Q: CAT 2024 Slot 3: If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8, and 8^f = 9, then what is the value of abcdef?

Answer: 2. a = log4 / log3, b = log5 / log4, c = log6 / log5, d = log7 / log6, e = log8 / log7, f = log9 / log8 abcdef = log9 / log3 = log base 3 of 9 = 2

Q: CAT 2024 Slot 3: $ a $ and $ b $ are natural numbers, then the value of $ a + b $ is: \[ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} \]

Answer: 8. Surd Algebra Question $ a $ and $ b $ are natural numbers, then the value of $ a + b $ is: \[ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} \] ✅ Solution: Expanding the left-hand side: \[ (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2 \] Match with the right-hand side: \[ a^2 + 3b^2 = 52 \quad \text{(1)} \\ 2ab = 30 \Rightarrow ab = 15 \quad \text{(2)} \] If $ ab = 15 $ and $ a^2 + 3b^2 = 52 $, where $ a $ and $ b $ are natural numbers, find $ a + b $. Solution using Hit and Trial: We try natural number pairs such that $ ab = 15 $. $ a $ $ b $ $ a^2 + 3b^2 $ Result 1 15 \( 1^2 + 3

Q: CAT 2024 Slot 2: Logarithm Expression Problem If $ a, b, c $ are positive real numbers such that: \[ a > 10 \geq b \geq c \quad \text{and} \quad \frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3} \] Then the greatest possible integer value of $ a $ is?

Answer: 14. Greatest Integer Value of a Problem: If $ a, b, c $ are positive real numbers such that: $ a > 10 \geq b \geq c $, and \[ \frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3}, \] then find the greatest possible integer value of $ a $. Step 1: Change of Base Using $ \log_b x = \frac{\log_k x}{\log_k b} $: \[ \log_8(a + b) = \frac{\log_2(a + b)}{\log_2 8} = \frac{\log_2(a + b)}{3}, \quad \log_{27}(a - b) = \frac{\log_3(a - b)}{3} \] So the given equation becomes: \[ \frac{1}{3} \cdot \frac{\log_2(a + b)}{\log_2 c} + \frac{1}{3} \cdot \frac{\log_3(a - b)}{\log_

Q: CAT 2023 Slot 3: Let $ n $ be any natural number such that $ 5^{n-1} < 3^{n+1} $. The least integer value of $ m $ that satisfies $ 3^{n+1} < 2^{n+m} $ for all such $ n $ is?

Answer: C · 5. The least value of $ m $ is 5. For $ n = 5 $, we need $ 3^6 < 2^{5 + m} $. This gives $ 729 < 2^{10} $, so $ m = 5 $.

Q: CAT 2022 Slot 2: The number of distinct integer values of n satisfying 4−log^2n/3−log4^n<0, is

Answer: 47. 4−logn23−logn4 16, it is negative. For any value of n 64, it is negative. For a fraction to be negative, the numerator and the denominator must be of opposite signs. In this case, n should be between 16 and 64. The number of values that n can take = 63 - 16 = 47.

Question 1

CAT 2025 Slot 3: If 12¹²^x × 4²⁴^(x +)¹² × 5²^y = 8⁴^z × 20¹²^x × 243^(3x-6) , where x, y and z are natural numbers, then x+y+z equals_____

Accepted Answer

Answer: 112. 12¹²^x × 4²⁴^x ^+ ¹² × 5²^y = 8⁴^z × 20¹²^x × 243^(3x-6) 12 = 2² × 3 , 4 = 2² , 8 = 2³ , 20 = 2² × 5 , 243 = 3⁵ Thus expression can be written as : 2^(24x) × 3^(12x) × 2^(48x+24) × 5^(2y) = 2^(12z) × 2²⁴^x× 5^(12x) × 3^(15x - 30) Cancels out the same powers and comparing the powers 12x = 15x - 30 , x = 10 2y = 12x , 2y = 12(10) , y = 60 12 z = 48x + 24 , 12z = 504 , z = 42 x + y + z = 10 + 60 + 42 = 112 Answer 112

Question 2

CAT 2025 Slot 2: The sum of digits of the number (625)⁶⁵ × (128)³⁶ is ______

Accepted Answer

Answer: 25. Final answer 25 Given: (625)⁶⁵ × (128)³⁶ Express the bases as powers of 5 and 2: 625 = 5⁴ and 128 = 2⁷ (625)⁶⁵ × (128)³⁶ = 5^(4×65) × 2^(7×36) = 5^260 × 2^252 = 5^8 × 5^252 × 2^252 = 5^8 × 10^252 Now 5⁸ = 390625 So the full number is: 390625 followed by 252 zeros. Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 = 25 Answer: 25

Question 3

CAT 2025 Slot 2: If log₆₄(x²) + log₈(√y) + 3 log₅₁₂(√(y)z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is

Accepted Answer

Answer: B · 48. Given log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4 x, y, z > 0 Step 1: Convert all logs to the same base (base 8) 64 = 8²; 512 = 8³ Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z) Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4 Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096 Apply the Arithme

Question 4

CAT 2025 Slot 1: The number of distinct integers n for which log_(1/4) of (n² − 7n + 11) > 0 is _____.

Accepted Answer

Answer: 0. log_(1/4) of (n² − 7n + 11) > 0 is _____. Using DPAC Strategy: Cycle n² − 7n + 11 < (1/4)^0 → n² − 7n + 11 < 1 Now we need values of n which satisfy 0 < n² − 7n + 11 < 1 Compute E(n) = n² − 7n + 11 for nearby integers: n = 1 → E = 5 (not in (0,1)) n = 2 → E = 1 (equals 1, excluded) n = 3 → E = −1 (negative, excluded) n = 4 → E = −1 (negative, excluded) n = 5 → E = 1 (equals 1, excluded) For n ≤ 0 or n ≥ 6, E ≥ 5 (too large). Therefore no integer n satisfies 0 < E < 1. Final answer: 0. Note: Official OG says answer is 2. Which is challenged by Cetking

Question 5

CAT 2024 Slot 3: If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8, and 8^f = 9, then what is the value of abcdef?

Accepted Answer

Answer: 2. a = log4 / log3, b = log5 / log4, c = log6 / log5, d = log7 / log6, e = log8 / log7, f = log9 / log8 abcdef = log9 / log3 = log base 3 of 9 = 2

Question 6

CAT 2024 Slot 3: $ a $ and $ b $ are natural numbers, then the value of $ a + b $ is: $$ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} $$

Accepted Answer

Answer: 8. Surd Algebra Question $ a $ and $ b $ are natural numbers, then the value of $ a + b $ is: $$ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} $$ ✅ Solution: Expanding the left-hand side: $$ (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2 $$ Match with the right-hand side: $$ a^2 + 3b^2 = 52 \quad \text{(1)} \ 2ab = 30 \Rightarrow ab = 15 \quad \text{(2)} $$ If $ ab = 15 $ and $ a^2 + 3b^2 = 52 $, where $ a $ and $ b $ are natural numbers, find $ a + b $. Solution using Hit and Trial: We try natural number pairs such that $ ab = 15 $. $ a $ $ b $ $ a^2 + 3b^2 $ Result 1 15 \( 1^2 + 3

Question 7

CAT 2024 Slot 2: Logarithm Expression Problem If $ a, b, c $ are positive real numbers such that: $$ a > 10 \geq b \geq c \quad \text{and} \quad \frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3} $$ Then the greatest possible integer value of $ a $ is?

Accepted Answer

Answer: 14. Greatest Integer Value of a Problem: If $ a, b, c $ are positive real numbers such that: $ a > 10 \geq b \geq c $, and $$ \frac{\log_8(a + b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3}, $$ then find the greatest possible integer value of $ a $. Step 1: Change of Base Using $ \log_b x = \frac{\log_k x}{\log_k b} $: $$ \log_8(a + b) = \frac{\log_2(a + b)}{\log_2 8} = \frac{\log_2(a + b)}{3}, \quad \log_{27}(a - b) = \frac{\log_3(a - b)}{3} $$ So the given equation becomes: \[ \frac{1}{3} \cdot \frac{\log_2(a + b)}{\log_2 c} + \frac{1}{3} \cdot \frac{\log_3(a - b)}{\log_

Question 8

CAT 2024 Slot 1: Logarithm Equation Question If $ x $ is a positive real number such that $ 4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13 $, then the greatest integer not exceeding $ x $ is:

Accepted Answer

Answer: 31. Logarithmic Equation - Greatest Integer Not Exceeding x Greatest Integer Not Exceeding $ x $ We are given the equation: $$ 4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13 $$ Step 1: Convert all logs to base 10 $$ \log_{100} x = \frac{\log_{10} x}{\log_{10} 100} = \frac{\log_{10} x}{2} \quad \text{and} \quad \log_{1000} x = \frac{\log_{10} x}{\log_{10} 1000} = \frac{\log_{10} x}{3} $$ Step 2: Substitute and simplify $$ 4 \log_{10} x + 4 \cdot \frac{\log_{10} x}{2} + 8 \cdot \frac{\log_{10} x}{3} = 4 \log x + 2 \log x + \frac{8}{3} \log x $$ \[ = \left(6 + \frac{8}{3}\right) \log x = \fr

Question 9

CAT 2023 Slot 3: Let $ n $ be any natural number such that $ 5^{n-1} < 3^{n+1} $. The least integer value of $ m $ that satisfies $ 3^{n+1} < 2^{n+m} $ for all such $ n $ is?

Accepted Answer

Answer: C · 5. The least value of $ m $ is 5. For $ n = 5 $, we need $ 3^6 < 2^{5 + m} $. This gives $ 729 < 2^{10} $, so $ m = 5 $.

Question 10

CAT 2022 Slot 2: The number of distinct integer values of n satisfying 4−log^2n/3−log4^n<0, is

Accepted Answer

Answer: 47. 4−logn23−logn4 16, it is negative. For any value of n 64, it is negative. For a fraction to be negative, the numerator and the denominator must be of opposite signs. In this case, n should be between 16 and 64. The number of values that n can take = 63 - 16 = 47.

Question 11

CAT 2021 Slot 3: For a real number a, if log15^a+log32^a/(log15a)(log32a)= = 4 then a must lie in the range

Accepted Answer

Answer: A · 4 < a < 5. loga15^a+loga32^a/(loga15^a×loga32^a) = 4 1/loga32^a+1/loga15^15 = 4 log32^a+log15^a = 4 log(32×15)^a = 4 a^4 = 32 × 15 = 480 We know that, 44 = 256 and 54 = 625 256 < 480 < 625 44 < a4 < 54 4 < a < 5

Question 12

CAT 2021 Slot 2: If log2^[3 + log3^{4 + log4^(x - 1)}] - 2 = 0 then 4x equals

Accepted Answer

Answer: 5. log_2[3+log_3{4+log_4(x-1)}]- 2= 0 log_2[3 + log_3{4 + log_4(x - 1)}] = 2 3 + log_3{4 + log_4(x - 1)} = 4 ( ∵ log_aN = x ⇒ N = a^x ) log_3{4 + log_4(x - 1)} = 1 4 + log_4(x - 1) = 3 log_4(x - 1) = -1 x - 1 = 1414 4x = 5

Question 13

CAT 2020 Slot 3: If x^1 = -1 and x^m = x^m + 1 + (m + 1) for every positive integer m, then x^100 equals

Accepted Answer

Answer: A · -5050. x_m = x_m + 1 + (m + 1) Let's say k = m + 1 The equation now becomes: x_k - 1 = x_k + k x_k = x_k - 1 - k x_k = - k + x_k - 1 Now the vizualization of this equation becomes far simpler... x_1 = -1 x_2 = -2 -1 x_3 = -3 -2 -1 x_4 = -4 -3 -2 -1 x_100 = -100 -99 -98 ... -3 -2 -1 x_100 = -(100 +99 +98 ... +3 +2 +1) x_100 = -(100(100+1)2100(100+1)2) = -5050