Problems on Rates and Unit Conversions

Date: 06/21/99 at 08:41:30
From: Peter Ravenscroft
Subject: Rates

Dear Doctor Math,

Peter has trouble on these questions on rates, so could you please 
send him the answers and how to set them up? 

Thank you,
Peter's mother


1. Convert these rates as directed
   A) $8.25/2.5 kg to $/kg
   B) $2.36/320 ml to $/ml
   C) 7 litres/$6.23 to $/litre

2. Which is the lower rate?
   A) $8.23 for 2.5 kg or $13.60 for 4 kg
   B) $1.75 /250 ml or $2.36/320 ml
   C) $421/week or $22,468/year

3. Convert these speeds to km/hour
   A) 18 m/sec
   B) 48 m/sec
   C) 30 m/min
   D) 105 m/min

4. Convert these rates as follows 
   A) 3.7 km/hr to m/sec
   B) 200 m/19.72 sec to m/sec and m/min
   C) $12.50/100 grams to $/kg and cents/g

5. mins converted to hours and minutes, convert the following:
   A) 259 mins (to hours and mins)
   B) 876 secs (to min and secs)
   C) 967 hours (to days and hours)
   D) 5871 mins (to hours, mins and secs)

6. A large fish aquarium measures 130 cm wide, 226 cm long and 40 cm 
   high. It is to be half filled with specially treated water, which 
   can be filtered into the tank at the rate of two hundred ml/min. 
   How long will it take to fill the tank to the correct level?

7. Copper has a density of 8.9 grms/cm^3 and steel 7.8 grams/cm^3. How 
   much would two steel and three copper balls weigh if the diameter 
   of each ball were 140 mm?

Date: 06/22/99 at 12:49:15
From: Doctor Rick
Subject: Re: Rates

Hi, Peter and Mom.

>1. Convert these rates as directed
>   A) $8.25/2.5 kg to $/kg
>   B) $2.36/320 ml to $/ml
>   C) 7 litres/$6.23 to $/litre

You can think of ratio problems as equivalent fraction problems, 
except that the numbers have units attached, and the numerator and 
denominator do nothave to be whole numbers. So, in (a),

  ? dollars    $8.25
  --------- = ------
     1 kg     2.5 kg

               $8.25          8.25
  ? dollars = ------ * 1 kg = ---- dollars
              2.5 kg           2.5

All you need to do is to work out this division. Note that in (c), you 
can't write

  ? dollars   7 litres
  --------- = --------
   1 litre      $6.23

You need to invert the righthand fraction so the units are in the same 
places in both. Then you will have

  ? dollars = $6.23 / 7


>2. Which is the lower rate?
>   A) $8.23 for 2.5 kg or $13.60 for 4 kg
>   B) $1.75 /250 ml or $2.36/320 ml
>   C) $421/week or $22,468/year

In (a) and (b), the units are the same in both rates, so the problems 
are the same as comparing fractions. You do this by making the 
denominators the same, and comparing the numerators. You can make the 
denominators both equal to 1 kg (in (a)), in other words, find each 
rate in dollars per kilogram, and compare them:

  $8.23 / 2.5 kg = 8.23/2.5 dollars/kg = 3.292 dollars/kg

  $13.60 / 4 kg = ? dollars/kg

Which is less?


>3. Convert these speeds to km/hour
>   A) 18 m/sec
>   B) 48 m/sec
>   C) 30 m/min
>   D) 105 m/min

Now we need to convert units. You need to do this in 2(c) also, but I 
will show you how here. What I like to do is to write a fraction that 
is equal to 1, using the units I want to convert between. In (a), 60 
sec = 1 minute, and 60 min = 1 hour, and 1000 m = 1 km, so I have 3 
fractions:

  60 sec   60 min   1000 m
  ------ = ------ = ------ = 1
   1 min    1 hr     1 km

Write the rate as a fraction, then multiply it by fractions so that 
the units cancel out (same units in the numerator and denominator) 
until only the units I want (km/hour) are left:

      /                //      ///
   18 m    1 km    60 sec   60 min   18 * 60 * 60 m
  ----- * ------ = ------ * ------ = -------------- = 64.8 km/hr
  1 sec   1000 m   1 min    1 hr        1000 hr
    //         /     ///


>4. Convert these rates as follows 
>   A) 3.7 km/hr to m/sec
>   B) 200 m/19.72 sec to m/sec and m/min
>   C) $12.50/100 grams to $/kg and cents/g

These are the same kind of problem as in 3.


>5. mins converted to hours and minutes, convert the following:
>   A) 259 mins (to hours and mins)
>   B) 876 secs (to min and secs)
>   C) 967 hours (to days and hours)
>   D) 5871 mins (to hours, mins and secs)

These problems are a lot like converting an improper fraction to a 
mixed number:

  259     19
  --- = 4 --
   60     60

Since 1/60 hour = 1 minute, 259 min = 259/60 hour. Divide 259 by 60, 
giving 4 with a remainder of 19; the 19 is the minutes you have left 
-- 4 hours, 19 minutes.

In (c), 24 hours = 1 day, so the improper fraction is 967/24 days.

In (d), should it read "5871 seconds"? If so, you do two conversions 
in a row: first convert seconds to minutes and seconds; then leave the 
seconds alone and convert the minutes to hours and minutes.


>6. A large fish aquarium measures 130 cm wide, 226 cm long and 40 cm 
>   high. It is to be half filled with specially treated water, which 
>   can be filtered into the tank at the rate of two hundred ml/min. 
>   How long will it take to fill the tank to the correct level?

This problem has two steps. Before you can use the rate, you need to 
convert the measurements given into a volume in millilitres. If you 
multiply the width, length, and height in centimetres (and divide this 
by 2 since the tank will be half filled), you will get a volume in 
cubic centimetres (cm^3). It just happens that 1 ml = 1 cm^3, so 
you've got what you need. Then

            1 min
  ___ ml * ------ = ? min
           200 ml

Notice that I inverted the rate so that the millilitres would cancel. 
The general rate equation is

   [distance, volume, etc.] = rate * time

We are using the form

   time = volume / rate


>7. Copper has a density of 8.9 grms/cm^3 and steel 7.8 grams/cm^3. 
>   How much would two steel and three copper balls weigh if the 
>   diameter of each ball were 140 mm?

Again you have to figure the volume first; do you know the formula for 
the volume of a sphere? If the diameter is in millimetres, you will 
get a volume in cubic millimetres (mm^3) which you will need to 
convert to cm^3.

Remember that 1 cm^3 = (1 cm)(1 cm)(1 cm), and use the conversion 
method I showed you above.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/