Let d > c > b > a. If c is twice as far from a as it is from d, and b is twice as far from c as it is from a, then (d – b) / (d – a) = A. 2/9 B. 1/3 C. 2/3 D. 7/9 E. 3/2

The correct answer is D. Geometrical and algebraic solutions are below.

Since the exercise mentions a number line, draw one. Since d > a, label the left end of the number line with d and the right end with a. The values of b and c will be between a and d:

Start with the relationship between c, a, and d. Since c is twice as far from a as it is from d, the segment between a and d is divided into 3 pieces – two pieces between c and a and one piece between d and c. Each piece is 1/3 of d to a.

The relationship between b, c, and a says that b is twice as far from c as it is from a. This means the segment between a and c is divided into 3 pieces.

The segment between c and a was 2/3 of the segment between d and a. Dividing 2/3 into 3 pieces makes each piece (2/3) / 3 = 2/9.

So d – b (the length of the segment between d and b) makes 7/9 of d – a (the length of the segment between d and a).

This question can also be solved algebraically. It might seem more difficult but it can be fast if you are good at fractions and simplifying expressions.

"c is twice as far from a as it is from d" means that |c – a| = 2|c – d| but since we know that d > c > b > a we can avoid using absolute values: c – a = 2(d – c)

In much the same way we transform "b is twice as far from c as it is from a" into algebra: c – b = 2(b – a)

Remember, that the question asks us to find: (d – b) / (d – a).

From formula c – a = 2(d – c) we can find that d = (3c – a) / 2

From formula c – b = 2(b – a) we can find that b = (c + 2a) / 3

Let us plug both formulas into expression we need to calculate: (d – b) / (d – a) = [(3c – a)/2 – (c + 2a)/3] / [(3c – a)/2 – a] = = [(7/6) × (c – a)] / [(3/2) × (c – a)]] = = (7/6) / (3/2) = 7/9

I don't quite understand the problem. I don't understand why 6 is twice as far from 0 as it is from 9. How is "if c is twice as far from a as it is from d, and b is twice as far from c as it is from a" interpret in the equation?

I don't quite understand the problem. I don't understand why 6 is twice as far from 0 as it is from 9.

All real numbers form a number line. Each number corresponds to a point. The distance between such points is the same as the distance between the numbers. Looking at the number line you can see that the distance between some real numbers a and b equals |a – b|.

Quote:

I don't understand why 6 is twice as far from 0 as it is from 9.

The distance between 6 and 0 is |6 – 0| = 6. The distance between 6 and 9 is |6 – 9| = 3. Therefore the distance between 6 and 0 is twice larger than the distance between 6 and 9.

Quote:

How is "if c is twice as far from a as it is from d, and b is twice as far from c as it is from a" interpret in the equation?

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