Proportions in mathematics has to do with the method of representing the comparisons of two or more quantities and it usually contains units.
There are three types of proportions in mathematics, which are:
Direct or Simple proportion:
This type of proportion compares two quantities in a way that when one quantity increases, the other quantity also increases and when it decreases, the other decreases as well.
Inverse Proportion:
This type of proportion compares two quantities in such a way that when one quantity increases, the other decreases and vise versa.
Complex Proportion:
This is the combination of either two or more Direct or Inverse Proportions, it may as well contain both direct and inverse Proportions with more than two quantities.
QUESTIONS AND SOLUTIONS:
Q1: 9 men do a piece of work in 28 days.
(a) How long will it take 14 men to do that same work?
(b) How many men will do it in 56 days?
Solution:
(1a) 9 men = 28 days
One man can do the work in (28 x 9) days = 252 days.
Therefore: 14 men can finish the work in (252/14) days = 18 days
(b) How many men will do it in 56 days?
Solution:
Let the total number of men that can do the work in 56 days be x.
9 men do it in 28 days
x men do it in 56 days
That is:
9 men = 28 days
x men = 56 days
Cross multiply
28x = 9 x 56
28x = 504
Divide both sides by the coefficient of x
x = 504/28
x = 18 men.
Q2: P and Q are partners in a venture. P contributed #20,000 for nine months and Q contributed #50,000 for one year. Find each person’s share of profit of #6,300.
Solution:
P = #20,000 for nine months
Q = #50,000 for one year(12 months)
Ratio at which they contributed the money P:Q = 9:12 = 3:4
Total income = #6,300
Add the ratio together :-
3 + 4 = 7
P is collecting 3/7 x #6,300 = #2,700
Q is collecting 4/7 x #6,300 = #3,600
Q3: A factory worker is paid #269.50 in 7 days working 11 hours a day. How much must be paid to him for 26 days working 9 hours a day?
Solution:
Let the amount he will be paid in 26 days working 9 hours be x.
Total duration of his work that fetched him #269.5 = number of days multiply by the hours he worked
= 7 x 11 = 77 hours
77 hours = #269.5
Total duration of his work that will fetch him #x = number of days multiply by the hours he worked
= 26 x 9 = 234 hours
234 hours = #x
So,
77 hours = #269.5
234 hours = #x
Cross multiply
77x = 269.5 x 234
77x = 63,063
x = #819.00
Q4: 24 men can plough 42 hectares of land in 14 days. How many days will it take 16 men to plough 64 hectares?
Solution:
If 24 men can plough 42 hectares of land in 14 days,
one man can plough the same 42 hectares in (14 x 24)/42
= 336/42
= 8 days.
Therefore 16 men can plough 64 hectares in 8 x (64/16)
= 8 x 4 = 32 days.
Q5: A motorist drives 80km per hour in 5 hours and another 60km per hour in 4 hours. Find the average speed for the whole journey.
Solution:
Speed = Distance/Time
For first journey:
80 = D/5
Distance = D = 80 x 5
= 400km
For second journey:
D = 60 x 4 = 240km
Therefore, average speed
= (D1 + D2)/(T1 + T2)
= (400 + 240)/(5 + 4)
= 640/9
= 71.11km per hour
Q6: Nneka is half as old as Joke and Joke is half as old as Zainab. The sum of their ages is 336 years. Find their ages.
Solution:
Let Zainab’s age be “y”
Joke’s age = y/2
Nneka’s age = y/2 ÷ 2
= y/2 x 1/2 = y/4
The sum of their ages
= y + y/2 + y/4 = 336
Solve for “y”
7y = 1344
Divide both sides by the coefficient of y
y = 1344/7
y = 192 years = Zainab’s age
Joke’s age = 192/2 = 96 years
Nneka’s age = 96/2 = 48 years.
Q7: If 10 students consume 1185kg of meat in 21 days, find how much 16 students will consume in 14 days if consumption takes the same rate.
Solution:
Let the amount that 16 students consume in 14 days be x.
10 students consume 1185kg of meat in 21 days
1 student consumes (10 x 21)/1185 days.
16 students will consume xkg in 14 days
So,
[(21 x 10)/1185] X (x/16) = 14
210x = 1185 x 16 x 14
210x = 265,440
x = 265,440/210
x = 1264kg
Q8: The cost of feeding 24 goats for 12 days is #360. Find the number of days that 20 goats will be fed with #400.
Solution:
Let the number of days be y.
If the cost of feeding 24 goats for 12 days is #360,
The cost of feeding 1 goat for 1 day
= (12 x 24)/ 360 = 288/360.
And the cost of feeding 20 goats in y days = 400
Again, the cost of feeding 1 goat for one day = (20 x y)/400
= 20y/400 = y/20
Equate both of them together and solve for y.
y/20 = 288/360
360y = 5,760
y = 16 days
Q9: Divide #738 in the ratio 3/2 : 2 : 10/3
Solution:
First add up the ratio
3/2 + 2 + 10/3 = 41/6
For portion 3/2
3/2 ÷ 41/6 x 738
= 3/2 x 6/41 x 738 = (3 x 3 x 738)/41
= 6642/41
= #162
For portion 2
2 ÷ 41/6 x 738
= 2 x 6/41 x 738 = (2 x 6 x 738)/41
= 8856/41
= #216
For portion 10/3
10/3 ÷ 41/6 x 738
= 10/3 x 6/41 x 738 = (10 x 2 x 738)/41
= 14760/41
= #360
Q10: 5kg of ginger costing #24 per kg is diluted with 10kg of garlic costing #18 per kg. Find the cost of the mixture per kg.
Solution:
Cost of 5kg of ginger = 5 x #24 = #120
Cost of 10kg of garlic = 10 x #18 = #180
Total cost of the two = #120 + #180 = #300
Total kg of the two = 5kg + 10kg = 15kg
Therefore, the cost of the mixture per kg = #300/15kg
= #20 per kg