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Mira walked to the school at a speed of 3 mph.
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21 Jul 2019, 05:48
21 Jul 2019, 05:48
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Question Stats:
70% (01:06) correct
29% (01:12) wrong based on 61 sessions
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Mira walked to the school at a speed of 3 mph. Once she reached the school, she realized that she forgot to bring her lunch box, so she rushed back home at a speed of 6 mph. She then walked back to school at a speed of 4 mph. All the time she used the same route.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Mira's average speed over the entire journey |
4 mph |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: Mira walked to the school at a speed of 3 mph.
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22 Jul 2019, 16:33
22 Jul 2019, 16:33
5
While these questions can be solved algebraically, it is often easier just to make up a distance. In this case, since we're using the numbers 3, 4, and 6, a distance of 12 will work nicely as it can be evenly divided by all three rates.
Average speed = total distance / total time
Total distance is just 3 * 12 = 36.
We'll need to calculate the time for each of the three trips using the \(d=rt\) formula, which we can rewrite as \(t=\frac{d}{r}\).
\(t1 = 12/3 = 4\)
\(t2 = 12/6 = 2\)
\(t3 = 12/4 = 3\)
That leaves us with \(\frac{36}{(4+2+3)}\) or 4, so the answer is C.
Average speed = total distance / total time
Total distance is just 3 * 12 = 36.
We'll need to calculate the time for each of the three trips using the \(d=rt\) formula, which we can rewrite as \(t=\frac{d}{r}\).
\(t1 = 12/3 = 4\)
\(t2 = 12/6 = 2\)
\(t3 = 12/4 = 3\)
That leaves us with \(\frac{36}{(4+2+3)}\) or 4, so the answer is C.
Show: :: algebra without assigning a distance; more complex, but useful to know for harder questions
Average speed = total distance / total time.
Total distance = 3d
We'll use the same formula as above to calculate individual times: \(t=\frac{d}{r}\). We'll get all those fractions to have a common denominator.
\(t1 = \frac{d}{3} = \frac{4d}{12}\)
\(t2 = \frac{d}{6} = \frac{2d}{12}\)
\(t3 = \frac{d}{4} = \frac{3d}{12}\)
\(total time = \frac{9d}{12}\)
\(average speed = \frac{3d}{(9d/12)} = 3d * \frac{12}{9d} = 4\)
Total distance = 3d
We'll use the same formula as above to calculate individual times: \(t=\frac{d}{r}\). We'll get all those fractions to have a common denominator.
\(t1 = \frac{d}{3} = \frac{4d}{12}\)
\(t2 = \frac{d}{6} = \frac{2d}{12}\)
\(t3 = \frac{d}{4} = \frac{3d}{12}\)
\(total time = \frac{9d}{12}\)
\(average speed = \frac{3d}{(9d/12)} = 3d * \frac{12}{9d} = 4\)
General Discussion
Re: Quant Comparison - Time speed
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21 Jul 2019, 13:56
21 Jul 2019, 13:56
1
let time spent walking to sch=p ; distance covered=speed * time=3p
time spent rushing back home=q ; distance covered=6q
time spent going back to sch=r ; distance covered=4r
since distance covered is same , then total distance =3* (any of the distance above)
choosing the distace covered when walkig to sch
total distance=3*3p=9p
avrg. spd= total distance /total time
avrg.spd=9p/p+q+r......(1)
however, 3p=6q=4r
q=0.5p, r=0.75p
therefore, avrg.spd=9p/p+0.5p+0.75p
avrg.spd=4mph
ANS: C
time spent rushing back home=q ; distance covered=6q
time spent going back to sch=r ; distance covered=4r
since distance covered is same , then total distance =3* (any of the distance above)
choosing the distace covered when walkig to sch
total distance=3*3p=9p
avrg. spd= total distance /total time
avrg.spd=9p/p+q+r......(1)
however, 3p=6q=4r
q=0.5p, r=0.75p
therefore, avrg.spd=9p/p+0.5p+0.75p
avrg.spd=4mph
ANS: C
