<p>Hey guys!</p>
<p>Can anyone help me with this problem-
If a,b,c are number such that a/b=3 and b/c=7 , then a+b/b+c is equal to which of the following?</p>
<p>(A)7/2
(B) 7/8
(C) 3/7
(D) 1/7
(E) 21</p>
<p>There’s probably a mathematical way to solve this which someone will explain later, but I would just Plug and Chug. I would say b is 7 making c =1 and a=21. Then I would plug this into the formula and get 3.5 or 7/2</p>
<p>given: a/b=3
a=3b
therefore, a+b=3b+b= 4b
given: b/c=7
therefore, b=7c
therefore, c=b/7
therefore, b+c=b+ b/7 = 8b/7</p>
<p>solve, (a+b) / (b+c)
= 4b / (8b/7)
= 4 / (8/7)
= 4 * (7/8)
= 7/2</p>
<p>If a,b,c are numbers such that a/b=3 and b/c=7 , then (a+b)/(b+c) is equal to which of the following? …</p>
<p>There are lots of b’s in the question. It pays to be observant of mathematical relationship in the SAT.</p>
<p>Divide both the numerator and denominator by b. You get (1 + a/b)/(1 +c/b) which is 4/(8/7) or 7/2</p>
<p>The fastest way is simply to assume a = 21, b = 7, c = 1 (or something like that).</p>
<p>Or you could express everything in terms of b: a = 3b, c = b/7, replace into (a+b)/(b+c). The first solution’s probably the best if you’re dealing with SAT-level math.</p>