Ratio word problem total cost of magazine and number of customers who bought the magazine
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During January of this year, Utopia Magazine sales reached a total of £ 488800. The ratio of normal price customer to discounted price customer was 5:2. How many customers in total bought this magazine?
I have tired to work this out by multiplying the total with each ratio and dividing by 7 but i dont seem to get the answer correct, can any one help with this problem?
algebra-precalculus
asked Dec 19 '18 at 23:00
italyitaly
519414
$endgroup$
add a comment |
$begingroup$
During January of this year, Utopia Magazine sales reached a total of £ 488800. The ratio of normal price customer to discounted price customer was 5:2. How many customers in total bought this magazine?
I have tired to work this out by multiplying the total with each ratio and dividing by 7 but i dont seem to get the answer correct, can any one help with this problem?
algebra-precalculus
asked Dec 19 '18 at 23:00
italyitaly
519414
$endgroup$
add a comment |
$begingroup$
During January of this year, Utopia Magazine sales reached a total of £ 488800. The ratio of normal price customer to discounted price customer was 5:2. How many customers in total bought this magazine?
I have tired to work this out by multiplying the total with each ratio and dividing by 7 but i dont seem to get the answer correct, can any one help with this problem?
algebra-precalculus
asked Dec 19 '18 at 23:00
italyitaly
519414
$endgroup$
During January of this year, Utopia Magazine sales reached a total of £ 488800. The ratio of normal price customer to discounted price customer was 5:2. How many customers in total bought this magazine?
I have tired to work this out by multiplying the total with each ratio and dividing by 7 but i dont seem to get the answer correct, can any one help with this problem?
algebra-precalculus
algebra-precalculus
asked Dec 19 '18 at 23:00
italyitaly
519414
asked Dec 19 '18 at 23:00
italyitaly
519414
asked Dec 19 '18 at 23:00
italyitaly
519414
asked Dec 19 '18 at 23:00
italyitaly
519414
asked Dec 19 '18 at 23:00
italyitaly
519414
519414
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $n$ be the number of normal price customers and $d$ be the number of discounted price customers. Then you have that
$$2n=5d$$
and
$$2.8n+2.1d=488,800$$
since $2.1=2.8-0.25cdot 2.8$. Can you solve this system of equations for $n$ and $d$, and then calculate the total number of customers $n+d$?
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Let $n$ be the number of normal price customers and $d$ be the number of discounted price customers. Then you have that
$$2n=5d$$
and
$$2.8n+2.1d=488,800$$
since $2.1=2.8-0.25cdot 2.8$. Can you solve this system of equations for $n$ and $d$, and then calculate the total number of customers $n+d$?
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
add a comment |
$begingroup$
Let $n$ be the number of normal price customers and $d$ be the number of discounted price customers. Then you have that
$$2n=5d$$
and
$$2.8n+2.1d=488,800$$
since $2.1=2.8-0.25cdot 2.8$. Can you solve this system of equations for $n$ and $d$, and then calculate the total number of customers $n+d$?
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
$endgroup$
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
add a comment |
$begingroup$
Let $n$ be the number of normal price customers and $d$ be the number of discounted price customers. Then you have that
$$2n=5d$$
and
$$2.8n+2.1d=488,800$$
since $2.1=2.8-0.25cdot 2.8$. Can you solve this system of equations for $n$ and $d$, and then calculate the total number of customers $n+d$?
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
$endgroup$
Let $n$ be the number of normal price customers and $d$ be the number of discounted price customers. Then you have that
$$2n=5d$$
and
$$2.8n+2.1d=488,800$$
since $2.1=2.8-0.25cdot 2.8$. Can you solve this system of equations for $n$ and $d$, and then calculate the total number of customers $n+d$?
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
answered Dec 19 '18 at 23:02
FrpzzdFrpzzd
23k841110
23k841110
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
add a comment |
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
$begingroup$
why does 2n=5d? That doesnt make sense to me. the number of discount customers is not equal to the number of Normal price customers
$endgroup$
– italy
Dec 19 '18 at 23:05
1
1
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
The ratio of $n$ to $d$ is $5:2$, so there are $5/2$ times as many $n$ as there are $d$, and $n=(5/2)d$, or $2n=5d$.
$endgroup$
– Frpzzd
Dec 19 '18 at 23:06
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
oh i see, thank you!
$endgroup$
– italy
Dec 19 '18 at 23:07
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
$begingroup$
@italy No problem! If this is helpful, don't forget to $checkmark$ and $uparrow$! :D
$endgroup$
– Frpzzd
Dec 19 '18 at 23:08
add a comment |
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