CAT 2025 QA Slot 2 Questions & Solutions

Key idea: Put \(A = m+2n\) and \(B = 2m+n\). Then \(AB = 27\), and since \(A+B = 3(m+n)\), the sum \(A+B\) must be a multiple of 3.

Solving back: \(m = \dfrac{2B-A}{3}\), \(n = \dfrac{2A-B}{3}\).

Integer factor pairs of 27: \((\pm1,\pm27), (\pm27,\pm1), (\pm3,\pm9), (\pm9,\pm3)\). Pairs with 1 and 27 give \(A+B=\pm28\), not a multiple of 3, so only the pairs with 3 and 9 work:

• \((A,B)=(3,9)\): \(m=5, n=-1\), so \(2m-3n = 13\)
• \((A,B)=(9,3)\): \(m=-1, n=5\), so \(2m-3n = -17\)
• \((A,B)=(-3,-9)\): \(m=-5, n=1\), so \(2m-3n = -13\)
• \((A,B)=(-9,-3)\): \(m=1, n=-5\), so \(2m-3n = 17\)

The largest value is 17.

Hence, the answer is 17.

Key idea: Convert every term to base 2, then use AM-GM.

Since \(64 = 2^6\), \(8 = 2^3\) and \(512 = 2^9\):

• \(\log_{64} x^2 = \dfrac{2}{6}\log_2 x = \dfrac{1}{3}\log_2 x\)
• \(\log_8 \sqrt{y} = \dfrac{1}{3}\cdot\dfrac{1}{2}\log_2 y = \dfrac{1}{6}\log_2 y\)
• \(3\log_{512}(\sqrt{y}\,z) = \dfrac{3}{9}\left(\dfrac{1}{2}\log_2 y + \log_2 z\right) = \dfrac{1}{6}\log_2 y + \dfrac{1}{3}\log_2 z\)

Adding: \(\dfrac{1}{3}\left(\log_2 x + \log_2 y + \log_2 z\right) = 4\), so \(\log_2(xyz) = 12\), i.e. \(xyz = 2^{12} = 4096\).

By AM-GM, \(x+y+z \ge 3\sqrt[3]{xyz} = 3 \times 16 = 48\), with equality at \(x=y=z=16\).

Hence, the answer is (B) 48.

Key idea: Turn every average into a total.

1. Days 1-7 total \(= 7 \times 60 = 420\).
2. Days 1-8 total \(= 8 \times 63 = 504\), so day 8 \(= 504 - 420 = 84\).
3. Day 9 \(= 84 - 11 = 73\), so days 1-9 total \(= 504 + 73 = 577\).
4. Days 2-9 total \(= 8 \times 66 = 528\).
5. Day 1 \(= 577 - 528 = 49\).

Hence, the answer is 49.

Key idea: For a composite \(f(g(x))\), \(x\) must lie in the domain of \(g\) and \(g(x)\) must lie in the domain of \(f\).

\(f\) is undefined at \(x = \tfrac{1}{2}\); \(g\) is undefined at \(x = 1\).

\(f(g(x))\): need \(x \ne 1\) and \(g(x) \ne \tfrac12\). Solving \(\dfrac{x}{x-1} = \dfrac12\) gives \(2x = x - 1\), i.e. \(x = -1\). So exclude \(1\) and \(-1\).

\(g(f(x))\): need \(x \ne \tfrac12\) and \(f(x) \ne 1\). Solving \(\dfrac{x}{2x-1} = 1\) gives \(x = 2x - 1\), i.e. \(x = 1\). So exclude \(\tfrac12\) and \(1\).

\(h\) needs both parts defined, so the excluded values are \(-1, \tfrac12, 1\).

Hence, the answer is (A) \(-1, \tfrac12\), and 1.

Key idea: Substitute \(t = 3^{x^2+2x-3}\) to get a quadratic.

Then \(9^{x^2+2x-3} = t^2\) and \(3^{x^2+2x-2} = 3t\). The equation becomes

• \(t = 3\): \(x^2 + 2x - 3 = 1\), i.e. \(x^2 + 2x - 4 = 0\). Discriminant \(4 + 16 > 0\), two real roots, product \(-4\).
• \(t = 9\): \(x^2 + 2x - 3 = 2\), i.e. \(x^2 + 2x - 5 = 0\). Discriminant \(4 + 20 > 0\), two real roots, product \(-5\).

The four roots are distinct, and their product is \((-4)(-5) = 20\).

Hence, the answer is (B) 20.

Key idea: In a regular hexagon, BC is parallel to AD, and the segment joining the midpoints of AB and CD is the midline of trapezium ABCD.

Take side \(= 1\). Then \(AD = 2\) (a long diagonal) and the distance between BC and AD is \(\dfrac{\sqrt3}{2}\).

PQ is the midline of trapezium ABCD, so \(PQ = \dfrac{BC + AD}{2} = \dfrac{1+2}{2} = \dfrac32\), and it lies halfway between BC and AD, so the height of PBCQ is \(\dfrac{\sqrt3}{4}\).

Ratio \(= 5 : 24\).

Hence, the answer is (A) 5 : 24.

Key idea: The expression is smallest when \(b, c, d\) are as close to \(a\) as possible.

Write \(b = a+x\), \(c = a+y\), \(d = a+z\) with integers \(x, y, z\). The condition becomes \(4a + x + y + z = 46\) and the expression is \(x^2 + y^2 + z^2\).

• Value 0 needs \(x=y=z=0\), so \(4a = 46\): not an integer.
• Value 1 needs exactly one of \(x,y,z\) to be \(\pm1\), so \(4a = 45\) or \(47\): impossible.
• Value 2: take \(a = 11\), \(x = y = 1\), \(z = 0\), i.e. \((a,b,c,d) = (11,12,12,11)\). Sum \(= 46\) and the expression \(= 1 + 1 + 0 = 2\).

Hence, the answer is 2.

Key idea: Take the three terms as \(\dfrac{a}{r}, a, ar\). Then the sum of pairwise products factorises neatly.

So \(52a = 624\), giving \(a = 12\).

Then \(\dfrac{12}{r} + 12 + 12r = 52\), so \(12r^2 - 40r + 12 = 0\), i.e. \(3r^2 - 10r + 3 = 0\), giving \(r = 3\) or \(\tfrac13\).

The GP is decreasing and infinite, so \(r = \tfrac13\): the terms are \(36, 12, 4, \ldots\)

Hence, the answer is (C) 54.

Key idea: Express \(b\) in terms of \(a\) and \(c\), then try small values of \(a\).

From \(3ac = 8a + 8b\): \(b = \dfrac{a(3c-8)}{8}\). For \(b \ge 1\) we need \(3c > 8\), so \(c \ge 3\).

• \(a = 1\): \(8 \mid 3c - 8\) forces \(8 \mid c\); smallest \(c = 8\), \(b = 2\). Value \(= 3 + 4 + 8 = 15\).
• \(a = 2\): \(b = \dfrac{3c-8}{4}\) needs \(4 \mid c\); \(c = 4\) gives \(b = 1\). The numbers \(2, 1, 4\) are distinct. Value \(= 6 + 2 + 4 = 12\).
• \(a \ge 3\): \(3a \ge 9\), while \(2b + c \ge 2 + 3 = 5\), so the value is at least 14.

Check: \(3 \cdot 2 \cdot 4 = 24 = 8(2+1)\). The minimum is 12.

Hence, the answer is 12.

Key idea: The centre of a circle lies on the bisector of the angle between two tangents drawn from an external point.

Let AB touch the circle at S. From A, the tangents AQ and AS are equal, so OA bisects \(\angle QOS\). Likewise, OB bisects \(\angle SOR\). Hence

In quadrilateral PQOR, \(\angle OQP = \angle ORP = 90^\circ\), so \(\angle QOR = 180^\circ - \angle P\).

So \(\angle AOB = 90^\circ - \dfrac{\angle P}{2}\). Setting \(50 = 90 - \dfrac{\angle P}{2}\) gives \(\angle APB = 80^\circ\).

Tip: Remember \(\angle AOB = 90^\circ - \tfrac{P}{2}\) for this configuration, where the circle is the excircle of triangle PAB.

Hence, the answer is 80.

Key idea: Work modulo 3. A divisor of the form \(3r+1\) leaves remainder 1 when divided by 3.

A divisor \(2^a 3^b 5^c 7^d\) must have \(b = 0\), otherwise it is a multiple of 3.

Modulo 3: \(2 \equiv -1\), \(5 \equiv -1\), \(7 \equiv 1\). So the divisor \(\equiv (-1)^{a+c}\), and we need \(a + c\) to be even.

• \(a \in \{0,\ldots,6\}\): 4 even values, 3 odd values.
• \(c \in \{0,\ldots,3\}\): 2 even values, 2 odd values.
• \(a, c\) both even: \(4 \times 2 = 8\); both odd: \(3 \times 2 = 6\). Total 14.
• \(d \in \{0,1,2\}\): 3 choices, with no effect on the remainder.

Number of divisors \(= 14 \times 3 = 42\).

Hence, the answer is (A) 42.

Key idea: Assign efficiencies in units of work per day.

Let Chandan do 1 unit/day. Then Bipin does 2 and Ankita does 4.

• Ankita works all 60 days: \(4 \times 60 = 240\) units.
• Chandan works all 60 days: \(1 \times 60 = 60\) units.
• Bipin works 20 days: \(2 \times 20 = 40\) units.

Total work \(= 240 + 60 + 40 = 340\) units. Chandan alone needs \(340 \div 1 = 340\) days.

Hence, the answer is 340.

Key idea: Put everything in terms of one variable using the expenditure ratio.

Let the expenditures be \(14k\) (Lakshmi) and \(21k\) (Meenakshi), keeping the ratio \(2:3\) and making 21 divisible by 7.

1. Lakshmi's income \(= \dfrac{6}{7} \times 21k = 18k\), so her saving \(= 18k - 14k = 4k\).
2. The savings ratio is \(4:9\), so Meenakshi saves \(9k\).
3. Meenakshi's income \(= 21k + 9k = 30k\).

Ratio of incomes \(= 18k : 30k = 3 : 5\).

Hence, the answer is (B) 3 : 5.

Key idea: Split into cases at \(x = -9\), where \(x + 9\) changes sign.

Case 1: \(x \ge -9\). \(|x+9| = x + 9\), so the expression is \(x^2 + x - x - 9 = x^2 - 9 > 0\), giving \(|x| > 3\). Combined with \(x \ge -9\): \(x \in [-9, -3) \cup (3, \infty)\).

Case 2: \(x < -9\). \(|x+9| = -x - 9\), so the expression is \(x^2 + 2x + 9 = (x+1)^2 + 8\), which is always positive. So every \(x < -9\) works.

Combining: \((-\infty, -3) \cup (3, \infty)\).

Shortcut: Test \(x = -10\): \(100 - 1 - 10 = 89 > 0\), which rules out options (C) and (D). Test \(x = -5\): \(25 - 4 - 5 = 16 > 0\), which rules out (A).

Hence, the answer is (B) \((-\infty, -3) \cup (3, \infty)\).

Key idea: Use the two scenarios to find the marked price, then equate profit % with discount %.

1. At 20% profit, the selling price is \(1.2 \times 1650 = 1980\). With double the discount, profit is 110, so the selling price is \(1760\).
2. If \(M\) is the marked price and \(d\) the discount fraction: \(M(1-d) = 1980\) and \(M(1-2d) = 1760\). Subtracting gives \(Md = 220\), so \(M = 2200\) (and \(d = 10\%\)).
3. Let the required discount be \(r\). Then the profit % is also \(r\): \[2200(1 - r) = 1650(1 + r) \;\Rightarrow\; 550 = 3850r \;\Rightarrow\; r = \frac17 \approx 14.29\%.\]

Hence, the answer is (B) 14.

Key idea: Eliminate the \(x^2\) term to find the common root, then use the sums of roots.

Let \(\alpha\) be the common root, \(\beta\) the other root of the first equation and \(\gamma\) the other root of the second. By sums of roots, \(\alpha + \beta = \dfrac53\) and \(\alpha + \gamma = 1\), so

Since \(\alpha\) satisfies both equations, take (first) \(- \tfrac32\) (second):

Therefore \(\beta + \gamma = \dfrac83 - p + \dfrac32 q\).

Hence, the answer is (D) \(\frac83 - p + \frac32 q\).

Key idea: Find the price per kg of pure coffee and of pure cocoa from the two given mixtures.

Let coffee cost \(c\) and cocoa cost \(k\) rupees per kg.

Subtracting: \(0.2(c - k) = 80\), so \(c - k = 400\). Substituting \(c = k + 400\) into the first equation: \(k + 64 = 240\), so \(k = 176\) and \(c = 576\).

If the new mixture is a fraction \(f\) coffee: \(176 + 400f = 376\), so \(f = \tfrac12\).

Coffee in 10 kg \(= 5\) kg.

Shortcut: Each extra 20% of coffee adds Rs 80, so every 1% adds Rs 4. Going from Rs 240 to Rs 376 adds Rs 136, i.e. 34% more coffee: \(16\% + 34\% = 50\%\).

Hence, the answer is (D) 5.

Key idea: Track the outstanding balance year by year, or equate the loan to the present value of the instalments.

With rate \(r\), after year 1 the balance is \(1000(1+r) - 530\), and after year 2 this grows to exactly 594:

Try \(r = 8\%\): \(1000 \times 1.08 - 530 = 550\), and \(550 \times 1.08 = 594\). This matches exactly.

(Algebraically, with \(R = 1+r\): \(1000R^2 - 530R - 594 = 0\), so \(R = \dfrac{530 + 1690}{2000} = 1.08\).)

Hence, the answer is (A) 8.

Key idea: Use Sneha's data to find the stream speed, then apply it to Rita.

Let the stream speed be \(v\) km/h. For Sneha, 48 minutes \(= 0.8\) h:

Rita rows upstream at \(5 - 1 = 4\) km/h and downstream at \(5 + 1 = 6\) km/h. If she goes \(d\) km upstream and comes back, in 100 minutes \(= \tfrac53\) h:

Total distance covered \(= 2d = 8\) km.

Hence, the answer is 8.

Key idea: Pair 2s with 5s to make powers of 10, which only add trailing zeros.

\(625^{65} = 5^{260}\) and \(128^{36} = 2^{252}\). So

The trailing zeros add nothing, so the digit sum is \(3 + 9 + 0 + 6 + 2 + 5 = 25\).

Hence, the answer is 25.

Key idea: Use the parallel line DF twice, first in triangle ADF and then in triangle CBE.

Given: \(AT : TD = 3 : 1\) and \(BT : TE = 4 : 1\). Let \(TE = 1\), so \(BE = 5\).

1. In \(\triangle ADF\), TE is parallel to DF (T on AD, E on AF). So \(\dfrac{TE}{DF} = \dfrac{AT}{AD} = \dfrac34\), giving \(DF = \dfrac43\).
2. In \(\triangle CBE\), DF is parallel to BE (D on CB, F on CE). So \(\dfrac{CD}{CB} = \dfrac{DF}{BE} = \dfrac{4/3}{5} = \dfrac{4}{15}\).
3. So \(CD : CB = 4 : 15\), and \(BD : CD = (15 - 4) : 4 = 11 : 4\).

Hence, the answer is (C) 11 : 4.

Key idea: Take the total as 100 and work through each share.

• Pinu \(= 20\% \text{ of } 100 = 20\); remaining \(= 80\).
• Meena \(= 40\% \text{ of } 80 = 32\).
• Seema \(= 20\%\) less than Pinu \(= 0.8 \times 20 = 16\).
• Rinu \(= 100 - 20 - 32 - 16 = 32\).

Pinu : Rinu \(= 20 : 32 = 5 : 8\).

Hence, the answer is (B) 5 : 8.