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CAT 2021 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 2021 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. 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 2021 Slot 1 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Q45MCQTime, Speed & Distance

Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is
  1. 184
  2. 192
  3. 190
  4. 180
Answer and solution

Answer: (C) 190

The faster train passes a lamp post by covering its own length, so its speed is 16012=403\frac{160}{12} = \frac{40}{3} m/s. 6 km/hr is 6×518=536 \times \frac{5}{18} = \frac{5}{3} m/s, so the slower train runs at 403−53=353\frac{40}{3} - \frac{5}{3} = \frac{35}{3} m/s. In opposite directions, the relative speed is 403+353=25\frac{40}{3} + \frac{35}{3} = 25 m/s. To cross each other, the trains cover the sum of their lengths: 160+x=25×14=350160 + x = 25 \times 14 = 350, so x=190x = 190 m. Each other option would need a different combined length (180 gives 340 m, 184 gives 344 m, 192 gives 352 m), so none fits the 14 seconds. Hence, option C (190).

Q46MCQPolygons & Circles

If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is
  1. 464\sqrt{6}
  2. 666\sqrt{6}
  3. 6\sqrt{6}
  4. 262\sqrt{6}
Answer and solution

Answer: (D) 262\sqrt{6}

Let the side of the hexagon be xx. A regular hexagon is six equilateral triangles of side xx, so its area is 6×34x2=332x26 \times \frac{\sqrt{3}}{4}x^2 = \frac{3\sqrt{3}}{2}x^2. The equilateral triangle of side 12 has area 34×122=363\frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3}. Equating the areas: 332x2=363\frac{3\sqrt{3}}{2}x^2 = 36\sqrt{3}, so x2=24x^2 = 24 and x=24=26x = \sqrt{24} = 2\sqrt{6}. Option A (464\sqrt{6}) would give x2=96x^2 = 96, a hexagon four times the triangle's area; options B and C give x2=216x^2 = 216 and x2=6x^2 = 6. Hence, option D (262\sqrt{6}).

Q47MCQPolygons & Circles

Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is
  1. 13\sqrt{13}
  2. 14\sqrt{14}
  3. 12\sqrt{12}
  4. 15\sqrt{15}
Answer and solution

Answer: (A) 13\sqrt{13}

A regular hexagon splits into six equilateral triangles of side 2 cm that meet at the centre. So the long diagonal AD passes through the centre and AD=2+2=4AD = 2 + 2 = 4 cm. Solution figure for question 47, CAT 2021 Slot 1 AD is an axis of symmetry, so it bisects the 120∘120^\circ interior angle at D. Hence ∠ADC=60∘\angle ADC = 60^\circ. T is the midpoint of CD, so DT=1DT = 1 cm. Solution figure for question 47, CAT 2021 Slot 1 In triangle ADT, by the cosine rule: AT2=AD2+DT2−2⋅AD⋅DTcos⁡60∘=16+1−2⋅4⋅1⋅12=13AT^2 = AD^2 + DT^2 - 2 \cdot AD \cdot DT \cos 60^\circ = 16 + 1 - 2 \cdot 4 \cdot 1 \cdot \frac{1}{2} = 13 So AT=13AT = \sqrt{13} cm. Option C (12\sqrt{12}) is the trap: using C instead of T, that is DT=2DT = 2, gives AC2=16+4−8=12AC^2 = 16 + 4 - 8 = 12. Hence, option A (13\sqrt{13}).

Q48MCQTime & Work

Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is
  1. 5
  2. 8
  3. 6
  4. 7
Answer and solution

Answer: (D) 7

Let the total amount of work be 60 units. Then Anu, Vinu, and Manu do 4, 5, and 3 units of work per day respectively. On the 1st day, Anu and Vinu work. Work done on the 1st day = 9 units On the 2nd day, Manu and Vinu work. Work done on the 2nd day = 8 units This cycle goes on. And in 6 days, the work completed is 9+8+9+8+9+8 = 51 units. On the 7th day, again Anu and Vinu work and complete the remaining 9 units of work. Thus, the number of days taken is 7 days.

Q49MCQLinear Equations

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is
  1. 3:2
  2. 11:7
  3. 11:3
  4. 7:3
Answer and solution

Answer: (C) 11:3

Let Neeta, Geeta and Sita earn nn, gg and ss a day. Given: n+g=6sn + g = 6s (i) and s+n=2gs + n = 2g (ii). Subtracting (i) from (ii): s−g=2g−6ss - g = 2g - 6s, so 7s=3g7s = 3g. Take g=7ag = 7a and s=3as = 3a. From (i), n=6s−g=18a−7a=11an = 6s - g = 18a - 7a = 11a. Check (ii): s+n=14a=2gs + n = 14a = 2g. So Neeta earns the most (11a11a) and Sita the least (3a3a), and the ratio is 11:311 : 3. Option D (7:37 : 3) is the trap: it compares Geeta with Sita, but Geeta is not the top earner. Hence, option C (11:3).

Q50MCQAverages, Mixtures & Alligations

Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to
  1. 26
  2. 18
  3. 16
  4. 20
Answer and solution

Answer: (B) 18

Take the fixed amount as ₹100 for each of the first three months, so ₹50 for each of the last two. Onion bought: 10010=10\frac{100}{10} = 10, 10020=5\frac{100}{20} = 5, 10025=4\frac{100}{25} = 4, 5025=2\frac{50}{25} = 2 and 5050=1\frac{50}{50} = 1 kg, a total of 22 kg. Money spent: 100×3+50×2=400100 \times 3 + 50 \times 2 = 400. Average expense per kg =40022≈18.18= \frac{400}{22} \approx 18.18, closest to 18. Option A (26) is the trap: it is the simple average of the five prices, 10+20+25+25+505=26\frac{10 + 20 + 25 + 25 + 50}{5} = 26, which ignores that more onion is bought when the price is low. Hence, option B (18).

Q51TITAPermutations & Combinations

The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

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

Answer and solution

Answer: 47

Every group contains 3 and 5, so the other members come from 1, 2, 4, 6, 7 and 8, and at least one is needed to reach three numbers. Split by 7 and 8, which cannot both appear. Neither 7 nor 8: any non-empty subset of {1,2,4,6}\{1, 2, 4, 6\}, so 24−1=152^4 - 1 = 15 groups. 7 but not 8: the group already has three numbers (3, 5, 7), and any subset of {1,2,4,6}\{1, 2, 4, 6\} may be added, so 24=162^4 = 16 groups. 8 but not 7: likewise, 16 groups. Total =15+16+16=47= 15 + 16 + 16 = 47. Check by size: adding kk of the six other numbers and removing the choices with both 7 and 8 gives 6+(15−1)+(20−4)+(15−6)+(6−4)+(1−1)=476 + (15 - 1) + (20 - 4) + (15 - 6) + (6 - 4) + (1 - 1) = 47. The answer is 47.

Q52TITAProfit, Loss & Discount

Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is

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Answer and solution

Answer: 1000

Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n+W, where W refers to the wage. Then SP in the first case = 12× 100+11× (n−100) Given profit is 300 in this case: 1200+11n-1100-8n-W=300 =>3n-W = 200 In second case: 1200+9n-900-8n-W=-300 (Loss). => W-n = 600. Adding the two equations: 2n = 800 n = 400. Thus W = 600 + 400 = 1000

Q53MCQLinear Equations

A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is
  1. 11
  2. 13
  3. 10
  4. 12
Answer and solution

Answer: (B) 13

Let an apple, an orange and a mango cost aa, oo and mm. First two baskets: 2a+4o+6m=a+4o+8m2a + 4o + 6m = a + 4o + 8m, so a=2ma = 2m. Second and third baskets: a+4o+8m=8o+7ma + 4o + 8m = 8o + 7m. With a=2ma = 2m this is 10m−7m=4o10m - 7m = 4o, so 4o=3m4o = 3m. Cost of the first basket: 2a+4o+6m=4m+3m+6m=13m2a + 4o + 6m = 4m + 3m + 6m = 13m. The third basket agrees: 8o+7m=6m+7m=13m8o + 7m = 6m + 7m = 13m. So a basket of 13 mangoes costs the same as each of the other baskets; 12 mangoes would cost only 12m12m, and 11 or 10 even less. Hence, option B (13).

Q54MCQInequalities & Modulus

The number of integers n that satisfy the inequalities ∣n−60∣<∣n−100∣<∣n−20∣ is
  1. 21
  2. 19
  3. 18
  4. 20
Answer and solution

Answer: (B) 19

∣n−a∣|n - a| is the distance from nn to aa on the number line. ∣n−100∣<∣n−20∣|n - 100| < |n - 20|: nn is closer to 100 than to 20, so n>60n > 60, the midpoint of 20 and 100. ∣n−60∣<∣n−100∣|n - 60| < |n - 100|: nn is closer to 60 than to 100, so n<80n < 80, the midpoint of 60 and 100. Both hold for 60<n<8060 < n < 80, that is n=61,62,…,79n = 61, 62, \ldots, 79: 19 integers. The endpoints fail: at n=60n = 60, ∣n−100∣=∣n−20∣=40|n - 100| = |n - 20| = 40, and at n=80n = 80, ∣n−60∣=∣n−100∣=20|n - 60| = |n - 100| = 20. Counting either one gives the trap answer 20 (option D). Hence, option B (19).

Q55TITADigits & Base Systems

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

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Answer and solution

Answer: 70

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits. Then (100c+10b+a)-(100a+10b+c)=198 99c-99a=198 c-a = 2. Now, a can take the values 1-7. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a single-digit number. Thus, there can be 7 cases. B can take the value of any digit from 0-9, as it does not affect the answer. Hence, the total cases will be 7× 10=70.

Q56MCQSimple & Compound Interest

Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to
  1. 929.48
  2. 934.65
  3. 931.72
  4. 926.84
Answer and solution

Answer: (C) 931.72

Let the principal be PP and the rate rr. The interest in any year is rr times the amount at the start of that year. Year 2 interest =Pr(1+r)=806.25= Pr(1 + r) = 806.25 and year 3 interest =Pr(1+r)2=866.72= Pr(1 + r)^2 = 866.72. Dividing: 1+r=866.72806.25=1.0751 + r = \frac{866.72}{806.25} = 1.075, so r=7.5%r = 7.5\%. Each year's interest is 1.075 times the previous year's, so year 4 interest =866.72×1.075≈931.72= 866.72 \times 1.075 \approx 931.72. Adding the same rupee increase again (866.72+60.47=927.19866.72 + 60.47 = 927.19) is the trap: compound interest grows by a fixed ratio, not a fixed amount. Hence, option C (931.72).

Q57TITAAverages, Mixtures & Alligations

Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was

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Answer and solution

Answer: 35

Let hospital A admit xx patients, so hospital B admits x+21x + 21. Average recovery days: A =200x= \frac{200}{x} and B =152x+21= \frac{152}{x + 21}, with 200x−152x+21=3\frac{200}{x} - \frac{152}{x + 21} = 3. Multiplying by x(x+21)x(x + 21): 200(x+21)−152x=3x(x+21)200(x + 21) - 152x = 3x(x + 21), so 48x+4200=3x2+63x48x + 4200 = 3x^2 + 63x. This gives 3x2+15x−4200=03x^2 + 15x - 4200 = 0, or x2+5x−1400=0x^2 + 5x - 1400 = 0, so (x+40)(x−35)=0(x + 40)(x - 35) = 0. As x>0x > 0, x=35x = 35. Check: A's average is 20035≈5.71\frac{200}{35} \approx 5.71 days and B's is 15256≈2.71\frac{152}{56} \approx 2.71 days, which differ by 3. The answer is 35.

Q58TITATriangles & Lines

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC=90∘\angle ABC = 90^\circ. If BC = 10 inches then the area of the triangle in square inches is

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Answer and solution

Answer: 120

The incircle has diameter 88, so its radius is r=4r = 4 inches. Solution figure for question 58, CAT 2021 Slot 1 For a triangle right-angled at B, the two tangent lengths from B both equal rr, which gives r=AB+BC−AC2r = \dfrac{AB + BC - AC}{2}. So AB+10−AC=8AB + 10 - AC = 8, i.e. AC=AB+2AC = AB + 2. By Pythagoras, AB2+102=AC2=(AB+2)2=AB2+4AB+4AB^2 + 10^2 = AC^2 = (AB + 2)^2 = AB^2 + 4AB + 4. So 4AB=964AB = 96, giving AB=24AB = 24 inches and AC=26AC = 26 inches. Area =12×AB×BC=12×24×10=120= \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times 24 \times 10 = 120 square inches. Check: the semi-perimeter is 24+10+262=30\dfrac{24 + 10 + 26}{2} = 30, and r=area30=12030=4r = \dfrac{\text{area}}{30} = \dfrac{120}{30} = 4, as required. The answer is 120.

Q59MCQInequalities & Modulus

f(x)=x2+2x−15x2−7x−18f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18} is negative if and only if
  1. −5<x<−2-5 < x < -2 or 3<x<93 < x < 9
  2. x<−5x < -5 or −2<x<3-2 < x < 3
  3. −2<x<3-2 < x < 3 or x>9x > 9
  4. x<−5x < -5 or 3<x<93 < x < 9
Answer and solution

Answer: (A) −5<x<−2-5 < x < -2 or 3<x<93 < x < 9

Factorise: f(x)=(x+5)(x−3)(x−9)(x+2)f(x) = \frac{(x + 5)(x - 3)}{(x - 9)(x + 2)}. The sign can change only at −5-5, −2-2, 33 and 99, and x=−2x = -2 and x=9x = 9 are excluded. x>9x > 9: all four factors are positive, so f(x)>0f(x) > 0. 3<x<93 < x < 9: only x−9x - 9 is negative, so f(x)<0f(x) < 0. −2<x<3-2 < x < 3: x−3x - 3 and x−9x - 9 are negative, so f(x)>0f(x) > 0. −5<x<−2-5 < x < -2: three factors are negative, so f(x)<0f(x) < 0. x<−5x < -5: all four are negative, so f(x)>0f(x) > 0. So f(x)<0f(x) < 0 exactly when −5<x<−2-5 < x < -2 or 3<x<93 < x < 9. Option D fails because for x<−5x < -5 the four negative factors make f(x)f(x) positive. Hence, option A (−5<x<−2-5 < x < -2 or 3<x<93 < x < 9).

Q60TITALogarithms

If 5−log⁡101+x+4log⁡101−x=log⁡1011−x25 - \log_{10}\sqrt{1+x} + 4\log_{10}\sqrt{1-x} = \log_{10}\dfrac{1}{\sqrt{1-x^2}}, then 100x100x equals

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

Answer and solution

Answer: 99

Write u=log⁡101+xu = \log_{10}\sqrt{1 + x} and v=log⁡101−xv = \log_{10}\sqrt{1 - x}. The right side is log⁡1011−x2=−log⁡10(1+x 1−x)=−u−v\log_{10}\frac{1}{\sqrt{1 - x^2}} = -\log_{10}\left(\sqrt{1 + x}\,\sqrt{1 - x}\right) = -u - v. So the equation is 5−u+4v=−u−v5 - u + 4v = -u - v, which gives 5=−5v5 = -5v, so v=−1v = -1. Then 1−x=10−1=110\sqrt{1 - x} = 10^{-1} = \frac{1}{10}, so 1−x=11001 - x = \frac{1}{100} and x=99100x = \frac{99}{100}. This keeps 1+x1 + x and 1−x1 - x positive, so every logarithm is defined, and 100x=99100x = 99. The answer is 99.

Q61TITATime & Work

Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

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

Answer and solution

Answer: 32

Take the project as 48 units (the LCM of 12, 16 and 24 months). Let Amar, Akbar and Anthony do mm, kk and nn units a month. Amar and Akbar: m+k=4812=4m + k = \frac{48}{12} = 4. Akbar and Anthony: k+n=4816=3k + n = \frac{48}{16} = 3. Anthony and Amar: m+n=4824=2m + n = \frac{48}{24} = 2. Adding: 2(m+k+n)=92(m + k + n) = 9, so m+k+n=4.5m + k + n = 4.5. Then m=4.5−3=1.5m = 4.5 - 3 = 1.5, k=4.5−2=2.5k = 4.5 - 2 = 2.5 and n=4.5−4=0.5n = 4.5 - 4 = 0.5. Akbar is the fastest and Anthony the slowest, so Amar is neither. Alone he needs 481.5=32\frac{48}{1.5} = 32 months. The answer is 32.

Q62MCQSequences & Series

If x0=1x_0 = 1, x1=2x_1 = 2, and xn+2=1+xn+1xnx_{n+2} = \frac{1 + x_{n+1}}{x_n}, n=0,1,2,3,…n = 0, 1, 2, 3, \ldots, then x2021x_{2021} is equal to
  1. 4
  2. 1
  3. 3
  4. 2
Answer and solution

Answer: (D) 2

Compute terms with xn+2=1+xn+1xnx_{n+2} = \frac{1 + x_{n+1}}{x_n}: x2=1+21=3x_2 = \frac{1 + 2}{1} = 3, x3=1+32=2x_3 = \frac{1 + 3}{2} = 2, x4=1+23=1x_4 = \frac{1 + 2}{3} = 1, x5=1+12=1x_5 = \frac{1 + 1}{2} = 1, x6=1+11=2x_6 = \frac{1 + 1}{1} = 2. Since x5=x0x_5 = x_0 and x6=x1x_6 = x_1, and each term depends only on the two before it, the sequence repeats every 5 terms: 1,2,3,2,1,…1, 2, 3, 2, 1, \ldots 2021=5×404+12021 = 5 \times 404 + 1, so x2021=x1=2x_{2021} = x_1 = 2. Option B (1) is the off-by-one trap: it is x2020=x0x_{2020} = x_0. The value 4 never occurs. Hence, option D (2).

Q63TITAAverages, Mixtures & Alligations

The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

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

Answer and solution

Answer: 200

Bottle 1 holds 800 cc at 33% and bottle 2 holds 800 cc at 17%. Some solution from bottle 1 is replaced by the same volume from bottle 2, so bottle 1 still holds 800 cc, now at 21%. By alligation, the 33% and 17% solutions are in the ratio (21−17):(33−21)=4:12=1:3(21 - 17) : (33 - 21) = 4 : 12 = 1 : 3. So of the 800 cc in bottle 1, 14\frac{1}{4} (200 cc) is the original solution and 34\frac{3}{4} (600 cc) came from bottle 2. Check: 200×33+600×17800=6600+10200800=21\frac{200 \times 33 + 600 \times 17}{800} = \frac{6600 + 10200}{800} = 21, as required. Bottle 2 gave away 600 cc of its 800 cc, so 800−600=200800 - 600 = 200 cc is left. The answer is 200.

Q64MCQPercentages

Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to
  1. 144
  2. 127
  3. 135
  4. 124
Answer and solution

Answer: (B) 127

Let the small box cost ₹100 and hold SS grams; the large box then costs ₹200 and holds LL grams. The large box's price per gram is 12% less: 200L=0.88×100S\frac{200}{L} = 0.88 \times \frac{100}{S}. So LS=20088=2511\frac{L}{S} = \frac{200}{88} = \frac{25}{11}. The large box holds more by (2511−1)×100=1411×100≈127.3%\left(\frac{25}{11} - 1\right) \times 100 = \frac{14}{11} \times 100 \approx 127.3\%, nearest to 127. Option D (124) is the trap: it treats a 12% lower price per gram as 12% more grams per rupee, giving 2×1.12=2.242 \times 1.12 = 2.24, a 124% excess. Hence, option B (127).

Q65MCQSequences & Series

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to
  1. 6119
  2. 6090
  3. 4941
  4. 7471
Answer and solution

Answer: (A) 6119

Group kk has 2k−12k - 1 numbers: 1, 3, 5, and so on. The first kk groups hold 1+3+⋯+(2k−1)=k21 + 3 + \cdots + (2k - 1) = k^2 numbers, so group kk ends at k2k^2. Group 14 ends at 142=19614^2 = 196, so group 15 runs from 197 to 152=22515^2 = 225, which is 2×15−1=292 \times 15 - 1 = 29 numbers. Sum =29×(197+225)2=29×211=6119= \frac{29 \times (197 + 225)}{2} = 29 \times 211 = 6119. Option B (6090) is the off-by-one trap: it is 29×21029 \times 210, the sum of 196 to 224, starting the group at 14214^2 instead of 142+114^2 + 1. Hence, option A (6119).

Q66MCQQuadratic & Polynomial Equations

If rr is a constant such that ∣x2−4x−13∣=r|x^2 - 4x - 13| = r has exactly three distinct real roots, then the value of rr is
  1. 17
  2. 21
  3. 15
  4. 18
Answer and solution

Answer: (A) 17

Let g(x)=x2−4x−13=(x−2)2−17g(x) = x^2 - 4x - 13 = (x - 2)^2 - 17. Its minimum is −17-17 at x=2x = 2, and it is negative between its two roots. Taking ∣g(x)∣|g(x)| reflects that negative part upward, so between the roots the graph rises to a peak of 17 at x=2x = 2. In the figure, C is the minimum of gg and C' is the peak of ∣g∣|g|. Solution figure for question 66, CAT 2021 Slot 1 A horizontal line y=ry = r with r>0r > 0 meets this graph four times if r<17r < 17, three times if r=17r = 17 (twice on the outer arms, once at the peak) and twice if r>17r > 17. Algebraically, g(x)=rg(x) = r gives (x−2)2=17+r(x - 2)^2 = 17 + r, which always has two roots, and g(x)=−rg(x) = -r gives (x−2)2=17−r(x - 2)^2 = 17 - r, which has a single root only when r=17r = 17. So only r=17r = 17 gives exactly three roots. Options B and D (21 and 18) give only 2 roots, and option C (15) gives 4. Hence, option A (17).