Concept: Ratio, Proportion & Partnership

Solution:
Required ratio is $\frac{15}{25}$.

This can further be simplified by eliminating the common factor from both the numbers. Here, the common factor is 5.

Hence, the simplified ratio = $\frac{3}{5}$

Example : 7 is what part of 8?

Solution:
Required part is the ratio of 7 and 8.

∴ part = $\frac{7}{8}$.

It means 7 is $\frac{7}{8}$^th of 8.

Example : $\frac{3}{48}$ is what part of $\frac{1}{12}$?

Solution:
Required part is the ratio of $\frac{3}{48}$ and $\frac{1}{12}$

∴ part = $\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$48$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}}=\frac{3}{48}\times \frac{12}{1}=\frac{3}{4}$ part.

It means $\frac{3}{48}$ is $\frac{3}{4}$^th of $\frac{1}{12}$.

Example:
In a ratio which is equal to 3 : 7, if the antecedent is 33, what is the consequent?

Solution:
$\frac{3}{7}=\frac{33}{x}$

⇒ x = 77

Example:
Find a fraction which bears the same ratio to $\frac{3}{7}$ that $\frac{1}{5}$ does to $\frac{7}{15}$?

Solution:
Let the number be x

Then, x ∶ 3/7 = 1/5 ∶ 7/15

⇒ $\frac{x}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}$

⇒ $x\times \frac{7}{15}=\frac{1}{5}\times \frac{3}{7}$

⇒ $x=\frac{1}{5}\times \frac{3}{7}\times \frac{15}{7}$ = $\frac{9}{49}$

Example : Divide Rs. 72 in the ratio 4 : 5.

Solution:
There are total 4 + 5 = 9 parts out of which first part has 4 parts which the other has 5 parts.

First part = $\frac{\mathrm{First} \mathrm{ratio} \mathrm{term}}{\mathrm{Sum} \mathrm{of} \mathrm{ratio} \mathrm{terms}}$ × Total amount = $\frac{4}{4+5}\times 72$ = Rs. 32

and, second part = $\frac{\mathrm{Second} \mathrm{ratio} \mathrm{term}}{\mathrm{Sum} \mathrm{of} \mathrm{ratio} \mathrm{terms}}$ × Total amount = $\frac{5}{4+5}\times 72$ = Rs. 40

Example : Amit and Atul have money in the ratio of 3 : 5. What is the amount with them respectively.

Solution:
Here, we know the ratio of the amounts with them i.e. 3 : 5.

Hence, the amount with Amit = 3x 

and the amount with Atul = 5x.

Now, there is no additional information given to calculate the value of x. Hence, we cannot determine the amount with each of them.

Hence, the answer cannot be determined.

Alternately,

Amount with Amit = $\frac{3}{8}$^th of the total amount.

Since we do not know the total amount, we cannot figure out the exact value of the amount with Amit.

Example : Amit and Atul have money in the ratio of 3 : 5. The total amount with them is Rs. 160. What is the amount with them respectively.

Solution:
Here, we know the ratio of the amounts with them i.e. 3 : 5.

Hence, the amount with Amit = 3x 

and the amount with Atul = 5x.

Now, we know the total amount with them is Rs. 160.

Hence, 3x + 5x = 160

⇒ x = 20

Now, we can calculate the amounts with each of them.

Amount with Amit = 3x = 60.

Amount with Atul = 5x = 100.

Alternately,

Amount with Amit = $\frac{3}{8}$^th of the total amount =$\frac{3}{8}$ × 160 = 60

Amount with Atul = $\frac{5}{8}$^th of the total amount =$\frac{5}{8}$ × 160 = 100

Example : The ratio between two numbers is 12 : 13. If each number is reduced by 20, the ratio becomes 2 : 3. Find the numbers.

Solution:
Let the numbers be 12x and 13x

∴ $\frac{12x-20}{13x-20}=\frac{2}{3}$

⇒ x = 2    

∴ Numbers are 24, 26.

Example : If bc : ac : ab = 1 : 2 : 3, find a/bc ∶b/ca

Solution: $\frac{bc}{ac}=\frac{1}{2}$⇒ a ∶ b = 2 ∶ 1.

∴ $\frac{a}{bc}:\frac{b}{ca}=\frac{a}{bc}\times \frac{ca}{b}=\frac{a^2}{b^2}=\frac{4}{1}$

Example : A person distributes his pens among four friends A, B, C and D in the ratio $\frac{1}{3}$ : $\frac{1}{4}$ ∶ $\frac{1}{5}$ ∶ $\frac{1}{6}$. What is the minimum number of pens that the person should have?

Solution:
LCM of 3, 4, 5 and 6 is 60. Hence, we multiply all terms with 60.

So, the pens are distributed among A, B, C and D in the ratio 1/3 × 60 ∶ 1/4 × 60 : 1/5 × 60 ∶ 1/6 × 60

i.e. 20 : 15 : 12 : 10

∴ Total number of pens = 20x + 15x + 12x + 10x = 57x

For minimum number of pens, x = 1

∴ The person should have at least 57 pens.

Example : A bag contains 25 paise, 10 paise and 5 paise coins in the ratio 1 : 2 : 3. If their total value is Rs. 30, the number of 5 paise coins is :

Solution:
Let the number of coins be x, 2x and 3x

Value of 'x' 25 paise coins = 25x, that of '2x' 10 paise coins = 2x × 10 and that of '3x' 5 paise coins = 3x × 5.

Since, the total value is Rs. 30.

⇒ x × 25 + 2x × 10 + 3x × 5 = 30 × 100 (converting all the amounts in paise)

⇒ 60x = 3000    

i.e. x = 50

∴ Number of 5 paise coins are 3x = 150.

Example : Two numbers are in the ratio 3 : 7 and if 6 be added to each of them, they are in the ratio 5 : 9, then find the numbers.

Solution:
Let the numbers be 3x and 7x

∴ $\frac{3x+6}{7x+6}=\frac{5}{9}$ or 27x + 54 = 35x + 30

or 8x = 24 i.e., x = 3

Numbers are 3 × 3; 7 × 3 i.e., 9 and 21.

Example : A profit of Rs. 84 is divided between A and B in the ratio of $\frac{1}{3}$ ∶ $\frac{1}{4}$. What will be the difference of their profits?

Solution:
$\frac{1}{3}$ ∶ $\frac{1}{4}$ is the same as $\frac{1}{3}$ × 12 ∶ $\frac{1}{4}$ × 12 i.e., 4 : 3

Share of A = 4/7 × Rs. 84 = Rs. 48

Share of B = 3/7 × Rs. 84 = Rs. 36

Difference = Rs. 48 – Rs. 36 = Rs. 12

Example : The ratio of ages of Meera and Meena is 4 : 3. The sum of their ages is 28 years. The ratio of their ages after 8 years will be :

Solution:
Age of Meena = 4/7 × 28 = 16 years

Age of Meera = 3/7 × 28 = 12 years

After 8 years their respective ages will be 24 years and 20 years which shall be in the ratio 24 : 20 or 6 : 5

Example : A man divided Rs. 2,60,000 amongst his three daughters such that four times the amount received by the first, thrice the amount received by the second and twice the amount received the third are all equal. Find the amount received by the third daughter.

Solution:
Let the amounts received by first, second and third daughter by x, y and z respectively.

Hence, 4x = 3y = 2z.

We divided all the expression by LCM of coefficients i.e., LCM(4, 3, 2) = 12

⇒ x/3 = y/4 = z/6

Now assuming, x/3 = y/4 = z/6 = k
⇒ x = 3k, y = 4k and z = 6k

∴ x : y : z = 3k : 4k : 6k = 3 : 4 : 6

∴ Third daughter’s share = 6/13 × 2,60,000 = 12,000.

Example : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ and $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{3}{4}$. Find a : b : c.

Solution:
Here, calculating combined ratio is very simple.

For every 2 parts of a, there are 3 parts of b and for every 3 parts of b, there are 4 parts of a.

Hence, for every 2 parts of a, there are 3 parts of b and 4 parts of c.

∴ a : b : c = 2 : 3 : 4

Example : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ and $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$. Find a : b : c.

Solution:
Here, since the common term ‘b’ has different values, we will first have to make its value same in both the ratios,

i.e. is equal to LCM of the two values.

⇒ $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ = $\frac{8}{12}$ (Multiplying 4 in both numerator and denominator)

⇒ $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$ = $\frac{12}{15}$ (Multiplying 3 in both numerator and denominator)

Now, for every 8 parts of a, there are 12 parts of b and 15 parts of c.

∴ a : b : c = 8 : 12 : 15

Example : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$, $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$, $\frac{\mathrm{c}}{\mathrm{d}}$ = $\frac{1}{2}$ and $\frac{\mathrm{d}}{\mathrm{e}}$ = $\frac{2}{1}$. Find a : b : c : d : e.

Solution: Here, a : b : c : d : e = (n₁ × n₂ × n₃ × n₄) : (d₁ × n₂ × n₃ × n₄) : (d₁ × d₂ × n₃ × n₄) : (d₁ × d₂ × d₃ × n₄) : (d₁ × d₂ × d₃ × d₄)

⇒ a : b : c : d : e = 2 × 4 × 1 × 2 : 3 × 4 × 1 × 2 : 3 × 5 × 1 × 2 : 3 × 5 × 2 × 2 : 3 × 5 × 2 × 1.

⇒ a : b : c : d : e = 16 : 24 : 30 : 60 : 30.

⇒ a : b : c : d : e = 8 : 12 : 15 : 30 : 15

Example : If 0.75 : x :: 5 : 8 then find x.

Solution:
Product of extremes = product of mean terms

0.75 × 8 = x × 5

⇒ x = 1.2

Example : The ratio of number of boys and girls in a school in 4 : 3. If there are 480 boys in the school, find the number of girls in the school.

Solution:
4 : 3 = 480 : x

⇒ (3 × 480)/4 = 360 girls

Example : The mean proportion of 0.32 and 0.02 is what?

Solution:
If x be the required mean proportional then,

0.32 : x :: x : 0.02

⇒ x² = 0.32 × 0.02

⇒ x = 0.08

Example : Find the third proportional to 16 and 24.

Solution:
16 : 24 :: 24 : x

⇒ x = (24 × 24)/16 = 36

Example : The sum of the squares of 3 numbers is 532 and the ratio of the first to the second as also of the second to the third is 3 : 2. What is the second number?

Solution: $\frac{First number}{Second number}=\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$ and $\frac{Second number}{Third number}=\frac{3}{2}\times \frac{2}{2}=\frac{6}{4}$

[Make the second number same in both the ratios, i.e. 6]

∴ First : Second : Third = 9 : 6 : 4

∴ (9x)² + (6x)² + (4x)² = 532 (given)

⇒ 133 x 2 = 532

⇒ x = 2

Second number is 6x = 12

Example : Rs. 3000 are divided amongst A, B and C so that if Rs. 20, Rs. 40 and Rs. 60 be taken from their shares respectively, they will have money in the ratio 3 : 4 : 5. Find the share of C.

Solution:
Total decrease = Rs. 20 + Rs. 40 + Rs. 60 = Rs. 120

Remaining sum = Rs. 3000 – Rs. 120 = Rs. 2880

C’s share = 5/12 × Rs. 2880 = Rs. 5 × 240 = Rs. 1200

∴ Share of C out of Rs. 3000 = Rs. 1200 + Rs. 60 = Rs. 1260

Example : Rs. 4400 are divided among A, B and C so that A receives 3/8 as much as B and C together. What is the share of A?

Solution:
when (B + C) get Rs. 1, A gets Rs. 3/8

∴ A’s share : (B + C)’s share = 3/8 : 1 = 3∶8

∴ A’s share = Rs. $\frac{3}{11}$ × 4400 = Rs. 1200

Example : The ratio of the first and second class fares between the two stations is 4 : 1 and that the number of passengers travelling by first and second class is 1 : 40. If Rs. 1100 is collected as fare, the amount collected from first class passengers is :

Solution:
Ratio of amounts collected from 1st and 2nd class = (4 × 1) : (1 × 40) = 1 : 10

∴ Amount collected from 1st class passengers = $\frac{1}{11}$ × Rs. 1100 = Rs. 100

Example : One year ago the ratio between Laxman’s and Gopal’s salary was 3 : 4. The ratio of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Laxman now is :

Solution:
Let the salaries of Laxman and Gopal one year before be a, b respectively and now c, d respectively.

⇒ a : b = 3 : 4

a : c = 4 : 5

b : d = 2 : 3

It is given that c + d = 4160

⇒ $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{3}{4}$; $\frac{\mathrm{a}}{\mathrm{c}}$ = $\frac{4}{5}$ and $\frac{\mathrm{b}}{\mathrm{d}}$ = 2/3

Let a be 3x and b be 4x.

Hence, c = 15x/4 and d = 6x

⇒ c ∶ d = $\frac{{\displaystyle \raisebox{1ex}{$15x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{6x}=\frac{5}{8}$

∵ c + d = 4160

∴ c = $\frac{5}{13}$ × 4160 = Rs. 1600.

Example : Ram and Shyam enter into a partnership. Ram invests Rs. 24,000 and Shyam invests Rs. 20,000. Find the share of each in a Profit of Rs. 66,000 after one year.

Solution:
Ratio of investments of Ram and Shyam = 24,000 : 20,000 = 6 : 5

Ram’s share in a profit of Rs. 66,000 = Rs. $\frac{6}{11}$ × 66000 = Rs. 36,000

Shyam’s share = Rs. $\frac{5}{11}$ × 66000 = Rs. 30,000

Example : Ram and Shyam enter into a partnership with same investment. Ram invests for 8 months and leaves the business. Find the ratio of share of each after one year.

Solution:
Ratio of profits of Ram and Shyam = I × 8 : I × 12 = 2 : 3

Example : Ram, Mohinder and Jatinder enter into a partnership. Ram invests Rs. 24,000, Mohinder Rs. 20,000 and Jatinder Rs. 30,000. Find the share of each in a profit of Rs. 74,000 after one year.

Solution:
Ratio of investments of Ram, Mohinder & Jatinder = 24,000 : 20,000 : 30,000 = 12 : 10 : 15

Ram’s share in a profit of Rs. 74,000 = Rs. 12/37 × 74000 = Rs. 24,000

Mohinder’s share = Rs. 10/37 × 74000 = Rs. 20,000

Jatinder’s share = 15/37 × 74000 = Rs. 30,000.

Example : Ram, Mohinder and Jatinder enter into a partnership. Ram invests Rs. 24,000 for 8 months, Mohinder Rs. 20,000 for 12 months and Jatinder Rs. 30,000 for 6 months. Find the share of each in a profit of Rs. 51,000 after one year.

Solution:
Ratio of profits of Ram, Mohinder & Jatinder = 24,000 × 8 : 20,000 × 12 : 30,000 × 6 = 16 : 20 : 15

Ram’s share in a profit of Rs. 51,000 = Rs. 16/51 × 51000 = Rs. 16,000

Mohinder’s share = Rs. 20/51 × 51000 = Rs. 20,000

Jatinder’s share = 15/51 × 51000 = Rs. 15,000.

Example : Rajkumar and Sandeep entered into a partnership investing Rs. 48,000 and Rs. 54,000 respectively. After 3 months Jugal Kishore also joined the business with a capital of Rs. 40,000. Find the share of each in a profit of Rs. 55,000 after a year.

Solution:
Ratio of profits of Raj Kumar, Sandeep and Jugal Kishore :
48000 × 12 : 54000 × 12 : 40000 × 9 = 8 : 9 : 5

Raj Kumar’s share in the profit = Rs. $\frac{8}{22}$ × 55,000 = Rs. 20,000

Sandeep’s share in the profit = Rs. $\frac{9}{22}$ × 55,000 = Rs. 22,500

Jugal Kishore’s share = Rs. $\frac{5}{22}$ × 55000 = Rs. 12,500

Example : P, Q and R contract a work for Rs. 1100. P and Q together are to do $\frac{7}{11}$ of the work. Find R’s share of the amount contracted for.

Solution:
R’s share of work = 1 - $\frac{7}{11}$ = $\frac{4}{11}$ of the total work.

Since R does 4/11^th of the total work, he would receive 4/11^th of the total amount as well.

∴ R’s share in total money = $\frac{4}{11}$ × 1100 = Rs. 400.

Example : In a partnership, A invests 1/6 of the capital for 1/6 of the time, B invests 1/3 of the capital for 1/3 of the time and C, the rest of the capital for whole time. Find A’s share of the total profit of Rs. 2,300.

Solution:
Capital of C = 1 - 1/6 - 1/3 = 1/2

Let the total time be 12 months

∴ A’s profit : B’s profit : C’s profit

= $\frac{1}{6}\times \left(\frac{1}{6}\times 12\right)$  : $\frac{1}{3}\times \left(\frac{1}{3}\times 12\right)$ : $\left(\frac{1}{2}\times 12\right)$ = $\frac{1}{3} : \frac{4}{3} : 6$ = 1 : 4 : 18

∴ Share of A = $\frac{1}{1+4+18}$ × 2300 = Rs. 100.