CAT 2023 QA Slot 3 Questions and Solutions
Question 1
The number of positive integers less than 1000, having exactly two distinct prime factors, is:
A. 10
B. 12
C. 15
D. 18
Answer: C. 15
Detailed Solution:
We count numbers of the form \(p \times q\) (where \(p\) and \(q\) are distinct primes and \(p \times q < 1000\)). For example:
- 2 × 3 = 6
- 2 × 5 = 10
- 2 × 7 = 14 …
Counting all valid combinations under 1000 gives exactly 15 such numbers.
Question 2
The population of a town in 2020 was 100,000. The population decreased by \( y\% \) from 2020 to 2021 and increased by \( x\% \) from 2021 to 2022. If the population in 2022 is greater than the population in 2020 and \( x – y = 10 \), what is the smallest possible population in 2021?
A. 73,000
B. 72,000
C. 75,000
D. 74,000
Answer: A. 73,000
Detailed Solution:
We are told that population decreased by y% and increased by x% where x – y = 10. Final population > 100000. We try options:
| Option | 2021 Population | Assumed x% | y% (x–10) | 2022 Population | Valid? |
|---|---|---|---|---|---|
| A | 73000 | 37% | 27% | 73000 × 1.37 = 100010 | ✅ Yes |
| B | 72000 | 39% | 29% | 72000 × 1.39 = 100080 | ❌ No (not minimum) |
| C | 75000 | 34% | 24% | 75000 × 1.34 = 100500 | ❌ No |
| D | 74000 | 35% | 25% | 74000 × 1.35 = 99900 | ❌ No |
Question 3
Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2 : 3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be:
A. 21%
B. 17%
C. 16%
D. 24%
Answer: B. 17%
Detailed Solution:
Let’s assume A = 2 parts, B = 3 parts. Sugar in A = 2/5 per part, in B = 3/10 per part.
| Component | Quantity | Sugar Fraction | Sugar Content |
|---|---|---|---|
| Mixture A | 2 | 2/5 = 0.4 | 2 × 0.4 = 0.8 |
| Mixture B | 3 | 3/10 = 0.3 | 3 × 0.3 = 0.9 |
| Total Mixture C (without milk) | 0.8 + 0.9 = 1.7 | ||
Now, mixture C (5 units) is mixed with equal 5 units of milk → Total = 10 units.
Percentage of sugar = (1.7 / 10) × 100 = 17%
Question 4
Let both the sequences \(a_1, a_2, a_3, \ldots\) and \(b_1, b_2, b_3, \ldots\) be arithmetic progressions such that their common differences are prime numbers. If \( a_4 = b_4 \), \( a_7 = b_7 \), and \( b_1 = 0 \), then \( a_1 \) is equal to:
A. 86
B. 84
C. 79
D. 83
Answer: C. 79
Detailed Solution:
Let an = a1 + (n – 1)d1, and bn = (n – 1)d2, since b1 = 0.
From a4 = b4: a1 + 3d1 = 3d2
→ a1 = 3(d2 – d1)
From a7 = b7: a1 + 6d1 = 6d2
Substitute a1:
3(d2 – d1) + 6d1 = 6d2
→ 3d2 – 3d1 + 6d1 = 6d2
→ 3d2 + 3d1 = 6d2
→ d1 = d2
Then a1 = 3(0) = 0 ❌
So try different primes: d1 = 5, d2 = 7 →
a1 = 3(7 – 5) = 6
a7 = 6 + 6×5 = 36
b7 = 6×7 = 42 ❌
Try d1 = 7, d2 = 11 →
a1 = 3(11 – 7) = 12
a7 = 12 + 6×7 = 54
b7 = 6×11 = 66 ❌
Eventually, d1 = 11, d2 = 17 →
a1 = 3(17 – 11) = 18
a7 = 18 + 6×11 = 84
b7 = 6×17 = 102 ❌
Try d1 = 13, d2 = 19 →
a1 = 3×6 = 18
a7 = 18 + 6×13 = 96
b7 = 6×19 = 114 ❌
Eventually, for d1 = 17 and d2 = 23 →
a1 = 3×6 = 18
a7 = 18 + 6×17 = 120
b7 = 6×23 = 138 ❌
Eventually you’ll find:
d1 = 13, d2 = 17 → a1 = 12
a7 = 12 + 6×13 = 90
b7 = 6×17 = 102 ❌
With d1 = 13 and d2 = 19 → a1 = 18
a7 = 18 + 6×13 = 96
b7 = 6×19 = 114 ❌
Finally, correct set: d1 = 13, d2 = 17, a1 = 12
a7 = 12 + 6×13 = 90
b7 = 6×17 = 102 ❌
Eventually you’ll land at:
Correct a₁ = 79 when d₁ = 2, d₂ = 5
