CAT 2023 QA Questions and Solutions
Question 1
In a right-angled triangle ∆ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ∆ABP, ∆ABQ, and ∆ABC are in arithmetic progression. If the area of ∆ABC is 1.5 times the area of ∆ABP, the length of PQ, in cm, is:
A. 2
B. 4
C. 5
D. 7
Answer: A. 2
Detailed Solution:
– Given AB = 5 cm and BC = 12 cm.
– The area of ∆ABC = \( \frac{1}{2} \times 5 \times 12 = 30 \, \text{cm}^2\).
– Since the area of ∆ABC is 1.5 times the area of ∆ABP, the area of ∆ABP = 20 cm² and the area of ∆ABQ = 30 cm².
– Therefore, the length of PQ = 2 cm.
Question 2
If \(\frac{\sqrt{5x + 9}}{\sqrt{5x – 9}} = 3(\sqrt{2} + 1)\), then find the value of \( \sqrt{10x + 9} \):
A. 3√7
B. 3√5
C. 4√3
D. 7√3
Answer: A. 3√7
Detailed Solution:
Given, \(\frac{\sqrt{5x + 9}}{\sqrt{5x – 9}} = 3(\sqrt{2} + 1)\)
– Simplifying, we get \( \sqrt{5x + 9} = 3(\sqrt{2} + 1) \sqrt{5x – 9} \).
– Solving, we find that \( x = 9 \).
– Therefore, \( \sqrt{10x + 9} = 3√7 \).
Question 3
A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with the same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16:9, then the ratio of cocoa in mixture P to that in mixture Q is:
A. 4 : 9
B. 1 : 3
C. 1 : 2
D. 5 : 9
Answer: D. 5 : 9
Detailed Solution:
Let the initial quantity of coffee in the jar be 100 kg, and r kg is replaced each time.
After first replacement, the quantity of coffee remaining = \( 100 \times \left( \frac{100 – r}{100} \right) = 100 – r \).
After second replacement, the quantity of coffee remaining = \( 100 \times \left( \frac{100 – r}{100} \right)^2 = 64 \).
Solving, r = 20 kg.
The required ratio = 5 : 9.
Question 4
Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to:
A. 49%
B. 42%
C. 45%
D. 47%
Answer: D. 47%
Detailed Solution:
Let the selling price be Rs. x.
Cost price of object A = \( \frac{x}{1.2} \) and for B = \( \frac{x}{0.9} \).
Assume x = 108 (LCM of 12 and 9).
Cost price of A = 90, and for B = 120.
New selling price of B for 10% profit = 132.
Profit percentage on A = \( \frac{(132 – 90)}{90} \times 100 = 46.67 \approx 47 \% \).
Question 5
If \( \log_2 (x + 12) = 4 \) and \( 3 \log_2 x = 1 \), then \( x + y \) equals:
A. 11
B. 68
C. 20
D. 10
Answer: D. 10
Detailed Solution:
Given, \( \log_2 (x + 12) = 4 \)
This implies \( x + 12 = 2^4 = 16 \).
Therefore, \( x = 16 – 12 = 4 \).
Now, \( 3 \log_2 x = 1 \) gives \( \log_2 x = \frac{1}{3} \).
\( \log_2 x = 2 \) => Value becomes x.
Question 6
The salaries of three friends Sita, Gita, and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25%, and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is:
A. 26%
B. 25%
C. 30%
D. 28%
Answer: A. 26%
Detailed Solution:
Let the initial salaries of Sita, Gita, and Mita be 5x, 6x, and 7x, respectively.
After the first year’s hike:
– Sita’s salary = 5x × 1.2 = 6x.
– Gita’s salary = 6x × 1.25 = 7.5x.
– Mita’s salary = 7x × 1.2 = 8.4x.
After the second year’s hike:
– Sita’s salary = 6x × 1.4 = 8.4x.
– Mita’s salary = 8.4x × 1.25 = 10.5x.
Since Gita’s salary is the average of Sita’s and Mita’s salaries after the second year:
\( Gita’s\, salary = \frac{8.4x + 10.5x}{2} = 9.45x \).
To calculate Gita’s percentage hike in the second year:
\( \frac{9.45x – 7.5x}{7.5x} \times 100 = \frac{1.95x}{7.5x} \times 100 \approx 26\% \).
