Quant Boosters  Sibanand Pattnaik  Set 1

First time when Dudul n gudul meet , together they cover then entire distance between Cuttack n Bhubaneswar.. 2nd time when they meet they covered thrice the distance. So I can say 3 * 10000  300 = 29.7 kms is the distance between 2 cities
Else s1/s2 = 10000/L  10000
And in second meet
S1/S2 = (L + 300)/2L  300
So equating 2 equations we can find L

Q14) How many whole number solutions exist for the equation x + y + z = 48 such that x < y < z?
a) 1225
b) 192
c) 200
d) 872

x + y + z = 48
total whole number solutions = n+r1 C r1 = 48+31 C 31 = 50C2
But 50C2 includes "ALL POSSIBLE WAYS IN WHICH 48 CAN BE DIVIDED INTO 3 GROUPS "
So broadly there are 3 type of cases in which 48 can be divided into 3 groups ..
CASE  1 > WHEN X , Y , Z are all equal i.e X = Y = Z = 16 .. so it can be done in only 1 way
CASE  2 > WHEN 2 GROUPS ARE SAME AND 1 GROUP is different ..
so cases like
0 0 48
1 1 46
2 2 44
.......
24 24 0so total 0 to 24 = (24  0 ) + 1 = 25 cases .. .. Just for your easy understanding i ve counted else you can directly do 2a + b = 48 so "a" can take values from 0 till 24 so 25 cases .
And now each of these 25 cases would have been arranged in 3!/2! ways ..why ?? because " 2 are same and 1 is different" .. so total 25 * 3 = 75 cases
BUT THESE 75 SOLUTIONS INCLUDE THAT "16 , 16 , 16" CASE AS WELL..so exclude that .. So finally 75  3 = 72 CASES ..
CASE  3 > WHEN ALL THE GROUPS ARE DIFFERENT
This includes cases like 3 , 24 , 21 ; i.e where X , Y ,Z are different ..
Now each these "ALL DIFFERENT ELEMENT " sets like ( 3 , 24 , 21 ) can be arranged among themselves in 3! i.e 6 ways ..But we need only ONE TYPE OUT OF THOSE 6 WAYs ..In simple words those 6 ways will include cases likeX < Y > Z ; X < Y < Z ; X > Y < Z ; X < Z < Y ; X > Z > Y ;X > Y > Z
BUT WE NEED ONLY "X > Y > Z" type cases so 1 out of 6 i.e 1/6 OF (ALL DIFFERENT CASES )
So X > Y > Z CASES = (TOTAL  CASE 2  CASE 1 )* 1/6
= (50c2  72 1)/6 = [192]

Q15) What is the probability of selecting a number is selected [100,999] such that sum of digits of number is 14.

a+b+c = 14
1st digit cannot be 0.
(a'+1)+b+c = 14
a'+b+c= 13 total = 15C2=105
a' cannot be 9 or more since (a'+1) would be 10 or more then
> (9+a')+b+c = 13
a'+b+c= 4 > 6C2
b and c cannot be 10 or more.
a'+(b+10)+c = 13
a'+b+c= 3 > 5C2 and same for c
=> 1056C22*5C2 = 1051520 = 70
=> 70/900 = 7/90

Q16) The volume of a mixture of milk and water was increased by 68.75% by adding pure milk to it. If the ratio of milk n water in the resultant solution is 3:1, find the ratio of milk n water in the original solution ?

68.75 = 11/ 16
let Vol 1= 16 then vol 2 = 27
We know water proportion initially * Volume 1 = Water proportion finally *Volume 2
=> water proportion initially * 16 = (1/4) * 27
=>water proportion initially = 27/64
so water : milk = 27: 37

Q17) How many pairs of factors of N = 360 will be coprime to each other?

360 = 36 * 10 = 2^3 * 3^2 * 5
ORDERED PAIRS = (2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) = 105
UNORDERED PAIR = [(2 * 3 + 1)(2 * 2 + 1) (2 * 1 + 1) 1 ]/2 = 52
As in this case variables are implicit so Its UNORDERED PAIRS .. so 52

Q18) How many ways can 360 be expressed as Product of 2 coprimes?

Let us suppose you have to express the number 360 as product of 2 co primes i,e a and b
So 360 = 2^3 * 3^2 *5
SO TOTAL 3 PRIMES Are there in 360 ..
We have to express as product of 2 coprimes "a" and "b"
So each of the primes i.e 2 , 3 and 5 has just 2 choices i.e either they can go "a" or to "b" .. so 2 * 2 * 2 = 8
there these 8 include cases like (1. 360 ) and (360,1) i.e "repeat cases
so 8/2 = 4
the formula is (2^n)/2 = 2^(n1)

Q19) Real madrid and Manchester united play a football match in which REAL beats MANCHESTER 4  2 ..In how many Ways can the goals be scored provided MANCHESTER never had a lead over REAL during the match?

if Manchester never had a lead means 1st goal was scored by REAL .. So score card at this juncture would ve been 1  0 ..The only way MANCHESTER could take lead is by scoring next 2 goals , i.e a pattern like RMMRRR.. Except this one case , MANCHESTER Can no way take lead as such ..
So (5!)/3! *2! = 10
NOW excluding this one case of RMMRRR ... We have 10  1 = 9

Q20) How many integers P are there between 0 and 10^100 such that units digit of P^3 is 1 ?

So , each number from 1  10 has a distinct unit digit for its cube ..
occurrence of each digit in cube is 1/10 times
So 0  10^100 it will be 10^100 * 1/10 = 10^99

Q21) 20 groups of five shooters each compete in an shooting championship. A shooter finishing in Nth position contributes N points to his team, and there are no ties. The team that wins will be the one that has the least score . Given that , the 1st position team's score is not the same as any other team, the number of winning scores that are possible is

The groupâ€™ scores must sum to 1 + 2 + . . . + 99 + 100 = 5050.
The winning group can be ATMOST 1/20 * 5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15.
However, not all scores between 15 and 252 inclusively are possible because all teams must have integer scores and no team can tie the winning team.
number of scores possible = 251  15 + 1 = 237

Q22) X varies inversely as y and x varies directly as the square of z. If y decreases by 43.75% and z decreases by 75%, by what percent does x change?
a. 88.88%
b. 50%
c. 45.45%
d. 54.54%
e. 44.44%

Here my approach would be
X varies inversely as Y so XY is Constant
X varies directly with z^2 so x/z^2 is a constant
so (XY)/Z^2 = constant
X = (constant * Z^2 )/Y
So multiplying factor of X = (1 * multiplying factor of Z^2 )multiplying factor of Y
=> multiplying factor of X = [1 * (1/4)^2] / (9/16)
=> multiplying factor of X = 1/9
As mulpliying factor of x is 1/9 so change is "x" is 8/9 .. i.e 11.11 * 8 = 88.88 %

Q23) Total positive integral solutions of 4x + 5y + 2z = 111?