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21 Jul 2022, 19:47
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67% (03:12) correct
33%
(03:23)
wrong
based on 24
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The ratio of cost of a notebook and a pen for the shopkeeper is 3:2. For every 5 notebooks sold, the shopkeeper gives 1 pen for free. The profit margin for each notebook is 50%. If the profit realized by the shopkeeper when a customer buys 10 notebooks is $231, what is the total cost (in $) of a notebook and a pen ?
A. 105
B. 110
C. 115
D. 120
E. 125
A. 105
B. 110
C. 115
D. 120
E. 125
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25 Jul 2022, 09:45
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CP of notebook- 3k
SP of notebook- 4.5k (3k*1.5)
CP of pen- 2k
When 10 notebooks is sold, two pens are also given by the Shopkeeper. So total cost price is equal to - 34k (3k*10+2k*2) and total selling price is 45k (4.5k*10)
Now profit is 231 , thus 45k-34k=231 so k=21.
To find cost of a pen and notebook. 5k=105 so 105 (A) is the correct answer
SP of notebook- 4.5k (3k*1.5)
CP of pen- 2k
When 10 notebooks is sold, two pens are also given by the Shopkeeper. So total cost price is equal to - 34k (3k*10+2k*2) and total selling price is 45k (4.5k*10)
Now profit is 231 , thus 45k-34k=231 so k=21.
To find cost of a pen and notebook. 5k=105 so 105 (A) is the correct answer
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25 Jul 2022, 16:38
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pintukr
The ratio of cost of a notebook and a pen for the shopkeeper is 3:2. For every 5 notebooks sold, the shopkeeper gives 1 pen for free. The profit margin for each notebook is 50%. If the profit realized by the shopkeeper when a customer buys 10 notebooks is $231, what is the total cost (in $) of a notebook and a pen ?
A. 105
B. 110
C. 115
D. 120
E. 125
A. 105
B. 110
C. 115
D. 120
E. 125
cn - cost of notebook
cp - cost of pen
also profit margin is 50% on notebook so sp = 1.5cn
also cn/cp = 3/2 or cp = 2/3 cn
now profit margin by selling 10 notebooks is 231, thus we have
1.5cn*10 - 10cn -2*(2/3cn)= 231 ---> here 2/3cn represents cost of 1 pen, as the shopkeeper gives away two pen on selling 10 notebooks
5-4/3cn =231
or cn =63
thus cp =2/3*(63) = 42
hence cost of pen, and notebook will be 63+ 42 = 105.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
