Question

10. The students in 3 classes
are in the ratio 3:4:5.If 20
students added in each class,
the ratio becomes 5:6:7. Find
the total no of students in all
the 3 classes now? A) 160 B)
170 C) 180 D) 200 E) None of
these​

Answer 1

None of the above

Step-by-step explanation:

Given:-

• The students in 3 classes are in the ratio 3:4:5.
• If 20 students added in each class,the ratio becomes 5:6:7

To find:-

The total number of students in all 3 classes now

Solution:-

Let

the number of students in each class=3x,4x,5x

• if 20 students added to each class then the new number of students =3x+20,4x+20,5x+20

ATQ,

$\\\qquad\quad\sf {:}\longrightarrow 3x+20:4x+20:5x+20=5:6:7$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {3x+20}{\dfrac {4x+20}{5x+20}}=\dfrac{5} {\dfrac {6}{7}}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 3x+20\times \dfrac {5x+20}{4x+20}=\dfrac {5\times 7}{6}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {(3x+20)(5x+20)}{4x+20}=\dfrac {35}{6}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {3x (5x+20)+20 (5x+20)}{4x+20}=\dfrac {35}{6}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {15x^2+60x+100x+400}{4x+20}=\dfrac {35}{6}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {15x^2+160x+400}{4x+20}=\dfrac {35}{6}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 6 (15x^2+160x+400)=35 (4x+20)$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 90x^2+960x+2400=140x+700$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 90x^2+960x-140x+2400-700=0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 90x^2+820x+1700=0$

• Here ,
• a=90,b=820,c=1700

use quadratic formula

$\boxed{\purple {\sf x=\dfrac {-b\underline{+}\sqrt {b^2-4ac}}{2a}}}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {-820\underline{+}\sqrt {(820)^2-4\times 90\times 1700}}{2\times 90}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {-820\underline{+}\sqrt {60400}}{180}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {-41}{9}+\dfrac {1}{9}\sqrt {151} \quad or\quad \dfrac {-41}{9}-\dfrac{1}{9}\sqrt {151}$

• ignoring negative value

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=3.2$

$\rule {250}{1}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=3x+20+4x+20+5x+20$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=3x+4x+5x+20+20+20$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=12x+60$

• Substitute the value of x

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=12 (3.2)+60$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=38.4+60$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=98.4$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Total\:number\:of\:students=99_{(Approximately)}$

$\\\\\therefore \sf Total\:Number \:of\:students\:is\:99.$