MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 5

Solve the following equations by substitution method.

$3y-2x=9; \, \, 2x+5y=15$

0%
$x = 0, y= 2$
0%
$x = 5, y= 2$
0%
$x = 1, y= 3$
0%
$x = 0, y= 3$

Explanation

$3y -2x =9; \, \, 2x +5y =15$

$y = \dfrac {9+2x}{3}$ Put it in second equation.

$2x + 5 \times \dfrac {9+2x}{3} =15$

$6x +45 +10x = 45$

$16x =0$

$\boxed{x =0}$

$y = \dfrac {9+2x}{3} = \dfrac {9+2 (0)}{3} = 3 \\ \boxed{y=3}$

Solve the following equations by substitution method.

$x=2y-1; \, \, y=2x-7$

0%
$x = 8, y= 3$
0%
$x = 2, y= 3$
0%
$x = 5, y= 3$
0%
$x = 1, y= 3$

Explanation

$x=2y-1; \, \, y=2x - 7$

$x = 2y-1$ Put it in second equation.

$y = 2 \times (2y -1) -7$

$y =4y -9$

$\boxed{y = 3}$

and

$x =2y -1 = 2 \times 3 -1 =5$

$\boxed{x=5}$

$\boxed{\,x = 5 \ and \ y = 3 \,}$

Solve the following equations by substitution method.

$3a-2b=-10; \, \, 2a+3b=2$

0%
$a = -2, b= 2$
0%
$a = -1, b= 2$
0%
$a= -4, b= 2$
0%
$a = -7, b= 2$

Explanation

$3a -2b =-10; \, \, 2a +3b =2$

$a = \dfrac {2b-10}{3} \Rightarrow$ Put it in second equation.

$2 \times \dfrac {2b-10}{3} +3b=2$

$4b - 20 +9b = 6$

$13b =26$

$\therefore b=2$

$a = \dfrac {2b-10}{3} = \dfrac {2(2)-10}{3} = -2$

$a = -2$ and

$b = 2$

Solve the following simultaneous equations by the method of equating coefficients.

$5x+7y=17;\,\, 7x+5y=19$

0%
$x = 1, y= 1$
0%
$x = 4, y= 1$
0%
$x = 3, y= 1$
0%
$x = 2, y= 1$

Explanation

$5x+7y=17$ ...(i)

$7x+5y=19$ ...(ii)

On multiplying (i) by 7 and (ii) by 5, we get

$35x+49y=119$

$\underline {\underset {-}35x\underset {-}{+}25y=\underset{-}95}$

$24y=24$

$\therefore y=1$

On putting

$y=1$ in (i), we get

$5x+7(1)=17$

$\Rightarrow x=2$

$\therefore x=2,y=1$

Solve the following equations by substitution method.

$5x + 3y = 21; 2x - y = 4$

0%
$x = 3, y = 2$
0%
$x =- 3, y = 2$
0%
$x = 3, y = -2$
0%
None of these

Explanation

$5x+3y=21$ .........(1)

$2x-y=4$ ...............(2)

$y=2x-4$

$5x+3(2x-4)=21$ .........putting value of

$y$ in (1)

$\Rightarrow 5x+6x-12=21$

$\Rightarrow 11x=33$ .......................

$\boxed{x=3}$

Putting value of

$x$ in (2)

$6-y=4$ ..........................

$\boxed{y=2}$

Solve the following simultaneous equations by the method of equating coefficients.

$\displaystyle \frac{x}{3}+5y=13; \, \, 2x+\frac{y}{2}=19$

0%
$x = 1, y= 2$
0%
$x = 9, y= 2$
0%
$x = 4, y= 2$
0%
$x = 8, y= 2$

Explanation

$\displaystyle \frac{x}{3}+5y=13$ ....(i)

$\displaystyle 2x+\frac{y}{2}=19$ .....(ii)

Multiplying equation (i) by

$3$ and equation (ii) by

$2$ . we get,

$x+15y=39$ ......(iii)

$4x+y=38$ ......(iv)

To make the coefficients of x equal multiplying equation (iii) by

$4$ . we get ,

$4x+60y=156$ .....(v)

Subtract equation (iii) from equation (iv). we get,

$4x+60y-(4x+y)=156-38$

$59y = 118 \Rightarrow y=2$

Substitute

$y=2$ in equation (iii). we get,

$x+15(2)=39 \Rightarrow x=9$

$\therefore x=9 , y=2$

Solve the following simultaneous equations by the method of equating coefficients.

$x-2y=-10; \, \, 3x-5y=-12$

0%
$x = 16, y= 18$
0%
$x = 32, y= 18$
0%
$x = 26, y= 18$
0%
$x = 22, y= 18$

Explanation

$x-2y=-10$ ...(i)

$3x-5y=-12$ ...(ii)

On multiplying (i) by 3, we get

$3x-6y=-30$ ...(iii)

now subtracting (ii) from (iii)

$3x-6y=-30$

$\underline {\underset {-}3x\underset {+}{-}5y=\underset{+}-12}$

$-y=-18$

$\therefore y=18$

On putting

$y=18$ in (ii), we get

$3x-5(18)=-12$

$3x-90=-12$

$3x=78$

$x=26$

$\therefore x=26,y=18$

Find the values of

$(x+y)$ and

$(x-y)$ without actually solving for

$x$ and

$y$ .

$8x-7y=11; \, \, 7x-8y=4$

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

Explanation

Rewriting the given equations, we get

$8x-7y=11$ ...(i)

$7x-8y=4$ ...(ii)

On adding (i) and (ii) , we get

$15x-15y=15$

$\Rightarrow x-y=1$

On subtracting (ii) from (i), we get

$x+y=7$

Find the values of

$(x+y)$ and

$(x-y)$ without actually solving for

$x$ and

$y$ .

$13x+15y=114 \quad and \quad15x+13y=110$

0%
$x+y= 9, x-y= -2$
0%
$x+y= 8, x-y= -2$
0%
$x+y= 3, x-y= -2$
0%
$x+y= 1, x-y= -2$

Explanation

$13x+15y=114$ ...(i)

$15x+13y=110$ ...(ii)

On adding (i) and (ii) , we get

$28x+28y=224$

$\Rightarrow x+y=\frac {224}{28} =8$

On subtracting (i) from (ii) , we get

$2x-2y=-4$

$\Rightarrow x-y=-\frac 42 =-2$ .

Find the values of

$(x+y)$ and

$(x-y)$ without actually solving for

$x$ and

$y$ .

$15x-12y=69; \, \, 12x-15y=39$

0%
$x+y = 10,x- y= 4$
0%
$x+y = 5,x- y= 4$
0%
$x+y = 13,x- y= 4$
0%
$x+y = 19,x- y= 4$

Explanation

$15x-12y=69$ ...(i)

$12x-15y=39$ ...(ii)

On adding (i) and (ii) , we get

$27x-27y=108$

$\Rightarrow x-y=4$

On subtracting (ii) from (i) , we get

$3x+3y=30$

$\Rightarrow x+y=10$

Find the values of

$(x+y)$ and

$(x-y)$ without actually solving for

$x$ and

$y$ .

$9x+11y=78; \, \, 11x+9y=82$

0%
$x+y = 2, x-y= 2$
0%
$x+y = 8, x-y= 2$
0%
$x+y = 7, x-y= 2$
0%
$x+y = 3, x-y= 2$

Explanation

$9x+11y=78$ ...(i)

$11x+9y=82$ ...(ii)

On adding (i) and (ii) , we get

$20x+20y=160$

$\Rightarrow x+y=8$

On subtracting (i) from (ii) , we get

$2x-2y=4$

$\Rightarrow x-y=2$

Solve the following simultaneous equations by the method of equating coefficients.

$4x+y=34; \, \, x+4y=16$

0%
$x = 8, y= 2$
0%
$x = 4, y= 2$
0%
$x = 9, y= 2$
0%
$x = 2, y= 2$

Explanation

$4x+y=34$ ...(i)

$x+4y=16$ ...(ii)

On multiplying (ii) by 4, we get

$4x+16y=64$ ...(iii)

now subtracting (i) from (iii)

$4x+16y=64$

$\underline {\underset {-}4x\underset {-}{+}y=\underset{-}34}$

$15y=30$

$\therefore y=2$

On putting

$y=2$ in (i), we get

$4x+2=34$

$x=8$

$\therefore x=8,y=2$

Solve the following simultaneous equations by the method of equating coefficients.

$\displaystyle \frac{2x-y}{3}+3x-2y=3; \, \, 4x-3y+\frac{5x-4y}{4}=1$

0%
$x = 4, y= 5$
0%
$x = 3, y= 5$
0%
$x = 7, y= 5$
0%
$x = 9, y= 5$

Explanation

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

$\displaystyle 4x-3y+\frac{5x-4y}{4}=1$ .....(ii)

Multiplying equation (i) by

$3$ and equation (ii) by

$4$ . we get,

$2x-y+9x-6y=9 \Rightarrow 11x-7y=9$ .......(iii)

$16x-12y+5x-4y=4 \Rightarrow 21x-16y=4$ ....(iv)

Multiplying equation (iii) by

$16$ and equation (iv) by

$7$ . we get,

$176x-112y=144$

$\underline{\underset {-}147x \underset {+}{-}112y=\underset {-}28}$

$29x=116 \Rightarrow x=4$

Substitute

$x=4$ in equation (iii). we get,

$11(4)-7y=9 \Rightarrow 7y=35 \Rightarrow y=5$

$\therefore x=4 , y=5$

Solve for

$x$ and

$y$ .

$5x+6y=7x+3y=3(x+6y-6)$

0%
$x =1 , y= 3$
0%
$x = 2, y= 6$
0%
$x = 3, y= 2$
0%
None of these

Explanation

Given,

$5x+6y=7x+3y=3(x+6y-6)$

By taking first and third we get,

$5x+6y = 3(x+6y-6)$

$\Rightarrow 5x+6y = 3x+18y-18$

$\Rightarrow 2x-12y=-18$ ....(i)

By taking last two we get,

$7x+3y=3(x+6y-6)$

$7x+3y=3x+18y-18$

$\Rightarrow 4x-15y=-18$ .....(ii)

On multiplying equation (i) by

$2$ . we get,

$4x-24y=-36$ ....(iii)

Subtracting (ii) from (iii). we get,

$4x-24y-(4x-15y)=-36-(-18)$

$\therefore -9y=-18 \Rightarrow y=2$

On substituting

$y=2$ in (i). we get,

$2x-12(2)=-18$

$\Rightarrow 2x=6$

$\therefore x=3$

$\therefore x=3 , y=2$

Solve the following simultaneous equations:

$\displaystyle \frac{x}{3}+\frac{y}{4}=4; \, \, \frac{5x}{6}-\frac{y}{8}=4$

0%
$x = 5, y= 8$
0%
$x = 2, y= 8$
0%
$x = 1, y= 8$
0%
$x = 6, y= 8$

Explanation

$\displaystyle \frac{x}{3}+\frac{y}{4}=4 \Rightarrow 4x+3y=48$ ....(i)

$\displaystyle \frac{5x}{6}-\frac{y}{8}=4 \Rightarrow 20x-3y=96$ ....(ii)

Since, coefficients of

$y$ are equal in both equations but with opposite sign.

So on adding (i) and (ii). we get,

$4x+3y+20x-3y=48+96$

$\Rightarrow 24x=144$

$\Rightarrow x=6$

On substituting

$x=6$ in (i). we get,

$4(6)+3y=48 \Rightarrow 3y=24 \Rightarrow y=8$

$\therefore x=6 , y=8$

Solve the following simultaneous equation.

$12x+17y=53; \, \, 17x+12y=63$

0%
$x = 3,\, y=1$
0%
$x = 2,\, y=1$
0%
$x = 1,\, y=1$
0%
$x = 9,\, y=1$

Explanation

$12x+17y=53$ ...(i)

$17x+12y=63$ ...(ii)
On adding (i) and (ii) , we get

$29x+29y=116$

$\Rightarrow x+y=4$ ....(iii)
On subtracting (i) from (ii) , we get

$5x-5y=10$

$\Rightarrow x-y=2$ ....(iv)
On adding (iii) and (iv) , we get

$2x=6$

$\Rightarrow x=3$
By putting

$x=3$ in (iii), we get

$3+y=4$

$\Rightarrow y=1$

Solve the following pair of equations

$64p-45q=289; \, \, 45p-64q=365$

0%
$p=0, \, q=-5$
0%
$p=4, \, q=-5$
0%
$p=1, \, q=-5$
0%
$p=2, \, q=-5$

Explanation

Solution:

$64p-45q=289$ .............(1)

$45p-64q=365$ .............(2)

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

$109p-109q=654$
i.e.,

$p-q=6$ .........(3)

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

$19p+19q=-76$
i.e.,

$p+q=-4$ .........(4)

Adding equation (3) and (4) we get,

$p=1$
and substituting

$p=1$ in equation (3) we get,

$q=-5$

Solve the following pair of equations

$31x-42y=51; \, \, 42x-31y=95$

0%
$x=3, \, y=1$
0%
$x=2, \, y=1$
0%
$x=1, \, y=1$
0%
$x=6, \, y=1$

Explanation

$31x-42y=51$ ...(i)

$42x-31y=95$ ...(ii)
On adding (i) and (ii) , we get

$73x-73y=146$

$\Rightarrow x-y=2$ ....(iii)
On subtracting (i) from (ii) , we get

$11x+11y=44$

$\Rightarrow x+y=4$ ....(iv)
On adding (iii) and (iv) , we get

$2x=6$

$\Rightarrow x=3$
By putting

$x=3$ in (iv), we get

$3+y=4$

$\Rightarrow y=1$

$\therefore x=3, y=1$

From the following figure, we can say:

$\displaystyle \frac{x}{3}+\frac{y}{4}=4; \, \, \frac{5x}{6}-\frac{y}{8}=4$

0%
$x=1, \, y=8$
0%
$x=7, \, y=8$
0%
$x=6, \, y=8$
0%
$x=3, \, y=8$

Explanation

Given ,

$\displaystyle \frac{x}{3}+\frac{y}{4}=4; \, \, \frac{5x}{6}-\frac{y}{8}=4$
Rewriting the given equations, we get

$4x+3y=48$ ...(i)

$20x-3y=96$ ...(ii)
On adding (i) and (ii), we get

$24x=144$

$\Rightarrow x=6$
On putting

$x=6$ in (i), we get

$4(6)+3y=48$

$\Rightarrow y=8$

$\therefore x=6,y=8$

Solve the following pair of equations

$\displaystyle \frac{2x}{15}+\frac{y}{5} =4; \, \, \frac{x}{3}=\frac{y}{2}$

0%
$x=15, \, y=10$
0%
$x=20, \, y=10$
0%
$x=5, \, y=10$
0%
$x=13, \, y=10$

Explanation

Given,

$\displaystyle \frac{2x}{15}+\frac{y}{5} =4; \, \, \frac{x}{3}=\frac{y}{2}$
Rewriting the given equations, we get

$2x+3y=60$ ...(i)

$2x-3y=0$ ...(ii)
On adding (i) and (ii) , we get

$4x=60$

$\Rightarrow x=15$
On putting

$x=15$ in (i), we get

$2(15)+3y=60$

$\Rightarrow y=10$

$\therefore x=15,y=10$

Solve the following pair of equations.

$35x-37y=68; \, \, 35y-37x+76=0$

0%
$x=2, \, y=1$
0%
$x=3, \, y=1$
0%
$x=6, \, y=1$
0%
$x=1, \, y=1$

Explanation

Rewriting the given equations, we get

$35x-37y=68$ ...(i)

$-37x+35y=-76 \Rightarrow 37x-35y=76$ ...(ii)
On adding (i) and (ii) , we get

$72x-72y=144$

$\Rightarrow x-y=2$ ....(iii)
On subtracting (i) from (ii) , we get

$2x+2y=8$

$\Rightarrow x+y=4$ ....(iv)
On adding (iii) and (iv) , we get

$2x=6$

$\Rightarrow x=3$
By putting

$x=3$ in (iv), we get

$3+y=4$

$\Rightarrow y=1$

Find the values of

$(m+n)$ and

$(m-n)$ without actually solving for

$m$ and

$n$ .

$17m+13n=133; \, \, 13m+17n=137$

0%
$m+n =5, m-n= -1$
0%
$m+n =1, m-n= -1$
0%
$m+n =2, m-n= -1$
0%
$m+n =9, m-n= -1$

Explanation

$17m+13n=133$ ...(i)

$13m+17n=137$ ...(ii)
On adding (i) and (ii) , we get

$30m+30n=270$

$\Rightarrow m+n=9$
On subtracting (ii) from (i) , we get

$4m-4n=-4$

$\Rightarrow m-n=-1$

Solve the following pair of equations.

$4(x+2)+5(y+2)=20$ and

$5x+4y=7$

0%
$x=7, \, y=-2$
0%
$x=4, \, y=-2$
0%
$x=3, \, y=-2$
0%
$x=2, \, y=-2$

Explanation

$4(x+2)+5(y+2)=20 \Rightarrow 4x+5y=2$ ....(i)

$5x+4y=7$ ........(ii)
On adding (i) and (ii) , we get

$9x+9y=9$

$\Rightarrow x+y=1$ ....(iii)
On subtracting (i) from (ii) , we get

$x-y=5$ ....(iv)
On adding (iii) and (iv) , we get

$2x=6$

$\Rightarrow x=3$
By putting

$x=3$ in (iv), we get

$3-y=5$

$\Rightarrow y=-2$

$\therefore x=3 , y=-2$

Solve the following pair of equations.

$25x-24y=197; \, \, 24x-25y=195$

0%
$x=1,\, y=-3$
0%
$x=9,\, y=-3$
0%
$x=5,\, y=-3$
0%
$x=3,\, y=-3$

Explanation

$25x-24y=197$ ...(i)

$24x-25y=195$ ...(ii)
On adding (i) and (ii) , we get

$49x-49y=392$

$\Rightarrow x-y=8$ ....(iii)
On subtracting (ii) from (i) , we get

$x+y=2$ ....(iv)
On adding (iii) and (iv) , we get

$2x=10$

$\Rightarrow x=5$
By putting

$x=5$ in (iv), we get

$5+y=2$

$\Rightarrow y=-3$

Solve the following pair of equations

$\displaystyle \frac{x}{8} -\frac{y}{7} = \frac{17}{28}; \, \, \frac{x}{7} -\frac{y}{8}=1$

0%
$x=5, \, y=8$
0%
$x=19, \, y=8$
0%
$x=12, \, y=8$
0%
$x=14, \, y=8$

Explanation

Given,

$\displaystyle \frac{x}{8} -\frac{y}{7} = \frac{17}{28}; \, \, \frac{x}{7} -\frac{y}{8}=1$
Rewriting the given equations, we get

$7x-8y=34$ ...(i)

$8x-7y=56$ ...(ii)
On multiplying (i) by 8 and (ii) by 7, we get

$56x-64y=272$

$\underline {\underset {-}56x\underset {+}{-}49y=\underset{-}392}$

$-15y=-120$

$\therefore y=8$
On putting

$y=8$ in (i), we get

$7x-8(8)=34$

$\Rightarrow 7x=34+64$

$\Rightarrow x=14$

$\therefore x=14,y=8$

Solve the following simultaneous equations:

$\displaystyle \frac{2x}{3}+\frac{y}{4}=12; \, \, \frac{4x}{3} - \frac{9y}{4}=2$

0%
$x=12, \, y=5$
0%
$x=15, \, y=8$
0%
$x=13, \, y=4$
0%
$x=14, \, y=9$

Explanation

Rewriting the given equations after taking LCM,

$8x + 3y = 144$ ...

$(1)$

$16x - 27y = 24$ ...

$(2)$

Multiplying

$(1)$ by

$2$

$16x + 6y = 288$ ...

$(3)$

$16x - 27y = 24$ ...

$(2)$

Subtracting

$(3)$ and

$(2)$ ,

$33y = 264$

$y = 8$

Substituting

$y = 8$ in

$(1)$ , we get

$8x + 24 = 144$

$8x = 120$

$x = 15$

$x = 15, y = 8$

Option

$B$

The ratio of the present ages of mother and son is

$12: 5$ . The mother's age at the time of the birth of the son was

$21$ years. Find their present ages.

0%
mother age $=$ $36$ years, son age $=$ $15$ years
0%
mother age $=$ $56$ years, son age $=$ $5$ years
0%
mother age $=$ $46$ years, son age $=$ $25$ years
0%
mother age $=$ $47$ years, son age $=$ $26$ years

Explanation

Let mother age be

$M$ and that of son age be

$S$

Therefore according to question,

$M = S+21$ ...(1)
and

$5M =12S$ ....(2)

Substitute equation (1) in equation (2), we get

$5\left(S+21\right) = 12S$

$\Rightarrow 5S+105 = 12S$

$\Rightarrow 7S= 105$

$\Rightarrow S= 15$

Using equation (1), we get

$M= 15+21 = 36$

Mother age is

$36$ years and Son age is

$15$ years.

A two-digit number is 3 more than six times the sum of its digits. If 18 is added to the number obtained by interchanged by interchanging the digits, we get the original number. Find the number.

0%
25
0%
45
0%
75
0%
None of these

Explanation

Let the tens and the units digits in the number be x and y, respectively.
So, the number may be written as

$10x+y$ .
According to the given condition.

$10x+y=6(x+y)+3 \Rightarrow 4x-5y=3$ .....(i)
If we interchange the digits of Original number then we get new number. i.e

$10y+x$
According to the given condition

$10y+x+18=10x+y$

$\Rightarrow 9x-9y = 18 \Rightarrow x-y=2$ .....(ii)
Now multiplying equation (ii) by

$4$ . we get,

$4x-4y=8$ ....(iii)
On subtracting (iii) from (i). we get,

$4x-5y-(4x-4y)=3-8$

$\therefore y=5$
On substituting

$y=5$ in (ii). we get,

$x-5=2 \Rightarrow x=7$

$\therefore$ number is

$10x+y = 10(7)+5=75$

From the following figure, we can say:

$\displaystyle \frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3}; \, \, \frac{3x}{2}+\frac{2y}{3}=13\frac{1}{3}$

0%
$x=1, \, y=4$
0%
$x=8, \, y=3$
0%
$x=6, \, y=5$
0%
None of these

Explanation

$\displaystyle \frac{2x}{3}+\frac{3y}{2}=8\frac{1}{3} \Rightarrow \frac {4x+9y}{6}=\frac {25}{3} \Rightarrow 4x+9y=50$ .......(i)

$\displaystyle \frac{3x}{2}+\frac{2y}{3}=13\frac{1}{3} \Rightarrow \frac {9x+4y}{6}=\frac {40}{3} \Rightarrow 9x+4y=80$ ......(ii)
Adding (i) and (ii). we get,

$13x+13y=130$

$\therefore x+y=10$ ......(iii)
On subtracting (i) from (ii). we get,

$5x-5y=30$

$\therefore x-y=6$ ......(iv)
Again adding (iii) and (iv). we get,

$2x=16 \Rightarrow x=8$
On substituting

$x=8$ in (iv). we get,

$8-y=6$

$\therefore y=2$

$\therefore x=8 , y=2$

Point

$A$ and

$B$ are

$70\ km$ apart on a highway. A car starts from

$A$ and another car starts from

$B$ at the same time. If they travel in the same direction, they meet in

$7$ hours, but if they travel towards each other they meet in one hour. What are their speeds?

0%
$30\ km/hr$ and $40\ km/hr$
0%
$36\ km/hr$ and $40\ km/hr$
0%
$19\ km/hr$ and $20\ km/hr$
0%
$40\ km/hr$ and $50\ km/hr$

Explanation

Let car speed from point

$A=x\, km/hr$ .
and car speed from point

$B=y\, km/hr$ .

Distance = speed

$\times$ time

$\therefore$ Distance from

$A = x \times 7 = 7x$
Distance from

$B = y \times 7 = 7y$
If they travel in the same direction, they meet in

$7$ hours. Point

$A$ and

$B$ are

$70 km$ apart.

$\Rightarrow 7x-7y=70$

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

$(i)$

if they travel towards each other they meet in one hour. Point

$A$ and

$B$ are

$70 km$ apart.

$x+y=70$ ...

$(ii)$

Adding

$(i)$ and

$(ii)$ . we get

$2x=80$

$\therefore x=40$

substituting

$x=40$ in

$(i)$ . we get

$40-y=10$

$\therefore y=30$

$\therefore x=40 km/hr$ and

$y=30 km/hr$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.