CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

By selling two transistors for Rs. $$600$$ each, a shopkeeper gains $$20$$ percent on one transistor and loses $$20$$ Percent on the other,
Find: (i) C.P.of each transistor.
(ii) total C P and total S.P. of both the transistors
(iii) profit or loss percent on the whole.


Solution

Let $$ CP_{1}$$ & $$ CP_{2}$$  be the cost price of $$ 1 ^{st}$$ transistors and $$ 2^{nd}$$ transistors respectively
selling price of each transistors, $$ SP_{1} = RS 600$$
Gain of one transistors = 20%
$$ \Rightarrow \frac{SP_{1}-CP_{1}}{CP_{1}}\times 100 =20.$$
$$\Rightarrow  \frac{600-CP_{1}}{CP_{1}}\times 100 = 20.$$
$$ \Rightarrow 600 - CP_{1} = \frac{20}{100}CP_{1}$$
$$\Rightarrow  600 = \frac{20}{100} CP_{1}+CP_{1}$$
$$\Rightarrow  \frac{120}{100}CP_{1}=600$$
$$\Rightarrow CP_{1} = \frac{600\times 100}{120}$$
$$\Rightarrow CP_{1} = Rs 500$$
and loss of other transistors = 20 %
$$\Rightarrow \frac{CP_{2}-SP_{1}}{CP^{1}} \times 100 = 20$$
$$\Rightarrow CP_{2}-600 = \frac{20}{100}CP_{2}$$
$$ \Rightarrow  CP_{2}\frac{-20}{100}CP_{2}=600$$
$$ \Rightarrow \frac{80}{100}CP_{2}=600$$
$$ \Rightarrow  CP_{2}= \frac{600\times 100}{80}$$
=RS 750
cost price of $$ 1^{st}$$ transistors = Rs 1350
Cost price of $$ 2^{nd}$$  transistors = Rs 750.
Total cost price ,CP = Rs (600+750) = 1350
Total selling price SP = Rs (600+600) = Rs 1200.
loss on the whole = $$ \frac{CP-SP}{CP}\times 100$$
$$ \frac{1350-1200}{1350}\times 100$$
$$ \frac{150}{1350}\times 100$$
= 11.11 % (approx.)

1193998_1340343_ans_67f837eb43764edba16b17165ec194c4.jpg

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image