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Answer 1

Given :-

• Metal A is 20 times heavier than iron.
• Metal B is 13 times heavier than iron.

To find :-

• In which ratio these two metals be mixed so that the mixture is 17 times heavier than iron ?

Solution :-

Let, 1 gm of Metal A be mixed with y gm of metal B to give (1 + y) gm of mixture.

Now,

→ 1A = 20 Iron

→ 1B = 13 Iron

→ Mixture = 17 Iron

And, 1 gm gold + y gm copper = (1 + y) gm mixture

So,

→ 20A * 1 iron + 13B * y iron = (1 + y ) * 17 iron

→ 20 + 13y = 17 + 17y

→ 20 - 17 = 17y - 13y

→ 3 = 4y

→ y = 3/4

Hence, the required ratio is 1 : 3/4 = 4 : 3 (Ans.) (Option D).