Quant Boosters  Hemant Malhotra  Set 1

a is a two digit number and b three digit number, so ab is five digit number.
If a = 12 and b = 123 and we are writing 12123 then we can also write it as 12 x 1000 + 123
so 1000 * a + b = 9 * (a) * b
1000/b + 1/a = 9
so 1000/b will be approx 9 so 1000/9 = 112 (approx value)
so b = 112 and a = 14
so sum = 126

Q2) In a ∆ABC, AD & BE are medians and G is the centroid, ∠AGE = 30° , AD=12cm & BE = 18cm. Find the area of the triangle?

∠AGE = 30
G will bisect BE in 2:1 ratio so GE = 4 and in same way AG = 12
so area of AGE = 1/2 * AG * GE * sin30 = 1/2 * 12 * 4 * 1/2 = 12
so area of ABC will be 6 times of area of AGE =12 * 6 = 72
[Note  Median divide triangle into 6 equal areas]

Q3) For prime numbers a and b, a + b = 102 and a > b. What is the least possible value of a  b ?

Question is not important but approach is important here
Minimum will be when a and b are as close as possible so
a = 51 + m and b = 51  m
where 51 + m and 51  m both are primes
at m = 8, 59 and 43 both are prime so minimum difference will be 16

Q4) At Ram's birthday party, the ratio of people who ate ice cream to people who ate cake was 3 : 2. People who ate both ice cream and cake were included in both categories. If 120 people were at the party, what is the maximum number of people who could have eaten both ice cream and cake?

Those who ate both will be maximum when all who ate cake also ate icecream
n(A U B) = n(A) + n(B)  n (A intersection B)
so 120 = 3x + 2x  2x
so x = 40
so Those who ate both = 2x = 80

Q5) How many ordered triples (a, b, c) satisfy the below equation where a, b and c are real numbers.
(a^2 − 1)^2 + (b^2 −4)^2 + (c^2 − 9)^2 = 0

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