CAT 2024 Slot 1 QA paper with solutions

Solution

Count functions from 6 elements onto 3 elements by inclusion–exclusion:

$3^6 - \binom{3}{1}2^6 + \binom{3}{2}1^6 = 729 - 192 + 3 = 540$.

Solution

Subtract the second equation from the first (after dividing both by 4):

$x^2 + y^2 + z^2 - x + y + z = -\dfrac{3}{4}$.

Complete squares: $\left(x - \dfrac{1}{2}\right)^2 + \left(y + \dfrac{1}{2}\right)^2 + \left(z + \dfrac{1}{2}\right)^2 = 0$.

So $x = \dfrac{1}{2}$, $y = z = -\dfrac{1}{2}$, and $a = 4\left(\dfrac{3}{4}\right) = 3$.

Solution

Let the common root be $r$. Adding the first two equations and subtracting the third:

$r^2 + 9 + 17 - 35 = 0 \Rightarrow r^2 = 9 \Rightarrow r = -3$ (negative).

$9 - 3m + 9 = 0 \Rightarrow m = 6$; $9 - 3n + 17 = 0 \Rightarrow n = \dfrac{26}{3}$.

$2m + 3n = 12 + 26 = 38$.

Solution

Let the first term be $a$ and the common difference $d$.

$2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) \Rightarrow 2a + 4d = 0 \Rightarrow a = -2d$.

$x_5 = a + 4d = 2d = -4 \Rightarrow d = -2$, $a = 4$.

$x_{100} = 4 + 99(-2) = -194$.

Solution

Renu needs $15 \times 4 = 60$ hours; Seema needs $8 \times 5 = 40$ hours.

Let Seema work $d$ days at 4 hours/day; Renu works $2d$ days at 2 hours/day.

$\dfrac{2 \times 2d}{60} + \dfrac{4d}{40} = \dfrac{d}{15} + \dfrac{d}{10} = \dfrac{d}{6} = 1 \Rightarrow d = 6$.

Solution

$10 \equiv 3 \pmod 7$ and $3^6 \equiv 1 \pmod 7$.

$100 = 6 \times 16 + 4$, so $10^{100} \equiv 3^4 = 81 \equiv 4 \pmod 7$.

Remainder $= 4$.

Solution

$32768 = 8^5$. Write everything as powers of 8:

$8^{-k} \cdot 8^{-5/3} = 8^{-1} \cdot 8^{-5/k} \Rightarrow k + \dfrac{5}{3} = 1 + \dfrac{5}{k}$.

$3k^2 + 2k - 15 = 0$ (two real roots).

Sum of values $= -\dfrac{2}{3}$.

Solution

Group by the value of $\lfloor\sqrt{n}\rfloor$:

1 for $n = 1$–3 (3 terms), 2 for 4–8 (5), 3 for 9–15 (7), 4 for 16–24 (9), 5 for 25–35 (11), 6 for 36–48 (13), 7 for 49–50 (2).

Sum $= 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$.

Solution

Take September incomes as $800, 600, 500$.

Rent $= 15\%(800) + 12\%(600) + 18\%(500) = 120 + 72 + 90 = 282$.

October incomes: $800 \times 1.10 = 880$, $600 \times 1.12 = 672$, $500 \times 1.15 = 575$. Total $= 2127$.

Rent is unchanged, so the share $= \dfrac{282}{2127} \times 100 \approx 13.26\%$.

Solution

Each digit appears 6 times in each place, so the sum $= 6 \times 1111 \times (a + b + c + d) = 6666S$.

$6666S = 153310 + n$ with $1 \leq n \leq 9$: $6666 \times 23 = 153318$, so $S = 23$, $n = 8$.

$a + b + c + d + n = 31$.

Solution

Triangle ADE is right-angled at D with $AD = 45$ and $DE = 28$.

$AE = \sqrt{45^2 + 28^2} = \sqrt{2809} = 53$.

Inradius of a right triangle $= \dfrac{45 + 28 - 53}{2} = 10$.

Solution

$x^2 + y^2 + 4x - 4y = (x + 2)^2 + (y - 2)^2 - 8$, so the condition is $4 \leq (x + 2)^2 + (y - 2)^2 \leq 8$: a ring with radii 2 and $2\sqrt{2}$ centred at $(-2, 2)$.

The line $y = x + 4$ passes through the centre, so it cuts the ring in half.

Area $= \dfrac{1}{2}\pi(8 - 4) = 2\pi$.

Solution

Let $L = \log_{10} x$. Then $\log_{100} x = \dfrac{L}{2}$ and $\log_{1000} x = \dfrac{L}{3}$.

$4L + 2L + \dfrac{8L}{3} = \dfrac{26L}{3} = 13 \Rightarrow L = 1.5$.

$x = 10^{1.5} = 10\sqrt{10} \approx 31.6$. Greatest integer $= 31$.

Solution

Let the cost be $c$; selling price $= 1.4c$.

New cost $= 0.6c$ and new selling price $= 1.4c - 5$, which gives $50\%$ profit:

$1.4c - 5 = 1.5 \times 0.6c = 0.9c \Rightarrow 0.5c = 5 \Rightarrow c = 10$.

Original selling price $= 1.4 \times 10 = 14$.

Solution

Each step keeps $\dfrac{1}{3}$ of the milk. After 4 steps, milk $= \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$ of the glass.

Water $= \dfrac{80}{81}$.

Milk : water $= 1 : 80$.

Solution

Apples $= 5k$, mangoes $= 2k$, oranges $= 187 - 7k$.

Unsold apples $= 5k - 75$; unsold oranges $= \dfrac{187 - 7k}{2}$.

$\dfrac{5k - 75}{(187 - 7k)/2} = \dfrac{3}{2} \Rightarrow 4(5k - 75) = 3(187 - 7k) \Rightarrow 41k = 861 \Rightarrow k = 21$.

Apples 105, mangoes 42, oranges 40. Unsold: $30 + 16 + 20 = 66$.

Solution

Let the speeds be $a$ (Anil) and $b$ (Sunil). Meeting in 1.5 h: $a + b = 30$.

Anil takes 1.25 h longer for the whole 45 km: $\dfrac{45}{a} - \dfrac{45}{30 - a} = 1.25$.

$36(30 - 2a) = a(30 - a) \Rightarrow a^2 - 102a + 1080 = 0 \Rightarrow a = 12$ (the other root, 90, is too large).

Anil's speed $= 12$ km/h.

Solution

Let the numbers be $a, b, c, d$.

$\dfrac{a + b}{2} = a + 1 \Rightarrow b = a + 2$; the average of two $= a + 1$.

Average of three $= a + 3 \Rightarrow c = 3(a + 3) - (2a + 2) = a + 7$.

Average of four $= a + 6 \Rightarrow d = 4(a + 6) - (3a + 9) = a + 15$.

Numbers: $a, a + 2, a + 7, a + 15$. Largest − smallest $= 15$.

Solution

Let the period in bank A be $n$ years. Interest from A $= 10000 \times 0.05n = 500n$.

Amount moved to B $= 10000 + 500n$; interest from B $= (10000 + 500n) \times 0.06 \times 5 = 3000 + 150n$.

$\dfrac{500n}{3000 + 150n} = \dfrac{10}{13} \Rightarrow 6500n = 30000 + 1500n \Rightarrow n = 6$.

Solution

Work backwards. The third customer takes half the remainder $+ 3$ and nothing is left: $\dfrac{R_3}{2} = 3 \Rightarrow R_3 = 6$.

Before the second: $\dfrac{R_2}{2} - 3 = 6 \Rightarrow R_2 = 18$.

Initially: $\dfrac{Q}{2} - 3 = 18 \Rightarrow Q = 42$ kg.

Solution

$29 - 12\sqrt{5} = (2\sqrt{5} - 3)^2$, so the positive root is $-3 + 2\sqrt{5}$.

$2\sqrt{5}$ can be written as $b\sqrt{n}$ with $b = 2, n = 5$ or $b = 1, n = 20$.

$a + b + n$ is $4$ or $18$. Maximum $= 18$.

Solution

Let the edges be $a, b, c$: $4(a + b + c) = 144 \Rightarrow a + b + c = 36$, and $2(ab + bc + ca) = 846$.

$a^2 + b^2 + c^2 = 36^2 - 846 = 450$, so the space diagonal $= \sqrt{450} = 15\sqrt{2}$ = sphere's diameter.

$R = \dfrac{15\sqrt{2}}{2}$. Volume $= \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi \times \dfrac{3375 \times 2\sqrt{2}}{8} = 1125\sqrt{2}\,\pi$.