Question 1

CAT 2025 Slot 3: If (x²+ 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is:

Accepted Answer

Answer: B · 44853√3. Given x² + 1/x² = 25 , x > 0 Using identity (a + b)² = a² + b² + 2ab I. x + 1/x: (x + 1/x)² = x² + 2 + 1/x² = 25 + 2 = 27 So, x + 1/x = 3√3 II. x³ + 1/x³: x³ + 1/x³ = (x + 1/x) * (x² + 1/x²) - (x + 1/x) x³ + 1/x³ = (3√3) * 25 - 3√3 = 72√3 III.x⁴ + 1/x⁴: x⁴ + 1/x⁴ = (x² + 1/x²)² - 2 x⁴ + 1/x⁴ = 25² - 2 = 625 - 2 = 623 IV.x⁷ + 1/x⁷: x⁷ + 1/x⁷ = (x³ + 1/x³) * (x⁴ + 1/x⁴) - (x + 1/x) x⁷ + 1/x⁷ = (72√3) * 623 - 3√3 = 44853√3

Question 2

CAT 2025 Slot 3: The ratio of the number of coins in boxes A and B was 17:7. After 108 coins were shifted from box A to box B, this ratio became 37:20. The number of coins to be further shifted from box A to box B to make this ratio 1:1 is:

Accepted Answer

Answer: 272. Correct Answer: 272 Let the number of coins in Box A be 17x and in Box B be 7x. After shifting 108 coins: New number of coins in A = 17x - 108, in B = 7x + 108. The ratio becomes 37:20, so we set up the equation: (17x - 108) / (7x + 108) = 37 / 20 Solving this gives x = 76. Coins in A = 17 * 76 = 1292, and in B = 7 * 76 = 532. After shifting 108 coins: A has 1184 coins, and B has 640 coins. To make the ratio 1:1: Set up the equation for the 1:1 ratio: (1184 - y) / (640 + y) = 1 Solving gives y = 272. Answer 272

Question 3

CAT 2025 Slot 3: The monthly sales of a product from January to April were 120, 135, 150 and 165 units respectively. The cost per unit of the product was Rs 240 per unit and a fixed marked price was used for the product in all the four months. However, 20%, 10% and 5% discount on the marked price

Accepted Answer

Answer: B · 525. Correct Answer: 2 Given: Units sold: January (120), February (135), March (150), April (165) Cost per unit: Rs 240 Discounts: 20% (January), 10% (February), 5% (March), 0% (April) Total profit: Rs 138,825 Let M be the marked price. Revenue for Each Month January: 120 × 0.80M = 96M February: 135 × 0.90M = 121.5M March: 150 × 0.95M = 142.5M April: 165 × M = 165M Total Revenue = 96M + 121.5M + 142.5M + 165M = 525M Total Cost = (120 + 135 + 150 + 165) × 240 = 136800 Profit = Revenue - Cost 138825 = 525M - 136800 525M = 275625 M = 275625 / 525 = 525 The marked price M = Rs 525, so the correct answe

Question 4

CAT 2025 Slot 3: The sum of all possible real values of x for which log x-3 (x² - 9) = log x-3 (x + 1) + 2 is :

Accepted Answer

Answer: C · (3+√33)/2. log (x-3) ((x²−9)/(x+1)) = 2 By definition of log : (x²−9)/(x+1) = (x−3)² Now equating : (x²−9)/(x+1) = x²−6x+9 x²−9 = (x²−6x+9)(x+1) Solve the cubic equation: x³−6x²+3x+18 = 0 Factor the cubic equation: (x−3)(x²−3x−6) = 0 Solve x²−3x−6 = 0 using the quadratic formula: x = (3 ± √33)/2 Sum of solutions: Sum = (3 + √33)/2 Option 3 is the answer

Question 5

CAT 2025 Slot 3: In an arithmetic progression, if the sum of fourth, seventh and tenth term is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is :

Accepted Answer

Answer: 65. Given : Sum of 4th, 7th, and 10th terms = 99 Sum of first 14 terms = 497 Use the formulas for AP (a + 3d) + (a + 6d) + (a + 9d) = 99 This simplifies to: a + 6d = 33 (Equation 1) (14/2) × [2a + 13d] = 497 2a + 13d = 71 (Equation 2) Solve the system of equations From Equation 1: a = 33 - 6d Substitute this into Equation 2: 2(33 - 6d) + 13d = 71 Simplifying gives: d = 5, a = 3 The first 5 terms are: T1 = 3, T2 = 8, T3 = 13, T4 = 18, T5 = 23 Sum of the first 5 terms: 3 + 8 + 13 + 18 + 23 = 65 Answer: 65

Question 6

CAT 2025 Slot 3: Let p, q and r be three natural numbers such that their sum is 900, and r is a perfect square whose values lies between 150 and 500.If p is less than 0.3q and  not more than 0.7q, then the sum of the maximum and minimum possible values of p is: _____

Accepted Answer

Answer: 397. Correct Answer: 397 Give p + q + r = 900 , 0.3q ≤ p ≤ 0.7q r  lies between 150 to 500 and is a perfect square So r = 169 , 196 , 225 , 256 , 289 , 324 , 361 , 400 , 441 , 484 For maximum p (0.7q) , r should be minimum, r = 169 0.7q + 1q + 169 = 900 , q = 430 so p = 0.7 × 430 = 301 For minimum p (0.3q) , r should be maximum r = 484 0.3q + 1q + 484 = 900 , q = 320 , so p = 0.3 × 320 = 96 Pmax + P min = 301 + 96 = 397 Answer 397

Question 7

CAT 2025 Slot 2: If a, b, c and d are integers such that their sum is 46, then the minimum possible value of (a − b)² + (a − c)² + (a − d)² is ______

Accepted Answer

Answer: 2. Minimum possible value is 2 Since a, b, c, d must be integers, they must be the integers closest to 11.5, which are 11 and 12. To satisfy the sum constraint a + b + c + d = 46, we can take values 11, 11, 12, 12. Expression becomes: (11 − 12)² + (11 − 12)² + (11 − 11)² = 1 + 1 + 0 = 2

Question 8

CAT 2025 Slot 1: Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 96

Accepted Answer

Answer: A · 86400. Work in minutes and meters: total time = 3 hr = 180 min, total distance = 224 km = 224000 m. Times are 30,40,50,60 (sum 180 using AP). If speeds are v, v+d, v+2d, v+3d then total distance = v·180 + d·(40 + 2·50 + 3·60) = v·180 + d·320. With v = 960 m/min, v·180 = 960×180 = 172800, so the extra distance to be made up by the d-terms = 224000 − 172800 = 51200. So d = 51200/320 = 160 m/min ⇒ v4 = 960 + 3·160 = 1440 m/min ⇒ distance in 4th part = 1440 × 60 = 86,400 m.

Question 9

CAT 2025 Slot 1: Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B th

Accepted Answer

Answer: 15. Let the trader hold x shares of stock B. Then stock C shares = 20 − x. Prices: A = 120 B = 90 C = 150 Portfolio value equation: 10(120) + x(90) + (20 − x)(150) = 3300 1200 + 90x + 3000 − 150x = 3300 1200 + 3000 + (90x − 150x) = 3300 4200 − 60x = 3300 60x = 900 x = 15 Answer: 15 shares of stock B.

Question 10

CAT 2025 Slot 1: If a -6b + 6c = 4 and 6a + 3b - 3c = 50, where a, b, and c are real numbers, the value of 2a + 3b - 3c is

Accepted Answer

Answer: D · 18. Step 1 From eq(2): 6a + 3b − 3c = 50 → take one-third → 2a + b − c = 50/3 Step 2 Compare target with this: Target = (2a + b − c) + 2(b − c) Step 3 Find (b − c) from eq(1): a − 6b + 6c = 4 → rewrite → a = 4 + 6(b − c) Step 4 Substitute into step 1: 2(4 + 6x) + x = 50/3 8 + 12x + x = 50/3 13x = 26/3 → x = 2/3 So (b − c) = 2/3 Step 5 Plug in: Target = 50/3 + 2*(2/3) = 18

Question 11

CAT 2025 Slot 1: If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____

Accepted Answer

Answer: 60. Let the diagonals be d1 and d2. Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792. In a rhombus, diagonals bisect each other at right angles. So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2. Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184. Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2). Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600. So |d1 − d2| = √3600 = 60. Final Answer: 60

Question 12

CAT 2024 Slot 3: For some constant real numbers p, k and a, consider the system of linear equations: px - 4y = 2 and 3x + ky = a. A necessary condition for the system to have no solution for (x,y) is:

Accepted Answer

Answer: pk = -12. Parallel: p/3 = -4/k Not coincident: 2a + k ≠ 0

Question 13

CAT 2024 Slot 3: Rajesh owns 20 hectares, Vimal owns 30 hectares with wheat and mustard in 5:3 ratio. Total wheat:mustard is 11:9. Find Rajesh's ratio.

Accepted Answer

Answer: 7:9. Rajesh wheat =9.5, mustard=10.5 → ratio 19:21 → closest 7:9

Question 14

CAT 2024 Slot 3: Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value after 5 years is 25:36, then the minimum number of years required for the value to exceed Rs 20000 is _____

Accepted Answer

Answer: 9. (1 + r)^3 / (1 + r)^5 = 25/36 → (1 + r)^2 = 36/25 = 1.44 → 1 + r =1.2 → r=20% 4000*(1.2)^n > 20000 → (1.2)^n>5 n=8:4.3, n=9:5.16 → n=9

Question 15

CAT 2024 Slot 2: 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, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up s

Accepted Answer

Answer: 940. Fruit Seller Problem Problem: A fruit seller has a stock of mangoes, bananas and apples, with at least one fruit of each type. Initially, mangoes make up 40% of the stock. He sells: Half of the mangoes 96 bananas 40% of the apples By the end of the day, he has sold 50% of his total stock. Find the smallest possible total number of fruits at the beginning of the day. Solution: Let total fruits be $ x $. Then: Mangoes = $ 0.4x $ Let bananas = $ b $ Then apples = $ x - 0.4x - b = 0.6x - b $ Fruits sold: Mangoes sold = $ \frac{1}{2} \times 0.4x = 0.2x $ Bananas sold = 96 Apples sold = \(

Question 16

CAT 2024 Slot 2: Find the value of x? $$ \left( x + 6\sqrt{2} \right)^{\frac{1}{2}} - \left( x - 6\sqrt{2} \right)^{\frac{1}{2}} = 2\sqrt{2}, $$

Accepted Answer

Answer: 11. Square Root Equation Solution Solution Explanation using Squaring We are given the equation: $$ \sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}} = 2\sqrt{2} $$ ✅ Step 1: Square both sides Using the identity $ (a - b)^2 = a^2 + b^2 - 2ab $: $$ (\sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}})^2 = (2\sqrt{2})^2 $$ $$ = (x + 6\sqrt{2}) + (x - 6\sqrt{2}) - 2\sqrt{(x + 6\sqrt{2})(x - 6\sqrt{2})} $$ $$ = 2x - 2\sqrt{x^2 - 72} = 8 $$ Divide both sides by 2: $$ x - \sqrt{x^2 - 72} = 4 $$ ✅ Step 2: Square both sides again $$ \sqrt{x^2 - 72} = x - 4 $$ $$ \Rightarrow x^2 - 72 = (x - 4)^2 = x^2 - 8x + 16 $$ \[

Question 17

CAT 2024 Slot 2: Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest

Accepted Answer

Answer: 20812. Compound and Simple Interest Problem Problem: Anil invests ₹22000 for 6 years in a scheme offering 4% interest per annum, compounded half-yearly. Sunil invests a certain amount in the same scheme for 5 years and then reinvests the entire amount for 1 year at 10% simple interest. If both receive the same amount at the end of 6 years, find Sunil's initial investment. Solution: Anil’s Investment: Compound interest formula: $$ A = P \left(1 + \frac{r}{100}\right)^n $$ - Principal $ P = 22000 $ - Rate per half-year = $ \frac{4}{2} = 2\% $ - Time = $ 6 \times 2 = 12 $ half-years \[ A_{\text{An

Question 18

CAT 2024 Slot 1: A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop

Accepted Answer

Answer: D · 42. Step 1: Understand the problem Let the initial quantity of grains be $ x $ kg. Step 2: Simplify after the first customer The first customer takes $ \frac{x}{2} + 3 $: Remaining grains = $ x - \left(\frac{x}{2} + 3\right) = \frac{x}{2} - 3 $. Step 3: Simplify after the second customer The second customer takes $ \frac{1}{2}\left(\frac{x}{2} - 3\right) + 3 = \frac{x}{4} + \frac{3}{2} $. Remaining grains = $ \frac{x}{2} - 3 - \left(\frac{x}{4} + \frac{3}{2}\right) = \frac{x}{4} - \frac{9}{2} $. Step 4: Simplify after the third customer The third customer takes \( \frac{1}{2}\left(\frac{

Question 19

CAT 2023 Slot 3: Let $ a $ and $ b $ be two sequences such that $ a_n = 13 + 6(n-1) $ and $ b_n = 15 + 7(n-1) $ for all natural numbers $ n $. Then, the largest three-digit integer that is common to both these sequences is:

Accepted Answer

Answer: C · 967. Using properties of arithmetic sequences, we find that the common terms of the two sequences form another arithmetic sequence. Calculating, the largest three-digit common term is 967.

Question 20

CAT 2023 Slot 3: A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process is repeated until the volume of milk becomes less than that of water. The smallest number of times this process is repeated to achieve this is:

Accepted Answer

Answer: C · 7. Using the formula for repeated removal and replacement, we find that the volume of milk becomes less than water after 7 repetitions.

Question 21

CAT 2023 Slot 3: If a person D joins a group of three people, the average weight of the group reduces by $ x $ kg. If another person E joins instead of D, the average increases by $ 2x $ kg. If E’s weight is 12 kg more than D’s, then the value of $ x $ is:

Accepted Answer

Answer: A · 1. Setting up equations for average changes, we solve and find that $ x = 1 $.

Question 22

CAT 2023 Slot 3: In a fruit market, there are initially 40% mangoes, 30% bananas, and 30% apples. If 50% of the mangoes, 30 bananas, and 40% of the apples are sold, what is the minimum number of fruits required initially to maintain a balance?

Accepted Answer

Answer: C · 150. By setting up the initial percentages and solving for minimum quantities, the required minimum initial number of fruits is 150.

Question 23

CAT 2023 Slot 3: Let $ 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:

Accepted Answer

Answer: A · 79. Using the given conditions and properties of arithmetic progressions, we find that $ a_1 = 79 $.

Question 24

CAT 2023 Slot 2: For some positive real number $ x $, if $ \log_{\sqrt{3}}(x) + \log_x(25) = \log_x(0.008) $, then the value of $ x $ is:

Accepted Answer

Answer: A · 7. Given $ \log_{\sqrt{3}}(x) + \log_x(25) = \log_x(0.008) $. Using properties of logarithms, solve step-by-step to get $ x = 7 $.

Question 25

CAT 2023 Slot 2: In a right-angled triangle, if the base is 12 cm and the height is 5 cm, then the length of the hypotenuse is:

Accepted Answer

Answer: A · 13. Using Pythagoras theorem: $ \sqrt{12^2 + 5^2} = 13 $.

Question 26

CAT 2023 Slot 2: Minu buys a pair of sunglasses at Rs. 1000 and sells to Kanu at 20% profit. Then, Kanu sells it back to Minu at 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs. 500, then the percentage of profit

Accepted Answer

Answer: B · 31.25%. After all transactions, Minu's final profit percentage is calculated using: $ \frac{500}{1600} \times 100 = 31.25\% $.

Question 27

CAT 2023 Slot 2: If $ \log_2(x + 12) = 4 $ and $ 3 \log_2 x = 1 $, then $ x + y $ equals:

Accepted Answer

Answer: D · 10. Solving the equations step-by-step, we get $ x = 4 $ and $ y = 6 $. Thus, $ x + y = 10 $.

Question 28

CAT 2023 Slot 2: A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the $ n $-th day exceeds one million,

Accepted Answer

Answer: D · 19. Let the number of organisms on day 1 be 2. Using the recurrence relation, the count on day $ n $ will exceed one million when $ n = 19 $.

Question 29

CAT 2023 Slot 2: In a lab experiment, the number of organisms increases exponentially. On day 1, the number of organisms is 2. What is the number on day 19?

Accepted Answer

Answer: D · 8,192. Using the exponential growth formula, the count on day 19 is calculated as $2^{18} = 8,192$.

Question 30

CAT 2023 Slot 2: The sum of all possible values of $x$ satisfying the equation $2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$, is

Accepted Answer

Answer: C · 1/2. Divide throughout by $2^{2x^{2}+x+16}$ to get $2^{2x^{2}-x-16}+2^{-2x^{2}+x+14}=1$. Put $t=2x^{2}-x-16$, so the second term is $2^{-t-2}$ and $2^{t}+2^{-t-2}=1$. With $y=2^{t}$ this is $y+\frac{1}{4y}=1$, i.e. $4y^{2}-4y+1=0$, i.e. $(2y-1)^{2}=0$, so $y=\frac{1}{2}$ and $t=-1$. Then $2x^{2}-x-16=-1$, i.e. $2x^{2}-x-15=0$, i.e. $(2x+5)(x-3)=0$, giving $x=3$ or $x=-\frac{5}{2}$. Sum $=3-\frac{5}{2}=\frac{1}{2}$.

Question 31

CAT 2023 Slot 2: A sum of money is shared equally among n people, giving each Rs 352. If instead two people get Rs 506 each and the rest is shared equally among the other people, each of those receives at most Rs 330. What is the maximum possible value of n?

Accepted Answer

Answer: 16. Total $T=352n$. In the second case the remaining $(n-2)$ people get at most Rs 330 each, so $352n \le 2(506)+330(n-2)$. Hence $352n \le 1012+330n-660$, i.e. $22n \le 352$, so $n \le 16$. Maximum $n=16$.

Question 32

CAT 2023 Slot 2: If $p^{2}+q^{2}-29=2pq-20=52-2pq$, then the difference between the maximum and minimum possible value of $p^{3}-q^{3}$ is

Accepted Answer

Answer: D · 378. From $2pq-20=52-2pq$ we get $4pq=72$, so $pq=18$. From $p^{2}+q^{2}-29=2pq-20$ we get $p^{2}+q^{2}-2pq=9$, i.e. $(p-q)^{2}=9$, so $p-q=\pm 3$. Using $p^{3}-q^{3}=(p-q)^{3}+3pq(p-q)$, if $p-q=3$ then $27+162=189$, and if $p-q=-3$ then $-27-162=-189$. Difference $=189-(-189)=378$.

Question 33

CAT 2023 Slot 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:

Accepted Answer

Answer: A · 2. - 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 34

CAT 2023 Slot 1: If $\frac{\sqrt{5x + 9}}{\sqrt{5x - 9}} = 3(\sqrt{2} + 1)$, then find the value of $ \sqrt{10x + 9} $:

Accepted Answer

Answer: A · 3√7. 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 35

CAT 2023 Slot 1: 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 co

Accepted Answer

Answer: D · 5 : 9. 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 36

CAT 2023 Slot 1: 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 m

Accepted Answer

Answer: D · 47%. 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 37

CAT 2023 Slot 1: If $ \log_2 (x + 12) = 4 $ and $ 3 \log_2 x = 1 $, then $ x + y $ equals:

Accepted Answer

Answer: D · 10. 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 38

CAT 2023 Slot 1: 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 G

Accepted Answer

Answer: A · 26%. 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 2

Question 39

CAT 2022 Slot 2: Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is __

Accepted Answer

Answer: 20. Let the capital be $C$. Annual interest rate on the whole capital $= \frac{1}{5}(6) + \frac{1}{3}(10) + \frac{7}{15}(1) = 1.2 + 3.33 + 0.47 = 5\%$. Interest each year is $0.05C$. We need $0.05C 	imes n \ge C$, so $n \ge 20$. Answer: 20 years.

Question 40

CAT 2022 Slot 2: Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his mont

Accepted Answer

Answer: B · 4400. Total saving required = 550*12 = 6600 Total expenses = 9*3500 + 3*3700 = 42600 So, total earning = 42600+6600 = 49200 If earning from 10th month = x Then 9*4000 + 3x = 49200 x = 4400 Let's break down the given information step by step: The total saving required is calculated as 550 multiplied by 12, resulting in 6600. The total expenses are calculated by multiplying 9 by 3500 (for the first 9 months) and 3 by 3700 (for the last 3 months), which amounts to 42600. Adding the total expenses to the total saving required gives us the total earning: 42600 + 6600 = 49200. Now, let's consider the earn