Problem Solving

A certain industrial machine requires 15 quarts (480 ounces) of oil every day, and it is given a mixture of oil A and oil B. Oil A consists of 5% special additives while oil B consists of 8% special additives. If the machine needs exactly 30 ounces of the special additives to ensure smooth operation, then how much of oil A, in ounces, should be in the mixture?

a 260

b 280

c 290

d 300

e 320

PLEASE EXPLAIN....

two equations, two unknowns:

You know that A+B=480
form an equation for the actual number of ounces of special additive, which should sum up to 30 ounces
5%A + 8%B = 30

5A/100 + 8b/100 = 30
multiply by 100
5A+8B = 30*100
Divide by 8 to Isolate B

5/8A + B = 30*100/8

Now subtract this equation from the top equation (A+B=480) to get
A-5/8A + B-B = 480 - 30*100/8
3/8A = 480 - 30*100/8
Multiply times 8
3A = 480*8 - 30*100
divide by 3
A = 160*8 - 10*100 = 1280-1000 =280.
Answer should be B.

You could reverse Plug in the answer choices here as well:
If we assume that answer choice B is correct, i.e. that A=280, then we have 5%*280 = 14 ounces additive from A
The remaining 200 ounces are B at 8% additive, or 8/100*200 = 8*2 = 16 ounce additive from B, which together with the 14 ounces from A brings you back to the required 30 ounces - which means that B checks out as the right answer.

Thanks. The question somehow got me all puzzled, but you seem to have made it easy.

well you can do it by plugging values from options.....

First take the middle value, say oil a is 290 ounces. So 5% of additives will be 14.5 (290*.05). to make it 30 ounces of additive we now require more 15.5 ounces of additives from oil B. so the remaining 190 (480-290) should give us 15.5. 190*.08=15.2....we fall short. so minimize oil A and maximize oil B
take value 280 for oil A...so additive from oil A= 280*.05= 14
additive from oil B required is 16... 200*.08 (480-280) = 16
14+16 =30

therefore answer is B.