-Step Solution:
Let race distance = D.
(D-30)/D * (D-20)/D = (D-48)/D
(D-30)(D-20) = D(D-48) → D² – 50D + 600 = D² – 48D → 2D = 600 → D = 300m (Statement I True).
Ratio A:B = 300:270 = 10:9. Ratio B:C = 300:280 = 15:14.
Combined Ratio A:B:C = 10:9:8.4. Not 50:45:40 (Statement II False).
Let race distance = D.
(D-30)/D * (D-20)/D = (D-48)/D
(D-30)(D-20) = D(D-48) → D² – 50D + 600 = D² – 48D → 2D = 600 → D = 300m (Statement I True).
Ratio A:B = 300:270 = 10:9. Ratio B:C = 300:280 = 15:14.
Combined Ratio A:B:C = 10:9:8.4. Not 50:45:40 (Statement II False).
14. Anand’s income is 140 more than Biren’s income and Chandu’s income is 80 more than Deepak’s. If the ratio of Anand’s & Chandu’s income is 2:3 and the ratio of Biren’s and Deepak’s income is 1:2, then the incomes are:
Correct Option: C
Step-by-Step Solution:
Biren = x, Anand = x + 140.
Deepak = y, Chandu = y + 80.
Ratio Biren:Deepak = 1:2 → y = 2x.
Ratio Anand:Chandu = (x+140)/(2x+80) = 2/3.
3x + 420 = 4x + 160 → x = 260.
Biren=260, Anand=400, Deepak=520, Chandu=600.
Biren = x, Anand = x + 140.
Deepak = y, Chandu = y + 80.
Ratio Biren:Deepak = 1:2 → y = 2x.
Ratio Anand:Chandu = (x+140)/(2x+80) = 2/3.
3x + 420 = 4x + 160 → x = 260.
Biren=260, Anand=400, Deepak=520, Chandu=600.
15. Arrange the following in descending order (Refer to official question paper for expressions).
Correct Option: C
Solution follows the official CMAT comparison logic for the given surds/powers.
16. A fair coin is tossed repeatedly. If tail appears on first four tosses, then the probability of head appearing on fifth toss equals:
Correct Option: B (1/2)
Logic: Each toss of a fair coin is an independent event. The outcome of previous tosses has no influence on the current toss. Probability remains 1/2.
17. Equal amounts of 1,000 lent for 3 years @ 30% SI and 30% CI. By how much percent is the CI greater than the SI?
Correct Option: B
Step-by-Step Solution:
SI = (1000 * 30 * 3) / 100 = 900.
CI Amount = 1000 * (1.3)³ = 2197.
CI = 2197 – 1000 = 1197.
Difference = 1197 – 900 = 297.
% Greater = (297/900) * 100 = 33%.
SI = (1000 * 30 * 3) / 100 = 900.
CI Amount = 1000 * (1.3)³ = 2197.
CI = 2197 – 1000 = 1197.
Difference = 1197 – 900 = 297.
% Greater = (297/900) * 100 = 33%.
18. A merchant earns 25% profit. 25% of his consignment was stolen. He sold the rest by increasing SP by 20%. New profit/loss?
Correct Option: C
Step-by-Step Solution:
Let CP = 100. Original SP = 125.
Stolen 25% → Remaining 75 units.
New SP per unit = 1.25 * 1.20 = 1.5.
Total New SP = 75 * 1.5 = 112.5.
Profit = 112.5 – 100 = 12.5%.
Let CP = 100. Original SP = 125.
Stolen 25% → Remaining 75 units.
New SP per unit = 1.25 * 1.20 = 1.5.
Total New SP = 75 * 1.5 = 112.5.
Profit = 112.5 – 100 = 12.5%.
19. Statement I: The perimeter of a triangle is greater than the sum of its three medians.
Statement II: In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.
Statement II: In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.
Correct Option: A
Explanation:
Statement I: Correct geometric property.
Statement II: Correct as per triangle inequality theorem (Sum of two sides > 2 * median).
Statement I: Correct geometric property.
Statement II: Correct as per triangle inequality theorem (Sum of two sides > 2 * median).
20. Match List I (Quadratic Equations) with List II (Roots):
| Equation | Roots |
|---|---|
| (A) x² + 17x + 52 = 0 | (II) -13, -4 |
| (B) x² – 17x + 52 = 0 | (IV) 13, 4 |
Correct Option: D
Matching roots based on sum and product of roots. Option D aligns with official matching.
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