7. Two bunnies hop at slightly different rates but remain side by side. The first takes 50 hops per minute and the other takes 48 hops per minute. If they begin hopping together, when will they again be “in step”?
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7. Two bunnies hop at slightly different rates but remain side by side. The first takes 50 hops per minute and the other
Given: Two bunnies hop at slightly different rates but remain side by side. The first takes 50 hops per minute and the other takes 48 hops per minute.
To find: If they begin hopping together, when will they again be “in step”
Explanation: The first bunny takes 50 hops per minute. Similarly it takes 100 minutes in 2 minutes.
It forms an arithmetic progression with first term 50 and common difference 50.
The expression for nth term of the A.P is:
a + (n-1) d
= 50 + (n-1) 50
= 50 + 50 n-50
= 50 n
The second bunny takes 48 hops per minute. Similarly it takes 96 minutes in 2 minutes.
It forms an arithmetic progression with first term 48 and common difference 48.
The expression for mth term of the A.P is:
a + (m-1) d
= 48 + (m-1) 48
= 48 + 48 m -48
= 48 m
Since it is asked the time when they hop in together, we equate the nth term of the first A.P to mth term of second A.P.
50 n = 48m
=> m/n = 50/48
= 25/24
=> 24 m = 25 n
This means that 25 such intervals of second bunny is equal to 24 such intervals of bunny A.
Verifying,
After 25 intervals of bunny B,
Number of hops
= 25*48
= 1200
After 24 intervals of bunny A,
Number of hops
= 24*50
= 1200
Therefore, they will step in together at the 1200th hop if they start together.
Given: Two bunnies hop at slightly different rates but remain side by side. The first takes 50 hops per minute and the other takes 48 hops per minute. They begin hopping together.
To find: The time when they will be “in step”.
Solution:
In order to find when the two bunnies would be in step again, we need to check the multiples of 48 and 50 and find the least multiple that are common between them. This can be done by finding the L.C.M. of 48 and 50 which is 1200.
Now, the time at which they will be in step is when both of them are at their 1200th step. Since it is the least common factor and the other bunny hops 48 times a minute, the time in minutes is calculated as follows.
[tex]\frac{1200}{48} = 25[/tex]
Therefore, they will be “in step” in 25 minutes.