## APTITUDE-Problems on RATIO AND PROPORTION

**Important Facts: **

**1.Ratio :** The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio, a:b, we call a as the first term of antecedent and b, the second term consequent.

Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9

**3Rule**: The multiplication or division of each term of 9 ratio by the same non-zero number does not affect the ratio.

**4.Proportion:** The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms.

Product of means=product of extremes

Thus, a:b::c:d => (b*c)=(a*d)

**5.Fourth proportional:** If a:b::c:d, then d is called the fourth proportional to a,b and c.

**6.Third proportional: **If a:b::b:c, then c is called third proportional to a and b.

**7.Mean proportional:** Mean proportional between a and b is SQRT(a*b).

**COMPARISION OF RATIOS:**

We say that (a:b)>(c:d) => (a/b)>(c/d)

**8.Compounded ratio:** The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf).

**9.Duplicate Ratio:** If (a:b) is (a2: b2 )

**10.Sub-duplicate ratio of **(a:b) is (SQRT(a):SQRT(b))

**11.Triplicate ratio of **(a:b) is (a3: b3 )

12.Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).

13.If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (component and dividend o)

** VARIATION: **

14.we say that x is directly proportional to y, if x=ky for some constant k and we write.

15.We say that x is inversely proportional to y, if xy=k for some constant and we write.

16. X is inversely proportional to y.

If a/b=c/d=e/f=g/h=k then

k=(a+c+e+g)/(b+d+f+h)

If a1/b1,a2/b2, a3/b3…………..an/bn are unequal

fractions then the ratio.

**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 MBAâ€™s and 5 CAâ€™s such that 3 CAâ€™s get as much as 2 MBAâ€™s and 3 Engâ€™s as much as 2 CAâ€™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.

**DIFFICULT 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 1400-441.66=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 twenty-five 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=(income-expenditure)=4x-2y=

Rs 1000

Amount of money saved by B=3x-y=Rs 1000

this can be even written as 6x-2y=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, Â¼:1/5:1/7=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

9x-2x=28 years

7x=28=>x=4

therefore saritha’s age =2*4=8 years

SIMPLE PROBLEMS

5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d is

Why is it multiplied by 3/4? (b:c=4:5=4*3/4:5*3/4)

Why is it multiplied by 15/24? (c:d=6:7=6*15/24:7*15/24)

There is a rule here that I can’t recollect, so could you explain those above steps?

Dear u can solve dis Q by Like dis ..

Let us consider

a:b=p:q

b:c=r:s

c:d=t:u

now here’s the formulae

a:b:c:d=prt:qrt:qst:qsu

now acc to ur question

ans is 48:72:90:105

if still u have prob

contact:- ash.vipin25@gmail.com

please sending aptitude question my address.

send me logical tricks ..

plz

please give me solution for, if x^1/p = y^1/q = z^1/r, then what is the value of p+q+r (x^1/p means x raised to the power 1/p and so on)

Please solve the problem for me. The question is find x:y for x(2x+ y)=15y2

x(2x + y) = 15y^2

=> 2x^2 + xy – 15y^2 = 0

=>2(x/y)^2 + x/y – 15 = 0 (dividing the previous eqn by y^2)

=> 2p^2 + p – 15 = 0 (say, x/y = p)

now this is a simple quadratic eqn, whose roots can be found as follows:

p=(-1+ sqrt(1+4*2*15))/4 = 30

or

p=(-1- sqrt(1+4*2*15))/4 = -61/2

since a ratio can’t be negative, the ans. is p=x:y=30:1

The salaries of A,B,C are in the ratio 2:3:5. If the increment of 15%, 10% , and 20% are allowed respectively in their salaries what would be the new ratio in their salaries.

2 * 1.15 = 2.3

3 * 1.1 = 3.3

5 * 1.2 = 6

thus the new ratio is:

2.3:3.3:6

or

23:33:60

Ramesh, Suresh and Harish can do a piece of work in 15 days, 10 days and 6 days respectively. How many days will they take to complete the work if they work together

total work = L.C.M.(15,10,6) = 60

their efficiencies are(in work per day)

in the ratio = 1/15 : 1/10 : 1/6 = 4 : 6 : 10

therefore if they work together,time taken will be

=60/(4+6+10) = 3 days

i learnt many things, thanks

why the ratio 2:5:3:4 is converted in to 5:6:8 in the second difficult problem

2.5 : 3 : 4

is same as

5 : 6: 8

the second one does not contain any decimal sign and easier to read and implement .

the 2nd expression is got when each term in the 1st expression is multiplied by 2..when each term in a ratio is multiplied by the same integer, the ratio doesn’t change

Re:MALLIRAJAN

the ratio is 2.5:3:4 not 2:5:3:4 writen mistake

dear saurav,

always while you deal with time and work calculate the individual efficiencies of the persons/ day…

for example ramesh does 1/15 of the work in one day so

Efficiency of ramesh is 1/15=6.66 %

Efficiency of ramesh is 1/10=10 %

Efficiency of ramesh is 1/6= 16.66 %

so all three does (6.66+10+16.66 )% in one day ..that is 32.5 % approx..

so 100 % work is done in 3.2 days approx…

please send me some more problems of ratio

very good posts!

Machines X & Y produce 8000 articles in 4 and 8 hours respectively. If they work alternatively for one hour each. X starting first, in what time 10000 articles can be produced?…please solve this

Please will be see that arthimetic Problems in Ratio and Proportion Problems

if a:b=3:2andb:c=3:5,then a:b:c is?

thoda hard thoda easy but I liked it

cost of 99 chocolates & 101 biscuits is Rs299 and that of 101 chocolates &99 biscuits is Rs301. find cost of 3 chocalates?

slove this one dont understand? —

x=3^1/3+3^1/3 , then 3x^3-9x is ___

if a sphere made of alloy copper and tin are in the ratio 86:14 .In a sphere made of another alloy copper and zinc are in the ratio 58:42. find the ratio of tin and zinc in the sphere made by melting the two sphere.

The income of A,B,C are in ratio 7:9:10 and there expenditure in the ratio 8:9:10.If A saves 1/4th of his income,find the proportion of their savings

hi frnds could u plz ans this

A,B and S were playing with 70 rs each.At the end the ratio of money between A and S was 1:2 and 4:1 between S and B.What is S’s gain in rs?

A:S & S:B = 1:2 & 4:1

MAKE IT A:S:B

i.e

2:4:1

now,2x+4x+x= 70*3

=>x=3o

‘S’ is having 4*30=120.

so ‘S’ gained 120-70 = rs50

Thhe ratio between the age of meera and priya is 3:4 respectively.Ten years ago the ratio between their was 4:7 respectively. what will be meera’s age after 5 years.

if the present ratio of the ages of a husband and wife arin the ratio of 5:4. what would be the ratio of their ages 20 years ago.

need the solution soon

by scoring a 100 in his 20th innings, Mr Tendulkar increased his earlier avg by 2. What was his average for d first?

THE RATIO COMPUNDED OF4:9AND THE DUPLICATE RATIO OF 3:4IS

a purse contains 1 rupee , 2 rupee, 5 rupee notes in the ratio 2:3:5.if the total amount in purse 66 find the number of 5 rupee coins.

the ratio compounded of 4:9 the duplicate ratio of 4:5,triplate ratio of 1:3,sub duplicate ratio of 81:256 and sub triplate ratio of 125:512is:

If m:n= 3:2 then (4m+5n) (4m-5n) is equal to. pls someone solve this