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Solutions Section III

101. y = log10 x and y = x-1 would intersect when log10 x = x

-1 or 101/x = x or xx = 10. There will be only one

value for this equation, hence [2].

102. For the conventional arithmetic series, we have a +2d+a+14d = a+5d+a+10d+a+12da +11d = 0, which is12th element of the series, hence [3].

103. Since from 50 questions, student scores 32, he has lost 18 marks on non-attempted or wrong questions. For

minimum wrong attempts, non-attempts should be maximum. Say he does not attempt 12 questions, in that

case he loses 2 marks from the correct attempts. Thus we get 34 correct + 12 non-attempted = 46 questionsLast four should even out, hence he must attempt at least 3 wrong to get 32 marks. Question is best

attempted through trial-&-error, and not by options. Answer is [3].

104. Since surface area of B is 300% higher than that of A, it is 4 times that of A. Now, as area is four times,

radius would be twice, and hence volume would be 8 times. Thus we get volume of A is 1/8 times thevolume of B or 87.5% lower than that of B, hence [4]. Remember, Ratio of volume = (Ratio of linear

measures)3, and Ratio of Area = (Ratio of linear measures)2.

105. Options [1], [3], and [4] may be possible, but [2] is not possible since there are 27 people and maximum

different number of acquaintances possible are only 26, hence [2].

106. Between 100 & 200, there are 14 multiples of 7, hence 7 of them should be even. Similarly, there are 11

multiples of 9 of which 6 would be even since first multiple of 9 i.e.108 is even. Thus there should be

7+6=13 such numbers, but a number multiple of both 7 & 9, i.e. 126 should be subtracted once, hence 12

even numbers would be multiple of 7 or 9 or both. Also from 100 to 200, there are 51 even integers, hence51-12 = 39 even integers will not be multiples of 7 or 9 or both, hence [3].

107. Going by options, student can check out for any one of the variable sum of the coefficients becomes zero.

Only option [1] satisfies he condition.

108. f(x) = | x- 2|+|2.5 -x |+|3.6-x| is a positive function, since each term is the absolute value. This would attain

the minimum value only when one of term is zero, and absolute value of other two is minimum. Also, 2.5

being the middle term between 2 and 3.6, x = 2.5 will give minimum absolute values for other two terms.Thus [2]. Also. You can checkout the options but None of the above may some time create trouble.

109. g(x) =max(5 - x, x+2) will attain minimum value when 5 x = x + 2 or x = 1.5. When x = 1.5, g(x) = 3.5,

hence [4].

110. When M is represented in base 2, base 3 and base 5 notation, we get last digit as 1, therefore M = 1+LCM

of 2, 3 & 5. Thus only numbers among options are [1] 31 and [4] 91. Now, converting 31 in different base

as given we get 31 = 1115 = 10113 = 111112, thus cannot be the answer since only two of the numbers

should begin with 1, hence [4].

111. Such problems can be solved by constraints or by logic. If only standard bags are made, factory can make

175 bags with 200 hours idle hours on machine B; and can make profit of 3500. If only deluxe bags are

made factory can make 125 bags with 75 idle hours on machine A; and can make profit of 3750. Now,

75+5x = 4y, and 10x = 6y or 5x = 3y or y = 75.where x = deluxe bags less than 125, y = standard bags.

Alternatively, 4x+5y < 700 and 6x+10y < 1250, solving we get x = 80, y = 75, hence [1].

112. Since the fastest runner runs at twice the speed the slowest runner, when the fastest first time meets the

slowest, slowest must have completed 1 circle, whereas the fastest must have completed 2 circles. Thus the

fastest covers 2000m in 5 minutes, and 4000m in 10 minutes, hence [3].

113. Since at the end of 9 years, the number is again same, purchase and sale of goats is done in equal number.

Now, Shepard buys p% of 912, and sells q% of (100+p)% of 912. q must be less than p as q is

calculated on higher value. Thus the answer is [3].114. G + B = M

T = 2(3+6+8+2) 1 = 37 = 19+G1+B1+M1 G B

G1+B1 = 16+M1M1 = 1.Question 114 cannot be determined, hence [4]. G1 3 B1

115. M1 = 1, hence [2].

6

8 2

14

M1

M

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116. If we take the simplest polygon, a square, it has for convex corners. If we make one corner a concave, we

get polygon of six sides with five convex and one concave corner. We can check similarly, we shall always

have 4 less concave corners than the number of convex corners. Thus polygon with 25 convex corners will

have 21 concave corners, hence [3].

117. p2 + q2 = (p+q)2 2pq = (-2)2 2(--1)[Sum of the roots, product of the roots]

= 2-2+6 = (-1)2+5, hence the minimum value = 5. Thus the answer is [4].

118. a, b, b, c, c, c, is a series in which 1st letter comes once, second comes twice, and so on. Therefore nshould be just more than 288 to give the sequential number of the letter at 288 th term. n(n+1)/2 > 288 or

n(n+1) > 576 n = 24. 24th

letter of alphabet is x, hence [4].119. Since rea is 4 times, the radius should be twice. Since OA = 2r, and OP = r, B

BAO = 300, which gives BAC = 600 and ABC an Equilateral triangle

with side AB = 23 r. Also 4r2 = 12 or r = (3/). Area of triangle ABC P

= 3/4(233/))2 = 93/, hence [3].A

C120. For minimum possible value of a2+b2+c2+d2, a+b+c+d = 4m+1 should be minimum, i.e. 5 (for m = 1). Let

integers be 1, 1, 1, 2, then a2+b2+c2+d2 = 7. This is true for option [2], since m = 1. Thus [2]. Maximum

possible value cannot be determined.

121. To find fraction of area, we can construct the diagram as shown in the figure, andjust count how many such triangles we can fit in the hexagon. Area of the triangle is1/12 the area of hexagon, hence [1].

122. 2x x 1 = 0 2x = x + 1. x = 0 & 1 satisfy this equation, hence two roots, or option [3]. Since there areonly two non-negative integral roots, and since a fractional value of x will give irrational value on LHS and

rational value on RHS, we can surely say that there are only two roots.

123. If r is the radius of area grazed by horses tied at P and at R, 2r is the

radius of the plot. Therefore area of the plot = 2r2. Area grazed by

horses tied at P & R = r2. In triangle PRS, PR = 2r, PS = r+r', and heightSO = 2r-r'. (r+r')2 = r2 + (2r-r')24r2 = 8rr' r = 2r'. Thus area grazedby horse tied at S will be (r/2)2. Thus fraction of area left un-grazed will

be (2r2 r2 r2/4)/2r2 = 3/8 = ~36%, hence (3).

124. 6-8-10 gives us a right triangle, hence the length perpendicular to hypotenuse BD = 68/10 = 4.8. AP =

1.2, and CQ = 3.2 AP:QC = 1.2:3.2 = 3:8, hence [4].125. Parallel lines AB and PQ give two similar triangles, ABD and PQD, hence AB/PQ = BD/QD. Similarly,

CD/PQ = BD/BQ. Dividing, we get AB/CD = BQ/QD = 3/1 [given AB/CD = 3/1]. BQ/(BQ+QD) =

(Componendo-dividendo rule of ratio) CD/PQ = BD/BQ = 4/3 = 1:0.75, hence [2].126. If log3 2, log3 (2

x -5), log3 (2x -7/2) are in arithmetic progression, then 2, (2x -5), and (2x -7/2) must be in

geometric progression. Therefore 2(2x 7/2) = (2x 5)2 or 2x+1 7 = 22x 102x + 25 22x 122x +36 = 4

(expressing so as to make perfect square), we get (2x

-6)2

= 4 or 2x

6 = 2 or x = 3., hence [4]. Substitutingthe values will lead you to wrong answer as both x = 2 and 3 satisfy the condition. But value x = 2 givesnegative value of 2x 5, log of which is not defined.

127. [mAOD - mBOC] = mBCD, [x y] = y or x = 3y, hence [1].

128. since x < y, and z < y, also x0, we can have following table, for y = 9, 89; for y = 8, 78,for y = 2,12. Therefore total numbers = 72+56+42+30+20+12+6+2 = 240, hence [3]. P

129. Since APB = 600, and AP = PB, we have equilateral triangle APB, hence AP = AB = b.

Also OP = p, and AO = b/2. Now applying Pythagoras theorem to AOP, hwe get b2 = h2+b2/2 or b2 = 2h2, hence [2]. D C

O

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A B

130. Drawing lines as shown in figure, we get a right triangle of sides r, r-10, r-20.

This gives us the pythagorian triplet 30-40-50, hence r = 50. Alternatively,

r2=(r-10)2 + (r-20)2, solving we get r = 10 or 50. 10 not being possible, we get

r=50, hence [3].

131. Series we get is 1, 3, 6, 10, where each nth term is sum of integers up to n.

Hence we have n = 8436 or n(n+1)(n+2)/6 = 8436 or n(n+1)(n+2) = 68436

n(n+1)(n+2) = 6341937 = 363738, hence n = 36, option [3]. Or we

can substitute option values in n(n+1)(n+2)/6 = 8436 to get the answer.132. When product is unity we get number as a+1/a, or all numbers would be unity, in any case sum will always

be greater than or equal to the number of values, hence the option [3].

133. Since -1 < v < 1, -2 < u < -0.5 and -2 < z < -0.5 and w = vz/u, z and u both always being positive, in case of

both minimum and maximum value, we must take highest absolute value of z/u = (-2)/(-0.5) = 4. Now

minimum value of w will be (-1)(4) = -4, and the maximum value will be (1)(4) = 4, hence [2].

Questions 134 to 136: From the information given we can draw figure as given N1

here. Also speed along OR = 30 km/hr, speed along IR = 20km/hr, and speed

along cord roads = 155km/hr. Now, if radius of inner ring road is ri, then radius of N2

outer ring road, ro = 2ri, and the length of cord road will be 5 ri.

134. Sum of lengths of all cord roads = 45ri, length of outer ring road = 4ri W1 W2 E2 E1

Thus the ratio =

5:, hence [3].135. If Amit traves on minor arc S1E1 and the cord road E1N2, the total distance

traveled by Amit is ri+5ri = ri(+5). Now he covers this distance in 90

minutes, hence ri/30 + 5ri/155 = 1.5 or 3ri/30 = 1.5 or ri = 15 thus radius of outer ringroad = 2ri = 30km, hence [3].

136. From above question, ri = 15, thus time taken for travel N1W2 will be 515/155 = 1hour, and time forcovering inner semi-circle = 15/20 = hour, thus total time taken will be 60+45 = 105 minutes, hence

[4].

137. 1 box containing green balls can be in 6 ways, 2 consecutively numbered boxes containing green balls can

be in 5 ways: 1-2, 2-3, 3-4, 4-5, 5-6, similarly three consecutive boxes in 4 ways and so on till six

consecutive boxes in 1 way, thus 1+2+3+4+5+6 = 21, hence [2].138. If 12 points are all connected with independent routes, we need 12C2 = 66 maximum number of edges.

Minimum edges will be required when all the points are collinear, then minimum 11 edges would berequired, hence [1].

139. Terms in set T are in arithmetic progression with first term 3 and common difference 8. Number of terms

therein are 3+(n-1)8 = 467 or (n-1) = 464/8 = 58 or n = 59. Thus there are 29 pairs of two numbers that

add up to 470. Therefore in set there can be maximum of 30 values, 29 from each pair and the middle term,

235 such that no two terms will give the sum of 470, hence [4].

140. When two curves intersect, value of y will be same and so equations can be equated with each other.

Equating, we get x3 + x2 + 5 = x2 + x + 5 or x3 = x or x3 x = 0 or x(x+1)(x-1) = 0, thus we get three values

of x: 0, -1, +1 graphs intersect in three distinct points, hence [4].141. If j = n, 2n-n = 1 student answered n questions wrongly; if j = (n-1), then 21 = 2 students answered n-1

questions wrongly, thus we get the series 1 + 2 + 4 + 8 + +2n-1 = 4095 (2n 1)/(2-1) = 4095 or 2n =4096, thus n = 12, hence [1].

142. Expanding {x2(y+z) + y2(x+z) + z2(x+y)}/xyz, we get {x/z + x/y + y/z + y/x + z/y + z/x} which gives threepairs of the form a+1/a which is always > 2, hence the expression will always be greater than 6, hence [3].

143. n such that (n-1)! Is not divisible by n will always be a prime number. Between 12 and 40 there are 13, 17,

19, 23, 29, 31, 37 Seven prime numbers, hence [2].

144. 244 < b111611 < b11. b being even does not answer the question, b > 16 alone gives the answer, hence [1].145. Statement A can give the sum and the product of the roots, and hence values of b and c respectively.

Statement B with value x = - can give the value of b = c, hence both independently are sufficient, hence[2].

146. AB not being diameter does not give any information, and hence statement A is not sufficient. Also from

statement B, data is inconsistent because arc passing through A, B & E cannot be minor A B

arc the circle, hence [4].

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147. {1/a2 + 1/a4 + 1/a6 + } > {1/a + 1/a3 + 1/a5 + } (1/a2)/[1-1/a2] > (1/a)/[1-1/a2] E 1/(a2-1) > a/(a2-1). (A) For -3 < a < 3, when a < -1, denominators being positive cancel out, and inequality

holds true, for -1 < a < 1, denominator becomes negative and hence can be cancelled by reversing

inequality, and thus inequality does not hold, also for a >1 inequality does not hold, hence not sufficient.

(B) Roots of equation 4x2 4x + l = 0 are + . In equality does not hold for either of the values & - hence (B) is sufficient, hence [1]. A

148. (A) Since AD = 1 cm, we get DB = 1; and DF = I cm and perimeter of DBF = 3 cmgives BF = 1 and hence BC = 2, and AC = 2, thus we can calculate area ofDEF.

(B) Since perimeter of

ABC is 6 and all sides are 2 each, we get sides of

DEFD E

as 2 each and hence calculate the area of the triangle. Thus answer is [2].

149. The three buy three bottles, one at 520 and other two at 30% discount or atRs.364 each thus total sum is 520+2364 = 1248, and average cost of each B F C

bottle Rs.416. Now, R pays 2 Euros or 92 Bahts, M pays 4 Euros and 27 Bahts or 211 Bahts, and S pays

1248-303 = 945 Bahts equivalent in US dollars. Thus R has to pay S, 416 92 = 324 Bahts, hence [4].

150. M has to pay S, 416 211 = 205 Bahts = 5 US dollars, hence [3].