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# Thomas was riding his bike at a constant rate to the store, which....

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Director
Joined: 11 Feb 2015
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Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 08:45
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1
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Difficulty:

25% (medium)

Question Stats:

81% (02:29) correct 19% (02:50) wrong based on 73 sessions

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Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

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Posts: 7202
Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 09:31
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1
CAMANISHPARMAR wrote:
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

Ways to do it..
i) choices
since speed,s, and s+2 leave a difference of 1 hour, both should divide 60 evenly and leave a difference 1..

A) 8 mph
8 does not divide evenly into 60 but 8+2 does....eliminate

B) 10 mph
Both 10 and 12 divide evenly in to 60, and difference is 60/10-60/12=6-5=1..YES

C) 12 mph
12+2 does not divide evenly into 60 but 12 does....eliminate

D) 15 mph
15+2 does not divide evenly into 60 but 15 does....eliminate

E) 18 mph
18 does not divide evenly into 60 but 18+2 does....eliminate

so B

II) algebraic

time taken at s speed = $$\frac{60}{s}$$
time taken at s+2 speed = $$\frac{60}{s+2}$$
so $$\frac{60}{s}-\frac{60}{s+2}=1.......60s+120-60s=s^2+2s......s^2+2s-120+0.......(s+12)(s-10)=0$$
so s = -12 or 10 but s cannot be negative so s=10

B
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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Updated on: 21 Sep 2018, 10:02
1
CAMANISHPARMAR wrote:
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

Distance $$60$$

Speed = $$s+2$$

Time = $$t-1$$

Distace = speed * time

1st Eq. $$60 = (s+2) (t-1)$$

2nd Eq. $$60 = s*t$$ ---- > $$S = \frac{60}{t}$$ (plug in the first one)

$$60 = (\frac{60}{t}+2) (t-1)$$

$$t60 = (60+2t) (t-1)$$ (multiply both parts by t )

$$t60 = 60t -60+2t-2$$

$$t60 = -62t +62t$$

niks18 why cant I get an answer from here ? any idea ?

Originally posted by dave13 on 21 Sep 2018, 09:51.
Last edited by dave13 on 21 Sep 2018, 10:02, edited 1 time in total.
Manager
Joined: 20 Feb 2017
Posts: 164
Location: India
Concentration: Operations, Strategy
WE: Engineering (Other)
Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 09:59
1
let the speed be x
so ,
60/x = 60/x+2 +1
from the option put x=10, then the equation solves. Hence B
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Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 10:00
2
dave13 wrote:
CAMANISHPARMAR wrote:
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

Distance $$60$$

Speed = $$s+2$$

Time = $$t-1$$

Distace = speed * time

1st Eq. $$60 = (s+2) (t-1)$$

2nd Eq. $$60 = s*t$$ ---- > $$S = \frac{60}{t}$$ (plug in the first one)

$$60 = (\frac{60}{t}+2) (t-1)$$ niks18 why cant I get an answer from here ? any idea ?

Hi dave13

You did not solve the equation. you will get an answer. It will be a quadratic equation where you will get two value of t. reject the negative value because time cannot be negative.
Instead of substituting t for s you should substitute s for t because you need speed here.
Math Expert
Joined: 02 Aug 2009
Posts: 7202
Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 10:01
2
dave13 wrote:
CAMANISHPARMAR wrote:
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

Distance $$60$$

Speed = $$s+2$$

Time = $$t-1$$

Distace = speed * time

1st Eq. $$60 = (s+2) (t-1)$$

2nd Eq. $$60 = s*t$$ ---- > $$S = \frac{60}{t}$$ (plug in the first one)

$$60 = (\frac{60}{t}+2) (t-1)$$ niks18 why cant I get an answer from here ? any idea ?

You will get your answer, but you will get time which will have to be converted in speed

$$60 = (\frac{60}{t}+2) (t-1)$$ ...

$$60 = (\frac{60+2t}{t}) (t-1)$$ .......$$60t=(60+2t)(t-1)=60t-60+2t^2-2t......2t^2-2t-60=0......t^2-t-30=0.......(t-6)(t+5)=0$$
So t is 6 or -5
But t cannot be negative, so t=6
Speed =60/6=10
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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21 Sep 2018, 10:15
1
CAMANISHPARMAR wrote:
Thomas was riding his bike at a constant rate to the store, which is 60 miles away. If Thomas had ridden his bike 2 miles per hour faster than he actually did, he would have saved one hour. How fast did he actually ride to the store?

A) 8 mph

B) 10 mph

C) 12 mph

D) 15 mph

E) 18 mph

$$\frac{60}{s} - \frac{60}{( s + 2)} = 1$$

Now, try to use the answer options to reach the correct Answer...

$$\frac{60}{10} - \frac{60}{( 10 + 2)} = 1$$, Answer must be (B)
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Re: Thomas was riding his bike at a constant rate to the store, which.... &nbs [#permalink] 21 Sep 2018, 10:15
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