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# If the city B lies along a certain route between A and C, what is the

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If the city B lies along a certain route between A and C, what is the  [#permalink]

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01 Jul 2017, 10:23
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If the city B lies along a certain route between A and C, what is the distance from A to B?

(1) Travelling from C to A at an average rate of 45 m.p.h. takes 25 minutes longer than at an average rate of 50 m.p.h.

(2) Traveling along the route at a constant 25 miles per gallon of gas, a car will use twice as much gas between A and B as it will use between B and C

Kudos for correct solution.

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If the city B lies along a certain route between A and C, what is the  [#permalink]

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01 Jul 2017, 11:21
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ydmuley wrote:
If the city B lies along a certain route between A and C, what is the distance from A to B?

(1) Travelling from C to A at an average rate of 45 m.p.h. takes 25 minutes longer than at an average rate of 50 m.p.h.

(2) Traveling along the route at a constant 25 miles per gallon of gas, a car will use twice as much gas between A and B as it will use between B and C

Kudos for correct solution.

Let Distance from $$A$$ to $$C$$ $$= D$$. Distance from $$A$$ to $$B$$ $$= D_1$$ and Distance from $$B$$ to $$C$$ $$= D_2$$

We need to find $$D_1$$.

(1)Travelling from C to A at an average rate of 45 m.p.h. takes 25 minutes longer than at an average rate of 50 m.p.h.
Let time taken with speed 50 m.p.h $$= t$$

$$D = 50*t$$ ----------- (i)

$$D = 45(t+\frac{5}{12})$$ -----------(ii) --------- ($$25$$ mins $$= \frac{5}{12} hr$$)

Equating (i) and (ii);

$$50t = 45(t+\frac{5}{12}) =45t + \frac{45*5}{12}$$

$$50t - 45t = \frac{15*5}{4}$$ $$=>$$ $$5t = \frac{15*5}{4}$$

$$t = \frac{15}{4}$$ hours

$$D = 50 * \frac{15}{4} = 187.5$$ miles

Does not have any information about distance from A to B or B to C. Hence I is Not Sufficient.

(2) Traveling along the route at a constant 25 miles per gallon of gas, a car will use twice as much gas between A and B as it will use between B and C. ( twice as much gas means twice the time.) Let time taken to travel from $$B$$ to $$C = t_1$$

$$D_1 = 25*2t_1 => 25 = \frac{D_1}{2t_1}$$ ---------- (iii)

$$D_2 = 25*t_1 => 25 = \frac{D_2}{t_1}$$ ----------- (iv)

Equating (iii) and (iv), we get; $$\frac{D_1}{2t_1} = \frac{D_2}{t_1}$$$$=>$$ $$D_2 = \frac{D_1}{2}$$

Cannot find Distance from A to B. Hence II is Not Sufficient.

Combining (1) and (2);

$$D_1 + D_2 = D$$

$$D_1 + \frac{D_1}{2} = D$$

$$\frac{2D_1 + D_1}{2} = D$$

$$3D_1 = 2D$$

$$D_1 = \frac{2}{3}D$$

$$D_1 = \frac{2}{3} * 187.5 = 125$$ miles .

If the city B lies along a certain route between A and C, what is the   [#permalink] 01 Jul 2017, 11:21
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