 Compound Proportions
 Continued Proportions
 Proportional Parts
 Rates of Work and Mixtures
 Past KCSE Questions on the Topic.
Compound Proportions
 The proportion involving two or more quantities is called compound proportion. Any four quantities a , b , c and d are in proportion if;
a = c
b d
Example
Find the value of a that makes 2, 5, a and 25 to be in proportion;
Solution
Since 2, 5, a, and 25 are in proportion
2 = a
5 25
5a = 2 x 25
a = 2 x 25
5
a = 10
Continued Proportions
 In continued proportion, all the ratios between different quantities are the same; but always remember that the relationship exists between two quantities for example:
P : Q Q : R R : S
10: 5 16 : 8 4 : 2  Note that in the example, the ratio between different quantities i.e. P:Q, Q:R and R:S are the same i.e. 2:1 when simplified.
 Continued proportion is very important when determining the net worth of individuals who own the same business or even calculating the amounts of profit that different individual owners of a company or business should take home.
Proportional Parts
In general, if n is to be divided in the ratio a: b: c, then the parts of n proportional to a, b, c are ^{a}/_{a+b+c} × n, ^{b}/_{a+b+c} ×n and ^{c}/_{a+b+c}×n respectively
Example
Omondi, Joel, cheroot shared sh 27,000 in the ratio 2:3:4 respectively. How much did each get?
Solution
The parts of sh 27,000 proportional to 2, 3, 4 are
^{2}/_{9} ×27,000 = sh 6000 →Omondi
^{3}/_{9} ×27,000 = sh 6000 →Joel
^{4}/_{9} ×27,000 = sh 6000 →Cheroot
Example
Three people – John, Debby and Dave contributed ksh 119, 000 to start a company. If the ratio of the contribution of John to Debby was 12:6 and the contribution of Debby to Dave was 8:4, determine the amount in dollars that every partner contributed.
Solution
Ratio of John to Debby’s contribution = 12:6 = 2:1
Ratio of Debby to Dave’s contribution = 8:4 = 2:1
As you can see, the ratio of the contribution of John to Debby and that of Debby to Dave is in continued proportion.
Hence John = Debby= 2
Debby Dave 1
To determine the ratio of the contribution between the three members, we do the calculation as follows:
John : Debby : Dave
12 : 6
8 : 4
We multiply the upper ratio by 8 and the lower ratio by 6, thus the resulting ratio will be:
John : Debby : Dave
96 : 48 : 24
= 4 : 2 : 1
The total ratio = 7
The contribution of the different members can then be found as follows:
John ^{4}/_{7} x ksh 119, 000 = ksh 68,000
Debby ^{2}/_{7} x ksh 119, 000 = ksh 34,000
Dave ^{1}/_{7} x ksh 119, 000 = ksh 17,000
John contributed ksh 68, 000 to the company while Debby contributed ksh 34, 000 and Dave contributed ksh 17, 000
Example 2
You are presented with three numbers which are in continued proportion. If the sum of the three numbers is 38 and the product of the first number and the third number is 144, find the three numbers.
Solution
Let us assume that the three numbers in continued proportion or Geometric Proportion are a, ar and ar^{2} where a is the first number and r is the rate.
a+ar+ar^{2 }= 38 ………………………….. (1)
The product of the 1 st and 3rd is
a × ar^{2 }= 144
Or
(ar)^{2} = 144………………………………..(2)
If we find the square root of (ar) 2, then we will have found the second number:
√[(ar)^{2 }]= √144
ar = 12
Since the value of the second number is 12, it then implies that the sum of the first and the third number is 26.
We now proceed and look for two numbers whose sum is 26 and product is 144.
Clearly, the numbers are 8 and 18.
Thus, the three numbers that we were looking for are 8, 12 and 18.
Let us work backwards and try to prove whether this is actually true:
8 + 12 + 18 = 18
What about the product of the first and the third number?
8 × 18 = 144
What about the continued proportion
a = ar = 2
ar ar^{2} 3
The numbers are in continued proportion
Example
Given that x: y =2:3, Find the ratio (5x – 4y): (x + y).
Solution
Since x: y =2: 3
^{x}/_{2} = ^{y}/_{3}= k,
x = 2k and y = 3k
(5x – 4y): (x + y) = (10k – 12k) : (2k + 3 k)
= −2k: 5k
= − 2: 5
Example
If^{ a}/_{b }= ^{c}/_{d} ,show that a − 3b = c − 3d.
b−3a d − 3c
Solution
^{a}/_{b}= ^{c}/_{d}→ ^{a}/_{c} = ^{b}/_{d}
^{a}/_{b} = ^{b}/_{d} = k
a = kc and b = kd
Substituting kc for a and kd for b in the expression a − 3b
b − 3a
kc − 3kd = k(c − 3d)
kd − 3kd k(d − 3c)
Therefore expression a − 3b = c − 3d
b−3a d − 3c
Rates of Work and Mixtures
Examples
195 men working 10 hour a day can finish a job in 20 days. How many men employed to finish the job in 15 days if they work 13 hours a day.
Solution:
Let x be the no. of men required
Days hours Men
20 10 195
15 13 x
20 x 1 0 x 1 95 = 15 ×13 × x
x = 20 ×10 ×195 = 200 men
15 ×13
Example
Tap P can fill a tank in 2 hrs, and tap Q can fill the same tank in 4 hrs. Tap R can empty the tank in 3 hrs.
 If tap R is closed, how long would it take taps P and Q to fill the tank?
 Calculate how long it would take to fill the tank when the three taps P, Q and R. are left running?
Solution
 Tap P fills ^{1}/_{2} of the tank in 1 h.
Tap Q fills ^{1}/_{4} of the tank in 1 h.
Tap R empties ^{1}/_{3} of the tank in 1 h.
In one hour, P and Q fill ^{1}/_{2} + ^{1}/_{4 }= ^{3}/_{4} of the tank
Therefore ^{3}/_{4} of the tank is filled in 1 h.
Time taken to fill the tank(^{4}/_{4}) = (^{4}/_{4 }÷ ^{3}/_{4})h
= ^{4}/_{3} h  In 1h, P and Q fill ^{3}/_{4} of tank while R empties ^{1}/_{3 }of the tank.
When all taps are open, (^{1}/_{2} + ^{1}/_{4 }− ^{1}/_{3} = ^{5}/_{12}) of the tank is filled in 1 hour.
^{5}/_{12} of tank is filled in 1 hour.
Therefore time required to fill the tank ^{12}/_{12} = (^{12}/_{12} ÷ ^{5}/_{12}) × 1 h
= 2^{2}/_{5} h
Example
In what proportion should grades of sugars costing sh. 45 and sh. 50 per kilogram be mixed in order to produce a blend worth sh. 48 per kilogram?
Solution
Method 1
Let n kilograms of the grade costing sh. 45 per kg be mixed with 1 kilogram of grade costing sh. 50 per kg.
Total cost of the two blends is sh. b(45n+50)
The mass of the mixture is (n +1) kg
Therefore total cost of the mixture is (n +1)48
45n + 50 = 48 (n +1)
45n + 50 = 48 n + 48
50 = 3n + 48
2 = 3n
n =^{2}/_{3}
The two grades are mixed in the proportion ^{2}/_{3} :1 = 2 :3
Method 2
Let x kg of grade costing sh 45 per kg be mixed with y kg of grade costing sh.50 per kg. The total cost will be sh. (45x + 50 y)
Cost per kg of the mixture is sh. 45x+50y
x+y
45x+50y = 48
x+y
45x + 50y = 48(x + y)
45x + 50y = 48x + 48y
2y = 3x
=^{x}/_{y} = ^{2}/_{3}
The proportion is x : y = 2:3
Past KCSE Questions on the Topic.
 Akinyi bought and beans from a wholesaler. She then mixed the maize and beans the ratio 4:3 she brought the maize as Kshs. 12 per kg and the beans 4 per kg. If she was to make a profit of 30% what should be the selling price of 1 kg of the mixture?
 A rectangular tank of base 2.4 m by 2.8 m and a height of 3m contains 3,600 liters of water initially. Water flows into the tank at the rate of 0.5 litres per second
Calculate the time in hours and minutes, required to fill the tank  A company is to construct a parking bay whose area is 135m^{2}. It is to be covered with concrete slab of uniform thickness of 0.15. To make the slab cement. Ballast and sand are to be mixed so that their masses are in the ratio 1 : 4 : 4. The mass of m^{3 }of dry slab is 2, 500kg.
Calculate
 The volume of the slab
 The mass of the dry slab
 The mass of cement to be used
 If one bag of the cement is 50 kg, find the number of bags to be purchased
 If a lorry carries 7 tonnes of sand, calculate the number of lorries of sand to be purchased.

 The mass of a mixture A of beans and maize is 72 kg. The ratio of beans to maize is 3:5 respectively
 Find the mass of maize in the mixture
 A second mixture of B of beans and maize of mass 98 kg in mixed with A. The final ratio of beans to maize is 8:9 respectively. Find the ratio of beans to maize in B
 A retailer bought 49 kg of grade 1 rice at Kshs. 65 per kilogram and 60 kg of grade II rice at Kshs 27.50 per kilogram. He mixed the tow types of rice.
 Find the buying price of one kilogram of the mixture
 He packed the mixture into 2 kg packets
 If he intends to make a 20% profit find the selling price per packet
 He sold 8 packets and then reduced the price by 10% in order to attract customers. Find the new selling price per packet.
 After selling ^{1}/_{3} of the remainder at reduced price, he raised the price so as to realize the original goal of 20% profit overall. Find the selling price per packet of the remaining rice.
 A trader sells a bag of beans for Kshs 1, 200. He mixed beans and maize in the ration 3 : 2. Find how much the trader should he sell a bag of the mixture to realize the same profit?
 Pipe A can fill an empty water tank in 3 hours while, pipe B can fill the same tank in 6 hours, when the tank is full it can be emptied by pipe C in 8 hours. Pipes A and B are opened at the same time when the tank is empty.
If one hour later, pipe C is also opened, find the total time taken to fill the tank  A solution whose volume is 80 litres is made 40% of water and 60% of alcohol. When litres of water are added, the percentage of alcohol drops to 40%
 Find the value of x
 Thirty litres of water is added to the new solution. Calculate the percentage
 If 5 litres of the solution in (b) is added to 2 litres of the original solution, calculate in the simplest form, the ratio of water to that of alcohol in the resulting solution
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