Question

a man bought to radio set for Rs 624 he sold one at a loss of 14% and the other at a profit of 14% if the selling price of each a same determine the cost price of each​

Answer 1

let x be the cp for one radio and y be that for next.

$x + y = 624 \: \: \: \: \: \: \: \: \: \: eqn \: no \: 1$

total sp=

$(1 + \frac{14}{100}) \times x+ (1 - \frac{14}{100}) \times y = \frac{114}{100} \times x + \frac{86}{100} \times y .$

given,

$\frac{114}{100} \times x = \frac{86}{100} \times y$

$114x = 86y \: \: \: \: \: eqno \: 2$

substitute eqn no 2 in 1

we get ,

114x+x=624

x= 624/115=Rs5.6

y=114x/86=Rs7.20

Answer 2

The cost price of first would be Rs. 355.68 and the cost price of second would be Rs. 268.32.

Step-by-step explanation:

Let the cost price of first radio set be 'x'.

Let the cost  price of second radio set be '624-x'

Loss % on first radio set = 14%

Profit % on second radio set =14%

So, According to question, it becomes,

$\dfrac{100-14}{100}x=\dfrac{100+14}{100}\times (624-x)\\\\\dfrac{86}{100}x=\dfrac{114}{100}(624-x)\\\\86x=114(624-x)\\\\86x+114x=71136\\\\200x=\dfrac{71136}\\\\x=\dfrac{71136}{200}\\\\x=355.68$

Hence, the cost price of first would be Rs. 355.68 and the cost price of second would be 624-355.68 = Rs. 268.32.

# learn more:

One merchant buy two radio set for 1000rs and one sell it 20% loss and other 20% profit. if the selling price of radio are same, then what is the cost price of each radio?

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