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CAT 2022 Slot 3 — 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 3, 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.

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Quantitative Ability

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

Q45MCQInequalities & Modulus

If c=16xy+49yxc = \dfrac{16x}{y} + \dfrac{49y}{x} for some non-zero real numbers xx and yy, then cc cannot take the value
  1. 60
  2. -50
  3. -70
  4. -60
Answer and solution

Answer: (B) -50

Let t=xyt = \frac{x}{y}, which can be any non-zero real number. Then c=16t+49tc = 16t + \frac{49}{t}. If t>0t > 0, by AM–GM: 16t+49t≥216t⋅49t=2×28=5616t + \frac{49}{t} \ge 2\sqrt{16t \cdot \frac{49}{t}} = 2 \times 28 = 56. If t<0t < 0, write t=−ut = -u with u>0u > 0: c=−(16u+49u)≤−56c = -\left(16u + \frac{49}{u}\right) \le -56. Both bounds are reached, at t=±74t = \pm\frac{7}{4}, and every value beyond them is possible too. So cc can take any value in (−∞,−56]∪[56,∞)(-\infty, -56] \cup [56, \infty), but nothing strictly between −56-56 and 5656. Of the options, 60, −60-60 and −70-70 lie in this range, while −50-50 does not. Hence, option B (−50-50).

Q46MCQQuadratic & Polynomial Equations

Suppose kk is any integer such that the equation 2x2+kx+5=02x^2 + kx + 5 = 0 has no real roots and the equation x2+(k−5)x+1=0x^2 + (k - 5)x + 1 = 0 has two distinct real roots for xx. Then, the number of possible values of kk is
  1. 9
  2. 7
  3. 8
  4. 13
Answer and solution

Answer: (A) 9

For 2x2+kx+5=02x^2 + kx + 5 = 0 to have no real roots, its discriminant must be negative: k2−40<0k^2 - 40 < 0, so −40<k<40-\sqrt{40} < k < \sqrt{40}. As 40≈6.32\sqrt{40} \approx 6.32, the integer values are −6,−5,…,6-6, -5, \dots, 6. For x2+(k−5)x+1=0x^2 + (k - 5)x + 1 = 0 to have two distinct real roots, its discriminant must be positive: (k−5)2−4>0(k - 5)^2 - 4 > 0, that is, k2−10k+21>0k^2 - 10k + 21 > 0, or (k−3)(k−7)>0(k - 3)(k - 7) > 0. So k<3k < 3 or k>7k > 7. Both conditions hold for k=−6,−5,−4,−3,−2,−1,0,1,2k = -6, -5, -4, -3, -2, -1, 0, 1, 2, which is 9 values. Option D (13) counts every integer from −6-6 to 66 and ignores the second condition, which rules out 3, 4, 5 and 6. Hence, option A (9).

Q47TITAIndices & Surds

If (75)3x−y=8752401\left(\sqrt\dfrac{7}{5}\right)^{3x - y} = \dfrac{875}{2401} and (4ab)6x−y=(2ab)y−6x\left(\dfrac{4a}{b}\right)^{6x - y} = \left(\dfrac{2a}{b}\right)^{y - 6x}, for all non-zero real values of aa and bb, then the value of x+yx + y is

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

Answer and solution

Answer: 14

First equation. The left side is (75)3x−y=(75)3x−y2\left(\sqrt{\dfrac{7}{5}}\right)^{3x - y} = \left(\dfrac{7}{5}\right)^{\frac{3x - y}{2}}. The right side is 8752401=7×1257×343=125343=(57)3=(75)−3\dfrac{875}{2401} = \dfrac{7 \times 125}{7 \times 343} = \dfrac{125}{343} = \left(\dfrac{5}{7}\right)^{3} = \left(\dfrac{7}{5}\right)^{-3}. So 3x−y2=−3\dfrac{3x - y}{2} = -3, giving 3x−y=−63x - y = -6. Second equation. Let n=6x−yn = 6x - y. Then (4ab)n=(2ab)−n\left(\dfrac{4a}{b}\right)^{n} = \left(\dfrac{2a}{b}\right)^{-n}, so (4ab⋅2ab)n=(8a2b2)n=1\left(\dfrac{4a}{b} \cdot \dfrac{2a}{b}\right)^{n} = \left(\dfrac{8a^2}{b^2}\right)^{n} = 1. This must hold for every non-zero aa and bb, which is possible only if n=0n = 0. So y=6xy = 6x. Substituting, 3x−6x=−63x - 6x = -6, so x=2x = 2 and y=12y = 12. Therefore x+y=2+12=14x + y = 2 + 12 = 14. The answer is 14.

Q48MCQAverages, Mixtures & Alligations

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is
  1. 23
  2. 24
  3. 23.5
  4. 22.5
Answer and solution

Answer: (D) 22.5

Let the six numbers in increasing order be a<b<c<d<e<fa < b < c < d < e < f. Then a+b=2×14=28a + b = 2 \times 14 = 28 and e+f=2×28=56e + f = 2 \times 28 = 56. The total is 28+c+d+5628 + c + d + 56, so cc and dd should be as large as possible. Both are less than ee, so ee should be as large as possible too. Since e<fe < f and e+f=56e + f = 56, ee cannot be 28 (then ff would also be 28). The largest choice is e=27e = 27, f=29f = 29. Then cc and dd are distinct and below 27, so at best d=26d = 26 and c=25c = 25. The two smallest can be, say, 13 and 15, which stay below 25. Maximum total =28+25+26+56=135= 28 + 25 + 26 + 56 = 135, so the maximum average is 1356=22.5\dfrac{135}{6} = 22.5. Option A (23) would need a total of 138, that is c+d=54c + d = 54, which is impossible when c+d≤51c + d \le 51. Hence, option D (22.5).

Q49TITATriangles & Lines

Suppose the medians BDBD and CECE of a triangle ABCABC intersect at a point OO. If area of triangle ABCABC is 108 sq. cm., then, the area of the triangle EODEOD, in sq. cm., is

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

Answer and solution

Answer: 9

Area of ABD : Area of BDC = 1:1 Therefore, area of ABD = 54 Area of ADE : Area of EDB = 1:1 Therefore, area of ADE = 27 O is the centroid and it divides the medians in the ratio of 2:1 Area of BEO : Area of EOD = 2:1 Area of EOD = 9
Solution figure for question 49, CAT 2022 Slot 3

Q50MCQQuadratic & Polynomial Equations

If (3+22)(3 + 2\sqrt{2}) is a root of the equation ax2+bx+c=0ax^2 + bx + c = 0 and (4+23)(4 + 2\sqrt{3}) is a root of the equation ay2+my+n=0ay^2 + my + n = 0 where aa, bb, cc, mm and nn are integers, then the value of (bm+c−2bn)\left(\dfrac{b}{m} + \dfrac{c - 2b}{n}\right) is
  1. 0
  2. 1
  3. 3
  4. 4
Answer and solution

Answer: (D) 4

The coefficients are integers, so the irrational roots come in conjugate pairs. First equation: the roots are 3+223 + 2\sqrt{2} and 3−223 - 2\sqrt{2}. Their sum is 6 and their product is 9−8=19 - 8 = 1. So −ba=6-\dfrac{b}{a} = 6 and ca=1\dfrac{c}{a} = 1, giving b=−6ab = -6a and c=ac = a. Second equation: the roots are 4+234 + 2\sqrt{3} and 4−234 - 2\sqrt{3}. Their sum is 8 and their product is 16−12=416 - 12 = 4. So m=−8am = -8a and n=4an = 4a. Now bm=−6a−8a=34\dfrac{b}{m} = \dfrac{-6a}{-8a} = \dfrac{3}{4} and c−2bn=a+12a4a=134\dfrac{c - 2b}{n} = \dfrac{a + 12a}{4a} = \dfrac{13}{4}. Adding, 34+134=164=4\dfrac{3}{4} + \dfrac{13}{4} = \dfrac{16}{4} = 4. Option B (1) is what you get if the −2b-2b term is dropped: 34+14=1\dfrac{3}{4} + \dfrac{1}{4} = 1. Hence, option D (4).

Q51MCQTime & Work

A group of NN people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of NN is
  1. 150
  2. 23
  3. 36
  4. 140
Answer and solution

Answer: (D) 140

Take the work one person does in one hour as 1 unit. In the first phase, NN people work 7 hours a day for 10 days, doing 70N70N units. This is 35% of the project, so the total work is 70N0.35=200N\dfrac{70N}{0.35} = 200N units, and the work left is 200N−70N=130N200N - 70N = 130N units. In the second phase, N−10N - 10 people work 10 hours a day for 14 days, doing 140(N−10)140(N - 10) units. So 140(N−10)=130N140(N - 10) = 130N, which gives 140N−1400=130N140N - 1400 = 130N, so 10N=140010N = 1400 and N=140N = 140. Check: the first phase does 70×140=980070 \times 140 = 9800 units, which is 35% of 28000. The second does 130×140=18200130 \times 140 = 18200 units, the remaining 65%. Option A (150) does not fit: 140 people working 140 hours would do 19600 units, but only 130×150=19500130 \times 150 = 19500 would be left. Hence, option D (140).

Q52MCQAverages, Mixtures & Alligations

A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio
  1. 1 : 1
  2. 10 : 13
  3. 3 : 10
  4. 10 : 3
Answer and solution

Answer: (A) 1 : 1

After the first transfer, the glass has 350 cc of milk, and the cup has 150 cc of milk and 500 cc of water, 650 cc in all. Milk and water in the cup are in the ratio 150:500=3:10150 : 500 = 3 : 10. The 150 cc taken back from the cup keeps this ratio. It carries 1013×150=150013\dfrac{10}{13} \times 150 = \dfrac{1500}{13} cc of water into the glass, along with 313×150=45013\dfrac{3}{13} \times 150 = \dfrac{450}{13} cc of milk. Water in the glass =150013= \dfrac{1500}{13} cc. Milk left in the cup =150−45013=150013= 150 - \dfrac{450}{13} = \dfrac{1500}{13} cc. These are equal, so the ratio is 1:11 : 1. This is expected: each container ends with 500 cc, so the water that entered the glass is exactly matched by the milk that stayed in the cup. Option D (10 : 3) is the ratio of water to milk inside the cup's mixture, not the ratio asked for. Hence, option A (1 : 1).

Q53MCQSimple & Compound Interest

Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at xx% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been
  1. 700
  2. 800
  3. 900
  4. 1000
Answer and solution

Answer: (B) 800

Interest from bank A =8000×5.5%=440= 8000 \times 5.5\% = 440. Interest from bank B =5000×5.6%=280= 5000 \times 5.6\% = 280. The rest, 20000−8000−5000=700020000 - 8000 - 5000 = 7000, is in bank C and earns 7000×x100=70x7000 \times \dfrac{x}{100} = 70x. The total interest is 5% of 20000, which is 1000. So 440+280+70x=1000440 + 280 + 70x = 1000, giving 70x=28070x = 280 and x=4x = 4. If all ₹20,000 were in bank C at 4%, the annual interest would be 20000×4%=80020000 \times 4\% = 800 rupees. Option D (1000) is her actual income from the three investments together, not the income from bank C alone. Hence, option B (800).

Q54MCQTime, Speed & Distance

Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance traveled, in km, by the slower car when it meets the other car while traveling towards each other, is
  1. 100
  2. 90
  3. 120
  4. 150
Answer and solution

Answer: (B) 90

When the cars travel towards each other, they meet after 1.5 hours. The slower car moves at 60 km/hr throughout, so it covers 60×1.5=9060 \times 1.5 = 90 km before they meet. The 10.5-hour figure is needed only to find the faster car's speed. Let it be vv km/hr. The gap between the cars is 1.5(v+60)=10.5(v−60)1.5(v + 60) = 10.5(v - 60), which gives 9v=7209v = 720, so v=80v = 80 and the gap is 210 km. Option C (120) is the distance covered by the faster car, 80×1.5=12080 \times 1.5 = 120 km. Together, 90+120=21090 + 120 = 210 km, which checks the gap. Hence, option B (90).

Q55MCQAverages, Mixtures & Alligations

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is
  1. 2222
  2. 2442
  3. 2592
  4. 3333
Answer and solution

Answer: (A) 2222

The digits are 1, 1, 2 and 4. The number of distinct arrangements is 4!2!=12\dfrac{4!}{2!} = 12. In any one place, each digit appears in proportion to how often it occurs among the four digits, that is, 2 : 1 : 1 for 1, 2 and 4. So in the units place, 1 appears 6 times, 2 appears 3 times and 4 appears 3 times, and the units digits add up to 6(1)+3(2)+3(4)=246(1) + 3(2) + 3(4) = 24. The same holds for the tens, hundreds and thousands places. So the sum of all 12 numbers is 24×(1+10+100+1000)=24×111124 \times (1 + 10 + 100 + 1000) = 24 \times 1111. Mean =24×111112=2×1111=2222= \dfrac{24 \times 1111}{12} = 2 \times 1111 = 2222. A quicker view: the average digit in every place is 1+1+2+44=2\dfrac{1 + 1 + 2 + 4}{4} = 2, so the mean is 2222. Every other option would need an average digit other than 2 in some place, which cannot happen. Hence, option A (2222).

Q56MCQPolygons & Circles

The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is
  1. 3
  2. 4
  3. 6
  4. 5
Answer and solution

Answer: (D) 5

In any quadrilateral, each side must be shorter than the sum of the other three. Let the fourth side be dd cm, a positive integer. For side dd: d<1+2+4=7d < 1 + 2 + 4 = 7. For the side of 4 cm: 4<1+2+d4 < 1 + 2 + d, so d>1d > 1. The conditions for the sides of 1 cm and 2 cm hold automatically, since 4+d4 + d is already larger than either. So 1<d<71 < d < 7, giving d=2,3,4,5,6d = 2, 3, 4, 5, 6: five possible lengths. Option C (6) wrongly includes d=1d = 1. Then 1+2+1=41 + 2 + 1 = 4 only equals the longest side, so the four sides would lie flat along a line instead of forming a quadrilateral. Hence, option D (5).

Q57TITASequences & Series

The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ... is

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

Answer and solution

Answer: 548

The progression has first term 38 and common difference 17. Its nnth term is 38+17(n−1)=17n+21=17(n+1)+438 + 17(n - 1) = 17n + 21 = 17(n + 1) + 4, so every term has the form 17k+417k + 4. Smallest 3-digit term: 17k+4≥10017k + 4 \ge 100 needs k≥6k \ge 6, giving 17×6+4=10617 \times 6 + 4 = 106. Largest 3-digit term: 17k+4≤99917k + 4 \le 999 needs k≤58k \le 58, giving 17×58+4=99017 \times 58 + 4 = 990. So the 3-digit terms are 106,123,140,…,990106, 123, 140, \dots, 990, and there are 58−6+1=5358 - 6 + 1 = 53 of them. In an arithmetic progression, the average of the terms equals the average of the first and last terms: 106+9902=10962=548\dfrac{106 + 990}{2} = \dfrac{1096}{2} = 548. The answer is 548.

Q58TITAAverages, Mixtures & Alligations

In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

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

Answer and solution

Answer: 63

Let section A have aa students, so section B has a+10a + 10. Combined average =32a+60(a+10)a+(a+10)=92a+6002a+10=46a+300a+5= \dfrac{32a + 60(a + 10)}{a + (a + 10)} = \dfrac{92a + 600}{2a + 10} = \dfrac{46a + 300}{a + 5}. Since 46a+300=46(a+5)+7046a + 300 = 46(a + 5) + 70, this equals 46+70a+546 + \dfrac{70}{a + 5}. It is an integer only when a+5a + 5 divides 70. The divisors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70. As a≥1a \ge 1, a+5≥6a + 5 \ge 6, so a+5a + 5 can be 7, 10, 14, 35 or 70, giving a=2,5,9,30,65a = 2, 5, 9, 30, 65. The maximum is 65 and the minimum is 2, so the difference is 65−2=6365 - 2 = 63. The answer is 63.

Q59MCQFunctions & Graphs

Let rr be a real number and f(x)={2x−rif x≥rrif x<rf(x) = \begin{cases} 2x - r & \text{if } x \geq r \\ r & \text{if } x < r \end{cases}. Then, the equation f(x)=f(f(x))f(x) = f(f(x)) holds for all real values of xx where
  1. x>rx > r
  2. x≤rx \leq r
  3. x≠rx \neq r
  4. x≥rx \geq r
Answer and solution

Answer: (B) x≤rx \leq r

Check the two parts of the definition separately. If x<rx < r: f(x)=rf(x) = r. Then f(f(x))=f(r)f(f(x)) = f(r), and since r≥rr \ge r, f(r)=2r−r=rf(r) = 2r - r = r. So f(f(x))=r=f(x)f(f(x)) = r = f(x), and the equation holds for every x<rx < r. If x≥rx \ge r: f(x)=2x−rf(x) = 2x - r. Since x≥rx \ge r, 2x−r≥r2x - r \ge r, so f(f(x))=2(2x−r)−r=4x−3rf(f(x)) = 2(2x - r) - r = 4x - 3r. The equation 2x−r=4x−3r2x - r = 4x - 3r gives x=rx = r. So among x≥rx \ge r, it holds only at x=rx = r. Together, the equation holds exactly when x≤rx \le r. Option D (x≥rx \geq r) fails because for any x>rx > r, f(f(x))=4x−3rf(f(x)) = 4x - 3r is larger than f(x)=2x−rf(x) = 2x - r. Option A (x>rx > r) fails for the same reason. Hence, option B (x≤rx \leq r).

Q60MCQPolygons & Circles

In a triangle ABCABC, AB=AC=8AB = AC = 8 cm. A circle drawn with BCBC as diameter passes through AA. Another circle drawn with center at AA passes through BB and CC. Then the area, in sq. cm, of the overlapping region between the two circles is
  1. 16π16\pi
  2. 16(π−1)16(\pi - 1)
  3. 32(π−1)32(\pi - 1)
  4. 32π32\pi
Answer and solution

Answer: (C) 32(π−1)32(\pi - 1)

Solution figure for question 60, CAT 2022 Slot 3 Call the circle with diameter BCBC C2, and the circle centred at AA C1. Since AA lies on C2, ∠BAC=90∘\angle BAC = 90^\circ (angle in a semicircle). With AB=AC=8AB = AC = 8, BC=82BC = 8\sqrt{2}, so C2 has radius 424\sqrt{2} and area π(42)2=32π\pi(4\sqrt{2})^2 = 32\pi. C1 passes through BB and CC, so its radius is AB=8AB = 8. BCBC is a common chord of the two circles and splits the overlap into two parts. On AA's side of BCBC: the half of C2 there lies wholly inside C1, since none of its points is more than 8 from AA. Its area is 12×32π=16π\frac{1}{2} \times 32\pi = 16\pi. On the other side: the overlap is the segment of C1 cut off by chord BCBC, which lies inside C2's other half. As ∠BAC=90∘\angle BAC = 90^\circ, it equals a quarter of C1 minus triangle ABCABC: 14π(8)2−12(8)(8)=16π−32\frac{1}{4}\pi(8)^2 - \frac{1}{2}(8)(8) = 16\pi - 32. Overlap =16π+(16π−32)=32π−32=32(π−1)= 16\pi + (16\pi - 32) = 32\pi - 32 = 32(\pi - 1). Option A (16π16\pi) counts only the half of C2 and misses the segment of C1. Hence, option C (32(π−1)32(\pi - 1)).

Q61TITAHCF & LCM

A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

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

Answer and solution

Answer: 150

Leaving 4 students over when divided by 9, 10, 12 or 25 means the number of students minus 4 is a multiple of each of them, and so of LCM(9,10,12,25)=900\text{LCM}(9, 10, 12, 25) = 900. So the number of students is 900k+4900k + 4. It is less than 5000, so kk can be 0 to 5, giving 4, 904, 1804, 2704, 3604 or 4504. It must also be divisible by 11. Checking: 4 is not; 904=11×82+2904 = 11 \times 82 + 2; 1804=11×1641804 = 11 \times 164; 2704=11×245+92704 = 11 \times 245 + 9; 3604=11×327+73604 = 11 \times 327 + 7; 4504=11×409+54504 = 11 \times 409 + 5. Only 1804 works. With 1804 students, 1804=12×150+41804 = 12 \times 150 + 4, so at most 150 teams of 12 can be formed, with 4 students left out. The answer is 150.

Q62MCQInequalities & Modulus

The minimum possible value of x2−6x+103−x\dfrac{x^2 - 6x + 10}{3 - x}, for x<3x < 3, is
  1. −12-\dfrac{1}{2}
  2. 2
  3. 12\dfrac{1}{2}
  4. -2
Answer and solution

Answer: (B) 2

Since x<3x < 3, put y=3−xy = 3 - x, so y>0y > 0. The numerator is x2−6x+10=(x−3)2+1=y2+1x^2 - 6x + 10 = (x - 3)^2 + 1 = y^2 + 1. So the expression becomes y2+1y=y+1y\dfrac{y^2 + 1}{y} = y + \dfrac{1}{y}. For y>0y > 0, AM–GM gives y+1y≥2y⋅1y=2y + \dfrac{1}{y} \ge 2\sqrt{y \cdot \dfrac{1}{y}} = 2, with equality when y=1y = 1, that is, at x=2x = 2. Check: at x=2x = 2, the expression is 4−12+103−2=2\dfrac{4 - 12 + 10}{3 - 2} = 2. Option D (−2) is the value at x=4x = 4, where y=−1y = -1. That is outside the range x<3x < 3, and for x<3x < 3 the expression is always positive. Hence, option B (2).

Q63TITALinear Equations

A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is

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

Answer and solution

Answer: 12

Let the numbers of ₹100, ₹250 and ₹500 cheques be xx, yy and zz. Count: x+y+z=100x + y + z = 100. Amount: 100x+250y+500z=15250100x + 250y + 500z = 15250. Dividing by 50, 2x+5y+10z=3052x + 5y + 10z = 305. Subtracting twice the first equation: 3y+8z=1053y + 8z = 105. So 3y=105−8z3y = 105 - 8z. For yy to be a whole number, 8z8z must be a multiple of 3, so zz is a multiple of 3. Also 8z≤1058z \le 105, so z≤13z \le 13. The largest multiple of 3 up to 13 is 12. Then 3y=105−96=93y = 105 - 96 = 9, so y=3y = 3 and x=100−12−3=85x = 100 - 12 - 3 = 85. Check: 85×100+3×250+12×500=8500+750+6000=1525085 \times 100 + 3 \times 250 + 12 \times 500 = 8500 + 750 + 6000 = 15250. z=13z = 13 fails because 105−104=1105 - 104 = 1 is not a multiple of 3. The answer is 12.

Q64TITATime, Speed & Distance

Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

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Answer and solution

Answer: 60

Let the speed of Moody be 'x' steps/sec and that of the escalator be 'y' steps/sec. In 30 seconds, Moody will finish riding the escalator when going in the same direction. Thus, total steps = 30(x+y) If Moody's speed becomes twice, the time becomes 20 seconds. Thus, total steps = 20(2x+y) Or 30x + 30y = 40x + 20y Or x = y So, total steps = 60y. Time taken by only escalator= 60y/y = 60s.

Q65MCQTime, Speed & Distance

Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be
  1. 4
  2. 12
  3. 8
  4. 6
Answer and solution

Answer: (B) 12

Solution figure for question 65, CAT 2022 Slot 3 Let O be the port, S the slower ship and F the faster one. They start at the corners of an equilateral triangle of side 24 km and sail straight along the sides SO and FO, so the angle at O stays 60∘60^\circ. When S has travelled 8 km, OS=24−8=16OS = 24 - 8 = 16 km. The triangle OSF is now right-angled, so the right angle is at S or at F. If it were at S, then OS=OFcos⁡60∘OS = OF \cos 60^\circ, so OF=32OF = 32 km, more than the starting 24 km. That is impossible. So the right angle is at F, and OF=OScos⁡60∘=16×12=8OF = OS \cos 60^\circ = 16 \times \frac{1}{2} = 8 km. F has travelled 24−8=1624 - 8 = 16 km. In the same time S covers 8 km and F covers 16 km, so F is twice as fast. F reaches the port after 24 km. In that time S covers 242=12\frac{24}{2} = 12 km, so it is 24−12=1224 - 12 = 12 km from the port. Option C (8) is F's distance from the port at the right-angle moment, not S's distance at the end. Hence, option B (12).

Q66TITATime & Work

Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is

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

Answer and solution

Answer: 11

Let Bob do 3 units of work a day. Alex is twice as fast as Bob, so he does 6 units a day. Alex is thrice as fast as Cole, so Cole does 2 units a day. Bob alone takes 40 days, so the job is 40×3=12040 \times 3 = 120 units. The roster does 6+3=96 + 3 = 9 units on day 1 (Alex and Bob), 3+2=53 + 2 = 5 units on day 2 (Bob and Cole) and 2+6=82 + 6 = 8 units on day 3 (Cole and Alex). Each 3-day cycle therefore does 22 units. Five cycles, 15 days, do 5×22=1105 \times 22 = 110 units, leaving 10. Day 16 (Alex and Bob) does 9 units, leaving 1, and day 17 (Bob and Cole) finishes the job. Alex works 2 days in every cycle, so 10 days in the first 15. He also works on day 16 but not on day 17. Alex works 10+1=1110 + 1 = 11 days. The answer is 11.