Quant Boosters  Maneesh  Set 4

n[2] = 500200/2 +1 = 151
n[5] = 61
n[7] = 43
n[10] = 31
n[14] = 21
n[35] = 9
n[70] = 5151 + 61 + 43  31  21  9 + 5 = 260  61 = 199
coprime to 70 > 301  199 = 102

Q26) Find the number of trailing zeros in 80C16

80!/64! * 16!
power of 2 = 2^78/2^63*2^15 => 0
Hence 0 trailing zeros

Q27) 4003 students line up in a row from left to right. Starting from the left, every 5th student is given a piece of quant book ; starting from the right, every 6th student is given a va book.
a) how many students get only quant book ?
b) how many students get only VA book ?
c) How many students get both the book ?
d) how many student got none of the book ?

solution by sagar gupta
5,10,15............4000 : 5a+5
2,8,14..............3998 : 6a +2quant books : 3995/5 + 1 = 800
verbal books : 3996/6 + 1 = 6675a+5 = 6b+2
5a = 6b  3
a = b + ( b3)/5
b=3 : 20
b=8 : 50
30c+20 form : both20,50.....4000
3980/30 + 1= 133only quant = 800  133 =667
only verbal = 534
both = 133
none = 2669

Q28) 17 numbers  2^0,2^1,2^2,.......2^16 were written on the board.You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboard when there is only one number left?

(solution by nitin gupta sir)
Let’s take 4 numbers , 2^0, 2^1,2^2,2^3 which are basically 1,2,4,8
Now, step 1 : 12 = 1,
Now the three numbers are 1 ,4,8
Step 2 : 14 = 5
Now the two numbers are 5 & 8
Step 3 : 8 – (5) = 13
So the largest number is 13
Similarly for 5 numbers , 2^0, 2^1,2^2,2^3 ,2^4 which are basically 1,2,4,8,16
Step 1: (12), 4,8,16
Step 2: (12 4), 8,16
Step 3: (12 48),16
Step 4: 16 – (1248) = 29
So for 2^0,2^1,2^2,.......2^16
Ans : 2^16 – (2^0 – 2^1 – 2^2 – 2^3 2^4 ……..2^15)
Rearranging this : 2^16 +2^15 + 2^14 + ……+ 2^1  2^0
= 2^17 – 3 = 131069

Q29) Consider the series : 0, 0, 2, 8, 22, 52, 114 ....
Find the sum of tenth and eleventh term of this series. (A10 + A11)

General term = 2^x  2x
A(10) = 2^10  20 = 1004
A(11) = 2^11  22 = 2026
Sum = 1004 + 2026 = 3030

Q30) The quantity of water (in ml) needed to reduce 9 ml saving lotion containing 50% alcohol to a lotion containing 30% alcohol is

4.5/(9+x) = 3/10
27 + 3x = 45
x = 6ml

Here
nf(x)= f(x^n)F(343sqrt3)= 3f(7)+1/2f(3)
F(7) and f(3) can be found using given values.