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

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

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New post 21 Sep 2018, 09:45
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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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Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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New post 21 Sep 2018, 10:31
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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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New post Updated on: 21 Sep 2018, 11:02
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, 10:51.
Last edited by dave13 on 21 Sep 2018, 11:02, edited 1 time in total.
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Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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New post 21 Sep 2018, 10:59
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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New post 21 Sep 2018, 11:00
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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.
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Re: Thomas was riding his bike at a constant rate to the store, which....  [#permalink]

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New post 21 Sep 2018, 11:01
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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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New post 21 Sep 2018, 11:15
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, 11:15
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