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# Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order

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Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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05 Mar 2019, 07:13
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25% (medium)

Question Stats:

76% (01:30) correct 24% (01:40) wrong based on 42 sessions

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Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order to keep the distance travelled same, by what percentage should the time ‘t’ be increased if the speed drops down by 20%?

A) 10%
B) 20%
C) 25%
D) 30%
E) 35%
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Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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05 Mar 2019, 10:02
Speed decrease by 20%, then time should be increased by $$\frac{20}{(100-20)}*100$$ = 25%

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Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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05 Mar 2019, 22:36
mangamma wrote:
Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order to keep the distance travelled same, by what percentage should the time ‘t’ be increased if the speed drops down by 20%?

A) 10%
B) 20%
C) 25%
D) 30%
E) 35%

Instead of solving, we can make use of numbers to arrive at the answer.
Initial speed=20mph
Let us assume t=4 hours
Then distance travelled=80miles

New speed=16mph
time taken to travel 80 miles=5 hours

i.e. t has increased from 4 to 5, which is 25%

Ans:C
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Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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06 Mar 2019, 12:42
Algebraic solution -->

S1 = 20mph for t1 hours - Therefore, the distance traveled (d) = 20*t1

Now we have been told that the new speed (S2) is 20% less than S1 --> S2 = S1*0.8 = 20*0.8 = 20*(4/5) = 16mph

if distance needs to remain the same, we can then write the new equation only in terms of S1, t1, S2, and t2.

S1*t1 = S2*t2
20*t1 = 16*t2
(20/16)*t1 = t2
(5/4)*t1 = t2
1.25t1 = t2
Therefore, t2 is 25% more than t1.
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Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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06 Mar 2019, 12:54
mangamma wrote:
Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order to keep the distance travelled same, by what percentage should the time ‘t’ be increased if the speed drops down by 20%?

A) 10%
B) 20%
C) 25%
D) 30%
E) 35%

The speed decreased by 20% = 4/5 of the regular speed.
Rate and time have a RECIPROCAL relationship.
If Jane travels at 4/5 of her regular speed, she will take 5/4 of her regular time -- an increase of 1/4 = 25%.

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Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order  [#permalink]

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31 Mar 2019, 21:20
Plug in numbers for the same.
Speed = 20
assuming distance = 100 , thus time taken is 5.
Now new speed is 16, while distance is yet 100 New time is 6.25
which is 5 + 0.25(5) = 25%
Re: Speed of Jane is 20 mph, with which he travels for ‘t’ hours. In order   [#permalink] 31 Mar 2019, 21:20
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