About Soroban Expertise I have a systematic and orderly way to organize the facts in a word problem, which (usually) leads clearly to the necessary equation. I think I can help with all types of word problems.
Experience 38 years of teaching college-level math, mostly at a two-year college.
Education/Credentials BS and MS in mathematics, SUNY Albany
Expert: Soroban Date: 2/10/2008 Subject: Word problems
Question Hi, I need help with the following word problems. Thank you.
1. If it takes 3 men 56 minutes to fill a tech 4’ x 6’ x 5’, and two of the men work twice as rapidly as the third, the number of minutes that it will take the two faster men alone to fill this trench is____
2. The dimensions of an office are 25 feet by 15 feet. It is to be fitted with desks 4 feet by 3 feet. The distance between the front of one
desk and the rear of another would be 3 feet, and the distance between the sides of 2 desks should be 4 feet. Assuming that no desk is placed closer than 1 ft. from any wall, the optimum number that can be placed
in the office is________
3. Two pieces of meat that together weighed 40lbs were sold for the same sum. What did the $1.20 piece weigh if they were worth $ 1.80 and $ 1.20 a pound?
4. A department vehicle has completed the first 5 miles of a 10-mile trip in 10 minutes. To complete the entire trip at an average rate of 45 miles per hour, the vehicle must travel the remaining 5 miles in how many minutes?
Thank you,
Daniel
Answer Hello, Daniel!
I can help you with MOST of these . . .
>> 1. If it takes 3 men 56 minutes to fill a trench 4’ x 6’ x 5’,
>> and two of the men work twice as rapidly as the third,
>> the number of minutes that it will take the two faster men alone to fill this trench is____
Let's say the first man (A) can do the job in X minutes.
In one minute, he can do 1/X of the job.
In 56 minutes, he can do 56/X of the job.
Then the second man (B) can do the job in X minutes.
In one minute, he can do 1/X of the job.
In 56 minutes, he can do 56/X of the job.
And the third man (C) can do the job in 2X minutes. (It takes him twice as long.)
In one minute, he can do 1/2X of the job.
In 56 minutes, he do 56/2X = 28/X of the job.
Working together for 56 minutes, they can do the entire job (1 job).
There is our equation: 56/X + 56/X + 28/X = 1
Multiply through by X: 56 + 56 + 28 = X --> X = 140
So: A takes 140 minutes, B takes 140 minutes, C takes 280 minutes
(working alone).
A does the job in 140 minutes.
In one minute, he does 1/140 of the job.
In M minutes, he does M/140 of the job.
B does the job in 140 minutes.
In one minute, he does 1/140 of the job.
In M minutes, he does M/140 of the job.
Together, working for M minutes, they can do the entire job (1 job).
There is our new equation: M/140 + M/140 = 1
Multiply by 140: M + M = 140 --> 2M = 140 --> M = 70
Therefore, A and B, working together, can do the job in 70 minutes.
>> 2. The dimensions of an office are 25 feet by 15 feet.
>> It is to be fitted with desks 4 feet by 3 feet.
>> The distance between the front of one desk and the rear of another is 3 feet,
>> and the distance between the sides of 2 desks should be 4 feet.
>> Assuming that no desk is placed closer than 1 ft. from any wall,
>> the optimum number that can be placed in the office is________
Sorry, this one will require me to make a number of sketches.
I'll have to work on it later.
>> 3. Two pieces of meat that together weighed 40lbs were sold for the same sum.
>> What did the $1.20 piece weigh if they were worth $ 1.80 and $ 1.20 a pound?
Let X = number of pounds of the $1.20/lb meat
Its cost is: 1.2X dollars
Then 40 - X = number of pounds of the $1.80 meat.
Its cost is: 1.8(40 - X)
>> 4. A car completed the first 5 miles of a 10-mile trip in 10 minutes.
>> To complete the entire trip at an average rate of 45 miles per hour,
>> the vehicle must travel the remaining 5 miles in how many minutes?
We can work this out without a lot of Algegra (if that's allowed).
We will use: [Distance] = [Speed] x [Time}
and its two variations: T = D/S and S = D/T
To average 45 miles per hour, that means: (45 miles) in (1 hour).
To cover 10 miles at 45 mph, it will take: 10/45 = 2/9 of an hour.
The car has already used up 10 minutes, or 1/6 of an hour.
It has: 2/9 - 1/6 = 1/18 of an hour left to cover the last 5 miles.
