An alloy contains a mixture of two metals X and Y in the ratio of 2 3 The second alloy contains a mi...........

Quantitative Aptitude (MTS) Mixture and Alligation

An alloy contains a mixture of two metals X and Y in the ratio of 2 : 3. The second alloy contains a mixture of the same metals,

X and Y, in the ratio 7 3. In what ratio should the first and the second alloys be mixed so as to make a new alloy containing 50%

of metal X?

This questions was previously asked in

SSC MTS 12th July 2022 Shift-2

3 : 4

3 : 1

5 : 6

2 : 1

- First alloy (A1) has metals X and Y in the ratio 2:3.

- Therefore, in A1, X is 2/(2+3) = 2/5 or 40%.

- Second alloy (A2) has metals X and Y in the ratio 7:3.

- Therefore, in A2, X is 7/(7+3) = 7/10 or 70%.

- We want a new alloy (A3) where X is 50%.

- Use the rule of alligation:

- Difference between A1 X% and A3 X% = 50 - 40 = 10

- Difference between A2 X% and A3 X% = 70 - 50 = 20

- Therefore, mix A1 and A2 in the ratio of 20:10, which simplifies to 2:1.

?? The correct answer is Option 4: 2 : 1.

Submit Error