One computer can upload 100 megabytes worth of data in 6

Question

One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?

(A) 6
(B) 7
(C) 9
(D) 11
(E) 13

(A) 6
(B) 7
(C) 9
(D) 11
(E) 13

Bunuel · Accepted Answer

Since the first computer can upload 100 megabytes worth of data in 6 seconds then in 6*7=42 seconds it can upload 7*100=700 megabytes worth of data, hence the second compute in 42 seconds uploads 1300-700=600 megabytes worth of data. The second computer can upload 600/6=100 megabytes of data in 42/6=7 seconds.

Answer: B.

mbaiseasy · Answer

Setup the rate equations. This always help.

rate-first=\frac{1}{A}=\frac{100}{6sec} 
 rate-together=\frac{1}{A}+\frac{1}{B}=\frac{1300}{42}

rate-together - rate-first = rate-second

\frac{1}{A}+\frac{1}{B}-\frac{1}{A}=\frac{1300}{42}-\frac{100}{6}=\frac{600}{42}=\frac{100}{7secs}

Answer: B

AbeinOhio · Answer

You are on the right track..

So Computer B's rate = 1300/42-100/6 = 600/42

So computer B can upload 600MB in 42 seconds.. divide numerator and denominator by 6 and you get 100/7 so 2nd computer takes 7 seconds to upload 100MB

LalaB · Answer

A needs 6sec for 100 ; A+B need 42 sec for 1300 or 100*42/1300=42/13 sec for 100 so now we have rate of A for 100 is 1/6 rate of B for 100 is 1/x rate of (a+b) for 100 is 13/42 1/6+1/x=13/42 x=7 sec hope it helps

galiya · Answer

RT=W
(100/6+X)*42=1300 --> 42X=600 --> X=600/42 rate of the second machine
600/42*t=100 --> t=7 sec. needs the second machine to upload data

megafan · Answer

One computer can upload 100 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 100 megabytes of data?

(A) 6
(B) 7
(C) 9
(D) 11
(E) 13

(A) 6
(B) 7
(C) 9
(D) 11
(E) 13

Source: Gmat Hacks 1800

GyanOne · Answer

The given computer can upload 100 MB of data in 6 seconds
=> In 42 seconds, it will upload 100 MB x 7 = 700 MB of data

In 42 seconds, the two computers working together upload 1300 MB of data
=> The other computer uploads 600 MB (1300-700 MB) of data in 42 seconds
=> The other computer uploads 100 MB of data in 42/6 = 7 seconds

Option B

Bunuel · Answer

Merging similar topics. Please refer to the solutions above.

TGC · Answer

Shouldn't be a 700 level problem.

Rate (1 computer) = 100/6
Rate ( 2 computers) = 1300/42

Rate (Computer in question) = 1300/42 - 100/6 = 100/7

Rate * Time = Work

100/7 * Time =100

Time =7

LogicGuru1 · Answer

Well, you have to understand the fundamental difference between time and rate. TIME is not an additive quantity but RATE is. Meaning you can add rates directly to get a valid new rate. (This never happens in time; remember those tricky average speed problems

) So here first computer's rate (R1) is 100/6 ------eq1 and rate of both computer (R1+R2)= 1300/42 ------eq2 so if you subtract eq 2 from eq 1 you will get the rate of second computer alone 1300/42-100/6= 700/42 R2 is 700/42 R2*T= work R2*T=100 700/42*T=100 T=6 ANSWER IS A

generis · Answer

This problem's solution is very quick if you realize that 6 seconds is a multiple of 42 seconds. Just multiply M1 rate's numerator and denominator by 7.

Machine 1, M1 =  \frac{100Mb}{6secs}

M1 + M2 =  \frac{1300Mb}{42secs}

M1 =  \frac{100Mb}{6secs}  *  \frac{7}{7} = \frac{700Mb}{42secs}

So M2 can do  \frac{(1300 - 700)}{42}  =  \frac{600Mb}{42secs}

\frac{600}{42}  =  \frac{100Mb}{x}

4,200 = 600x
x = 7

Answer B

Aitolkyn18 · Answer

If A=100/6, and A+B = 1300/42, to find out B
A multiple 7 times, as 42/6 = 7, then A=700/42, A+B = 700+ (1300-700)/42
B= 600/42, to upload 100 Mb (600/42)/6 = 100/7, the unswer is 7 sec

One computer can upload 100 megabytes worth of data in 6

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

One computer can upload 100 megabytes worth of data in 6

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards