CAT 2023 Slot 2 QA paper with solutions

Solution

$144^{145} = 2^{580} \times 3^{290}$.

Make $n$ as large and $m$ as small as possible: $m = 1$, $a = 3^{290}$, $b = 2$, $n = 580$.

$n - m = 579$.

Solution

$\dfrac{x + 3}{y - 3} - \dfrac{x}{y} > 0 \Rightarrow \dfrac{3(x + y)}{y(y - 3)} > 0$.

Option B: $y < 0$ makes $y(y - 3) > 0$, and $-x < y$ means $x + y > 0$. The fraction is positive ✓.

(The other options do not guarantee the sign.)

Solution

$k$ divides $(3m + 4n) - 2(m + 2n) = m$.

Then $k$ also divides $(m + 2n) - m = 2n$.

So $k$ is a common divisor of $m$ and $2n$.

Solution

Let $u = 2^{2x^2}$ and $v = 2^{x + 15}$. Then $2^{4x^2} = u^2$, $2^{2x^2 + x + 16} = 2uv$, $2^{2x + 30} = v^2$.

$u^2 - 2uv + v^2 = 0 \Rightarrow u = v \Rightarrow 2x^2 = x + 15$.

$2x^2 - x - 15 = 0$: sum of roots $= \dfrac{1}{2}$.

Solution

Exactly two factors other than 1 and itself means exactly 4 factors: $p^3$ or $pq$.

$p^3 < 50$: 8, 27 (2 numbers).

$pq < 50$: $2 \times \{3, 5, 7, 11, 13, 17, 19, 23\}$ (8), $3 \times \{5, 7, 11, 13\}$ (4), $5 \times 7$ (1).

Total $= 2 + 13 = 15$.

Solution

$\dfrac{\log_x 25}{\log_x 0.008} = \log_{0.008} 25 = \dfrac{\log 5^2}{\log 5^{-3}} = -\dfrac{2}{3}$.

So $\log_{\sqrt{3}} x = \dfrac{16}{3} + \dfrac{2}{3} = 6 \Rightarrow x = (\sqrt{3})^6 = 27$.

$\log_3(3x^2) = 1 + 2\log_3 27 = 1 + 6 = 7$.

Solution

$(x - 1)^2 + 2kx + 11 = x^2 + (2k - 2)x + 12$. No real roots: $(2k - 2)^2 < 48 \Rightarrow (k - 1)^2 < 12$.

$k - 1 \leq 3$, so the largest integer $k = 4$.

$\dfrac{4}{4y} + 9y = \dfrac{1}{y} + 9y \geq 2\sqrt{9} = 6$ (AM–GM), at $y = \dfrac{1}{3}$.

Least value $= 6$.

Solution

Let A fill in $a$ hours, B empty in $a - 1$, C fill in $c$.

All three: $\dfrac{1}{a} - \dfrac{1}{a - 1} + \dfrac{1}{c} = \dfrac{1}{2}$.

B and C for 1 hour, then C for $1.25$ hours: $\dfrac{2.25}{c} - \dfrac{1}{a - 1} = 1$.

Substituting $\dfrac{1}{c}$ from the first: $0.125 = \dfrac{2.25}{a} - \dfrac{1.25}{a - 1} \Rightarrow a^2 - 9a + 18 = 0 \Rightarrow a = 3$ or $6$.

$a < 5 \Rightarrow a = 3$. Then $\dfrac{1}{c} = \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{2}{3} \Rightarrow c = 1.5$ hours $= 90$ minutes.

Solution

Rate per half-year $= 4\%$.

After 1 year: $200000 \times 1.04^2 = 216320$. After repaying 10320, outstanding $= 206000$.

After 2 more years (4 half-years): $206000 \times 1.04^4 \approx 206000 \times 1.16986 \approx 240990.9$.

Total paid $= 10320 + 240990.9 = 251310.9$.

Interest $\approx 251311 - 200000 = 51311$.

Solution

Ravi and Ashok approach at $40 + 50 = 90$ km/h $= 25$ m/s, so they meet in $\dfrac{225}{25} = 9$ s.

In 9 s Ravi moves $\dfrac{40 \times 1000}{3600} \times 9 = 100$ m.

Vijay must cover $54 + 100 = 154$ m in 9 s: $\dfrac{154}{9}$ m/s $= \dfrac{154}{9} \times 3.6 = 61.6$ km/h.

Solution

Minu buys at 1000 and sells to Kanu at $20\%$ profit: 1200 (profit 200).

Kanu sells it back at a $20\%$ loss: $0.8 \times 1200 = 960$.

Let Minu sell to Tanu at $x$. Total profit: $200 + (x - 960) = 500 \Rightarrow x = 1260$.

Profit on the last sale $= \dfrac{1260 - 960}{960} \times 100 = 31.25\%$.

Solution

Price $= kw^2$. The four distinct integer weights add to 18.

Highest total: make one piece as big as possible: $1, 2, 3, 12$ gives $\sum w^2 = 158$.

Lowest total: keep pieces close: $3, 4, 5, 6$ gives $\sum w^2 = 86$.

$k(158 - 86) = 288000 \Rightarrow k = 4000$.

Original price $= 4000 \times 18^2 = 1296000$.

Solution

Take 100 employees and total salary $= 600$.

Manufacturing: 20 employees share $\dfrac{1}{6} \times 600 = 100$, so average $= 5$.

Others: 80 employees share $500$, so average $= 6.25$.

Ratio $= 5 : 6.25 = 4 : 5$.

Solution

Total money $= 352n$.

After two people take 506 each, the rest get $\dfrac{352n - 1012}{n - 2} \leq 330$.

$352n - 1012 \leq 330n - 660 \Rightarrow 22n \leq 352 \Rightarrow n \leq 16$.

Maximum $n = 16$.

Solution

Selling price $= 0.9 \times 1.25 = 1.125 \times$ average cost, so profit $= 12.5\%$ of total cost.

Total cost $= \dfrac{51000}{0.125} = 408000$.

$1000w + 1125b = 408000 \Rightarrow 8w + 9b = 3264$.

For the most shirts, buy as many cheaper (white) shirts as possible. Since $3264$ is divisible by 8, $b$ must be a multiple of 8; take $b = 8$.

$w = \dfrac{3264 - 72}{8} = 399$. Total $= 399 + 8 = 407$.

Solution

After $n$ replacements, milk $= 40 \times 0.9^n$. We need this below 20: $0.9^n < 0.5$.

$0.9^6 \approx 0.531$, $0.9^7 \approx 0.478$.

Smallest $n = 7$.

Solution

A triangle on a diameter is right-angled. Let the legs be $ak$ and $bk$; the hypotenuse is $2r$.

$k^2(a^2 + b^2) = 4r^2$.

Area $= \dfrac{1}{2}abk^2 = \dfrac{1}{2}ab \times \dfrac{4r^2}{a^2 + b^2} = \dfrac{2abr^2}{a^2 + b^2}$.

Solution

Rectangle area $= 54$. In a GP with third term $= 4 \times$ first, the ratio is 2: areas are $x, 2x, 4x$.

$7x = 54$, and area of ABP $= \dfrac{9 \times BP}{2}$, area of APQ $= \dfrac{9 \times PQ}{2}$, so $PQ = 2BP$.

$BP = \dfrac{2 \times 54/7}{9} = \dfrac{12}{7}$, $PQ = \dfrac{24}{7}$, $QC = 6 - \dfrac{36}{7} = \dfrac{6}{7}$.

$BP : PQ : QC = 2 : 4 : 1$.

Solution

Between $x = 0$ and $x = 5$, $|x - 5| = 5 - x$, so $|x - y| = 7 - x$.

This gives the lines $y = 7$ and $y = 2x - 7$.

At $x = 0$ the region runs from $y = -7$ to $7$ (length 14); at $x = 5$ from $y = 3$ to $7$ (length 4).

Area $= \dfrac{1}{2}(14 + 4) \times 5 = 45$.

Solution

Let the common differences be primes $p$ (for $a$) and $q$ (for $b$). $b_2 = 0 \Rightarrow b_1 = -q$.

$a_5 = b_9$: $a_1 + 4p = 7q$. $a_{19} = b_{19}$: $a_1 + 18p = 17q$.

Subtracting: $14p = 10q \Rightarrow 7p = 5q \Rightarrow p = 5$, $q = 7$.

$a_1 = 49 - 20 = 29$; $a_{11} = 29 + 10 \times 5 = 79$.

Solution

$2pq - 20 = 52 - 2pq \Rightarrow pq = 18$. Then $p^2 + q^2 - 29 = 16 \Rightarrow p^2 + q^2 = 45$.

$(p - q)^2 = 45 - 36 = 9 \Rightarrow p - q = \pm 3$.

$p^3 - q^3 = (p - q)(p^2 + pq + q^2) = \pm 3 \times 63 = \pm 189$.

Difference $= 189 - (-189) = 378$.

Solution

$a_n = 6n + 7$: all terms $\equiv 1 \pmod 6$ (from 13). $b_n = 7n + 8$: all terms $\equiv 1 \pmod 7$ (from 15).

Common terms are $\equiv 1 \pmod{42}$: $43, 85, \dots$

Largest three-digit: $42 \times 23 + 1 = 967$.