# Quant Boosters by Hemant Malhotra - Set 8

• Two vessels contain a mixture of spirit and water. in the first vessel the ratio of spirit to water is 8:3 and in the second vessel the ratio of spirit to water is 5:1. a 35 litre cask is filled from these vessels so as to contain a mixture of spirit and water is in the ratio of 4:1. how many litres are taken from the first vessel

8/11----------------------5/6
------------- * 4/5

240-----------------------275
-------------264

(275-264)---------(264-240)

so 11:24
so 11 from 1st vessel

A dishonest seller uses a weight of 800g in place of 1 kg and adds 20 % impurities in sugar ...what would be his profit percentage if he claims to sell sugar at cost price

Method1- he is already making a profit of 20% by giving 800 for 1000
Then he is mixing 20% impurity that means 160/640 * 100.
The profit would be 25%. Then apply successive % tage. 25+20+20 * 25/100 = 50% profit.

Method2- let the price be 1 rs per gram :-
let x gram fo sugar he uses , then wid 20% x impurity , such that the total is 800 so x+1.2x= 800
x= 800/1.2 =666.66
also he sell it at 1000g = 1000
thus profit = 1000-666.66 = 333.33
50% profit

Method3- uses a weight of 800gm so multiplying factor =8/10
added 20% impurities , so multiplying factor is 12/10
. now in both the cases it is profit so =1 *10/8 * 12/10 = 1.5 so 50% profit

a, b, c, d are 4 consecutive digits such that a > b > c > d.
How many numbers abcd are there such that abcd is divisible by 37

abcd = x * 10^3 + (x-1) * 10^2 +(x-2) * 10 +(x-3) mod 37
= x + 26(x-1) + 10(x-2) + x-3
= 38x -49 mod 37
= x-12 mod 37
where x can take value from 9 - 3
so x-12 will not be div by 37 so OA=0

What will be the minimum possible value of |36^a - 5^b|, where a and b are natural numbers

Unit digit of 36^a - 5^b is always 1 or 9
36^a- 5^b is divided by 4 remainder is 3
so Lowest such number is 11 (which is possible for a = 1, b = 2)

If r, s, t, and u denote the roots of the polynomial f(x) = x^4 + 3x^3 + 3x + 2, find the value of
1/r^2 + 1/s^2 + 1/t^2 + 1/u^2

OA= 9/4

Method 1.. find every value sum and all then find ans

Method2- Transformation Method1-
x^4+3x^3+3x+2=0
put x=sqrtx
x^2+3xsqrtx+3sqrtx+2=0
x^2+2=-3x*Sqrtx-3sqrtx
so x^4+4+4x^2=9x^3+9x+18x^2
so x^4-9x^3-14x^2-9x+4=0
now reverse of this
1/x^4-9/x^3-14/x^2-9/x+4=0
so 4x^4-9x^3+...............=0
so sum of roots=9/4

If A is the greatest integral multiple of 36 such that all of its digits are even and no two of the digits are the same. Find the remainder when A is divided by 1000.

36=9*4
so number should be div by 9 also
Combination 8,6,4,0 is possible.
max no. Will be 8640 ....remainder is 640

Raju likes the number 17. He likes monic quadratics (they are his favourite) with integer coefficients such that 17 is a root of the quadratic and the roots differ by no more than 17. Compute the sum of the coefficients of all of Raju's favorite polynomials. (A monic quadratic is a quadratic polynomial whose x² term has a coefficient of 1.)

any quadratic can be represented in form of (x-a)(x-b)
now we need sum of coefficient which can be find by putting x=1
so (1-a) (1-b)
now one root is 17 and difference between a and b < =17
b could take integral values from to 34
so sum will be
-16 * (1-0)+(-16 * (1-1))+........(1-34))
so -16 * (1-0+1-1+1-2+1-3+1-4....1-34)
=16 * 16 * 35=8960

A man is running in a railway tunnel. When he has covered exactly two-fifth of the length of the
tunnel he sees a train approaching the tunnel from behind. He realizes that irrespective of the
direction in which he runs he will be hit by the train at one of the ends of the tunnel. If the speed of
the man is 10 km/hr, find the speed of the train.
(a) 60 km/hr
(b) 50 km/hr
(c) 40 km/hr
(d) 30 km/hr

|train______a_______|2b_____man________3b____|
Sman /Strain = 3b/ 5b+a = 2b / a
a = 10b
so Sman/ Strain = 1/5
so 5 * 10=50

There are 100 tokens numbered 1 t0 100. In how many ways we can draw two tokens simultaneously so that their sum is greater than 100.

Method1- For 100, 99 numbers can be taken ( 1 to 99 )
For 99, 97 numbers can be taken ( 2 to 98 )
For 98, 95 numbers can be taken ( 3 to 97 )
....
...
For 51, 1 number can be taken ( only 50 )
So find out the sum of ( 99+97+95+....3+1) = 2500

Method2- 101c2 - 50}/2 = 2500

If all the roots of the equation 30x^3- ax2 + bx - 1001 = 0, where a, b > 0, are rational and improper fractions, what is the maximum possible value of (a + b)

Proper roots will be 1001/2, 1/3 and 1/5 (for maximum a + b)
30x^3 - ax^2 + bx - 1001 = 30(x - 1001/2)(x - 1/3)(x - 1/5)
now put x = -1
a + b = -1031 + 30(1003)(4)(6)/30
so a + b = 23041
for improper it should be a =347 and b = 1142 ... and the roots be ( 7/5 , 11/3 and 13/2 ) ..
1489
OA= for proper = 23041 and for improper 1489

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