Quant Boosters  Sagar Gupta, CAT Quant  99.2 Percentile  Set 4

Author : Sagar Gupta  MBA student at Symbiosis Institute of Operations Management. 99.2 Percentile in CAT 2015 (Quant)
If x, y and z are whole numbers such that x ≥ y,then how many solutions are possible for the equation x + y + z = 36?
(a) 361
(b) 323
(c) 382
(d) 342[ 38c2  19 ] /2 = 342 , when x > y and x < y
when x > = y
342 + 19 = 361In a group if 80% drink tea and 60% drink coffee,what is the maximum percentage of people drinking either tea /coffee but not both?
I + II = 100
I + 2 II = 140
II = 40
I = 601 unit of A is made by mixing 4 units of B and 5 units of C. 1 unit of B is made by mixing 1 unit of X, 4 units of Y and 1 unit of Z. 1 unit of C is made by mixing 2 units of X, 6 units of Y and 1 unit of Z. The weight of 1 unit each of X, Y and Z is 5 kgs, 3 kgs and 8 kgs respectively. What is the total weight of Y required to make 1400 kgs of A?
(1) 630 kgs
(2) 720 kgs
(3) 690 kgs
(4) 870 kgs
(5) 570 kgsA = 4B + 5C
B = X + 4Y + Z
C = 2X + 6Y + Z
X = 5 , Y = 3 , Z = 8
B = 5 + 12 + 8 =25
C = 10 + 18 + 8 = 36
A = 100 + 180 = 280
280 of A with 46Y or 46*3 = 138
690 kgs for 1400 kgs of AFind HCF of 2^100  1, 2^120 1 ?
HCF ( 2^a 1 , 2^b 1 ) = 2 ^( HCF a, b)  1
= 2^20  1Find last two digits of 15 * 37 * 63 * 51 * 97 * 17
15 * 37 * 63 * 51 * 97 * 17 mod 100
15 * 37 * (37) * 51 * (3) * 17 mod 100
45 * 37^2 * 17^2 * 3 mod 100
45 * 89 * 03 * 69
35A five digit number divisible by 3 is to be formed using numbers 0, 1,2,3,4,5 without repetition . The total number of ways in which this can be done is....
A. 216
B. 240
C. 600
D. 31251 + 2 + 3 + 4 + 5 =15
> 5! ways
0 + 1 + 2 + 4 + 5 =12
>5!  [ 5! /5 ] =120  24 =96
total 120 +96 =216The total number of ways of selecting two numbers from the set {1,2,3,4,......,30}, so that their sum is divisible by 3, is
a. 95
b. 145
c. 190
d. None of the above.3k, 3k + 1, 3k + 2
divisible by 3:
case 1 : 3k + 3k
10 * 9 = 90 ways
case 2 : (3k+2) + (3k+1)
10 * 10 = 100 ways
total =190 waysTrain X started from point A at 9:00 am with a speed 72km/hr towards station Y. 2 hrs after train Y starts from point B towards X at a speed of 90 km/hr. They cross each other at 1:30 pm. But owing to signal problems at 12:00 noon the speeds of both the trains is reduced by same quantity such that they now cross each other at 4:30 pm. Calculate the new speed of train that started from point A.
Train X starts at 9:00 am @72 km/hr
at 11:00 am, it is 144 kms from A
A...144 kms...A'......d...........B
Now Y starts from B at 11:00am @90 km/hr
A' and B is the position at 11:00 am
:
They should have met at 1:30pm
72 * 2.5 + 90 * 2.5 = d =405 kms
:
But they meet at 4:30 pm
from 11:00 to 12:00 , they travel at normal speed
Distance reduced = 72+90=162
Distance between them at 12:00 = 243 kms
Now they reduce their speed by k
(72k) * 4.5 + (90k) * 4.5 =243
k=54
7254 =18km/hrabcd is a 4 digit perfect square, then if each digit is increased by 3 the number is still a perfect square. Find abcd.
abcd = x^2
1000a +100b +10c +d = x^2
1000(a+3) +100(b+3) +10(c+3) +d+3 = y^2
y^2x^2=3000 +300 + 30 + 3
y^2  x^2 = 3 [ 1000 +100 +10 +1 ]
y^2  x^2 =3333
3333=101*33
y+x =101
yx =33
y = 67
x = 34
x^2 =1156What is the remainder when 72 to the power 7202 is divided by 625?
E[625]=500
7202 mod 500 = 2
72^2 mod 625 = 5184 mod 625 = 184