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Problem Solving

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by [email protected] » Fri Sep 19, 2014 10:34 am
Hi RiyaR,

This question is a real "pattern-matcher's" question. You don't have to do a lot of math to solve it, but you do have to think about how fractions can be equal to one another....

We're told that B and C are positive.

We're also told that 1/B = B/C = C/8

Let's start with that last fraction....C/8. We're told that the other two fractions are also equal to C/8 so they both need to have denominators that are either FACTORS of 8 or MULTIPLES of 8. Since the answer choices are relatively small, we're probably going to be dealing with FACTORS of 8....

So, what divides evenly into 8? 1, 2, 4, 8.....

B/C = C/8

If C = 8, then B = 8, but then 1/B would not equal the other two fractions, so C CANNOT = 8
If C = 4, then B = 2.....and we have 1/2 = 2/4 = 4/8. This is exactly what we're looking for.

The question asks us for B+C.....2+4 = 6

Final Answer: B

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by Brent@GMATPrepNow » Fri Sep 19, 2014 12:32 pm
RiyaR wrote:If b and c are positive numbers and 1/b = b/c = c/8 , then b + c =
(A) 4
(B) 6
(C) 7
(D) 8
(E) 9
Here's an algebraic solution.

Begin with 1/b = b/c = c/8
The lowest common denominator of b, c and 8 is 8bc.
So, multiply all three sides by 8bc to get: 8c = 8b² = bc²

Let's focus on 8c = 8b²
Divide both sides by 8 to get: c = b²

Now, let's focus on 8c = bc²
Divide both sides by c to get: 8 = bc

So, we have the following system:
c = b²
8 = bc

Take the red equation, and replace c with b² to get: 8 = b(b² )
Simplify: 8 = b³
Solve: b = 2

Now that we know that b = 2, we can use the fact that 1/b = c/8 to get 1/2 = c/8
This means that c = 4

So, b + c = 2 + 4 = [spoiler]6 = B[/spoiler]

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by GMATGuruNY » Sat Sep 20, 2014 3:04 am
RiyaR wrote:17. If b and c are positive numbers and 1/b=b/c=c/8 , then b + c =
(A) 4
(B) 6
(C) 7
(D) 8
(E) 9
Given that the answer choices are integers, b and c are almost certainly integers themselves.

1/b = c/8.
bc = 8.

8 has only two factor pairs:
1 and 8
2 and 4.

Since 1/b = b/c, c = b².
Implication:
c is a PERFECT SQUARE.
In the factor pairs for 8, only 1 and 4 are perfect squares.
If c=1 and b=8, then c ≠ b².
Thus:
c=4 and b=2, implying that b+c = 2+4 = 6.

The correct answer is B.
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by Matt@VeritasPrep » Mon Sep 22, 2014 9:16 am
I'd proceed more or less the same way GMATGuru did: by assuming that the answers are friendly, familiar integers.

Since 1/b = c/8, it's probably the case that 1/b is in reduced terms and c/8 is not.

c/8 has a few reducible possibilities: 2/8, 4/8, and 8/8.

If c/8 = 2/8, then 1/b = 1/4. But then b/c = 4/2, which is not the same as 1/4 or 2/8.

If c/8 = 4/8, then 1/b = 1/2. Then b/c = 2/4, which IS the same as 1/2 and 4/8. Success!

Hence b = 2, c = 4, and b + c = 6.
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by Vipul1991 » Sat Aug 12, 2017 10:26 pm
1/b=b/c=c/8
Therefore, from eqn (1)
1/b=b/c => b^2=c
And from eqn (2)
b/c=c/8 => c^2=8b

from (1) & (2) we have,

Replacing c in terms of b,
b^4=8b
=> b^3=8 ; b=2

Replacing b=2 in eqn (1) we have c=4
Hence b+c=6
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Re: Problem Solving

by Scott@TargetTestPrep » Sun Feb 16, 2020 7:46 am
RiyaR wrote: ↑
Fri Sep 19, 2014 10:15 am
 17. If b and c are positive numbers and 1/b=b/c=c/8 , then b + c =
(A) 4
(B) 6
(C) 7
(D) 8
(E) 9
We see that the equation can be re-expressed as three separate equations: 1/b = b/c, b/c = c/8, and 1/b = c/8.

Using the first equation, we have:

1/b = b/c

b^2 = c

Using the third equation, we have:

1/b = c/8

bc = 8

Since c = b^2, we have:

b(b^2) = 8

b^3 = 8

b = 2

So c = 2^2 = 4, and therefore, b + c = 2 + 4 = 6.

Answer: B

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