# Quant Boosters - Sagar Gupta - Set 3

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Compilation of posts from Sagar Gupta - 99.2 Percentile in CAT 2015 (Quant)

• Q1) Let a secret three digit number be cba. If the sum of cab + bac + bca + abc + acb = 2536, what is c+b+a ?

• 222a +222b+222c -a-10b-100c = 2536
221a + 212b + 122c = 2536
122(a+b+c) + (99a + 90b) = 2536
2536 mod 9 = 7
99a + 90b mod 9 = 0
122(a+b+c) mod 9 = 7
5(a+b+c) mod 9 = 7
so : a+b+c = 14 as 70 mod 9 = 7

• Q2) The average height of a group of n persons is 90. Two persons with heights 60 and 65 respectively leave the group. A third person with height between 92 and 97 joins the group. The new average height of the group is a prime number. Find the possible height of the person who joined the group if it is known that the new average height is less than 120?
a. 92
b. 93
c. 97
d. 94
e. CBD

• a) 90n -125 + 92 = 90n - 33 / n-1 = 33 + 57n / n-1 => n=4 / n=20 / n=58 -> new avg age => 109 or 93 or 90 : 109 possible
b) 90n - 125 + 93 = 90n - 32/n-1 = 32 + 58n/n-1 => n=59 -> new avg age => 91 : not possible
c)90n - 125 + 97 = 90n - 28/n-1 = 28 + 62n/n-1 => n=32 -> new avg age = 92 : not possible
d) 90n -125 + 94 = 90n -31 / n-1 = 31 + 59n/n-1 => n=60 -> new avg age = 91 : not possible
only 109 is possible : which comes from option A

• Q3) There are 20 people in your applicant pool, including 5 pairs of identical twins. If you hire 5 people randomly, what are the chances you will hire at least 1 pair of identical twins?

• 10C5 + 10C4 * 10C1 + 10C3 * (10C2 - 5 ) + 10C2 * ( 10C3 - 5 * 8 ) + 10C1 * ( 10C4 - 5C2 - 5C1 * (8C2-4 ) ) + 10C0 * ( 10C5 - 5C2 * 6C1 - 5C1 * (8C3 -4 * 6 )
252 + 10 * 210 + 120 * ( 40) + 45 * (80) + 10(210 -10-120 ) + ( 192 - 160 ) = 11584

• Q4) Find the number of integer solutions for |x| + 2*|y| + |z| = 4

• |x| + 2|y| + |z| = 4
y=0 => |x| +|z| = 4 : 16
y=+/-1 =>|x| + |z| =2 : 8
y=+/-2 => |x| + |z| = 0 : 1
16 + 2 [ 8 + 1 ] = 34

• Q5) Find the minimum value of x^2 * y^2 - y^2 * z^2 - z^2 * x^2 where x^2 + y^2 + z^2 = 5 and x, y & z are real numbers

• since u need minimum value...put x=0
-y^2 * z^2 y^2+z^2 = 5
min when y=z = root ( 5/2 )
-y^2 z^2 = -25/4 = OA

• Q6) Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?

• a0 , b0 , c0 , d0 , e0
a0 : a1 to a24 + sara
b0 : b1 to b24+ sara
c0 : c1 to c24 + sara
d0 : d1 to d24 + sara
e0 : e1 to e24+ sara
2 friends of sara are connected to each other : say a-b
24 * 5 + 3 = 123
all five friends are mutual friends
each has 20 distinct friends and give people a0,b0,c0,d0,e0 : 20 * 5 + 5 =105

• Q7) A man sells two houses at the rate of Rs. 1.995 lakh each. On one he gains 5% and on the other, he loses 5%. His gain or loss percent in the whole transaction is ?

• General Formula : if equal rate of interest ( one + and other - ) , the total loss % is given by r^2 / 100.
cp = a
sp = y
y = a ( 1 + r/100 ) => a = y / ( 1 + r/100 )
cp = b
sp = y
y = b ( 1 - r/100 ) => b = y / ( 1 - r/100 )
cp = ( 2y / 1 - r^2 / 10000 )
sp = 2y
2y ( 1 - r^2 / 10000 ) = 2y ( 1 + x / 100 )
( 1 - r^2 / 10000 ) = 1 - x/100
r^2 / 10000 = x/100
x = r^2 / 100 = loss %

• Q8) Given N = 35 x 36 x 37 .......... x 67, what is the remainder left when N is divided by 289?

• 35 x 36 x 37 .......... x 67 mod 289
take out 17 common
( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17
1 * 2 *....16 * 3 * -1 * -2 * ... -16 mod 17
16! * 16 ! * 3 mod 17
16 * 16 * 3 = 1 * 3 mod 17 = 3
17 * 3 = 51
35 x 36 x 37 .......... x 67 mod 289
take out 17 common
( 35 * 36 *...50 * 3 * 52 * 53 *....67 ) mod 17
1 * 2 *....16 * 3 * -1 * -2 * ... -16 mod 17
16! * 16 ! * 3 mod 17
16 * 16 * 3 = 1 * 3 mod 17 = 3
17 * 3 = 51

• Q9) Find the number of ordered pairs of integers for : x^2 + y^2 - xy = 727

• x^2 + y^2 - xy = 727
x^2 - yx + (y^2 - 727 ) = 0
D = y^2 - 4(y^2 - 727)
D = 2908 - 3y^2
2907-3y^2+1 should be a perfect square
3(969-y^2)+1 should be a perfect square
clearly y=31 , which gives 2 possible x = 13 and x= 18
values can be interchanged here :
y =13 will give x=31 , x=18
y=18 will also give x=31 , x=13
same will be possible when all these values are of negative sign
so total 3 * 2 * 2 = 12 values

• Q10) Remainder of 1776^1816! divided by 2000 is ?

61

63

61

61

61

42

61

63