Quant Boosters  Swetabh Kumar  Set 8

Q) If Ratio of Cost Price and Mark Price is 5:8 and that of % profit on sale to % discount is 2:3 then find % of Discount.
profit on sale = profit on SP = Profit/SP *100%
CP=50, MP=80.
3 * (SP50) * 100/SP = 2 * D (D=discount%)
Also, SP = (1D/100) * 80 = 80 * (100D)/100
so 3 * {80(100D)/100 50 } * 100 = 2D * {80 * (100D)}/100
150 { 752D} = D {2002D}
D^2  250D + 5625 = 0
D= (250200)/2 = 25%Q) Find the product of the irrational roots of the equation (2x – 1) (2x 3) (2x – 5) (2x – 7) = 9.
(2x1)(2x7)(2x3)(2x5) = 9
(4x^216x+7) (4x^216x+15)=9
Let 4x^216x=t
(t+7)(t+15)=9
t^2+22t+96=0
t=16, t=6
so 4x^216x+16=0 and 4x^216x+6=0
x^24x+4=0 2x^28x+3=0
1st gives x=2 and second eqn gives both irrational roots. so c/a= 3/2Q) For how many ordered triplet (x,y,z) of positive integers less than 10 is the product xyz divisible by 20?
20=5 * 2^2 so one 5 should be there and rest we have to distribute
(5,2,2)....(5,2,8) so 3+6 * 3 = 21
(5,4,4)...(5,4,8) so 3+2 * 6 = 15
(5,6,6,), (5,6,8) = 3+6=9
(5,8,8)= 3 so till here= 48.
ab (5,4,1/3/5/7/9) = 3+6 * 4 = 27
and (5,8,1/3/5/7/9) = 3+6 * 4 = 27
so 54 extra.
hence 48+54=102Q) The points (3,7), (6,2) and (2,k) are the vertices of a triangle. For how many values of k is the triangle a right triangle ?
A=(3,7), B(6,2), C(2,k)
AB^2 = 9+25 = 34
AC^2 = 1+(7k)^2 = k^214k+50
BC^2 = 16+(2k)^2 = k^24k+20
if AB is hyp. 2k^218k+70 = 34
k^29k+18= 0
2 real values here.
if AC is hyp: k^24k+54 = k^214k+50
10k=4 so k=0.4 so one value
if BC is hyp, k^214k+84=k^24k+20 so k=6.4 so one value.
so (2+1+1)= 4 valuesQ) Choti has 11 different toys and Bade has 8 different toys.Find the number of ways in which they can exchange their toys so that each keeps their initial number of toys?
Goods have to be "exchanged". 19C8 or 19C11 is the total possibilities, but since they have to be "exchanged" so it is 19C81.
(original configuration not to be counted)Q) In how many ways can seven balls of different colours be put into 4 identical boxes such that none of the box is empty?
(4,1,1,1): 7C4 * 3 * 2 * 4 = 840
(3,1,1,2): 7C3 * 4C2 * 2C1 * 4!/2! = 5040
(2,2,2,1): 7C2 * 5C2 * 3C2 * 4 = 2520
so(840+5040+2520)/4! = 350Q) A team is planning to participate in a Dahihandi competition. On average, they succeed in breaking the handi on 2 out of every 11 attempts. How many attempts will they have to make to have a fair chance of succeeding in breaking the handi?
2/11 + 9/11 * 2/11 + 9/11 * 9/11 * 2/11..n terms.
r=9/11
{ 1(9/11)^n} > = 0.5
(9/11)^n < = 0.5
so n=4
PS: Fair chance means >=50% probabilityQ) A person has just sufficient money to buy either 30 guavas, 50 plums or 70 peaches. He spends 20% of the money on travelling, and buys 14 peaches, 'x' guavas and 'y' plums using rest of the money. If x, y > 0, what is the minimum value of the sum of x and y?
30a =50b = 70c =5T
4T = 14c + xa + yb
4T = 14 * 5T/70 + x * 5T/30 + y * 5T/50
4 = 1+ x/6 + y/10
x/6+y/10 = 3
5x+3y = 90
so min. sum when x=15, y=5 so 20Q) How many ordered pairs of integers (a,b), are there such that their product is a positive integer less than 100?
In the above question, if (a,b) is not different from (b,a), then how many such pairs are possible?a) when both are +ve:
1 * 1 ... 1 * 99 and reverse = 99 * 2  1 = 197
2 * 2...2 * 49 and reverse = 48 * 2  1 = 95
3 * 3....3 * 33 = 31 * 2  1 = 61
4 * 4...4 * 24 = 21 * 2  1 = 41
5 * 5...5 * 19 = 15 * 2  1 = 29
6 * 6....6 * 16 = 11 * 2  1 = 21
7 * 7...7 * 14 = 8 * 2  1 = 15
8 * 8 .... 8 * 12 = 5 * 2  1 = 9
9 * 9...9 * 11 = 3 * 2  1 = 5
so total = 473
same when both are ve so 473 * 2 = 946b) Unordered = 99+48+31+21+15+11+8+5+3= 241 so 241*2= 482
Q) Find the number of positive solutions for 2x + 3y more than 60 and 2y + 3x less than 60
Draw 3x+2y = 60 and 2x+3y = 60
find the common region as per question
a triangle will be formed with cordinates (12,12 ) ( 0,20 ) ( 0,30 )
area of this triangle = 60
use picks theorem : A = i + b/2  1
61 = i + b/2
calculate the number of boundary pts of triangle boundary , total 20 boundary pts will be there
9 on the vertical line between 0,20 and 0,30
3 on the line between 0,20 and 12,12
5 on the line between 0,30 and 12,12
3 vertices
total 20 boundary points
61  20/2 = 51 integral points