Maths MCQ

Let us assume that the number to be subtracted is $ x$
Then, $(15 - x) : (19 - x) = 15 : 3$
$\Rightarrow\frac{15-\text{x}}{19-\text{x}}=\frac{3}{4}$
$\Rightarrow 60 - 4x = 57 - 3x$
$\Rightarrow x = 3$

Amount $= Rs. 420$
And ratio $ = 3 : 4$
Sum of ratios $= 3 + 4 = 7$
A’s share $=\frac{420\times3}{7}=\text{Rs. }60\times3=\text{Rs. }180$

$2\text{A}=3\text{B}=4\text{C}=\text{x}$
$\therefore\text{A}=\frac{\text{x}}{2},\text{B}=\frac{\text{x}}{3},\text{C}=\frac{\text{x}}{4}$
$\therefore\text{A : B : C}=\frac{\text{x}}{2}:\frac{\text{x}}{3}:\frac{\text{x}}{4}=\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$
$\frac{6\ :\ 4\ :\ 3}{12}=6:4:3$

$2\text{A}=3\text{B}$
$\Rightarrow\frac{\text{A}}{\text{B}}=\frac{3}{2}$
$4\text{B}=5\text{C}$
$\Rightarrow\frac{\text{B}}{\text{C}}=\frac{5}{4}$
Multiplying, We get
$\frac{\text{A}}{\text{B}}\times\frac{\text{B}}{\text{C}}=\frac{3}{2}\times\frac{5}{4}$
$\Rightarrow\frac{\text{A}}{\text{C}}=\frac{15}{8}$
$\therefore\text{A : C}=15:8$

$\frac{\text{A}}{3}=\frac{\text{B}}{4}=\frac{\text{C}}{5}=1$ (Suppose)
$\text{A}=3,\text{B}=4,\text{C} =5$
$\text{A : B : C}=3:4:5$

Ratio in two numbers $= 3 : 5$
Let first number $= 3x$
Then second number $= 5x$
According to the condition,
$\frac{3\text{x}+10}{5\text{x}+10}=\frac{5}{7}$
$($By cross multiplication$)$
$25x + 50 = 21x + 70$
$25x - 21x = 70 - 50$
$4x = 20$
$x = 5$
First number $= 3 \times 5 = 15$
and second number $= 5 \times 5 = 25$
Sum of numbers $= 15 + 25 = 40$

Suppose that the present ages of $A$ and $B$ are $5x$ yrs and $2x$ yrs, respectively.
After 5 years, the ages of $A$ and $B$ will be $(5x + 5)$ yrs and $(2x + 5)$ yrs, respectively
Then, $(5x+ 5) : (2x + 5) = 15 : 7$
$\Rightarrow\frac{5\text{x}+5}{2\text{x}+5}=\frac{15}{7}$
Cross multiplying; we get:
$\Rightarrow 35x + 35 = 30x + 75$
$\Rightarrow 5x = 40$
$\Rightarrow x = 8$
Then, the present age of $A$ is $5 \times 8 = 40$ years.

$(3\text{a}+5\text{b}):(3\text{a}-5\text{b})=5:1$
$\Rightarrow\frac{3\text{a}+5\text{b}}{3\text{a}-5\text{b}}=\frac{5}{1}$
$3\text{a}+5\text{b}=5(3\text{a}-5\text{b})$
$\Rightarrow3\text{a}+5\text{b}=15\text{a}-25\text{b}$
$\Rightarrow3\text{a}-15\text{a}=-25\text{b}-5\text{b}$
$-12\text{a}=-30\text{b}$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{-30}{-12}=\frac{5}{2}$
$\therefore\text{a : b}=5:2$

Let x be subtracted from each of the term
$\frac{15-\text{x}}{19-\text{x}}=\frac{3}{4}$
$\Rightarrow 4 (15 - x) = 3 (19 - x)$
$\Rightarrow 60 - 4x = 57 - 3x$
$\Rightarrow -4x + 3x = 57 - 60$
$\Rightarrow -x = -3$
$x = 3$
Required number $= 3$

$\text{A}=\frac{2}{3}\text{B}$
$\text{C}=\frac{4}{3}\text{B}$
$\therefore\text{A : B : C}=\frac{2}{3}\text{ B : B : }\frac{4}{3}\text{ B}$
$=2 : 3 : 4$

Let x be the number of bosy.
Then, $8 : 5 = x : 160$
$\Rightarrow\frac{8}{5}=\frac{\text{x}}{160}$
$\Rightarrow\text{x}=\frac{8\times160}{5}=256$
$\therefore$ Total strength of the school $= 256 + 160 = 416$

Let $x$ be the third proportional to $9$ and $12$ then
$9 : 12 :: x : 12$
$\Rightarrow 9 \times x = 12 \times 12$
$\Rightarrow\text{x}=\frac{12\times12}{9}=\frac{144}{9}=16$
Third proportional $= 16$

$\frac{2}{3}$ and $\frac{4}{7}$
$\text{LCM}$ of $3$ and $7 = 21$
$\therefore\frac{2}{3}=\frac{2\times7}{3\times7}=\frac{14}{21}$
$\frac{4}{7}=\frac{4\times3}{7\times3}=\frac{12}{21}$
It is clear that $\frac{14}{21}$ or $\frac{2}{3}$ is greater.

$\frac{1}{\text{x}}:\frac{1}{\text{y}}:\frac{1}{\text{z}}=2:3:5$
Let $\frac{1}{\text{x}}=2\text{k},\frac{1}{\text{y}}=3\text{k},\frac{1}{\text{z}}=5\text{k}$
$\Rightarrow\text{x}=\frac{1}{2\text{k}},\text{y}=\frac{1}{3\text{k}},\text{z}=\frac{1}{5\text{z}}$
$\therefore\text{x : y : z}=\frac{1}{2\text{k}}:\frac{1}{3\text{k}}:\frac{1}{5\text{k}}$
$=\frac{15\ :\ 10\ :\ 6}{30\text{k}}$
$\therefore$ Ratio of $x : y : z = 15 : 10 : 6$