What percentage of steel workers in the country earn over $37,000 anually and has a net worth over $200,000?

1) 55% of all steel workers in the country have an annual income over $37,000

2) 12% of all steel workers in the country with an annual income over $37,000 also have a networth above $200,000

I chose B, but the answer is C "both statements together are suff". Can you please show algebraically how you arrive at the answer?
thanks.

Hey TJ,

This one is probably even more conceptual than algebraic, but the algebra helps, too. The question looks for the percentage of the total that meet two criteria - an annual income > $37K and a net worth of over $200K.

Let's call T the total, I the group that meets the income requirement, N the group that meets the net work requirement, and B the group that meets both (we won't need all of those variables in the end, but we can set up that way). With that in mind, we need to be able to solve for:

B/T * 100 (both/total expressed as a percentage)

Statement 1 tells us that 55% of the total meet the income requirement. Algebraically, that's:

1) 55/100 (T) = I

We don't have any information about the "both" group, so this statement is not sufficient.

Statement 2 tells us that, of those who meet the income requirement, 12% also meet the net worth requirement, and therefore both requirements. Algebraically, that's:

2) 12/100 (I) = B

We don't have any information about the "total" group, so this statement is not sufficient.


Together:

55/100 (T) = I

12/100 (I) = B

so

12/100 [(55/100) T] = B

or, conceptually, 12% of 55% of the total represents the "both" group, so we can multiply .12 * .55 (ugly math, but we don't need to do it since we know we'll get a number) to get the percentage.

I hope that helps...

got it. thanks.