Question 7 & 8

Question 7: One hundred cats can eat 100 mice in 100 days time, then in how many days four cats will eat four mice ?


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This is a  situation of direct and indirect proportion i-e as the days increase, more cats require more mice and few cats will  require more mice, and more cats will finish eating more mice in few days.

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 !

Question 8: A man completes a 10 hour journey. He travels 21 km/h in the first half of the journey and at rate of 24 km/h in second half. Find the total kilometers of the journey.

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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+y’ 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


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;


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.


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 

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