Question

Two radios were bought at the same cost. One was sold at a profit of 12% and other at a loss of 17%. If the difference in the selling price of two radios was 116. Find the C.P. of each radio

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Answer 1

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$\sf{let~ the~ cost ~of ~radios~ be ~x.}$

$\sf{\longmapsto SP ~of ~radio~ -1}$

$\bold{\longmapsto SP=cp(\frac{100+p\%}{100})}$

$\bold{\longmapsto x(\frac{100+12}{100})}$

$\bold{\longmapsto \frac{112x}{100}}$

$\bold{\longmapsto SP~ of ~radio ~-2}$

$\bold{\longmapsto x(\frac{100+17}{100})}$

$\bold{\longmapsto \frac{117x}{100}}$

given, SP of radio-1 - sp of radio-2 =125

$\bold{\longmapsto \frac{117x}{100} - \frac{112x}{100} = 125\frac{5x}{100}=125}$

$\bold{\longmapsto 5x=125 × 100}$

$\bold{\longmapsto 5x=12500}$

$\bold{\longmapsto x=\frac{12500}{5}}$

$\bold{\longmapsto x=Rs.2500}$

$\sf{\longmapsto CP of radios=2500}$

$\sf{SP ~of ~radio-1=2500(\frac{100+12}{100})}$

$\bold{\longmapsto 2500 × \frac{112}{100}}$

$\bold{\longmapsto Rs.2800}$

$\sf{SP~ of~ radio-2=2500(\frac{100+17}{100})}$

$\bold{\longmapsto 2500 ×\frac{ 117}{100}}$

$\bold{\longmapsto Rs.2925}$

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Answer 2

Answer:

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$\sf{let~ the~ cost ~of ~radios~ be ~x.}$

$\sf{\longmapsto SP ~of ~radio~ -1}⟼SP of radio −1$

$\bold{\longmapsto SP=cp(\frac{100+p\%}{100}}$

$\{\longmapsto x(\frac{100+12}{100}$

$\bold{\longmapsto \frac{112x}{100}}$

$\bold{\longmapsto x(\frac{100+17}{100}}$

$\bold{\longmapsto \frac{117x}{100}}⟼$

given, SP of radio-1 - sp of radio-2 =125

$\bold{\longmapsto \frac{117x}{100} - \frac{112x}{100} = 125\frac{5x}{100}=125}⟼$

$\bold{\longmapsto 5x=125 × 100}⟼5x=125×100$

$\bold{\longmapsto 5x=12500}⟼5x=12500$

$\bold{\longmapsto x=\frac{12500}{5}}$

$\bold{\longmapsto x=Rs.2500}⟼x=Rs.2500$

$\sf{\longmapsto CP of radios=2500}⟼CPofradios=2500$

$\sf{SP ~of ~radio-1=2500(\frac{100+12}{100}}$

$\bold{\longmapsto 2500 × \frac{112}{100}}⟼2500×$

$\bold{\longmapsto Rs.2800}$

$\sf{SP~ of~ radio-2=2500(\frac{100+17}{100}}$

$\bold{\longmapsto 2500 ×\frac{ 117}{100}}⟼2500×$

$\bold{\longmapsto Rs.2925}$

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