Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?
A-150
B-144
C-120
D-90
E-36
I know this is very simple prob.I am getting ans B but OA is A
Help
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Harold's Time
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sandipgumtya
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The distance before and after will be the same.sandipgumtya wrote:Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?
A-150
B-144
C-120
D-90
E-36
I know this is very simple prob.I am getting ans B but OA is A
Help
speed = distance/time
distance = speed x time
therefore speed before x time before = new speed x new time
S X T = 1.2S X T/1.2
T/1.2 = 180/1.2 = 1800/12 = 150
ANS= A
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20% faster = 120% of last year's speed = 6/5 of last year's speed.sandipgumtya wrote:Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?
A-150
B-144
C-120
D-90
E-36
Rate and time are RECIPROCALS.
6/5 of last year's speed implies 5/6 of last year's time:
(5/6)(3 hours) = (5/6)(180 minutes) = 150 minutes.
The correct answer is A.
Alternate approach:
Let last year's speed = 10 feet per minute.
Since last year Harold traveled at a speed of 10 feet per minute for 180 minutes, the total distance = rt = 10*180 = 1800 feet.
If the speed is increased by 20% to 12 feet per minute, the time to travel 1800 feet = d/r = 1800/12 = 150 minutes.
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Hello..
My approach,
1. rate x time = distance
2. Convert 3 hours to 180 mins as all answer options are minutes
3. LY - r x 180 = d
4. TY - (120/100)r x t = d
5. Solve #4 after substituting value of d from #3
6. (12/10)t = 180 --> t = 150
Bullzi
My approach,
1. rate x time = distance
2. Convert 3 hours to 180 mins as all answer options are minutes
3. LY - r x 180 = d
4. TY - (120/100)r x t = d
5. Solve #4 after substituting value of d from #3
6. (12/10)t = 180 --> t = 150
Bullzi
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sandipgumtya
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Hi sandipgumtya,
We're told that the SPEED is 20% faster; with your math, you're attempting to manipulate the TIME, which is why that result is incorrect.
GMAT assassins aren't born, they're made,
Rich
We're told that the SPEED is 20% faster; with your math, you're attempting to manipulate the TIME, which is why that result is incorrect.
GMAT assassins aren't born, they're made,
Rich
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sandipgumtya
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As speed and time are inversely proportional,20% faster means 20% lesser time.Is this concept wrong. Pl correct me in detail.
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Hi sandipgymtya,
Percent Change is calculated based on the ORIGINAL numbers involved. While speed and time ARE related, their respective percent changes are NOT the same.
Assuming that the distance stays the same, Increasing a speed by 20% decreases the time by 16 1/6% (not 20%).
Here's why:
Original speed and time:
D = (R)(T)
20% increase in speed:
D = (6/5)(R)(5/6)(T)
Since the distance is the same, the (6/5) is 'cancelled out' by the (5/6) - the speed increased by 1/5 (or 20%), but the time decreased by 1/6 (or 16 1/6%).
GMAT assassins aren't born, they're made,
Rich
Percent Change is calculated based on the ORIGINAL numbers involved. While speed and time ARE related, their respective percent changes are NOT the same.
Assuming that the distance stays the same, Increasing a speed by 20% decreases the time by 16 1/6% (not 20%).
Here's why:
Original speed and time:
D = (R)(T)
20% increase in speed:
D = (6/5)(R)(5/6)(T)
Since the distance is the same, the (6/5) is 'cancelled out' by the (5/6) - the speed increased by 1/5 (or 20%), but the time decreased by 1/6 (or 16 1/6%).
GMAT assassins aren't born, they're made,
Rich
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sandipgumtya
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GMATGuruNY
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As noted above:sandipgumtya wrote:As speed and time are inversely proportional,20% faster means 20% lesser time.Is this concept wrong. Pl correct me in detail.
20% faster = (120/100)r = (6/5)r.
Since rate and time are reciprocals:
(6/5)r --> (5/6)t --> 1/6 less time.
Other examples:
50% faster = (150/100)r = (3/2)r.
Since rate and time are reciprocals:
(3/2)r --> (2/3)t --> 1/3 less time.
20% slower = (80/100)r = (4/5)r.
Since rate and time are reciprocals:
(4/5)r --> (5/4)t = 1/4 more time.
50% slower = (50/100)r = (1/2)r.
Since rate and time are reciprocals:
(1/2)r --> 2t --> twice the time.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Recall that rate x time = job (or rate = job/time, or time = job/rate). Since Harold's time to complete the event was 3 hours, we can say that his rate was 1 job per 3 hours; i.e., his rate is 1/3. Since he's increasing his speed by 20%, his new rate is:sandipgumtya wrote:Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?
A-150
B-144
C-120
D-90
E-36
(1/3) x 1.2
(1/3) x 12/10 = 4/10 = 2/5
Since his new rate is 2/5, his new time is 1/(2/5) = 5/2 = 2.5 hours. Thus he can complete the event in 2.5 x 60 = 150 minutes.
Answer: A
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