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31 May 2022, 08:30
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
88% (01:45) correct
12%
(02:00)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
At the beginning of the school year, Caitlin paid $22 for four pens and three notebooks. Two months later, she decided to buy six more of the same pens and five more of the same notebooks before the price changed. If she spent $35 on these additional pens and notebooks, what was the cost of one notebook?
A. $2
B. $2.50
C. $3
D. $4
E. $6
A. $2
B. $2.50
C. $3
D. $4
E. $6
31 May 2022, 22:49
Kudos
Bookmarks
4 Pen and 3 Notebook cost $22
And, 6 Pen and 5 Notebook cost $35
4P + 3N = 22 ......(1)
Therefore, 3(4P + 3N) = 3*22
12P + 9N = 66 ......(2)
6P + 5N = 35....... (3)
Therefore, 2(6P + 5N) = 3*35
12P + 10N = 70 ......(4)
Solving, (2) and (4) simultaneously we get,
N =4
Answer: D
And, 6 Pen and 5 Notebook cost $35
4P + 3N = 22 ......(1)
Therefore, 3(4P + 3N) = 3*22
12P + 9N = 66 ......(2)
6P + 5N = 35....... (3)
Therefore, 2(6P + 5N) = 3*35
12P + 10N = 70 ......(4)
Solving, (2) and (4) simultaneously we get,
N =4
Answer: D
01 Jun 2022, 07:51
Kudos
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Let the cost of one pen be $p and that of one notebook be $q.
At the beginning of the school year, Caitlin paid $22 to buy four pens and three notebooks.
Therefore, 4p + 3q = 22. Let’s call this equation as equation (1)
Two months later, she decided to buy six more of the same pens and five more of the same notebooks, at the same price, spending $35 on this purchase.
Hence, 6p + 5q = 35. Let’s call this equation as equation (2)
There are two unknowns and two independent equations in these unknowns. The two equations can be solved simultaneously for p and q.
Mutliplying equation (1) by 3 and equation (2) by 2, we have,
12p + 9q = 66
12p + 10q = 70
Subtracting the equations, we have, q = 4. This is the cost of one notebook, which is what we are trying to find out.
There is no need to calculate the value of p.
The correct answer option is D.
At the beginning of the school year, Caitlin paid $22 to buy four pens and three notebooks.
Therefore, 4p + 3q = 22. Let’s call this equation as equation (1)
Two months later, she decided to buy six more of the same pens and five more of the same notebooks, at the same price, spending $35 on this purchase.
Hence, 6p + 5q = 35. Let’s call this equation as equation (2)
There are two unknowns and two independent equations in these unknowns. The two equations can be solved simultaneously for p and q.
Mutliplying equation (1) by 3 and equation (2) by 2, we have,
12p + 9q = 66
12p + 10q = 70
Subtracting the equations, we have, q = 4. This is the cost of one notebook, which is what we are trying to find out.
There is no need to calculate the value of p.
The correct answer option is D.
