The average cost of a few cups was $8. A f t e r a n o t h e r c u p t h a t c o s t s$ 32 was added, the average cost became $12. How many cups were there in the end?

Can i have the steps to solve it.

Thank you😀

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Solution

HCost of few cups = x
initially number of cups = n
(since 1 cup is added)
Number of cups in the end = n+1
Average = (total cost) ÷ (total number of cups)
Therefore,
(x÷n) = 8
x = 8n {eqn no.1}

[(x+32)÷(n+1)] = 12
(x+32) = 12(n+1)
x = [12(n+1)-32] {eqn no.2}
LHS of eqn no.1 = LHS of eqn no.2
Therefore,
8n = [12(n+1)-32]
8n = 12n + 12 - 32
12n - 8n = 32 - 12
4n = 20
n = 20 ÷ 4
n = 5
Therefore number of cups in the end is (n+1) = 6

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The average cost of a few cups was 8.Afteranothercupthatcosts32 was added, the average cost became $12. How many cups were there in the end? Can i have the steps to solve it. Thank you😀

The average cost of a few cups was $8. A f t e r a n o t h e r c u p t h a t c o s t s$ 32 was added, the average cost became $12. How many cups were there in the end?

Can i have the steps to solve it.

Thank you😀