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CAT 2024 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.

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Quantitative Ability

CAT 2024 Slot 3 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q47MCQSet Theory

In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are
  1. 72 and 88, respectively
  2. 75 and 96, respectively
  3. 72 and 80, respectively
  4. 75 and 90, respectively
Answer and solution

Answer: (C) 72 and 80, respectively

Girls GG satisfy 110≤G≤150110 \le G \le 150 (44% and 60% of 250), and boys B=250−GB = 250 - G. Swimming: 0.5B+0.8G0.5B + 0.8G. Running: 0.7B+0.6G0.7B + 0.6G. Every student chose at least one, so both == swimming ++ running −250=1.2B+1.4G−250- 250 = 1.2B + 1.4G - 250. Substituting B=250−GB = 250 - G: both =300−1.2G+1.4G−250=50+0.2G= 300 - 1.2G + 1.4G - 250 = 50 + 0.2G. This rises with GG, so the minimum is at G=110G = 110: 50+22=7250 + 22 = 72, and the maximum is at G=150G = 150: 50+30=8050 + 30 = 80. Both give whole numbers of students (B=140B = 140 and B=100B = 100). Option A (72 and 88) has the right minimum, but 88 would need G=190G = 190, above the 60% cap of 150. Hence, option C (72 and 80, respectively).

Q48MCQIndices & Surds

If (a+b3)2=52+303(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}, where a and b are natural numbers, then a+ba + b equals
  1. 7
  2. 8
  3. 9
  4. 10
Answer and solution

Answer: (B) 8

(a+b3)2=a2+3b2+2ab3(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}. As aa and bb are natural numbers, match the parts with and without 3\sqrt{3} in 52+30352 + 30\sqrt{3}: a2+3b2=52a^2 + 3b^2 = 52 and 2ab=302ab = 30, i.e. ab=15ab = 15. With ab=15ab = 15, (a,b)(a, b) is one of (1,15)(1, 15), (3,5)(3, 5), (5,3)(5, 3), (15,1)(15, 1). Only (5,3)(5, 3) gives a2+3b2=25+27=52a^2 + 3b^2 = 25 + 27 = 52. Check: (5+33)2=25+303+27=52+303(5 + 3\sqrt{3})^2 = 25 + 30\sqrt{3} + 27 = 52 + 30\sqrt{3}. So a+b=5+3=8a + b = 5 + 3 = 8. The other options are impossible: every pair with ab=15ab = 15 has a+b=16a + b = 16 or 88, so 77, 99 and 1010 cannot occur. The pair (3,5)(3, 5) also sums to 88 but gives 9+75=849 + 75 = 84, not 5252. Hence, option B (8).

Q49TITAAverages, Mixtures & Alligations

The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is

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

Answer and solution

Answer: 70

Let the numbers be x<y<zx < y < z, so x+y+z=3×28=84x + y + z = 3 \times 28 = 84. After the changes the numbers are x+7x + 7, yy and z−10z - 10, in the same order. Their sum is 84+7−10=8184 + 7 - 10 = 81, so the new mean is 27. The new mean is 2 more than the middle number, so y=25y = 25 and x+z=84−25=59x + z = 84 - 25 = 59. The new difference between the largest and smallest is (z−10)−(x+7)=64(z - 10) - (x + 7) = 64, so z−x=81z - x = 81. Adding the two equations: 2z=1402z = 140, so z=70z = 70 and x=−11x = -11. Check: the new numbers −4-4, 25 and 60 keep the same order. The answer is 70.

Q50MCQRemainders

If 106810^{68} is divided by 13, the remainder is
  1. 5
  2. 8
  3. 9
  4. 4
Answer and solution

Answer: (C) 9

Work modulo 13. Since 13×77=100113 \times 77 = 1001, we have 103=1000≡−1(mod13)10^3 = 1000 \equiv -1 \pmod{13}, and so 106≡110^6 \equiv 1. 68=6×11+268 = 6 \times 11 + 2, so 1068=(106)11×102≡1×100(mod13)10^{68} = (10^6)^{11} \times 10^2 \equiv 1 \times 100 \pmod{13}. 100=13×7+9100 = 13 \times 7 + 9, so the remainder is 99.

Q51MCQTime & Work

Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is
  1. 1/20
  2. 3/10
  3. 1/5
  4. 3/20
Answer and solution

Answer: (B) 3/10

Take the job as 60 units. Sam does 60÷20=360 \div 20 = 3 units a day. Mohit is twice as fast, 6 units a day. Mohit is thrice as fast as Ayna, so Ayna does 2 units a day. Day 1 (Sam + Mohit): 9. Day 2 (Sam + Ayna): 5. Day 3 (Mohit + Ayna): 8. One 3-day cycle does 22 units. Two cycles (6 days) do 44 units, leaving 16. Day 7 does 9, leaving 7; day 8 does 5, leaving 2; Mohit and Ayna finish the last 2 units on day 9. Sam worked on days 1, 2, 4, 5, 7 and 8: 6×3=186 \times 3 = 18 units, which is 1860=310\dfrac{18}{60} = \dfrac{3}{10} of the job. Option D (3/20) is half of this; it would count only one of Sam's two working days in each cycle. Hence, option B (3/10).

Q52MCQPolygons & Circles

A circular plot of land is divided into two regions by a chord of length 10310\sqrt{3} meters such that the chord subtends an angle of 120∘120^\circ at the center. Then, the area, in square meters, of the smaller region is
  1. 20(4π3+3)20\left(\frac{4\pi}{3} + \sqrt{3}\right)
  2. 25(4π3+3)25\left(\frac{4\pi}{3} + \sqrt{3}\right)
  3. 20(4π3−3)20\left(\frac{4\pi}{3} - \sqrt{3}\right)
  4. 25(4π3−3)25\left(\frac{4\pi}{3} - \sqrt{3}\right)
Answer and solution

Answer: (D) 25(4π3−3)25\left(\frac{4\pi}{3} - \sqrt{3}\right)

The chord subtends 120∘120^\circ at the centre, so its length is 2Rsin⁡60∘=R32R\sin 60^\circ = R\sqrt{3}. R3=103R\sqrt{3} = 10\sqrt{3}, so R=10R = 10 m. The smaller region is the minor segment: the sector minus the triangle formed by the two radii and the chord. Sector: 120360×π×102=100π3\dfrac{120}{360} \times \pi \times 10^2 = \dfrac{100\pi}{3}. Triangle: 12×102×sin⁡120∘=12×100×32=253\dfrac12 \times 10^2 \times \sin 120^\circ = \dfrac12 \times 100 \times \dfrac{\sqrt3}{2} = 25\sqrt{3}. Segment: 100π3−253=25(4π3−3)\dfrac{100\pi}{3} - 25\sqrt{3} = 25\left(\dfrac{4\pi}{3} - \sqrt{3}\right) square metres. Option B adds the triangle instead of subtracting it, giving the sector plus the triangle rather than the segment. Options A and C use 20 in place of 25. Hence, option D (25(4π3−3)25\left(\frac{4\pi}{3} - \sqrt{3}\right)).

Q53MCQSequences & Series

Consider the sequence t1=1,t2=−1t_1 = 1, t_2 = -1 and tn=(n−3n−1)tn−2t_n = \left(\frac{n-3}{n-1}\right)t_{n-2} for n≥3n \ge 3. Then, the value of the sum 1t2+1t4+1t6+.......+1t2022+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + ....... + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}, is
  1. -1024144
  2. -1022121
  3. -1023132
  4. -1026169
Answer and solution

Answer: (A) -1024144

Only even-numbered terms are needed. With n=2kn = 2k, the rule gives t2k=2k−32k−1 t2k−2t_{2k} = \dfrac{2k - 3}{2k - 1}\, t_{2k-2}. t2=−1t_2 = -1, t4=13t2=−13t_4 = \dfrac13 t_2 = -\dfrac13, t6=35t4=−15t_6 = \dfrac35 t_4 = -\dfrac15, and in general t2k=−12k−1t_{2k} = -\dfrac{1}{2k - 1}. Each step keeps this form, since 2k−32k−1×(−12k−3)=−12k−1\dfrac{2k-3}{2k-1} \times \left(-\dfrac{1}{2k-3}\right) = -\dfrac{1}{2k-1}. So 1t2k=−(2k−1)\dfrac{1}{t_{2k}} = -(2k - 1), and the sum is −(1+3+5+⋯+2023)-(1 + 3 + 5 + \dots + 2023). The last term, 2023, is 2k−12k - 1 with k=1012k = 1012, so there are 1012 odd numbers. The sum of the first mm odd numbers is m2m^2, so the total is −10122=−1024144-1012^2 = -1024144. Options B, C and D are −10112-1011^2, −1011×1012-1011 \times 1012 and −10132-1013^2; they come from miscounting the terms. Hence, option A (-1024144).

Q54TITAInequalities & Modulus

The number of distinct real values of x, satisfying the equation max(x, 2) - min(x, 2) = |x + 2| - |x - 2|, is

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

Answer and solution

Answer: 2

Solution figure for question 54, CAT 2024 Slot 3

Note that max⁡(x,2)−min⁡(x,2)=∣x−2∣\max(x, 2) - \min(x, 2) = |x - 2|.

So the equation becomes: ∣x−2∣=∣x+2∣−∣x−2∣|x - 2| = |x + 2| - |x - 2|, which gives 2∣x−2∣=∣x+2∣2|x - 2| = |x + 2|.

Case 1: x≥2x \geq 2: 2(x−2)=x+2⇒x=62(x-2) = x+2 \Rightarrow x = 6.

Case 2: −2≤x<2-2 \leq x < 2: 2(2−x)=x+2⇒4−2x=x+2⇒x=232(2-x) = x+2 \Rightarrow 4-2x = x+2 \Rightarrow x = \frac{2}{3}.

Case 3: x<−2x < -2: 2(2−x)=−(x+2)⇒4−2x=−x−2⇒x=62(2-x) = -(x+2) \Rightarrow 4-2x = -x-2 \Rightarrow x = 6, which contradicts x<−2x < -2.

So there are exactly 2 distinct real values of xx that satisfy the equation.

Q55TITASimple & Compound Interest

Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is

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Answer and solution

Answer: 9

The value after nn years is 4000(1+r)n4000(1 + r)^n. The ratio of the values after 3 and 5 years is (1+r)3(1+r)5=1(1+r)2=2536\dfrac{(1+r)^3}{(1+r)^5} = \dfrac{1}{(1+r)^2} = \dfrac{25}{36}, so (1+r)2=3625(1+r)^2 = \dfrac{36}{25} and 1+r=65=1.21 + r = \dfrac65 = 1.2. We need 4000×1.2n>200004000 \times 1.2^n > 20000, that is, 1.2n>51.2^n > 5. Powers of 1.2: 1.22=1.441.2^2 = 1.44, 1.24=1.442=2.07361.2^4 = 1.44^2 = 2.0736, 1.28=2.07362≈4.301.2^8 = 2.0736^2 \approx 4.30, and 1.29≈4.30×1.2≈5.161.2^9 \approx 4.30 \times 1.2 \approx 5.16. So 8 years is not enough (4000×4.30≈172004000 \times 4.30 \approx 17200), and 9 years is (4000×5.16≈206404000 \times 5.16 \approx 20640). The answer is 9.

Q56MCQQuadratic & Polynomial Equations

The sum of all distinct real values of x that satisfy the equation 10x+410x=81210^x + \frac{4}{10^x} = \frac{81}{2}, is
  1. 2log⁡1022\log_{10} 2
  2. 4log⁡1024\log_{10} 2
  3. log⁡102\log_{10} 2
  4. 3log⁡1023\log_{10} 2
Answer and solution

Answer: (A) 2log⁡1022\log_{10} 2

Let y=10xy = 10^x, which is positive for every real xx. The equation becomes y+4y=812y + \dfrac{4}{y} = \dfrac{81}{2}, i.e. 2y2−81y+8=02y^2 - 81y + 8 = 0. Its discriminant 812−6481^2 - 64 is positive, and the roots have sum 812>0\dfrac{81}{2} > 0 and product 82=4>0\dfrac{8}{2} = 4 > 0. So both roots are positive and distinct, and each gives one real x=log⁡10yx = \log_{10} y; the two xx values are distinct. Sum of the xx values =log⁡10y1+log⁡10y2=log⁡10(y1y2)=log⁡104=2log⁡102= \log_{10} y_1 + \log_{10} y_2 = \log_{10}(y_1 y_2) = \log_{10} 4 = 2\log_{10} 2. Option D (3log⁡102=log⁡1083\log_{10} 2 = \log_{10} 8) comes from taking the product of the roots as 88 without dividing by the leading coefficient 22. Hence, option A (2log⁡1022\log_{10} 2).

Q57MCQTime, Speed & Distance

A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is
  1. 720
  2. 800
  3. 780
  4. 640
Answer and solution

Answer: (A) 720

Let the speed be SS km/h and the time TT hours, so the distance is STST. Faster: (S+6)(T−4)=ST(S + 6)(T - 4) = ST gives 6T−4S=246T - 4S = 24. Slower: (S−6)(T+6)=ST(S - 6)(T + 6) = ST gives 6S−6T=366S - 6T = 36. Adding the two equations: 2S=602S = 60, so S=30S = 30. Then 6T=24+120=1446T = 24 + 120 = 144, so T=24T = 24. Distance =30×24=720= 30 \times 24 = 720 km. Check: at 36 km/h it takes 20 hours (4 fewer), and at 24 km/h it takes 30 hours (6 more). The two equations fix SS and TT, so no other option can satisfy both conditions. Hence, option A (720).

Q58TITALogarithms

If 3a=4,4b=5,5c=6,6d=7,7e=83^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8 and 8f=98^f = 9, then the value of the product abcdef is

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

Answer and solution

Answer: 2

Substitute each equation into the next. 3a=43^a = 4. Raise both sides to the power bb: 3ab=4b=53^{ab} = 4^b = 5. Raise to the power cc: 3abc=5c=63^{abc} = 5^c = 6. Continuing the same way: 3abcd=6d=73^{abcd} = 6^d = 7, then 3abcde=7e=83^{abcde} = 7^e = 8, then 3abcdef=8f=93^{abcdef} = 8^f = 9. So 3abcdef=9=323^{abcdef} = 9 = 3^2, which gives abcdef=2abcdef = 2. The same result follows from logarithms: a=log⁡34a = \log_3 4, b=log⁡45b = \log_4 5, and so on, and by the change-of-base rule the product log⁡4log⁡3⋅log⁡5log⁡4⋯log⁡9log⁡8\dfrac{\log 4}{\log 3} \cdot \dfrac{\log 5}{\log 4} \cdots \dfrac{\log 9}{\log 8} telescopes to log⁡39=2\log_3 9 = 2. The answer is 2.

Q59MCQProfit, Loss & Discount

Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is
  1. 10
  2. 25
  3. 15
  4. 20
Answer and solution

Answer: (A) 10

Ravi saves Rs 15 through a 10% discount, so 10% of the marked price is 15 and the marked price is Rs 150. The marked price gives a 20% profit, so 1.2×CP=1501.2 \times \text{CP} = 150 and the cost price is Rs 125. Ravi pays 150−15=135150 - 15 = 135, so Gopi's profit is 135−125=10135 - 125 = 10 rupees. Option C (15) is the discount, not the profit, and option B (25) is the profit Gopi would make at the full marked price. Hence, option A (10).

Q60TITAAverages, Mixtures & Alligations

A certain amount of water was poured into a 300 litre container and the remaining portion of the container was filled with milk. Then an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again. If the resulting solution contains 72% milk, then the amount of water, in litres, that was initially poured into the container was

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

Answer and solution

Answer: 30

Let xx litres of water be poured in first, so the milk is 300−x300 - x litres and the milk fraction is 300−x300\dfrac{300 - x}{300}. Taking out 2x2x litres leaves 300−2x300 - 2x litres in the same ratio. Refilling with water adds no milk, so the milk left is 300−x300×(300−2x)=72%×300=216\dfrac{300 - x}{300} \times (300 - 2x) = 72\% \times 300 = 216. So (300−x)(300−2x)=64800(300 - x)(300 - 2x) = 64800, that is, 2x2−900x+90000=648002x^2 - 900x + 90000 = 64800, or x2−450x+12600=0x^2 - 450x + 12600 = 0. This factors as (x−30)(x−420)=0(x - 30)(x - 420) = 0. Since 2x2x litres were taken from a 300-litre container, x≤150x \le 150, so x=30x = 30. Check: 270 litres of milk; removing 60 litres leaves 240×0.9=216240 \times 0.9 = 216 litres of milk, which is 72% of 300. The answer is 30.

Q61MCQPolygons & Circles

A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is
  1. 72(2 + √2)
  2. 36(1 + √2)
  3. 72(1 + √2)
  4. 36(2 + √2)
Answer and solution

Answer: (D) 36(2 + √2)

A regular octagon has interior angles of 135∘135^\circ. The square's side AC skips one vertex, so it is the third side of triangle ABC, with AB=BC=6AB = BC = 6 and ∠ABC=135∘\angle ABC = 135^\circ. Solution figure for question 61, CAT 2024 Slot 3 By the cosine rule, AC2=62+62−2⋅6⋅6cos⁡135∘=72+72⋅22=72+362AC^2 = 6^2 + 6^2 - 2 \cdot 6 \cdot 6 \cos 135^\circ = 72 + 72 \cdot \dfrac{\sqrt2}{2} = 72 + 36\sqrt2. The area of square ACEG is AC2=72+362=36(2+2)AC^2 = 72 + 36\sqrt2 = 36(2 + \sqrt2) sq cm. Option A, 72(2+2)72(2 + \sqrt2), is twice this. Option C, 72(1+2)72(1 + \sqrt2), is the area of the whole octagon, 2(1+2)s22(1 + \sqrt2)s^2 with s=6s = 6. Hence, option D (36(2 + √2)).

Q62TITAInequalities & Modulus

The number of distinct integer solutions (x, y) of the equation |x + y| + |x - y| = 2, is

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

Answer and solution

Answer: 8

For any real xx and yy, ∣x+y∣+∣x−y∣=2max⁡(∣x∣,∣y∣)|x + y| + |x - y| = 2\max(|x|, |y|). If ∣x∣≥∣y∣|x| \ge |y|, then x+yx + y and x−yx - y both have the sign of xx (or are 0), so their absolute values add to ∣2x∣=2∣x∣|2x| = 2|x|. The case ∣y∣≥∣x∣|y| \ge |x| is the same with the roles swapped. So the equation says max⁡(∣x∣,∣y∣)=1\max(|x|, |y|) = 1. Integer pairs with ∣x∣≤1|x| \le 1 and ∣y∣≤1|y| \le 1 form a 3×33 \times 3 grid of 9 points. Removing (0,0)(0, 0), where the maximum is 0, leaves 8: (±1,0)(\pm1, 0), (0,±1)(0, \pm1), (1,1)(1, 1), (1,−1)(1, -1), (−1,1)(-1, 1), (−1,−1)(-1, -1). Check one: (1,1)(1, 1) gives ∣2∣+∣0∣=2|2| + |0| = 2. The answer is 8.

Q63MCQFunctions & Graphs

For any non-zero real number x, let f(x) + 2f(1/x) = 3x. Then, the sum of all possible values of x for which f(x) = 3, is
  1. 3
  2. -2
  3. -3
  4. 2
Answer and solution

Answer: (C) -3

Replacing xx by 1x\dfrac1x in f(x)+2f(1x)=3xf(x) + 2f\left(\dfrac1x\right) = 3x gives f(1x)+2f(x)=3xf\left(\dfrac1x\right) + 2f(x) = \dfrac3x. Double this second equation, 2f(1x)+4f(x)=6x2f\left(\dfrac1x\right) + 4f(x) = \dfrac6x, and subtract the first: 3f(x)=6x−3x3f(x) = \dfrac6x - 3x, so f(x)=2x−xf(x) = \dfrac2x - x. Setting f(x)=3f(x) = 3: 2x−x=3\dfrac2x - x = 3, so 2−x2=3x2 - x^2 = 3x, that is, x2+3x−2=0x^2 + 3x - 2 = 0. Its discriminant is 9+8=17>09 + 8 = 17 > 0, so it has two real roots, and neither is 0 because the constant term is −2-2. Their sum is −3-3. Option A (3) has the wrong sign; it comes from writing the quadratic as x2−3x−2=0x^2 - 3x - 2 = 0. Hence, option C (-3).

Q64MCQLinear Equations

For some constant real numbers p, k and a, consider the following system of linear equations in x and y: px - 4y = 2 3x + ky = a A necessary condition for the system to have no solution for (x, y), is
  1. ap + 6 = 0
  2. 2a + k ≠ 0
  3. ap - 6 = 0
  4. kp + 12 ≠ 0
Answer and solution

Answer: (B) 2a + k ≠ 0

The system has no solution when the two lines are parallel but distinct: the xx and yy coefficients are in proportion, but the constants are not. So we need p3=−4k\dfrac{p}{3} = \dfrac{-4}{k}, that is, kp=−12kp = -12, and −4k≠2a\dfrac{-4}{k} \ne \dfrac{2}{a}. The second condition is −4a≠2k-4a \ne 2k, that is, 2a+k≠02a + k \ne 0. If 2a+k=02a + k = 0 as well, the second equation is a multiple of the first and there are infinitely many solutions. So 2a+k≠02a + k \ne 0 holds whenever there is no solution: it is a necessary condition. Option D is the strongest distractor, but kp+12≠0kp + 12 \ne 0 is the condition for a unique solution; no solution needs kp+12=0kp + 12 = 0. With kp=−12kp = -12, option C (ap−6=0ap - 6 = 0) is the same as 2a+k=02a + k = 0, which gives infinitely many solutions, and option A (ap+6=0ap + 6 = 0) is only one of many no-solution cases, so it is not necessary. Hence, option B (2a + k ≠ 0).

Q65MCQRatios, Proportions & Partnership

Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of 5 : 3. If the total cultivation area of wheat and mustard are in the ratio 11 : 9, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is
  1. 4 : 3
  2. 7 : 9
  3. 3 : 7
  4. 1 : 1
Answer and solution

Answer: (B) 7 : 9

Total land =20+30=50= 20 + 30 = 50 hectares. Wheat : mustard is 11:911 : 9 overall, so wheat =1120×50=27.5= \dfrac{11}{20} \times 50 = 27.5 ha and mustard =22.5= 22.5 ha. Vimal's 30 ha are in the ratio 5:35 : 3: wheat =58×30=18.75= \dfrac58 \times 30 = 18.75 ha and mustard =11.25= 11.25 ha. Rajesh's wheat =27.5−18.75=8.75= 27.5 - 18.75 = 8.75 ha and mustard =22.5−11.25=11.25= 22.5 - 11.25 = 11.25 ha, which add to his 20 ha. Ratio =8.75:11.25=875:1125=7:9= 8.75 : 11.25 = 875 : 1125 = 7 : 9. Option A (4 : 3) would give Rajesh more wheat than mustard, but overall wheat is 55% of the land while Vimal's is 62.5%, so Rajesh's wheat share must be below 55%. Hence, option B (7 : 9).

Q66TITATriangles & Lines

The midpoints of sides AB, BC, and AC in ΔABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ΔABC is 1440 sq cm, then the area, in sq cm, of ΔXYZ is

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

Answer and solution

Answer: 90

M, N and P are the midpoints of AB, BC and CA, so triangle MNP joins the midpoints of triangle ABC. Its area is 14×1440=360\dfrac14 \times 1440 = 360 sq cm. Solution figure for question 66, CAT 2024 Slot 3 Now locate X. MP joins the midpoints of AB and AC, so MP is parallel to BC. The median AN runs from A to the midpoint of BC, and a line from A cuts every segment parallel to BC in the same ratio, so AN meets MP at its midpoint. So X is the midpoint of MP. In the same way, the median BP meets MN (parallel to AC) at its midpoint Y, and the median CM meets NP (parallel to AB) at its midpoint Z. So triangle XYZ joins the midpoints of the sides of triangle MNP, and its area is 14×360=90\dfrac14 \times 360 = 90 sq cm. The answer is 90.

Q67TITAPermutations & Combinations

The number of all positive integers up to 500 with non-repeating digits is

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

Answer and solution

Answer: 378

Count the numbers from 1 to 500 whose digits are all different. One-digit numbers (1 to 9): 9. Two-digit numbers: the first digit is 1 to 9 (9 ways), and the second is any of the other 9 digits, including 0: 9×9=819 \times 9 = 81. Three-digit numbers from 100 to 499: the first digit is 1, 2, 3 or 4 (4 ways), the second is any of the 9 remaining digits, and the third any of the 8 left: 4×9×8=2884 \times 9 \times 8 = 288. 500 repeats the digit 0, so it is not counted. Total =9+81+288=378= 9 + 81 + 288 = 378. The answer is 378.

Q68MCQPercentages

After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
  1. 30
  2. 27.5
  3. 25
  4. 20
Answer and solution

Answer: (C) 25

Let the first increase be r%r\%, so the second is 2r%2r\%. Writing a=r100a = \dfrac{r}{100}: (1+a)(1+2a)=1.875=158(1 + a)(1 + 2a) = 1.875 = \dfrac{15}{8}. Expanding: 1+3a+2a2=1581 + 3a + 2a^2 = \dfrac{15}{8}, so 16a2+24a−7=016a^2 + 24a - 7 = 0, which factors as (4a−1)(4a+7)=0(4a - 1)(4a + 7) = 0. The positive root is a=14a = \dfrac14, so r=25r = 25. Check: 1.25×1.50=1.8751.25 \times 1.50 = 1.875. Option A (30) would give 1.30×1.60=2.081.30 \times 1.60 = 2.08, which is too large, and option D (20) gives 1.20×1.40=1.681.20 \times 1.40 = 1.68, which is too small. Hence, option C (25).