quant questions 151-200 and solutions

A Problem Booklet for IIM-CAT 2008

School of Management

Quantitative Aptitude (Volume III)

A Sample Booklet( not for sale)

Rahul Kumar JhaMar-Nov 2008

Problem Set

151) Let a, b, c be distinct such that a + 1/b = b + 1/c = c + 1/a, then abcequals

a) 1 b) 0 c) -1 d) greater than 1 e) Less than -1

152) Find tan 9− tan 27− tan 63 + tan 81 ( all angles in degrees)

a) 2 b) 3 c) 4 d) 5 e) none of these

153) Let tan θ = tan 10− tan 50 + tan 70 then θ is ?( all angles in degress)

a) 15 b) 30 c) 45 d) 60 e) 75

154)9x + 16y = n find n, where n is the greatest number such that bothof (x, y) are not positive?

a)119 b) 135 c) 128 d) 144 e) 169

155) Find the number of triangles formed when all the diagonals of a pen-tagon are joined?


156)Let r be a root of x2 + 5x + 7 = 0. Compute (r− 1)(r + 2)(r + 6)(r + 3)

a) 0 b) 6 c) -6 d) 13 e) -13

157)Find the number of positive solution to 1x2−10x−29

+ 1x2−10x−45

= 2x2−10x−69

a)0 b) 1 c) 2 d) 3 e) 4

158)Find the number of quadratic polynomials ax2 + bx + c such that:

a) a, b, c are distinct.

1

b) a, b, c ∈ {1, 2, 3, ...2008}c) x + 1 divides ax2 + bx + c

a) 2013018 b) 2013021 c) 2014024 d) 2018040 e) noneof these

159) Suppose K be the number of integers n such that 2n+1n2 is also an integer.

Then K isa) 0 b) 1 c) 2 d) 3 e) none of these

160) Let a, b be positive integers such that a1 =√

ab, b1 = a+b2

, a2 =√a1b1, b2 = a1+b1

2... thus is general for all n, an =

√an−1bn−1, bn = an−1+bn−1

2

Let X = |bn − an|, and Y = |bn + an| then

a) X ≤ |b−a|2n+1 b) X ≤ |b−a|

2n c) Y ≤ |b−a|2n+1 d) Y ≤ |b−a|

2n

e) none of these

161) Triangle ABC has base AB of length k and C is such that < CAB =2 < CBA < 120◦. Then the locus of C is :

a) Straight line b) circle d) ellipse d) Parabola e) noneof these

162) For how many primes p is p2 + 3p− 1 also prime?


163) There are two ants on opposite corners of a cube. On each move,they can travel along an edge to an adjacent vertex. If the probability thatthey both return to their starting position after 4 moves is m

n, where m and

n are relatively prime integers, find m + n. (NOTE:They do not stop if theycollide.)


164) Find the sum 1 + 47

+ 949

+ 16343

+ · · ·

2

a) 2749

b) 4927

c) 243343

d) 343243

e) none of these

165) In a certain town, every girl is acquainted with more boys then girls.Furthermore, all the girls acquainted with a given boy are also acquaintedwith each other.Thena) The number of boys is always greater than the number of girls.b) The number of girls is always greater than the number of boys .c) There are at least as many boys as girls in the town .d) There are at least as many girls as boys in the town.e) none of these

166) Find the number of pairs of positive integers (x, y) such that x6 =y2 + 127


167) Let m > 1 be a positive integer then find the number of pairs (x, y)of positive integers such that x2 − y2 = m3


168)Let N be a positive integer. Then which of the following is(are) true?

a) If N is divisible by 4, then N can be expressed as the sum of two ormore consecutive odd integers.b) If N is a prime number, then N cannot be expressed as the sum of two ormore consecutive odd integers.c) If N is twice some odd integer, then N cannot be expressed as the sum oftwo or more consecutive odd integers.d) At least two of the foregoinge) All of the foregoing

169) Given natural number N, denote as f(N) the following sum of N in-tegers:f(N) = [N/1] + [N/2] + [N/3] + ... + [N/N] (where [x] denotes the greatestinteger that does not exceed x.). Then find f(128)-f(127):


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170) Let n and m be positive integers. An n ∗ m rectangle is tiled withunit squares. Let r(n, m) denote the number of rectangles formed by theedge of these unit squares. Thus, for example, r(2, 1) = 3. Find r(11, 12)


171) Find the number of real positive integral roots of x(5−x)x+1

(x +

5− x

x + 1

)= 6

a) 0 b) 1 c) 2 d) 3 e) 4

172) Let p1, p2, p3 and p4 be distinct prime numbers satisfying :2p1 + 3p2 + 5p3 + 7p4 = 16211p1 + 7p2 + 5p3 + 4p4 = 162Then p1p2p3p4 = k find the number of possible values of k?


173) How many ways to fill the 4X4 board by nonnegative integers, suchthat sum of the numbers of each row and each column is 3?

a) 2006 b) 2007 c) 2008 d) 2009 e) 2010

174)Let H be a set of 2000 nonzero real numbers. How many negativeelements should H have in order to maximize the number of four-elementsubsets of H with a negative product of elements?

a) 1039 b) 961 c) either a or b d) neither a nor b e) none of these

175) Let p(x) = x2 + bx + c , where b and c are integers. If p(x) is afactor of both x4 + 6x2 + 25 and 3x4 + 4x2 + 28x + 5 , what is p(1)?


176) Two different teams with four students each were training for an in-ternational math contest. The students were given an assessment exam andtheir scores were ranked. What is the probability that the four best scores

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come from only one of the teams assuming that no two students got the samescore?

a) 1/2 b) 1/16 c) 1/35 d) 1/288 e) 1/576

177) Let A = {a1, a2..ak} be any set of k composite numbers such that1 ≤ ai ≤ 120 for all i such that1 ≤ i ≤ k. Find the least value of k suchthat there exists at least one pair (ai, aj), 1 ≤ i, j ≤ k in A which is not coprime ?

a) 3 b) 4 c) 5 d) 6 e) 7

178) A polynomial P has four roots 1/4, 1/2, 2, 4. The product of the rootsis 1, and P (1) = 1. Find P (0).

a) 0 b) 1/9 c) -1/9 d) 8/9 e) -8/9

179) Find the sum of all positive integers m such that m2 + 25m + 19 isa perfect square?

a) 125 b) 130 c) 159 d) 164 e) 198

180) Given that m, a, and b are positive integers such thatm + 5a + 7b = 84 andm + 7a + 10b = 114 then find9m + 23a + 30b =?


181)Let there be two lines AC and BC such that < ACB = 90◦. Wedraw circles C1, C2, ..Cn with radii R1, R2, ..Rn, (Ri > Rj, if i > j). Allthese circles are tangent to the lines AC and BC, and any two circlesCp, Cp+1(1 ≤ p ≤ n−1) are tangent to each other. Find R1+R2+R3+...+Rn

when n = N where N is very large

a)√

2+12

R1 b) (√

2 + 1)R1 c)√

2+12

RN d) (√

2 + 1)RN e)none of these

5

182) a, b, c, d be real numbers in G.P. If u, v, w satisfy the equations u +2v + 3w = 6; 4u + 5v + 6w = 12; 6u + 9v = 4 then roots of the equations(1/u + 1/v + 1/w)x2 + [(b− c)2 + (c− a)2 + (d− b)2]x + u + v + w = 0 and20x2 + [10(a− d)2]x− 9 = 0 are

a) equal to each other b) reciprocal to each other c) at leastone of the foregoing d) all of the foregoing e) none of the foregoing

Directions for problem 183 and 184:a and b are positive integers such that gcd(a, b) = 1 andab

+ 14a9b

is an integer.

183) What is the value of b?a) 1 b) 3 c) 9 d) 6 e) can not be uniquely determined

184) How many pairs (a, b) exist?


185)A boy spends Rs. 81 in buying some pens and pencils. If a pen costsRs.7 and a pencil Rs 3, What is the ratio of pens to pencils when the maxi-mum number of pens are purchased such that no extra money is given to theshopkeeper?

a) 3 : 2 b) 2 : 1 c) 5 : 4 d) 7 : 2 e) none of these

186) Find the sum 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12


187) Arrange in ascending order x = 6013 , y = 2 + 7

13 , 3,4

a) 3,x,y,4 b) 3,y,x,4 c) 3,x,4,y d) 3,y,4,x e) cannot bedetermined

188) Let a, b, c be complex numbers such that:a2

b+c+ b2

a+c+ c2

a+b= 0 Then

6

ab+c

+ ba+c

+ ca+b

equals?

a) 0 b) 1 c) -3 d) a, b and c e) b and c

189)Which is/are true?

I) log(√

3−√

5 +√

3 +√

5) = 1/2II) 3 < 6 log 2 + log 31 < 4III)if a = log7(2

√2 + 1)+log2(

√2− 1) then log7(2

√2− 1)+log2(

√2 + 1) =

2− a

a) I,II b) II,III c) I,III d) I,II and III e) none of these

190)If A, B and C denote the angles of a triangle then find the maximumvalue of sin2 A + sin B sin C cos A

a) 1 b) 8/9 c) 9/8 d) 4/9 e) 9/4

Directions for problem 191-200:Each question is followed by 2 state-ments X and Y. Answer each question using the following instructionsChoose A if the question can be answered using X aloneChoose B if the question can be answered using Y aloneChoose C if the question can be answered using either X or (exclusive) YChoose D if the question can be answered using X and Y togetherChoose E if the question can not be answered using X and Y also

191) What are the ages of three brothers?X: The product of their ages is 21Y: The Sum of their ages is not divisible by 3

192) Is x = y?X:(x + y)( 1

x+ 1

y) = 4

Y:(x− 50)2 = (y − 50)2

193)What is the value of m and n?X: n is an even number, m is an odd number, m.nY: mn = 30

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194) In a family, Kamla, her brother, her daughter and her son live. Theyplay a game of cards. Is Kamla the best player?

X: The worst player’s twin and the best player are of the opposite sex.Y: The worst player and the best player are of the same age

195) a + b + c equals?

X: a, b and c are in A.PY: a2 + b2 + c2 = 83, where a, b and c ε N .

196) Let p(x) = x2 + 40. Then for any two positive integers i and j wherei > j, is p(i) + p(j) a composite number?

X: p(i)p(j) is not a composite numberY: p(2i) + p(2j) is a composite number

197)What is the length of the side AB of triangle ABC?

(X) AB <= AC = 2, and area of triangle ABC is 2(Y) Exactly two sides have integer length

198) Is p = q, if pqrst = 4?

X: r = s = t Y: Three of p, q, r, s and t are integers.

199)Find the age of three sisters?X: The product of their ages is 72,The sum is 14Y: the eldest studies in class 3

200)Can we measure 15 minutes using two given sand glasses?

X:One sand glass takes 7 minute and other takes 11 minutes for the sandto go through to the bottom.Y:One sand glass takes 4 minutes and other takes 10 minutes for the sandto go through to the bottom.

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Answer Key

151) a) 152) c) 153) d) 154) a) 155) c)156) d) 157) b) 158) c) 159) c) 160) b)161) d) 162) b) 163) d) 164) b) 165) c)166) b) 167) e) 168) e) 169) d) 170) b)171) e) 172) b) 173) c) 174) c) 175) b)176) c) 177) c) 178) d) 179) d) 180) b)181) c) 182) b) 183) b) 184) c) 185) a)186) c) 187) b) 188) e) 189) d) 190) c)191) e) 192) a) 193) d) 194) d) 195) d)196) a) 197) a) 198) d) 199) d) 200) a)

Appropriate care has been taken to keep this document free. But, if you findany errors, please report to the author at [email protected]. Kindlynote the solutions of these problems are discussed in full athttp://www.pagalguy.com/forum/quantitative-questions-and-answers/27454-official-quant-thread-for-cat08.htmlThis document is not for sale! Good Luck !

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Short Hints

152) use tan 90− x = cot x and then tan θ + cot θ = 2sin 2θ

154) ax + by = n if gcd(a, b) = 1 then the greatest n such that both (x, y)are not positive is ab− b− a

155) 10C3 − 5.2 = 120− 10 = 110

156)(r2 + 5r − 6)(r2 + 5r + 6)= (r2 + 5r + 7− 13)(r2 + 5r + 7− 1)= (−13)(−1) = 13

157) put x2 − 10x = y will reduce to a quadratic in y and we get finallyx = 13,−3 so positive x is 13

158) solution is finding the solns of a + c = b now b = 3 gives (2, 1), (1, 2).b = 4 gives (3, 1), (1, 3) so 2 solution. then for each b we have solutions as2(d b

2e−1) so summing we get 2+2+4+4+· · ·+2006+2006 = 2·1003·1004 =

2014024

160) X = |bn−an| = | bn−1+an−1

2−√

an−1bn−1| = | (√

bn−1−√

an−1)2

2| ≤ | bn−1−an−1

2|

now use the inequality n times we get X ≤ |b−a|2n

161)can be easily proved take one vertex at origin other at (k,0) and thirdas (x,y) use the formula for tan 2θ = 2 tan θ

1−tan2 θ

162) p2 + 3p− 1 = 6k + 3p = 3(2k + p) ( for all primes p ≥ 5so cases remaining are p = 2, 3clearly p = 2 gives p2 + 3p− 1 = 9and p = 3 gives p2 + 3p− 1 = 17so only prime is p = 3

163)At each step each ant can go three ways, so total no of ways 34.34 = 81.81now each ant can come back to original in 6 ways along one face ABCD be

10

the square, then ABCDA, ADCBA,ABABA,ADADA,ABCBA,ADCDAnow we have 3 faces so we get 6.3 = 18same for the other ant with C as initial point so m/n = 18.18/81.81 = 4/81m + n = 85

164) Tn = n2

7n−1

now Let S = 1 + 4/7 + 9/49 + 16/343 + ............S/7 = 1/7 + 4/49 + 9/343 + ........S − S/7 = 6S/7 = 1 + 3/7 + 5/49 + 7/343 + .............6S/49 = 1/7 + 3/49 + 5/343 + ...........6S/7− 6S/49 = 36S/49 = 1 + 2/7 + 2/49 + 2/343 + .......or, 36S/49 = 1 + 2(1/7 + 1/49 + 1/343 + ......) = 1 + 2(1/6) = 4/3

or, S = 49/27

165)If n girls know a certain boy, any one girl in the group would have(n − 1) girl friends; this would imply that every girl (in the group) wouldhave to know at least n boys, so that they would know more boys than girls.Now, if every girl knew the same boys, there would be n boys and n girls (theminimum of boys). If there are girls that do not know each other through aspecific boy, there would be (n + m) boys, and n girls, withn + m > n .

166) Note that 127 is prime . now (x3 + y)(x3 − y) = 127.1 as x, y arepositive integers x3 + y = 127, x3 − y = 1 so we get one pair (4, 63)

167) let x = m+m2

2and y = m2−m

2then clearly for each m we have one

solution. so infinite solutions

168) n=4k, n=(2k-1)+(2k+1); b) if n is the sum of m consecutive odd in-tegers, m—n. as we are given that n is prime, we have m=n, so n is beingexpressed as the sum of n consecutive odd integers. In this case, all primes(except for 2) can be expressed as the sum of consecutive odd integers: 3=-1+1+3, 5=-3+-1+1+3+5, etc. (follow the pattern); c)the sum of an oddnumber of odd integers will be odd, thus it cannot :equiv:2 mod 4. So, wehave that the number of odd integers is even, meaning that the mean fallsright in between two terms of the sequence, i.e. it is even. However, thesum of this arithmetic series will be the number of terms times the mean,i.e. an even number of terms times an even mean will be :equiv:0 mod 4.

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contradictionSo all three are correct

169)the only time in which one of the terms in the expansion of f(128) will bediff. than in one of the terms of f(127) will be in the [N/1] term, and in anysubsequent terms of the form [N/d], where d—128. To see why, observe thatfor any term of the form [N/k], k not dividing 128 and not equal to 1, 128 and127 will have the same integer quotients upon division by k and thus thatterm will be the same. Ok, finally, note that for each term of the form [N/d],k—128, the difference between [128/d] and [127/d] will be 1, as [128/d] willhave exactly one more factor of d. So essentially what the question is askingfor is the number of divisors of 128, that is (7+1)=8

170) r(n,m)=n(n+1)m(m+1)/4. To see why, partition these rectangles upaccording to their width, and assume that there are A rectangles that canbe formed from a 1*m column. Well then there will be n*A rectangles ofwidth 1, (n-1)*A rectangles of width 2, (n-2)*A rectangles of width 3 ... 1*Arectangle of width n. When we add these up, we get n(n+1)A/2. Ok, nowwe need to find A. We partition the rectangles that can be formed from a1*m column similarly, except with respect to their length. There will be (ina similar fashion) m rectangles of length 1, m-1 rectangles of length 2, ..., 1rectangle of length m, i.e. A=m(m+1)/2. Putting all this together gives theformula at the beginning.

171) on solving we get x=2,1 as roots so 2 solutions .

172) subtract the two equations gives 9p1 + 4p2 = 3p4 clearly 3 dividesp2 hence p2 = 3 this give 3p1 + 4 = p4 solving which gives (p1, p4) =(5, 19), (11, 37).. the second and further solutions wont work as it will leadto negative value of p3 . hence only one set!p.s: there is no further need but we can proceed to find p3 = 2 :)

174) Let H has p positive elements and n negative elements.then p + n =2000 = s now pC3.n +n C3.p are the number of subsets of H with negativeproduct. further we getpn6

(p2 + n2 − 3p− 3n + 4) = pn6

((p + n)2 − 2pn− 3(p + n) + 4) = pn6

(−2pn +(s2 − 3s + 4)) let pn = x we get a quadratic in x we get x = 1039 or 961

12

175) x4 + 6x2 + 25 can be easily factorized as (x2 − 2x + 5)(x2 + 2x + 5)so one of them is p(x) further investigation proves p(x) = x2 − 2x + 5

176) Suppose the teams are A and B. there are two favorable ways. Allfrom A or all from B. Trick here is to find the total no of ways which comesout to be 8C4. I leave it to you to figure it out :)

177) 121 = 112 > 120 so the prime factors are 2, 3, 5, 7 which make thecomposite numbers of A. By pigeon− hole principle there must be at max4 such composites which will not be co-prime. So the required minimum k is 5

178) P (x) = a(4x− 1)(2x− 1)(x− 2)(x− 4)

179) let m2 + 25m + 19 = k2 where k is a positive integer.⇒ 4(m2 + 25m + 19) = 4k2

⇒ 4m2 + 100m + 76 = r2 (r = 2k)⇒ 4m2 + 100m + 76 + 549 = r2 + 549⇒ (2m + 25)2 − r2 = 549⇒ (2m + 25 + r)(2m + 25− r) = 549 = 32.61solving we get m = 5, 34, 125

180) m + 5a + 7b = 84 and m + 7a + 10b = 114gives 2a + 3b = 30this givesm + a + b = 24so9m+23a+30b = (m+5a+7b)+(m+7a+10b)+7(m+a+ b)+2(2a+3b) =84 + 114 + 7 ∗ 24 + 2 ∗ 30 = 198 + 168 + 60 = 426

181) Any circle Ci will form a square of the side Ri so OiC =√

2Ri us-ing this we can form a relation that 0i+1C = Ri+1 + Ri + Oi ⇒

√2Ri+1 =√

2RiRi+1 +Ri use this and sum of an infinite GP we get the answer in termsof RN

182) solve for u, v, w we get u = −1/3, v = 2/3 and w = 5/3 givesu + v + w = 2 and (1/u + 1/v + 1/w) = −9/10 and also use the condi-

13

tion of gp then (a− d)2 = (b− c)2 + (c− a)2 + (d− b)2 so the two equationshave reciprocal roots

183) Let ab

+ 14a9b

= k where k is an integer⇒ 9a2 + 14b2 = 9abk clearly 9 divides RHS and 9a2 so 9 divides b2 so b is ofthe form 3m or 9t where gcd(m, 3) = 1 andgcd(t, 9) = 1 but if b = 9t , RHSwill be divisible by 81 and LHS by 9 so b = 3

184) b = 3 so we get that only 1, 2, 7, 14 i.e the divisors of 14 should di-vide a else it would not divide 14b2 so a = 1, 2, 7, 14 hence 4 pairs

185)7x+3y = 81, 3 divides RHS and 3y so 3 divides x. So we get x = 3, 6, 9we need max x, so x = 9 which gives y = 6 so ratio is 3 : 2

186) series is just the sum of numbers whose reciprocals have only 2 and3 as prime divisors. So sum = (1 + 1/2 + 1/4...)(1 + 1/3 + 1/9..) =[1/(1− 1/2)][1/(1− 1/3)] = 3

187)Just prove which is larger?(4(m + n))(1/3) or m(1/3) + n(1/3) (becauseif we let m = 8, n = 7 we get our case) for positive reals m, n. Then letm = a3, n = b3 and try to see which is bigger. we will get x > y

188)(

ab+c

+ ba+c

+ ca+b

)(a + b + c) = a(b+c)

b+c+ b(a+c)

a+c+ c(a+b)

a+b= a + b + c so

either the sum is −3 or 1

189) II and III are obvious. what remains is I.√

3−√

5 +√

3 +√

5 =√6−2

√5+√

6+2√

5√2

=

√(√

5−1)2+√

(√

5+1)2√2

= 2√

5√2

=√

10 take log we get 1/2

190) sin2 A+sin B sin C cos A = a2

4R2 + b2R· c2R· b2+c2−a2

2bc= a2+b2+c2

8R2 = 9R2−OH2

8R2 ≤9R2

8R2 = 98

191)easy !

193) 15,2 easy!

194) we cannot comment anything specific by using either alone. If we useboth together, we look at this

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If Kamla is worst player , the best player is female , but this defies othercondition that the best player and worst player are of same age

now if her brother is worst player best player is then her son. this meanskatrina her son and her brother are of same age. not possible again

if her son is worst player, the best player is male so he must be katrina’sbrothernow if we assume that her son and her brother are of same age this canhappen.

=> option (D)

195) easy 3,5,7

196) p(i)p(j) is not a composite number=> i2 − j2 is a prime as i, j are positive integers and i > j, (i2 − j2) can’t be1=> (i + j)(i − j) = prime so i − j = 1 let p be the prime so i = (p + 1)/2,j = (p− 1)/2clearly p is not 2 hence all p is oddp(i) + p(j) = 80 + (p2 + 1)/2now p2 = 6k +1 ( can be easily proved) [ p(p−1)(p+1) is divisible by 6 nowp is a prime so p2 − 1 = 6k, p2 = 6k + 1, for any prime p greater than 3]therefore p(i) + p(j) = 80 + (p2 + 1)/2 becomes80 + (6k + 2)/2 = 81 + 3k = 3(27 + k)so not a prime =¿ can be answered by using X

now, p(2i) + p(2j) is a composite number4(i2 + j2 + 20) is composite

now i and j can be anythingcan’t make any conclusions

=> choice (A) is the right answer

197) Area = ∗ (AB).(AC).sinA => 2 = ∗ AB.2.sinA => ABsinA = 2.But AB <= 2andsinA <= 1 => AB = 2.

15

=> choice (A) is the right answer199) 8.3.3 and 6.6.2 so can’t say with X also can’t say with Y. But Y sayseldest so combining the two, we get eldest in case 1, so ages are 8,3,3. optiond200) 15 = 11.2− 7 so if we start the two timers together when the 7 minutetimer runs out, reverse it, now when the 11 minute timer runs out, the 7minute timer will show 4, reverse it. we got 11+4 so A other two timers areeven and 15 is odd

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quant questions 151-200 and solutions

Documents

consecutive odd integers

odd integers

log2

positive integers

unit squares

lines ac

triangle abc

composite numbers