Quant Boosters  Hemant Malhotra  Set 4

p + q + r = 7 so p + q = 7  r
q + r = 7  p
and p + r = 7  q
so we need (7p)(7q)(7r)
means we need equation whose roots are 7p,7q,7r or
x^37x^26x+5=(xp)((xq))(xr))
put x=7
so 7^37 * 7^26 * 7+5 =  37

Q4) If three medians of the triangle are of the lengths 9, 12 and 15, find the area of the triangle.

Method 1 – Direct formula
Area of triangle using medians, A = 4/3 ( Sqrt(s(sp)(sq)(sr))
p, q and r are medians and s = (p+q+r)/3Method2 ABC is the triangle, AD, BE, CF are medians and G centroid
extend GD to a point M such that GD = DM, then CM = BG
so, triangle MCG; has sides 6, 8 and 10
area = 24
area of ABC = 24 * 3 = 72.Method 3
4 * (a^2+b^2+c^2)=3 * (9^2+12^2+15^2)
so a^2+b^2+c^2=3 * (81+225+144)/4
now we have a^2+b^2+c^2
apply a^2+b^2=2*(AD^2+(c^2/4))

Q5) Given 3x^2 + x = 1, find the value of 6x^3  x^2  3x + 2015.

6x^3+2x^2 = (2x)(3x^2+x) = 2x
and (3x^2+x) = 1
so 6x^3x^23x+2015 = 2x12x+2015 = 2014.

Q6) Let a, b and c be the roots of x^3  2x^2 + 2x + 4 = 0.
Find the value of 1/(a^2 + b^2) + 1/(b^2 + c^2) + 1/(c^2 + a^2) + 1/4

a + b + c = 2
ab + bc + ac = 2
abc = 4
here a+b+c=ab+bc+ca so a^2+b^2+c^2=0
question reduces to
(1/a^2 + 1/b^2 + 1/c^2) + 1/4
now we can easily find the value by solving thisAnother method : x^32x^2+2x+4=0
roots are a,b,c equation whose roots are a^2,b^2,c^2
will be (x^3/2)2x+2sqrtx+4=0
x*sqrtx2x+2sqrtx+4=
x^3+4x+4x^2=4x^2+1616x
so x^3+20x12=0
now roots of this equation are a^2,b^2,c^2
so equation whose roots are 1/a^2,1/b^2,1/c^2 will be reverse of this
so 1/x^3+20/x12=0
12x^3+20x^2+1=0
so 12x^320x^21=0
so 1/a^2 + 1/b^2 + 1/c^2 = 5/4
so 5/4 + 1/4 = 1

Q7) From 100 litres of milk, 10 litres of water is added and then 20 litres of solution is removed. Next 30 litres of the water is added and 20 litres of solution is removed. Find the amount of milk, in litres, in the solution now.

Method to tackle These Type Of Questions
always take initial as those part which is not added,
here milk as it is never added
here in first step 10 litre of water is added so solution=100+10=110
in the 1st time 20 is removed out off 110 so
leftout part is 90/110 so our multiplying factor is 90/110
no 30 is added to 90 so 120 solution
in the 2nd case 20 is removed from 120 so left out is 100/120
so amount of milk= 100 * 90/110 * 100/120=750/11
PS Always remember Our concentration should be on total solution and removed part .... Added part will only increase the value of solution

Q8) A shopkeeper claims to sell his items at a loss of 10% but instead he secretly makes a profit of 20% by using a false weight. what is the effective profit percentage he gets on every 100gm he sells?

Method1 Approach= He claims to sell 120 when he actually sells 100 (using false weights)
Now, he claims to sell them at a loss of 10%
He makes 120*0.9 = 108
so 8% profitMethod2 = Multiplying factor
100 * 9/10 * 12/10=108 so 8%Method3 2010+(20*(10)/100=8%

Q9) A dishonest trader, at the time of selling and purchasing, weighs 10% less and 20% more per kilogram respectively. Find the percentage profit earned by treachery. (Assuming he sells at Cost Price)
a) 30%
b) 20%
c) 25%
d) 33.33%

Method1 Let's say original weight is 100 now he is taking 20% extra so total 120 and selling 10% less so 90
So (12090/90) * 100
= 30/90 * 100= 33.33%Method2
multiplying factor
1 * 10/9 * 6/5=1.33 so 33%

Q10) A beaker contains milk and water in the ratio 5 : 3. If a person removes 6 L of this mixture and replaces it with pure water, the ratio of milk to water becomes 1 : 1. What is the amount of milk present in beaker initially

Always remember one thing
Final =Initial * fraction left 1 * fraction left 2 * fraction left 3 * (so on)
Answer = 75/4

Q11) 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

8/115/6
 * 4/5240275
264(275264)(264240)
so 11:24
so 11 from 1st vessel

Q12) 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 = 1000666.66 = 333.33
50% profitMethod3 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

Q13) 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