Hello,
Please give me a proper equation for this.
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute, and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than gear P?
Answer: 12
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Word translation - circular gears
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GMATGuruNY
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When elements COMPETE, calculate the DIFFERENCE between the rates.Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than geat P ?
A. 6
B. 8
C. 10
D.12
E.15
Q's rate - P's rate = 40-10 = 30 revolutions per minute.
Implication:
Every minute Q makes 30 more revolutions than P.
Time for Q to make 6 more revolutions = (number of revolutions)/(rate difference) = 6/30 = 1/5 of a minute = 12 seconds.
The correct answer is D.
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Brent@GMATPrepNow
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We can solve it using equivalent ratios.szDave wrote:Hello,
Please give me a proper equation for this.
Circular gears P and Q start rotating at the same time at constant speeds. Gear P makes 10 revolutions per minute, and gear Q makes 40 revolutions per minute. How many seconds after the gears start rotating will gear Q have made exactly 6 more revolutions than gear P?
Answer: 12
We know that, for every 60 seconds, Q makes 30 more revolutions than P does.
We want to determine how many second it will take Q to make 6 more revolutions than P does.
In our heads, we can see that it will take 1/5 the time (i.e., 12 seconds), but let's use equivalent ratios.
The ratio will be: (# of seconds)/(# of extra revolutions Q makes).
So, we get: 60/30 = x/6
Cross multiply to get 30x = (60)(6)
x = 12
In other words, it will take 12 seconds.
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Brent
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If you want proper equation and not the quicker solution like above then,
10 rev / min = 1/6 rev per sec.
40 rev/min = 2/3 rev per sec.
Now say after x sec the condition is met, then it means
X*(2/3) = X*(1/6) + 6
4X = X + 36
10 rev / min = 1/6 rev per sec.
40 rev/min = 2/3 rev per sec.
Now say after x sec the condition is met, then it means
X*(2/3) = X*(1/6) + 6
4X = X + 36
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sana.noor
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my approach:
rate of p = 1/6 rev per sec
rate of q = 2/3 rev per sec
RT=D
for P--> 1/6 (t) = X rev...t = 6x---1
for Q---> 2/3 rev (t) = X + 6 .....t = (3x + 18)/2-----2
keeping the two equation equal
6x = (3x + 18)/2
solve for x = 2
putting the value of x in either equaion will give you the same answer 12 seconds..
rate of p = 1/6 rev per sec
rate of q = 2/3 rev per sec
RT=D
for P--> 1/6 (t) = X rev...t = 6x---1
for Q---> 2/3 rev (t) = X + 6 .....t = (3x + 18)/2-----2
keeping the two equation equal
6x = (3x + 18)/2
solve for x = 2
putting the value of x in either equaion will give you the same answer 12 seconds..
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