Math, asked by sagarsarangi, 1 year ago

Anil sold 2 washing machines at 12000 each. On one he gains 25% and the other he loses 25%. How much does he gain or loss in the while transaction and find the overall gain or loss %

Answers

Answered by rustyattacker03629
24
S. P. Of 1st washing machine = ₹12000
Gain = 25%
C. P. Of 1st washing machine,
= S. P. x 100 / 100 + gain%
= 12000 x 100 / 100 +25%
= 12000 x 100 / 125
= ₹9600

Now, S. P. Of 2nd washing machine = ₹12000
Loss = 25 %
C. P. Of washing machine,
= S. P. x 100 / 100+ loss%
= 12000 x 100 / 100 - 25 %
= 12000 x 100 / 75
= ₹16000

Now, Total S. P. Of washing machines = S. P. Of 1st washing machine + S. P. Of 2nd washing machine = ₹12000 + ₹12000
= ₹24000

Now, Total C. P. Of washing machines = C. P. Of 1st washing machine + S. P. Of 2nd washing machine = ₹9600 + ₹ 16000
= ₹25600


Here, C. P. > S. P.
Therefore, loss = C. P. - S. P.
= ₹25600 - 24000
= ₹1600

Now, loss%= (loss / C. P.) x 100
= (1600 / 25600) x 100
= 6.25%
Answered by Anonymous
7

\huge{\underline{\underline {\mathtt{\purple{H}\pink{O}\green{L}\blue{A}\red{!}\orange{!}}}}}

Question

Anil sold two washing machine at 12000 each.on one he a profit of 25% and another a loss of 25% .find his overall loss or profit in the whole transaction and find the overall gain or loss %

\LARGE{\underline{\underline {\mathtt{\purple{A}\pink{N}\green{S}\blue{W}\red{E}\orange{R}}}}}

\huge\fbox\red{6.25%}

Solution

For 1st machine,

x +  \frac{x}{4}  = 12000

 \frac{5x}{4}  = 12000

x = 9600

Hence the profit on 1st machine = ₹2400

For 2nd machine, y -  \frac{y}{4}  = 12000

 \frac{3y}{4}  = 12000

Hence loss on 2nd machine = ₹4000

Hence total loss/ profit = 4000 - 2400 = ₹1600

= (1600/25600) ×100

= 6.25%

Hope it helps uh!!

Mark as Brainliest

Do thank and follow

Similar questions