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

Q45TITARatios, Proportions & Partnership

In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is

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Answer and solution

Answer: 43200

Literate males : literate females =2:3=2:3, so literate females =3600×32=5400=3600\times\frac{3}{2}=5400. Now write the illiterate counts in terms of the totals and use their ratio. Let there be 5y5y males and 4y4y females. Then illiterate males =5y−3600=5y-3600 and illiterate females =4y−5400=4y-5400, and 5y−36004y−5400=43\frac{5y-3600}{4y-5400}=\frac{4}{3} 15y−10800=16y−2160015y-10800=16y-21600, so y=10800y=10800. Females =4y=43200=4y=43200. Check: males =54000=54000, so illiterate males =50400=50400 and illiterate females =43200−5400=37800=43200-5400=37800, in the ratio 4:34:3. The answer is 43200.

Q46MCQAverages, Mixtures & Alligations

The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is
  1. 1 : 4
  2. 1 : 2
  3. 4 : 1
  4. 3 : 1
Answer and solution

Answer: (C) 4 : 1

Let there be nn original students with average weight xx kg, and mm new students with average x+3x+3 kg. The new average is x+0.6x+0.6 kg (600 g): nx+m(x+3)n+m=x+0.6\frac{nx+m(x+3)}{n+m}=x+0.6 nx+mx+3m=nx+mx+0.6n+0.6mnx+mx+3m=nx+mx+0.6n+0.6m 2.4m=0.6n2.4m=0.6n, so n=4mn=4m. So n:m=4:1n:m=4:1. Option A (1 : 4) reverses this: the new students are 3 kg heavier yet raise the average by only 0.6 kg, so they must be few. Hence, option C (4 : 1).

Q47MCQSequences & Series

For any natural number nn, suppose the sum of the first nn terms of an arithmetic progression is (n+2n2)(n + 2n^2). If the nthn^{th} term of the progression is divisible by 9, then the smallest possible value of nn is
  1. 9
  2. 4
  3. 7
  4. 8
Answer and solution

Answer: (C) 7

It is given, Sn=2n2+nS_n = 2n^2 + n Sn−1=2(n−1)2+(n−1)=2n2−3n+1S_{n-1} = 2(n-1)^2 + (n-1) = 2n^2 - 3n + 1 Tn=Sn−Sn−1=2n2+n−(2n2−3n+1)=4n−1T_n = S_n - S_{n-1} = 2n^2 + n - (2n^2 - 3n + 1) = 4n - 1 Tn=4n−1T_n = 4n - 1 The terms are 3, 7, 11, 15, 19, 23, 27,...... 27 is the first term in the series divisible by 9. 27 is the 7th term. Therefore, the least possible value of n is 7.

Q48MCQFunctions & Graphs

Let 0≤a≤x≤1000 \leq a \leq x \leq 100 and f(x)=∣x−a∣+∣x−100∣+∣x−a−50∣f(x) = |x - a| + |x - 100| + |x - a - 50|. Then the maximum value of f(x)f(x) becomes 100 when aa is equal to
  1. 25
  2. 100
  3. 50
  4. 0
Answer and solution

Answer: (C) 50

For a≤x≤100a \le x \le 100: ∣x−a∣=x−a|x - a| = x - a and ∣x−100∣=100−x|x - 100| = 100 - x. These add to 100−a100 - a, so f(x)=100−a+∣x−(a+50)∣f(x) = 100 - a + |x - (a + 50)|. As xx runs over [a,100][a, 100], x−(a+50)x - (a + 50) runs from −50-50 to 50−a50 - a. Its largest absolute value is max⁡(50,∣50−a∣)\max(50, |50 - a|), which is 5050 because 0≤a≤1000 \le a \le 100 keeps ∣50−a∣≤50|50 - a| \le 50. It is reached at x=ax = a. So the maximum of ff is 100−a+50=150−a100 - a + 50 = 150 - a. Setting 150−a=100150 - a = 100 gives a=50a = 50. Check: with a=50a = 50, f(50)=0+50+50=100f(50) = 0 + 50 + 50 = 100, and no xx in [50,100][50, 100] gives more. The other options give a different maximum: a=25a = 25 gives 125125 and a=0a = 0 gives 150150. With a=100a = 100, xx can only be 100100, and f(100)=0+0+50=50f(100) = 0 + 0 + 50 = 50. Hence, option C (50).

Q49MCQTime, Speed & Distance

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is
  1. 6
  2. 15
  3. 10
  4. 12
Answer and solution

Answer: (B) 15

Let the trains meet at M, a distance xx from X, on a track XY of length DD. Let the speeds of A and B be aa and bb. Solution figure for question 49, CAT 2022 Slot 1 Train A takes 10 minutes for the whole trip: D=10aD=10a. After meeting, B covers xx in 9 minutes: x=9bx=9b. Up to the meeting both run for the same time: xa=D−xb\frac{x}{a}=\frac{D-x}{b}. Substituting, 9ba=10a−9bb\frac{9b}{a}=\frac{10a-9b}{b}, so 9b2=10a2−9ab9b^2=10a^2-9ab, i.e. 10a2−9ab−9b2=010a^2-9ab-9b^2=0, i.e. (2a−3b)(5a+3b)=0(2a-3b)(5a+3b)=0. Speeds are positive, so a=3b2a=\frac{3b}{2} and D=10a=15bD=10a=15b. So B takes Db=15\frac{D}{b}=15 minutes from Y to X. Option C (10) is train A's time, not B's. Hence, option B (15).

Q50TITAMensuration

A trapezium ABCDABCD has side ADAD parallel to BCBC, ∠BAD=90°\angle BAD = 90°, BC=3BC = 3 cm and AD=8AD = 8 cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is

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Answer and solution

Answer: 66

Since AD∥BCAD\parallel BC and ∠BAD=90°\angle BAD=90°, ABAB is the height; call it yy. Drop CE⊥ADCE\perp AD: AE=BC=3AE=BC=3, so ED=8−3=5ED=8-3=5 and CE=yCE=y. Solution figure for question 50, CAT 2022 Slot 1 In right triangle CEDCED, the legs are yy and 5, so CD=y2+25CD=\sqrt{y^2+25}. The perimeter is 36: 3+8+y+y2+25=363+8+y+\sqrt{y^2+25}=36, so y2+25=25−y\sqrt{y^2+25}=25-y. Squaring, y2+25=625−50y+y2y^2+25=625-50y+y^2, so 50y=60050y=600 and y=12y=12 (then CD=13CD=13, and 3+8+12+13=363+8+12+13=36). Area of the trapezium =12×(sum of parallel sides)×height=12(3+8)×12=66=\frac{1}{2}\times(\text{sum of parallel sides})\times\text{height}=\frac{1}{2}(3+8)\times12=66. The answer is 66.

Q51MCQProfit, Loss & Discount

Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is
  1. 1680
  2. 1176
  3. 2520
  4. 1440
Answer and solution

Answer: (A) 1680

Let 7C=30P=9A=630k7C=30P=9A=630k, where CC, PP, AA are the costs per kg of cashews, peanuts and almonds. Then C=90kC=90k, P=21kP=21k, A=70kA=70k. Cost of 4 kg cashews, 14 kg peanuts and 6 kg almonds =360k+294k+420k=1074k=360k+294k+420k=1074k. The 24 kg mixture is marked to give ₹1752 profit, so its full marked price is M=1074k+1752M=1074k+1752. She sells 4 kg at the marked price and 20 kg at 80% of it: 424M+2024(0.8M)=5M6\frac{4}{24}M+\frac{20}{24}(0.8M)=\frac{5M}{6}. Profit is ₹744: 56(1074k+1752)−1074k=744\frac{5}{6}(1074k+1752)-1074k=744 895k+1460−1074k=744895k+1460-1074k=744, so 179k=716179k=716 and k=4k=4. Almonds cost 420k=1680420k=1680 rupees. Options D (1440) and B (1176) are the amounts spent on cashews (360k360k) and peanuts (294k294k). Hence, option A (1680).

Q52TITAProperties of Numbers

Let AA be the largest positive integer that divides all the numbers of the form 3k+4k+5k3^k + 4^k + 5^k, and BB be the largest positive integer that divides all the numbers of the form 4k+3(4k)+4k+24^k + 3(4^k) + 4^{k+2}, where kk is any positive integer. Then (A+B)(A + B) equals

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Answer and solution

Answer: 82

A number that divides every value must divide the first few values, so start there. AA: at k=1k=1, 3+4+5=123+4+5=12; at k=2k=2, 9+16+25=509+16+25=50. So AA divides gcd⁡(12,50)=2\gcd(12,50)=2. Also 3k3^k and 5k5^k are odd and 4k4^k is even, so every value is even. Hence A=2A=2. BB: 4k+3⋅4k+4k+2=4⋅4k+16⋅4k=20⋅4k=5⋅4k+14^k+3\cdot4^k+4^{k+2}=4\cdot4^k+16\cdot4^k=20\cdot4^k=5\cdot4^{k+1}. At k=1k=1 this is 80, and for every k≥1k\ge1, 5⋅4k+1=80⋅4k−15\cdot4^{k+1}=80\cdot4^{k-1} is a multiple of 80. So B=80B=80. A+B=2+80=82A+B=2+80=82. The answer is 82.

Q53MCQQuadratic & Polynomial Equations

Let aa, bb, cc be non-zero real numbers such that b2<4acb^2 < 4ac, and f(x)=ax2+bx+cf(x) = ax^2 + bx + c. If the set SS consists of all integers mm such that f(m)<0f(m) < 0, then the set SS must necessarily be
  1. the set of all positive integers
  2. the set of all integers
  3. either the empty set or the set of all integers
  4. the empty set
Answer and solution

Answer: (C) either the empty set or the set of all integers

Since b2<4acb^2<4ac, the discriminant b2−4acb^2-4ac is negative, so f(x)=ax2+bx+cf(x)=ax^2+bx+c has no real root and never changes sign. If a>0a>0, f(x)>0f(x)>0 for every xx, so no integer mm has f(m)<0f(m)<0 and SS is empty. If a<0a<0, f(x)<0f(x)<0 for every xx, so every integer is in SS. The sign of aa is not given, and both cases fit b2<4acb^2<4ac (for example b=1b=1 with a=c=1a=c=1, or with a=c=−1a=c=-1). So SS is either empty or all integers. Option D (the empty set) covers only the case a>0a>0. Hence, option C (either the empty set or the set of all integers).

Q54TITAPermutations & Combinations

The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is

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Answer and solution

Answer: 84

No child gets an odd number and each gets some, so each child gets an even number, at least 2. Let the children get 2a2a, 2b2b, 2c2c and 2d2d balloons, with a,b,c,d≥1a,b,c,d\ge1: 2a+2b+2c+2d=202a+2b+2c+2d=20, so a+b+c+d=10a+b+c+d=10. The balloons are identical but the children are not, so we count ordered solutions. The number of ways to write 10 as an ordered sum of 4 positive integers is (10−14−1)=(93)=9⋅8⋅76=84\binom{10-1}{4-1}=\binom{9}{3}=\frac{9\cdot8\cdot7}{6}=84. The answer is 84.

Q55MCQQuadratic & Polynomial Equations

Let aa and bb be natural numbers. If a2+ab+a=14a^2 + ab + a = 14 and b2+ab+b=28b^2 + ab + b = 28, then (2a+b)(2a + b) equals
  1. 8
  2. 7
  3. 10
  4. 9
Answer and solution

Answer: (A) 8

Take out the common factor in each equation: a(a+b+1)=14a(a+b+1)=14 and b(a+b+1)=28b(a+b+1)=28. Both share the factor a+b+1a+b+1, which is positive, so dividing the equations gives ab=1428=12\frac{a}{b}=\frac{14}{28}=\frac{1}{2}, i.e. b=2ab=2a. Substituting in the first: a(3a+1)=14a(3a+1)=14, i.e. 3a2+a−14=03a^2+a-14=0, i.e. (3a+7)(a−2)=0(3a+7)(a-2)=0. aa is a natural number, so a=2a=2 and b=4b=4. Check: 4+8+2=144+8+2=14 and 16+8+4=2816+8+4=28. 2a+b=4+4=82a+b=4+4=8. Option B (7) is a+b+1a+b+1, not 2a+b2a+b. Hence, option A (8).

Q56TITAProfit, Loss & Discount

Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹308.32 per kg and makes an overall profit of 64%. Then, Amal's cost price for syrup, in rupees per kg, is

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Answer and solution

Answer: 160

Let juice cost 10c10c per kg; syrup is 20% cheaper, so 8c8c per kg. Total cost =110×8c+120×10c=880c+1200c=2080c=110\times8c+120\times10c=880c+1200c=2080c. 10 kg syrup at 10% profit sells for 1.1×80c=88c1.1\times80c=88c; 20 kg juice at 20% profit sells for 1.2×200c=240c1.2\times200c=240c. The remaining 100 kg syrup and 100 kg juice, 200 kg, sell at ₹308.32 per kg: 200×308.32=61664200\times308.32=61664. Overall profit is 64%: 61664+88c+240c=1.64×2080c61664+88c+240c=1.64\times2080c 61664+328c=3411.2c61664+328c=3411.2c, so 3083.2c=616643083.2c=61664 and c=20c=20. Syrup costs 8c=1608c=160 rupees per kg. The answer is 160.

Q57MCQPolygons & Circles

All the vertices of a rectangle lie on a circle of radius RR. If the perimeter of the rectangle is PP, then the area of the rectangle is
  1. P216−R2\frac{P^2}{16} - R^2
  2. P28−2R2\frac{P^2}{8} - 2R^2
  3. P22−2PR\frac{P^2}{2} - 2PR
  4. P28−R22\frac{P^2}{8} - \frac{R^2}{2}
Answer and solution

Answer: (B) P28−2R2\frac{P^2}{8} - 2R^2

Let the sides be ll and bb. All four vertices lie on the circle, so each diagonal is a diameter (the figure labels the radius rr): l2+b2=(2R)2=4R2l^2+b^2=(2R)^2=4R^2. Solution figure for question 57, CAT 2022 Slot 1 The perimeter gives 2(l+b)=P2(l+b)=P, so l+b=P2l+b=\frac{P}{2}. Squaring: l2+b2+2lb=P24l^2+b^2+2lb=\frac{P^2}{4} 4R2+2lb=P244R^2+2lb=\frac{P^2}{4} lb=P28−2R2lb=\frac{P^2}{8}-2R^2. This is the area of the rectangle. Option A (P216−R2\frac{P^2}{16}-R^2) is exactly half of it, so it fails. Hence, option B (P28−2R2\frac{P^2}{8} - 2R^2).

Q58MCQAverages, Mixtures & Alligations

The average of three integers is 13. When a natural number nn is included, the average of these four integers remains an odd integer. The minimum possible value of nn is
  1. 3
  2. 4
  3. 5
  4. 1
Answer and solution

Answer: (C) 5

The three integers add to 3×13=393\times13=39. With nn included, the average is 39+n4\frac{39+n}{4}, which must be an odd integer. So 39+n=4k39+n=4k with kk odd. Since n≥1n\ge1, 39+n≥4039+n\ge40, so k≥10k\ge10; the smallest odd such kk is 11. 39+n=4439+n=44, so n=5n=5. Option D (n=1n=1) gives an average of 404=10\frac{40}{4}=10, which is even; options A and B give 424\frac{42}{4} and 434\frac{43}{4}, which are not integers. Hence, option C (5).

Q59MCQAverages, Mixtures & Alligations

A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is
  1. 1 : 4
  2. 1 : 5
  3. 1 : 6
  4. 1 : 7
Answer and solution

Answer: (D) 1 : 7

Take 1 part of the original mixture: it holds 12\frac{1}{2} part lemon juice and 12\frac{1}{2} part sugar syrup. Add 3 parts of sugar syrup. Lemon juice =12=\frac{1}{2} part. Sugar syrup =12+3=72=\frac{1}{2}+3=\frac{7}{2} parts. Ratio =12:72=1:7=\frac{1}{2}:\frac{7}{2}=1:7, so lemon juice is 18\frac{1}{8} of the new mixture. Option C (1 : 6) counts only the added syrup and forgets the half part already in the mixture. Hence, option D (1 : 7).

Q60MCQInequalities & Modulus

The largest real value of aa for which the equation ∣x+a∣+∣x−1∣=2|x + a| + |x - 1| = 2 has an infinite number of solutions for xx is
  1. -1
  2. 0
  3. 1
  4. 2
Answer and solution

Answer: (C) 1

∣x+a∣+∣x−1∣|x+a|+|x-1| is the sum of the distances from xx to −a-a and to 11 on the number line. This sum is at least the distance between −a-a and 11, which is ∣1+a∣|1+a|, and equals it for every xx between them. If ∣1+a∣=2|1+a|=2, every xx between −a-a and 11 is a solution: infinitely many. If ∣1+a∣<2|1+a|<2 there are exactly two solutions, and if ∣1+a∣>2|1+a|>2 there are none. So 1+a=±21+a=\pm2, giving a=1a=1 or a=−3a=-3. The largest is a=1a=1, when every xx from −1-1 to 11 works. Option D (a=2a=2) puts −2-2 and 11 a distance 3 apart, so the sum is never 2. Hence, option C (1).

Q61MCQSet Theory

In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is
  1. 47
  2. 53
  3. 52
  4. 48
Answer and solution

Answer: (A) 47

Let nn, ss, dd, tt be the numbers liking none, exactly one, exactly two and all three drinks. n+s+d+t=100n+s+d+t=100 and s+2d+3t=73+80+52=205s+2d+3t=73+80+52=205. Subtracting, d+2t−n=105d+2t-n=105. Maximum: t≤52t\le52, as only 52 like lemonade. t=52t=52 works with d=1d=1, n=0n=0, s=47s=47 (20 coffee only, 27 tea only, 1 coffee and tea). Minimum: d=100−t−s−n≤100−td=100-t-s-n\le100-t, so 105=d+2t−n≤100+t105=d+2t-n\le100+t, giving t≥5t\ge5. t=5t=5 works with d=95d=95 and s=n=0s=n=0 (48 coffee and tea, 20 coffee and lemonade, 27 tea and lemonade). Difference =52−5=47=52-5=47. Option C (52) is the maximum alone, not the difference. Hence, option A (47).

Q62MCQCoordinate Geometry

Let ABCDABCD be a parallelogram such that the coordinates of its three vertices AA, BB, CC are (1,1)(1, 1), (3,4)(3, 4) and (−2,8)(-2, 8), respectively. Then, the coordinates of the vertex DD are
  1. (0,11)(0, 11)
  2. (4,5)(4, 5)
  3. (−3,4)(-3, 4)
  4. (−4,5)(-4, 5)
Answer and solution

Answer: (D) (−4,5)(-4, 5)

In a parallelogram the diagonals bisect each other, so diagonals AC and BD have the same midpoint. Midpoint of AC =(1+(−2)2,1+82)=(−12,92)= \left(\frac{1 + (-2)}{2}, \frac{1 + 8}{2}\right) = \left(-\frac{1}{2}, \frac{9}{2}\right). Let D=(x,y)D = (x, y). Midpoint of BD =(x+32,y+42)= \left(\frac{x + 3}{2}, \frac{y + 4}{2}\right). Equating: x+3=−1x + 3 = -1, so x=−4x = -4; and y+4=9y + 4 = 9, so y=5y = 5. So D=(−4,5)D = (-4, 5), which is the same as A+C−BA + C - B. The trap (0,11)(0, 11) comes from treating BC as a diagonal, but in ABCD the diagonals are AC and BD. Hence, option D ((−4,5)(-4, 5)).

Q63TITAProperties of Numbers

For natural numbers xx, yy, and zz, if xy+yz=19xy + yz = 19 and yz+xz=51yz + xz = 51, then the minimum possible value of xyzxyz is

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Answer and solution

Answer: 34

It is given, y(x + z) = 19 y cannot be 19. If y = 19, x + z = 1 which is not possible when both x and z are natural numbers. Therefore, y = 1 and x + z = 19 It is given, z(x + y) = 51 z can take values 3 and 17 Case 1: If z = 3, y = 1 and x = 16 xyz = 3*1*16 = 48 Case 2: If z = 17, y = 1 and x = 2 xyz = 17*1*2 = 34 Minimum value xyz can take is 34.

Q64MCQSimple & Compound Interest

Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is
  1. 37.5%
  2. 62.5%
  3. 60%
  4. 40%
Answer and solution

Answer: (A) 37.5%

Let the two parts be xx and yy. The simple interest is equal: x×15×4100=y×12×3100\frac{x\times15\times4}{100}=\frac{y\times12\times3}{100} 60x=36y60x=36y, so 5x=3y5x=3y and x:y=3:5x:y=3:5. So the savings split into 8 equal shares, 3 in the first part. First part =33+5×100%=37.5%=\frac{3}{3+5}\times100\%=37.5\%. This makes sense: the first part earns at a higher rate for longer, so it must be the smaller part. Option B (62.5%) is the second part's share. Hence, option A (37.5%).

Q65TITARatios, Proportions & Partnership

Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is

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Answer and solution

Answer: 111

The ratio 3:53:5 means the counts are 3a3a ahead of Pinky and 5a5a behind her, for some whole number aa. Including Pinky, the queue has 3a+5a+1=8a+13a+5a+1=8a+1 people. The queue has fewer than 300 people, so 8a+1<3008a+1<300, giving 8a<2998a<299 and a<37.375a<37.375. The largest whole value is a=37a=37. Then 3a=3×37=1113a=3\times37=111 people stand ahead of her (with 185 behind, 297 in all). The answer is 111.

Q66TITAProperties of Numbers

For any real number xx, let [x][x] be the largest integer less than or equal to xx. If ∑n=1N[15+n25]=25\sum_{n=1}^{N}\left[\frac{1}{5} + \frac{n}{25}\right] = 25, then NN is

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Answer and solution

Answer: 44

[15+n25]=[n+525]\left[\frac{1}{5}+\frac{n}{25}\right]=\left[\frac{n+5}{25}\right]. For n=1n=1 to 1919, n+5n+5 runs from 6 to 24, so each term is 0. For n=20n=20 to 4444, n+5n+5 runs from 25 to 49, so each term is 1. At n=45n=45, n+5=50n+5=50 and the term becomes 2. So for 20≤N≤4420\le N\le44 the sum counts the terms from n=20n=20 to n=Nn=N, which is N−19N-19. Setting N−19=25N-19=25 gives N=44N=44. Going on to N=45N=45 would make the sum 27. The answer is 44.