CAT 2025 Slot 1 QA paper with solutions

Solution

Minimum of $f$: at $x = 2c$, $f = 8c - 4c^2$. Maximum of $g$: at $x = \dfrac{3c}{2}$, $g = \dfrac{9c^2}{4} - 2c$.

$8c - 4c^2 > \dfrac{9c^2}{4} - 2c \Rightarrow 10c > \dfrac{25c^2}{4} \Rightarrow 0 < c < 1.6$.

Of the options, only $c = \dfrac{1}{2}$ works.

Solution

Part 1: 960 m/min for 30 min $= 28800$ m.

Times are in AP with total 180 min: $4(30) + 6d = 180 \Rightarrow d = 10$, so times are 30, 40, 50, 60 min.

Speeds are $960, 960 + e, 960 + 2e, 960 + 3e$ (m/min).

$960 \times 180 + e(40 + 100 + 180) = 224000 \Rightarrow 320e = 51200 \Rightarrow e = 160$.

Part 4: $60 \times (960 + 480) = 86400$ m.

Solution

Allowed digits (non-zero, not perfect squares): 2, 3, 5, 6, 7, 8. Primes among them: 2, 3, 5, 7; non-primes: 6, 8.

Exactly one prime means the digits are 6, 8 and one prime.

Smallest such number: digits 2, 6, 8 give $N = 268$.

$268 = 2^2 \times 67$, so the number of factors $= 3 \times 2 = 6$.

Solution

For $x \in [k, k + 1)$, $[x]^2 = k^2$, so we need $[x^2] = k^2$, i.e. $x^2 < k^2 + 1$.

$k = 3$: $x \in [3, \sqrt{10})$. $k = 4$: $x \in [4, \sqrt{17})$. $k = 5$: $x \in [5, \sqrt{26})$. Also $x = 6$.

Option A, $(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}$, lies entirely inside these intervals.

The others include points such as $\sqrt{10}$, $\sqrt{17}$ or values above $\sqrt{17}$, which fail.

Solution

Let B and C be the numbers of shares, with $B + C = 20$.

$10 \times 120 + 90B + 150C = 3300 \Rightarrow 90B + 150(20 - B) = 2100$.

$3000 - 60B = 2100 \Rightarrow B = 15$.

Solution

Left side $= 3^{1^2 + 2^2 + \dots + 20^2} = 3^{2870}$.

Right side $= 3^{21 + 22 + \dots + (20 + m)} = 3^{20m + \frac{m(m + 1)}{2}}$.

Need $20m + \dfrac{m(m + 1)}{2} > 2870$. $m = 57$: $1140 + 1653 = 2793$ ✗. $m = 58$: $1160 + 1711 = 2871$ ✓.

Smallest $m = 58$.

Solution

The base $\dfrac{1}{4}$ is less than 1, so $\log_{1/4}(t) > 0$ means $0 < t < 1$.

We need $0 < n^2 - 7n + 11 < 1$. The upper bound gives $n^2 - 7n + 10 < 0 \Rightarrow 2 < n < 5$, so $n = 3$ or $4$.

But $n = 3$ gives $-1$ and $n = 4$ gives $-1$, which are not positive.

No integer works: $0$.

Solution

We need $y \geq 3$, $x > y$ and $x + y \leq 13$.

$y = 3$: $x = 4$ to $10$ (7 pairs). $y = 4$: $x = 5$ to $9$ (5). $y = 5$: $x = 6$ to $8$ (3). $y = 6$: $x = 7$ (1). $y \geq 7$: none.

Total $= 7 + 5 + 3 + 1 = 16$.

Solution

SI for the extra $3.5$ years $= 18960 - 13920 = 5040$, so SI per year $= 1440$.

Principal $= 13920 - 3 \times 1440 = 9600$. Rate $= \dfrac{1440}{9600} = 15\%$.

Half-yearly compounding for 2 years: 4 periods at $7.5\%$.

$9600 \times 1.075^4 = 9600 \times 1.33547 \approx 12820.5$.

Interest $\approx 3220.5 \approx 3221$.

Solution

Acid at start $= 30\%$ of 200 $= 60$ L.

Replace $20\%$ with water: acid $= 60 \times 0.8 = 48$ L.

Replace $10\%$ with acid: acid $= 48 \times 0.9 + 20 = 63.2$ L.

Replace $15\%$ with water: acid $= 63.2 \times 0.85 = 53.72$ L.

Percentage $= \dfrac{53.72}{200} \times 100 \approx 26.9\% \approx 27\%$.

Solution

Let the number of girls be $g$ and boys be $b$ ($b > 10$).

Girls remaining $= 0.6g$, boys remaining $= 0.4b$.

Given $0.6g - 0.4b = 8 \Rightarrow 3g - 2b = 40$.

Since $0.6g$ and $0.4b$ must be whole numbers, both $g$ and $b$ are multiples of 5.

Try $b = 15$: $3g = 70$ ✗. $b = 20$: $3g = 80$ ✗. $b = 25$: $3g = 90 \Rightarrow g = 30$ ✓.

Check: $18$ girls and $10$ boys remain, difference $8$ ✓.

Minimum total $= 30 + 25 = 55$.

Solution

Sandwich choices: $5 \times 4 \times 2 = 40$.

Sauces (none, one or two of 6): $1 + 6 + 15 = 22$.

Total $= 40 \times 22 = 880$.

Solution

There are 29 odd numbers from 1 to 57, with total $29^2 = 841$.

Let $k$ be the $j$-th odd number ($k = 2j - 1$). Sum below $= (j - 1)^2$; sum above $= 841 - j^2$.

$(j - 1)^2 = 841 - j^2 \Rightarrow j^2 - j - 420 = 0 \Rightarrow j = 21$.

$k = 2(21) - 1 = 41$.

Solution

Cost per complete job: Arun $= 2160 \times 24 = 51840$, Varun $= 2400 \times 21 = 50400$, Tarun $= 2160 \times 15 = 32400$.

Tarun is cheapest, so use him on all 10 days: he finishes $\dfrac{10}{15} = \dfrac{2}{3}$ of the job for $21600$.

The remaining $\dfrac{1}{3}$ goes to the next cheapest, Varun: $\dfrac{1}{3} \times 21 = 7$ full days, costing $7 \times 2400 = 16800$.

Minimum total $= 21600 + 16800 = 38400$.

Solution

Let investment in gold $= G$. Then bonds $= 0.25G$ and stocks $= 100000 - 1.25G$.

Total return: $0.10(100000 - 1.25G) + 0.06(0.25G) + 0.08G = 8200$

$10000 - 0.125G + 0.015G + 0.08G = 8200 \Rightarrow 0.03G = 1800 \Rightarrow G = 60000$.

Bonds $= 0.25 \times 60000 = 15000$.

Gain from bonds $= 6\%$ of $15000 = 900$.

Solution

Let $u = b - c$. The equations become $a - 6u = 4$ and $6a + 3u = 50$.

From the first, $a = 4 + 6u$. Then $24 + 39u = 50 \Rightarrow u = \dfrac{2}{3}$, $a = 8$.

$2a + 3b - 3c = 2a + 3u = 16 + 2 = 18$.

Solution

In parallelogram PQRS, $S = P + R - Q = (-3 + 9 - 1, -2 + 1 + 5) = (5, 4)$.

Line through $Q(1, -5)$ and $S(5, 4)$ has slope $\dfrac{9}{4}$.

At $y = 0$: $x = 1 + \dfrac{5 \times 4}{9} = \dfrac{29}{9}$.

Solution

$CP = CQ = 6\sqrt{2}$ and $\angle PQC = 45^\circ$, so triangle $CPQ$ is right-angled at C: $PQ = 12$, and the distance from C to PQ $= 6$.

The distances are in ratio $3 : 2$, so SR is 4 cm from C: half of SR $= \sqrt{72 - 16} = 2\sqrt{14}$, so $SR = 4\sqrt{14}$.

The chords lie on opposite sides of C, so they are $6 + 4 = 10$ cm apart.

Area $= \dfrac{1}{2}(12 + 4\sqrt{14}) \times 10 = 20(3 + \sqrt{14})$.

Solution

Let the shifts be $13k$ and $9k$.

$\dfrac{13k - 21}{9k + 21} = \dfrac{19}{14} \Rightarrow 182k - 294 = 171k + 399 \Rightarrow k = 63$.

After the move: morning $= 798$, afternoon $= 588$.

New students $3m$ and $8m$: $\dfrac{798 + 3m}{588 + 8m} = \dfrac{5}{4} \Rightarrow 3192 + 12m = 2940 + 40m \Rightarrow m = 9$.

New students $= 11m = 99$.

Solution

Diagonals $d_1, d_2$: area $= \dfrac{d_1 d_2}{2} = 396 \Rightarrow d_1 d_2 = 792$.

Half-diagonals and the side form a right triangle: $d_1^2 + d_2^2 = 4 \times 36^2 = 5184$.

$(d_1 - d_2)^2 = 5184 - 1584 = 3600 \Rightarrow |d_1 - d_2| = 60$.

Solution

Integer roots with sum 5 and product $k \geq 0$: $(0, 5), (1, 4), (2, 3)$.

$k = 0, 4, 6$: $3$ values.

Solution

Let the marked price be $M$. Discounted price per chair $= 0.78M$.

The customer pays for 12 chairs and gets 13: revenue $= 12 \times 0.78M = 9.36M$.

Cost of 13 chairs $= 13 \times 100 = 1300$; a $26\%$ profit means revenue $= 1.26 \times 1300 = 1638$.

$9.36M = 1638 \Rightarrow M = 175$.