Quant Boosters  Maneesh  Set 4

in one attempt net value of distance boy climb is 3–2 = 1m
in 6 attempts boy climbs = 6 m
in the 7th attempt, he climbs = 6+3–2 = 7 m
in the eighth attempt he reaches 7+3 = 10 m
Hence answer 8 attempts

Q20) What is the remainder when (32^32)^32 is divided by 9?

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 twodigit 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 > (a1)^2 = (b+7)^2
a1 = b+7 and a1= b7 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 possiblecase 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 = 924case 2) 3 even and one odd
4 * 1 = 4 waystotal 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 > ((364)/4)+1 = 9

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

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)