Ratio and Mixture Problems - Class XII

Let the numerator be $x+5$ and denominator be $x$

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are ($140%$ of $5x$), ($150%$ of  $7x$) and ($175%$ of $8x$).
 $\rightarrow \left ( \dfrac{140}{100} x\times  5x\right ),\left ( \dfrac{150}{100} x\times 7x\right ), and \left ( \dfrac{175}{100} \times  8x\right )$
$\rightarrow 7x, \dfrac{21x}{2}$ and $14x$
$\therefore $ The required ratio$ = 7x:$ $\dfrac{21x}{2}$$: 14x$
$\rightarrow$ $14x : 21x : 28x$
$\rightarrow$ $2 : 3 : 4$

Population after two years,
$=8000\times \dfrac{110}{100}\times \dfrac{120}{100}=10560$

Let  the number of bacteria is $a$. The number of bacteria doubles every $12$   minutes, which means that it doubles $5$ times in an hour.
Therefore,
$(2)(2)(2)(2)(2)a =a\times2^5$
                            $ = 32 a$  at the end of the hour.
Thus, the ratio of the number of bacteria at the end of 1 hour to the number of bacteria at the beginning of that hour is $\cfrac{32a}{a} = 32:1$.

Given that cost of two TV sets is $Rs.10,000$ each

Let past salaries be 3x and 5x. Present salaries are (3/2)*3x = 4.5x and (5/4)*5x = 6.25x. Total = 10.75x = 8600. x = 800. Ram's present salary = 4.5 * 800 = 3600.

It is given that $x$ and $y$ are inversely proportional

Let initial quantities be 7x and 2x. After removing 18 litres of mixture, the remaining quantities are proportional, and adding 18 litres of pure liquid B changes the ratio to 2:3. Solving the algebraic equation yields 490/17 litres for liquid A.

Let the numbers be $2x$ and $3x $

Given, sides of the triangle $=10cm, 10cm, 12cm$

 Let the present ages of Amit and his father be 2x and 5x years respectively 
Four years hence 
Amit's Age = ( 2x+ 4) years
Father's age = (5x + 4) years
Given $\displaystyle \frac{2x+4}{5x+4}=\frac{5}{11}\Rightarrow (2x+4)=5(5x+4)$

Let the two numbers be x and y Then $\displaystyle x^{2}-y^{2}=80%$ of $\displaystyle (x^{2}+y^{2})$
$\displaystyle \Rightarrow x^{2}-y^{2}=\frac{4}{5}(x^{2}+y^{2})\Rightarrow x^{2}-\frac{4}{5}x^{2}=\frac{4}{5}y^{2}+y^{2}$
$\displaystyle \Rightarrow \frac{1}{5}x^{2}=\frac{9}{5}y^{2}\Rightarrow \frac{x^{2}}{y^{2}}=\frac{9}{1}\Rightarrow \frac{x}{y}=\frac{3}{1}\Rightarrow x:y=3:1$

$a =3k$
$b=k$
$c= 5k-4k =k$
$d =6k-5k =k$
$\frac {a}{b+2c+3d}=\frac {3k}{k+2k+3k}=\frac {1}{2}$

Solution:

Let the whole number is X.
Now, according to question,
(6-X) / (7-X) < 16/21
21 *(6-X) < 16 *(7-X)
126 - 21X < 112 - 16X
126 - 112 < -16X + 21X
14 < 5X
5X > 14
X > 2.8
So, Least such whole number would be 3.

Length of ribbon originally $=30cm$
Let the original length be $5x$ and reduced length be $3x$.
But $5x=30cm$
$\Longrightarrow x=\dfrac{30}{5}cm=6cm$
Therefore, reduced length $=3\times6cm=18cm$

Multiplying any fraction by a value greater than 1 will increase its value.

Let the new quantity be x
$\therefore \dfrac {\text {Original quantity}}{\text {New quantity}}$
$\dfrac {245}{x} = \dfrac {7}{4}$
$\therefore x = 140$
Hence, on decreasing $245$ by $4 : 7$ we get $140$ kg.

$3 : 4$ is the ratio of the new quantity to the original quantity.
Let the new number be x.
$\therefore \dfrac {x}{60} = \dfrac {3}{4}$
$\therefore x = 45$
$\therefore$ The decreased quantity is $45$

New price $= 20$
Old price $= 15$
Multiplying ratio
$\dfrac {\text {new price}}{\text {old price}} = \dfrac {20}{15} = \dfrac {4}{5}$

Let the new price of pencil box be x
$\dfrac {\text {Original price}}{\text {New price}}$
$\therefore \dfrac {70}{x} = \dfrac {14}{3}$
$\therefore x = 65$
Hence the reduced price of pencil box is Rs $65$.

Final quantity $= 63$
Original quantity $= 90$
Multiplying ratio
$\dfrac {\text {Final quantities}}{\text {Original quantity}} = \dfrac {63}{90} = \dfrac {7}{10}$

Value=$540\times \dfrac{2}{3}=360$

$\cfrac {456}{x} = \dfrac {8}{9}$

Original quantity $= 750$
Final quantity $= 1000$

Multiplying ratio:
$\dfrac {\text {final quantity}}{\text {original quantity}} = \dfrac {1000}{750} = \dfrac {4}{3}$

Increasing Rs $40$ in the ratio $5.4$ implies that the ratio of the new quantity to the old quantity is $5 : 4$.
$\therefore$ Let the increased quantity be x
$\therefore \dfrac {\text {New amount}}{\text {original amount}} = \dfrac {x}{40} = \dfrac {5}{4}$
$\therefore x = 50$
that is the increased quantity is $50$

Let the new salary be $x$

$3$ pounds of peat moss is to $5$ pounds of compost is the composition in $8$ pounds potting soil.

Let the numbers are $x$ and $y$ whose ratio is $1 : 2$

Let initial investments by $A, B$ and $C$ are $3x, 5x, 7x$

It is given that,
$\displaystyle \dfrac{l _1}{l _2} , = , \dfrac{5}{3} , and , \dfrac{s _1}{s _2} , = , \dfrac{6}{5} , \Rightarrow , \dfrac{t _1}{t _2} , = , \frac{s _1}{l _2} , = , \dfrac{l _1}{l _2} , \times , \dfrac{s _2}{s _1} , = , \dfrac{5}{3} , \times , \dfrac{5}{6} , = , \dfrac{25}{18}$
Hence, $t _1 : t _2$ = $25 : 18$

Let the volume of spirit and water be x and 3x Then, total volume = 4x. Resultant  volume of solution = 1.25 $\times$ 4x = 5x
Therefore, increase in volume = $5x - 4x = x$
So, the new ratio of spirit of water $2x : 3x = 2:3$
It is to be noted that increase in volume is due to addition of spirit only.

Given that expense on Wheat, Meat and Vegetable =12x + 17x + 3x = 32x
New expense on wheat, Meat and Vegetable
= 1.2 $\times 12x + 1.3 \times 17x + 1.5 \times 3x $
= 14.4x + 22.1x + 4.5x = 41x
Percentage increase in expense = $\frac{9}{32} \times 100 = 28\frac{1}{8}$%