Find the unit price of each item which was paid as whole
$begingroup$
I am trying to help my daughter in her math and there is this question I can't quite get my head around, The sum is:
Three friends go into a book shop. Salma buys a cook book and a novel, she pays $ 20.75. Isla buys the same novel and a dictionary. Her bill comes to $ 26.65. Josh buys a cook book and a Dictionary and pays $ 30.90.
What is the price of each book?
problem-solving word-problem
edited Dec 10 '18 at 8:27
Arthur
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
$endgroup$
add a comment |
$begingroup$
I am trying to help my daughter in her math and there is this question I can't quite get my head around, The sum is:
Three friends go into a book shop. Salma buys a cook book and a novel, she pays $ 20.75. Isla buys the same novel and a dictionary. Her bill comes to $ 26.65. Josh buys a cook book and a Dictionary and pays $ 30.90.
What is the price of each book?
problem-solving word-problem
edited Dec 10 '18 at 8:27
Arthur
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
$endgroup$
add a comment |
$begingroup$
I am trying to help my daughter in her math and there is this question I can't quite get my head around, The sum is:
Three friends go into a book shop. Salma buys a cook book and a novel, she pays $ 20.75. Isla buys the same novel and a dictionary. Her bill comes to $ 26.65. Josh buys a cook book and a Dictionary and pays $ 30.90.
What is the price of each book?
problem-solving word-problem
edited Dec 10 '18 at 8:27
Arthur
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
$endgroup$
I am trying to help my daughter in her math and there is this question I can't quite get my head around, The sum is:
Three friends go into a book shop. Salma buys a cook book and a novel, she pays $ 20.75. Isla buys the same novel and a dictionary. Her bill comes to $ 26.65. Josh buys a cook book and a Dictionary and pays $ 30.90.
What is the price of each book?
problem-solving word-problem
problem-solving word-problem
edited Dec 10 '18 at 8:27
Arthur
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
edited Dec 10 '18 at 8:27
Arthur
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
edited Dec 10 '18 at 8:27
Arthur
115k7116198
edited Dec 10 '18 at 8:27
Arthur
115k7116198
edited Dec 10 '18 at 8:27
Arthur
115k7116198
115k7116198
asked Dec 10 '18 at 8:25
PremPrem
31
asked Dec 10 '18 at 8:25
PremPrem
31
asked Dec 10 '18 at 8:25
PremPrem
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Setting $C=$ price of cook book; $N=$price of novel and $D=$price of dictionary, we get the equations
$$
left{
begin{array}{crccc}
C& + N & & = &20.75\
& N & + D & = & 26.65 \
C &&+D &=& 30.90
end{array}
right.
$$
This is fairly easy to solve. Adding the first and last equations together, we obtain
$$
left{
begin{array}{crccc}
2C& + N & +D & = &51.65\
& N & + D & = & 26.65 \
end{array}
right.
$$
From which we can easily see (by subtracting the equations from each other) that
$$
2C = 25.00 qquad Rightarrow qquad C = 12.50
$$
now we can plug in this value to the original last equation, and we obtain $D = 30.90 - 12.50 = 18.40$. Similarly, plugging this result in the second equation of the original set, we get $N=26.65 - 18.40 = 8.25$
The final answers are therefore
$$
C = 12.50 qquad N = 8.25 qquad D = 18.40
$$
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
$endgroup$
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
add a comment |
$begingroup$
I'll try to put it in a way, so you may explain it to your daughter.
Let's denote $C$ price for cook book, $N$ novel, $D$ dictionary.
So you have
$C + N = 20.75$, $N + D = 20.65$ and $C + D = 30.90$.
Now the cost of buying a cook book and selling a dictionary $ C - D = C + N - (N + D) = 20.75 -20.65 = 0.10$. And finally buying two cookbooks: $2C = 2C - D + D = C + D + (C - D) = 30.90 + 0.10 = 31.00$. So $C = 15.50$. The rest should by easy.
answered Dec 10 '18 at 8:38
denklodenklo
4357
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Setting $C=$ price of cook book; $N=$price of novel and $D=$price of dictionary, we get the equations
$$
left{
begin{array}{crccc}
C& + N & & = &20.75\
& N & + D & = & 26.65 \
C &&+D &=& 30.90
end{array}
right.
$$
This is fairly easy to solve. Adding the first and last equations together, we obtain
$$
left{
begin{array}{crccc}
2C& + N & +D & = &51.65\
& N & + D & = & 26.65 \
end{array}
right.
$$
From which we can easily see (by subtracting the equations from each other) that
$$
2C = 25.00 qquad Rightarrow qquad C = 12.50
$$
now we can plug in this value to the original last equation, and we obtain $D = 30.90 - 12.50 = 18.40$. Similarly, plugging this result in the second equation of the original set, we get $N=26.65 - 18.40 = 8.25$
The final answers are therefore
$$
C = 12.50 qquad N = 8.25 qquad D = 18.40
$$
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
$endgroup$
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
add a comment |
$begingroup$
Hint: Setting $C=$ price of cook book; $N=$price of novel and $D=$price of dictionary, we get the equations
$$
left{
begin{array}{crccc}
C& + N & & = &20.75\
& N & + D & = & 26.65 \
C &&+D &=& 30.90
end{array}
right.
$$
This is fairly easy to solve. Adding the first and last equations together, we obtain
$$
left{
begin{array}{crccc}
2C& + N & +D & = &51.65\
& N & + D & = & 26.65 \
end{array}
right.
$$
From which we can easily see (by subtracting the equations from each other) that
$$
2C = 25.00 qquad Rightarrow qquad C = 12.50
$$
now we can plug in this value to the original last equation, and we obtain $D = 30.90 - 12.50 = 18.40$. Similarly, plugging this result in the second equation of the original set, we get $N=26.65 - 18.40 = 8.25$
The final answers are therefore
$$
C = 12.50 qquad N = 8.25 qquad D = 18.40
$$
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
$endgroup$
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
add a comment |
$begingroup$
Hint: Setting $C=$ price of cook book; $N=$price of novel and $D=$price of dictionary, we get the equations
$$
left{
begin{array}{crccc}
C& + N & & = &20.75\
& N & + D & = & 26.65 \
C &&+D &=& 30.90
end{array}
right.
$$
This is fairly easy to solve. Adding the first and last equations together, we obtain
$$
left{
begin{array}{crccc}
2C& + N & +D & = &51.65\
& N & + D & = & 26.65 \
end{array}
right.
$$
From which we can easily see (by subtracting the equations from each other) that
$$
2C = 25.00 qquad Rightarrow qquad C = 12.50
$$
now we can plug in this value to the original last equation, and we obtain $D = 30.90 - 12.50 = 18.40$. Similarly, plugging this result in the second equation of the original set, we get $N=26.65 - 18.40 = 8.25$
The final answers are therefore
$$
C = 12.50 qquad N = 8.25 qquad D = 18.40
$$
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
$endgroup$
Hint: Setting $C=$ price of cook book; $N=$price of novel and $D=$price of dictionary, we get the equations
$$
left{
begin{array}{crccc}
C& + N & & = &20.75\
& N & + D & = & 26.65 \
C &&+D &=& 30.90
end{array}
right.
$$
This is fairly easy to solve. Adding the first and last equations together, we obtain
$$
left{
begin{array}{crccc}
2C& + N & +D & = &51.65\
& N & + D & = & 26.65 \
end{array}
right.
$$
From which we can easily see (by subtracting the equations from each other) that
$$
2C = 25.00 qquad Rightarrow qquad C = 12.50
$$
now we can plug in this value to the original last equation, and we obtain $D = 30.90 - 12.50 = 18.40$. Similarly, plugging this result in the second equation of the original set, we get $N=26.65 - 18.40 = 8.25$
The final answers are therefore
$$
C = 12.50 qquad N = 8.25 qquad D = 18.40
$$
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
edited Dec 10 '18 at 8:50
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
answered Dec 10 '18 at 8:34
Matti P.Matti P.
2,0381414
2,0381414
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
add a comment |
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
Actually I don't know how to work out the answer. Also can you advise where I can find similar sums online?
$endgroup$
– Prem
Dec 10 '18 at 8:37
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
About finding similar stuff online: I don't have any recommendations right now, but I would advise googling "solving systems of linear equations".
$endgroup$
– Matti P.
Dec 10 '18 at 9:03
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
$begingroup$
Matti P. Thank you so much for the prompt response, more than the answer I was keen to know about the process to achieving the answer. This helps a great deal. Gold bless.
$endgroup$
– Prem
Dec 10 '18 at 9:20
add a comment |
$begingroup$
I'll try to put it in a way, so you may explain it to your daughter.
Let's denote $C$ price for cook book, $N$ novel, $D$ dictionary.
So you have
$C + N = 20.75$, $N + D = 20.65$ and $C + D = 30.90$.
Now the cost of buying a cook book and selling a dictionary $ C - D = C + N - (N + D) = 20.75 -20.65 = 0.10$. And finally buying two cookbooks: $2C = 2C - D + D = C + D + (C - D) = 30.90 + 0.10 = 31.00$. So $C = 15.50$. The rest should by easy.
answered Dec 10 '18 at 8:38
denklodenklo
4357
$endgroup$
add a comment |
$begingroup$
I'll try to put it in a way, so you may explain it to your daughter.
Let's denote $C$ price for cook book, $N$ novel, $D$ dictionary.
So you have
$C + N = 20.75$, $N + D = 20.65$ and $C + D = 30.90$.
Now the cost of buying a cook book and selling a dictionary $ C - D = C + N - (N + D) = 20.75 -20.65 = 0.10$. And finally buying two cookbooks: $2C = 2C - D + D = C + D + (C - D) = 30.90 + 0.10 = 31.00$. So $C = 15.50$. The rest should by easy.
answered Dec 10 '18 at 8:38
denklodenklo
4357
$endgroup$
add a comment |
$begingroup$
I'll try to put it in a way, so you may explain it to your daughter.
Let's denote $C$ price for cook book, $N$ novel, $D$ dictionary.
So you have
$C + N = 20.75$, $N + D = 20.65$ and $C + D = 30.90$.
Now the cost of buying a cook book and selling a dictionary $ C - D = C + N - (N + D) = 20.75 -20.65 = 0.10$. And finally buying two cookbooks: $2C = 2C - D + D = C + D + (C - D) = 30.90 + 0.10 = 31.00$. So $C = 15.50$. The rest should by easy.
answered Dec 10 '18 at 8:38
denklodenklo
4357
$endgroup$
I'll try to put it in a way, so you may explain it to your daughter.
Let's denote $C$ price for cook book, $N$ novel, $D$ dictionary.
So you have
$C + N = 20.75$, $N + D = 20.65$ and $C + D = 30.90$.
Now the cost of buying a cook book and selling a dictionary $ C - D = C + N - (N + D) = 20.75 -20.65 = 0.10$. And finally buying two cookbooks: $2C = 2C - D + D = C + D + (C - D) = 30.90 + 0.10 = 31.00$. So $C = 15.50$. The rest should by easy.
answered Dec 10 '18 at 8:38
denklodenklo
4357
answered Dec 10 '18 at 8:38
denklodenklo
4357
answered Dec 10 '18 at 8:38
denklodenklo
4357
answered Dec 10 '18 at 8:38
denklodenklo
4357
4357
add a comment |
add a comment |
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