Quant Boosters  Hemant Malhotra  Set 1

all values will be equal to zero
a^2  1 = 0 so a = +/ 1
b^2  4 = 0 so b = +/ 4
c^2  9 = 0 so c = +/ 3
so 2 * 2 * 2 = 8

Q6) How many subsets A of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} have the property that no two elements of A sum to 11

Pairs that makes 11 = (1 ,10), (2 ,9), (3 8), (4 7), (5 6)
we have 5 pairs and out of each pair only one or none can go to subset A
so for each pair we have 3 possibilities.
so number of subsets of A satisfying the condition = 3^5 = 243

Q7) The centroid Of a triangle is at (1, 1) While its orthocentre at (5, 3). The circumcentre of the triangle could be at
a) (1, 3)
b) (8/3, 0)
c) (0, 8/3)
d) (7/3, 1/3)

Centroid bisect ortho and circum in 2:1 ratio
distance between (1,1) and (5,3) is sqrt(16+16)=sqrt32
so distance between centroid and circum should be sqrt32/2
check by options
sqrt(4 + 4) = sqrt8
so OA = A

Q8) A person travels equal distance at a speed of 3 kmph, 4kmph and 5 kmph and takes 1 hour 34 minutes to complete the journey. Find his average speed for the whole journey
a) 3.83 kmph
b) 4.62 kmph
c) 4 kmph
d) 4.11 kmph

Basic approach to solve 180/47 = 3.8 approx
Quick approach  Average speed of journey will be equal to HM always (when equal distance with different speeds)
HM < AM
AM = (3 + 4 + 5)/3 = 4
so HM < 4
only 1st option satisfies.

Q9) Find the range of T = 2Cos^4θ + Sin^2θ + 3
a) [1/2, 5]
b) [2, 3]
c) [31/8, 5]
d) [31/7, 4]

OA  C
Method 1 
Change in cos or sin
2cos^4 x + 1  cos^2x + 3
let cos^2x = y
2y^2  y + 4 = T
so dT/dy = 4y  1 = 0
so y = 1/4
so 2/16  1/4 + 4
1/8  1/4 + 4
1  2 + 32/8 = 31/8 min value
and max will be 5 when x = 0Method 2  Exam Condition Approach
max value is 5 at x = 0 so 2 and 4 ruled out
now square of any thing will not be negative so min will be 3 in (worst cases ignoring sin and cos)
so only option possible = C

Q10) In the figure given below, if O is the centre of the circle and < BAO = 54 then find < OAD + < OCD
a) 108
b) 126
c) 112
d) 120

ABCD is cyclic quadrilateral
so BAD + BCD = 180
so OAD + OCD = 180  BAO = 180  54 = 126

Q11) How many integral solutions does the equation x^y = xy has
a) 1
b) 2
c) 3
c) 4
d) More than 4

x^y = xy
let y = 1
then x = x
we can put any integral value of x here so more than 4

Q12) Find the minimum number of coins required to pay the amounts of 67 paise, Rs 1.03 and 83 paise to three persons A, B and C, respectively, using only coins of the denominations of 2 paise, 5 paise, 10 paise, 25 paise and 50 paise.
a) 17
b) 20
c) 18
d) 19
e) 16

(Attack on highest coefficient )
2a + 5b + 10c + 25d + 50e = 67
now we need min so e = 1
2a + 5b + 10c + 25d = 17
so c = 1, b = 1, a = 1
so we need min 4 coins
now 2a + 5b + 10c + 25d + 50e = 103
so e = 1 and d = 1, c = 2, b = 0, a = 4 so 8 coins
now 2a + 5b + 10c + 25d + 50e = 83
so e = 1, d = 1
so 2a + 5b + 10c = 8 so a = 4
so 6 coins
so min coins needed = 4 + 8 + 6 = 18

Q13) For how many positive integer values of ‘x’ is x – 1 – 2 – 3 – 4 < 5

x – 1 – 2 – 3 – 4 < 5
so x  1 < 5 + 4 + 3 + 2
so x1 < 14
so 14 < x  1 < 14
so 13 < x < 15
so max 14 and min = 12 so 1 to 14
14 positive values

Q14) Because of an error , a printing machine cannot print numbers 6 and 8 . this machine is used to print page numbers on a volume of an encyclopedia. What number is printed on 503rd page of encyclopedia

0 to 9 means we are working on base 10. Now 2 values are missing so we are working on base 8
numbers used 0, 1, 2, 3, 4, 5, 7 and 9
so 6 will be printed as 7 and 7 as 9 here
now convert 503 in base 8 so 767
but 7 will be printed as 9 and 6 as 7
so OA = 979

Q15) A G.P with the first term as 'a' and common ratio 'r' > 1, is such that the average of the first n terms is 13 and the average of the first 'n + 1' terms is 30. If a = r = n, then what is the second term of the G.P.?