MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 9

Solve the following pair of linear equations by the substitution method

$s-t=3, \dfrac {s}{3}+\dfrac {t}{2}=6$

0%
$s = 2, t = 7$
0%
$s = 9, t = 6$
0%
$s = 3, t = 1$
0%
$s = 6, t = -9$

Explanation

$s-t=3$ .....(i)

$\frac {s}{3}+\frac {t}{2}=6$ .....(ii)
From (i)

$s = t + 3$ ...(iii)
Substituting s from (iii) in (ii), we get

$\frac {t+3}{3}+\frac {t}{2}=6$

$\Rightarrow 2(t + 3) + 3t = 36$

$\Rightarrow 5t + 6 = 36$

$\Rightarrow t = 6$
From (iii),

$s =6 +3 = 9$ ,
Hence,

$s = 9, t = 6$

Solve the following pair of linear equations by the substitution method

$x + y = 14, x - y = 4$

0%
$x = 8, y = 5$
0%
$x = 6, y = 9$
0%
$x = 7, y = 10$
0%
None of these

Explanation

$x + y = 14$ ....(i)

$x - y = 4$ .....(ii)
From (ii)

$y=x - 4$ ...(iii)
Substituting y from (iii) in (i), we get

$x + x - 4 = 14$

$\Rightarrow 2x = 18$

$\Rightarrow x =9$
Substituting

$x = 9$ in (iii), we get

$y = 9 - 4 = 5$ ,
i.e,

$y = 5$

$x = 9, y = 5$

Solve the following pair of linear equations by the substitution method.

$0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3$

0%
$x=1, y=-2$
0%
$x=6, y=-7$
0%
$x=5, y=1$
0%
$x=2, y=3$

Explanation

The given equations are

$\Rightarrow .2x+.3y=1.3........eq1$

$\Rightarrow x=\frac{1.3-.3y}{.2}.......eq2$

and

$\Rightarrow .4x+.5y=2.3.......eq3$
Substitute the value of x in eq3

$\Rightarrow .4\times\left (\frac{1.3-.3y}{.2} \right)+.5y=2.3$

$\Rightarrow 2.6-.6y+.5y=2.3$

$\Rightarrow -.1y=-.3 \Rightarrow y=3$
Substitute

$y=3$ in eq 2

$\Rightarrow x=\frac{1.3-.9}{.2}\Rightarrow x=2$
Hence x=2 and y=3

The sum of the numerator and denominator of a fraction is

$4$ more than twice the numerator. If the numerator and denominator are increased by

$3$ , they are in the ratio

$2 : 3$ . Determine the fraction

0%
$\dfrac {1}{7}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{7}$
0%
$\dfrac {5}{9}$

Form the pair of linear equations for the following problem and find their solution by substitution method.

The larger of two supplementary angles exceeds the smaller by

$18$ degrees.

0%
$x + y = 100$ and $x - y = 18$ , $x=90$ and $\,y=16$
0%
$x + y = 60$ and $x - y = 18$ , $x=95$ and $\,\,y=70$
0%
$x + y = 180$ and $x - y = 18$ , $x=99$ and $\,\,y=81$
0%
None of these

Explanation

Let the supplementary angles be

$x$ and

$y$ ,

$(x> y)$

$\therefore$ Sum of both angles is

$180^0$

$\Rightarrow x + y = 180$ ...

$(1)$

The larger angle exceeds the smaller by

$18^0$

$\Rightarrow x - y = 18$ ...

$(2)$

$\Rightarrow y=x -18$

Put value of

$y$ in (1), we get

$2x=198$

$\Rightarrow x = 99$

Now substituting the value of

$x$ in equation

$(1)$

$\Rightarrow 99+y = 180$

$\Rightarrow y = 180-99$

$\Rightarrow y = 81$

Hence,

$x = 99^0$ and

$y = 81^0$

The pair of linear equations for the given problem are

$x + y = 180$ and

$x - y = 18$ .

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.

$27$ for a book kept for seven days, while Susy paid Rs.

$21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

0%
Fixed charge: Rs. $12$ , Charge for each extra day: Rs. $9$
0%
Fixed charge: Rs. $15$ , Charge for each extra day: Rs. $3$
0%
Fixed charge: Rs. $18$ , Charge for each extra day: Rs. $3$
0%
Data insufficient

Explanation

Let the fixed charge for

$3$ days

$=x$ and additional charge

$=y$
According to the question

$\Rightarrow x+4y=27....eq1$

$\Rightarrow x+2y=21....eq2$
Subtract

$eq1$ and

$eq2$

$\Rightarrow \left ( 3x+4y=27 \right )-\left ( 3x+2y=21 \right )\Rightarrow 2y=6\Rightarrow y=3$
put

$y=3$ in

$eq1$

$\Rightarrow x+4\times3=27\Rightarrow x=15$
Fixed charges

$=$ Rs.

$15$
The charge for each extra day

$=$ Rs.

$3$

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs.What are the fixed charges and the charge per kilometer? How much does a person have to pay for traveling a distance of 25 km?

0%
Fixed charge is Rs. $5$ and the charge per kilometer is Rs. $10$ For $25$ km person have to pay Rs. $255$ .
0%
Fixed charge is Rs. $10$ . and the charge per kilometer is Rs. $5$ . For $25$ km person have to pay Rs $135$ .
0%
Fixed charge is Rs. $15$ and the charge per kilometer is Rs. $5$ . For $25$ km person have to pay Rs. $140$ .
0%
Fixed charge is
Rs. $50$ . and the charge per kilometer is Rs $20$ . For $25$ km person have to pay Rs $50$

Explanation

Let fixed charge be Rs x and charge per km be Rs y.

$\Rightarrow x + 10y = 105 ...(1)$

$\Rightarrow x + 15y = 155 ..(2)$ .

Now From eq 2

$\Rightarrow 15y = 155-x$

$\Rightarrow y = \dfrac{155-x}{15}...(3)$

Substituting y from eq3 in eq1

$\Rightarrow x+\dfrac{10\left ( 155-x \right )}{15} = 105$

$\Rightarrow 15x+1550-10x =1575$

$\Rightarrow 5x = 1575-1550$

$\Rightarrow 5x = 25$

$\Rightarrow x = 5$

Substituting x in eq2

$\Rightarrow 5+15y = 155$

$\Rightarrow 15y = 155-5$

$\Rightarrow 15y = 150$

$\Rightarrow y = 10$

Hence

$x= 5$ and

$y = 10$

Fixed charge is Rs.

55 and the charge per kilometer is Rs.

1010

For

2525 km person have to pay =

$5+10 \times 25= 255 Rs.$

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Meena went to a bank to withdraw Rs.

$2000.$ She asked the cashier to give her Rs.

$50$ and Rs.

$100$ notes only. Meena got

$25$ notes in all. Find how many notes of Rs.

$50$ and Rs.

$100$ she received

0%
$15$ notes of Rs. $50$ and $18$ notes of Rs. $100.$
0%
$12$ notes of Rs. $50$ and $18$ notes of Rs. $100$ .
0%
$8$ notes of Rs. $50$ and $25$ notes of Rs. $100.$
0%
$10$ notes of Rs. $50$ and $15$ notes of Rs. $100$ .

Explanation

Let

$x$ and

$y$ be the number of '

$50$ and '

$100$ notes.respectively.

Then,

$x +y=25$ ....(1)
and

$50x + 100y = 2000$ ...(2)
Now multiplying (1) from

$100$ and (2) from (1) subtracting (2) from (1) will get

$\Rightarrow 100\left ( x +y \right )-1\left ( 50x -100y \right ) = 100\times 25 -1\times 2000$

$\Rightarrow 100x+100y-50x-100y = 2500-2000$

$\Rightarrow 50x = 500$

$\Rightarrow x = 10$
Now putting

$x$ in (1)

$\Rightarrow 10+y = 25$

$\Rightarrow y = 25-10$

$\Rightarrow y = 15$
Hence Meena Received

$10$ notes of Rs.

$50$ and

$15$ notes of Rs.

$100$ .

Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p. How many boys are there ?

0%
$52$
0%
$54$
0%
$68$
0%
$62$

Explanation

Let number of boys be

$x$ and number of girls be

$y$
Since, Total number of children's are

$150$

$\therefore$

$x+y=150$ ....(1)
Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p

$\therefore$ Out of

$Rs. 58$ Boys will get Rs.

$0.25 x$ and girls will get Rs.

$0.5y$ .

$\therefore$

$\dfrac{25}{100}x+\dfrac{50}{100}y=58$

$\therefore$

$x+2y=232$ ...(2)
Subtract equation (1) from equation (2), we get

$y = 82$

And put

$y = 82$ in equation (1)

$x=68$ and

$y=82$
Number of boys are

$68$

A fraction becomes

$\dfrac {9}{11}$ , if

$2$ is added to both the numerator and the denominator. If

$3$ is added to both the numerator and the denominator it becomes

$\dfrac {5}{6}$ . Find the fraction.

0%
$\dfrac{7}{9}$
0%
$\dfrac{5}{7}$
0%
$\dfrac{9}{11}$
0%
None of these

Explanation

Let $\dfrac {x}{y}$ be the fraction, where $x$ and $y$ are positive integers.

$\Rightarrow \dfrac {x+2}{y+2}=\dfrac {9}{11}$ ..(1)
$\Rightarrow \dfrac {x+3}{y+3}=\dfrac {5}{6}$ ..(2)

From (1) we get
$\Rightarrow 11x+22 = 9y+18$
$\Rightarrow 11x-9y = -4$ ...(3)

From (2) we get
$\Rightarrow 6x +18 = 5y +15$
$\Rightarrow 6x-5y = -3$ ...(4)

From (4) will get
$y = \dfrac{3+6x}{5}$

Now substituting $y$ in (3),

$\Rightarrow 11x-\dfrac{9\left ( 3+6x \right )}{5} = -4$

$\Rightarrow 55x-27-54x = -20$

$\Rightarrow 55x-54x = -20+27$

$\Rightarrow x = 7$

Now substituting $x$ in (4)

$\Rightarrow 6\times 7 -5y = -3$

$\Rightarrow 42-5y = -3$

$\Rightarrow -5y = -3 - 42$

$\Rightarrow -5y = -45$

$\Rightarrow y = 9$

Hence, $x = 7$ and $y = 9$ .

And fraction $\dfrac{x}{y}=\dfrac{7}{9}$

Form the pair of linear equations in the following problem, and find their solutions (if they exist) by the elimination method:

The sum of the digits of a two-digit number is

$9$ . Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

0%
Number is $18$
0%
Number is $36$
0%
Number is $81$
0%
None of these

Explanation

Let the unit's digit of the number

$=x$
Ten's digit of the number

$=y$

$\therefore$ Number

$=10y+x$ or

$yx$ .
Sum of two digits are

$9$ .

$\therefore x+y=9$ ........(1)

The number obtained by reversing the order of digit

$\Rightarrow 10x+y$ or

$xy$ .
According to the second condition, we have

$9\left(10y+x \right)=2\left(10x+y \right)$

$\Rightarrow 90y+9x=20x+2y$

$\Rightarrow 88y-11x=0$
Divide by

$11$ ,

$\Rightarrow 8y-x=0.......$ (2)

Adding both the equation (1) and (2) we get,

$\Rightarrow 9y=9$

$\Rightarrow y=1$

Put

$y=1$ in eq. (1) we get,

$\Rightarrow x+1=9 \Rightarrow x=8$
Hence, the original number is

$10y+x=10\times1+8=18$ or

$yx=18$ .

Five years hence, the age of jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

0%
Present age of Jacob: $50$ years and Present age of Jacob's son: $15$ years
0%
Present age of Jacob: $50$ . years and Present age of Jacob's son: $10$ years
0%
Present age of Jacob: $40$ years and Present age of Jacob's son: $10$ years
0%
Present age of Jacob: $35$ years and Present age of Jacob's son: $15$ years

Explanation

Let

$x$ be the present age of Jacob's son and

$y$ be the present age of Jacob. Then as per the question

$(y + 5) =3(x + 5)\ ......(1)$

$(y- 5) = 7 (x - 5)\ ......(2)$

From equation

$(1)$ ,

$\Rightarrow y +5 = 3x +15$

$\Rightarrow 3x-y = -10\ ......(3)$

From equation

$(2)$ ,

$\Rightarrow y-5 = 7x-35$

$\Rightarrow 7x-y = 30\ .......(4)$

from equation

$(3)$ we get

$\Rightarrow -y = -10 -3x$

$\Rightarrow y = 10 +3x$

Now Substitute

$y$ in

$(4)$

$\Rightarrow 7x-\left ( 10+3x \right ) = 30$

$\Rightarrow 7x -10-3x = 30$

$\Rightarrow 4x = 30+10$

$\Rightarrow 4x = 40$

$\Rightarrow x = 10$

Now substituting

$x$ in (1) will give

$\Rightarrow 3\times 30-y = -10$

$\Rightarrow 30 - y = -10$

$\Rightarrow -y = -40$

$\Rightarrow y = 40$

Hence, Jacob's son's current age

$=$

$10$ and Jacob's Current age

$=$

$40$ .

Two years ago, a father was five times as old as his son. Two years later, his age will be

$8$ more than

$3$ times the age of the son. Find the present age of father.

0%
$22$ years
0%
$60$ years
0%
$39$ years
0%
$42$ years

Explanation

Let the present age of the father be

$x$ and that of the son be

$y$

Two years ago their ages were

$(x-2)$ and

$(y-2)$ respectively

Then, according to question, we have

$(x-2)=5(y-2)$

$\Rightarrow x-2=5y-10$

$\Rightarrow x-5y=-8......(1)$

Two years later their ages will be

$(x+2)$ and

$(y+2)$ respectively

Then again,

$(x+2)=3(y+2)+8$

$\Rightarrow x+2=3y+6+8$

$\Rightarrow x-3y=12.....(2)$

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

$\Rightarrow (x-3y)-( x-5y)=12-(-8)$

$\Rightarrow x-3y- x+5y=12+8$

$\Rightarrow 2y=20$

$\Rightarrow y=10$

Putting

$y = 10$ in equation (2), we get

$x-3\times 10=12$

$\Rightarrow x=12+30$

$\Rightarrow x=42$
The present ages of father and son are

$42$ years and

$10$ years respectively.

So,

$Op-D$

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Five years ago Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

0%
Nuri is $40$ years old and Sonu is $10$ years old.
0%
Nuri is $50$ years old and Sonu is $20$ years old.
0%
Nuri is $70$ years old and Sonu is $30$ years old.
0%
Nuri is $60$ years old and Sonu is $10$ years old.

Explanation

Let present age of Nuri=

$x$ years
Let present age of sonu=

$y$ years
Five years ago
Nuri was

$\left(x-5 \right) years$
Sonu was

$\left(y-5 \right) years$
According to the first condition

$\Rightarrow \left(x-5 \right)=3\left(y-5 \right)$

$\Rightarrow x-5=3y-15 \Rightarrow x-3y=-10.........eq1$
Ten years later
Nuri will be

$\left(x+10 \right) years$
Sonu will be

$\left(y+10 \right)$

$\Rightarrow \left(x+10 \right)=2\left(y+10 \right)$

$\Rightarrow x+10=2y+20 \Rightarrow x+2y=10.....eq2$
Subtracting

$eq 1 and eq2$

$\Rightarrow -y=-2- \Rightarrow y=20$
put y=20 in

$eq1$

$x-3y=10 \Rightarrow x=10-60 \Rightarrow x=50$
Hence, present age of Nuri= 50 years
Let present age of sonu=20 years

The sum of the digits of a two-digit number is

$5$ . The digit obtained by increasing the digit in tens' place by unity is one-eighth of the number. Then the number is

0%
less than $30$
0%
lies between $30$ and $40$
0%
more than $37$
0%
lies between $40$ and $50$

Explanation

$x+y=5$
and

$x+1=\dfrac {1}{8}(10x+y)$ .........(1)
or

$8x+8=10x+y$
or

$2x+y=8$ .........(2)
Subtract equation (1) from equation (2)

$2x+y=8$

$+ x+y=5$

$\underline {(-) (-) (-)}$

$x=3$

$\therefore y=2$
Number is

$32$ .

A fraction becomes

$\dfrac {1}{3}$ when

$1$ is subtracted from the numerator and it becomes

$\dfrac {1}{4}$ when

$8$ is added to its denominator. Find the fraction.

0%
$\dfrac{1}{4}$
0%
$\dfrac{4}{17}$
0%
$\dfrac{5}{12}$
0%
None of these

Explanation

Let the fraction be

$\dfrac {x}{y}$

Now as per the question we will have

$\dfrac {x-1}{y} = \dfrac {1}{3}..(1)$

$\dfrac {x}{y+8} = \dfrac {1}{4}..(2)$

From eq1 will get

$3x-y = 3...(3)$

From eq2 will get

$4x-y = 8...(4)$

Now subtract eq4 from eq3

$\Rightarrow 3x-y-4x+y = 3-8$

$\Rightarrow -x = -5$

$\Rightarrow x = 5$

Now put x in eq 4 will get

$\Rightarrow 4 \times 5-y = 8$

$\Rightarrow 20-y =8$

$\Rightarrow -y = 8-20$

$\Rightarrow -y = -12$

$\Rightarrow y = 12$

Hence fraction will be

$\dfrac {5}{12}$

The difference between two numbers is

$7$ and their sum is

$35$ . What will be their product?

0%
$324$
0%
$294$
0%
$79$
0%
$245$

Explanation

$x + y = 35$

$x - y = 7$

Add both the equations

$\displaystyle \Rightarrow$

$2x = 42$

$\displaystyle \Rightarrow$

$x = 21$

$\displaystyle \therefore$

$21 + y = 35$

$\displaystyle \Rightarrow$

$y = 14$

$\displaystyle \therefore$

$x$

$\displaystyle \times$

$y = 21$

$\displaystyle \times$

$14 = 294$

$\displaystyle \frac{x+y}{x-y}=\frac{5}{3}\: and\: \frac{x}{\left ( y+2 \right )}=2$ the value of (x , y) is

0%
(4, 1)
0%
(2, 8)
0%
(1, 4)
0%
(8, 2)

Explanation

$\dfrac{x}{\left ( y+2 \right )}=2$

$x=2y+4$ .....

$(1)$

$\displaystyle \dfrac{x+y}{x-y}=\frac{5}{3}$ .....

$(2)$
From eqs

$(1)$ and

$(2)$

$\dfrac { 2y+4+y }{ 2y+4-y } =\dfrac { 5 }{ 3 }$

$\dfrac { 3y+4 }{ y+4 } =\dfrac { 5 }{ 3 }$

$9y+12=5y+20$

$4y=8$

$y=2$
and

$x=2y+4=2\times 2+4=8$
Hence,

$(x , y)=(8,2)$

Sum of two numbers is

$35$ and their difference is

$13$ . Then the numbers are

0%
$10, 25$
0%
$22, 13$
0%
$24, 11$
0%
$20, 15$

Explanation

${\textbf{Step -1: Framing equations}}$

$\text{Without loss of generality, suppose,}$ $x$ ${\text{and}}$ $y$ ${\text{be two numbers and }} x > y.$

${\text{Given that, sum of two numbers is 35}}{\text{.}}$

$\Rightarrow x + y = 35$ $\ldots \left( 1 \right)$

${\text{Difference between these two numbers is 13}}{\text{.}}$

$\Rightarrow x - y = 13$ $\ldots \left( 2 \right)$

${\textbf{Step -2: Adding equation }}\left( \mathbf 1 \right)$ ${\textbf{and equation }}\left(\mathbf 2 \right)\textbf .$

$\Rightarrow x + y + \left( {x - y} \right) = 35 + 13$

$\Rightarrow 2x = 48$

$\Rightarrow x = 24$

${\text{Substituting the value of}}$ $x$ ${\text{in equation }}\left( 1 \right)$ ${\text{we get,}}$

$\Rightarrow 24 + y = 35$

$\Rightarrow y = 35 - 24$

$\Rightarrow y = 11$

${\text{Hence, these two numbers are 24 and 11}}{\text{.}}$

${\textbf{ Hence, option (C) is correct answer.}}$

In covering a distance of

$30$ km, Abhay takes

$2$ hours more than Sameer. If Abhay doubles his speed, then he would take

$1$ hour less than Sameer. What is Abhay's speed? (in km/hr)

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Let Abhay's speed be

$x$ km/hr and Sameer's speed by

$y$ km/hr
Then

$\displaystyle \frac{30}{x}-\frac{30}{y}=2$ ..... (i)

and

$\displaystyle \frac{30}{y}-\frac{30}{2x}=1$ ......(ii)
Adding equation (i) and (ii), we get

$\displaystyle \frac{30}{x}-\frac{30}{2x}=3$

$\Rightarrow \displaystyle \frac{30}{2x}=3$

$\Rightarrow 2x=10$

$\Rightarrow x=5$ km/hr

If Rs.

$50$ is distributed amount

$150$ children giving

$50$ p to each boy and

$25$ p to each girl. Then the number of boys is

0%
$25$
0%
$40$
0%
$36$
0%
$50$

Explanation

Let number of boys

$= x$
Number of girls

$= y$
Then,

$x + y = 150$ .....(1)
and

$0.5x + 0.25 y = 50$

$\Rightarrow 2x + y = 200$ ......(2)
Subtracting (1) from (2), we get

$x = 50$
Thus, number of boys are

$50$ .

The electricity bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed. When in a certain month 540 units are consumed, the bill is Rs.In another month 620 units are consumed and the bill is Rs.In yet another month 500 units are consumed then, the bill for that month would be

0%
Rs. 1560
0%
Rs. 1680
0%
Rs. 1840
0%
Rs. 1950

Explanation

Let the fixed amount be Rs.

$x$ and the cost of each unit be Rs.

$y$ . Then,

$540y + x = 1800$ ...(i) and

$620y + x = 2040$ ...(ii)
On subtracting (i) from (ii), we get

$80y = 240 \Rightarrow y = 3$
Putting y = 3 in (i), we get :

$540 \times 3 + x = 1800 \Rightarrow x = \left(1800 - 1620\right) = 180$

$\therefore$ Fixed charges

$= Rs. 180,$ Charge per unit

$= Rs. 3$
Total charges for consuming

$500$ units

$= Rs.\left (180 + 500 \times 3\right) = Rs. 1680$

Albert buys

$4$ horses and

$9$ cows for Rs.

$13400$ . If he sells the horses at

$10\%$ profit and the cows at

$20\%$ profit, then he earns a total profit of Rs.

$1880$ . The cost of a horse is

0%
Rs. $1000$
0%
Rs. $2000$
0%
Rs. $2500$
0%
Rs. $3000$

The real numbers

$x$ and

$y$ are such that

$\displaystyle x+\frac{2}{y}=\frac{8}{3}$ and

$\displaystyle y+\frac{2}{x}=3$ .

The value of

$xy$ , is

0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{16}{9}$
0%
$2$
0%
$4$

Explanation

$\displaystyle x+\frac{2}{y}=\frac{8}{3}$ ...(i)

$\displaystyle y+\frac{2}{x}=3\Rightarrow y=\frac{3x-2}{x}$ ...(ii)

From (i)

$\displaystyle x+\frac{2x}{3x-2}=\frac{8}{3}$

$\Rightarrow \displaystyle 3(3x^2-2x+2x)=8(3x-2)$

$\Rightarrow \displaystyle 9x^2-24x+16=0$

$\Rightarrow \displaystyle (3x-4)^2=0$

$\Rightarrow \displaystyle x=\frac43$

From (ii), we get

$y=\dfrac{3\times\frac43-2}{\frac43}=\dfrac32$
Now,

$xy=\dfrac43\times\dfrac32=2$
Hence, option C is correct.

Mr. Joshi has

$430$ cabbage-plants which he wants to plant out. Some

$25$ to a row and the rest

$20$ to a row. If there are to be

$18$ rows in all how many rows of

$25$ will there be?

0%
$10$
0%
$14$
0%
$8$
0%
$12$

Explanation

Let '

$x$ ' rows of

$25$ and '

$y$ ' rows of

$20$ .

Given,

$25x + 20y = 430$

$5x+4y=86$ ....(1)

Also

$x + y = 18$ ....(2)

Multiply equation (1) by

$4$ , we get

$4x+4y=72$ ....(3)

Subtract equations (1) and (3), we get

$x=14$

Hence, answer is

$14$ .

$\displaystyle r+s=t$ and

$\displaystyle x+t=y-2s$ , then which of the following must be true ?

0%
$\displaystyle r+2s-x=y-t$
0%
$\displaystyle r+2s-t=x+y$
0%
$\displaystyle x+r=y+s$
0%
$\displaystyle x+r=y-3s$

Explanation

$\displaystyle r+s=t$ ---eq1 and

$\displaystyle x+t=y-2s$ ----eq2
putting the value of t in eq 2

$\displaystyle x+r+s=y-2s$

$\displaystyle x+r=y-3s$
Answer (D)

$\displaystyle x+r=y-3s$

10 years ago the age of the father was 5 times that of the son. 20 years hence the age of the father will be twice that of the son. The present age of the father (in years) is

0%
$40$
0%
$45$
0%
$60$
0%
$70$

Explanation

Let the present age of father be

$x$ years. and that of son be

$y$ years
10 years ago the age of the father was 5 times that of the son.

$\therefore$

$x-10=5(y-10)$

$\therefore$

$x-5y=-40$ ...(1)
20 years hence, the age of the father will be twice that of the son.

$\therefore$

$x+20=2(y+20)$

$\therefore$

$x-2y=20$ ...(2)
Subtracting (1) and (2) we get ,

$y=20$ and

$x=60$

$\therefore$ The present age of father is

$60$ years

What values of x and y satisfies the equations

$\displaystyle \frac{x}{6}+\frac{y}{15}=4$ and

$\displaystyle \frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$

0%
$(6,15)$
0%
$(18,15)$
0%
$(15,18)$
0%
$(12,30)$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$\frac { x }{ 3 } -\frac { y }{ 12 } =4\frac { 3 }{ 4 }$ or

$4x-y=57$ and write it as

$y=4x-57...........(1)$

Substitute the value of

$y$ in Equation

$\frac { x }{ 6 } +\frac { y }{ 15 } =4$ or

$15x+6y=360$ . We get

$15x+6(4x-57)=360$

i.e.

$15x+24x-342=360$

i.e.

$39x=702$

i.e.

$x=18$

Therefore,

$x=18$

Substituting this value of

$x$ in equation 1, we get

$y=72-57$

i.e.

$y=15$

Hence, the solution is

$x = 18, y =15$ .

A certain two digits number is equal to five times the sum of its digits. If

$9$ were added to the number, its digits would be reversed. The sum of the digits of the number is:

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Let the ones and tens digit of the number be

$y$ and

$x.$

According to question, we have

$10x+y=5\left( x+y \right)$

$\Rightarrow 5x-4y=0\quad\quad\quad\dots(i)$

And,

$10x+y+9=10y+x$

$\Rightarrow 9x-9y=-9$

$\Rightarrow x-y=-1\quad\quad\quad\dots(ii)$

Multiply

$(ii)$ by

$-4$ we get,

$\Rightarrow -4x+4y=4\quad\quad\quad\dots(iii)$

Add equations

$(i)$ and

$(iii),$

$\left( {5x - 4y} \right) + \left( { - 4x + 4y} \right) = 0 + 4$

$\Rightarrow 5x - 4x - 4y + 4y = 4$

$\Rightarrow x = 4$

Substitute

$x=4$ in equation

$(ii),$

$4 - y = - 1$

$\Rightarrow - y = - 5$

$\Rightarrow y=5$

Thus, the sum of the digits of the number is equal to

$x+y=9.$ .

The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father are respectively,

0%
$4$ and $24$
0%
$5$ and $30$
0%
$6$ and $36$
0%
$3$ and $24$

Explanation

Let the present age of father's be

$x$ years and present age of son's be

$y$ years.
According to the problem

$x=6y$
After

$4$ years

$x+4 = 4(y+4)$
Hence we get two equations

$x=6y$ .........

$(1)$

$x+4 = 4(y+4)$ .....

$(2)$
Simplifying eq

$(2)$

$x+4 = 4y+16$

$x-4y = 12$
Put

$x=6y$ in eq

$(2)$ , we get

$6y-4y = 12$

$2y=12$

$y=6$ years
and

$x=6y=36$ years
Present age of son

$=6$ years
and present age of father

$=36$ years

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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