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CAT 2023 Slot 1 — 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 2023 Slot 1, 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. Free — sign in and you come straight back to this paper.

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Quantitative Ability

CAT 2023 Slot 1 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45MCQProperties of Numbers

Let nn be the least positive integer such that 168168 is a factor of 1134n1134^n. If mm is the least positive integer such that 1134n1134^n is a factor of 168m168^m, then m+nm + n equals
  1. 9
  2. 15
  3. 12
  4. 24
Answer and solution

Answer: (B) 15

Factorise: 1134=2×34×71134 = 2 \times 3^4 \times 7 and 168=23×3×7168 = 2^3 \times 3 \times 7. For 168 to divide 1134n=2n×34n×7n1134^n = 2^n \times 3^{4n} \times 7^n, the power of 2 needs n≥3n \ge 3 (the powers of 3 and 7 are then enough). So n=3n = 3. Then 11343=23×312×731134^3 = 2^3 \times 3^{12} \times 7^3. For this to divide 168m=23m×3m×7m168^m = 2^{3m} \times 3^m \times 7^m, we need 3m≥33m \ge 3, m≥12m \ge 12 and m≥3m \ge 3. The power of 3 decides it: m=12m = 12. So m+n=12+3=15m + n = 12 + 3 = 15. Option C (12) is mm alone; the question asks for m+nm + n. Hence, option B (15).

Q46MCQLogarithms

If xx and yy are positive real numbers such that log⁡x(x2+12)=4\log_x(x^2 + 12) = 4 and 3log⁡yx=13\log_y x = 1, then x+yx + y equals
  1. 20
  2. 68
  3. 10
  4. 11
Answer and solution

Answer: (C) 10

log⁡x(x2+12)=4\log_x(x^2 + 12) = 4 means x4=x2+12x^4 = x^2 + 12, so x4−x2−12=0x^4 - x^2 - 12 = 0. This factorises as (x2−4)(x2+3)=0(x^2 - 4)(x^2 + 3) = 0. Since x2+3>0x^2 + 3 > 0, x2=4x^2 = 4, and as xx is positive, x=2x = 2. 3log⁡yx=13\log_y x = 1 gives log⁡yx=13\log_y x = \frac{1}{3}, so x=y1/3x = y^{1/3} and y=x3=8y = x^3 = 8. So x+y=2+8=10x + y = 2 + 8 = 10. Option B (68) comes from stopping at x2=4x^2 = 4 and taking x=4x = 4, which gives y=64y = 64; but x2=4x^2 = 4 means x=2x = 2. Hence, option C (10).

Q47MCQIndices & Surds

If 5x+9+5x−9=3(2+2)\sqrt{5x+9} + \sqrt{5x-9} = 3(2 + \sqrt{2}), then 10x+9\sqrt{10x+9} is equal to
  1. 3313\sqrt{31}
  2. 454\sqrt{5}
  3. 373\sqrt{7}
  4. 272\sqrt{7}
Answer and solution

Answer: (C) 373\sqrt{7}

Let a=5x+9a = \sqrt{5x + 9} and b=5x−9b = \sqrt{5x - 9}. Then a+b=3(2+2)=6+32a + b = 3(2 + \sqrt{2}) = 6 + 3\sqrt{2} and a2−b2=(5x+9)−(5x−9)=18a^2 - b^2 = (5x + 9) - (5x - 9) = 18. So a−b=186+32=18(6−32)36−18=6−32a - b = \frac{18}{6 + 3\sqrt{2}} = \frac{18(6 - 3\sqrt{2})}{36 - 18} = 6 - 3\sqrt{2}. Adding, 2a=122a = 12, so a=6a = 6 and b=32b = 3\sqrt{2}. 5x+9=365x + 9 = 36 gives 5x=275x = 27, so x=275x = \frac{27}{5}. Check: 5x−9=185x - 9 = 18 and 18=32\sqrt{18} = 3\sqrt{2}. Then 10x+9=54+9=6310x + 9 = 54 + 9 = 63, and 63=37\sqrt{63} = 3\sqrt{7}. Option A (3313\sqrt{31}) is the trap of reading 5x=275x = 27 as x=27x = 27: 270+9=279=331\sqrt{270 + 9} = \sqrt{279} = 3\sqrt{31}. Hence, option C (373\sqrt{7}).

Q48MCQQuadratic & Polynomial Equations

If xx and yy are real numbers such that x2+(x−2y−1)2=−4y(x+y)x^2 + (x - 2y - 1)^2 = -4y(x + y), then the value x−2yx - 2y is
  1. 0
  2. 1
  3. -1
  4. 2
Answer and solution

Answer: (B) 1

Move everything to one side: x2+4xy+4y2+(x−2y−1)2=0x^2 + 4xy + 4y^2 + (x - 2y - 1)^2 = 0. The first three terms are (x+2y)2(x + 2y)^2, so (x+2y)2+(x−2y−1)2=0(x + 2y)^2 + (x - 2y - 1)^2 = 0. A sum of two squares of real numbers is 0 only when both are 0. So x+2y=0x + 2y = 0 and x−2y−1=0x - 2y - 1 = 0, which gives x−2y=1x - 2y = 1. These give x=12x = \frac{1}{2} and y=−14y = -\frac{1}{4}, so such real numbers exist. Option A (0) is the value of x+2yx + 2y, not of x−2yx - 2y. Hence, option B (1).

Q49TITAInequalities & Modulus

The number of integer solutions of equation 2∣x∣(x2+1)=5x22|x|(x^2 + 1) = 5x^2 is

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

Answer and solution

Answer: 3

Consider the sign of xx. x=0x = 0: both sides are 0, so it is a solution. x>0x > 0: ∣x∣=x|x| = x, and dividing by xx gives 2(x2+1)=5x2(x^2 + 1) = 5x, so 2x2−5x+2=02x^2 - 5x + 2 = 0, i.e. (x−2)(2x−1)=0(x - 2)(2x - 1) = 0. So x=2x = 2 or x=12x = \frac{1}{2}; only x=2x = 2 is an integer. x<0x < 0: ∣x∣=−x|x| = -x, and dividing by −x-x gives 2(x2+1)=−5x2(x^2 + 1) = -5x, so 2x2+5x+2=02x^2 + 5x + 2 = 0, i.e. (x+2)(2x+1)=0(x + 2)(2x + 1) = 0. So x=−2x = -2 or x=−12x = -\frac{1}{2}; only x=−2x = -2 is an integer. The integer solutions are −2-2, 0 and 2. The answer is 3.

Q50TITAQuadratic & Polynomial Equations

The equation x3+(2r+1)x2+(4r−1)x+2=0x^3 + (2r + 1)x^2 + (4r - 1)x + 2 = 0 has −2-2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of rr is

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

Answer and solution

Answer: 2

Since −2-2 is a root, (x+2)(x + 2) is a factor. Dividing it out (synthetic division by −2-2): x3+(2r+1)x2+(4r−1)x+2=(x+2)(x2+(2r−1)x+1)x^3 + (2r+1)x^2 + (4r-1)x + 2 = (x + 2)\left(x^2 + (2r - 1)x + 1\right) Check: expanding gives x2x^2-coefficient (2r−1)+2=2r+1(2r - 1) + 2 = 2r + 1 and xx-coefficient 1+2(2r−1)=4r−11 + 2(2r - 1) = 4r - 1. The other two roots are real when the quadratic's discriminant is non-negative: (2r−1)2−4≥0(2r - 1)^2 - 4 \ge 0, so ∣2r−1∣≥2|2r - 1| \ge 2, i.e. r≥32r \ge \frac{3}{2} or r≤−12r \le -\frac{1}{2}. r=0r = 0 and r=1r = 1 give a negative discriminant, so the smallest non-negative integer is r=2r = 2. The answer is 2.

Q51TITAQuadratic & Polynomial Equations

Let α\alpha and β\beta be the two distinct roots of the equation 2x2−6x+k=02x^2 - 6x + k = 0, such that (α+β)(\alpha + \beta) and αβ\alpha\beta are the distinct roots of the equation x2+px+p=0x^2 + px + p = 0. Then, the value of 8(k−p)8(k - p) is

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

Answer and solution

Answer: 6

From 2x2−6x+k=02x^2 - 6x + k = 0: α+β=3\alpha + \beta = 3 and αβ=k2\alpha\beta = \dfrac{k}{2}. These two numbers are the roots of x2+px+p=0x^2 + px + p = 0, so sum: 3+k2=−p3 + \dfrac{k}{2} = -p, and product: 3⋅k2=p3 \cdot \dfrac{k}{2} = p. Adding the two equations: 3+k2+3k2=0⇒3+2k=0⇒k=−323 + \dfrac{k}{2} + \dfrac{3k}{2} = 0 \Rightarrow 3 + 2k = 0 \Rightarrow k = -\dfrac32, and then p=−94p = -\dfrac94. 8(k−p)=8(−32+94)=8×34=68(k - p) = 8\left(-\dfrac32 + \dfrac94\right) = 8 \times \dfrac34 = 6. (Check: 2x2−6x−32=02x^2 - 6x - \tfrac32 = 0 has distinct real roots, and 3≠−343 \ne -\tfrac34.)

Q52MCQAverages, Mixtures & Alligations

A mixture P is formed by removing a certain amount of coffee from a coffee jar and replacing the same amount with cocoa powder. The same amount is again removed from mixture P and replaced with same amount of cocoa powder to form a new mixture Q. If the ratio of coffee and cocoa in the mixture Q is 16:916 : 9, then the ratio of cocoa in mixture P to that in mixture Q is
  1. 1:31 : 3
  2. 1:21 : 2
  3. 5:95 : 9
  4. 4:94 : 9
Answer and solution

Answer: (C) 5:95 : 9

Take 25 units of coffee at the start, so mixture Q has 16 units of coffee and 9 of cocoa. Let xx units be removed and replaced with cocoa each time. Each replacement keeps a fraction 1−x251 - \frac{x}{25} of the coffee, so after two replacements: 25(1−x25)2=1625\left(1 - \frac{x}{25}\right)^2 = 16, so 1−x25=451 - \frac{x}{25} = \frac{4}{5} and x=5x = 5. Mixture P: the first removal takes out 5 units of pure coffee and adds 5 units of cocoa, so P has 5 units of cocoa. Mixture Q has 9 units of cocoa. So the ratio is 5:95 : 9. Option D (4:94 : 9) uses the 4 in 45\frac{4}{5}, which is P's share of coffee, not its cocoa. Hence, option C (5:95 : 9).

Q53MCQClocks

The minor angle between the hours hand and minutes hand of a clock was observed at 8:48 am. The minimum duration, in minutes, after 8:48 am when this angle increases by 50% is
  1. 3611\dfrac{36}{11}
  2. 2
  3. 4
  4. 2411\dfrac{24}{11}
Answer and solution

Answer: (D) 2411\dfrac{24}{11}

At 8:48 the hour hand is 8×30+48×0.5=264∘8 \times 30 + 48 \times 0.5 = 264^\circ from 12, and the minute hand is 48×6=288∘48 \times 6 = 288^\circ. The minor angle is 288∘−264∘=24∘288^\circ - 264^\circ = 24^\circ, with the minute hand ahead. A 50% increase makes it 36∘36^\circ, so the angle must grow by 12∘12^\circ. The minute hand gains on the hour hand at 6−0.5=5.5∘6 - 0.5 = 5.5^\circ per minute, and since it is ahead, the angle starts growing at once. Time needed: 125.5=2411\frac{12}{5.5} = \frac{24}{11} minutes. Option B (2) comes from 126\frac{12}{6}, which ignores the hour hand's movement. Hence, option D (2411\dfrac{24}{11}).

Q54MCQProfit, Loss & Discount

Gita sells two objects A and B at the same price such that she makes a profit of 20% on object A and a loss of 10% on object B. If she increases the selling price such that objects A and B are still sold at an equal price and a profit of 10% is made on object B, then the profit made on object A will be nearest to
  1. 42%
  2. 45%
  3. 47%
  4. 49%
Answer and solution

Answer: (C) 47%

Let the common selling price be pp. A is sold at a 20% profit, so its cost is p1.2=5p6\frac{p}{1.2} = \frac{5p}{6}. B is sold at a 10% loss, so its cost is p0.9=10p9\frac{p}{0.9} = \frac{10p}{9}. The new common price gives a 10% profit on B: 1.1×10p9=11p91.1 \times \frac{10p}{9} = \frac{11p}{9}. Profit on A =11p9−5p6=22p−15p18=7p18= \frac{11p}{9} - \frac{5p}{6} = \frac{22p - 15p}{18} = \frac{7p}{18}. Profit % on A =7p/185p/6×100=715×100≈46.67%= \frac{7p/18}{5p/6} \times 100 = \frac{7}{15} \times 100 \approx 46.67\%. This is nearest to 47%; options B (45%) and D (49%) are each more than 1.6 points away. Hence, option C (47%).

Q55MCQTime, Speed & Distance

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of xx km till then, where xx is a whole number and is palindromic, i.e., xx remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was
  1. 110
  2. 90
  3. 100
  4. 80
Answer and solution

Answer: (C) 100

The trip lasted 8 hours at no more than 110 km/h, so the distance was at most 8×110=8808 \times 110 = 880 km. The starting reading xx was therefore at least 26862−880=2598226862 - 880 = 25982. To make the average speed as large as possible, xx must be the smallest palindrome from 25982 up. Five-digit palindromes starting with 2 have the form 2 a b a 22\,a\,b\,a\,2. With a=5a = 5 the largest is 25952, below 25982. With a=6a = 6 the smallest is 26062. So x=26062x = 26062, the distance is 26862−26062=80026862 - 26062 = 800 km, and the average speed is 8008=100\frac{800}{8} = 100 km/h. Option A (110) would need a starting reading of 25982, which is not a palindrome. Hence, option C (100).

Q56MCQAverages, Mixtures & Alligations

In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is
  1. 21
  2. 20
  3. 22
  4. 19
Answer and solution

Answer: (A) 21

Let each girl score gg and each boy bb. Then 4g+6b=10×24=2404g + 6b = 10 \times 24 = 240, so 6b=240−4g6b = 240 - 4g. The total for 2 girls and 6 boys is 2g+6b=2g+240−4g=240−2g2g + 6b = 2g + 240 - 4g = 240 - 2g. g≥bg \ge b means 6g≥240−4g6g \ge 240 - 4g, so g≥24g \ge 24. g≤2bg \le 2b means 3g≤240−4g3g \le 240 - 4g, so g≤2407≈34.29g \le \frac{240}{7} \approx 34.29. So 240−2g240 - 2g runs from 240−4807≈171.43240 - \frac{480}{7} \approx 171.43 up to 240−48=192240 - 48 = 192, both ends included. The integers in this range are 172 to 192, which is 192−172+1=21192 - 172 + 1 = 21 values. Option B (20) forgets to count both ends; 192 itself is reached when g=b=24g = b = 24. Hence, option A (21).

Q57MCQPercentages

The salaries of three friends Sita, Gita and Mita are initially in the ratio 5:6:75 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is
  1. 25%
  2. 28%
  3. 26%
  4. 30%
Answer and solution

Answer: (C) 26%

Let the starting salaries be 5p5p, 6p6p and 7p7p. After the first year's hikes of 20%, 25% and 20%: Sita =1.2×5p=6p= 1.2 \times 5p = 6p, Gita =1.25×6p=7.5p= 1.25 \times 6p = 7.5p, Mita =1.2×7p=8.4p= 1.2 \times 7p = 8.4p. In the second year, Sita =1.4×6p=8.4p= 1.4 \times 6p = 8.4p and Mita =1.25×8.4p=10.5p= 1.25 \times 8.4p = 10.5p. Let Gita's new salary be gg. It equals the mean of the three: 3g=8.4p+g+10.5p3g = 8.4p + g + 10.5p, so 2g=18.9p2g = 18.9p and g=9.45pg = 9.45p. Gita's hike =9.45p−7.5p7.5p×100=1.957.5×100=26%= \frac{9.45p - 7.5p}{7.5p} \times 100 = \frac{1.95}{7.5} \times 100 = 26\%. Option A (25%) would give Gita 1.25×7.5p=9.375p1.25 \times 7.5p = 9.375p, which is not the mean 9.45p9.45p. Hence, option C (26%).

Q58TITATime, Speed & Distance

Arvind travels from town A to town B, and Surbhi from town B to town A, both starting at the same time along the same route. After meeting each other, Arvind takes 6 hours to reach town B while Surbhi takes 24 hours to reach town A. If Arvind travelled at a speed of 54 km/h, then the distance, in km, between town A and town B is

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

Answer and solution

Answer: 972

Let Arvind's speed be a=54a = 54 km/h and Surbhi's be ss, and let them meet after tt hours. After meeting, Arvind covers in 6 hours the stretch Surbhi covered in tt hours: 6a=st6a = st. Surbhi covers in 24 hours the stretch Arvind covered in tt hours: 24s=at24s = at. Multiplying, 144as=ast2144as = ast^2, so t2=144t^2 = 144 and t=12t = 12. Then s=6at=6×5412=27s = \frac{6a}{t} = \frac{6 \times 54}{12} = 27 km/h. Distance =(a+s)t=(54+27)×12=972= (a + s)t = (54 + 27) \times 12 = 972 km. Check: Arvind's whole trip takes 12+6=1812 + 6 = 18 hours at 54 km/h, which is 972 km. The answer is 972.

Q59TITASimple & Compound Interest

Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received 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 same, then the initial investment made by Sunil, in rupees, is

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

Answer and solution

Answer: 20808

4% a year compounded half-yearly is 2% per half-year. Anil: 12 half-years, so his amount is 22000×(1.02)1222000 \times (1.02)^{12}. Sunil invests SS: 10 half-years, then 10% simple interest for one year, so his amount is S×(1.02)10×1.1S \times (1.02)^{10} \times 1.1. Setting them equal: S×(1.02)10×1.1=22000×(1.02)12S \times (1.02)^{10} \times 1.1 = 22000 \times (1.02)^{12} S=22000×(1.02)21.1=20000×1.0404=20808S = \frac{22000 \times (1.02)^2}{1.1} = 20000 \times 1.0404 = 20808. The answer is 20808.

Q60TITATime & Work

The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is

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

Answer and solution

Answer: 27

The daily amounts of work are in HP, so the times each takes alone are in AP. Let Amal take TT days; Kamal takes 2T2T, so Sunil takes the middle term T+2T2=1.5T\dfrac{T + 2T}{2} = 1.5T. Work done: 4T+91.5T+162T=4+6+8T=18T=1\dfrac{4}{T} + \dfrac{9}{1.5T} + \dfrac{16}{2T} = \dfrac{4 + 6 + 8}{T} = \dfrac{18}{T} = 1, so T=18T = 18. Sunil alone takes 1.5×18=271.5 \times 18 = 27 days.

Q61MCQPolygons & Circles

Let CC be the circle x2+y2+4x−6y−3=0x^2 + y^2 + 4x - 6y - 3 = 0 and LL be the locus of the point of intersection of a pair of tangents to CC with the angle between the two tangents equal to 60°60°. Then, the point at which LL touches the line x=6x = 6 is
  1. (6,6)(6, 6)
  2. (6,3)(6, 3)
  3. (6,8)(6, 8)
  4. (6,4)(6, 4)
Answer and solution

Answer: (B) (6,3)(6, 3)

Write the circle as (x+2)2+(y−3)2=4+9+3=16(x + 2)^2 + (y - 3)^2 = 4 + 9 + 3 = 16: centre (−2,3)(-2, 3), radius 4. Let the tangents meet at (h,k)(h, k), at distance dd from the centre. The line to the centre bisects the 60∘60^\circ angle, so it makes 30∘30^\circ with each tangent. In the right triangle with the radius to the point of contact, sin⁡30∘=4d\sin 30^\circ = \frac{4}{d}, so d=8d = 8. So LL is the circle (h+2)2+(k−3)2=64(h + 2)^2 + (k - 3)^2 = 64, with centre (−2,3)(-2, 3) and radius 8. Its rightmost point has h=−2+8=6h = -2 + 8 = 6, so LL touches the line x=6x = 6 there, level with the centre: k=3k = 3. Check: at h=6h = 6, 64+(k−3)2=6464 + (k - 3)^2 = 64 gives k=3k = 3 only. The other options are not on LL: for (6,6)(6, 6), 64+9≠6464 + 9 \ne 64. Hence, option B ((6,3)(6, 3)).

Q62MCQPolygons & Circles

A quadrilateral ABCDABCD is inscribed in a circle such that AB:CD=2:1AB : CD = 2 : 1 and BC:AD=5:4BC : AD = 5 : 4. If ACAC and BDBD intersect at the point EE, then AE:CEAE : CE equals
  1. 2:12 : 1
  2. 1:21 : 2
  3. 8:58 : 5
  4. 5:85 : 8
Answer and solution

Answer: (C) 8:58 : 5

ABCD is cyclic and its diagonals meet at E. Solution figure for question 62, CAT 2023 Slot 1 Angles in the same segment are equal: ∠DAC=∠DBC\angle DAC = \angle DBC and ∠ADB=∠ACB\angle ADB = \angle ACB. So triangles AED and BEC are similar, with A matching B and D matching C. Hence AEBE=ADBC=45\frac{AE}{BE} = \frac{AD}{BC} = \frac{4}{5}. Likewise ∠BAC=∠BDC\angle BAC = \angle BDC and ∠ABD=∠ACD\angle ABD = \angle ACD, so triangles AEB and DEC are similar, with A matching D and B matching C. Hence BECE=ABDC=21\frac{BE}{CE} = \frac{AB}{DC} = \frac{2}{1}. Multiplying, AECE=45×2=85\frac{AE}{CE} = \frac{4}{5} \times 2 = \frac{8}{5}. Option D (5:85 : 8) is CE:AECE : AE, the same ratio read the wrong way round. Hence, option C (8:58 : 5).

Q63TITATriangles & Lines

In a right-angled triangle ΔABC\Delta ABC, the altitude ABAB is 5 cm, and the base BCBC is 12 cm. PP and QQ are two points on BCBC such that the areas of ΔABP\Delta ABP, ΔABQ\Delta ABQ and ΔABC\Delta ABC are in arithmetic progression. If the area of ΔABC\Delta ABC is 1.5 times the area of ΔABP\Delta ABP, the length of PQPQ, in cm, is

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

Answer and solution

Answer: 2

Given that ABC is a right-angled triangle with AB = 5 and BC = 12 => Area of the triangle = 0.5 * 5 * 12 = 30. Let us assume BP = p, BQ = q => Area of ABP = 0.5 * 5 * p = 2.5p => Area of ABQ = 0.5 * 5 * q = 2.5q Given the area of ABC is 1.5 times that of ABP => 30 = 1.5 * 2.5p => 20 = 2.5p => p = 8. Given Areas of ABP, ABQ and ABC are in A.P. => 2 * 2.5q = 2.5 * 8 + 30 => 5q = 50 => q = 10. PQ = BQ - BP = q - p = 10 - 8 = 2.

Q64MCQSequences & Series

For some positive and distinct real numbers xx, yy and zz, if 1y+z\dfrac{1}{\sqrt{y}+\sqrt{z}} is the arithmetic mean of 1x+z\dfrac{1}{\sqrt{x}+\sqrt{z}} and 1x+y\dfrac{1}{\sqrt{x}+\sqrt{y}}, then the relationship which will always hold true, is
  1. x\sqrt{x}, z\sqrt{z} and y\sqrt{y} are in arithmetic progression
  2. yy, xx and zz are in arithmetic progression
  3. xx, yy and zz are in arithmetic progression
  4. x\sqrt{x}, y\sqrt{y} and z\sqrt{z} are in arithmetic progression
Answer and solution

Answer: (B) yy, xx and zz are in arithmetic progression

Write p=xp = \sqrt{x}, q=yq = \sqrt{y} and r=zr = \sqrt{z}. The condition says 2q+r=1p+r+1p+q=2p+q+r(p+r)(p+q)\dfrac{2}{q + r} = \dfrac{1}{p + r} + \dfrac{1}{p + q} = \dfrac{2p + q + r}{(p + r)(p + q)}. Cross-multiplying: 2(p+r)(p+q)=(q+r)(2p+q+r)2(p + r)(p + q) = (q + r)(2p + q + r). The left side is 2p2+2pq+2pr+2qr2p^2 + 2pq + 2pr + 2qr, and the right side is 2pq+2pr+q2+2qr+r22pq + 2pr + q^2 + 2qr + r^2. Cancelling the common terms leaves 2p2=q2+r22p^2 = q^2 + r^2, i.e. 2x=y+z2x = y + z. So xx is the arithmetic mean of yy and zz: yy, xx and zz are in AP. Options A and D are the traps: the relation holds for xx, yy, zz themselves, not for their square roots. For example, y=1y = 1, x=5x = 5, z=9z = 9 satisfy it, but 11, 5\sqrt{5} and 33 are not in AP in any order. Option C would need 2y=x+z2y = x + z instead. Hence, option B (yy, xx and zz are in arithmetic progression).

Q65MCQPermutations & Combinations

The number of all natural numbers up to 1000 with non-repeating digits is
  1. 504
  2. 648
  3. 738
  4. 585
Answer and solution

Answer: (C) 738

Count the numbers from 1 to 1000 whose digits are all different. 1-digit: 1 to 9, so 9. 2-digit: 9 choices for the first digit (1 to 9), then 9 for the second (0 to 9 except the first): 9×9=819 \times 9 = 81. 3-digit: 9 choices for the first digit, 9 for the second and 8 for the third: 9×9×8=6489 \times 9 \times 8 = 648. 1000 repeats the digit 0, so it is not counted. Total =9+81+648=738= 9 + 81 + 648 = 738. Option B (648) counts only the 3-digit numbers. Hence, option C (738).

Q66TITASequences & Series

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nthn^{th} day exceeds one million, then the lowest possible value of nn is

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

Answer and solution

Answer: 19

Let TnT_n be the count on day nn: T1=2T_1 = 2 and Tn=2Tn−1+3T_n = 2T_{n-1} + 3. Add 3 to both sides: Tn+3=2(Tn−1+3)T_n + 3 = 2(T_{n-1} + 3). So Tn+3T_n + 3 doubles each day, starting from T1+3=5T_1 + 3 = 5: Tn+3=5×2n−1T_n + 3 = 5 \times 2^{n-1}, so Tn=5×2n−1−3T_n = 5 \times 2^{n-1} - 3. Check: T2=10−3=7T_2 = 10 - 3 = 7 and T3=20−3=17T_3 = 20 - 3 = 17, matching 2×2+32 \times 2 + 3 and 2×7+32 \times 7 + 3. We need 5×2n−1−3>10000005 \times 2^{n-1} - 3 > 1000000, i.e. 2n−1>200000.62^{n-1} > 200000.6. 217=1310722^{17} = 131072 is too small and 218=2621442^{18} = 262144 is enough, so n−1=18n - 1 = 18. Check: T18=5×131072−3=655357T_{18} = 5 \times 131072 - 3 = 655357, below a million; T19=5×262144−3=1310717T_{19} = 5 \times 262144 - 3 = 1310717, above it. The answer is 19.