Question 53 & 54

Past Exam Question

This is a  Situation of direct proportion i-e as the days increase, more cats require more mice and few cats will  require few mice.

Formula here is ;

\frac{x}{100}=\frac{4}{100}\times \frac{100}{4}

\Rightarrow \frac{x}{100}=1

\Rightarrow x=100

So, 4 cats will also require 100 days to eat 4 mice !

Here the technique of distance formula is to be used. Distance is calculated by multiplying speed with time taken. Here we have to calculate the distance.

Let the total distance is ‘x’ Kms  from point A to C!

Let distance  covered from point A to B in first half  i-e in 5 hours is X kms  and distance covered  from B to C  in  second half i-e  next 5 hours is Y kms

Case 1 –  Distance for First Half (5 hours) :

We are given that in first half he travels with speed of  21 kms per hour.

We know Distance = Speed x Time

i-e

S= V \times T

                                                                                                                                  S is distance, V is speed and T is time; Put values of S, V and T in the relation, we get;

Here,

S= X,   V=21 km/h and T=5 hours

X=21 \times 5= 105 kms

Case 2 – Distance For second half ( Next 5 hours):

We use same formula. Here Time is again 5 hours, but for second half speed will be  24 kms per hour.

Here,

S= Y,   V=24 km/h and T=5 hours

Y=24 \times 5= 120 kms

So, total distance is  X+Y i-e 105+120 = 225 kms