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+r-1 C r-1 = 48+3-1 C 3-1 = 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 0

so 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 like

X < 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
=> 105-6C2-2*5C2 = 105-15-20 = 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 co-primes "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^(n-1)

• 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 ?

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