CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 7 - MCQExams.com

On simplifying $$ \dfrac {(x-2)}{5} = \dfrac {(1-y)}{4} $$, we get $$ 4x + 5y = 13 $$---- (1)
And equation $$ 26x + 3y = -4
$$ ---- (2)
Multiplying equation  $$ (1) $$ with $$ 3 $$ we get, $$ 12x + 15y = 39  $$ ----- equation $$ (3) $$ 
Multiplying equation  $$ (2) $$ with $$ 5 $$ we get, $$ 130x + 15y = - 20  $$ ----- equation $$ (4) $$ 

subtracting equation $$ (3) $$

from $$ (4) $$, we get $$ 118x = -59 => x = - \dfrac {1}{2} $$

Substituting $$ x = - \dfrac {1}{2} $$ in the equation $$ (1) $$, we get $$ 4 (-\dfrac {1}{2}) + 5y = 13 $$

 $$\implies  y = 3 $$

Rewriting the given equations, we get
$$7x-10y=4$$                              ...(i)
$$12x+18y=1$$                           ...(ii)
On multiplying (i) by 12 and (ii) by 7 and subtracting, we get
$$84x-120y=48$$
$$\underline {\underset {-}84x\underset {-}{+}126y=\underset{-}7}$$
             $$-246y=41$$
$$\therefore y=\dfrac{-41}{246}$$
On putting $$y=\dfrac{-41}{246}$$ in (i), we get
$$7x-10\times\dfrac{-41}{246}=4\Rightarrow x=\dfrac{82}{246}$$           
Now, $$4x+6y=4\times\dfrac{82}{246}+6\times\dfrac{-41}{246}=\dfrac{82}{246}=\dfrac13$$ and
$$8y-x=8\times \dfrac{-41}{246}-\dfrac{82}{246}=\dfrac{-810}{246}=\dfrac{-5}{3}$$

Multiplying eq(2) with $$ 4 $$ we get, 

$$  \dfrac {100}{3x-2y}  - \dfrac {34}{3x+4y} = 18 $$    ...(3)

Adding eq(1) and eq(3) we get, 

$$  \dfrac {115}{3x-2y}   = 23=> 3x - 2y = 5 $$       ...(4)

Substituting $$ 3x - 2y = 5 $$ in the eq(1), we get,

$$  \dfrac {34}{3x+4y}  + \dfrac {15}{5} = 5 => \dfrac {34}{3x+4y} = 2 $$

$$=> 3x +4y = 17  $$                   ...(5)

Subtracting eq(4) from eq(5) we get, 

$$ 6y = 12 => y = 2 $$

Substituting $$ y = 2 $$ in the eq(5) we get,

$$ 3x + 4(2) = 17 => x = 3 $$

As per the statement, "If A gives 10 pencils to B, then B will have twice as many as A":

$$ \implies 2(x - 10) = y + 10$$

           $$ 2x - y = 30 $$ --- (1)
Also, as per the statement, "if B gives 10 pencils to A, then they will have the same number of pencils" 

$$\implies x + 10 = y - 10 $$

           $$x - y = -20  $$ --- (2)

Subtracting equation $$ (2) $$ from $$ (1) $$, we get:

$$2x-y-x+y=30+20$$

 $$  x = 50$$

Substituting $$ x = 50 $$ in equation $$ (2) $$, we get:

 $$ 50 -y = -20$$

$$\implies y = 70 $$

So, A has $$50$$ pencils and B has $$70$$ pencils.

Multiplying equation  $$

(1) $$ with $$ 3 $$ we get, $$ 3x + 3y = 552  $$ ----- equation $$

(3) $$

Adding equations $$2 $$ and $$ 3 $$, we get $$ 10x = 636 => x = 63.6 $$

Substituting

$$ x = 63.6 $$ in the equation $$ (2) $$, we get $$ 63.6 + y = 184 =>y = 120.4

$$

Thus , the parts are $$ 63.6 ; 120.4 $$

Multiplying equation $$ (2) $$ with $$ 1.05 $$ we get, $$ 1.1235x + 1.1025y = 1223.25 $$ ----- equation $$ (4) $$

Subtracting equation $$(4) $$ from $$ (3) $$, we get $$ 0.0424y = 25.44 \Rightarrow y = 600 $$

Substituting $$ y = 600 $$ in the equation $$ (2) $$, we get $$ 1.07x + 1.05(600) = 1165 \Rightarrow x = 500 $$

Hence, cost price of $$A$$ is Rs. $$ 500 $$ and of $$B$$ is Rs. $$ 600 $$

Substituting $$ y = 18 $$ in the equation $$ (1) $$, we get $$ x + 18 = 40 => x = 22 $$

Hence $$ 22 kg $$ and $$ 18 kg $$ of two types of sweets were bought. 

Substituting $$ x = 600 $$ in the equation $$ (1) $$, we get $$ 600 + y = 1250 \Rightarrow y = 650 $$

Let father's age $$=x$$, son's age $$=y$$    

 As per question,

$$x+y=65\Rightarrow y=65-x $$         .....(1)                                                

$$2\left( x-y \right )=50\Rightarrow2x-2y=50 $$       .......(2)
$$\Rightarrow 2x-2\left (65-x  \right )=50$$    $$[$$ substituting value of $$y$$ from equation (1) in equation (2) $$]$$

$$\Rightarrow2x-130+2x=50$$
$$\Rightarrow x=45 $$

So father's age $$=45 \text{ years}$$