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KiRa [710]
3 years ago
9

A coin bank that excepts only nickels and dimes contains $9.15. There are three more than twice as many nickels as there are dim

es. How many of each coin are in the bank?
Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0

45 dimes and 93 nickels were in bank

<em><u>Solution:</u></em>

Let "n" be the number of nickels

Let "d" be the number of dimes

We know that,

value of 1 nickel = $ 0.05

value of 1 dime = $ 0.10

<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>

Number of nickels = 3 + 2(number of dimes)

n = 3 + 2d ---- eqn 1

<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>

number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15

n \times 0.05 + d \times 0.10 = 9.15

0.05n + 0.10d = 9.15  ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Substitute eqn 1 in eqn 2

0.05(3 + 2d) + 0.10d = 9.15

0.15 + 0.1d + 0.10d = 9.15

0.2d = 9

<h3>d = 45</h3>

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

<h3>n = 93</h3>

Thus 45 dimes and 93 nickels were in bank

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Answer:

a) <em>The equation</em> (10s + 8w) <em>represents </em><em>the </em><em>calories </em><em>Bridget </em><em>ate </em><em>on </em><em>Monday </em><em>and </em><em>the </em><em>equation</em> (20s + w) <em>represents </em><em>the </em><em>calories</em><em> </em><em>she </em><em>ate</em><em> </em><em>the </em><em>next </em><em>day.</em>

<em>b)</em><em> </em><em>The </em><em>number </em><em>of </em><em>calories</em><em> </em><em>in </em><em>each </em><em>strawberry</em><em> </em><em>is </em>4 <em>and </em><em>the </em><em>number </em><em>of </em><em>calories </em><em>in </em><em>each </em><em>vanilla</em><em> </em><em>wafer</em><em> cookie</em><em> </em><em>is </em>19. The solution is s= 4 and w = 19.

Step-by-step explanation:

For part A, Bridget ate 10 strawberries and 8 vanilla wafer cookies on Monday. Since the the number of calories in a strawberry is <em>s</em> and the number of calories in a vanilla wafer cookie is <em>w </em>, the number of calories Bridget ate on Monday is <em>10s + 8w</em><em>.</em><em> </em>The next day, Bridget ate 20 strawberries and 1 vanilla wafer cookie. Hence, the number of calories Bridget ate on the next day is 20s<em> + w</em>.

For part B,

we will create two different simultaneous equations.

Equation 1: 10s + 8w = 192

Equation 2: 20s + w = 99

We need to find one of the terms first to solve the other term. For this case, I will solve for w first.

Multiply the first equation by 2.

Equation 3: 20s + 16w = 192*2 = 384.

Now, subtract equation 2 from this new equation.

Equation 4:

(20s + 16w) - (20s + w) = 384 - 99

20s + 16w - 20s - w = 285

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This leaves only w left and we can solve w.

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Now, we can solve for s using equation 2.

20s + 19 = 99

20s = 99-19 = 80

Hence,

s = 80/20 = 4

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