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# Had Frank driven at an averagespeed of 40 miles per hour, he would hav

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Had Frank driven at an averagespeed of 40 miles per hour, he would hav  [#permalink]

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10 Mar 2019, 10:20
1
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Difficulty:

35% (medium)

Question Stats:

64% (01:18) correct 36% (01:38) wrong based on 55 sessions

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Had Frank driven at an averagespeed of 40 miles per hour, he would have reached his office 10 minutes earlier than he usually did. At what average speed should Frank drive to reach his office 20 minutes earlier than he usually did?

(1) Frank usually takes 5/6 hours to reach his office

(2) The distance to his office is 80/3 miles
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Re: Had Frank driven at an averagespeed of 40 miles per hour, he would hav  [#permalink]

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10 Mar 2019, 10:42
2
From question. 40(t-10) = d and d = x(t-20)

From statement 1:

Time taken is 56*60 min. Enough to find d from 1st equation and eventually find the time required.
Hence Sufficient.

From statement 2:

d = 803 miles.
Again, time taken can be calculated from 1st equation. Hence we can find the speed.
Sufficient.

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Re: Had Frank driven at an averagespeed of 40 miles per hour, he would hav  [#permalink]

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30 Mar 2019, 06:46
1
mangamma wrote:
Had Frank driven at an averagespeed of 40 miles per hour, he would have reached his office 10 minutes earlier than he usually did. At what average speed should Frank drive to reach his office 20 minutes earlier than he usually did?

(1) Frank usually takes 56 hours to reach his office

(2) The distance to his office is 803 miles

A)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.

B)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.

C)BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

E)Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Typo error in statments....
1. 5/6 not 56
2. 80/3 not 803
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Re: Had Frank driven at an averagespeed of 40 miles per hour, he would hav  [#permalink]

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30 Mar 2019, 06:48
kiran120680 wrote:
mangamma wrote:
Had Frank driven at an averagespeed of 40 miles per hour, he would have reached his office 10 minutes earlier than he usually did. At what average speed should Frank drive to reach his office 20 minutes earlier than he usually did?

(1) Frank usually takes 56 hours to reach his office

(2) The distance to his office is 803 miles

A)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.

B)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.

C)BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

E)Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Typo error in statments....
1. 5/6 not 56
2. 80/3 not 803

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Edited. Thank you.
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Re: Had Frank driven at an averagespeed of 40 miles per hour, he would hav  [#permalink]

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31 Mar 2019, 07:39
$$S_{initial}$$ = x miles per hour ; $$t_{initial}$$ = y mins ; Distance = d miles
$$S_1$$ = 40 miles per hour ; $$t_1$$ = y - 10 mins ; Distance = d miles
$$S_2$$ = $$S_2$$ miles per hour ; $$t_2$$ = y - 20 mins ; Distance = d miles

We have to find $$S_2$$

$$\frac{40}{S_2}$$ =$$\frac{y - 20}{y - 10}$$ ..................(1)

Statement 1) $$t_{initial}$$ = y mins = $$\frac{5}{6}*60$$ = 50 mins

Substituting "y" in eq (1), we can find $$S_2$$. SUFFICIENT.

Statement 2) Distance = d miles = $$\frac{80}{3}$$ miles

calculating the time taken by traveling at 40 miles per hour

40 = $$\frac{80}{3(y - 10)}$$

We will get y value and then we can substitute in eq (1) to get $$S_2$$. SUFFICIENT.

OPTION: D
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Re: Had Frank driven at an averagespeed of 40 miles per hour, he would hav   [#permalink] 31 Mar 2019, 07:39
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