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The time it took car A to travel 400 miles was 2 hours less than the

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Joined: 26 Feb 2011
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The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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10 Mar 2011, 10:42
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The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20
B. 30
C. 40
D. 50
E. 80
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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10 Mar 2011, 11:02
Let B's average speed be b

$$\frac{400}{b+10} = \frac{400}{b}-2$$

$$\frac{400}{b+10} = \frac{400}{b}-2$$

$$(b+50)(b-40)=0$$

b=40 miles/h

Ans: "C"
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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10 Mar 2011, 11:08
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The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20
B. 30
C. 40
D. 50
E. 80

Let the rate of B be $$x$$ miles per hour, then the rate of A would be $$x+10$$ miles per hour.

We also know that the time of B equals to time of A + 2: $$\frac{400}{x}=\frac{400}{x+10}+2$$, from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that $$\frac{400}{40}=10=\frac{400}{40+10}+2=8+2$$

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The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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31 Dec 2013, 11:44
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20
B. 30
C. 40
D. 50
E. 80

Agree with Bunuel, in these problems one will find it easier to plug in answer choices, that is, backsolving

One gets

400/B+10 - 400/B = 2

So beginning with C one gets that 10 - 8 = 2

So bingo!

Cheers!
J
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The time it took car A to travel 400 miles was 2 hours less  [#permalink]

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31 Dec 2013, 14:49
Bunuel wrote:
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20
B. 30
C. 40
D. 50
E. 80

Let the rate of B be $$x$$ miles per hour, then the rate of A would be $$x+10$$ miles per hour.

We also know that the time of B equals to time of A + 2: $$\frac{400}{x}=\frac{400}{x+10}+2$$, from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that $$\frac{400}{40}=10=\frac{400}{40+10}+2=8+2$$

I'm struggling with this one for some odd reason:

The setup for my work was as follows
Car B: Rate = R + 10, Time = T + 2, Distance = 400
Car A: Rate = R, Time = T, Distance = 400

I then substituted B's Rate (400/T) into Ys equation --> giving me --> (400 / T) + 10 = (400 / T - 2)
After solving this out, I got T = 20, therefore Car B's average speed is 20mph.

Can either of you help me out with where I went wrong here. Much appreciated. Just over a month from my first GMAT experience!
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The time it took car A to travel 400 miles was 2 hours less  [#permalink]

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01 Jan 2014, 05:57
bparrish89 wrote:
Bunuel wrote:
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20
B. 30
C. 40
D. 50
E. 80

Let the rate of B be $$x$$ miles per hour, then the rate of A would be $$x+10$$ miles per hour.

We also know that the time of B equals to time of A + 2: $$\frac{400}{x}=\frac{400}{x+10}+2$$, from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that $$\frac{400}{40}=10=\frac{400}{40+10}+2=8+2$$

I'm struggling with this one for some odd reason:

The setup for my work was as follows
Car B: Rate = R + 10, Time = T + 2, Distance = 400
Car A: Rate = R, Time = T, Distance = 400

I then substituted B's Rate (400/T) into Ys equation --> giving me --> (400 / T) + 10 = (400 / T - 2)
After solving this out, I got T = 20, therefore Car B's average speed is 20mph.

Can either of you help me out with where I went wrong here. Much appreciated. Just over a month from my first GMAT experience!

We are given that Car As average speed was 10 miles per hour greater than that of car B, thus if you say that the rate of A is R miles per hour, then the rate of B is R-10 miles per hour (not R+10).

The equation should be 400/T -10 = 400/(T+2) --> T=8 --> B's rate = 400/(T+2) = 40.

Hope it's clear.
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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22 Jun 2014, 22:13
To be academic with use GMAT strategy)

R * T = D
A x-2= 400
B x = 400

Question is what is the rate of B. I agree that backsolving is the best strategy and start with C. If R of B=40 then x=10 and R of A=50 and this is 10 more than B. So answer is C
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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22 Jun 2014, 22:46
1
.......................... A................. B

Time .................. t .................. t+2

Speed ............... s+10 ............... s

t(s+10) = s(t+2) = 400

s = 5t

Substitute value of t in any of the formed equation:

$$\frac{s}{5} (s+10) = 400$$

$$s^2 + 10s - 2000 = 0$$

Speed of B (s) = 40

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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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03 Jul 2014, 10:45
The first step should be mentioned here: Translate the task into numbers and variables !

Here you have a lot of different possibilities just 2 examples:

I. (r+10) * (t-2) =400
II. r * t = 400

OR:

I. (r+10) * t = 400
II. r * (t+2) = 400

As you can see, no we have 2 equations with 2 variables... tons of different possibilities to solve this as well:
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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11 Aug 2018, 14:00
Bunuel wrote:
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car As average speed was 10 miles per hour greater than that of car B, what was car Bs average speed in miles per hour?
A/ 20
B/ 30
C/ 40
D/ 50
E/ 80

Let the rate of B be $$x$$ miles per hour, then the rate of A would be $$x+10$$ miles per hour.

We also know that the time of B equals to time of A + 2: $$\frac{400}{x}=\frac{400}{x+10}+2$$, from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that $$\frac{400}{40}=10=\frac{400}{40+10}+2=8+2$$

Hi Bunuel,
This question needs an edit.
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A ‘s average speed was 10 mi greater than that of average speed of Car B , the what was Car B’s average speed in miles per hour?

Probus
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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11 Aug 2018, 14:23
Probus wrote:
Bunuel wrote:
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car As average speed was 10 miles per hour greater than that of car B, what was car B`s average speed in miles per hour?
A/ 20
B/ 30
C/ 40
D/ 50
E/ 80

Let the rate of B be $$x$$ miles per hour, then the rate of A would be $$x+10$$ miles per hour.

We also know that the time of B equals to time of A + 2: $$\frac{400}{x}=\frac{400}{x+10}+2$$, from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that $$\frac{400}{40}=10=\frac{400}{40+10}+2=8+2$$

Hi Bunuel,
This question needs an edit.
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A ‘s average speed was 10 mi greater than that of average speed of Car B , the what was Car B’s average speed in miles per hour?

Probus

_______________
Edited. Thank you.
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Re: The time it took car A to travel 400 miles was 2 hours less than the  [#permalink]

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11 Apr 2019, 11:12
Hi swerve,

We're told that the time it took car A to travel 400 miles was 2 hours LESS than the time it took car B to travel the same distance and car A's average speed was 10 miles per hour GREATER than that of car B. We're asked for car B's average speed in miles per hour. This question can be approached in a couple of different ways. Since it's essentially about basic arithmetic (re: two 'pairs' of numbers that have a product of 400), you would likely find it fastest to TEST THE ANSWERS.

To start, the difference in the time that the two cars traveled is exactly 2 hours and we know that Car B was going exactly 10 mph slower than Car A. Based on the prompt and the answer choices, we're clearly dealing with nice 'round' numbers, so let's start with an answer that divides evenly into 400...

Let's TEST Answer C: 40 miles/hour

IF.... Car B is traveling 40 mph, then it takes 400/40 = 10 hours to complete that drive.
We're told that Car A's speed is 10 mph greater than Car B's speed, so....
Car A is traveling 50 mph... and that would take 400/50 = 8 hours to complete that drive.
Here, the difference in travel time is exactly 2 hours - and this matches what we were told - so this MUST be the answer.

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Re: The time it took car A to travel 400 miles was 2 hours less than the   [#permalink] 11 Apr 2019, 11:12
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