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A lump of two metals weighing 18 g is worth Rs. 87 but if their weight is interchanged, it would be worth Rs. 78.60. If the price of one metal be Rs. 6.70 per gram, find the weight of the other metal in the mixture.
Answer & Solution
Correct Answer:
Option
A
Cost of (18g of 1st metal + 18g of 2nd metal) = Rs. 165.60
Cost of (1g of 1st metal + 1g metal of 2nd metal) = Rs. 9.20
Hence cost of 1g of 2nd metal,
= 9.20 - 6.70
= Rs. 2.5
Mean price = Rs. $$\frac{{87}}{{18}}$$
Now, $$\frac{{{\text{quantity}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{metal}}}}{{{\text{quantity}}\,{\text{of}}\,{2^{{\text{nd}}}}\,{\text{metal}}}}$$ $$ = $$ $$\frac{{14}}{6}:\frac{{56}}{{30}} = 5:4$$
Quantity of 2nd metal = $$\frac{{18 \times 4}}{9}$$ = 8g
Now, $$\frac{{{\text{quantity}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{metal}}}}{{{\text{quantity}}\,{\text{of}}\,{2^{{\text{nd}}}}\,{\text{metal}}}}$$ $$ = $$ $$\frac{{14}}{6}:\frac{{56}}{{30}} = 5:4$$
Quantity of 2nd metal = $$\frac{{18 \times 4}}{9}$$ = 8g
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Loginprice of second metal per gram=p
weight of one metal in mixture=x
weight of second metal in mixture=18-x
before interchange: 6.7*x+p(18-x)=87
after interchange:6.7(18-x)+px=78.6
after solving: p=2.5 ,x=10
so weight of second metal in mixture=8