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CBSE 10th Standard Maths Subject Pair of Linear Equation in Two Variables Ncert Exemplar 4 Marks Questions 2021

Published on: 26/05/2021

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Questions:

Generic 5

5 x 5 = 25

Form the pair of linear equation in the following problems and find their solutions graphically.
The cost of 4 pens and 4 pencil boxes is Rs.100. Threetimes the cost of a pen is Rs.15 more than the cost of a pencil box. Find the cost of a pen and a pencil box.

For which values of p and q, will the following pair of linear equations have infinitely many solutions?
4x+5y=2, (2p+71)x+(p+8q)y=2q-p+1

Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs.1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would received Rs.20 more as annual interest. How much money did she invest in each scheme?

Reduce the following pair of equations into a pair of linear equations and solve them
\(\frac { 2xy }{ x+y } =\frac { 3 }{ 2 } ,\quad \frac { xy }{ 2x-y } =\frac { -3 }{ 10 } ;\quad x+y\neq 0,\quad 2x-y\neq 0\)

Find the solution of the pair of equations \(\frac { x }{ 10 } +\frac { y }{ 5 } -1=0\) and \(\frac { x }{ 8 } +\frac { y }{ 6 } =15\) and find \(\lambda ,\) if \(y=\lambda x+5.\)

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Answers Key:

Generic 5

5 x 5 = 25

Cost of one pen = Rs.10, cost of one pencil box = Rs.15

p=-1, q=2

Rs.12000 in scheme A and Rs.10000 in scheme B

The given system of equations is
\(\frac { 2xy }{ x+y } =\frac { 3 }{ 2 } \) and \(\frac { xy }{ 2x-y } =\frac { -3 }{ 10 } \)
\(\Rightarrow\) \(\frac { x+y }{ 2xy } =\frac { 4 }{ 3 } \) and \(\frac { 2x-y }{ xy } =\frac { -10 }{ 3 } \)
\(\frac { 1 }{ y } +\frac { 1 }{ x } =\frac { 4 }{ 3 } \) and \(\frac { 2 }{ y } -\frac { 1 }{ x } =\frac { -10 }{ 3 } \)
Put \(\frac { 1 }{ x } =u\) and \(\frac { 1 }{ y } =v\) then the system of equations becomes u+v=\(\frac { 4 }{3 } \) and -u+2v=\(\frac { -10 }{3 } \)
Now, solve these equation.
x=\(\frac {1}{2}\) and y=\(\frac {-3}{2}\)

By solving both equations. Find the values of x and y and then put these values in \(y=\lambda x+5\) to get required values of \(\lambda .\)
x=340, y=-165, \(\lambda =-\frac { 1 }{ 2 } \)