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CAT 2024 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.

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Quantitative Ability

CAT 2024 Slot 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q47TITAProfit, Loss & Discount

Bina incurs 19% loss when she sells a product at Rs. 4860 to Shyam, who in turn sells this product to Hari. If Bina would have sold this product to Shyam at the purchase price of Hari, she would have obtained 17% profit. Then, the profit, in rupees, made by Shyam is

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

Answer and solution

Answer: 2160

Let the cost price of the item be C

We are given that Bina sells this at 19% loss or at (1 - 0.19)C = 0.81C at 4860

This gives us the value of C at Rs. 6000

If Bina had sold this at 17% profit, the selling price would have been 1.17× 6000 = 7020

So Shyam bought the product at 4860 and sold it to Hari at 7020

Giving the profit made by Shyam to be 7020−4860 = 2160

Therefore, 2160 is the correct answer.

Q48TITATriangles & Lines

The coordinates of the three vertices of a triangle are: (1,2)(1, 2), (7,2)(7, 2), and (1,10)(1, 10). Then the radius of the incircle of the triangle is

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

Answer and solution

Answer: 2

Solution figure for question 48, CAT 2024 Slot 2 The points (1,2)(1, 2) and (7,2)(7, 2) lie on a horizontal line, and (1,2)(1, 2) and (1,10)(1, 10) on a vertical line, so the triangle has a right angle at (1,2)(1, 2). Legs: 7−1=67 - 1 = 6 and 10−2=810 - 2 = 8. Hypotenuse: 62+82=10\sqrt{6^2 + 8^2} = 10. Area =12×6×8=24= \dfrac{1}{2} \times 6 \times 8 = 24. Semi-perimeter s=6+8+102=12s = \dfrac{6 + 8 + 10}{2} = 12. The inradius rr of a triangle satisfies r×s=Arear \times s = \text{Area}, so r=2412=2r = \dfrac{24}{12} = 2. The answer is 2.

Q49TITAPercentages

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

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

Answer and solution

Answer: 340

Let the stock be SS fruits, with aa apples. Mangoes are 40% of the stock, 2S5\dfrac{2S}{5}. Fruits sold: half the mangoes, S5\dfrac{S}{5}; 96 bananas; and 40% of the apples, 2a5\dfrac{2a}{5}. This is half the stock: S5+96+2a5=S2\dfrac{S}{5} + 96 + \dfrac{2a}{5} = \dfrac{S}{2} Multiplying by 10: 2S+960+4a=5S2S + 960 + 4a = 5S, so S=320+4a3S = 320 + \dfrac{4a}{3}. For SS to be a whole number, aa must be a multiple of 3. For 2a5\dfrac{2a}{5} apples to be sold, aa must be a multiple of 5. So aa is a multiple of 15, and the smallest positive value is a=15a = 15. Then S=320+20=340S = 320 + 20 = 340. Check: 136 mangoes, 15 apples and 189 bananas; sold 68+96+6=17068 + 96 + 6 = 170, half of 340. The answer is 340.

Q50TITALogarithms

If aa, bb and cc are positive real numbers such that a>10≥b≥ca > 10 \geq b \geq c and log⁡8(a+b)log⁡2c+log⁡27(a−b)log⁡3c=23\dfrac{\log_8(a+b)}{\log_2 c} + \dfrac{\log_{27}(a-b)}{\log_3 c} = \dfrac{2}{3}, then the greatest possible integer value of aa is

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Answer and solution

Answer: 14

log⁡8(a+b)log⁡2c=13log⁡2(a+b)log⁡2c=13log⁡c(a+b)\dfrac{\log_8(a+b)}{\log_2 c} = \dfrac{\tfrac{1}{3}\log_2(a+b)}{\log_2 c} = \dfrac{1}{3}\log_c(a+b), and in the same way log⁡27(a−b)log⁡3c=13log⁡c(a−b)\dfrac{\log_{27}(a-b)}{\log_3 c} = \dfrac{1}{3}\log_c(a-b). So 13log⁡c((a+b)(a−b))=23\dfrac{1}{3}\log_c\big((a+b)(a-b)\big) = \dfrac{2}{3}, which gives a2−b2=c2a^2 - b^2 = c^2, i.e. a2=b2+c2a^2 = b^2 + c^2. With c≤b≤10c \le b \le 10: a2≤100+100=200a^2 \le 100 + 100 = 200, so a≤200≈14.14a \le \sqrt{200} \approx 14.14. a=14a = 14 is possible: b=10b = 10, c=96≈9.8c = \sqrt{96} \approx 9.8 satisfies 10≥b≥c10 \ge b \ge c, and c≠1c \ne 1. The greatest integer value of aa is 1414.

Q51MCQFunctions & Graphs

A function ff maps the set of natural numbers to whole numbers, such that f(xy)=f(x)f(y)+f(x)+f(y)f(xy) = f(x)f(y) + f(x) + f(y) for all xx, yy and f(p)=1f(p) = 1 for every prime number pp. Then, the value of f(160000)f(160000) is
  1. 4095
  2. 8191
  3. 2047
  4. 1023
Answer and solution

Answer: (A) 4095

First, 160000=28×54160000 = 2^8 \times 5^4. Add 1 to both sides of the rule: f(xy)+1=f(x)f(y)+f(x)+f(y)+1=(f(x)+1)(f(y)+1)f(xy) + 1 = f(x)f(y) + f(x) + f(y) + 1 = (f(x) + 1)(f(y) + 1). So g(n)=f(n)+1g(n) = f(n) + 1 satisfies g(xy)=g(x)g(y)g(xy) = g(x)g(y), with g(p)=1+1=2g(p) = 1 + 1 = 2 for every prime pp. Hence g(pk)=2kg(p^k) = 2^k; for example f(p2)=3f(p^2) = 3 and f(p3)=7f(p^3) = 7. g(160000)=g(28)×g(54)=28×24=212=4096g(160000) = g(2^8) \times g(5^4) = 2^8 \times 2^4 = 2^{12} = 4096. So f(160000)=4096−1=4095f(160000) = 4096 - 1 = 4095. Option B (8191=213−18191 = 2^{13} - 1) would need 13 prime factors, but 160000160000 has 8+4=128 + 4 = 12 counted with repetition. Hence, option A (4095).

Q52MCQQuadratic & Polynomial Equations

The roots α\alpha, β\beta of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 0, satisfy 1α2+1β2=15\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = 15. The value of (α3+β3)2(\alpha^3 + \beta^3)^2 is
  1. 16
  2. 4
  3. 1
  4. 9
Answer and solution

Answer: (B) 4

For 3x2+λx−1=03x^2 + \lambda x - 1 = 0: α+β=−λ3\alpha + \beta = -\dfrac{\lambda}{3} and αβ=−13\alpha\beta = -\dfrac{1}{3}. 1α2+1β2=(α+β)2−2αβ(αβ)2=9(λ29+23)=λ2+6\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 9\left(\dfrac{\lambda^2}{9} + \dfrac{2}{3}\right) = \lambda^2 + 6. Setting this equal to 1515 gives λ2=9\lambda^2 = 9, so (α+β)2=λ29=1(\alpha + \beta)^2 = \dfrac{\lambda^2}{9} = 1. Write s=α+βs = \alpha + \beta, so s=±1s = \pm 1. Then α3+β3=s3−3αβ s=s3+s=s(s2+1)=±2\alpha^3 + \beta^3 = s^3 - 3\alpha\beta\,s = s^3 + s = s(s^2 + 1) = \pm 2. So (α3+β3)2=4(\alpha^3 + \beta^3)^2 = 4, whichever sign λ\lambda has. Option D (9) is λ2\lambda^2 and option C (1) is (α+β)2(\alpha + \beta)^2; neither is the quantity asked for. Hence, option B (4).

Q53MCQRatios, Proportions & Partnership

When Rajesh's age was same as the present age of Garima, the ratio of their ages was 3:23 : 2. When Garima's age becomes the same as the present age of Rajesh, the ratio of the ages of Rajesh and Garima will become
  1. 3:23 : 2
  2. 4:34 : 3
  3. 5:45 : 4
  4. 2:12 : 1
Answer and solution

Answer: (C) 5:45 : 4

Let Garima's present age be GG and Rajesh's R=G+xR = G + x, where xx is the age gap. xx years ago, Rajesh was GG (Garima's present age) and Garima was G−xG - x: GG−x=32⇒2G=3G−3x⇒G=3x\dfrac{G}{G - x} = \dfrac{3}{2} \Rightarrow 2G = 3G - 3x \Rightarrow G = 3x, so R=4xR = 4x. In xx years, Garima will be 4x4x (Rajesh's present age) and Rajesh will be 5x5x. Ratio =5x:4x=5:4= 5x : 4x = 5 : 4. Option B (4:34 : 3) is their present ratio, not the future one. Hence, option C (5:45 : 4).

Q54MCQPolygons & Circles

Three circles of equal radii touch (but not cross) each other externally. Two other circles, XX and YY, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of XX is more than that of YY, the ratio of the radii of XX and YY is
  1. 7+43:17 + 4\sqrt{3} : 1
  2. 4+23:14 + 2\sqrt{3} : 1
  3. 4+3:14 + \sqrt{3} : 1
  4. 2+3:12 + \sqrt{3} : 1
Answer and solution

Answer: (A) 7+43:17 + 4\sqrt{3} : 1

Let each of the three equal circles have radius RR. Their centres form an equilateral triangle of side 2R2R, and both X and Y are centred at its centre O. Solution figure for question 54, CAT 2024 Slot 2 In the right triangle below, the side from a centre to a touching point is RR, the hypotenuse to O is R+rR + r, and the angle at the centre is 30∘30^\circ. So cos⁡30∘=RR+r\cos 30^\circ = \dfrac{R}{R + r}, giving R+r=2R3R + r = \dfrac{2R}{\sqrt{3}}. Solution figure for question 54, CAT 2024 Slot 2 Small circle Y: r=2R3−R=(2−3)R3r = \dfrac{2R}{\sqrt{3}} - R = \dfrac{(2 - \sqrt{3})R}{\sqrt{3}}. Large circle X reaches one radius beyond each centre: its radius is 2R3+R=(2+3)R3\dfrac{2R}{\sqrt{3}} + R = \dfrac{(2 + \sqrt{3})R}{\sqrt{3}}. Ratio =2+32−3=(2+3)24−3=7+43= \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{(2 + \sqrt{3})^2}{4 - 3} = 7 + 4\sqrt{3}. This is about 13.9; options B, C and D give only about 7.5, 5.7 and 3.7. Hence, option A (7+43:17 + 4\sqrt{3} : 1).

Q55MCQTriangles & Lines

ABCDABCD is a trapezium in which ABAB is parallel to CDCD. The sides ADAD and BCBC when extended, intersect at point EE. If AB=2AB = 2 cm, CD=1CD = 1 cm, and perimeter of ABCDABCD is 6 cm, then the perimeter, in cm, of △AEB\triangle AEB is
  1. 8
  2. 10
  3. 9
  4. 7
Answer and solution

Answer: (A) 8

Solution figure for question 55, CAT 2024 Slot 2 The perimeter of ABCDABCD is 6 cm, with AB=2AB = 2 and CD=1CD = 1, so AD+BC=3AD + BC = 3. Since CD∥ABCD \parallel AB, triangles EDCEDC and EABEAB are similar, with ratio CD:AB=1:2CD : AB = 1 : 2. So ED=12EAED = \dfrac{1}{2}EA, which makes D the midpoint of EAEA; likewise C is the midpoint of EBEB. Hence EA=2ADEA = 2AD and EB=2BCEB = 2BC. Perimeter of △AEB=EA+EB+AB=2(AD+BC)+2=2×3+2=8\triangle AEB = EA + EB + AB = 2(AD + BC) + 2 = 2 \times 3 + 2 = 8 cm. This holds however the 3 cm is split between ADAD and BCBC, so no other value, such as 7 or 9, is possible. Hence, option A (8).

Q56MCQAverages, Mixtures & Alligations

A company has 40 employees whose names are listed in a certain order. In the year 2022, the average bonus of the first 30 employees was Rs. 40000, of the last 30 employees was Rs. 60000, and of the first 10 and last 10 employees together was Rs. 50000. Next year, the average bonus of the first 10 employees increased by 100%, of the last 10 employees increased by 200% and of the remaining employees was unchanged. Then, the average bonus, in rupees, of all the 40 employees together in the year 2023 was
  1. 95000
  2. 90000
  3. 80000
  4. 85000
Answer and solution

Answer: (A) 95000

Split the 40 employees into four groups of 10 with average bonuses a,b,c,da, b, c, d, in order. First 30: a+b+c3=40000\dfrac{a + b + c}{3} = 40000, so a+b+c=120000a + b + c = 120000. Last 30: b+c+d3=60000\dfrac{b + c + d}{3} = 60000, so b+c+d=180000b + c + d = 180000. First and last 10: a+d2=50000\dfrac{a + d}{2} = 50000, so a+d=100000a + d = 100000. Adding the first two: a+d+2(b+c)=300000a + d + 2(b + c) = 300000, so b+c=100000b + c = 100000. Then a=20000a = 20000 and d=80000d = 80000. In 2023, aa doubles to 40000 and dd triples to 240000, while b+cb + c stays 100000. Average of all 40 =40000+100000+2400004=3800004=95000= \dfrac{40000 + 100000 + 240000}{4} = \dfrac{380000}{4} = 95000. Swapping the two increases (tripling aa, doubling dd) would give 80000, option C. Hence, option A (95000).

Q57TITATime & Work

Amal and Vimal together can complete a task in 150 days, while Vimal and Sunil together can complete the same task in 100 days. Amal starts working on the task and works for 75 days, then Vimal takes over and works for 135 days. Finally, Sunil takes over and completes the remaining task in 45 days. If Amal had started the task alone and worked on all days, Vimal had worked on every second day, and Sunil had worked on every third day, then the number of days required to complete the task would have been

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

Answer and solution

Answer: 139

Let Amal, Vimal and Sunil do aa, vv and ss units of work per day, and let the task be TT units. 150a+150v=T150a + 150v = T …(1) 100v+100s=T100v + 100s = T …(2) 75a+135v+45s=T75a + 135v + 45s = T …(3) Adding (1) and (2): 150a+250v+100s=2T150a + 250v + 100s = 2T. Doubling (3): 150a+270v+90s=2T150a + 270v + 90s = 2T. Subtracting: 10s=20v10s = 20v, so s=2vs = 2v. From (2), T=100v+200v=300vT = 100v + 200v = 300v; from (1), 150a=150v150a = 150v, so a=va = v. New plan: Amal works every day, Vimal every second day and Sunil every third day. In each 6-day block, Amal works 6 days (6v6v), Vimal 3 days (3v3v) and Sunil 2 days (2×2v=4v2 \times 2v = 4v), a total of 13v13v. 23 blocks take 23×6=13823 \times 6 = 138 days and complete 299v299v. The last vv is done by Amal on day 139. The answer is 139.

Q58MCQInequalities & Modulus

All the values of xx satisfying the inequality 1x+5≤12x−3\dfrac{1}{x+5} \leq \dfrac{1}{2x-3} are
  1. x<−5x < -5 or 32<x≤8\dfrac{3}{2} < x \leq 8
  2. −5<x<32-5 < x < \dfrac{3}{2} or x>32x > \dfrac{3}{2}
  3. x<−5x < -5 or x>32x > \dfrac{3}{2}
  4. −5<x<32-5 < x < \dfrac{3}{2} or 32<x≤8\dfrac{3}{2} < x \leq 8
Answer and solution

Answer: (A) x<−5x < -5 or 32<x≤8\dfrac{3}{2} < x \leq 8

The critical points are x=−5x = -5 and x=32x = \dfrac{3}{2}, where a denominator is zero. For x>32x > \dfrac{3}{2}: both denominators are positive, so 1x+5≤12x−3\dfrac{1}{x+5} \le \dfrac{1}{2x-3} becomes 2x−3≤x+52x - 3 \le x + 5, i.e. x≤8x \le 8. Valid: 32<x≤8\dfrac{3}{2} < x \le 8. For −5<x<32-5 < x < \dfrac{3}{2}: the left side is positive and the right side negative, so the inequality never holds. For x<−5x < -5: both denominators are negative, so their product is positive and again 2x−3≤x+52x - 3 \le x + 5, i.e. x≤8x \le 8, which every x<−5x < -5 satisfies. So x<−5x < -5 or 32<x≤8\dfrac{3}{2} < x \le 8. Option C drops the upper limit 8, but x=13x = 13 gives 118≤123\dfrac{1}{18} \le \dfrac{1}{23}, which is false. Hence, option A (x<−5x < -5 or 32<x≤8\dfrac{3}{2} < x \leq 8).

Q59MCQSimple & Compound Interest

Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
  1. 20860
  2. 20640
  3. 20480
  4. 20808
Answer and solution

Answer: (D) 20808

Anil: 4% a year compounded half-yearly is 2% per half-year, for 12 half-years. He receives 22000×(1.02)1222000 \times (1.02)^{12}. Sunil invests XX. After 5 years (10 half-years) he has X(1.02)10X(1.02)^{10}; one more year at 10% simple interest multiplies this by 1.1. Equating the two amounts: X(1.02)10×1.1=22000×(1.02)12X(1.02)^{10} \times 1.1 = 22000 \times (1.02)^{12} X=22000×(1.02)21.1=20000×1.0404=20808X = \dfrac{22000 \times (1.02)^2}{1.1} = 20000 \times 1.0404 = 20808 The result is exact, so the nearby values 20860 (option A) and 20640 (option B) do not satisfy the equation. Hence, option D (20808).

Q60MCQTime, Speed & Distance

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is
  1. 7 : 30 pm
  2. 7 : 00 pm
  3. 9 : 00 pm
  4. 10 : 30 pm
Answer and solution

Answer: (A) 7 : 30 pm

Let the scheduled travel time be tt hours and the route length DD km. Day 1: at 60 km/h the bus takes t+3.5t + 3.5 hours, so D=60(t+3.5)D = 60(t + 3.5). Day 2: it covers 23D\dfrac{2}{3}D in t3\dfrac{t}{3} hours, so to arrive on time it covers the remaining D3\dfrac{D}{3} in 2t3\dfrac{2t}{3} hours at 40 km/h. So D3=40×2t3\dfrac{D}{3} = 40 \times \dfrac{2t}{3}, giving D=80tD = 80t. Equating: 60t+210=80t60t + 210 = 80t, so t=10.5t = 10.5 hours. Starting at 9 am, the bus is scheduled to arrive 10.5 hours later, at 7:30 pm. Option B (7 : 00 pm) would mean t=10t = 10, but then 60×13.5=81060 \times 13.5 = 810 km does not equal 80×10=80080 \times 10 = 800 km. Hence, option A (7 : 30 pm).

Q61MCQIndices & Surds

If mm and nn are natural numbers such that n>1n > 1, and mn=225×340m^n = 2^{25} \times 3^{40}, then m−nm - n equals
  1. 209932
  2. 209937
  3. 209942
  4. 209947
Answer and solution

Answer: (D) 209947

Since mn=225×340m^n = 2^{25} \times 3^{40} with mm a natural number, nn must divide both exponents, 25 and 40. Their only common factor greater than 1 is 5, so n=5n = 5. Then m5=(25)5×(38)5m^5 = (2^5)^5 \times (3^8)^5, so m=25×38=32×6561=209952m = 2^5 \times 3^8 = 32 \times 6561 = 209952. m−n=209952−5=209947m - n = 209952 - 5 = 209947. Option C (209942) would need n=10n = 10, but 10 does not divide 25; the other options need values of nn that are just as impossible. Hence, option D (209947).

Q62MCQRemainders

When 33333^{333} is divided by 11, the remainder is
  1. 5
  2. 10
  3. 1
  4. 6
Answer and solution

Answer: (A) 5

Look for a power of 3 that leaves remainder 1 when divided by 11. 35=243=11×22+13^5 = 243 = 11 \times 22 + 1, so 353^5 leaves remainder 1. Write 333=5×66+3333 = 5 \times 66 + 3: 3333=(35)66×333^{333} = (3^5)^{66} \times 3^3 (35)66(3^5)^{66} leaves remainder 166=11^{66} = 1, and 33=27=11×2+53^3 = 27 = 11 \times 2 + 5 leaves remainder 5. So 33333^{333} leaves remainder 1×5=51 \times 5 = 5. Option C (1) would be the remainder only if 333 were a multiple of 5; the leftover 333^3 changes it to 5. Hence, option A (5).

Q63TITAQuadratic & Polynomial Equations

If xx and yy are real numbers such that 4x2+4y2−4xy−6y+3=04x^2 + 4y^2 - 4xy - 6y + 3 = 0, then the value of (4x+5y)(4x + 5y) is

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Answer and solution

Answer: 7

Complete the squares. The terms 4x2−4xy4x^2 - 4xy need a y2y^2 to form (2x−y)2=4x2−4xy+y2(2x - y)^2 = 4x^2 - 4xy + y^2. Taking that y2y^2 from 4y24y^2 leaves 3y23y^2: 4x2+4y2−4xy−6y+3=(2x−y)2+3y2−6y+3=(2x−y)2+3(y−1)24x^2 + 4y^2 - 4xy - 6y + 3 = (2x - y)^2 + 3y^2 - 6y + 3 = (2x - y)^2 + 3(y - 1)^2 So (2x−y)2+3(y−1)2=0(2x - y)^2 + 3(y - 1)^2 = 0. A sum of squares of real numbers is zero only if each square is zero. So y−1=0y - 1 = 0, giving y=1y = 1, and 2x−y=02x - y = 0, giving x=12x = \dfrac{1}{2}. Then 4x+5y=4×12+5×1=2+5=74x + 5y = 4 \times \dfrac{1}{2} + 5 \times 1 = 2 + 5 = 7. The answer is 7.

Q64TITAIndices & Surds

If x+62−x−62=22\sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}} = 2\sqrt{2}, then xx equals

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Answer and solution

Answer: 11

Let u=x+62u = \sqrt{x + 6\sqrt{2}} and v=x−62v = \sqrt{x - 6\sqrt{2}}, so u−v=22u - v = 2\sqrt{2}. Then u2−v2=(x+62)−(x−62)=122u^2 - v^2 = (x + 6\sqrt{2}) - (x - 6\sqrt{2}) = 12\sqrt{2}. Since u2−v2=(u+v)(u−v)u^2 - v^2 = (u + v)(u - v), we get (u+v)×22=122(u + v) \times 2\sqrt{2} = 12\sqrt{2}, so u+v=6u + v = 6. From u+v=6u + v = 6 and u−v=22u - v = 2\sqrt{2}: u=3+2u = 3 + \sqrt{2} and v=3−2v = 3 - \sqrt{2}, both positive, as square roots must be. Then x=u2−62=(9+62+2)−62=11x = u^2 - 6\sqrt{2} = (9 + 6\sqrt{2} + 2) - 6\sqrt{2} = 11. Check: 11−62=(3−2)211 - 6\sqrt{2} = (3 - \sqrt{2})^2, so 11+62−11−62=(3+2)−(3−2)=22\sqrt{11 + 6\sqrt{2}} - \sqrt{11 - 6\sqrt{2}} = (3 + \sqrt{2}) - (3 - \sqrt{2}) = 2\sqrt{2}. The answer is 11.

Q65TITAPermutations & Combinations

PP, QQ, RR and SS are four towns. One can travel between PP and QQ along 3 direct paths, between QQ and SS along 4 direct paths, and between PP and RR along 4 direct paths. There is no direct path between PP and SS, while there are few direct paths between QQ and RR, and between RR and SS. One can travel from PP to SS either via QQ, or via RR, or via QQ followed by RR, respectively, in exactly 62 possible ways. One can also travel from QQ to RR either directly, or via PP, or via SS, in exactly 27 possible ways. Then, the number of direct paths between QQ and RR is

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Answer and solution

Answer: 7

Solution figure for question 65, CAT 2024 Slot 2 Let bb be the number of direct paths between Q and R, and aa the number between R and S. P to S in 62 ways: via Q, 3×4=123 \times 4 = 12; via R, 4a4a; via Q then R, 3×b×a=3ab3 \times b \times a = 3ab. So 12+4a+3ab=6212 + 4a + 3ab = 62, or 4a+3ab=504a + 3ab = 50 …(1) Q to R in 27 ways: directly, bb; via P, 3×4=123 \times 4 = 12; via S, 4a4a. So b+12+4a=27b + 12 + 4a = 27, or 4a+b=154a + b = 15 …(2) Subtracting (2) from (1): 3ab−b=353ab - b = 35, so b(3a−1)=35b(3a - 1) = 35. So bb is a divisor of 35: b=1b = 1: a=12a = 12, but 4a+b=49≠154a + b = 49 \ne 15. b=5b = 5: 3a−1=73a - 1 = 7, so aa is not a whole number. b=7b = 7: 3a−1=53a - 1 = 5, so a=2a = 2, and 4(2)+7=154(2) + 7 = 15 fits (2). b=35b = 35: 3a−1=13a - 1 = 1, so aa is not a whole number. So there are 7 direct paths between Q and R. The answer is 7.

Q66MCQInequalities & Modulus

If xx and yy satisfy the equations ∣x∣+x+y=15|x| + x + y = 15 and x+∣y∣−y=20x + |y| - y = 20, then (x−y)(x - y) equals
  1. 20
  2. 15
  3. 5
  4. 10
Answer and solution

Answer: (B) 15

Split into cases by the signs of xx and yy. x≥0x \ge 0, y≥0y \ge 0: the equations become 2x+y=152x + y = 15 and x=20x = 20, giving y=−25y = -25, which contradicts y≥0y \ge 0. x<0x < 0, y≥0y \ge 0: they become y=15y = 15 and x=20x = 20, contradicting x<0x < 0. x<0x < 0, y<0y < 0: the first becomes y=15y = 15, contradicting y<0y < 0. x≥0x \ge 0, y<0y < 0: they become 2x+y=152x + y = 15 and x−2y=20x - 2y = 20. From the first, y=15−2xy = 15 - 2x; substituting, x−30+4x=20x - 30 + 4x = 20, so x=10x = 10 and y=−5y = -5, which fits. So x−y=10−(−5)=15x - y = 10 - (-5) = 15. Option C (5) is x+yx + y and option D (10) is xx alone; neither is x−yx - y. Hence, option B (15).

Q67MCQAverages, Mixtures & Alligations

A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution was
  1. 5:35 : 3
  2. 3:53 : 5
  3. 5:45 : 4
  4. 4:54 : 5
Answer and solution

Answer: (B) 3:53 : 5

After 2 litres of water are added, the solution is 50% acid; let it hold TT litres of acid and TT litres of water. Adding 15 litres of acid makes it 80% acid: T+152T+15=45\dfrac{T + 15}{2T + 15} = \dfrac{4}{5} 5T+75=8T+605T + 75 = 8T + 60, so T=5T = 5. So after the water was added there were 5 litres of acid and 5 litres of water. Before the 2 litres of water were added, there were 5 litres of acid and 3 litres of water. Water : acid =3:5= 3 : 5. Option A (5:35 : 3) is the acid-to-water ratio, the reverse of what is asked. Hence, option B (3:53 : 5).

Q68MCQSequences & Series

The sum of the infinite series 15(15−17)+(15)2((15)2−(17)2)+(15)3((15)3−(17)3)+⋯\dfrac{1}{5}\left(\dfrac{1}{5} - \dfrac{1}{7}\right) + \left(\dfrac{1}{5}\right)^2\left(\left(\dfrac{1}{5}\right)^2 - \left(\dfrac{1}{7}\right)^2\right) + \left(\dfrac{1}{5}\right)^3\left(\left(\dfrac{1}{5}\right)^3 - \left(\dfrac{1}{7}\right)^3\right) + \cdots is equal to
  1. 7/816
  2. 5/408
  3. 7/408
  4. 5/816
Answer and solution

Answer: (B) 5/408

The nn-th term is (15)n((15)n−(17)n)=(125)n−(135)n\left(\dfrac{1}{5}\right)^n\left(\left(\dfrac{1}{5}\right)^n - \left(\dfrac{1}{7}\right)^n\right) = \left(\dfrac{1}{25}\right)^n - \left(\dfrac{1}{35}\right)^n. So the series splits into two infinite geometric series: 125+(125)2+⋯=1/251−1/25=124\dfrac{1}{25} + \left(\dfrac{1}{25}\right)^2 + \cdots = \dfrac{1/25}{1 - 1/25} = \dfrac{1}{24} 135+(135)2+⋯=1/351−1/35=134\dfrac{1}{35} + \left(\dfrac{1}{35}\right)^2 + \cdots = \dfrac{1/35}{1 - 1/35} = \dfrac{1}{34} Sum =124−134=34−24816=10816=5408= \dfrac{1}{24} - \dfrac{1}{34} = \dfrac{34 - 24}{816} = \dfrac{10}{816} = \dfrac{5}{408}. Option D (5816\dfrac{5}{816}) is half of this value: it comes from halving the numerator 1010 without halving the denominator 816816. Hence, option B (5/408).