CAT 2025 QA Slot 1 Questions & Solutions

Key idea: \(b\) and \(c\) appear only as \(b-c\) in every expression, so treat \(t=b-c\) as a single unknown.

1. The equations become \(a-6t=4\) and \(6a+3t=50\).
2. From the first, \(a=4+6t\). Substituting: \(6(4+6t)+3t=50 \Rightarrow 24+39t=50 \Rightarrow t=\frac{2}{3}\).
3. Then \(a=4+6\cdot\frac{2}{3}=8\).
4. Required value: \(2a+3(b-c)=2(8)+3\cdot\frac{2}{3}=16+2=18\).

Tip: When three unknowns appear with only two equations, look for a grouping (here \(b-c\)) that reduces the count of unknowns.

Hence, the answer is (A) 18.

Key idea: Multiply the independent choices; the sauce choice is "0, 1 or 2 sauces", counted with combinations.

1. Sandwich type, bread and size: \(5\times4\times2=40\) ways.
2. Sauces (up to 2 of 6, order irrelevant): \(\binom{6}{0}+\binom{6}{1}+\binom{6}{2}=1+6+15=22\) ways.
3. Total orders: \(40\times22=880\).

Tip: "Optionally" and "up to" mean the no-sauce case (\(\binom{6}{0}=1\)) must be included.

Hence, the answer is (A) 880.

Key idea: For a base between 0 and 1, \(\log_b y>0\) exactly when \(0<y<1\).

1. We need \(0<n^2-7n+11<1\).
2. Upper bound: \(n^2-7n+10<0 \Rightarrow (n-2)(n-5)<0 \Rightarrow 2<n<5\), so \(n=3\) or \(4\).
3. Check the lower bound: \(n=3\) gives \(9-21+11=-1\); \(n=4\) gives \(16-28+11=-1\). Both are negative, so the logarithm is not even defined.
4. No integer satisfies both conditions.

Tip: The quadratic has its minimum at \(n=3.5\) with value \(-0.25\), and it is negative for all \(n\) in \((2.38, 4.62)\), which swallows every candidate.

Hence, the answer is (B) 0.

Key idea: The sum of the first \(m\) odd numbers is \(m^2\).

1. \(\{1,3,\ldots,57\}\) has 29 terms with total sum \(29^2=841\).
2. Let \(k=2m-1\) be the \(m\)-th odd number. Sum of elements below \(k\) is \((m-1)^2\); sum of elements above \(k\) is \(841-m^2\).
3. Set them equal: \((m-1)^2=841-m^2 \Rightarrow 2m^2-2m-840=0 \Rightarrow m^2-m-420=0 \Rightarrow (m-21)(m+20)=0\).
4. So \(m=21\) and \(k=2(21)-1=41\).

Check: below 41 the sum is \(20^2=400\); above 41 it is \(841-441=400\).

Hence, the answer is (A) 41.

Key idea: First list the allowed digits, then build the smallest number.

1. Non-zero digits that are not perfect squares: remove 1, 4, 9, leaving \(\{2,3,5,6,7,8\}\).
2. Primes among them: 2, 3, 5, 7. Non-primes: 6, 8.
3. Exactly one digit is prime, so the other two must be the only non-primes, 6 and 8 (digits are distinct).
4. Digits are {prime, 6, 8}. For the minimum, take the smallest prime 2 in the hundreds place: \(N=268\).
5. \(268=2^2\times67\), so the number of factors is \((2+1)(1+1)=6\).

Hence, the answer is 6.

Key idea: The difference between the two amounts is simple interest for the extra time, which gives the yearly interest, then the principal and rate.

1. Interest for \(6.5-3=3.5\) years \(=18960-13920=5040\), so interest per year \(=1440\).
2. Principal \(=13920-3\times1440=9600\); rate \(=\frac{1440}{9600}=15\%\) p.a.
3. Compounded half-yearly for 2 years: 4 periods at \(7.5\%\).
4. \(1.075^2=1.155625\), and \(1.155625^2\approx1.335469\).
5. Interest \(=9600\times0.335469\approx3220.5\).

Hence, the answer is (D) 3221.

Key idea: If the roots are integers \(p, q\), then \(p+q=5\) and \(pq=k\ge0\).

1. \(pq\ge0\) means both roots have the same sign (or one is zero). Two negative numbers cannot add up to 5, so both roots are non-negative.
2. Non-negative pairs with sum 5: \((0,5),(1,4),(2,3)\).
3. These give \(k=0,4,6\).

Tip: Equivalently, the discriminant \(25-4k\) must be a perfect square; \(k=0,4,6\) give 25, 9, 1.

Hence, the answer is 3.

Key idea: The extreme value of a quadratic occurs at its vertex.

1. \(f(x)=(x-2c)^2+8c-4c^2\), so \(\min f=8c-4c^2\).
2. \(g(x)=-\left(x-\frac{3c}{2}\right)^2+\frac{9c^2}{4}-2c\), so \(\max g=\frac{9c^2}{4}-2c\).
3. Require \(8c-4c^2>\frac{9c^2}{4}-2c \Rightarrow 10c>\frac{25c^2}{4} \Rightarrow 25c^2-40c<0 \Rightarrow c(5c-8)<0\).
4. So \(0<c<\frac{8}{5}\). Among the options only \(c=\frac{1}{2}\) lies in this interval.

Hence, the answer is (D) \(\frac{1}{2}\).

Key idea: Track the litres of acid; the total volume stays 200 L throughout.

1. Initially acid \(=30\%\) of 200 \(=60\) L.
2. Replace 20% with water: acid \(=60\times0.8=48\) L.
3. Replace 10% (20 L) with acid: 4.8 L acid removed, 20 L acid added, so acid \(=48-4.8+20=63.2\) L.
4. Replace 15% with water: acid \(=63.2\times0.85=53.72\) L.
5. Percentage \(=\frac{53.72}{200}\times100=26.86\%\approx27\%\).

Tip: Replacing a fraction \(f\) with water simply multiplies the acid by \((1-f)\); only the "replace with acid" step needs an addition.

Hence, the answer is (D) 27.

Key idea: Use the first part to fix the starting terms of both progressions, then use total time and total distance.

1. First part: speed \(960\) m/min \(=57.6\) km/h, time \(30\) min \(=\frac12\) h, distance 28.8 km.
2. Times in AP: \(\frac12,\frac12+d,\frac12+2d,\frac12+3d\) with sum \(2+6d=3\Rightarrow d=\frac16\). Times: \(\frac12,\frac23,\frac56,1\) h.
3. Speeds in AP: \(57.6+je\) for \(j=0,1,2,3\). Total distance:
   \(57.6\left(\frac12+\frac23+\frac56+1\right)+e\left(\frac23+\frac{10}{6}+3\right)=224\)
   \(172.8+\frac{16}{3}e=224\Rightarrow e=9.6\).
4. Fourth part: speed \(57.6+28.8=86.4\) km/h for 1 h, i.e. 86.4 km \(=86400\) m.

Check: distances 28.8, 44.8, 64, 86.4 km add up to 224 km.

Hence, the answer is (D) 86400.

Key idea: Use the first change to find the actual numbers, then the second change to find the new joiners.

1. Let the shifts have \(13x\) and \(9x\) students. After the transfer: \(\frac{13x-21}{9x+21}=\frac{19}{14}\).
2. \(182x-294=171x+399\Rightarrow11x=693\Rightarrow x=63\). Now morning \(=798\), afternoon \(=588\) (ratio \(42\cdot19:42\cdot14\)).
3. New students \(3y\) and \(8y\): \(\frac{798+3y}{588+8y}=\frac54\Rightarrow3192+12y=2940+40y\Rightarrow y=9\).
4. New students \(=11y=99\).

Hence, the answer is (D) 99.

Key idea: \(CP=CQ\) (radii), so \(\angle PQC=45^\circ\) makes \(\triangle PCQ\) a right isosceles triangle.

1. \(\angle PCQ=90^\circ\), so \(PQ=r\sqrt2=6\sqrt2\cdot\sqrt2=12\), and its distance from \(C\) is \(\frac{PQ}{2}=6\).
2. Distances are in ratio \(3:2\), so \(SR\) is at distance 4 from \(C\). Half of \(SR=\sqrt{72-16}=\sqrt{56}=2\sqrt{14}\), so \(SR=4\sqrt{14}\).
3. The chords lie on opposite sides of the centre, so the distance between them is \(6+4=10\).
4. \(PQRS\) is a trapezium: area \(=\frac12(12+4\sqrt{14})\times10=60+20\sqrt{14}=20(3+\sqrt{14})\).

Hence, the answer is (C) \(20(3+\sqrt{14})\).

Key idea: In a parallelogram the diagonals bisect each other, so \(P+R=Q+S\).

1. \(S=P+R-Q=(-3+9-1,\,-2+1+5)=(5,4)\).
2. Line through \(Q(1,-5)\) and \(S(5,4)\) has slope \(\frac{4-(-5)}{5-1}=\frac94\): \(y+5=\frac94(x-1)\).
3. Put \(y=0\): \(5=\frac94(x-1)\Rightarrow x-1=\frac{20}{9}\Rightarrow x=\frac{29}{9}\).

Hence, the answer is (B) \(\frac{29}{9}\).

Key idea: All the head-counts (initial and remaining) must be whole numbers.

1. Let there be \(b\) boys and \(g\) girls. Remaining: \(0.6g\) girls and \(0.4b\) boys, with \(0.6g=0.4b+8\), i.e. \(3g=2b+40\).
2. \(0.4b\) and \(0.6g\) are integers, so \(b\) and \(g\) are multiples of 5. Also \(b>10\).
3. The total \(b+g=b+\frac{2b+40}{3}\) increases with \(b\), so test the smallest valid \(b\):
   • \(b=15\): \(g=\frac{70}{3}\), not an integer.
   • \(b=20\): \(g=\frac{80}{3}\), not an integer.
   • \(b=25\): \(g=30\), a multiple of 5. Valid.
4. Check: 18 girls and 10 boys remain; \(18=10+8\).

Minimum total \(=25+30=55\).

Hence, the answer is 55.

Key idea: The diagonals of a rhombus bisect each other at right angles, giving \(d_1^2+d_2^2=4a^2\) and area \(=\frac12d_1d_2\).

1. \(d_1^2+d_2^2=4\times36^2=5184\).
2. \(\frac12d_1d_2=396\Rightarrow d_1d_2=792\), so \(2d_1d_2=1584\).
3. \((d_1-d_2)^2=5184-1584=3600\Rightarrow|d_1-d_2|=60\).

Hence, the answer is 60.

Key idea: Equate the money received for 13 chairs with cost plus 26% profit.

1. Let the marked price be \(M\). Discounted price per chair \(=0.78M\); the customer pays for 12 chairs: \(12\times0.78M=9.36M\).
2. Cost of 13 chairs \(=1300\); with 26% profit the revenue must be \(1300\times1.26=1638\).
3. \(9.36M=1638\Rightarrow M=175\).

Hence, the answer is 175.

Key idea: Fix \(y\) and count the integers \(x\) allowed between \(y+1\) and the upper limit set by \(x+y<14\), i.e. \(x\le13-y\).

Total \(=7+5+3+1=16\).

Tip: The counts fall by 2 each time \(y\) rises by 1, so the sum is a quick AP of odd numbers.

Hence, the answer is 16.

Key idea: Compare each worker's cost for the whole task (daily charge \(\times\) days needed alone) and use the cheapest workers as much as the deadline allows.

1. Cost of the full task: Arun \(2160\times24=51840\), Varun \(2400\times21=50400\), Tarun \(2160\times15=32400\).
2. Tarun is cheapest per unit of work but alone needs 15 days, more than 10. So use him all 10 days: he does \(\frac{10}{15}=\frac23\) of the task for \(10\times2160=21600\).
3. The remaining \(\frac13\) goes to the next cheapest, Varun: \(\frac13\times21=7\) full days (no partial-day waste), costing \(7\times2400=16800\). This fits within 10 days alongside Tarun.
4. Total \(=21600+16800=38400\).

Hence, the answer is (D) 38400.

Key idea: Everything is a power of 3, so compare exponents.

1. LHS \(=\prod_{k=1}^{20}(3^k)^k=3^{\sum k^2}\), and \(\sum_{k=1}^{20}k^2=\frac{20\cdot21\cdot41}{6}=2870\).
2. RHS \(=3^{21+22+\cdots+(20+m)}\), with exponent \(\frac{(20+m)(21+m)}{2}-210\).
3. Need \(\frac{(20+m)(21+m)}{2}-210>2870\Rightarrow(20+m)(21+m)>6160\).
4. Let \(n=20+m\): \(77\times78=6006\) (too small), \(78\times79=6162>6160\). So \(n=78\), \(m=58\).

Hence, the answer is (C) 58.

Key idea: Remove the known stock A value, then solve two equations in B and C.

1. Value of A: \(10\times120=1200\), so B and C together are worth \(3300-1200=2100\).
2. Let \(b\) shares of B and \(20-b\) of C: \(90b+150(20-b)=2100\Rightarrow3000-60b=2100\Rightarrow b=15\).

Shortcut: If all 20 were C, value would be 3000; each B instead of C cuts 60, and we need to cut 900, so \(900/60=15\).

Hence, the answer is 15.

Key idea: Express everything in terms of the gold investment \(G\).

1. Bonds \(=0.25G\), stocks \(S=100000-1.25G\).
2. Returns: \(0.10S+0.06(0.25G)+0.08G=8200\Rightarrow10000-0.125G+0.015G+0.08G=8200\).
3. \(10000-0.03G=8200\Rightarrow G=60000\). Bonds \(=15000\), stocks \(=25000\).
4. Gain from bonds \(=6\%\times15000=900\).

Check: \(2500+900+4800=8200\).

Hence, the answer is 900.

Key idea: On each interval \([n,n+1)\), \([x]=n\), so we need \(n^2\le x^2<n^2+1\), i.e. \(n\le x<\sqrt{n^2+1}\).

1. \(n=3\): \(x\in[3,\sqrt{10})\).
2. \(n=4\): \(x\in[4,\sqrt{17})\).
3. \(n=5\): \(x\in[5,\sqrt{26})\).
4. \(x=6\): \([36]=36=6^2\), valid.

So \(S=[3,\sqrt{10})\cup[4,\sqrt{17})\cup[5,\sqrt{26})\cup\{6\}\).

Checking options: (A) contains points in \([\sqrt{17},\sqrt{18})\), not in \(S\). (B) and (C) include the endpoint \(\sqrt{10}\), where \([x^2]=10\ne9\). (D) \((3,\sqrt{10})\cup[5,\sqrt{26})\cup\{6\}\) lies entirely inside \(S\).

Tip: The right endpoints \(\sqrt{n^2+1}\) are always excluded; any option with a closed bracket there is out.

Hence, the answer is (D).