Question

200kg of an alloy of metal A and B in the ratio of 1 : 4 is mixed with a kgs of alloy of metal A and B in the ratio of 2 : 3. If percentage of metal A in mixture of two alloys is in between 30% and 32%. Find the maximum value of a can take.
1. 250
2. 300
3. 100
4. 200

Answer 1

Correct Answer - Option 2 : 300

Calculation:

Quantity of metal A in 200kg alloy = 1/5 × 200 = 40

Quantity of metal B in a kg alloy = 2/5 × a = 2a/5

% of tin in mixture of two alloys = (40 + 2a/5)/(200 + a) × 100

⇒ 30 ≤ (40 + 2a/5)/(200 + a) × 100 ≤ 32

Then,

⇒ 30 ≤ (200 + 2a)/(200 + a) × 20

⇒ 600 + 3a ≤ 400 + 4a

⇒ 200 ≤ a

And,

⇒ (200 + 2a)/(200 + a) × 20 ≤ 32

⇒ 4000 + 40a ≤ 6400 + 32a

⇒ a ≤ 300

Range in between values of a can take = 200 ≤ a ≤ 300.

∴ Maximum value a can take is 300.

Ratio of metal A and B in alloy 1 of m kg = a : b

Ratio of metal A and B in alloy 2 of n kg = c : d

Percentage of metal A in mixture of two alloys = (am/(a + b) + cn/(c + d)) × 100/(m + n)

Similarly, for % metal B in mixture of two alloys.