# Quant Boosters by Hemant Malhotra - Set 15

• Two cars run toward each other. Their speeds are 30kph and 40kph. At the moment they are 105 km apart, a bee flies at 50kph from the bumper of one car to the bumper of the other and shuttles back and forth until the two cars collide. Calculate the total distance traveled by the bee

105/(40+30) * 50
= 75

Find the sum of the series given below.
2 + 1 + 4 + 2 + 8 + 3 + 16 + 4 + 32 + 5 + … + 20th term.

(a) 2047
(b) 2092
(c) 2147
(d) 2101**

OA=D
2,4,8,16,32 ... IN gp SUM OF 10 TERMS OF GP
1,2,3,4,5 ..... IN ap SUM OF 10 TERMS OF AP
SO 2101

What is the maximum number of natural numbers that one can select from the first 99 natural numbers and be sure that no three of the selected numbers add up to 99?
(1) 33
(2) 67
(3) 49
(4) 66
(5) None of these

32+33+34 = 99 so for every number greater than 33 that will not give us sum=99 , so from 33 to 99 =67 numbers we can choose

If f(x) is cubic polynomial with coefficient of x^3 as 1. It is given that it has non-negative real roots and f(0) = -64. Find the largest possible value of f(-1).
a) -125
b) -65
c) -64
d) -120
e) None of these

If f(x) is cubic polynomial with coefficient of x^3 as 1 .. always consider this cubic polynomial and standard polynomial is ax^3+bx^2+cx+d=0 and in this when b=c=0 so x^3+d=0 and x^3-64=0 so x^3-64=0 (x-4)(x^2+16+4x)) so x=4 and x^2+4x+16=0 now roots of x^2+4x+16=0 is imaginary so this is not the case.
F(x) = x^3 - b * x^2 + c * x - 64
f(-1) = -1 - b - c -64
roots are 4,4,4
-1 -12 - 48 - 64 = -125

Ram starts a simple calculation. He multiplies the integers 1, 2, 3,…n taking two at a time and adds up the products. If n is 12, the final sum will be ?
a) 2717
b) 2600
c) 2742
d) 2718
e) None of These

(a + b)^2 - (a^2 + b^2) = 2 * (ab)
(a + b + c)^2 - (a^2 + b^2 + c^2) = 2 * (ab + bc + ca)
and so on
(1 + 2 + 3.........12)^2 - (1^2 + 2^2......12^2) = 2 * (required answer)
so required value=5434/2=2717

There are three runners viz , Nishant , Deepak and Mohit who jog on the same path. Nishant goes jogging every two days. Deepak goes jogging every four days. Mohit goes jogging every seven days. If its the first day that they started this routine, what is the total number of days that each person will jog by himself in the next seven weeks?
(a) 12
(b) 13
(c) 14
(d) 15
(e)16

Nishant = 25 days,
Deepak = 13 days,
Mohit = 7 days.
Nishant jogs on days 1,3,5,7,.....49, and Deepak jobs on days 1,5,9,.....49, and Mohit jogs on days 1, 8, 15, 22,...43.
when Deepak jogs, Nishant will always be jogging also, so Deepak jogs 0 days alone.
Nishant jogs 25-13=12 days without Deepak also jogging,
but Mohit jogs on 4 odd numbered days, 2 of which Deepak also jogs, giving us 12-4+2=10 days that Nishant jogs alone.
Nishant jogs on every odd numbered day, so Mohit jogs alone only on even numbered days.
Because there are 4 odd numbered days , there are 3 even numbered days, so Mohit jogs alone on 3 days.
so
Nishant: 10
Deepak: 0
Mohit: 3
so 10+0+3=13

The sum of all the possible values of integral a such that
(a^2-2a)^(a^2+47)=(a^2-2a)^ (16a-16) is
(a) 16
(b) 17
(c) 18
(d) 19
(e) none

a^2 - 2a will be equal to 0 or 1 or -1 then it’s possible
when a^2 - 2a = 0
so a=0,2
when a^2-2a=1 then not real values
a^2-2a=-1 so a=1
so 0,1,2 these three possible values
now when same base
((a^2+47)= (16a-16))
so a^2-16a+63=0
so a=7,9
so sum 0+1+2+7+9=19

Ten liters content of a milk water solution, which contain milk and water in the ratio 7:3 are removed and replaced with water to bring down the concentration of milk by 10% points. The amount (in liters) of water that needs to be added to the resulting solution in order to reduce the concentration of milk to 50% is?

When water= 7 liter extra
milk =7 liter decrease
then decrement of milk =10%
x * 10/100=7 so x=70 so total solution=70
now initially there was 49 liter milk and 21 liter water
but 7 liter milk reduced so 42 milk and 21+7=28 liter water
so (28+a)=(70+a) * 50% so a=14
If a and b are two real numbers such that a + b = 1, then find the maximum possible value of the product of (a^a x b^b) and (a^b and b^a)
a^(a+b) * b^(a+b)
a * b
if a+b=1 and we want max a * b so 1/4

Let {Vn} be a sequence such that V1 = 2, V2 = 1 and 2 Vn – 3 Vn-1 + Vn–2 = 0 for n > 2, then find the value of V9.

V1=2
V2=1
2Vn-3Vn-1+Vn-2=0 for n > 2
now put n=3
so 2V3-3V2+V1=0
so 2V3=3V2-V1
so 2V3=3-2=1 so V3=1/2
now put n=4
2V4-3V3+V2=0
2V4=3 * 1/2 -1=1/2 so V4=1/4
now check V1=2,V2=1,V3=1/2,V4=1/4 and So on so this is GP

sherlock homes and dr. watson have to travel from rajiv gandhi chowk to indira gandhi international airpot via the metro. they have enough coins of 1,5,10,25 paise. sherlock homes agrees to pay for dr. watson, only if he tells all the possible combination of coins that can be used to pay for the ticket.
how many combinations are possible, if the fare is 50 paise?
a 52
b 49
c 45
d 44

a+5b+10c+25d =50 ,
Method 1-
Put d=0,a+5b+10c=50 now solve this
Put d=1 then d = 2

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