CAT 2023 Slot 1 QA
Q1–10 of 22If $x$ and $y$ are positive real numbers such that $\log_x(x^2 + 12) = 4$ and $3\log_y x = 1$, then $x + y$ equals
log_x(x² + 12) = 4 and 3·log_y(x) = 1, find x + y Solution: log_x(x²+12)=4 means x²+12 = x⁴. Let t = x² (t>0). Then t+12=t² → t²-t-12=0 → (t-4)(t+3)=0 → t=4 (rejecting t=-3). So x²=4 → x=2 (x positive real). Now 3·log_y(x)=1 → log_y(x)=1/3 → x = y^(1/3) → 2 = y^(1/3) → y=8. x+y = 2+8 = 10. Answer: (d) 10
If $x$ and $y$ are real numbers such that $x^2 + (x - 2y - 1)^2 = -4y(x + y)$, then the value of $x - 2y$ is
Expand (x-2y-1)² = x²+4y²+1-4xy-2x+4y. LHS total = x² + x²+4y²+1-4xy-2x+4y = 2x²+4y²+1-4xy-2x+4y RHS = -4xy-4y² Bring RHS to left: 2x²+4y²+1-4xy-2x+4y+4xy+4y² = 0 → 2x² - 2x + 8y² + 4y + 1 = 0 Complete the square: 2(x-1/2)² - 1/2 + 8(y+1/4)² - 1/2 + 1 = 0 → 2(x-1/2)² + 8(y+1/4)² = 0 Since both squared terms are non-negative, both must equal 0: x = 1/2, y = -1/4 x - 2y = 1/2 - 2(-1/4) = 1/2 + 1/2 = 1 Answer: (b) 1
If $\sqrt{5x+9} + \sqrt{5x-9} = 3(2 + \sqrt{2})$, then $\sqrt{10x+9}$ is equal to
Let a=√(5x+9), b=√(5x-9). Given a+b = 6+3√2. a²-b² = (5x+9)-(5x-9) = 18 So a-b = 18/(a+b) = 18/(6+3√2) = 6/(2+√2) = 6(2-√2)/2 = 3(2-√2) = 6-3√2 Adding: a+b=6+3√2 and a-b=6-3√2 → 2a=12 → a=6 a² = 36 = 5x+9 → 5x=27 → x=27/5 10x+9 = 54+9 = 63 → √63 = 3√7 Answer: (d) 3√7
Let $n$ be the least positive integer such that 168 is a factor of $1134^n$. If $m$ is the least positive integer such that $1134^n$ is a factor of $168^m$, then $m + n$ equals
1134 = 2 × 3⁴ × 7, and 168 = 2³ × 3 × 7 For 168 | 1134ⁿ = 2ⁿ·3⁴ⁿ·7ⁿ, we need n≥3 (for 2³), n≥1 (for 3¹), n≥1 (for 7¹) → n=3 So 1134³ = 2³ × 3¹² × 7³ For 1134³ | 168ᵐ = 2³ᵐ·3ᵐ·7ᵐ, we need 3m≥3 (m≥1), m≥12, m≥3 → m=12 m+n = 12+3 = 15 Answer: (c) 15
The number of integer solutions of the equation $2|x|(x^2 + 1) = 5x^2$ is _______
x=0: LHS=0=RHS → works. For x≠0, let t=|x|>0: 2t(t²+1)=5t² → 2t³-5t²+2t=0 → t(2t²-5t+2)=0 2t²-5t+2=0 → t=(5±3)/4 → t=2 or t=0.5 Since t=|x| must be a positive integer, only t=2 is valid → x=2 or x=-2 Total solutions: x=0, 2, -2 Answer: 3
Let $\alpha$ and $\beta$ be the two distinct roots of the equation $2x^2 - 6x + k = 0$, such that $(\alpha + \beta)$ and $\alpha\beta$ are the distinct roots of the equation $x^2 + px + p = 0$. Then, the value of $8(k - p)$ is _______
α+β = 3, αβ = k/2 Sum of new roots: (α+β)+αβ = -p → 3 + k/2 = -p ...(i) Product of new roots: (α+β)(αβ) = p → 3(k/2) = p → p = 3k/2 ...(ii) Substitute (ii) into (i): 3 + k/2 = -3k/2 → 3 = -2k → k = -3/2 p = 3(-3/2)/2 = -9/4 k - p = -3/2 + 9/4 = 3/4 8(k-p) = 8 × 3/4 = 6 Answer: 6
The equation $x^3 + (2r+1)x^2 + (4r-1)x + 2 = 0$ has $-2$ as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of $r$ is _______
Since −2 is a root, factor out (x+2): x³+(2r+1)x²+(4r-1)x+2 = (x+2)(x²+(2r-1)x+1) The other two roots come from x²+(2r-1)x+1=0 For real roots, discriminant ≥ 0: (2r-1)² - 4 ≥ 0 → (2r-1)² ≥ 4 → 2r-1 ≥ 2 or 2r-1 ≤ -2 → r ≥ 3/2 or r ≤ -1/2 Minimum non-negative integer satisfying r ≥ 3/2 is r = 2 Answer: 2
Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of $x$ km till then, where $x$ is a whole number and is palindromic (i.e., $x$ remains unchanged when its digits are reversed). At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was
Distance travelled = 26862 - x, and since max speed is 110 km/h, distance ≤ 8×110 = 880 So x ≥ 26862 - 880 = 25982 To maximize distance (26862-x), minimize x, so find the smallest palindrome ≥ 25982. 5-digit palindromes of form "2b c b 2": for b=5, max value is 25952 (<25982); for b=6, minimum value is 26062 (≥25982) So smallest valid palindrome is x = 26062 Distance = 26862 - 26062 = 800 km Average speed = 800/8 = 100 km/h Answer: (b) 100
The minor angle between the hour hand and minute hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8:48 am when this angle increases by 50% is
Hour hand angle = 8×30 + 48×0.5 = 240+24 = 264° Minute hand angle = 48×6 = 288° Current angle = 288-264 = 24° Target angle = 24×1.5 = 36° (increase of 12°) Relative speed of minute hand over hour hand = 6-0.5 = 5.5°/min Time needed = 12/5.5 = 24/11 minutes Answer: (b) 24/11
In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
Let girl's marks = g, boy's marks = b. Total: 4g+6b=240 → 2g+3b=120 → g = 60-1.5b Condition: b ≤ g ≤ 2b b ≤ 60-1.5b → b ≤ 24 60-1.5b ≤ 2b → b ≥ 120/7 ≈ 17.14 So b ∈ [120/7, 24] Total of 2 girls + 6 boys, T = 2g+6b = 2(60-1.5b)+6b = 120+3b As b ranges over [120/7,24], T ranges over [120+360/7, 192] = [≈171.43, 192] Integer values of T: 172 to 192 inclusive → 192-172+1 = 21 values Answer: (b) 21