# Quant Boosters - Maneesh - Set 4

• 32^2 mod 9 = 1024 mod 9 = -2mod 9
(32^32)^32 mod 9 = (-2)^(16 * 32)mod 9 = 2^(512) mod 9
Since 2^3 mod 9 = -1
2^512 = 2^510 * 2^2
2^510 mod 9 = 2^(3 * 170) mod 9 = -1^170 = 1
2^2 mod 9 = 4 mod 9 = 4
2^512 mod 9 = 4 * 1 = 4
Hence answer 4

• Q21) If (a + 1/a)^2 = 3 then (a^3 + 1/a^3) = ?

• a+1/a = root3
(a+1/a)^3 = 3root3
a^3 + 1/a^3 + 3(a + 1/a) = 3 root3
a^3 + 1/a^3 + 3root3 = 3root 3
a^3 + 1/a^3 = 0

• Q22) If a and b are two-digit prime numbers such that a^2 - b^2 = 2a + 14b + 48. Find the largest possible value of a + b.

• solution by Ytiam Unata

a^2 - 2a + 1 = b^2 + 14b + 49 ---> (a-1)^2 = (b+7)^2
a-1 = b+7 and a-1= -b-7 or, a - b = 8 and a + b = -6.
89 and 97
sum = 186

• Q23) Mahesh wrote his class tests in 4 subjects, each with a maximum mark of 50. He scored only 51 marks in all but observed that the marks he scored in each subject is either odd or prime. If he did not get 1 mark in any subject, in how many ways could he have scored the total of 51 marks in the four subjects?
a) 924
b) 1018
c) 235
d) 928

• a + b + c + d = 51
3 odd one even
1 odd 3 evn possible

case 1) one even and 3 odd
a + b + c = 49
2A + 1 + 2B + 2C + 1 + 1 = 49 Where A,B,C =1,2,3...
2A + 2B + 2C = 46
A + B + C = 23
4 * 22c2 = 4 * 11 * 21 = 4 * 231 = 924

case 2) 3 even and one odd
4 * 1 = 4 ways

total 924 + 4 = 928

• Q24) What is the total number of positive integer solutions that satisfy the equation 4x + 3y = 120

• 4x + 3y = 120
when x = 3 y = 36
y = 4 x = 27
so values of x will be 3,6,9...27
values of y will be 4,8...36
answer --> ((36-4)/4)+1 = 9

• Q25) How many numbers between 200 and 500 (both inclusive) are coprime to 70 ?

• n[2] = 500-200/2 +1 = 151
n[5] = 61
n[7] = 43
n[10] = 31
n[14] = 21
n[35] = 9
n[70] = 5

151 + 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 +2

quant books : 3995/5 + 1 = 800
verbal books : 3996/6 + 1 = 667

5a+5 = 6b+2
5a = 6b - 3
a = b + ( b-3)/5
b=3 : 20
b=8 : 50
30c+20 form : both

20,50.....4000
3980/30 + 1= 133

only 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 : 1-2 = -1,
Now the three numbers are -1 ,4,8
Step 2 : -1-4 = -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: (1-2), 4,8,16
Step 2: (1-2 -4), 8,16
Step 3: (1-2 -4-8),16
Step 4: 16 – (1-2-4-8) = 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

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