Quant Boosters  Sibanand Pattnaik  Set 1

N must be a factor of (414 – 399) = 15
15 = 3 x 5
So 15 has 4 factors
So 4 possible values of N

Q11) When 101, 201 and 301 are divided by N, the remainder is 1 in each case. How many values can N take?

N is a factor of 100 = 2^2 x 5^2
So 100 has 9 factors
However N cannot be 1
So N can take 8 values

Q12) How many pairs of numbers can you have whose LCM is 144?
a) 22
b) 21
c) 18
d) 23

I've explained the ORERED part here. but this is un ordered
so (ORDERED  1 )/2 is ur answer

Q13) Dudul n Gudul make round trips between Cuttack n Bhubaneswar respectively ..They first met at 10kms from Bhubaneswar, they reached their respective destinations n returned back n on their way , they meet for the 2nd time at 300 metres from Cuttack ..Find the distance between Cuttack n Bhubaneswar ?

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 ?