CAT 2025 Slot 3 QA paper with solutions

Solution

Let the marked price be $M$.

Revenue $= M(120 \times 0.8 + 135 \times 0.9 + 150 \times 0.95 + 165) = M(96 + 121.5 + 142.5 + 165) = 525M$.

Units sold $= 570$, so cost $= 570 \times 240 = 136800$.

$525M - 136800 = 138825 \Rightarrow 525M = 275625 \Rightarrow M = 525$.

Solution

Rate of one member per hour: A $= \dfrac{1}{5 \times 40} = \dfrac{1}{200}$, B $= \dfrac{1}{400}$, C $= \dfrac{1}{40}$.

Starting group: $\dfrac{2}{200} + \dfrac{3}{400} + \dfrac{1}{40} = 0.01 + 0.0075 + 0.025 = 0.0425$ per hour.

In 23 hours: $0.9775$ done; $0.0225$ remains for the last hour.

Without the C member the group does $0.0175$ per hour, so it needs an extra $0.005$, i.e. $\dfrac{0.005}{0.0025} = 2$ more B members.

Solution

Let $s, a, c$ be the numbers of science, arts and commerce students.

$s + a + c = 1500$ and $1100s + 1000a + 800c = 1550000$.

Subtract $800 \times$ the first: $300s + 200a = 350000 \Rightarrow 3s + 2a = 3500$.

Since $s \leq a$: $3500 = 3s + 2a \geq 5s \Rightarrow s \leq 700$.

$s = 700$ works ($a = 700$, $c = 100$). Maximum $= 700$.

Solution

$a_4 + a_7 + a_{10} = 3a_7 = 99 \Rightarrow a_7 = 33$.

$S_{14} = 7(a_1 + a_{14}) = 7(a_7 + a_8) = 497 \Rightarrow a_8 = 38$, so $d = 5$ and $a_1 = 3$.

$S_5 = \dfrac{5}{2}(2 \times 3 + 4 \times 5) = 65$.

Solution

Let her walking speed be $w$; running speed $= 1.4w$.

BC walked in 3.5 h: BC $= 3.5w$. BA run in 3.5 h: AB $= 4.9w$.

Walk A to B: $4.9$ h. Run B to C: $\dfrac{3.5w}{1.4w} = 2.5$ h.

Total $= 7.4$ h $= 444$ minutes.

Solution

Let the number be $ABCD$. $A + B + C = 15$ and $B + C + D = 16 \Rightarrow D = A + 1$.

$C = D + 6$, so $D \leq 3$, and $A = D - 1 \geq 1$ gives $D = 2$ or $3$.

$D = 2$: $A = 1$, $C = 8$, $B = 6$: 1682. $D = 3$: $A = 2$, $C = 9$, $B = 4$: 2493.

Difference $= 2493 - 1682 = 811$.

Solution

Let the initial speed be $v$ and the distance left after the stop be $D$.

Making up 20 minutes: $\dfrac{D}{v} - \dfrac{D}{v + 3} = \dfrac{1}{3} \Rightarrow D = \dfrac{v(v + 3)}{9}$.

Making up 30 minutes: $\dfrac{D}{v} - \dfrac{D}{v + 5} = \dfrac{1}{2} \Rightarrow D = \dfrac{v(v + 5)}{10}$.

$\dfrac{v + 3}{9} = \dfrac{v + 5}{10} \Rightarrow v = 15$ km/h.

Solution

The rates are $4 : 9 : 36$, so all three together are $\dfrac{49}{4}$ times as fast as A.

Time $= 15 \times \dfrac{4}{49} = \dfrac{60}{49}$ hours $\approx 73.5$ minutes, nearest 73.

Solution

Let $t = x^2 + 3x$. Then $f(x) + 1 = t(t + 2) + 1 = (t + 1)^2$, so $|t + 1| = 9701$.

$t = 9700$: $x^2 + 3x - 9700 = 0$ has real roots with sum $-3$.

$t = -9702$: $x^2 + 3x + 9702 = 0$ has no real roots.

Sum of real roots $= -3$.

Solution

Let $y = f(x)$: $2yx^2 + (4y - 2)x + (3 - 6y) = 0$ must have a real solution.

Discriminant $\geq 0$: $(4y - 2)^2 - 8y(3 - 6y) \geq 0 \Rightarrow 64y^2 - 40y + 4 \geq 0 \Rightarrow (8y - 1)(2y - 1) \geq 0$.

$y \leq \dfrac{1}{8}$ or $y \geq \dfrac{1}{2}$.

Range: $\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty\right)$.

Solution

$10^{25} - 123 = \underbrace{99\ldots9}_{22}877$ (25 digits).

Adding $10^{50}$ puts a 1 in front with zeros in between.

Digit sum $= 1 + 22 \times 9 + 8 + 7 + 7 = 1 + 198 + 22 = 221$.

Solution

Altitude to BC $= \sqrt{50^2 - 40^2} = 30$. Area $= \dfrac{1}{2} \times 80 \times 30 = 1200$.

Altitudes to AB and AC $= \dfrac{2 \times 1200}{50} = 48$ each.

Sum $= 30 + 48 + 48 = 126$.

Solution

$\left(x + \dfrac{1}{x}\right)^2 = 25 + 2 = 27 \Rightarrow x + \dfrac{1}{x} = 3\sqrt{3}$ (as $x > 0$).

$x^3 + \dfrac{1}{x^3} = (3\sqrt{3})^3 - 3(3\sqrt{3}) = 81\sqrt{3} - 9\sqrt{3} = 72\sqrt{3}$.

$x^4 + \dfrac{1}{x^4} = 25^2 - 2 = 623$.

$x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right)\left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 72\sqrt{3} \times 623 - 3\sqrt{3} = 44853\sqrt{3}$.

Solution

Let P∩C $=$ C∩M $= x$ and P∩M $= 2x$ (each including those in all three, $t \geq 1$).

$150 = 75 + 111 + 40 - 4x + t \Rightarrow t = 4x - 76$.

Physics but not maths $= 75 - 2x$, largest when $x$ is smallest. $t \geq 1 \Rightarrow x \geq 20$ (then $t = 4$).

Check: only-C $= 40 - 40 + 4 = 4$, only-P $= 19$, only-M $= 55$; all valid.

Maximum $= 75 - 40 = 35$.

Solution

The base $x - 3$ must be positive and not 1: $x > 3$, $x \neq 4$.

$\log_{x-3}(x^2 - 9) = \log_{x-3}\left[(x + 1)(x - 3)^2\right]$, so $(x - 3)(x + 3) = (x + 1)(x - 3)^2$.

$x + 3 = (x + 1)(x - 3) \Rightarrow x^2 - 3x - 6 = 0 \Rightarrow x = \dfrac{3 \pm \sqrt{33}}{2}$.

Only $\dfrac{3 + \sqrt{33}}{2} \approx 4.37$ satisfies $x > 3$. Sum $= \dfrac{3 + \sqrt{33}}{2}$.

Solution

Total salary $= 30 \times 60000 = 1800000$.

A $5\%$ rise in the average means the total rises by $90000$.

This equals $20\%$ of the managers' total, so managers' total $= 450000$.

Engineers' total $= 1350000$; average $= \dfrac{1350000}{25} = 54000$.

Solution

The height AD equals the diameter: $AD = 6$.

A tangential quadrilateral has $AB + DC = AD + BC$, so with $AB = 3DC$: $BC = 4DC - 6$.

Also $BC^2 = 6^2 + (AB - DC)^2 = 36 + 4DC^2$.

$(4DC - 6)^2 = 36 + 4DC^2 \Rightarrow 12DC^2 = 48DC \Rightarrow DC = 4$, $AB = 12$.

Area $= \dfrac{1}{2}(12 + 4) \times 6 = 48$.

Solution

Write each side in primes.

Left: $2^{24x} 3^{12x} \cdot 2^{48x + 24} \cdot 5^{2y} = 2^{72x + 24}\, 3^{12x}\, 5^{2y}$.

Right: $2^{12z} \cdot 2^{24x} 5^{12x} \cdot 3^{15x - 30}$.

Powers of 3: $12x = 15x - 30 \Rightarrow x = 10$. Powers of 5: $2y = 120 \Rightarrow y = 60$.

Powers of 2: $744 = 12z + 240 \Rightarrow z = 42$.

$x + y + z = 112$.

Solution

Triangles ABD and AEB share $\angle BAD$, and $\angle ABD = \angle ACB = \angle AEB$ (since $AB = AC$).

So they are similar: $\dfrac{AB}{AE} = \dfrac{AD}{AB}$.

$AE = \dfrac{AB^2}{AD} = \dfrac{144}{8} = 18$.

Solution

Let $x$ litres be moved. B then has $x$ alcohol and 60 water.

Moving $x$ back carries $\dfrac{x^2}{60 + x}$ alcohol and $\dfrac{60x}{60 + x}$ water.

A: alcohol $= 60 - x + \dfrac{x^2}{60 + x} = \dfrac{3600}{60 + x}$, water $= \dfrac{60x}{60 + x}$.

$\dfrac{3600}{60x} = \dfrac{15}{4} \Rightarrow \dfrac{60}{x} = \dfrac{15}{4} \Rightarrow x = 16$.

Solution

Let A $= 17k$, B $= 7k$.

$\dfrac{17k - 108}{7k + 108} = \dfrac{37}{20} \Rightarrow 340k - 2160 = 259k + 3996 \Rightarrow k = 76$.

Total coins $= 24k = 1824$. After the first shift, A $= 1292 - 108 = 1184$.

For 1 : 1 each box needs 912. Coins to shift $= 1184 - 912 = 272$.

Solution

$p + q = 900 - r$ with $r \in \{169, 196, \dots, 484\}$.

$0.3q \leq p \leq 0.7q$ means $\dfrac{3}{13} \leq \dfrac{p}{p + q} \leq \dfrac{7}{17}$.

Maximum $p$: take $r = 169$, so $p + q = 731$ and $p \leq \dfrac{7}{17} \times 731 = 301$ ($q = 430$ ✓).

Minimum $p$: take $r = 484$, so $p + q = 416$ and $p \geq \dfrac{3}{13} \times 416 = 96$ ($q = 320$ ✓).

Sum $= 301 + 96 = 397$.