Question

A box contains only $2 coins and $10 coins and there

are 43 coins in total. If the total value of all the coins
in the box is greater than $278, at least how many
$10 coins are there in the box?

Answer 1

Answer: 25

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Work Shown:

x = number of $2 coins

y = number of $10 coins

x+y = 43 coins total

x = 43-y

2x = value of just the $2 coins only

10y = value of all the $10 coins only

2x+10y = total value of all coins mentioned

This value must be greater than $278, so,

2x+10y > 278

2( x ) + 10y > 278

2(43-y)+10y > 278

86-2y+10y > 278

8y+86 > 278

8y > 278 - 86

8y > 192

y > 192/8

y > 24

The smallest y can be is y = 25.

We need at least 25 coins worth $10 each.