CATin
  1. CATin
  2. CAT past papers
  3. CAT 2022 Slot 2
  4. QA

CAT 2022 Slot 2 — QA questions with answers

All 22 questions of the Quantitative Ability section (14 MCQs, 8 TITA). Try each one, then open its answer and solution.

CAT 2022 Slot 2, timed like the real exam (120 minutes, 40 a section) and scored the CAT way: +3 right, −1 for a wrong MCQ, 0 for a wrong TITA, with the full post-mock analysis. Older official papers are in CATin Pro, with 80+ mocks and the full planner. The last three years of papers are free.

Sit this paper as a timed mock with CATin Pro

Quantitative Ability

CAT 2022 Slot 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45TITASimple & Compound Interest

Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 20

Take the capital as 15x15x, so that each part is a whole multiple of xx. One-fifth, 3x3x, earns 6%; one-third, 5x5x, earns 10%; the remaining 15x−3x−5x=7x15x - 3x - 5x = 7x earns 1%. Simple interest each year: 6100(3x)+10100(5x)+1100(7x)=18x+50x+7x100=0.75x\frac{6}{100}(3x) + \frac{10}{100}(5x) + \frac{1}{100}(7x) = \frac{18x + 50x + 7x}{100} = 0.75x. After TT years the interest is 0.75xT0.75xT. We need 0.75xT≥15x0.75xT \ge 15x, so T≥150.75=20T \ge \frac{15}{0.75} = 20. After 19 years the interest is only 14.25x14.25x, less than the capital, so at least 20 years are needed. The answer is 20.

Q46TITAAverages, Mixtures & Alligations

The average of a non-decreasing sequence of NN numbers a1,a2,…,aNa_1, a_2, \ldots, a_N is 300. If a1a_1 is replaced by 6a16a_1, the new average becomes 400. Then, the number of possible values of a1a_1 is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 14

The sum of the NN numbers is 300N300N. Replacing a1a_1 by 6a16a_1 adds 5a15a_1 to the sum, and the new sum is 400N400N. So 5a1=100N5a_1 = 100N, which gives a1=20Na_1 = 20N. Upper limit: the sequence is non-decreasing, so a1a_1 is the smallest number and cannot exceed the average. Hence 20N≤30020N \le 300, so N≤15N \le 15. (At N=15N = 15, a1=300a_1 = 300 and every term is 300.) Lower limit: if N=1N = 1, the only number equals the average, so a1=300a_1 = 300, but a1=20Na_1 = 20N gives 20. So N≥2N \ge 2. Every NN from 2 to 15 works: take a1=20Na_1 = 20N and each of the other N−1N-1 numbers equal to 280NN−1\frac{280N}{N-1}, which is at least 20N20N exactly when N≤15N \le 15. So a1a_1 can be 40,60,…,30040, 60, \ldots, 300, which is 14 values. The answer is 14.

Q47TITAIndices & Surds

The number of integer solutions of the equation (x2−10)(x2−3x−10)=1\left(x^2 - 10\right)^{\left(x^2 - 3x - 10\right)} = 1 is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

A power aba^b equals 1 in three cases: the exponent is 0 (with a≠0a \ne 0), the base is 1, or the base is −1-1 with an even exponent. Case 1, exponent 0: x2−3x−10=0x^2 - 3x - 10 = 0 gives (x−5)(x+2)=0(x-5)(x+2) = 0, so x=5x = 5 or x=−2x = -2. The bases are 25−10=1525 - 10 = 15 and 4−10=−64 - 10 = -6, both non-zero, so both work. Case 2, base 1: x2−10=1x^2 - 10 = 1 gives x2=11x^2 = 11, which has no integer solution. Case 3, base −1-1: x2−10=−1x^2 - 10 = -1 gives x2=9x^2 = 9, so x=3x = 3 or x=−3x = -3. For x=3x = 3 the exponent is 9−9−10=−109 - 9 - 10 = -10, and (−1)−10=1(-1)^{-10} = 1. For x=−3x = -3 the exponent is 9+9−10=89 + 9 - 10 = 8, and (−1)8=1(-1)^{8} = 1. Both work, so the integer solutions are −3,−2,3,5-3, -2, 3, 5: four in all. The answer is 4.

Q48MCQAverages, Mixtures & Alligations

Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be
  1. 4400
  2. 4200
  3. 4300
  4. 4350
Answer and solution

Answer: (A) 4400

Yearly savings target: 550×12=6600550 \times 12 = 6600 rupees. Savings in the first nine months: 9×(4000−3500)=9×500=45009 \times (4000 - 3500) = 9 \times 500 = 4500 rupees. So the last three months must save 6600−4500=21006600 - 4500 = 2100 rupees, i.e. 21003=700\frac{2100}{3} = 700 rupees a month. From the tenth month his expense is ₹3700 a month, so his monthly earnings must be 3700+700=44003700 + 700 = 4400 rupees. The nearest option, ₹4350, would save only ₹650 a month, or ₹1950 over three months, which is ₹150 short of the target. Hence, option A (4400).

Q49MCQTriangles & Lines

In triangle ABCABC, altitudes ADAD and BEBE are drawn to the corresponding bases. If ∠BAC=45°\angle BAC = 45° and ∠ABC=θ\angle ABC = \theta, then ADBE\dfrac{AD}{BE} equals
  1. 2cos⁡θ\sqrt{2}\cos\theta
  2. (sin⁡θ+cos⁡θ)2\dfrac{\left({\sin\theta + \cos\theta}\right)}{\sqrt{2}}
  3. 1
  4. 2sin⁡θ\sqrt{2}\sin\theta
Answer and solution

Answer: (D) 2sin⁡θ\sqrt{2}\sin\theta

Draw the triangle with both altitudes: Solution figure for question 49, CAT 2022 Slot 2 Since BE⊥ACBE \perp AC, triangle ABEABE is right-angled at EE, and ∠BAE=∠BAC=45°\angle BAE = \angle BAC = 45°. So it is an isosceles right triangle: let AE=BE=xAE = BE = x; then AB=x2AB = x\sqrt{2}. Since AD⊥BCAD \perp BC, triangle ABDABD is right-angled at DD, with ∠ABD=θ\angle ABD = \theta. The side ADAD is opposite θ\theta, so AD=ABsin⁡θ=x2sin⁡θAD = AB\sin\theta = x\sqrt{2}\sin\theta. Therefore ADBE=x2sin⁡θx=2sin⁡θ\dfrac{AD}{BE} = \dfrac{x\sqrt{2}\sin\theta}{x} = \sqrt{2}\sin\theta. Option A, 2cos⁡θ\sqrt{2}\cos\theta, would come from using the side next to θ\theta, BD=ABcos⁡θBD = AB\cos\theta, instead of the opposite side ADAD. Hence, option D (2sin⁡θ\sqrt{2}\sin\theta).

Q50MCQFunctions & Graphs

Let f(x)f(x) be a quadratic polynomial in xx such that f(x)≥0f(x) \geq 0 for all real numbers xx. If f(2)=0f(2) = 0 and f(4)=6f(4) = 6, then f(−2)f(-2) is equal to
  1. 12
  2. 24
  3. 6
  4. 36
Answer and solution

Answer: (B) 24

Since f(x)≥0f(x) \ge 0 for every real xx and f(2)=0f(2) = 0, the graph touches the x-axis at x=2x = 2 without crossing it. If 2 were a simple root, ff would change sign there and take negative values. So 2 is a repeated root, and f(x)=a(x−2)2f(x) = a(x-2)^2 with a>0a > 0. From f(4)=6f(4) = 6: a(4−2)2=4a=6a(4-2)^2 = 4a = 6, so a=32a = \frac{3}{2}. Then f(−2)=32(−2−2)2=32×16=24f(-2) = \frac{3}{2}(-2-2)^2 = \frac{3}{2} \times 16 = 24. Option A, 12, doubles f(4)f(4) because −2-2 is twice as far from 2 as 4 is. But ff grows with the square of that distance, so the factor is 22=42^2 = 4, giving 4×6=244 \times 6 = 24. Hence, option B (24).

Q51MCQQuadratic & Polynomial Equations

Let rr and cc be real numbers. If rr and −r-r are roots of 5x3+cx2−10x+9=05x^3 + cx^2 - 10x + 9 = 0, then cc equals
  1. −92-\dfrac{9}{2}
  2. 92\dfrac{9}{2}
  3. −4-4
  4. 44
Answer and solution

Answer: (A) −92-\dfrac{9}{2}

Since rr and −r-r are both roots, substitute each: 5r3+cr2−10r+9=05r^3 + cr^2 - 10r + 9 = 0 ...(1) −5r3+cr2+10r+9=0-5r^3 + cr^2 + 10r + 9 = 0 ...(2) Adding (1) and (2): 2cr2+18=02cr^2 + 18 = 0, so cr2=−9cr^2 = -9. Subtracting (2) from (1): 10r3−20r=010r^3 - 20r = 0, so 10r(r2−2)=010r(r^2 - 2) = 0. If r=0r = 0, equation (1) would give 9=09 = 0, which is false. So r2=2r^2 = 2. Then 2c=−92c = -9, so c=−92c = -\frac{9}{2}. Check with the roots: the third root is p=−c5=910p = -\frac{c}{5} = \frac{9}{10}, since rr and −r-r cancel in the sum, and the product r(−r)p=−2×910=−95r(-r)p = -2 \times \frac{9}{10} = -\frac{9}{5}, as required. Option B, 92\frac{9}{2}, has the wrong sign: it would make cr2=9cr^2 = 9, not −9-9. Hence, option A (−92-\dfrac{9}{2}).

Q52MCQTime, Speed & Distance

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is
  1. 20
  2. 12
  3. 18
  4. 24
Answer and solution

Answer: (C) 18

The ships move south and west from the meeting point, at right angles, so their paths and the line joining them form a right triangle. Solution figure for question 52, CAT 2022 Slot 2 Let the slower speed be xx km/h and the faster x+6x + 6 km/h. In 2 hours they cover 2x2x and 2x+122x + 12 km. By Pythagoras: (2x)2+(2x+12)2=602(2x)^2 + (2x+12)^2 = 60^2. Dividing by 4: x2+(x+6)2=900x^2 + (x+6)^2 = 900, so 2x2+12x+36=9002x^2 + 12x + 36 = 900, i.e. x2+6x−432=0x^2 + 6x - 432 = 0. This factors as (x+24)(x−18)=0(x + 24)(x - 18) = 0. Speed is positive, so x=18x = 18. Check: 362+482=1296+2304=3600=60236^2 + 48^2 = 1296 + 2304 = 3600 = 60^2. Option D, 24, is the faster ship's speed, 18+618 + 6, not the slower one's. Hence, option C (18).

Q53TITAFunctions & Graphs

Suppose for all integers xx, there are two functions ff and gg such that f(x)+f(x−1)−1=0f(x) + f(x-1) - 1 = 0 and g(x)=x2g(x) = x^2. If f(x2−x)=5f(x^2 - x) = 5, then the value of the sum f(g(5))+g(f(5))f(g(5)) + g(f(5)) is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 12

The rule f(x)+f(x−1)−1=0f(x) + f(x-1) - 1 = 0 gives f(x)=1−f(x−1)f(x) = 1 - f(x-1) for every integer xx. Put x=1x = 1 in f(x2−x)=5f(x^2 - x) = 5: f(0)=5f(0) = 5. Then f(1)=1−5=−4f(1) = 1 - 5 = -4, f(2)=1−(−4)=5f(2) = 1 - (-4) = 5, f(3)=−4f(3) = -4, and so on. The values alternate: f(n)=5f(n) = 5 for even nn and f(n)=−4f(n) = -4 for odd nn. This agrees with f(x2−x)=5f(x^2 - x) = 5, since x2−x=x(x−1)x^2 - x = x(x-1) is always even. Now g(5)=25g(5) = 25, which is odd, so f(g(5))=f(25)=−4f(g(5)) = f(25) = -4. Also f(5)=−4f(5) = -4, since 5 is odd, so g(f(5))=(−4)2=16g(f(5)) = (-4)^2 = 16. Sum: −4+16=12-4 + 16 = 12. The answer is 12.

Q54TITALinear Equations

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 24

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z. It is given, x + y + z = 75 ...... (1) 3x - y + z = 97 ...... (2) (2)-(1) -> x - y = 11 (1)+(2) -> 2x + z = 86 z > x + y z > 75 - z z > 37.5 Minimum possible value of z is 38 2x + 38 = 86 2x = 48 x = 24 The maximum number of correct answers is 24.

Q55TITAPolygons & Circles

Regular polygons AA and BB have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of BB equals

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 10

Let AA have nn sides and BB have 2n2n sides. Each interior angle of a regular polygon with kk sides is (k−2)×180°k\frac{(k-2) \times 180°}{k}. So the angle of AA is (n−2)180°n\frac{(n-2)180°}{n}, and the angle of BB is (2n−2)180°2n=(n−1)180°n\frac{(2n-2)180°}{2n} = \frac{(n-1)180°}{n}. Their ratio is n−2n−1=34\frac{n-2}{n-1} = \frac{3}{4}. Cross-multiplying: 4n−8=3n−34n - 8 = 3n - 3, so n=5n = 5. Check: a regular pentagon's angle is 108°108° and a regular decagon's is 144°144°, and 108:144=3:4108 : 144 = 3 : 4. So BB has 2×5=102 \times 5 = 10 sides. The answer is 10.

Q56MCQRatios, Proportions & Partnership

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was
  1. 50240
  2. 40192
  3. 60288
  4. 62800
Answer and solution

Answer: (D) 62800

Let there be 100x100x registered voters. Votes cast: 80x80x. One candidate gets 30% of the votes cast: 0.3×80x=24x0.3 \times 80x = 24x. The remaining 80x−24x=56x80x - 24x = 56x votes are shared in the ratio 1 : 2 : 3, i.e. 56x6\frac{56x}{6}, 112x6\frac{112x}{6} and 168x6=28x\frac{168x}{6} = 28x. The first two are about 9.3x9.3x and 18.7x18.7x. So the winner has 28x28x votes and the runner-up has 24x24x. 28x−24x=4x=251228x - 24x = 4x = 2512, so x=628x = 628. Registered voters: 100x=62800100x = 62800. Option A, 50240, is the number of votes cast, 80x80x, not the number of registered voters. Hence, option D (62800).

Q57MCQSequences & Series

On day one, there are 100 particles in a laboratory experiment. On day nn, where n≥2n \geq 2, one out of every nn particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day mm, then mm equals
  1. 19
  2. 16
  3. 18
  4. 17
Answer and solution

Answer: (A) 19

On day nn, one particle in every nn produces a new one, so the count is multiplied by 1+1n=n+1n1 + \frac{1}{n} = \frac{n+1}{n}. Day 1: 100. Day 2: 100×32=150100 \times \frac{3}{2} = 150 (50 new). Day 3: 150×43=200150 \times \frac{4}{3} = 200 (50 new). Day 4: 200×54=250200 \times \frac{5}{4} = 250 (50 new). The pattern holds every day: if the count on day n−1n-1 is 50n50n, then on day nn one in nn of them produces, adding 50nn=50\frac{50n}{n} = 50. So the count on day mm is 100+50(m−1)=50(m+1)100 + 50(m-1) = 50(m+1). Setting 50(m+1)=100050(m+1) = 1000 gives m+1=20m + 1 = 20, so m=19m = 19. Option C, 18, is one day short: on day 18 there are only 50×19=95050 \times 19 = 950 particles. Hence, option A (19).

Q58MCQPermutations & Combinations

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is
  1. 1440
  2. 1200
  3. 1480
  4. 1420
Answer and solution

Answer: (A) 1440

Numbers with 1, 2 or 3 digits are below 2000, so only 4-, 5- and 6-digit numbers count. No digit repeats, and the first digit cannot be 0. 4-digit numbers: to exceed 2000 the first digit must be 2, 3, 4 or 5 (4 choices). The other three places are filled from the remaining 5 digits in 5×4×3=605 \times 4 \times 3 = 60 ways. Total 4×60=2404 \times 60 = 240. (2000 itself cannot be formed, as it repeats 0.) 5-digit numbers: the first digit is 1 to 5 (5 choices), and the rest can be filled in 5×4×3×2=1205 \times 4 \times 3 \times 2 = 120 ways. Total 5×120=6005 \times 120 = 600. 6-digit numbers: 5 choices for the first digit, then 5!=1205! = 120 ways for the rest. Total 600. Every 5- and 6-digit number exceeds 2000. Sum: 240+600+600=1440240 + 600 + 600 = 1440. Option B, 1200, counts only the 5- and 6-digit numbers and misses the 240 four-digit ones. Hence, option A (1440).

Q59MCQProperties of Numbers

For some natural number nn, assume that (15000)!(15000)! is divisible by (n!)!(n!)!. The largest possible value of nn is
  1. 4
  2. 7
  3. 6
  4. 5
Answer and solution

Answer: (B) 7

To find the largest possible value of n, we need to find the value of n such that n! is less than 15000. 7! = 5040 8! = 40320 > 15000 This implies 15000! is not divisible by 40320! Therefore, maximum value n can take is 7. The answer is option B.

Q60TITATime & Work

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 6

The times are in the ratio 5 : 8 : 10, so the work rates are in the ratio 15:18:110=8:5:4\frac{1}{5} : \frac{1}{8} : \frac{1}{10} = 8 : 5 : 4. Let Anu, Tanu and Manu do 8k8k, 5k5k and 4k4k units of work per hour. Together they finish in 4 days of 8 hours, i.e. 32 hours. So the job is 32×(8k+5k+4k)=32×17k=544k32 \times (8k + 5k + 4k) = 32 \times 17k = 544k units. Anu and Tanu work for 6 days at 6 hours 40 minutes a day: 6×623=406 \times 6\frac{2}{3} = 40 hours. They do 40×(8k+5k)=520k40 \times (8k + 5k) = 520k units. Remaining work: 544k−520k=24k544k - 520k = 24k units. Manu does 4k4k units an hour, so he needs 24k4k=6\frac{24k}{4k} = 6 hours. The answer is 6.

Q61MCQAverages, Mixtures & Alligations

There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is
  1. 4 : 5
  2. 6 : 5
  3. 5 : 4
  4. 5 : 6
Answer and solution

Answer: (D) 5 : 6

Take each container to hold 200 L, so the first starts with 100 L of sugar syrup and the second with 100 L of milk. Step 1: half of the first container (50 L of syrup) goes to the second. First: 50 L syrup. Second: 50 L syrup and 100 L milk. Step 2: half of the second container (75 L) goes back. It carries 25 L syrup and 50 L milk. First: 75 L syrup and 50 L milk. Second: 25 L syrup and 50 L milk. Step 3: half of the first container (62.5 L) goes to the second. It carries 37.5 L syrup and 25 L milk. Second: 25+37.5=62.525 + 37.5 = 62.5 L syrup and 50+25=7550 + 25 = 75 L milk. Solution figure for question 61, CAT 2022 Slot 2 Syrup : milk in the second container =62.5:75=5:6= 62.5 : 75 = 5 : 6. Option B (6 : 5) is the same ratio turned round: it is milk to syrup, not syrup to milk. Hence, option D (5 : 6).

Q62MCQSequences & Series

Consider the arithmetic progression 3, 7, 11, ... and let AnA_n denote the sum of the first nn terms of this progression. Then the value of 125∑n=125An\dfrac{1}{25}\sum_{n=1}^{25} A_n is
  1. 455
  2. 442
  3. 415
  4. 404
Answer and solution

Answer: (A) 455

The progression has first term a=3a = 3 and common difference d=4d = 4, so An=n2[2(3)+(n−1)4]=n2(4n+2)=n(2n+1)=2n2+nA_n = \frac{n}{2}\left[2(3) + (n-1)4\right] = \frac{n}{2}(4n + 2) = n(2n + 1) = 2n^2 + n. Add these for n=1n = 1 to 2525, using ∑n2=n(n+1)(2n+1)6\sum n^2 = \frac{n(n+1)(2n+1)}{6} and ∑n=n(n+1)2\sum n = \frac{n(n+1)}{2}: ∑n=125An=2⋅25⋅26⋅516+25⋅262=11050+325=11375\sum_{n=1}^{25} A_n = 2 \cdot \frac{25 \cdot 26 \cdot 51}{6} + \frac{25 \cdot 26}{2} = 11050 + 325 = 11375. Divide by 25: 125∑n=125An=1137525=442+13=455\frac{1}{25}\sum_{n=1}^{25} A_n = \frac{11375}{25} = 442 + 13 = 455. Option B (442) is the trap. It is only the 2∑n22\sum n^2 part divided by 25, 1105025\frac{11050}{25}; it leaves out the 32525=13\frac{325}{25} = 13 that comes from ∑n\sum n. Hence, option A (455).

Q63TITALogarithms

The number of distinct integer values of nn satisfying 4−log⁡2n3−log⁡4n<0\dfrac{4 - \log_2 n}{3 - \log_4 n} < 0, is

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 47

The logarithms need n>0n > 0. Let y=log⁡2ny = \log_2 n. Then log⁡4n=log⁡2nlog⁡24=y2\log_4 n = \frac{\log_2 n}{\log_2 4} = \frac{y}{2}, and the inequality becomes 4−y3−y2<0\frac{4 - y}{3 - \frac{y}{2}} < 0, that is, 2(4−y)6−y<0\frac{2(4 - y)}{6 - y} < 0. A fraction is negative when its top and bottom have opposite signs. For y<4y < 4 both 4−y4 - y and 6−y6 - y are positive, and for y>6y > 6 both are negative. So the fraction is negative only when 4<y<64 < y < 6 (at y=6y = 6 the fraction is undefined, and at y=4y = 4 it is zero). So 4<log⁡2n<64 < \log_2 n < 6, which gives 24<n<262^4 < n < 2^6, that is, 16<n<6416 < n < 64. The integers in this range are 17,18,…,6317, 18, \dots, 63, which is 63−17+1=4763 - 17 + 1 = 47 values. The answer is 47.

Q64MCQLinear Equations

If aa and bb are non-negative real numbers such that a+2b=6a + 2b = 6, then the average of the maximum and minimum possible values of (a+b)(a + b) is
  1. 3
  2. 4
  3. 3.5
  4. 4.5
Answer and solution

Answer: (D) 4.5

From a+2b=6a + 2b = 6 with a,b≥0a, b \ge 0, bb can run from 00 (when a=6a = 6) up to 33 (when a=0a = 0). Write a+b=(a+2b)−b=6−ba + b = (a + 2b) - b = 6 - b. So a+ba + b gets smaller as bb gets larger. Maximum: at b=0b = 0, a=6a = 6 and a+b=6a + b = 6. Minimum: at b=3b = 3, a=0a = 0 and a+b=3a + b = 3. Average of the maximum and minimum =6+32=4.5= \frac{6 + 3}{2} = 4.5. Option A (3) is only the minimum value of a+ba + b, not the average of the two extremes. Hence, option D (4.5).

Q65MCQAverages, Mixtures & Alligations

Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is
  1. 22
  2. 21
  3. 24
  4. 20
Answer and solution

Answer: (D) 20

The five marks add up to 5×38=1905 \times 38 = 190, and no two are equal. Exactly three students scored more than 32, so the other two, Amit and one more student, scored 32 or less. Amit's lowest mark: make the other four as high as possible. The three above 32 can be at most 50, 49 and 48, and the fourth student at most 32. They total 179179, so Amit scores at least 190−179=11190 - 179 = 11. Amit's highest mark: he has the least marks, so he must be below the other student who scored 32 or less. That student has at most 32, so Amit has at most 31. This works: with Amit on 31 and the fourth student on 32, the top three need 190−63=127190 - 63 = 127, for example 50, 44 and 33. Difference =31−11=20= 31 - 11 = 20. Option B (21) comes from taking Amit's highest mark as 32, which would leave no mark for the fourth student between Amit and 33. Hence, option D (20).

Q66MCQTriangles & Lines

The length of each side of an equilateral triangle ABCABC is 3 cm. Let DD be a point on BCBC such that the area of triangle ADCADC is half the area of triangle ABDABD. Then the length of ADAD, in cm, is
  1. 8\sqrt{8}
  2. 6\sqrt{6}
  3. 7\sqrt{7}
  4. 5\sqrt{5}
Answer and solution

Answer: (C) 7\sqrt{7}

Triangles ABD and ADC have the same height from A to BC, so their areas are in the ratio of their bases, BD:DCBD : DC. Area ADC is half of area ABD, so BD=2 DCBD = 2\,DC. Since BD+DC=3BD + DC = 3, we get BD=2BD = 2 and DC=1DC = 1. Solution figure for question 66, CAT 2022 Slot 2 In triangle ADC, AC=3AC = 3, DC=1DC = 1 and ∠ACD=60∘\angle ACD = 60^\circ. By the cosine rule, AD2=AC2+DC2−2(AC)(DC)cos⁡60∘=9+1−2(3)(1)(12)=7AD^2 = AC^2 + DC^2 - 2(AC)(DC)\cos 60^\circ = 9 + 1 - 2(3)(1)\left(\frac{1}{2}\right) = 7, so AD=7AD = \sqrt{7}. Options B (6\sqrt{6}) and D (5\sqrt{5}) are impossible for any point D on BC: they are shorter than the altitude 332=6.75\frac{3\sqrt{3}}{2} = \sqrt{6.75}, the shortest distance from A to BC. Option A (8\sqrt{8}) would need D about 0.38 cm from an end of BC, not 1 cm from C. Hence, option C (7\sqrt{7}).