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A tortoise makes a journey in two parts; it can either walk at 4cm/s or crawl at 3 cm/s. If the tortoise walks the first part and crawls the second, it takes 110 seconds. If it crawls the first part and walks the second, it takes 100 seconds. Find the lengths of the two parts of the journey.

2007-06-09 22:17:17 · 5 answers · asked by Kris 1 in Science & Mathematics Mathematics

5 answers

let A cm be the length of the 1st part, B be the 2nd part.
A/4 + B/3 = 110 <--- walks the first part
A = (1320-4B)/3
A/3 + B/4 = 100 <-------crawls the 1 st part
[(1320-4B)/3] / 3 + B/4 = 100
B = 240
A = 120

2007-06-09 22:28:59 · answer #1 · answered by Alex W 1 · 0 1

This is a distance/rate/time problem.

Let
d = total distance
x = length first part
d - x = length second part

Use the formula:

(distance 1)/(rate 1) + (distance 2)/(rate 2) = time

We have two equations and two unknowns.

x/4 + (d - x)/3 = 110
x/3 + (d - x)/4 = 100

Multiply both equations thru by 12

3x + 4(d - x) = 1320
4x + 3(d - x) = 1200

Simplify

-x + 4d = 1320
x + 3d = 1200

Add the two equations.

7d = 2520
d = 360

Plug into the first equation and solve for x.

-x + 4d = 1320
x = 4d - 1320 = 4*(360) - 1320 = 120

d - x = 360 - 120 = 240

The length of the parts are:

First part = 120 cm
Second part = 240 cm

2007-06-10 05:29:37 · answer #2 · answered by Northstar 7 · 0 0

the first part is x
and second part is y
so we can get the equation x/4 +y/3 =110
x/3+y/4=100
so we can get the x=120 y=240

2007-06-10 05:29:06 · answer #3 · answered by Xu D 2 · 0 0

let A cm be the length of the 1st part, B be the 2nd part.

A/4 + B/3 = 110 ----------walks the first part
A = (1320-4B)/3

A/3 + B/4 = 100 ------------crawls the 1 st part
[(1320-4B)/3] / 3 + B/4 = 100

B = 240
A = 120

2007-06-10 05:46:46 · answer #4 · answered by Krish 5 · 0 0

part1=120 cm
part2=240 cm

2007-06-10 05:26:03 · answer #5 · answered by yoovraj s 2 · 0 0

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