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gtnhenbr [62]
3 years ago
12

Pablo ran a concession stand last Saturday and made $79.80 from selling a total of 53 hot dogs and hamburgers. Each hot dog sold

for $1.40 and each hamburger sold for $1.80. Which system of equations can be used to determine the number of hot dogs, x, and hamburgers, y, that were sold?
A. 1.4x + 1.8y = 79.8
39x + 14y = 53

B. 0.7x + 0.9y = 79.8
x + y = 3.2

C. 1.4x + 1.8y = 79.8
x + y = 53

D. 2.8x + 3.6y = 3.2
x + y = 546
Mathematics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

C. 1.4x + 1.8y = 79.8

x + y = 53

Step-by-step explanation:

The problem statement gives rise to two equations, one for the amount of money made, and one for the number of items sold. If x and y represent the numbers of items, and if 53 items were sold, then one of the equations will be ...

x + y = 53

This is sufficient to let you choose the correct answer.

___

Since "x" items were sold for $1.40 and "y" items were sold for $1.80, the sales revenue will be the sum of products of price and quantity:

1.40x +1.80y = 79.80

This confirms the choice of answer.

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An adult ticket to a museum costs 3$ more than a children’s ticket. When 200 adult tickets and 100 children tickets are sold, th
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<h3><u>Solution:</u></h3>

Let "c" be the cost of one children ticket

Let "a" be the cost of one adult ticket

Given that adult ticket to a museum costs 3$ more than a children’s ticket

<em>Cost of one adult ticket = 3 + cost of one children ticket</em>

a = 3 + c  ------ eqn 1

<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>

200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100

200 \times a + 100 \times c = 2100

200a + 100c = 2100 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>

Substitute eqn 1 in eqn 2

200(3 + c) + 100c = 2100

600 + 200c + 100c = 2100

600 + 300c = 2100

300c = 1500

<h3>c = 5</h3>

Thus the cost of children’s ticket is $ 5

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The price, e, of an entertainment system at extreme electronics is $220 less than twice the price, u, of the same system at ultr
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Answer: The price of system at extreme electronics is $570 and the price of system at ultra electronics is $395.

Step-by-step explanation:

Let the price of the system at the ultra electronics be u

Let the price of the system at extreme electronics be e

According to question, we have

2u-e=220\\\\and\\\\e-u=175

Now, using the substitution method, we will solve the above system of equations.

e-u=175\\\\e=175+u

Now, put the value of u in the first equation :

2u-e=220\\\\2u-(175+u)=220\\\\2u-175-u=220\\\\u-175=220\\\\u=220+175\\\\u=\$395

Now, we put the value of u in the equation which is given by:

e=175+u\\\\e=175+395\\\\e=\$570

Hence, the price of system at extreme electronics is $570 and the price of system at ultra electronics is $395.

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Step-by-step explanation:

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