CAT 2024 Slot 2 QA paper with solutions

Solution

Bina's cost price $= \dfrac{4860}{0.81} = 6000$.

Hari's purchase price would give Bina $17\%$ profit: $1.17 \times 6000 = 7020$.

Shyam bought at 4860 and sold at 7020.

Shyam's profit $= 7020 - 4860 = 2160$.

Solution

The triangle is right-angled at $(1, 2)$ with legs 6 and 8, so the hypotenuse is 10.

Inradius of a right triangle $= \dfrac{6 + 8 - 10}{2} = 2$.

Solution

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

Fruits sold $= 0.2N + 96 + 0.4A = 0.5N \Rightarrow 0.4A = 0.3N - 96 \Rightarrow A = 0.75N - 240$.

Then $B = 0.6N - A = 240 - 0.15N$.

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

Whole numbers: half the mangoes ($0.2N$) needs $N$ to be a multiple of 5, and $A = 0.75N - 240$ needs $N$ to be a multiple of 4. So $N$ is a multiple of 20 (then $0.4A$ is also whole).

Smallest multiple of 20 greater than 320 is $340$.

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

Solution

$\dfrac{\log_8(a + b)}{\log_2 c} = \dfrac{1}{3}\log_c(a + b)$ and $\dfrac{\log_{27}(a - b)}{\log_3 c} = \dfrac{1}{3}\log_c(a - b)$.

So $\dfrac{1}{3}\log_c(a^2 - b^2) = \dfrac{2}{3} \Rightarrow a^2 - b^2 = c^2 \Rightarrow a^2 = b^2 + c^2$.

With $b, c \leq 10$: $a^2 \leq 200 \Rightarrow a \leq 14.1$.

$a = 14$ is possible (e.g. $b = c = \sqrt{98} \approx 9.9$). Greatest integer $a = 14$.

Solution

Let $g(x) = f(x) + 1$. Then $g(xy) = f(xy) + 1 = (f(x) + 1)(f(y) + 1) = g(x)g(y)$, and $g(p) = 2$ for every prime.

$160000 = 2^8 \times 5^4$, which has 12 prime factors (with repetition), so $g(160000) = 2^{12} = 4096$.

$f(160000) = 4096 - 1 = 4095$.

Solution

$\alpha\beta = -\dfrac{1}{3}$. $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = 9(\alpha^2 + \beta^2) = 15 \Rightarrow \alpha^2 + \beta^2 = \dfrac{5}{3}$.

$(\alpha + \beta)^2 = \dfrac{5}{3} - \dfrac{2}{3} = 1$.

$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = 2(\alpha + \beta)$.

$(\alpha^3 + \beta^3)^2 = 4(\alpha + \beta)^2 = 4$.

Solution

Let the age gap be $d$ (Rajesh is older).

$d$ years ago Rajesh was Garima's present age $G$ and Garima was $G - d$: $\dfrac{G}{G - d} = \dfrac{3}{2} \Rightarrow G = 3d$, so Rajesh $= 4d$.

In $d$ years Garima will be $4d$ and Rajesh $5d$.

Ratio $= 5 : 4$.

Solution

Take the three equal radii as 1. Their centres form an equilateral triangle of side 2, with circumradius $\dfrac{2}{\sqrt{3}}$.

Small circle Y (in the middle): radius $\dfrac{2}{\sqrt{3}} - 1$. Large circle X (around them): radius $\dfrac{2}{\sqrt{3}} + 1$.

Ratio $= \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}$.

Solution

Triangle EDC is similar to EAB with ratio $\dfrac{CD}{AB} = \dfrac{1}{2}$, so D and C are the midpoints of EA and EB.

So $EA = 2AD$ and $EB = 2BC$.

Perimeter of ABCD: $2 + 1 + AD + BC = 6 \Rightarrow AD + BC = 3$.

Perimeter of AEB $= 2 + 2(AD + BC) = 8$.

Solution

Let the bonus totals be $F$ (first 10), $M$ (middle 20) and $L$ (last 10).

$F + M = 30 \times 40000 = 1.2$ million; $M + L = 1.8$ million; $F + L = 20 \times 50000 = 1$ million.

Adding: $2(F + M + L) = 4$ million, so the total is 2 million: $M = 1$, $F = 0.2$, $L = 0.8$ (million).

2023: $2F + M + 3L = 0.4 + 1 + 2.4 = 3.8$ million.

Average $= \dfrac{3800000}{40} = 95000$.

Solution

Let the daily rates be $A, V, S$: $A + V = \dfrac{1}{150}$ and $V + S = \dfrac{1}{100}$.

$75A + 135V + 45S = 1 \Rightarrow 75(A + V) + 15V + 45(V + S) = 1 \Rightarrow 0.5 + 15V + 0.45 = 1$.

$V = \dfrac{1}{300}$, $A = \dfrac{1}{300}$, $S = \dfrac{1}{150}$. Take the job as 300 units: $A = 1$, $V = 1$, $S = 2$ per day.

New schedule over a 6-day cycle: A works 6 days, V 3 days, S 2 days: $6 + 3 + 4 = 13$ units.

23 cycles $= 138$ days give 299 units; on day 139 Amal does the last unit.

Answer: $139$ days.

Solution

$\dfrac{1}{x + 5} - \dfrac{1}{2x - 3} \leq 0 \Rightarrow \dfrac{x - 8}{(x + 5)(2x - 3)} \leq 0$.

Critical points: $-5$, $\dfrac{3}{2}$, $8$. Sign check:

$x < -5$: negative ✓. $-5 < x < \dfrac{3}{2}$: positive ✗. $\dfrac{3}{2} < x < 8$: negative ✓. $x = 8$: zero ✓. $x > 8$: positive ✗.

Answer: $x < -5$ or $\dfrac{3}{2} < x \leq 8$.

Solution

Half-yearly rate $= 2\%$. Anil: $22000 \times 1.02^{12}$.

Sunil (principal $P$): $P \times 1.02^{10} \times 1.10$.

Equal amounts: $P = \dfrac{22000 \times 1.02^{12}}{1.02^{10} \times 1.1} = 20000 \times 1.02^2 = 20000 \times 1.0404 = 20808$.

Solution

Let the scheduled travel time be $T$ and the distance $D$.

Day 1: $\dfrac{D}{60} = T + 3.5$.

Day 2: the last third of the route at 40 km/h takes $\dfrac{2T}{3}$: $\dfrac{D}{120} = \dfrac{2T}{3} \Rightarrow D = 80T$.

$\dfrac{80T}{60} = T + 3.5 \Rightarrow T = 10.5$ h.

Scheduled arrival $= 9$ am $+ 10.5$ h $= 7{:}30$ pm.

Solution

$n$ must divide both exponents 25 and 40, and $n > 1$, so $n = 5$.

$m = 2^5 \times 3^8 = 32 \times 6561 = 209952$.

$m - n = 209947$.

Solution

$3^5 = 243 = 22 \times 11 + 1$, so $3^5 \equiv 1 \pmod{11}$.

$333 = 5 \times 66 + 3 \Rightarrow 3^{333} \equiv 3^3 = 27 \equiv 5 \pmod{11}$.

Solution

Rewrite: $(2x - y)^2 + 3y^2 - 6y + 3 = (2x - y)^2 + 3(y - 1)^2 = 0$.

Both squares are zero: $y = 1$, $x = \dfrac{1}{2}$.

$4x + 5y = 2 + 5 = 7$.

Solution

Let $A = x + 6\sqrt{2}$, $B = x - 6\sqrt{2}$, so $A - B = 12\sqrt{2}$.

$\sqrt{A} - \sqrt{B} = 2\sqrt{2} \Rightarrow \sqrt{A} + \sqrt{B} = \dfrac{12\sqrt{2}}{2\sqrt{2}} = 6$.

$\sqrt{A} = 3 + \sqrt{2} \Rightarrow A = 11 + 6\sqrt{2}$.

$x = 11$.

Solution

Let Q–R have $x$ paths and R–S have $y$ paths.

P to S: $3 \times 4 + 4y + 3xy = 62 \Rightarrow y(4 + 3x) = 50$.

Q to R: $x + 3 \times 4 + 4y = 27 \Rightarrow x + 4y = 15$.

Try $y = 2$: $x = 7$, and $2(4 + 21) = 50$ ✓.

Direct Q–R paths $= 7$.

Solution

If $x < 0$: the first equation gives $y = 15$, then the second gives $x = 20$ ✗.

So $x \geq 0$: $2x + y = 15$. If $y \geq 0$, the second gives $x = 20$ and $y = -25$ ✗, so $y < 0$: $x - 2y = 20$.

Solving: $x = 10$, $y = -5$.

$x - y = 15$.

Solution

Let acid $= a$ and water $= w$.

After adding 2 L water: $a = 0.5(a + w + 2) \Rightarrow a = w + 2$.

After adding 15 L acid: $a + 15 = 0.8(a + w + 17) = 0.8(2a + 15)$.

$a + 15 = 1.6a + 12 \Rightarrow a = 5$, so $w = 3$.

Water : acid $= 3 : 5$.

Solution

The $n$-th term is $\left(\dfrac{1}{25}\right)^n - \left(\dfrac{1}{35}\right)^n$.

Sum $= \dfrac{1/25}{1 - 1/25} - \dfrac{1/35}{1 - 1/35} = \dfrac{1}{24} - \dfrac{1}{34} = \dfrac{10}{816} = \dfrac{5}{408}$.