CAT 2025 Slot 2 QA paper with solutions

Solution

Let $u = 3^{x^2 + 2x - 3}$. Then $9^{x^2 + 2x - 3} = u^2$ and $3^{x^2 + 2x - 2} = 3u$.

$u^2 - 12u + 27 = 0 \Rightarrow u = 3$ or $9$.

$x^2 + 2x - 3 = 1 \Rightarrow x^2 + 2x - 4 = 0$ (product of roots $-4$).

$x^2 + 2x - 3 = 2 \Rightarrow x^2 + 2x - 5 = 0$ (product of roots $-5$).

Product of all four values $= (-4)(-5) = 20$.

Solution

Sum of days 1–7 $= 7 \times 60 = 420$; sum of days 1–8 $= 8 \times 63 = 504$, so day 8 $= 84$.

Day 9 $= 84 - 11 = 73$, so the sum of days 1–9 $= 577$.

Sum of days 2–9 $= 8 \times 66 = 528$.

Day 1 $= 577 - 528 = 49$.

Solution

Case $x \geq -9$: $x^2 - (x + 9) + x = x^2 - 9 > 0 \Rightarrow x < -3$ or $x > 3$, giving $[-9, -3) \cup (3, \infty)$.

Case $x < -9$: $x^2 + x + 9 + x = x^2 + 2x + 9 > 0$ always (discriminant $< 0$).

Combined: $(-\infty, -3) \cup (3, \infty)$.

Solution

Selling price at the original discount $d$: $M(1 - d) = 1.2 \times 1650 = 1980$.

With the discount doubled: $M(1 - 2d) = 1650 + 110 = 1760$.

Subtracting: $Md = 220$, so $M = 1980 + 220 = 2200$.

Now find a discount $r$ where the profit $\%$ is also $r$:

$2200(1 - r) = 1650(1 + r) \Rightarrow 550 = 3850r \Rightarrow r = \dfrac{1}{7} \approx 14.3\%$.

Nearest: $14$.

Solution

Let $A = m + 2n$ and $B = 2m + n$ with $AB = 27$. Then $m = \dfrac{2B - A}{3}$, $n = \dfrac{2A - B}{3}$, and $2m - 3n = \dfrac{7B - 8A}{3}$.

Integer $m, n$ need $2B - A$ divisible by 3: $(A, B) = (3, 9), (9, 3), (-3, -9), (-9, -3)$.

Values of $2m - 3n$: $13, -17, -13, 17$.

Maximum $= 17$ (at $m = 1$, $n = -5$).

Solution

$625^{65} = 5^{260}$ and $128^{36} = 2^{252}$.

Product $= 10^{252} \times 5^8 = 390625 \times 10^{252}$.

Sum of digits $= 3 + 9 + 0 + 6 + 2 + 5 = 25$.

Solution

Let the common root be $\alpha$: $3\alpha^2 - 5\alpha + p = 0$ and $2\alpha^2 - 2\alpha + q = 0$.

$2 \times$ first $- 3 \times$ second: $-4\alpha + 2p - 3q = 0 \Rightarrow \alpha = \dfrac{2p - 3q}{4}$.

Other roots: $\dfrac{5}{3} - \alpha$ and $1 - \alpha$. Sum $= \dfrac{8}{3} - 2\alpha = \dfrac{8}{3} - p + \dfrac{3}{2}q$.

Solution

Read the equation as $\log_{64} x^2 + \log_8 \sqrt{y} + 3\log_{512}(\sqrt{y}\,z) = 4$ and convert to base 2:

$\dfrac{2}{6}\log_2 x + \dfrac{1}{6}\log_2 y + \dfrac{3}{9}\left(\dfrac{1}{2}\log_2 y + \log_2 z\right) = 4$.

$\dfrac{1}{3}(\log_2 x + \log_2 y + \log_2 z) = 4 \Rightarrow xyz = 2^{12} = 4096$.

By AM–GM, $x + y + z \geq 3\sqrt[3]{4096} = 3 \times 16 = 48$.

Solution

Let the stream speed be $c$. For Sneha: $\dfrac{14}{6 - c} - \dfrac{14}{6 + c} = 0.8$.

$\dfrac{28c}{36 - c^2} = 0.8 \Rightarrow c^2 + 35c - 36 = 0 \Rightarrow c = 1$ km/h.

Rita goes $d$ km and back: $\dfrac{d}{4} + \dfrac{d}{6} = \dfrac{100}{60} \Rightarrow \dfrac{5d}{12} = \dfrac{5}{3} \Rightarrow d = 4$.

Total distance $= 2d = 8$ km.

Solution

$3ac = 8(a + b) \Rightarrow b = \dfrac{a(3c - 8)}{8}$.

$a = 1$: $c = 8$, $b = 2$ gives $3 + 4 + 8 = 15$.

$a = 2$: $b = \dfrac{3c - 8}{4}$; $c = 4$ gives $b = 1$, and $3(2) + 2(1) + 4 = 12$ (check: $3 \times 2 \times 4 = 24 = 8 \times 3$ ✓).

Larger $a$ only increase the total (e.g. $a = 4$: 28; $a = 8$: 29).

Smallest value $= 12$.

Solution

$g$ is undefined at $x = 1$; $f$ is undefined at $x = \dfrac{1}{2}$.

$f(g(x))$ also fails where $g(x) = \dfrac{1}{2}$: $\dfrac{x}{x - 1} = \dfrac{1}{2} \Rightarrow x = -1$.

$g(f(x))$ also fails where $f(x) = 1$: $\dfrac{x}{2x - 1} = 1 \Rightarrow x = 1$.

Excluded values: $-1, \dfrac{1}{2}$ and $1$.

Solution

The loan equals the present value of the two instalments:

$1000 = \dfrac{530}{1 + r} + \dfrac{594}{(1 + r)^2}$.

Check $r = 8\%$: $\dfrac{530}{1.08} = 490.74$ and $\dfrac{594}{1.1664} = 509.26$; total $= 1000$ ✓.

Rate $= 8\%$.

Solution

For tangents from P with a third tangent AB, the angle at the centre is $\angle AOB = 90^\circ - \dfrac{\angle APB}{2}$.

$50 = 90 - \dfrac{\angle APB}{2} \Rightarrow \angle APB = 80^\circ$.

Solution

A divisor $\equiv 1 \pmod 3$ cannot contain 3, so it is $2^a 5^c 7^d$.

Mod 3: $2 \equiv -1$, $5 \equiv -1$, $7 \equiv 1$. We need $(-1)^{a + c} = 1$, i.e. $a + c$ even.

$a \in \{0..6\}$ (4 even, 3 odd), $c \in \{0..3\}$ (2 even, 2 odd): $4 \times 2 + 3 \times 2 = 14$ ways; $d$ has 3 choices.

Total $= 14 \times 3 = 42$.

Solution

Take side 2, centre at the origin: $B(1, \sqrt{3})$, $C(-1, \sqrt{3})$, $P(1.5, \tfrac{\sqrt{3}}{2})$, $Q(-1.5, \tfrac{\sqrt{3}}{2})$.

PBCQ has parallel sides $PQ = 3$ and $BC = 2$, height $\dfrac{\sqrt{3}}{2}$: area $= \dfrac{1}{2}(5)\dfrac{\sqrt{3}}{2} = \dfrac{5\sqrt{3}}{4}$.

Hexagon area $= \dfrac{3\sqrt{3}}{2} \times 4 = 6\sqrt{3}$.

Ratio $= \dfrac{5\sqrt{3}/4}{6\sqrt{3}} = 5 : 24$.

Solution

Write $b = a + p$, $c = a + q$, $d = a + r$ (integers). Then $4a + p + q + r = 46$.

Since 46 is not a multiple of 4, $p + q + r \neq 0$; in fact $p + q + r \equiv 2 \pmod 4$, so $|p + q + r| \geq 2$.

We minimise $p^2 + q^2 + r^2$: with $p + q + r = 2$, take $p = q = 1$, $r = 0$ (e.g. $a = 11$: 11, 12, 12, 11).

Minimum $= 1 + 1 + 0 = 2$.

Solution

Let expenditures be $2k$ (Lakshmi) and $3k$ (Meenakshi).

Lakshmi's income $= \dfrac{6}{7} \times 3k = \dfrac{18k}{7}$, so her savings $= \dfrac{18k}{7} - 2k = \dfrac{4k}{7}$.

Meenakshi's savings $= \dfrac{9}{4} \times \dfrac{4k}{7} = \dfrac{9k}{7}$; her income $= 3k + \dfrac{9k}{7} = \dfrac{30k}{7}$.

Ratio of incomes $= 18 : 30 = 3 : 5$.

Solution

Terms $a, ar, ar^2$: $a(1 + r + r^2) = 52$.

Pairwise products: $a^2 r(1 + r + r^2) = 624 \Rightarrow ar \times 52 = 624 \Rightarrow ar = 12$.

$\dfrac{12}{r} + 12 + 12r = 52 \Rightarrow 3r^2 - 10r + 3 = 0 \Rightarrow r = \dfrac{1}{3}$ (decreasing), $a = 36$.

Sum $= \dfrac{36}{1 - 1/3} = 54$.

Solution

Let coffee cost $c$ and cocoa cost $k$ per kg.

$0.16c + 0.84k = 240$ and $0.36c + 0.64k = 320$.

Subtracting: $0.2(c - k) = 80 \Rightarrow c - k = 400$, so $k = 240 - 0.16 \times 400 = 176$.

New mix with coffee fraction $p$: $k + 400p = 376 \Rightarrow p = 0.5$.

Coffee in 10 kg $= 5$ kg.

Solution

$AT : TD = 3 : 1$ and $BT : TE = 4 : 1$. Use mass points with mass 1 at A.

On AD: $1 \times 3 = m_D \times 1 \Rightarrow m_D = 3$, so $m_B + m_C = 3$.

On BE: $m_B \times 4 = m_E \times 1$, and $m_E = m_A + m_C = 1 + m_C$, so $1 + m_C = 4m_B$.

Solving: $m_B = 0.8$, $m_C = 2.2$.

$BD : DC = m_C : m_B = 2.2 : 0.8 = 11 : 4$.

Solution

Let Chandan's rate be $c$; Bipin $= 2c$, Ankita $= 4c$.

First 20 days (all three): $20 \times 7c = 140c$. Next 40 days (Ankita and Chandan): $40 \times 5c = 200c$.

Job $= 340c$, so Chandan alone takes $340$ days.

Solution

Let the total be $100$.

Pinu $= 20$. Remaining $= 80$, so Meena $= 40\%$ of $80 = 32$.

Seema gets $20\%$ less than Pinu $= 16$.

Rinu $= 100 - 20 - 32 - 16 = 32$.

Pinu : Rinu $= 20 : 32 = 5 : 8$.