# Quant Boosters - Swetabh Kumar - Set 6

• A guy throws a pair of dice. If he doesn't get the sum as even number, he throws the dice again and again until he gets it. And then he tosses the coin. What is the probability of him getting a head?

sum even = 1/2
so 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 * 1/2....
= 1/4 / (1/2) = 1/2
if A and B are independent, then P(A int B ) = P(A) * P(B ). That is why we multiplied probabilities here.

If Ratio of Cost Price and Mark Price is 5:8 and that of % profit on sale to % discount is 2:3 then find % of Discount.
a.8.33%
b. 16.66%
c. 25%
d. CBD

CP=50, MP=80.
3 * (SP-50) * 100/SP = 2 * D (D=discount%)
Also, SP = (1-D/100) * 80 = 80 * (100-D)/100
so 3 * {80(100-D)/100 -50 } * 100 = 2D * {80 * (100-D)}/100
150 { 75-2D} = D {200-2D}
D^2 - 250D + 5625 = 0
D= (250-200)/2 = 25%

Find the product of the irrational roots of the equation (2x – 1) (2x -3) (2x – 5) (2x – 7) = 9.
a) 4
b) 3/2
c) 3/4
d) 4/3

(2x-1)(2x-7)(2x-3)(2x-5) = 9
(4x^2-16x+7) (4x^2-16x+15)=9
Let 4x^2-16x=t
(t+7)(t+15)=9
t^2+22t+96=0
t=-16, t=-6
so 4x^2-16x+16=0 and 4x^2-16x+6=0
x^2-4x+4=0 2x^2-8x+3=0
1st gives x=2 and second eqn gives both irrational roots. so c/a= 3/2

How many solutions are possible for the inequality |x-1| + |x-6| < 5

Median of (1,6) = 3.5
x = 3.5 gives 5 as min value. so cannot be less than 5

If N be the number of consecutive zeros at the end of the decimal representation of the expression 1! × 2! × 3! × 4! x ... x 99! × 100! Find the remainder when N is divided by 1000?

1^100 * 2^99 * 3^98 ... 100^1
power of 5: (1+6+11....96)+(76+51+26+1) = 970+154=1124
so 1124 mod 1000=124

A and B started running from the same point and in opposite directions around a circular track of radius 24.5 m. A’s speed was twice that of B’s speed. They met each other aиer 14 seconds. What was A’s speed?

A=2x, B=x
length= 2pi * r = 49pi = 154 m.
so given, 154/3x = 14 x=11/3 so 2x=22/3

Find the approximate sum of the series 11/4, 31/8, 79/16 up to 10 terms?

(2+1-1/4) + (3+1-1/8) + (4+1-1/16)...10 terms
(2+3+4....11) + (1+1+1...10 times) - (1/4+1/8+1/16...10 terms of GP)
= 65 + 10 - 1/2{1-(1/2)^10} = 75- 1/2 {1023/1024} = 74.5 approx

If [x] denotes the greatest integral function, then find the value of [√1] + [√2] + [√3] +..........+ [√2004]

1 comes 3 times (from rt 1 to rt 3)
2 comes 5 times (from rt 4 to rt 8 )
3 comes 7 times (from rt 9 to rt 15) and so on.
so ever number N comes 2n+1 times. so N(2N+1)
till 43. After that the remaining 69 numbers with 44.

1 * 3 + 2 * 5 + 3 * 7..... 43 * 87 + 44 * 69
so 2n^2 + n sigma till 43 + 44*69
2 * 43 * 44 * 87/6 + 43 * 22 + 44 * 69 = 58850

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