Balanced High-Probability Paper

Disclaimer: This is a pattern-based predicted practice paper workflow output, not a leaked, exact, or official CAT paper.

Variant Blueprint

• Variant type: balanced
• Strategy: Maximize score with moderate difficulty.
• Target section mix: QA 22 from the frozen selected pool; VARC/DILR are outside the current candidate scope
• Expected overlap score: 0.849
• Risk score: 0.35
• Diversity score: 0.864

Questions

Q1. QA - Arithmetic / Profit and Loss

• Archetype/type: Profit-loss transaction problem
• Difficulty: Hard
• Score: 0.91

A shop buys 90 identical bags at Rs 640 each. It marks every bag at the same price. Of the 90 bags, 36 are sold at a 10% discount, 30 at a 20% discount, 12 at a 35% discount because of minor defects, and 12 remain unsold. If the shop still makes an overall profit of Rs 6600 on the entire purchase, what is the marked price of each bag?

A. Rs 900 B. Rs 960 C. Rs 1000 D. Rs 1080

Q2. QA - Arithmetic / Mixtures and Alligation

• Archetype/type: Mixture concentration problem
• Difficulty: Hard
• Score: 0.905

Solution A is $40%$ acid by volume and Solution B is $70%$ acid by volume. A chemist mixes $a$ litres of Solution A with $b$ litres of Solution B. After mixing, exactly $10$ litres of pure water evaporates from the combined mixture. The remaining mixture has volume $40$ litres and is $75%$ acid. Find the ratio $a:b$.

A. $1:1$ B. $1:2$ C. $2:3$ D. $3:5$

Q3. QA - Arithmetic / Percentages

• Archetype/type: Successive percentage change
• Difficulty: Hard
• Score: 0.892

A value is first increased by p%, then decreased by 20%, and then again increased by p%. The final value is exactly 90% of what the value would have been after only a single p% increase from the original. If p is positive, find p.

Q4. QA - Arithmetic / Time and Work

• Archetype/type: Work-efficiency problem
• Difficulty: Medium
• Score: 0.882

A can finish a job alone in 30 days, while B can finish it alone in 20 days. A starts the job alone. B joins after 6 days and leaves as soon as only one-eighth of the job is left. A then completes the remaining work alone. In how many days is the job completed?

A. 17.25 days B. 17.85 days C. 18.25 days D. 18.75 days

Q5. QA - Arithmetic / Ratio and Proportion

• Archetype/type: Ratio sharing problem
• Difficulty: Hard
• Score: 0.873

The monthly incomes of P and Q are in the ratio 8:11, while their monthly expenditures are in the ratio 5:7. Q saves Rs 3000 more than P. After P invests 20% of his saving and Q invests 25% of his saving, their remaining savings are equal. What is Q's monthly income in rupees?

Q6. QA - Geometry / Triangles

• Archetype/type: Triangle similarity problem
• Difficulty: Medium-Hard
• Score: 0.865

In triangle $ABC$, point $D$ lies on side $AB$ and point $E$ lies on side $AC$ such that $DE \parallel BC$. The line through $D$ parallel to $AC$ meets $BC$ at point $F$. If the area of triangle $ADE$ is $16$ and the area of quadrilateral $DECF$ is $40$, find the area of triangle $DBF$.

A. $20$ B. $24$ C. $25$ D. $36$

Q7. QA - Arithmetic / Averages

• Archetype/type: Weighted average problem
• Difficulty: Hard
• Score: 0.859

In a batch of $40$ candidates, the average score is $69.6$. The highest $5$ candidates have an average score of $92$, and the lowest $7$ candidates have an average score of $48$. The lowest $7$ candidates are replaced by $7$ new candidates. If the new average score of all $40$ candidates becomes $74.5$, find the average score of the $7$ new candidates.

Q8. QA - Algebra / Inequalities

• Archetype/type: Inequality optimization problem
• Difficulty: Hard
• Score: 0.855

For integer $k$ with $0 \le k \le 15$, how many values of $k$ make both roots of [x^2 + (k - 3)x + (k - 2) = 0] distinct negative real numbers?

A. $7$ B. $8$ C. $9$ D. $12$

Q9. QA - Algebra / Linear Equations

• Archetype/type: Integer feasibility under hidden divisibility constraint
• Difficulty: Hard
• Score: 0.854

A company has three departments X, Y, and Z. Every employee belongs to exactly one department. The number of employees in X is a positive multiple of 4. The number of employees in Y exceeds that in X by a positive multiple of 3. The number of employees in Z equals exactly half the sum of the employees in X and Y. If the total number of employees in the company is between 100 and 130, inclusive, how many distinct values can the total number of employees take?

Q10. QA - Algebra / Functions

• Archetype/type: Function value/domain problem
• Difficulty: Hard
• Score: 0.852

A function $f$ defined on positive integers satisfies $f(1) = 1$ and [f(n) = n - f(f(n - 1)) \quad \text{for } n \ge 2.] Find $f(20)$.

A. $10$ B. $11$ C. $12$ D. $13$

Q11. QA - Algebra / Logarithms

• Archetype/type: Base-change identity with algebraic cancellation
• Difficulty: Hard
• Score: 0.842

Let $,a = \log_{12} 18,$ and $,b = \log_{24} 54$. ; What is the value of [ \frac{1 - ab}{a - b};? ]

A. $3$ B. $5$ C. $7$ D. $-5$

Q12. QA - Geometry / Coordinate Geometry

• Archetype/type: Coordinate geometry line/slope/distance problem
• Difficulty: Hard
• Score: 0.839

In a parallelogram $ABCD$ with vertices taken in that order, $A=(1,2)$ and $B=(7,5)$. The vertex $C=(p,q)$ has integer coordinates and $p>10$. The diagonals intersect at a point lying on the line $x+2y=16$, and the area of the parallelogram is $42$. Find $p+q$.

Q13. QA - Arithmetic / Time Speed Distance

• Archetype/type: Relative speed problem
• Difficulty: Hard
• Score: 0.83

Train A crosses a 150 m platform in 20 seconds. It overtakes a 180 m long train B, moving in the same direction, in 45 seconds. If train A is 250 m long, what is the speed of train B?

A. 32.4 km/h B. 36 km/h C. 37.6 km/h D. 40 km/h

Q14. QA - Algebra / Progressions

• Archetype/type: AP/GP sequence problem
• Difficulty: Hard
• Score: 0.824

An arithmetic progression has positive integer first term $a$ and positive integer common difference $d$. The sum of the first $7$ terms equals the sum of the next $5$ terms, i.e.\ the sum of the $8$th to $12$th terms. If the $10$th term lies strictly between $50$ and $80$, how many ordered pairs $(a,d)$ are possible?

A. $1$ B. $2$ C. $3$ D. $4$

Q15. QA - Geometry / Mensuration

• Archetype/type: Mensuration volume/area problem
• Difficulty: Hard
• Score: 0.811

A right circular cone has base radius $10$ cm and height $24$ cm. It is cut by a plane parallel to the base so that the curved surface area of the smaller cone removed from the top is one-fourth of the curved surface area of the original cone. The remaining frustum is melted and recast into identical solid cylinders, each of radius $5$ cm and height $7$ cm. How many such cylinders can be made?

A. $3$ B. $4$ C. $5$ D. $6$

Q16. QA - Number System / Divisibility

• Archetype/type: Divisibility/factor-counting problem
• Difficulty: Hard
• Score: 0.808

Find the least positive integer $n$ such that $n$ leaves remainders $3$, $5$, and $7$ when divided by $5$, $7$, and $9$, respectively, and $n$ is divisible by $11$.

A. $943$ B. $1258$ C. $1573$ D. $1888$

Q17. QA - Algebra / Quadratic Equations

• Archetype/type: Roots and coefficients problem
• Difficulty: Hard
• Score: 0.801

For how many integer values of m does the equation x^2 - (m+5)x + 6m = 0 have two distinct positive integer roots, one of which is exactly twice the other?

A. 0 B. 1 C. 2 D. 3

Q18. QA - Arithmetic / Simple Interest Compound Interest

• Archetype/type: Interest-rate/time comparison problem
• Difficulty: Hard
• Score: 0.783

A sum of Rs.\ $64{,}000$ is invested in Scheme P at compound interest of $12.5%$ per annum, compounded annually, for two years. The entire amount obtained at the end of two years is then reinvested in Scheme Q at simple interest of $r%$ per annum for three years. The simple interest earned from Scheme Q is Rs.\ $7{,}300$ more than the compound interest earned from Scheme P. Find $r$.

Q19. QA - Number System / Digits

• Archetype/type: Digit manipulation problem
• Difficulty: Hard
• Score: 0.727

How many three-digit numbers have distinct non-zero digits and become larger by 198 when their digits are reversed?

A. 35 B. 42 C. 49 D. 56

Q20. QA - Arithmetic / Profit and Loss

• Archetype/type: Profit-loss transaction problem
• Difficulty: Medium
• Score: 0.908

A retailer buys 100 identical speakers at Rs 600 each and marks all of them at the same price. He sells 50 speakers at a discount of 10%, 30 speakers at a discount of 20%, and the remaining speakers, after minor carton damage, at a discount of 35%. If his overall profit is 9 1/3%, what is the marked price of each speaker?

A. Rs 760 B. Rs 780 C. Rs 800 D. Rs 840

Q21. QA - Arithmetic / Percentages

• Archetype/type: Successive percentage change
• Difficulty: Hard
• Score: 0.887

In January, revenue from product X was exactly 40% of a firm's total revenue. In February, total revenue increased by 25%, but revenue from X decreased by 10% because only some contracts were renewed. By what percentage did revenue from all other products increase from January to February?

Q22. QA - Arithmetic / Percentages

• Archetype/type: Successive percentage change
• Difficulty: Medium-Hard
• Score: 0.872

In an entrance test, $80%$ of candidates clear Section A. Of those who clear Section A, $75%$ clear Section B. Of those who clear both sections, $40%$ clear Section C. Among the candidates who do not clear all three sections, the candidates who failed only Section C (i.e., cleared A and B but failed C) form $\dfrac{9}{19}$ of that group. If the number of such candidates is $720$, find the total number of candidates who appeared.

Answer Key

Detailed Solutions

Q1

Step 1: Total cost of all 90 bags = 90 x 640 = Rs 57600. Step 2: Profit is calculated on the entire purchase, including the unsold bags, so required revenue = 57600 + 6600 = Rs 64200. Step 3: Let the marked price be M. Only 78 bags are sold, giving revenue 36 x 0.90M + 30 x 0.80M + 12 x 0.65M = 32.4M + 24M + 7.8M = 64.2M. Step 4: Therefore 64.2M = 64200, so M = Rs 1000. Step 5: Verification: revenue = 36 x 900 + 30 x 800 + 12 x 650 = 32400 + 24000 + 7800 = Rs 64200.

Q2

Step 1: After evaporation, the remaining volume is $40$ litres and the acid concentration is $75%$. Hence acid in the final mixture is [ 0.75 \times 40=30 \text{ litres}. ] Step 2: Evaporation removes only water, not acid. Therefore acid before evaporation was also $30$ litres. Step 3: Since $10$ litres of water evaporated and final volume is $40$ litres, the pre-evaporation mixed volume was [ 40+10=50 \text{ litres}. ] So [ a+b=50. ] Step 4: Acid balance before evaporation: [ 0.40a+0.70b=30. ] Step 5: Substitute $a=50-b$: [ 0.40(50-b)+0.70b=30, ] [ 20-0.40b+0.70b=30, ] [ 0.30b=10 \Rightarrow b=\frac{100}{3}. ] Then [ a=50-\frac{100}{3}=\frac{50}{3}. ] Step 6: Therefore [ a:b=\frac{50}{3}:\frac{100}{3}=1:2. ]

Q3

Step 1: Let the original value be 100. Step 2: After the three-stage change, the value is 100(1+p/100)(0.8)(1+p/100). Step 3: After only a single p% increase, the value would be 100(1+p/100). Step 4: The first value is 90% of the second, so 0.8(1+p/100)^2 = 0.9(1+p/100). Step 5: Since p is positive, divide by the common positive factor to get 0.8(1+p/100)=0.9, so p=12.5.

Q4

A's rate is 1/30 and B's rate is 1/20. In the first 6 days, A completes 6/30 = 1/5 of the job. B joins and they work together at rate 1/30 + 1/20 = 1/12 per day. B leaves when only 1/8 is left, so 7/8 of the job has been completed. The part done together is 7/8 - 1/5 = (35 - 8)/40 = 27/40. Time together = (27/40)/(1/12) = 81/10 = 8.1 days. The remaining 1/8 is done by A alone in (1/8)/(1/30) = 15/4 = 3.75 days. Total time = 6 + 8.1 + 3.75 = 17.85 days.

Q5

Step 1: Let incomes be 8x and 11x, and expenditures be 5y and 7y. Then P's saving is 8x - 5y and Q's saving is 11x - 7y. Step 2: Q saves Rs 3000 more, so 11x - 7y = 8x - 5y + 3000, giving 3x - 2y = 3000. Step 3: After investment, remaining savings are 80% of P's saving and 75% of Q's saving, so 0.8(8x - 5y) = 0.75(11x - 7y). Since Q's saving exceeds P's by 3000, this is also 0.8(SQ - 3000) = 0.75SQ. Hence SQ = 48000. Step 4: Therefore 11x - 7y = 48000 and 3x - 2y = 3000. From the second equation, y = 1.5x - 1500. Step 5: Substitute: 11x - 7(1.5x - 1500) = 48000, so 0.5x + 10500 = 48000, x = 75000. Q's income = 11x = Rs 825000.

Q6

Step 1: Let $\dfrac{AD}{AB} = k$. Since $DE \parallel BC$, triangle $ADE \sim ABC$ with linear ratio $k$, so $[ADE] = k^2 \cdot [ABC]$. Step 2: Since $DF \parallel AC$, triangle $DBF \sim ABC$ (same angle at $B$, with $DB/AB = 1-k$ and $BF/BC = 1-k$). Hence $[DBF] = (1-k)^2 \cdot [ABC]$. Step 3: Identify the hidden structure of $DECF$. Both $DE \parallel BC$ and $DF \parallel AC$. By the proportionality theorem, $DE = k \cdot BC$ and $FC = BC - BF = (1-(1-k)) \cdot BC = k \cdot BC$. So $DE = FC$ with $DE \parallel FC$, meaning $DECF$ is a \emph{parallelogram}, not just a trapezoid. Step 4: Since the three regions $[ADE]$, $[DECF]$, $[DBF]$ partition $[ABC]$: [[ABC] = [ADE] + [DBF] + [DECF] = k^2 T + (1-k)^2 T + 2k(1-k)T = T.] So $[DECF] = 2k(1-k),T$. Step 5: Now $[DECF]^2 = 4 k^2 (1-k)^2 T^2 = 4 \cdot [ADE] \cdot [DBF]$. Substituting: [40^2 = 4 \cdot 16 \cdot [DBF] ;\Rightarrow; [DBF] = \dfrac{1600}{64} = 25.]

Q7

Step 1: Convert the original average into the original total: [ 40\times 69.6=2784. ] Step 2: The total score of the lowest $7$ candidates is [ 7\times 48=336. ] Step 3: After replacing these $7$ candidates, the total score of the unchanged $33$ candidates is [ 2784-336=2448. ] Step 4: The new average of all $40$ candidates is $74.5$, so the new total score is [ 40\times 74.5=2980. ] Step 5: Therefore, the total score of the $7$ new candidates is [ 2980-2448=532. ] Step 6: Hence their average score is [ \frac{532}{7}=76. ]

Q8

Step 1: For both roots negative and \emph{distinct} real: \begin{itemize} \item Discriminant $> 0$ (strict, for distinct): $(k-3)^2 - 4(k-2) > 0$. \item Sum of roots negative: $-(k-3) < 0$, i.e.\ $k > 3$. \item Product of roots positive: $k - 2 > 0$, i.e.\ $k > 2$. \end{itemize} Step 2: Simplify discriminant: $(k-3)^2 - 4(k-2) = k^2 - 6k + 9 - 4k + 8 = k^2 - 10k + 17 > 0$. Step 3: Roots of $k^2 - 10k + 17 = 0$: $k = \dfrac{10 \pm \sqrt{100 - 68}}{2} = \dfrac{10 \pm \sqrt{32}}{2} = 5 \pm 2\sqrt{2}$. Approximately $5 + 2\sqrt{2} \approx 7.83$ and $5 - 2\sqrt{2} \approx 2.17$. Step 4: So $k^2 - 10k + 17 > 0$ iff $k < 5 - 2\sqrt{2}$ or $k > 5 + 2\sqrt{2}$. Step 5: Combine with $k > 3$ and $k > 2$: intersection is $k > 5 + 2\sqrt{2} \approx 7.83$ (the branch $k < 5 - 2\sqrt{2} \approx 2.17$ is killed by $k > 3$). Step 6: Integer $k$ with $7.83 < k \le 15$: $k \in {8, 9, 10, 11, 12, 13, 14, 15}$. Count $= 8$. Step 7: Verify endpoints. $k = 8$: $x^2 + 5x + 6 = (x+2)(x+3) = 0$, roots $-2, -3$, both negative and distinct (\checkmark). $k = 7$: $x^2 + 4x + 5 = 0$, discriminant $= 16 - 20 = -4 < 0$, complex roots (\times) (good).

Q9

Step 1: Let the number of employees in X be $4a$ where $a \ge 1$, and let Y exceed X by $3b$ where $b \ge 1$. Then $|Y| = 4a + 3b$. Step 2: The number of employees in Z is [ |Z| = \frac{|X| + |Y|}{2} = \frac{4a + 4a + 3b}{2} = \frac{8a + 3b}{2}. ] For $|Z|$ to be a positive integer, $8a + 3b$ must be even. Since $8a$ is always even, $3b$ must be even, which forces $b$ to be even. Write $b = 2k$ with $k \ge 1$. Step 3: With $b = 2k$, we get $|Z| = 4a + 3k$ and [ T = |X| + |Y| + |Z| = 4a + (4a + 6k) + (4a + 3k) = 12a + 9k = 3(4a + 3k). ] So $T$ is always a multiple of 3. Step 4: Impose $100 \le T \le 130$: [ 100 \le 3(4a + 3k) \le 130 \implies 34 \le 4a + 3k \le 43. ] Let $m = 4a + 3k$ where $a \ge 1$ and $k \ge 1$, so $m \ge 7$. For any integer $m \ge 7$, a representation $m = 4a + 3k$ with $a, k \ge 1$ exists: if $m \equiv 1 \pmod{3}$, take $a = 1$; if $m \equiv 2 \pmod{3}$, take $a = 2$; if $m \equiv 0 \pmod{3}$, take $a = 3$; and solve for $k$ accordingly. Each resulting $k$ is positive for $m \ge 7$. Step 5: The integers $m$ from 34 to 43 inclusive give $43 - 34 + 1 = 10$ values of $m$, hence 10 distinct totals $T \in {102, 105, 108, 111, 114, 117, 120, 123, 126, 129}$.

Q10

Step 1: Compute $f(n)$ step by step. $f(1) = 1$.\ $f(2) = 2 - f(f(1)) = 2 - f(1) = 1$.\ $f(3) = 3 - f(f(2)) = 3 - f(1) = 2$.\ $f(4) = 4 - f(f(3)) = 4 - f(2) = 3$.\ $f(5) = 5 - f(f(4)) = 5 - f(3) = 3$.\ $f(6) = 6 - f(f(5)) = 6 - f(3) = 4$.\ $f(7) = 7 - f(f(6)) = 7 - f(4) = 4$.\ $f(8) = 8 - f(f(7)) = 8 - f(4) = 5$. Step 2: Continue. $f(9) = 9 - f(f(8)) = 9 - f(5) = 6$.\ $f(10) = 10 - f(f(9)) = 10 - f(6) = 6$.\ $f(11) = 11 - f(f(10)) = 11 - f(6) = 7$.\ $f(12) = 12 - f(f(11)) = 12 - f(7) = 8$. Step 3: Continue. $f(13) = 13 - f(f(12)) = 13 - f(8) = 8$.\ $f(14) = 14 - f(f(13)) = 14 - f(8) = 9$.\ $f(15) = 15 - f(f(14)) = 15 - f(9) = 9$.\ $f(16) = 16 - f(f(15)) = 16 - f(9) = 10$. Step 4: Continue. $f(17) = 17 - f(f(16)) = 17 - f(10) = 11$.\ $f(18) = 18 - f(f(17)) = 18 - f(11) = 11$.\ $f(19) = 19 - f(f(18)) = 19 - f(11) = 12$.\ $f(20) = 20 - f(f(19)) = 20 - f(12) = 20 - 8 = 12$. Step 5: $f(20) = 12$.

Q11

Step 1: Set $p = \ln 2$ and $q = \ln 3$. Then [ a = \frac{\ln 18}{\ln 12} = \frac{p + 2q}{2p + q}, \qquad b = \frac{\ln 54}{\ln 24} = \frac{p + 3q}{3p + q}. ] Step 2: Compute $a - b$ over the common denominator $(2p+q)(3p+q)$: \begin{align*} a - b &= \frac{(p+2q)(3p+q) - (p+3q)(2p+q)}{(2p+q)(3p+q)}. \end{align*} Expanding the numerator: [ (3p^2 + 7pq + 2q^2) - (2p^2 + 7pq + 3q^2) = p^2 - q^2 = (p-q)(p+q). ] Step 3: Compute $ab$: [ ab = \frac{(p+2q)(p+3q)}{(2p+q)(3p+q)} = \frac{p^2 + 5pq + 6q^2}{6p^2 + 5pq + q^2}. ] Step 4: Compute $1 - ab$: [ 1 - ab = \frac{(6p^2 + 5pq + q^2) - (p^2 + 5pq + 6q^2)}{(2p+q)(3p+q)} = \frac{5p^2 - 5q^2}{(2p+q)(3p+q)} = \frac{5(p-q)(p+q)}{(2p+q)(3p+q)}. ] Step 5: Form the ratio: [ \frac{1 - ab}{a - b} = \frac{5(p-q)(p+q)}{(2p+q)(3p+q)} \cdot \frac{(2p+q)(3p+q)}{(p-q)(p+q)} = 5. ] Note that $p \neq q$ (since $\ln 2 \neq \ln 3$) and $p + q \neq 0$, so the cancellation is valid.

Q12

Step 1: In a parallelogram, the diagonals bisect each other. Hence the midpoint of $AC$ is [ \left(\frac{1+p}{2},\frac{2+q}{2}\right). ] Step 2: This point lies on $x+2y=16$, so [ \frac{1+p}{2}+2\left(\frac{2+q}{2}\right)=16. ] Step 3: Simplify: [ \frac{1+p}{2}+2+q=16. ] Multiplying by $2$, [ 1+p+4+2q=32, ] so [ p+2q=27. ] Step 4: The adjacent side vectors are [ \overrightarrow{AB}=(6,3),\qquad \overrightarrow{BC}=(p-7,q-5). ] The area of the parallelogram is the absolute value of their determinant: [ \left|6(q-5)-3(p-7)\right|=42. ] Step 5: Substitute $p=27-2q$: [ \left|6q-30-3(27-2q-7)\right|=42. ] This becomes [ \left|6q-30-3(20-2q)\right|=42, ] or [ |12q-90|=42. ] Step 6: Therefore, [ 12q-90=42 \quad \text{or} \quad 12q-90=-42. ] So [ q=11 \quad \text{or} \quad q=4. ] Step 7: From $p+2q=27$, these give [ (q,p)=(11,5) \quad \text{or} \quad (4,19). ] The condition $p>10$ selects [ p=19,\qquad q=4. ] Hence [ p+q=23. ]

Q13

Step 1: While crossing the platform, Train A covers its own length plus platform length = 250 + 150 = 400 m in 20 seconds. So speed of A = 20 m/s. Step 2: While overtaking Train B in the same direction, Train A must cover the sum of both train lengths relative to B: 250 + 180 = 430 m. Step 3: Relative speed = 430/45 = 86/9 m/s. Step 4: Speed of B = 20 - 86/9 = 94/9 m/s. Step 5: Convert to km/h: (94/9) x 18/5 = 188/5 = 37.6 km/h.

Q14

Step 1: Sum of the first $7$ terms is [ S_7=\frac{7}{2}(2a+6d)=7a+21d. ] Step 2: Sum of the $8$th to $12$th terms is [ (a+7d)+(a+8d)+(a+9d)+(a+10d)+(a+11d)=5a+45d. ] Step 3: Equating the two sums, [ 7a+21d=5a+45d \Rightarrow 2a=24d \Rightarrow a=12d. ] Step 4: The $10$th term is [ a+9d=12d+9d=21d. ] The condition $50<21d<80$ gives [ \frac{50}{21}<d<\frac{80}{21}. ] Step 5: Since $d$ is a positive integer, the only possible value is $d=3$. Then $a=36$. Hence the only ordered pair is $(36,3)$.

Q15

Step 1: The original cone has radius $10$ and height $24$, so its slant height is [ \sqrt{10^2+24^2}=26. ] Step 2: The curved surface area of a cone is proportional to $rl$. For similar cones, both $r$ and $l$ scale by the same linear factor, so curved surface area scales as the square of the linear factor. Step 3: Since the smaller cone has one-fourth the curved surface area of the original cone, its linear scale factor is [ \sqrt{\frac14}=\frac12. ] Step 4: Therefore the volume of the smaller cone is [ \left(\frac12\right)^3=\frac18 ] of the original cone's volume. Step 5: The original cone's volume is [ \frac13\pi(10)^2(24)=800\pi. ] So the frustum volume is [ 800\pi-\frac18(800\pi)=700\pi. ] Step 6: Each cylinder has volume [ \pi(5)^2(7)=175\pi. ] Therefore the number of cylinders is [ \frac{700\pi}{175\pi}=4. ]

Q16

Step 1: The given remainder conditions are [ n\equiv 3\pmod 5,\quad n\equiv 5\pmod 7,\quad n\equiv 7\pmod 9. ] Step 2: Each remainder is $2$ less than the corresponding divisor. Hence [ n+2 ] is divisible by $5$, $7$, and $9$. Step 3: Since [ \text{lcm}(5,7,9)=315, ] we can write [ n+2=315k, ] so [ n=315k-2. ] Step 4: Now impose divisibility by $11$: [ 315k-2\equiv 0\pmod {11}. ] Step 5: Since [ 315\equiv 7\pmod {11}, ] we get [ 7k-2\equiv 0\pmod {11}, ] or [ 7k\equiv 2\pmod {11}. ] Step 6: The inverse of $7$ modulo $11$ is $8$, because [ 7\cdot 8=56\equiv 1\pmod {11}. ] Therefore, [ k\equiv 2\cdot 8=16\equiv 5\pmod {11}. ] Step 7: The least positive value is $k=5$. Hence [ n=315(5)-2=1575-2=1573. ]

Q17

Step 1: Let the roots be a and 2a, where a is a positive integer. Step 2: Sum of roots = 3a = m + 5, and product = 2a^2 = 6m, so m = a^2/3. Step 3: Since m is an integer, a must be divisible by 3. Let a = 3t, then m = 3t^2. Step 4: Use the sum equation: 3a = m + 5 becomes 9t = 3t^2 + 5, or 3t^2 - 9t + 5 = 0. This has discriminant 81 - 60 = 21, not a perfect square, so no integer t works. Step 5: Therefore there are 0 integer values of m.

Q18

Step 1: Since $12.5% = \dfrac{1}{8}$, the two-year compound multiplier is [ \left(1+\frac{1}{8}\right)^2=\left(\frac{9}{8}\right)^2=\frac{81}{64}. ] Step 2: Amount from Scheme P after two years is [ 64000 \times \frac{81}{64}=81000. ] So the compound interest earned in Scheme P is [ 81000-64000=17000. ] Step 3: The simple interest earned in Scheme Q is Rs.\ $7{,}300$ more than this, hence [ SI_Q = 17000+7300=24300. ] Step 4: In Scheme Q, the principal is not Rs.\ $64{,}000$; it is the matured amount Rs.\ $81{,}000$. Thus [ 24300 = \frac{81000 \cdot r \cdot 3}{100}. ] Step 5: Solving, [ r=\frac{24300 \cdot 100}{81000 \cdot 3}=10. ] Hence the required rate is $10%$ per annum.

Q19

Step 1: Let the number be 100a+10b+c. Step 2: Reversed-original equals 99(c-a). Step 3: Given 198, c-a=2. Step 4: a can be 1 through 7. Step 5: For each, b has 7 non-zero choices excluding a and c, total 49.

Q20

Total cost price = 100 x 600 = Rs 60000. Let the common marked price be M. Revenue from the three groups is 50 x 0.90M + 30 x 0.80M + 20 x 0.65M = 45M + 24M + 13M = 82M. Overall profit is 9 1/3% = 28/3%, so required total selling price = 60000 x (1 + 7/75) = 60000 x 82/75 = Rs 65600. Hence 82M = 65600, so M = Rs 800. Verification: revenue = 50 x 720 + 30 x 640 + 20 x 520 = 36000 + 19200 + 10400 = Rs 65600.

Q21

Step 1: Let January total revenue be 100, so X revenue is 40 and other revenue is 60. Step 2: February total revenue is 125 after the 25% increase. Step 3: February X revenue is 90% of 40, which is 36. Step 4: Therefore February revenue from other products is 125 - 36 = 89. Step 5: The increase in other-products revenue is 29 on a base of 60, so the percentage increase is 2900/60 = 145/3%.

Q22

Step 1: Let total candidates $= N$. Clear A $= 0.80 N$. Clear A and B $= 0.75 \times 0.80 N = 0.60 N$. Step 2: Clear all three $= 0.40 \times 0.60 N = 0.24 N$. Hence candidates who do not clear all three $= N - 0.24N = 0.76N$. Step 3: Failed only C means cleared A and B but failed C, so this group is $0.60N - 0.24N = 0.36N$. Step 4: The given fraction is consistent because $0.36N/0.76N = 36/76 = 9/19$. Step 5: Given $0.36N = 720$, we get $N = 720/0.36 = 2000$.

Evidence Lineage