CAT 2024 Slot 3 QA paper with solutions

Solution

Girls $g$ is between $44\%$ and $60\%$ of 250: $110 \leq g \leq 150$; boys $= 250 - g$.

Swimming $= 0.5(250 - g) + 0.8g$, running $= 0.7(250 - g) + 0.6g$.

Both $=$ swimming $+$ running $- 250 = 50 + 0.2g$.

$g = 110$ gives 72 and $g = 150$ gives 80.

Solution

$(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}$.

$2ab = 30 \Rightarrow ab = 15$, and $a^2 + 3b^2 = 52$.

$a = 5, b = 3$: $25 + 27 = 52$ ✓.

$a + b = 8$.

Solution

Let the numbers be $a < b < c$ with $a + b + c = 84$.

New numbers $a + 7, b, c - 10$ have sum 81 and mean 27, which is $b + 2$, so $b = 25$.

$(c - 10) - (a + 7) = 64 \Rightarrow c - a = 81$, and $a + c = 84 - 25 = 59$.

$c = \dfrac{59 + 81}{2} = 70$.

Solution

$10^3 = 1000 \equiv -1 \pmod{13}$, so $10^6 \equiv 1$.

$68 = 6 \times 11 + 2 \Rightarrow 10^{68} \equiv 10^2 = 100 \equiv 9 \pmod{13}$.

Solution

Rates per day (job = 60 units): Sam $= 3$, Mohit $= 6$, Ayna $= 2$.

Days 1, 2, 3 give $9 + 5 + 8 = 22$ units. After 6 days: 44 units.

Day 7 (Sam + Mohit): 53. Day 8 (Sam + Ayna): 58. Day 9 (Mohit + Ayna) finishes the last 2 units.

Sam worked days 1, 2, 4, 5, 7, 8: $6 \times 3 = 18$ units.

Fraction $= \dfrac{18}{60} = \dfrac{3}{10}$.

Solution

Radius: $\dfrac{10\sqrt{3}}{2\sin 60^\circ} = 10$ m.

Smaller region $=$ sector $-$ triangle $= \dfrac{120}{360}\pi(100) - \dfrac{1}{2}(100)\sin 120^\circ$.

$= \dfrac{100\pi}{3} - 25\sqrt{3} = 25\left(\dfrac{4\pi}{3} - \sqrt{3}\right)$.

Solution

Even terms: $t_4 = \dfrac{1}{3}t_2$, $t_6 = \dfrac{3}{5}t_4$, …, so $t_{2k} = -\dfrac{1}{2k - 1}$.

$\dfrac{1}{t_{2k}} = -(2k - 1)$.

Sum for $k = 1$ to $1012$ $= -(1 + 3 + \dots + 2023) = -1012^2 = -1024144$.

Solution

Left side $= |x - 2|$. Right side: $4$ for $x \geq 2$; $2x$ for $-2 \leq x \leq 2$; $-4$ for $x \leq -2$.

$x \geq 2$: $x - 2 = 4 \Rightarrow x = 6$ ✓.

$-2 \leq x \leq 2$: $2 - x = 2x \Rightarrow x = \dfrac{2}{3}$ ✓.

$x \leq -2$: $|x - 2| = -4$ is impossible.

Number of solutions $= 2$.

Solution

$\dfrac{A_3}{A_5} = \dfrac{1}{(1 + r)^2} = \dfrac{25}{36} \Rightarrow 1 + r = 1.2$, i.e. $20\%$.

We need $4000 \times 1.2^n > 20000 \Rightarrow 1.2^n > 5$.

$1.2^8 \approx 4.30$ and $1.2^9 \approx 5.16$.

Minimum years $= 9$.

Solution

Let $t = 10^x$: $t + \dfrac{4}{t} = \dfrac{81}{2} \Rightarrow 2t^2 - 81t + 8 = 0$, with two positive roots.

Product of roots $t_1 t_2 = 4 \Rightarrow 10^{x_1 + x_2} = 4$.

$x_1 + x_2 = \log_{10} 4 = 2\log_{10} 2$.

Solution

$\dfrac{D}{v} - \dfrac{D}{v + 6} = 4 \Rightarrow \dfrac{6D}{v(v + 6)} = 4$.

$\dfrac{D}{v - 6} - \dfrac{D}{v} = 6 \Rightarrow \dfrac{6D}{v(v - 6)} = 6$.

Dividing: $\dfrac{v - 6}{v + 6} = \dfrac{2}{3} \Rightarrow v = 30$.

$D = \dfrac{4 \times 30 \times 36}{6} = 720$ km.

Solution

$a = \log_3 4$, $b = \log_4 5$, …, $f = \log_8 9$.

The chain telescopes: $abcdef = \log_3 9 = 2$.

Solution

A $10\%$ discount saves Rs 15, so the marked price $= 150$.

Marked price $= 1.2 \times$ cost $\Rightarrow$ cost $= 125$.

Ravi pays $150 - 15 = 135$.

Gopi's profit $= 135 - 125 = 10$.

Solution

Let the water be $w$ L, so milk $= 300 - w$.

Removing $2w$ L removes the fraction $\dfrac{2w}{300}$ of the milk.

$(300 - w)\left(1 - \dfrac{2w}{300}\right) = 0.72 \times 300 = 216$.

$(300 - w)(300 - 2w) = 64800 \Rightarrow w^2 - 450w + 12600 = 0 \Rightarrow w = 30$ or $420$.

$w = 30$ L (420 exceeds the container).

Solution

Each interior angle of a regular octagon is $135^\circ$. In triangle ABC:

$AC^2 = 6^2 + 6^2 - 2 \times 36 \cos 135^\circ = 72 + 36\sqrt{2}$.

Area of square ACEG $= AC^2 = 36(2 + \sqrt{2})$.

Solution

$|x + y| + |x - y| = 2\max(|x|, |y|)$, so we need $\max(|x|, |y|) = 1$.

These are the integer points on the boundary of the square with corners $(\pm 1, \pm 1)$:

$(\pm 1, 0), (0, \pm 1), (\pm 1, \pm 1)$. Total $= 8$.

Solution

Replace $x$ by $\dfrac{1}{x}$: $f\left(\dfrac{1}{x}\right) + 2f(x) = \dfrac{3}{x}$.

Solve the two equations: $3f(x) = \dfrac{6}{x} - 3x \Rightarrow f(x) = \dfrac{2}{x} - x$.

$f(x) = 3 \Rightarrow 2 - x^2 = 3x \Rightarrow x^2 + 3x - 2 = 0$.

Sum of the values $= -3$.

Solution

No solution means the lines are parallel but distinct: $\dfrac{p}{3} = \dfrac{-4}{k} \neq \dfrac{2}{a}$.

Parallel: $pk = -12$. Distinct: $\dfrac{-4}{k} \neq \dfrac{2}{a} \Rightarrow -4a \neq 2k \Rightarrow 2a + k \neq 0$.

So $2a + k \neq 0$ is a necessary condition.

Solution

Vimal's 30 ha in $5 : 3$: wheat $= 18.75$, mustard $= 11.25$.

Total 50 ha in $11 : 9$: wheat $= 27.5$, mustard $= 22.5$.

Rajesh: wheat $= 27.5 - 18.75 = 8.75$, mustard $= 22.5 - 11.25 = 11.25$.

Ratio $= 8.75 : 11.25 = 7 : 9$.

Solution

The median from A bisects MP (the midline), so X is the midpoint of MP. Similarly Y and Z are the midpoints of MN and NP.

So XYZ is the medial triangle of MNP, and MNP is the medial triangle of ABC.

Area $= \dfrac{1}{4} \times \dfrac{1}{4} \times 1440 = 90$.

Solution

1-digit: 9. 2-digit: $9 \times 9 = 81$. 3-digit up to 500: first digit 1–4, so $4 \times 9 \times 8 = 288$ (500 itself repeats 0).

Total $= 9 + 81 + 288 = 378$.

Solution

Let the first increment be $x$ (as a fraction); the second is $2x$.

$(1 + x)(1 + 2x) = 1.875 \Rightarrow 2x^2 + 3x - 0.875 = 0 \Rightarrow 16x^2 + 24x - 7 = 0$.

$x = \dfrac{-24 + \sqrt{576 + 448}}{32} = \dfrac{-24 + 32}{32} = 0.25$.

First increment $= 25\%$.