CAT 2023 Slot 3 QA paper with solutions

Solution

$x^8 + x^{-8} = 47 \Rightarrow x^4 + x^{-4} = 7 \Rightarrow x^2 + x^{-2} = 3 \Rightarrow x + x^{-1} = \sqrt{5}$.

$x^3 + x^{-3} = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$; $x^6 + x^{-6} = 20 - 2 = 18$.

$x^9 + x^{-9} = (x^3 + x^{-3})(x^6 - 1 + x^{-6}) = 2\sqrt{5} \times 17 = 34\sqrt{5}$.

Solution

$8^m = 2^{3m}$ and $8^n = 2^{3n}$. Powers of 2 strictly between them: $3n - 3m - 1 = 41 \Rightarrow n - m = 14$.

Smallest sum: $m = 1$, $n = 15$, so $n + m = 16$.

Solution

Infinitely many solutions: $\dfrac{a + 5}{1} = \dfrac{b^2 - 15}{1} = \dfrac{8b}{4} = 2b$.

$b^2 - 15 = 2b \Rightarrow b = 5$ or $b = -3$.

$b = 5 \Rightarrow a = 5$, $ab = 25$. $b = -3 \Rightarrow a = -11$, $ab = 33$.

Maximum $ab = 33$.

Solution

The terms are $\dfrac{1}{2}$, $\log_4(2^x - 9)$ and $\log_4\left(2^x + \dfrac{17}{2}\right)$.

AP: $2\log_4(t - 9) = \dfrac{1}{2} + \log_4\left(t + \dfrac{17}{2}\right)$ with $t = 2^x$, i.e. $(t - 9)^2 = 2\left(t + \dfrac{17}{2}\right) = 2t + 17$.

$t^2 - 20t + 64 = 0 \Rightarrow t = 16$ (as $t > 9$).

Common difference $= \log_4 7 - \dfrac{1}{2} = \log_4 7 - \log_4 2 = \log_4\dfrac{7}{2}$.

Solution

$5^{n-1} < 3^{n+1}$ holds for $n = 1, 2, 3, 4, 5$ (at $n = 5$: $625 < 729$) and fails from $n = 6$ ($3125 > 2187$).

We need $3^{n+1} < 2^{n+m}$ for each such $n$:

$n = 1$: $9 < 2^{1+m} \Rightarrow m \geq 3$. $n = 3$: $81 < 2^{3+m} \Rightarrow m \geq 4$. $n = 5$: $729 < 2^{5+m} \Rightarrow 5 + m \geq 10 \Rightarrow m \geq 5$.

Least $m = 5$.

Solution

15 factors: $p^{14}$ or $p^4 q^2$.

Smallest candidates: $2^4 \times 3^2 = 144$, $2^2 \times 3^4 = 324$, $2^4 \times 5^2 = 400$, …

Sum of the two smallest $= 144 + 324 = 468$.

Solution

Let $u, v$ be the reciprocals of the roots: $(u - v)^2 = \dfrac{1}{9}$ and $u^2 + v^2 = \dfrac{5}{9}$.

$2uv = \dfrac{4}{9} \Rightarrow uv = \dfrac{2}{9}$; $(u + v)^2 = 1 \Rightarrow u + v = \pm 1$.

So $u, v = \pm\dfrac{1}{3}, \pm\dfrac{2}{3}$, and the roots are $3, \dfrac{3}{2}$ or $-3, -\dfrac{3}{2}$.

Roots $3, 1.5$: $b = -4.5$, $c = 4.5$, $b + c = 0$. Roots $-3, -1.5$: $b = 4.5$, $c = 4.5$, $b + c = 9$.

Largest $= 9$.

Solution

Buying: he pays Rs 100 but gets 105 cm, so cost per cm $= \dfrac{100}{105}$.

Selling: price after $5\%$ discount $= 0.95 \times 110 = 104.5$ for what should be 100 cm, but he gives only 95 cm.

Revenue per cm actually given $= \dfrac{104.5}{95} = 1.1$.

Profit $= \dfrac{1.1}{100/105} - 1 = 1.1 \times 1.05 - 1 = 0.155 = 15.5\%$.

Solution

Let the daily rates be $R, K, G$ (Rakshita $= K$) and $T = R + K + G$.

$6T + 3K = 1 \Rightarrow K = \dfrac{1 - 6T}{3}$.

$T < \dfrac{1}{7} \Rightarrow K > \dfrac{1}{21}$, so Rakshita takes fewer than 21 days.

$R + G = T - K = \dfrac{9T - 1}{3} > \dfrac{1}{15} \Rightarrow T > \dfrac{2}{15} \Rightarrow K < \dfrac{1}{15}$, so she takes more than 15 days.

Her time lies strictly between 15 and 21 days, so it cannot be $21$.

Solution

Population in 2022 $= 100000\left(1 - \dfrac{y}{100}\right)\left(1 + \dfrac{x}{100}\right)$ with $x = y + 10$.

It must exceed 100000: $(100 - y)(110 + y) > 10000$

$\Rightarrow 11000 - 10y - y^2 > 10000 \Rightarrow y^2 + 10y - 1000 < 0$.

Root: $y = \dfrac{-10 + \sqrt{4100}}{2} \approx 27.02$, so the largest natural $y = 27$.

The 2021 population is lowest when $y$ is largest: $100000 \times 0.73 = 73000$.

Solution

Sugar in A $= \dfrac{2}{5} = 40\%$; sugar in B $= \dfrac{3}{10} = 30\%$.

C mixes A and B in $2 : 3$: sugar $= 0.4 \times 0.4 + 0.6 \times 0.3 = 0.34$.

Adding an equal amount of milk halves this: $17\%$.

Solution

Let the average of A, B, C be $a$ (total $3a$).

With D: $\dfrac{3a + D}{4} = a - x \Rightarrow D = a - 4x$.

With E: $\dfrac{3a + E}{4} = a + 2x \Rightarrow E = a + 8x$.

$E - D = 12x = 12 \Rightarrow x = 1$.

Solution

Take AB $= 12$ units. Boat 1: $b + r = 6$, $b - r = 4$, so $b = 5$, $r = 1$ (A to B is downstream).

Boat 2 with speed $s$: $\dfrac{12}{s - 1} + \dfrac{12}{s + 1} = 6 \Rightarrow s^2 - 4s - 1 = 0 \Rightarrow s = 2 + \sqrt{5} \approx 4.24$.

Boat 2 is slower. A to B (downstream): $\dfrac{12}{s + 1} = \dfrac{12}{3 + \sqrt{5}} = 3(3 - \sqrt{5})$ hours.

Solution

Let A collect $3x$ and B collect $4x$ per week.

A in 5 weeks: $15x$ is a multiple of 7, so $x$ is a multiple of 7.

B in 3 weeks: $12x$ is a multiple of 24, so $x$ is even.

Smallest $x = 14$, so A collects $3 \times 14 = 42$ per week.

Solution

Let the total stock be $N$. Mangoes $= 0.4N$; let apples $= A$ and bananas $= B$, with $A + B = 0.6N$.

Fruits sold $= 0.2N + 96 + 0.4A = 0.5N \Rightarrow A = 0.75N - 240$, so $B = 240 - 0.15N$.

Conditions: $A \geq 1 \Rightarrow N > 320$; $B \geq 96 \Rightarrow N \leq 960$.

For whole numbers, $N$ must be a multiple of 5 (half the mangoes) and of 4 (for $A$), i.e. a multiple of 20.

Smallest such $N > 320$ is $340$.

Check: $A = 15$, $B = 189$, mangoes $= 136$; sold $= 68 + 96 + 6 = 170 = 50\%$ ✓.

Solution

Let the daily rates be $G$ and $S$: $G + S = \dfrac{1}{20}$.

Losing $40\%$ of Gautam's work equals $50\%$ extra of Suhani's: $0.4G = 0.5S \Rightarrow S = 0.8G$.

$1.8G = \dfrac{1}{20} \Rightarrow G = \dfrac{1}{36}$.

Gautam is faster and takes $36$ days.

Solution

Let $\angle B = \angle C = \beta$, so $\angle A = 180^\circ - 2\beta$.

$\angle OAB = 90^\circ - \beta$ (AD bisects $\angle A$) and $\angle OBA = 90^\circ - \angle A = 2\beta - 90^\circ$.

In triangle AOB: $105 + (90 - \beta) + (2\beta - 90) = 180 \Rightarrow \beta = 75^\circ$, $\angle A = 30^\circ$.

$AD = AB\cos 15^\circ$ and $BE = AB\sin 30^\circ = \dfrac{AB}{2}$.

$\dfrac{AD}{BE} = 2\cos 15^\circ$.

Solution

Let the rectangle be $2x$ wide and $y$ tall with $x^2 + y^2 = 4$.

Area $= 2xy$ is largest when $x = y = \sqrt{2}$.

Sides: $2\sqrt{2}$ and $\sqrt{2}$. Ratio $= 2 : 1$.

Solution

Interior + exterior $= 180^\circ$ and interior $-$ exterior $= 120^\circ$, so exterior $= 30^\circ$ and $n = 12$.

Diagonals $= \dfrac{12 \times 9}{2} = 54$.

Solution

The series is $\sum_{n \geq 0} \left(\dfrac{1}{4}\right)^n \sum_{k=0}^{n} \left(\dfrac{1}{3}\right)^k$.

Swap the order: $\sum_{k \geq 0} \left(\dfrac{1}{3}\right)^k \sum_{n \geq k} \left(\dfrac{1}{4}\right)^n = \sum_{k} \left(\dfrac{1}{3}\right)^k \left(\dfrac{1}{4}\right)^k \cdot \dfrac{4}{3}$.

$= \dfrac{4}{3} \times \dfrac{1}{1 - 1/12} = \dfrac{4}{3} \times \dfrac{12}{11} = \dfrac{16}{11}$.

Solution

$a_n$: 54, 62, …, 846 (all $\equiv 6 \pmod 8$). $b_n$: 102, 106, …, 498 (all $\equiv 2 \pmod 4$).

Every $a_n$ is $\equiv 2 \pmod 4$, so the common terms are the $a_n$ between 102 and 498: 102, 110, …, 494.

That is 50 terms; sum $= \dfrac{50(102 + 494)}{2} = 14900$.

Solution

Let $u = 3x + 2y$ and $v = 2x - 5y$. Then $5u + 2v = 19x$, so $f(u, v) = 5u + 2v$.

$f(x, 2x) = 5x + 4x = 9x = 27 \Rightarrow x = 3$.