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

Published on: 26/05/2021

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Questions:

Generic 5

5 x 5 = 25

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?

Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour, if she travels 2 km by rickshaw and the remaining distance by bus.
On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 min longer. Find the speed of rickshaw and of the bus.

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 \(y=\lambda x+5.\)

Write a pair of linear equations which has the unique solution x=-1 and y=3. How many such pairs can you write?

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Answers Key:

Generic 5

5 x 5 = 25

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

10 km/h, 40 km/h

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, \(\lambda =-\frac { 1 }{ 2 } \)

3x+2y=3, x+y=2; Infinitely many pairs are possible.