# Quant Boosters - Sibanand Pattnaik - Set 1

• 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?

y needs to be odd, else LHS would become even.
say y = 2k-1 k is positive
=> 4x + 5 (2k-1) + 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 cases

HCF > 1
only possible case for HCF = Either 2 and 5 ..
case -1 (when HCF =2)
2x + 2y + 2z = 20
=> x + y + z = 10
9c2

Case -2 , (When HCF = 5)
5x + 5y + 5z = 20
=> x + y + z = 4
so 3c2

Therefore , 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 ?

• It is unordered distribution of a+b+c = 22
so total cases = 24c2
2 same n 1 different cases .. will be .. 0 0 22 ; 1 1 20 ....till 11 11 0 .
so total 0 - 11 that is 12 cases and each of them ll be arranged in 3!/2! i.e 3 ways
why ?? because they are of type A A B ..
so( 24c2 - 36 ) / 6 + 12 = 52

• Q30) X and Y have some chocolates with them which they wish to sell . The cost of each chocolate is equal to the number of total chocolates with both of them together initially. Together they sell all the chocolates and after that they start distributing the money collected in this particular fashion. First X takes a 10 rupee note, then Y takes a 10 rupee note and so on. In end it's turn of Y who didn't get any more 10 rupees. How much rupees Y get in his last turn?

• As X started the distribution part and took a 10 rupee note first and in the end also, he is able to take a 10 rupee note. That means Total amount, which needs to be a perfect square for n mangoes @ n rupees per mango, is odd multiple of 10 plus some more which is less than 10. That means ten's place digit of the perfect square is ODD. So certainly unit digit of perfect square is 6.

• @sibanand_pattnaik

hi,
there is a bit confusion i have.
100 + a^3 = -900 (a = -10)
so how would the answer change here?