Ratios and Proportions - Class VII

$\dfrac{36}{64}=\dfrac{36\div 4}{64\div 4} =\dfrac{9}{16}$

Using the property of ratios, if a/b = c/d = e/f, then each ratio is equal to (a+c+e)/(b+d+f). Adding the numerators gives 2(x+y+z) and adding the denominators gives (a+b)(x+y+z). Canceling (x+y+z) yields 2/(a+b).

$\dfrac{\text{number of cricket players}}{\text{total number of players}}=\dfrac{30}{30+20}=\dfrac{30}{50} =\dfrac{3}{5}$

$\dfrac{\text{length of red ribbon}}{\text{length of blue ribbon}}= \dfrac{80cm}{220cm} = \dfrac{8}{22} = \dfrac{4}{11}$

Given $a:b=3:5$, Let $a=3k$ and $b=5k$

Total population is 3540 and males are 2065. Females = 3540 - 2065 = 1475. The ratio of males to females is 2065/1475. Dividing both by 295 gives 7/5.

Let the age of the two brothers be $x$ and $y$ respectively

In this Question the sum of ratio of marbles must be 25.
A. $12+13=25$
B. $11+14=25$
C. $1+10=11$
D. $8+17=25$

Given, $A:B=2:3$, $B:C=4:5$ and $C:D=6:7$

For the ratio to be equal the product of mean =product of extremes
for ex   $\dfrac{a}{b}=\dfrac{c}{d}$
product of $a\times d=b\times c$
where
$a\times d$ $=$product of extreme.
and$b\times c$ $=$ product of means.

Given  $a:b = 3:4$ and $b:c = 8:9$
Then $\displaystyle \frac{a}{b}\times \frac{b}{c}=\frac{3}{4}\times \frac{8}{9}$

$A, :, B, = , \displaystyle {\frac{1}{2}, :, \frac{3}{8},  =, 4, :, 3,}$

To find a : b : c, b is made same in both the ratios. L.C.M of 4 and 5 is 20
$a, :, b, =, \displaystyle {\frac{3}{4}, =, \frac{3, \times, 5}{4, \times, 5}, =, \frac{15}{20}}$
$b, :, c, =, \displaystyle {\frac{5}{9}, =, \frac{5, \times, 4}{9, \times, 4}, =, \frac{20}{36}}$
$\therefore$   $ a : b : c = 15 : 20 : 36$

Given a : b = 8 : 9, b : c = 18 : 40
$\displaystyle {\frac{a}{c}, =, \frac{a}{c}, \times, \frac{b}{b}, =, \frac{a}{b}, \times, \frac{b}{c}, =, \frac{8}{9}, \times, \frac{18}{40}, =, \frac{2}{5}}$
a : c = 2 : 5

Let the third number be x.
Then first number=120% of x=$\frac{120}{100}\times x=\frac{6x}{5}$
Second number=150% of x$\frac{150}{100}\times x=\frac{3x}{2}$
$\therefore$Ratio of first two number=$\frac{6x}{5}:\frac{3x}{2}=12x:15x=4:5$

$\displaystyle \frac{M}{a}=m+n;\frac{N}{b}=m-n$
$\displaystyle \therefore \left ( \frac{M}{a}+\frac{N}{b} \right )\div \left ( \frac{M}{a}-\frac{N}{b} \right )=2m\div 2n=\frac{m}{n}$

We know,

Length = 2m = 200 cm
 Width = 28 cm
$\dfrac{Length}{ Width}$ = $\dfrac{200}{28}$

The three quantities $a,,b,,c$ are said to be in continued proportion if $a:b::b:c$
$\Longrightarrow\dfrac{a}{b}=\dfrac{b}{c}$
$\Longrightarrow a\times c=b^2$
$\Longrightarrow b^2=ac$

The $a:b:c=A:B:C$
can be represented by ratio representation is
$ratio=\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}$

$\dfrac{2}{9}=\dfrac{x}{18}$
$\Rightarrow;9x=36$
$\Rightarrow;x=4$

Consider the amount of pulse of price $Rs15$ $=x$

$\cfrac { ax+b }{ cx+d } \neq \cfrac { b }{ d } \Longrightarrow adx+bd\neq bcx+bd\Longrightarrow (ad-bc)x\neq 0\ \therefore x\neq 0\quad and\quad ad\neq bc$

Given 

The length of the rectangular field is $50 m$ and breadth of the field is $15 m$.

The ratio of breadth and length is,

${\rm{breadth}}:{\rm{length}} = 15:50=\dfrac{15}{50}=\dfrac{3}{10}$

$ = 3:10$

Since, ${\rm{Rs}}{\rm{.1}} = {\rm{100}}\;{\rm{paise}}$

The ratio is given as,

$r = \dfrac{{50\;{\rm{paise}}}}{{5 \times 100\;{\rm{paise}}}}$

$r = \dfrac{1}{{10}}$

Thus, the required ratio is $1:10$.

Reduced form of $5$ liters and $2500ml$ 

$Ratio=\dfrac{Present\ age\ of\ father}{Present\ age\ of\ son.}$

firstly year change into months

Reduced form of 7 min 20 sec, 5 min 6 sec is 

Reduced form of $3.kg$ and $1900gm$ in ratio

We have,

Number of boys $=32$

Number of girls $=18$

Then,

Total student in a class

$ =32+18 $

$ =50 $

The ratio of number of boys us

$ =\dfrac{Boys}{Total\,students} $

$ =\dfrac{32}{50} $

$ =\dfrac{16}{25} $

Hence, this is the answer.

To find A:D, multiply the ratios: (A/B) * (B/C) * (C/D) = (2/3) * (4/5) * (5/9). The B and C terms cancel out, leaving A/D = (2 * 4 * 5) / (3 * 5 * 9) = 40/135 = 8/27.

Given x/y = 3/5, let x = 3k and y = 5k. Substitute these into (2x+3y)/(5x+7y): (2(3k) + 3(5k)) / (5(3k) + 7(5k)) = (6k + 15k) / (15k + 35k) = 21k / 50k = 21/50.

$33\displaystyle\frac{1}{3}%$ of $A=1.5$ of $B=\displaystyle\frac{1}{8}$ of $C$ $\Rightarrow\displaystyle\frac{100}{3\times100}A=\displaystyle\frac{3}{2}B=\displaystyle\frac{1}{8}C;;\Rightarrow:\displaystyle\frac{1}{3}A=\displaystyle\frac{3}{2}B=\displaystyle\frac{1}{8}C=K:(say)$  $\Rightarrow:A=3K,,B=\displaystyle\frac{2}{3}K,,C=8K$  $\Rightarrow:A,\colon,B,\colon,C=3,\colon,\displaystyle\frac{2}{3},\colon,8=3\times3,\colon,\displaystyle\frac{2}{3}\times3,\colon,8\times3=9,\colon,2,\colon,24$

Given,

Given, $\dfrac{a}{b}=\dfrac{7}{9}$ and$\dfrac{b}{c}=\dfrac{3}{5}$
Then $\displaystyle \frac{a}{b}\times \frac{b}{c}=\frac{7}{9}\times \dfrac{3}{5}\Rightarrow \frac{a}{c}=\frac{7}{15}$
$\Rightarrow \dfrac{b}{c}=\dfrac{3}{5}=\dfrac{9}{15}$
$\Rightarrow a:b=7:9$ and $b:c=9:15$

$;A,\colon,B=\displaystyle\frac{1}{2} \colon\displaystyle\frac{1}{3}=\displaystyle\frac{1}{2}\times6\ \colon\displaystyle\frac{1}{3}\times6=3,\colon,2$

Let the required number be y
$a : b : : x : y$
$a \times y = b \times x$
$\displaystyle y=\frac{bx}{a}$

$\displaystyle \frac {a}{b} = \frac {2}{9} \div \frac {1}{3}=  \frac {2}{9} \times \frac {3}{1}, \frac {b}{c}= \frac {2}{7} \div \frac {5}{14} = \frac {2}{7} \times \frac {14}{5}= \frac {4}{5}$

$\displaystyle \frac {A}{B} = \frac {1}{2} \Rightarrow A = \frac {1}{2} B , \frac {B}{C} = \frac {3}{4} \Rightarrow C = \ \frac {4B}{3} $

Given, $3A = 4B = 6C$ 

Let the number of $50 p$ coins be $2x$, $1$ rupee coins be $3x$ and $2$ rupee coins be $4x$.
Total amount=240 

$A:B = \displaystyle \frac{1}{2} : \frac{3}{8} = \frac{1}{2} \div \frac{3}{8} = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3} = 4 : 3 = 8:6$

$\displaystyle \frac{A}{B} = \frac{4}{5} $ and $\displaystyle \frac{B}{C} = \frac{2}{3}$
To make both the values of B equal, take LCM of both values of B, i.e., 5 and 2 which is equal to 10.
$\therefore \displaystyle \frac{A}{B} = \frac{4}{5} = \frac{8}{10}$ and $\displaystyle \frac{B}{C} = \frac{2}{3} = \frac{10}{15}$
$\Rightarrow A : B : C =8 : 10 : 15$
$\Rightarrow A = 8x, B = 10 x, C = 15 x$
Given, $A = 800 \Rightarrow 8x = 800$
$\Rightarrow x = 100       \Rightarrow C = 1500$

Let $x = 3k$ and $y = k$
Then $\displaystyle \frac{x^{3}-y^{3}}{x^{3}+y^{3}}$

Speed of cycle which is $15km/hr$ to the speed of scooter which is $30km/hr = \cfrac {15 km}{30 km} =\cfrac 12 = 1:2$

Given that,  If $A:B=3:4,B:C=5:6$,then$A:C$

Given$ a:b=3:4$  and $b:c= 5:6$  

$a:b::b:c =3:4::5:6$ 

$a:c= 3/4*5/6=5/8$ 

Hence, this is the answer. 

Total time in office $ = 9:30  to  4:30  = 7  hours  $

Ratio of two quantities can be found when they are of same units.

So, $ 7  hours   = 7 \times 60 = 420  minutes   $
Hence, ratio $ = \dfrac {420  min }{20  min } = 21:1 $

Office hours
$= 10.30$ AM to $5.30$ AM=
 $i.e., 7 hrs = 420 min$.
Lunch time $= 30$ min
$420:30$
14:1

To find the ratio of two numbers, we have to consider their fraction. 

$b:c=\begin{pmatrix}6\times\dfrac{7}{6}\end{pmatrix}:\begin{pmatrix}11\times\dfrac{7}{6}\end{pmatrix}=7:\dfrac{77}{6}$
$a:b:c=5:7:\dfrac{77}{6}=30:42:77$

$Ratio 1 = 4 : 3$
$Ratio 2 = 5 : 6$
Multiplying ratio $1$ by antecedent of ratio $2$ and ratio $2$ by consequent of ratio $1$,
Ratio $1 = 20 : 15$
Ratio $2 = 15 : 18$
Thus, for two ratios $a : b$ and $b : c, a : b : c$ is called the continued ratio.
$\therefore 20 : 15 : 18$ is the continued ratio for $4 : 3$ and $5 : 6$.

$R _{1}$ = $2 : 5$
$R _{2}$ =$ 6 : 7$
Multiplying the LCM of the consequent of ratio $1$ and antecedent of ratio $2$, to both the ratios.
$R _1$ = $12 : 30$
$R _2$ = $30 : 35$
Thus, the continued ration for $2 : 5$ and $6 : 7$ is $12 : 30 : 35$

Let $A = 2x; B = 5x$ be the present ages of $A$ and $B$ respectively.
After $8$ years, their ages will be $(2x + 8)$ years and $(5x + 8)$ years respectively.
Therefore, $ (2x + 8) : (5x + 8) : : 1 : 2$
$\Rightarrow  (5x + 8)\times 1 = 2(2x + 8)$
$\Rightarrow x = 8$
Thus difference of their present ages is $5x - 2x = 3x$ i.e., $24$ years.

Let the two numbers be $2x$ and $3x.$ Thus,

$With\quad ratios,add\quad all\quad the\quad parts\quad together\quad to\quad get\quad a\quad total.In\quad this\quad case,it\quad is\quad 4red\quad and\quad 7blue.$

$A:B=7:5\B:C=9:11$

$\displaystyle\frac{1}{3}$ of Rs. $9.30=Rs. \left(\displaystyle\frac{1}{3}\times 9.30\right)=Rs. 3.10$
$0.6$ of Rs. $1.55=$Rs. $(0.6\times 1.55)=$ Rs. $0.93$
$\therefore$ Required ratio$=3.10:0.93=10:3$.

Cost of $1$ candy $=50$ paise
Cost of $1$ ice-cream $=Rs. 20=20\times 100$ paise
$=2000$ paise
$\therefore$ Required ratio $=50:2000=1:40$.