# Quant Boosters by VP - Set 1

• Increment of 2 zeroes as result of multiple of 25
Means if you check for 24! =>
24/5 => 4
25/5 => 5
5/5 => 1
So 5 +1 => 6
Same for 26!

So
98! And 100!
99! And 101!
Similarly
148! --> 150!
149! --> 151!
And
173! --> 175!
174! --> 176!

Remember you don't have to consider 125! (As it results into increment of 3 zeroes )
So 6 possibilities

• Q7) Which of the following statements is false?
a) (23!)^2 > 23^23
b) (20!∗19!∗18!) < 57!
c) (33!)^4 < 33^60
d) None of these

• B is clearly right since 20 + 19 + 18 = 57

A is right coz
Lhs=> 23!^2 = (23 * 1)(22 * 2)(21 * 3)...(2 * 22)(3 * 23)

However in rhs each terms is =>23 * 23 * 23...
Certainly lhs > rhs

Now in 3rd
33!^2 > 33^33
So 33!^2 * 33!^2 > 33^33 * 33^33
33!^4 > 33^66 hence 3rd is false

OA:C

• Q8) In a plane there are 37 straight lines of which 13 passes through Point A and 11 passes through Point B. Besides, no line passes through 3 points and no line passes through both A and B and no two are parallel. Find the total number of point of intersections of the straight lines

• Total number of points of intersection of 37 lines is 37c2
But 13 straight lines out of the given 37 straight lines pass through the same point A.
Therefore instead of getting (13c2) points, we get only one point A.
Similarly 11 straight lines out of the given 37 straight lines intersect at point B.
Therefore instead of getting (11c2) points, we get only one point B.
Hence the number of intersection points of the lines is = 37c2 - 13c2 -11c2 + 2 = 535

• Q9) Find the number of ways of selection 4 letters from the word EXAMINATION

• EXAMINATION has 11 letters in total
And out of which
E --> 1
X --> 1
M -->1
T ---> 1
O ---> 1
(A) --> 2
I --> 2
N --> 2

8 distinct letters.

1. 4 letters selected, which are all distinct: 8C4 = 70
2. 2 letters alike, and 2 distinct
(eg: AAEM)
= 3c1 x 7c2 = 63
3. 2 letters alike, and 2 letters alike
(eg: NNII)
= 3c2 = 3

So answer is, 70 + 63 + 3 = 136

• Q10) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

• Method:1 (Not recommended)
_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
3 0 _ _ : 49 added
3 1 _ _ : 49 added
3 3 _ _ : 49 added
3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
Total 294 more. Remaining 21
3 9 0 _ : 7
3 9 1 _ : 7
3 9 3 _ : 7
3939

Method:2
Base method approach
0 -> 0
1 -> 1
2 -> 3
3 -> 4
4 -> 7
5 -> 8
6 -> 9
1000th term : 1000 in base 7 = 2626 < - > 3939

• Q11) One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?

• Bus .......... A....... Vikram ...B
AB is a tunnel
And suppose person is at point C
AC : CB = 1:3
Now the time at which bus reaches A
Vikram would reach from point C to point A
And the time at which bus reaches to B
Vikram reaches from C to B
Let assume
Length of tunnel is 4km
So if Vikram moves 1 km forward towards B
Bus would reach at A
So remaining distance covered by Vikram is 2km
And bus covered the whole distance 4km(length of tunnel) in the same time with a speed of 80km/hr
So speed of Vikram = 80/2 => 40km/hr

• Q12) Find the ten's place digit of 625^246 - 441^128

• (a5)^even ends with 25
Last two digits
(41)^28
=> (81)^14
=> (61)^7
=> 21
Or (41)^28
=> (4 * 8)1
=> 21

So 25-21 = 04

digit at tenth place = 0

• Q13) 25 males and 12 females can complete a piece of work in 12 days. They worked together for 8 days and the women left the work.the remaining men completed the work in 6 days. Find out in how many days 15women can complete the entire work?

• They worked for 8 days
I.e completed 2/3 work
So Remaining (1/3) work is completed by men in 6 days
I.e males complete this work in 18 days
So 18 * X /(18+X) = 12
=> 3x /(18+X) = 2
=> 3x = 36 + 2x
=> X = 36
I.e 12 women complete this work in 36 days
So 15 women in 12 * 36/15
I.e 144/5 days

• Q14) X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day one man whose capacity to do the work is twice that of X, joined. On the third day another man whose capacity is thrice that of X, joined and the process continues till the work is completed. In how many days will the work be completed, if everyone works for four hours a day?

• Total work = 20 * 7 = 140 units
Efficiency of X => 1unit /hour

First day
1 * 4

Second day
Another man joined
Whose efficiency is 2
So total work on second day
(1+2) * 4

So basically
4 [ 1 + 1+2 + 1+2+3 + 1+2+3+4 + 1+2+3+4+5 ] = 140
So 5 days

• Q15) How many multiples of 18 are there which are less than 3500 and also 2 more than the square of a natural number

• 18k = t^2 + 2 ---(1)
Now t^2 mod 18
Min Possibility 0
and max possibility - 17
so for t^2 +2 to be divisible by 18
t^2 must give remainder as 16 .
so t must be of type 18k+ 4 or 18k - 4 ..
√3600 = 60
√3500 = 59 (approx)
So maximum value of t can be 59
so t = 4 (18 * 0 +4)
14 (18 * 1 - 4)
22 ,
32,
40,
50 ,
58(18 * 3 + 4 )
So 7

• Q16)

61

61

61

33

61

61

63