8. A shopkeeper carns a profit of 15% on selling a radio, whereas if it is sold for360 less, he suffers a loss of 15%. What is the cost price of the radio for theshopkeeper? About the author Quinn
Answer: Cp of radio is 1200 Step-by-step explanation: Let CP of radio be ‘x’ and SP of radio be ‘y’. Then , profit=SP-CP=y-x given that, profit%=15% We know that, profit%=(profit/CP)×100 15={(y-x)/x}×100 15x=100y-100x 115x=100y 23x=20y x=20y/23 If the shopkeeper sells the radio for Rs 360 less then the loss occured is of 15%. Now we have, SP=y-360 CP=x loss=CP-SP=x-(y-360) loss=x-y+360 loss%=15% we know that, loss%=(loss/CP)×100 15={(x-y+360)/x}×100 15x=100x-100y+36000 -85x=-100y+36000 Putting eq1 in above equation we get, -85{20y/23}=-100y+36000 -1700y=-2300y+828000 600y=828000 y=828000/600 y=1380 Now by putting value of y in eq1 we get, x=(20×1380)/23 x=1200 Hence CP of radio for the shopkeeper=Rs1200 Reply
Answer:
Cp of radio is 1200
Step-by-step explanation:
Let CP of radio be ‘x’ and SP of radio be ‘y’.
Then ,
profit=SP-CP=y-x
given that,
profit%=15%
We know that,
profit%=(profit/CP)×100
15={(y-x)/x}×100
15x=100y-100x
115x=100y
23x=20y
x=20y/23
If the shopkeeper sells the radio for Rs 360 less then the loss occured is of 15%.
Now we have,
SP=y-360
CP=x
loss=CP-SP=x-(y-360)
loss=x-y+360
loss%=15%
we know that,
loss%=(loss/CP)×100
15={(x-y+360)/x}×100
15x=100x-100y+36000
-85x=-100y+36000
Putting eq1 in above equation we get,
-85{20y/23}=-100y+36000
-1700y=-2300y+828000
600y=828000
y=828000/600
y=1380
Now by putting value of y in eq1 we get,
x=(20×1380)/23
x=1200
Hence CP of radio for the shopkeeper=Rs1200