Ratio and Proportions problems
Posted by
Ravi Kumar at Monday, November 24, 2008
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Ratio and Proportions problems
Ratio and Proportions problems
Simple Problems:
1.If a:b =5:9 and b:c=4:7 Find a:b:c?
Sol: a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9
a:b:c=5:9:63/9=20:36:63
2.Find the fourth proportion to 4,9,12
Sol: d is the fourth proportion to a,b,c
a:b=c:d
4:9=12:x
4x=9*12=>x=27
3.Find third proportion to 16,36
Sol: If a:b=b:c then c is the third proportion to a,b
16:36=36:x
16x=36*36
x=81
4.Find mean proportion between 0.08 and 0.18
Sol: mean proportion between a and b=square root of ab
mean proportion =square root of 0.08*0.18=0.12
5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is
Sol: a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4
c:d=6:7=6*15/24:7*15/24=15/4:35/8
a:b:c:d=2:3:15/4:35/8=16:24:30:35
6.2A=3B=4C then A:B:C?
Sol: let 2A=3B=4C=k then
A=k/2, B=k/3, C=k/4
A:B:C=k/2:k/3:k/4=6:4:3
7.15% of x=20% of y then x:y is
Sol: (15/100)*x=(20/100)*y
3x=4y
x:y=4:3
8.a/3=b/4=c/7 then (a+b+c)/c=
Sol: let a/3=b/4=c/7=k
(a+b+c)/c=(3k+4k+7k)/7k=2
9.Rs 3650 is divided among 4 engineers, 3 MBA’s and 5 CA’s
such that 3 CA’s get as much as 2 MBA’s and 3 Eng’s as
much as 2 CA’s .Find the share of an MBA.
Sol: 4E+3M+5C=3650
3C=2M, that is M=1.5C
3E=2C that is E=.66 C
Then, (4*0.66C)+(3*1.5C)+5C=3650
C=3650/12.166
C=300
M=1.5 and C=450.
Difficule Problems:
1.Three containers A,B and C are having mixtures of milk and
water in the ratio of 1:5 and 3:5 and 5:7 respectively. If
the capacities of the containers are in the ratio of all the
three containers are in the ratio 5:4:5, find the ratio of
milk to water, if the mixtures of all the three containers
are mixed together.
Sol: Assume that there are 500,400 and 500 liters respectively
in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of
the three containers.
Thus, the total milk is 441.66 liters. Hence, the amount of
water in the mixture is 1400441.66=958.33liters.
Hence, the ratio of milk to water is
441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.
2.A certain number of one rupee,fifty parse and twenty five
paise coins are in the ratio of 2:5:3:4, add up to Rs 210.
How many 50 paise coins were there?
Sol: the ratio of 2.5:3:4 can be written as 5:6:8
let us assume that there are 5 one rupee coins,6 fifty
paise coins and 8 twentyfive paise coins in all.
their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10
If the total is Rs 10,number of 50 paise coins are 6.
if the total is Rs 210, number of 50 paise coins would be
210*6/10=126.
3.The incomes of A and B are in the ratio of 4:3 and their
expenditure are in the ratio of 2:1 . if each one saves
Rs 1000,what are their incomes?
Sol: Ratio of incomes of A and B=4:3
Ratio of expenditures of A and B=2:1
Amount of money saved by A=Amount of money saved by B=Rs 1000
let the incomes of A and B be 4x and 3x respectively
let the expense of A and B be 2y and 1yrespectively
Amount of money saved by A=(incomeexpenditure)=4x2y=Rs 1000
Amount of money saved by B=3xy=Rs 1000
this can be even written as 6x2y=Rs 2000
now solve 1 and 3 to get
x=Rs 500
therefore income of A=4x=4*500=Rs 2000
income of B=3x=3*500=Rs 1500
4.A sum of Rs 1162 is divided among A,B and C. Such that 4
times A's share share is equal to 5 times B's share and 7
times C's share . What is the share of C?
Sol: 4 times of A's share =5 times of B's share=7 times of
C's share=1
therefore , the ratio of their share =1/4:1/5:1/7
LCM of 4,5,7=140
therefore, ¼:1/5:1/7=35:28:20
the ratio now can be written as 35:28:20
therefore C's share=(20/83)*1162=20*14=Rs 280.
5.The ratio of the present ages of saritha and her mother is
2:9, mother's age at the time of saritha's birth was 28 years,
what is saritha's present age?
Sol: ratio of ages of saritha and her mother =2:9
let the present age of saritha be 2x years.
then the mother's present age would be 9x years
Difference in their ages =28 years
9x2x=28 years
7x=28=>x=4
therefore saritha's age =2*4=8 years