Quant Boosters  Swetabh Kumar  Set 3

Q27) Two cars P and Q are moving at uniform speeds, 50 km/hr and 25 km/hr respectively, on two straight roads intersecting at right angle to each other. P passes the intersecting point of the roads when Q has still to travel 50 km to reach it. What is the shortest distance between the cars during the journey?
a. 20√5 km
b. 50 km
c. 25 km
d. 25√2 km

P's distance be Y, Q's be X
Y/50 = (X50)/25
Y=2X100
say Y=100, X=100
distance at any time=d
d^2 = (10050t)^2 + (10025t)^2
diff wrt t for minima
(10050t)*2 +(10025t) = 0
t=12/5
d^2 = 20^2 + 40^2 = 2000 so d= 20 rt 5

Q28) Two men are walking towards each other alongside a railway track. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout. How long after the train has passed the second man will the two men meet?
a. 89.7 minutes
b. 90 minutes
c. 90.3 seconds
d. 91 seconds

Train speed= T, men speed: v1, v2. length L.
so L/(Tv1)=20 and L/(T+v2) = 18
so 20T20v1=18T+18v2
T= 9v2+10v1 (1)
Ditsane between the men = 600 (T+v2)
Distance covered in these 10 mins = 600(v1+v2)
so reqd time = {600(T+v2)600(v1+v2)}/(v1+v2)
= (6000600)= 5400 sec= 90 mins ( using 1)

Q29) 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)

Q30) Find the number of positive solutions for 2x + 3y > 60 and 2y + 3x < 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

shouldn't it be 89.7 mins?
You might have missed 18 seconds both men have moved while from the moment the train meets the second man to the moment it passes him.
so reqd time = {600(T+v2)600(v1+v2)18(v1+v2)}/(v1+v2)
which gives 540018 seconds or 89.7 mins

We can do it another way too:
Put x= y,
F(2x) = f(x)^2
Put y=2x
F(3x)= f(x)^3
=>f(nx)=f(x)^nF(8)=F(2×4)=f(4)^2=1/9