Quant Boosters  Sibanand Pattnaik  Set 1

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 group’ 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?

Answer by Bruce Wayyne
y needs to be odd, else LHS would become even.
say y = 2k1 k is positive
=> 4x + 5 (2k1) + 2z = 111
2x + 5k +z = 58
Let k = 1,
2x+z= 53 so (1, 51)...(26, 1) so 26
k=2
2x+z=48 so (23, 2).....(1, 46) so 23
k=3
2x+z=43 so (1,41)....(21, 1) so 21
k=4
2x+z= 38 so (1, 36)....(18,2) so 18
k=5
2x+z= 33 so (1,31)....(16,1) so 16
Hence the pattern is +3, +2, +3, +2....so on.
so (26+21+16+11+6+1) + (23+18+ 13 + 8 + 3) = 146

Q24) 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
or
milk : water = 37: 27

Q25) HCF of 3 numbers is 6 and their sum is 120. How many such triplets exist ?

6x + 6y + 6z = 120
=> x + y + z = 20
19c2 casesHCF > 1
only possible case for HCF = Either 2 and 5 ..
case 1 (when HCF =2)
2x + 2y + 2z = 20
=> x + y + z = 10
9c2Case 2 , (When HCF = 5)
5x + 5y + 5z = 20
=> x + y + z = 4
so 3c2Therefore , required cases = 19c2  9c2  3c2 = 132 (ORDERED )
Now remove the 2 same 1 cases where HCF is 1
cases , 1 , 1 ,18 ; 3, 3 , 14 ; 7,7,6; 9,9,2 ... each of these arranged in 3!/2! ways so , 12 ways ..
so (132  12)/6 + 4 = 24
(for un ordered) ..

Q26) Ram bought a few mangoes and apples spending an amount of at most 2000.If each mango cost 4 and Apple 6 and Ram bought at least 1. Find the different possible amounts could have spent in purchasing fruits?

Sum of 3 number is even so either all three are even or 2 odd and 1 even ..
7^4 > 2002 .. so it must be less than this ... 5^4 is 3 digit number so we MUST TAKE 6^4 else how can you form a 4 digit number .. ??
that way 6^4 is included , now only 2 cases are possible
either the other 2 number ll be even or they have to be odd ... By common sense , we ignore 4 and 2 so it must be 3^4 and 5^4
so 6 + 5 + 3 = 14

Q27) If 1/a + 1/b + 1/c + 1/d = 2, where a , b , c , d are distinct natural numbers, what is the value of a + b + c + d ?

If sum of Factors of N excluding N is equal to N then N is called a perfect number ..
for Eg 28 = 1 + 2 + 4 + 7 + 14
496 = 1 +2 +4 +8 + 16 + 31 + 62 + 124 + 248
PERFECT NUMBERS show another property
The sum of the reciprocal of the factors of a perfect number INCLUDING THE NUMBER ITSELF = 2
Again i repeat "INCLUDING THE NUMBER ITSELF"
for eg : 28 is a perfect number
1+ 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 2

Q28) Find the largest value of a for which 100 + a^3 becomes perfectly divisible by 10 + a

by Factor theorem, if 100 + a^3 is divisible by 10 + a then substitute a = 10 in 100 + a^3 = 100  1000 = 900
Now 10 + a must be a factor of 900
so let K * (10 + a) = 900
Now to obtain the greatest value of "a", we have to put K as 1
so 10 + a = 900
=> a = 890

Q29) How many ways can 22 identical balls be distributed in 3 identical boxes ?