Quant Boosters by VP  Set 1

A.P. So differences will be equal
2P – P = q + P – 2P = 2P – 2q – 6 – q – P
P = q – P = P – 3q – 6
∴ q = 2P and 2P – 6 = 4q
On solving, we get P = – 1 and q = – 2
∴ A.P. is –1, –2, –3………….
∴ T99 = a + 98d = – 1 + 98 x –1 = – 99

Q24) What is the range of x satisfying x^2 + 2x – 15 > 0 and x^23x4 < 0
(a) 5 < x < 4
(b) 1 < x < 4
(c) 5 < x < 3
(d) 3 < x < 4

x^2 + 2x – 15 > 0 or (x + 5)(x  3) > 0
∴ x <  5 or x > 3
x^ 2 – 3x – 4 < 0
(x – 4) (x + 1) < 0
⇒ 1 < x < 4
∴ common solution is 3 < x < 4

Q25) There are 100 people in a room. Some or all of them shake hands. Probability that there will be two people in the room who have shaken hands with the same number of people?

No of people = 100
Possible no of handshakes = 0 to 99 => 100So it seems all 100 ppl can have different no of handshakes, but notice 0 and 99 both won't b possible. If out of 100 ppl, there is one person who shakes hand with 0 ppl(none) then can there be any person among them who will shake hand with 99 ppl(all except himself? Not possible since he won't b shaking hand with himself and that guy who didn't shake hand)
Hence it means 0 and 99 both won't coexist, instead any one of those two will come
Hence now we got 100 people and 99 different no of handshakes. So even if 99 people got those 99 different number, 100th person will get no of handshakes same as one of those 99, and hence there will be atlst two people with same no of handshakes no matter what. Hence it's a certain event and probability is 1.

Q26) If x + (1/x) = 1, find x^127 + (1/x)^127

(x + 1/x) = 1
x^2 + 1 = x
x^2 + 1  x = 0
Now try to observe
Where have you seen such terms
a^3 + b^3 = (a+b) (a^2 + b^2  ab)
So here we have to multiply by
(x+1) on both sides
(x+1) (x^2 x +1 ) = 0
=> (x^3 +1) = 0
x^3 = 1
Now x^127 => (x^3)^42 * x
=> (1)^42 * x
=> x
So x + (1/x) which is equal to 1

Q27) A natural number equals 75 times the average of its digits. How many three digit numbers satisfy this condition

If average is 1
Then number is 75 (which can't be possible )If average is 2
Then number is 150 ( as yes it's average is 2 also )If average is 3
Number is 75*3 => 225 ( yes average of the digits is also 3)Average is 4
Number is 300 ( but average of digits is not 4)Average is 5
Number is 375 ( average of digits is 5 ,so possibile )Average is 6
75 * 6 => 450 ( average of digits is not 6)Average is 7
75 * 7 => 575 ( average is not 7)Average is 8
75 × 8 => 600 ( again not possible )Average is 9
75 * 9 => 675 ( but average can't be 9)So.here Basically two concepts
 Table of 75
 Average of digits can't be more than 9
As it is a three digit number
If you take average as 10 then sum.of digits would be more than 27 which can't be possible
So three numbers are possible  150, 225 , 375

Q28) The regional passport office has a waiting list of 6665 applicants for passport. The list shows that there at least 6 males between any two females. The largest possible number of females in the waiting list is

First person is female
Then eight person is female
So one combination of
1F MMMMMM
Repeats
6665 /7 => 952(1/7)
So 952 combinations will have 952 female
And one more female in the last
So 952 + 1 = 953

Q29) In an arithmetic progression, the sum of the first N terms is T and the sum of the first 2N terms is 6T. If the sum of the first 3N terms is kT, then find k

Take N = 1
Sum of first term is T
Sum of first two terms is 6T
I.e first term = T
Second term = 5T
So third term has to be 9T
Sum of first three terms = T + 5T + 9T
=> 15T
So k = 15

Q30) A clock was correct at 2 p.m, but then it began to lose 30 minutes each hour. It now shows 6 pm, but it stopped 3 hours ago. What is the correct time now?

The clock loses 30 minutes per hour.
i.e 30 minutes of this faulty clock = 60 minutes of the correct clock
From 2 p.m to 6 p.m ,
total number of hours = 4 hours
4 hours of this faulty clock => 4 x 60/30= 8 hours of original clock
So, The correct time when the clock show 6 p.m = 6 p.m + 4 = 10 p.m
But The clock stopped 3 hours ago ,
So present time is 10 p.m + 3 hours = 1 a.m