How many mg of a metal containing 38 percent nickel must be combined with 5 mg of pure nickel to form an alloy containing 68 percent nickel? Which of the following equations could be used to solve this?

Answer:

the Answer is C

Step-by-step explanation:

The metal needs to be 16/3 mg in weight to get the required alloy.

We can use the third equation,

(a(0.38) + 5(1.00))/(a+5) = 68/100

to solve the problem.

A percentage is a ratio where a number is shown as a fraction with respect to 100 parts of it. To remove the percentage sign we divide the number by 100.

We are asked to mix a certain quantity of metal in mg containing 38 percent of nickel to 5 mg of pure nickel such that the new alloy contains 68 percent of nickel.

Let the quantity of metal taken to be 'a' mg.

∴ Nickel provided by the metal is 38% of a,

or, (38/100)*a

or, a(0.38)

We took 5 mg of pure nickel so it is 100 percent pure,

∴ Nickel in the alloy from pure nickel = 100% of 5

or, (100/100)*5

or, 5(1.00)

Total weight of the alloy = a + 5

Total nickel in the alloy = a(0.38) + 5(1.00)

We are said that the alloy constitutes 68 percent of nickel,

∴ (a(0.38) + 5(1.00))/(a+5) = 68%

or, (a(0.38) + 5(1.00))/(a+5) = 68/100

This is the equation needed to find the quantity of metal used to make the alloy. It is the same as option 3.

On solving the equation we get that the required quantity of the metal is 16/3 mg.

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