A equals b% of what number translates algebraicaly to

A = B/100 * x, where x is "what number"

so to solve for X we need another equation that relates A to B

in 1)
A=3B we can substitute in original to get an x value (suff)

in 2)
we can rearange equation to get 2A- 56A/14 = -6B which can give the form cA = dB where c and d are some constants which we don't care about but we need to know we can get the form above (suff)

coolgirl26 wrote:A equals to B % of what number?

1) A= 3B

2) 6B+2A= 56A/14


The answer is D but how and why



And what do they mean when they say A= B% of what number

please show me all your work and short cuts and understanding involvd

As m&m says:

A equals b% of what number translates algebraicaly to

A = B/100 * x, where x is "what number"

The question is asking us for the value of x, so let's collect all the other variables on one side:

A = x(B/100)
A/B = x/100

So, if we can solve for the specific relationship A/B, we can solve for x.

(1) A = 3B

or

A/B = 3

We can substitute this value for A/B into our original equation and solve for x: sufficient.

(2) 6B+2A= 56A/14

or

6B + 2A = 4A

6B = 2A

6/2 = A/B

We can substitute this value for A/B into our original equation and solve for x: sufficient.

Each statement is sufficient alone: choose (D).

Although m&m's statement about just needing another equation that "relates A to B" wasn't 100% accurate (since we do need the specific relationship A/B... e.g. knowing that A=B+2 would NOT have been sufficient), m&m does make another excellent point in his (her? their? do the yellow ones really count?) treatment of statement (2):

c and d are some constants which we don't care about but we need to know we can get the form above

For statement (2), as soon as you see that it can be rearranged to get A/B on one side, you don't need to do any further work.