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alexandr402 [8]

3 years ago

12

A pile of coins, consisting of quarters and half dollars, is worth $11.75. If there are 2 more quarters than half dollars, how m

any of each are there?

1 answer:

Black_prince [1.1K]3 years ago

4 0

The pile contains 17 quarters and 15 half-dollars.

Let <em>x</em> = the number of quarters and <em>y</em> = the number of half-dollars.

We have two equations:

(1) $0.25<em>x</em> + $0.50<em>y</em> = $11.75

(2) <em>x</em> = <em>y</em> +2

Substitute the value of <em>x</em> from Equation (2) into Equation (1).

0.25(<em>y</em>+2) + 0.50<em>y</em> = 11.75

0.25<em>y</em> + 0.50 + 0.50<em>y</em> = 11.75

0.75<em>y</em> = 11.75 – 0.50 = 11.25

<em>y</em> = 11.25/0.75 = 15

Substitute the value of <em>y</em> in Equation (2).

<em>x</em> = 15 + 2 = 17

The pile contains 17 quarters and 15 half-dollars.

<em>Check</em>: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.

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${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

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★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

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${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

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${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

${\underline{\rule{290pt}{2pt}}}$

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