3 Marks Question - Maths STD 10 Questions - Vidyadip

Questions

3 Marks Question

Take a timed test

86 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Find the value of k for which the following system of equations has a unique solution:

$kx + 2y = 5$

$3x + y = 1$

Answer

The given system of equation is
$kx + 2y - 5 = 0$
$3x + y - 1 = 0$
The given system of equation is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1= k, b_1 = 2, c_1 = -5$
And, $a_2 = 3, b_2 = 1, c_2 = -1$
For a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\therefore\frac{\text{k}}{3}\neq\frac{2}{1}$
$\Rightarrow\text{k}\neq6$
So, the given system of equations will have a unique for all real value of k othere then 6.

View full question & answer→

Question 23 Marks

Find the value of k for which the following system of equations has no solution:

$2x + ky = 11$

$5x - 7y = 5$

Answer

Given,
$2x + ky = 11$
$5x - 7y = 5$
To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For no solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{2}{5}=\frac{\text{k}}{-7}\neq\frac{11}{5}$
$\frac{2}{5}=\frac{\text{k}}{-7}$
$\text{k}=\frac{-14}{5}$
Hence for $\text{k}=\frac{-14}{5}$ the system of equations has no solution.

View full question & answer→

Question 33 Marks

Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.

Answer

Let the present age of father be x years and the present age of son be y years.
Then years ago Father's age = (x - 10) years
Son's age = (y - 10) years
Using the given information we get
x - 10 = 12(y - 10)
⇒ x - 10 = 12y - 120
⇒ x - 12y = -120 + 10
⇒ x - 12y = -110 .....(i)
Ten years hence, Father's age = (x + 10) years
Son's age = (y + 10) years
Using the given information we get
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y = 20 - 10
⇒ x - 2y = 10 ......(ii)
Subtracting equation (i) from equation (ii) we get
-2y + 12y = 10 + 110
⇒ 10y = 120
$\Rightarrow\text{y}=\frac{120}{10}$
⇒ y = 12
Putting y = 12 in equation (ii) we get
x - 2 × 12 = 10
⇒ x - 24 = 10
⇒ x = 10 + 24
⇒ x = 34
Hence, present age of father's is 34 years and present age of son is 12 years.

View full question & answer→

Question 43 Marks

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 Rs. 50 and Rs 100 she received.

Answer

Let the number of Rs. 50 notes and Rs. 100 notes be x and y respectively.
According to the question,
x + y = 25 ....(i)
50x + 100y = 2000 ....(ii)
Multiplying equation (i) by 50, we obtain,
50x + 50y = 1250
Subtracting equation (iii) from equation (ii) we obtain,
50y = 750
y = 15
Substituting the value of y in equation (i) we obtain,
x = 10
Hence, meena received 10 notes of Rs. 50 and 15 notes of Rs. 100
Concept insight: This problem talks about two types of notes, Rs. 50 notes and Rs. 100 notes. And the number of both these notes with meena is not known. So, we denote the number of Rs. 50 notes and Rs. 100 notes by variables x and y respectively. Now two liner equatoion can be formed by the given condition which can be solved by eliminatiing one of the variables.

View full question & answer→

Question 53 Marks

The cost of 4 pens and 4 pencil boxes is ₹ 100. Three times the cost of a pen is ₹ 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and pencil box.

Answer

Let the cost of a pen be ₹ x and the cost of a pencil box be ₹ y.
Then, by given condition.
4x + 4y = 100
⇒ x + y = 25 .....(i)
and 3x = y + 15
⇒ 3x - y = 15 .....(ii)
On adding eq. (i) and (ii) we get
4x = 40
⇒ x = 10
By substituting x = 10, in eq. (i) we get
y = 25 - 10
⇒ y = 15
Hence, the cost of a pen and a pencil box are ₹ 10 and ₹ 15, respectively.

View full question & answer→

Question 63 Marks

The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

Answer

Let numerator and denominator be x and y respectively.
Fraction $=\frac{\text{x}}{\text{y}}$
According to question,
x = y - 4 ....(i)
and 8(x - 2) = (y + 1)
⇒ 8x - 16 = y + 1
⇒ 8x - y - 17 = 0 .....(ii)
Putting x = (y - 4) in (2), we get
⇒ 8(y - 4) - y - 17 = 0
⇒ 8y - 32 - y - 17 = 0
⇒ 7y - 49 = 0
⇒ 7y = 49
$\Rightarrow\text{y}=\frac{49}{7}$
⇒ y = 7
Putting y = 7 in (i) we get
⇒ x = 7 - 4
⇒ x = 3
Thus, the fraction $=\frac{3}{7}$

View full question & answer→

Question 73 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$2x + 3y = 2$

$(k + 2)x + (2k + 1)y = 2(k - 1)$

Answer

The given equations are
$2x + 3y = 2 .....(i)$
$(k + 2)x + (2k + 1)y = 2(k - 1) .....(ii)$
The given equations are of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = 3, c_1 = -2$
and, $a_2 = (k + 2), b_2 = (2k + 1)$, and $c_2 = -2(k - 1)$
For infinitely many solutions
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{(\text{k}+2)}=\frac{3}{(2\text{k}+1)}=\frac{-2}{-2(\text{k}-1)}$
$\Rightarrow\frac{2}{(\text{k}+2)}=\frac{3}{(2\text{k}+1)}$
$\Rightarrow 2(2k + 1) = 3(k + 2)$
$\Rightarrow 4k + 2 = 3k + 6$
$\Rightarrow 4k - 3k = 6 - 2$
$\Rightarrow k = 4$
$\Rightarrow k = 4$
Thus, the given system of equations will have infinitely many solutions, if $k = 4$

View full question & answer→

Question 83 Marks

Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Answer

Let the present age of a man and his son be x and y years respectively.
6 years leave (later)
Age of man = (x + 6) years
Age of his son = (y + 6) years
3 years ago
Age of man = (x - 3) years
Age of his son = (y - 3) years
According to question
(x + 6) = 3(y + 6)
⇒ x + 6 = 3y + 18
⇒ x - 3y - 12 = 0 .....(i)
and, (x - 3) = 9(y - 3)
⇒ x - 3 = 9y - 27
⇒ x - 9y + 24 = 0 .....(ii)
Subtracting (ii) from (i) we get
⇒ 6y - 36 = 0
⇒ 6y = 36
$\Rightarrow\text{y}=\frac{36}{6}$
⇒ y = 6 years
Putting y = 6 in (i) we get
⇒ x - 3 × 6 - 12 = 0
⇒ x - 18 - 12 = 0
⇒ x - 30 = 0
⇒ x = 30 years
Thus, present ages of man and his son are 30 years and 6 years respectively.

View full question & answer→

Question 93 Marks

Find the values of a and b for which the following system of equations has infinitely many solutions:

$2x + 3y - 7 = 0$

$(a - 1)x + (a + 1)y = (3a - 1)$

Answer

The given equations are
$2x + 3y - 7 = 0 .....(i)$
$(a - 1)x + (a + 1)y = (3a - 1) .....(ii)$
The given equations are of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = 3, c_1 = -7$
And, $a_2 = (a - 1), b_2 = (a + 1) and c_2 = -(3a - 1)$
For infinite many solutions
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{(\text{a}-1)}=\frac{3}{(\text{a}+1)}=\frac{-7}{-(3\text{a}-1)}$
$\Rightarrow\frac{2}{(\text{a}-1)}=\frac{3}{(\text{a}+1)}$
$\Rightarrow 2a + 2 = 3a - 3$
$\Rightarrow 3a - 2a = 2 + 3$
$\Rightarrow a = 5$
Thus, $a = 5$

View full question & answer→

Question 103 Marks

Find the value of k for which the following system of equations has no solution:

$x + 2y = 0$

$2x + ky = 5$

Answer

The given system of equations may be written as,
$x + 2y = 0$
$2x + ky = 5$
The given system of equations is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 1, b_1 = 2, c_1 = 0$
and, $a_2 = 2, b_2 = k, c_2 = -5$
For a solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{2}{\text{k}}$
and, $\frac{\text{c}_1}{\text{c}_2}=\frac{0}{-5}$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\Rightarrow\frac{1}{2}=\frac{2}{\text{k}}$
$\Rightarrow\text{k}=4$
Hence, the given system of equations has no solutions, when k = 4

View full question & answer→

Question 113 Marks

The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to $\frac{1}{3}.$ Find the fraction.

Answer

Let the fraction be $\frac{\text{x}}{\text{y}}$
Then, according to the given conditions we have
x + y = 18 .... (i)
And, $\frac{\text{x}}{\text{y}+2}=\frac{1}{3}$
⇒ 3x = y + 2
⇒ 3x - y = 2 .....(ii)
Adding equation (i) and equation (ii) we get
x + 3x = 18 + 2
⇒ 4x = 20
$\text{x}=\frac{20}{4}=5$
Putting x = 5 in equation (i) we get
5 + y = 18
⇒ y = 18 - 5
⇒ y = 13
Hence, the fraction is $\frac{5}{13}$

View full question & answer→

Question 123 Marks

The sum of digits of a two number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.

Answer

Let the digits at unit tens place be x and y respectively number
= 10y + x
Sum of digits of number is 15
x + y = 15 .....(i)
When digits are reversed then number will be 10x + y
According to equation
10x + y = 10y + x + 9
9x - 9y = 9
x - y = 1 .....(ii)
Adding (i) and (ii) we get
⇒ 2x = 16
$\Rightarrow\text{x}=\frac{16}{2}$
⇒ x = 8
Putting x = 8 in (i) we get
⇒ 8 + y = 15
⇒ y = 7
Thus, the number will be (7 × 10 + 8) = 78

View full question & answer→

Question 133 Marks

The difference between two numbers is 26 and one number is three times the other. Find them.

Answer

Let x and y be two numbers
According to question
x - y = 26 ....(i)
and x = 3y .......(ii)
Putting x = 3y in (i) we get
⇒ 3y - y = 26
⇒ 2y = 26
⇒ y = 13
Putting y = 13 in (ii) we get
⇒ x = 3 × 13
⇒ x = 39
Thus number will be 39, 13.

View full question & answer→

Question 143 Marks

For what value of k the following system of equations will be inconsistent?

$4x + 6y = 11$

$2x + ky = 7$

Answer

The given equations are,
$4x + 6y = 11 ......(i)$
$2x + ky = 7 .......(ii)$
The given equations are, of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 4, b_1 = 6, c_1 = -11,$
$a_2 = 2, b_2 = k$ and $c_2 = -7$
For inconsistent solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{4}{2}=\frac{6}{\text{k}}\neq\frac{-11}{-7}$
$\Rightarrow\text{k}=\frac{6\times2}{4}$
$\Rightarrow\text{k}=\frac{12}{4}=3$
Thus, the given system of equations will be inconsistent for k = 3

View full question & answer→

Question 153 Marks

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer

We know that the sum of supplementary angles will be 180°
Let the longer supplementary angles will be 'y'
Then, x + y = 180° ....(i)
If larger of supplementary angles exceeds the smaller by 18 degree, According to the given condition. We have,
x = y + 18° ....(ii)
Substitute x = y + 18° in equation (i) we get
x + y = 180°
y + 18° + y = 180°
2y + 18° = 180°
2y = 180° - 18°
2y = 162°
$\text{y}=\frac{162^\circ}{2}$
y = 81°
Put y = 81° equation (ii) we get
x = y + 18°
x = 81° + 18°
x = 99°
Hence, the larger supplementary angle is 99°, The smaller supplementary angle is 81°

View full question & answer→

Question 163 Marks

A person rowing at the rate of 5km/ h in still water, takes thrice as much time in going 40km upstream as in going 40km downstream. Find the speed of the stream.

Answer

Let the speed of the stream be v km/ h.
Given that, a person rowing in still water = 5km/ h.
The speed of a person rowing in downstream = (5 + v) km/ h.
and the speed of a person Has rowing in upstream = (5 - v) km/ h.
Now, the person taken time to cover 40km downstream,
$\text{t}_1=\frac{40}{5+\text{v}}$
$\Big[\because\text{Speed}=\frac{\text{Distance}}{\text{Time}}\Big]$
and the person has taken time to cover 40km upstream,
$\text{t}_2=\frac{40}{5-\text{v}}\text{h.}$
By condition, $\text{t}_2=\text{t}_1\times3$
$\Rightarrow\frac{40}{5-\text{v}}=\frac{40}{5+\text{v}}\times3$
$\Rightarrow\frac{1}{5-\text{v}}=\frac{3}{5+\text{v}}$
$\Rightarrow5+\text{v}=15-3\text{v}$
$\Rightarrow4\text{v}=10$
$\therefore\text{v}=\frac{10}{4}=2.5\text{km/ h.}$
Hence, the speed of the stream is 2.5km/ h.

View full question & answer→

Question 173 Marks

If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes $\frac{6}{5}.$ And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes $\frac{2}{5}.$ Find the fraction.

Answer

Let the fraction be $\frac{\text{x}}{\text{y}}$
Then, according to the given conditions, we have
$\frac{2\text{x}}{\text{y}-5}=\frac{6}{5}$
⇒ 5 × 2x = 6(y - 5)
⇒ 10x = 6y - 30
⇒ 10x - 6y = -30
⇒ 2(5x - 3y) = -30
⇒ 5x - 3y = -15 ....(i)
and, $\frac{\text{x}+8}{2\text{y}}=\frac{2}{5}$
⇒ 5(x + 8) = 2 × 2y
⇒ 5x + 40 = 4y
⇒ 5x - 4y = -40 .....(ii)
Subtracting equation (ii) by equation (i) we get
-3y + 4y = -15 + 40
⇒ y = 25
Putting y = 25 in equation (i) we get
5x - 3 × 25 = -15
⇒ 5x - 75 = -15
⇒ 5x = -15 + 75
⇒ 5x = 60
$\Rightarrow\text{x}=\frac{60}{5}=12$
Hence, the fraction is $\frac{12}{25}$

View full question & answer→

Question 183 Marks

Solve the following systems of equations:
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=13,$
$\frac{5}{\text{x}}-\frac{4}{\text{y}}=-2.$

Answer

Let us write the given pair of equation as
$2\Big(\frac{1}{\text{x}}\Big)+3\Big(\frac{1}{\text{y}}\Big)=13\ .....(\text{i})$
$5\Big(\frac{1}{\text{x}}\Big)-4\Big(\frac{1}{\text{y}}\Big)=-2\ .....(\text{ii})$
These equation are not in the from ax + by + c = 0. However, if we substitute
$\frac{1}{\text{x}}=\text{p}$ and $\frac{1}{\text{y}}=\text{q}$ in equationss (i) and (ii) we get
$2\text{p}+3\text{q}=13$
$5\text{p}-4\text{q}=-2$
So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3.
you know that $\text{p}=\frac{1}{\text{x}}$ and $\text{q}=\frac{1}{\text{y}}.$
Substitute the values of p and q to get
$\frac{1}{\text{x}}=2$ i.e., $\text{x}=\frac{1}{2}$ and $\frac{1}{\text{y}}=3$ i.e., $\text{y}=\frac{1}{3}.$

View full question & answer→

Question 193 Marks

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.

Answer

Let the fraction be $\frac{\text{x}}{\text{y}}$
Then, according to the given conditions, we have
x + y = 2x +4
⇒ x + y - 2x = 4
⇒ -x + y = 4 ....(i)
And, $\frac{\text{x}+3}{\text{y}+3}=\frac{2}{3}$
⇒ 3(x + 3) = 2(y + 3)
⇒ 3x + 9 = 2y + 6
⇒ 3x - 2y = 6 - 9
⇒ 3x - 2y = -3 .....(ii)
Multiplying equation (i) by 3 we get
-3x + 3y = 12 .....(iii)
Adding equation (ii) and equation (iii) we get
-2y + 3y = -3 + 12
⇒ y = 9
Putting y = 9 in equation (i) we get
-x + 9 = 4
⇒ -x = 4 - 9
⇒ -x = -5
⇒ x = 5
Hence, the fraction is $\frac{5}{9}$

View full question & answer→

Question 203 Marks

Solve the following system of equations by the method of cross-multiplication:

2x - y = 6,

x - y = 2.

Answer

Given,
2x - y = 6
x - y = 2
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation,
2x - y - 6 = 0
x - y - 2 = 0
By cross-multiplication method we get,
$\Rightarrow\frac{\text{x}}{\big((-1)\times(-2)\big)-\big((-1)\times(-6)\big)}=\frac{-\text{y}}{(2)(-2)-(1)(-6)}\\=\frac{\text{1}}{\big((2)\times(-1)\big)-\big((1)\times(-1)\big)}$
$\Rightarrow\frac{\text{x}}{2-6}=\frac{-\text{y}}{-4+6}=\frac{1}{-2+1}$
$\Rightarrow\frac{\text{x}}{-4}=\frac{-\text{y}}{2}=\frac{1}{-1}$
$\Rightarrow\text{x}=\frac{-4}{-1}=4$
$\Rightarrow\text{y}=\frac{-2}{-1}=2$
Hence we get the value of x = 4 and y = 2.

View full question & answer→

Question 213 Marks

Father's age is three times the sum of age of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.

Answer

Let the present age of father be x years and the sum of the present age of two chidren be y years. Then,
x = 3y ......(i)
Five years hence, father's age = (x + 5) years
Sum of children's age = (y + 5 + 5)
= (y + 10) years
Using the given inform ation we get
x + 5 = 2(y + 10)
⇒ x + 5 = 2y + 20
⇒ x - 2y = 20 - 5
⇒ x - 2y = 15 .....(ii)
Substituting x = 3y in equation (ii) we get
3y - 2y = 15
⇒ y = 15
Putting y = 15 in equation (i) we get
x = 3 × 15 = 45
Hence, present age of father is 45 years.

View full question & answer→

Question 223 Marks

The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

Answer

Let the present age (in year) of father and his two children be x, y and z year, respectively.
Now by given condition x = 2(y + z) .......(i)
and after 20 years,
(x + 20) = (y + 20) + (z + 20)
⇒ y + z + 40 = x + 20
⇒ y + z = x - 20
On putting the value of (y + z) in eq. (i) and we get the present age of father
⇒ x = 2(x - 20)
⇒ x = 2x - 40
⇒ x = 40
Hence, the father’s age is 40 years.

View full question & answer→

Question 233 Marks

If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction.

Answer

Let numerator and denominator be x and y respectively.
fraction $=\frac{\text{x}}{\text{y}}$
According to question,
$\frac{\text{x}+1}{\text{y}-1}=1$
2x - x - 1 - 2 = 0
x - 3 = 0
x = 3
⇒ 2x = y + 1
⇒ 2x - y - 1 = 0 ....(ii)
Subtracting (i) from (ii) we get
⇒ 3 - y + 2 = 0
⇒ y = 5
Thus, fraction $=\frac{3}{5}$

View full question & answer→

Question 243 Marks

Find the values of a and b for which the following system of equations has infinitely many solutions:
$(2a - 1)x - 3y = 5$
$3x + (b - 2)y = 3$

Answer

Given
$(2a - 1)x - 3y = 5$
$3x + (b - 2)y = 3$
To find: To determine for what value of k the system of equation has infinitely many solutions,
We know that the system of equations,
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For infinitely many solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{2\text{a}-1}{3}=\frac{3}{-(\text{b}-2)}=\frac{5}{3}$
Coniser
$\frac{3}{-(\text{b}-2)}=\frac{5}{3}$
$-5\text{b}+10=9$
$-5\text{b}=-1$
$\text{b}=\frac{1}{5}$
Again consider
$\frac{2\text{a}-1}{3}=\frac{5}{3}$
$2\text{a}-1=5$
$2\text{a}=6$
$\text{a}=3$
Hence for a = 3 and $\text{b}=\frac{1}{5}$ the system of equation has infinitely many solution.

View full question & answer→

Question 253 Marks

Solve the following systems of equations:
$\frac{1}{7\text{x}}+\frac{1}{6\text{y}}=3,$
$\frac{1}{2\text{x}}-\frac{1}{3\text{y}}=5.$

Answer

Let $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$
$\frac{\text{u}}{7}+\frac{\text{v}}{6}=3$
$\Rightarrow6\text{u}+7\text{v}=126\ .....(\text{i})$
And $\frac{\text{u}}{2}-\frac{\text{v}}{3}=5$
$\Rightarrow3\text{u}-2\text{v}=30\ .....(\text{ii})$
multiplying (ii) by 2 we get,
$\Rightarrow6\text{u}-4\text{v}=60\ .....(\text{iii})$
Subtracting (iii) from (i) we get
$\Rightarrow11\text{v}=66$
$\Rightarrow\text{v}=6$
Putting v = 6 in (iii) we get
$\Rightarrow6\text{u}-4\times6=60$
$\Rightarrow6\text{u}=84$
$\Rightarrow\text{u}=\frac{84}{6}=14$
$\therefore\frac{1}{\text{x}}=\text{u}$
$\Rightarrow\text{x}=\frac{1}{\text{u}}$
$\Rightarrow\text{x}=\frac{1}{14}$
and $\frac{1}{\text{y}}=\text{v}$
$\Rightarrow\text{y}=\frac{1}{\text{v}}$
$\Rightarrow\text{y}=\frac{1}{6}$
Thus, $\text{x}=\frac{1}{14}$ and $\text{y}=\frac{1}{6}$

View full question & answer→

Question 263 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$2x + 3y = 7$

$(k + 1)x + (2k - 1)y = 4k + 1$

Answer

The given equations are
$2x + 3y = 7 ....(i)$
$(k + 1)x + (2k - 1)y = 4k + 1 ....(ii)$
The given equations are of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = 3, c_1 = -7$
and, $a_2 = (k + 1), b_2 = (2k - 1)$ and $c_2 = -(4k + 1)$
For infinitely many solutions
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{(\text{k}+1)}=\frac{3}{(2\text{k}-1)}=\frac{-7}{-(4\text{k}+1)}$
$\Rightarrow\frac{2}{(\text{k}+1)}=\frac{3}{(2\text{k}-1)}$
$\Rightarrow 2(2k - 1) = 3(k + 1)$
$\Rightarrow 4k - 2 = 3k + 3$
$\Rightarrow 4k - 3k = 3 + 2$
$\Rightarrow k = 5$
Thus, the given system of equations has infinitely many solutions, if $k =, 5$

View full question & answer→

Question 273 Marks

A father is three times as old as his son. After twelve years, his age will be twice as that of his son then. Find their present ages.

Answer

Let the present ages of a father and his son be x years and y years respectively.
Father is three times as old as his son
x = 3y ....(i)
After 12 years,
Age of father = (x + 12) years
Age of son = (y + 12) years
According to question
(x + 12) = 2(y + 12)
⇒ x + 12 = 2y + 24
⇒ 3y - 2y = 24 - 12
⇒ y = 12
Putting y = 12 in (i) we get
⇒ x = 3 × 12
⇒ x = 36
Thus, present ages of father and his son are 36 years and 12 years respectively.

View full question & answer→

Question 283 Marks

Solve the following system of equations by the method of cross-multiplication:

3x + 2y + 25 = 0,

2x + y + 10 = 0.

Answer

The given system of equation is
3x + 2y + 25 = 0 ....(i)
2x + y + 10 = 0 ....(ii)
By cross-multiplication, we get
$\Rightarrow\frac{\text{x}}{2\times10-25\times1}=\frac{\text{y}}{25\times2-3\times10}\\=\frac{1}{3\times1-2\times2}$
$\Rightarrow\frac{\text{x}}{20-25}=\frac{\text{y}}{50-30}=\frac{1}{3-4}$
$\Rightarrow\frac{\text{x}}{-5}=\frac{\text{y}}{20}=\frac{1}{-1}$
$\text{x}=\frac{-5}{-1}=5$ and $\text{y}=\frac{20}{-1}=-20$
Thus, x = 5 and y = -20.

View full question & answer→

Question 293 Marks

A fraction becomes $\frac{1}{3}$ if 1 is subtracted from both its numerator and denominator. It 1 is added to both the numerator and denominator, it becomes $\frac{1}{2}.$ Find the fraction.

Answer

Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is $\frac{\text{x}}{\text{y}}$
If 1 is subtracted from both numerator and the denominator, the fraction becomes $\frac{1}{3}$
Thus, we have
$\frac{\text{x}-1}{\text{y}-1}=\frac{1}{3}$
⇒ 3(x - 1) = y - 1
⇒ 3x - 3 = y - 1
⇒ 3x - y - 2 = 0
If 1 is added to both numerator and the denominator, the fraction becomes $\frac{1}{2}$
Thus, we have
$\frac{\text{x}+1}{\text{y}+1}=\frac{1}{2}$
⇒ 2(x + 1) = y + 1
⇒ 2x + 2 = y + 1
⇒ 2x - y + 1 = 0
So, we have two equations
3x - y - 2 = 0
2x - y + 1 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
$ \Rightarrow\frac{\text{x}}{(-1)\times1-(-1)\times(-2)}=\frac{-\text{y}}{3\times1-2\times(-2)}\\=\frac{1}{3\times(-1)-2\times(-1)}$
$\Rightarrow\frac{\text{x}}{-1-2}=\frac{-\text{y}}{3+4}=\frac{1}{-3+2}$
$\Rightarrow\frac{\text{x}}{-3}=\frac{-\text{y}}{7}=\frac{1}{-1}$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{y}}{7}=1$
$\Rightarrow\text{x}=3,\ \text{y}=7$
Hence, the fraction is $\frac{3}{7}$

View full question & answer→

Question 303 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$8x + 5y = 9$

$kx + 10y = 18$

Answer

Given
$8x + 5y = 9$
$kx + 10y = 18$
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$_
For infinitely many solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{8}{\text{k}}=\frac{5}{10}=\frac{9}{18}$
$\frac{8}{\text{k}}=\frac{5}{10}$
$\text{k}=\frac{8\times10}{5}$
$\text{k}=8\times2$
$\text{k}=16$
Hence, the given system of equations will have infinitely many solutions if k = 16

View full question & answer→

Question 313 Marks

Find the value of k for which the following system of equations has no solution:

kx + 2y = 3

12x + ky = 6

Answer

kx + 2y = 3
12x + ky = 6
For no solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\text{k}}{12}=\frac{2}{\text{k}}\neq\frac{3}{6}$
$\frac{\text{k}}{12}=\frac{3}{\text{k}}$
$\text{k}^2=36$
$\text{k}=\pm6$ i.e.,
$\text{k}=6,-6$
Also,
$\frac{3}{\text{k}}\neq\frac{3}{6}$
$\frac{3\times6}{3}\neq\text{k}$
$\text{k}\neq6$
k = -6 satisfies both the condition
Hence, k = -6

View full question & answer→

Question 323 Marks

Half the perimeter of a garden, whose length is 4 more than its width is 36m. Find the dimension of the garden.

Answer

Let the length and breadth of the garden be x m and y m respectively. Then,
x = y + 4 ....(i)
And, $\frac{1}{2}$ [perimeter of a garden] = 36
$=\frac{1}{2}[2(\text{x}+\text{y})]=36$ $\big[\because$ perimeter of rectangle $=2(\text{l}+\text{b})\big]$
⇒ x + y = 36
Substituting x = y + 4 in equation (ii) we get
y + 4 + y = 36
⇒ 2y = 36 - 4
⇒ 2y = 32
$\Rightarrow\text{y}=\frac{32}{2}$
⇒ y = 16
Putting y = 16 in equation (i) we get
⇒ x = 16 + 4
⇒ x = 20
Hence, the length and breadth of the garden are 20m and 16m respectively.

View full question & answer→

Question 333 Marks

The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Answer

Let x and y be the any two numbers
According to question,
x + y = 8 ....(i)
(x + y) = 4(x - y)
8 = 4(x - y) [from (i)]
2 = (x - y)
x - y = 2 .....(ii)
Adding (i) and (ii) we get
⇒ 2x = 10
$\Rightarrow\text{x}=\frac{10}{2}$
⇒ x = 5
Putting x = 5 in (i) we get
⇒ 5 + y = 8
⇒ y = 3
Thus, the numbers are 5 and 3

View full question & answer→

Question 343 Marks

Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

Intersecting lines.
Parallel lines.
Coincident lines.

Answer

We have, $2x + 3y - 8 = 0$
Let another equation of line is:
$4x + 9y - 4 = 0$
Here, $a_1 = 2, b_1 = 3, c_1 = -8$
$a_2= 4, b_2 = 9, c_2 = -4$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{9}=\frac{1}{3}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-4}=\frac{2}{1}$
$\therefore\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\therefore$ 2x + 3y - 8 = 0 and 4x + 9y - 4 = 0 intersect each other at one point.
Hence, required equation of line is 4x + 9y - 4 = 0
We have, $2x + 3y - 8 = 0$
Let another equation of line is:
$4x + 6y - 4 = 0$
Here, $a_1 = 2, b_1 = 3, c_1 = -8$
$a_2= 4, b_2 = 9, c_2 = -4$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-4}=\frac{2}{1}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ Lines are parallel to each other.
Hence, required equation of line is $4x + 6y - 4 = 0.$
Let another equation of line is:
$4x + 6y - 16 = 0$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-16}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ Lines are coincident to each other.

View full question & answer→

Question 353 Marks

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Answer

Let the two numbers be x and y.
Then, by the first condition, ratio of these two numbers = 5 : 6
x : y = 5 : 6
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac{5}{6}$
$\Rightarrow\text{y}=\frac{6\text{x}}{5}\ ....(\text{i})$
And by the second condition, then, 8 is subtracted form each of the numbers, then ratio become 4 : 5
$\frac{\text{x}-8}{\text{y}-8}=\frac{4}{5}$
⇒ 5x - 40 = 4y - 32
⇒ 5x - 4y = 8 .....(ii)
Now, put the value of y in eq. (ii) we get
$5\text{x}-4\Big(\frac{6\text{x}}{5}\Big)=8$
⇒ 25x - 24x = 40
⇒ x = 40
Put the value of x in eq. (i) we get
$\text{y}=\frac{6}{5}\times40$
⇒ y = 6 × 8
⇒ y = 48
Hence, the required numbers are 40 and 48.

View full question & answer→

Question 363 Marks

Reena has pend and pencils which together are 40 in number. If she has 5 more pencils and 5 less pens, the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.

Answer

Let the number of pens be x and that of pencils be y. Then,
x + y = 40 .....(i)
and, (y + 5) = 4(x - 5)
⇒ y + 5 = 4x - 20
⇒ 5 + 20 = 4x - y
⇒ 4x - y = 25 ......(ii)
Adding equation (i) and equation (ii) we get
x + 4x = 40 + 25
⇒ 5x = 65
$\Rightarrow\text{x}=\frac{65}{50}=13$
Putting x = 13 in equation (i) we get
13 + y = 40
⇒ y = 40 - 13 = 27
Hence, Reena has 13 pens and 27 pencils.

View full question & answer→

Question 373 Marks

A lending library has a fixed charge for the first three days and 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.

Answer

Let fixed charge be Rs. x and additional charge be Rs. y respectively.
According to question
x + 4y = 27 .....(i)
x + 2y = 21 ........(ii)
Subtracting (ii) from (i) we get
⇒ 2y = 6
$\Rightarrow\text{y}=\frac{6}{2}=3$
Putting y = 3 in eq. (i) we get
⇒ x + 3 × 4 = 27
⇒ x + 12 = 27
⇒ x = 27 - 12
⇒ x = 15
Thus, fixed charge = Rs. 15 and charge for each extra day = Rs. 3

View full question & answer→

Question 383 Marks

Find the values of k for which the system will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?

$2x + ky = 1$

$3x - 5y = 7$

Answer

The given equations are,
$2x + ky = 1 ....(i)$
$3x - 5y = 7 ......(ii)$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = k, c_1 = -1$
And, $a_2 = 3, b_2 = -5$, and $c_2 = -7$
For unique solution,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{3}\neq\frac{\text{k}}{-5}$
$\Rightarrow\text{k}\neq\frac{-10}{3}$
Thus, the given system of equations has unique solution if $\text{k}\neq\frac{-10}{3}$
For no solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}=\frac{-1}{-7}$
$\Rightarrow\frac{2}{3}=\frac{\text{k}}{-5}$ and $\frac{\text{k}}{-5}=\frac{1}{7}$
$\Rightarrow\text{k}=\frac{-10}{3}$ and $\text{k}=\frac{-5}{7}$

View full question & answer→

Question 393 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$2x + 3y - 5 = 0$

$6x + ky - 15 = 0$

Answer

The given equation are
$2x + 3y - 5 = 0 .....(i)$
$6x + ky - 15 = 0 ........(ii)$
The given equation are of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where,$ a_1 = 2, b_1 = 3, c_1 = -5$
$a_2 = 6, b_2 = k$ and$ c_2 = -15$
For infinitely many solutions
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{6}=\frac{3}{\text{k}}=\frac{-5}{-15}$
$\Rightarrow\frac{2}{6}=\frac{3}{\text{k}}$
$\Rightarrow\text{k}=\frac{3\times6}{2}$
$\Rightarrow\text{k}=9$
Thus, the given system of equations has infinitely many solutions if k = 9.

View full question & answer→

Question 403 Marks

Obtain the condition for the following system of linear equations to have a unique solution:

$ax + by = c$

$lx + my = n$

Answer

Given,
$ax + by = c$
$lx + my = n$
To find: To determine the condition for the system of equation to have a unique equation,
We know that the system of equations,
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Here,
$\Rightarrow\frac{\text{a}}{\text{l}}\neq\frac{\text{b}}{\text{m}}$
$\Rightarrow\text{am}\neq\text{bl}$
Hence for $\text{am}\neq\text{bl}$ the system of equations have unique solution.

View full question & answer→

Question 413 Marks

A father is three times as old as his son. In 12 years time, he will be twice as old as his son. Find the present ages of father and the son.

Answer

Let the present ages of a father and his son be x and y years respectively.
After 12 years
Age of father = (x + 12) years
Age of his son = (y + 12) years
According to question
x = 3y ....(i)
And (x + 12) = 2(y + 12)
⇒ x + 12 = 2y + 24
⇒ 3y + 12 = 2y + 24
⇒ 3y - 2y = 24 - 12
⇒ y = 12 years
Putting y = 12 in (i) we get
⇒ x = 3 × 12
⇒ x = 36 years
Thus, present age of father = 36 years.
and present age of his son = 12 years.

View full question & answer→

Question 423 Marks

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes $\frac{1}{2}.$ Find the fraction.

Answer

Let the fraction be $\frac{\text{x}}{\text{y}}$
Then, according to the given conditions, we have
x + y = 12 .....(i)
and $\frac{\text{x}}{\text{y}+3}=\frac{1}{2}$
⇒ 2x = y + 3
⇒ 2x - y = 3 .....(ii)
Adding equation (i) and equation (ii) we get
x + 2x = 12 + 3
⇒ 3x = 15
$\Rightarrow\text{x}=\frac{15}{3}$
⇒ x = 5
Putting x = 5 in equation (i) we get
5 + y = 12
⇒ y = 12 - 5
⇒ y = 7
Hence, the fraction is $\frac{5}{7}$

View full question & answer→

Question 433 Marks

Solve the following systems of equations:
${\text{x}}+2{\text{y}}=\frac{3}{2},$
$2\text{x}+\text{y}=\frac{3}{2}.$

Answer

The given system of equations is,
${\text{x}}+2{\text{y}}=\frac{3}{2}\ .....(\text{i})$
$2\text{x}+\text{y}=\frac{3}{2}\ .......(\text{ii})$
Let us eliminate y from the given equations. The co-efficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is. so, we make the co-efficients of y equal to 2 in the equations.
Multiplying (i) by 1 and (ii) by 2, we get
${\text{x}}+2{\text{y}}=\frac{3}{2}\ .....(\text{iii})$
$4\text{x}+2\text{y}=3\ .....(\text{iv})$
Subtracting (iii) from (iv), we get
$4\text{x}-\text{x}+2\text{y}-2\text{y}=3-\frac{3}{2}$
$\Rightarrow3\text{x}=\frac{6-3}{2}$
$\Rightarrow3\text{x}=\frac{3}{2}$
$\Rightarrow\text{x}=\frac{3}{2\times3}$
$\Rightarrow\text{x}=\frac{1}{2}$
Putting $\text{x}=\frac{1}{2},$ in equation (iv), we get
$4\times\frac{1}{2}+2\text{y}=3$
$\Rightarrow2+2\text{y}=3$
$\Rightarrow2\text{y}=3-2$
$\Rightarrow\text{y}=\frac{1}{2}$
Hence, solution of the given of equation is $\text{x}=\frac{1}{2},\text{y}=\frac{1}{2}.$

View full question & answer→

Question 443 Marks

Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$ have:
No solution?

Answer

The given pair of linear equations is
$\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$
Here,
$\text{a}_1=\lambda,\text{b}_1=1,\text{c}_1=-\lambda^2$
$\text{a}_2=1,\text{b}_2=\lambda,\text{c}_2=-1$
For no solution,
$\frac{\text{a}_1}{\text{a}_1}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\lambda}{1}=\frac{1}{\lambda}\neq\frac{-\lambda^2}{-1}$
$\Rightarrow\lambda^2-1=0$
$\Rightarrow(\lambda-1)(\lambda+1)=0$
$\Rightarrow\lambda=1,-1$
Here, we take only $\lambda=-1$ because at $\lambda=1$ the system of linear equations has infinitely many solutions.

View full question & answer→

Question 453 Marks

Gloria is walking along the path joining (-2, 3) and (2, -2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.

Answer

Given, Gloria is walking along the path joining (-2, 3) and (2, -2), while suresh is walking along the path joining (0, 5) and (4, 0).

View full question & answer→

Question 463 Marks

For what value of $\alpha,$ the system of equations will have no solution
$\alpha\text{x}+3\text{y}=\alpha-3$
$12\text{x}+\alpha\text{y}=\alpha$

Answer

The given system of equations may be written as,
$\alpha\text{x}+3\text{y}-(\alpha-3)=0$
$12\text{x}+\alpha\text{y}-\alpha=0$
The given system of equations is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $\text{a}_1=\alpha,\text{b}_1=3,\text{c}_1=-(\alpha-3)$
and, $\text{a}_2=12,\text{b}_2=\alpha,\text{c}_2=-\alpha$
For no solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\alpha}{12}=\frac{3}{\alpha}\neq\frac{-(\alpha-3)}{-\alpha}$
Now,
$\frac{3}{\alpha}\neq\frac{-(\alpha-3)}{-\alpha}$
$\Rightarrow\frac{3}{\alpha}\neq\frac{\alpha-3}{\alpha}$
$\Rightarrow3\neq\alpha-3$
$\Rightarrow3+3\neq\alpha$
$\Rightarrow6\neq\alpha$
$\Rightarrow\alpha\neq6$
and $\frac{\alpha}{12}=\frac{3}{\alpha}$
$\Rightarrow\alpha^2=36$
$\Rightarrow\alpha=\pm6$
$\Rightarrow\alpha=-6$ $\big[\therefore\alpha\neq6\big]$
Hence, the given system of equations will have no solution, if $\alpha=-6$

View full question & answer→

Question 473 Marks

Prove that there is a value of $(\text{c}\neq0)$ for which the system has infinitely many solutions. Find this value.

$6x + 3y = c - 3$

$12x + cy = c$

Answer

The given system of equations may be written as,
$6x + 3y - (c - 3) = 0$
$12x + cy - c = 0$
This is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 6, b_1 = 3, c_1 = -(c - 3)$
And, $a_2 = 12, b_2 = c, c_2 = -c$
For infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}=\frac{-(\text{c}-3)}{-\text{c}}$
$\Rightarrow\frac{6}{12}=\frac{3}{\text{c}}$ and $$$\frac{3}{\text{c}}=\frac{\text{c}-3}{\text{c}}$
$\Rightarrow6\text{c}=12\times3$ and $3=(\text{c}-3)$
$\Rightarrow\text{c}=\frac{36}{6}$ and $\text{c}-3=3$
$\Rightarrow\text{c}=6$ and $\text{c}=6$
Now,
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{12}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-(6-3)}{-6}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
Clearly, for this value of c, we have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Hence, the given system of equations has infinitely many solutions, if $c = 6.$

View full question & answer→

Question 483 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$kx - 2y + 6 = 0$

$4x - 3y + 9 = 0$

Answer

The given system of equations is
$kx - 2y + 6 = 0$
$4x - 3y + 9 = 0$
The system of equations is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = k, b_1 = -2, c_1 = 6$
and, $a_2 = 4, b_2 = -3, c_2 = 9$
For infinitely many solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\text{k}}{4}=\frac{-2}{-3}=\frac{6}{9}$
Now,
$\frac{\text{k}}{4}=\frac{6}{9}$
$\Rightarrow\frac{\text{k}}{4}=\frac{2}{3}$
$\Rightarrow\text{k}=\frac{2\times4}{3}$
$\Rightarrow\text{k}=\frac{8}{3}$
Hence, the given system of equations will have infinitely many solutions if $\text{k}=\frac{8}{3}$

View full question & answer→

Question 493 Marks

Ten years later, A will be twice as old as B and five years ago, A was three times as old as B. What are the present ages of A and B?

Answer

Let the age of A and B be x and y years respectively. Then,
Ten years later, the age of A will be (x + 10) years and, the age of B will be (y + 10) years.
$\therefore$ x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y = 20 - 10
⇒ x - 2y = 10 .....(i)
Five years ago A's age = (x - 5) years
B's ago = (y - 5) years
using the given inform ation we get
x - 5 = 3(y - 5)
⇒ x - 5 = 3y - 15
⇒ x - 3y = - 10 ....(ii)
Subtracting equation (ii) from equation (i) we get
-2y + 3y = 10 + 10
⇒ y = 20
Putting y = 20 in equation (i) we get
x - 2 × 20 = 10
⇒ x = 10 + 40
⇒ x = 50
Hence A's age = 50 years
B's age = 20 years

View full question & answer→

Question 503 Marks

Find the value of k for which the following system of equations has no solution:

$kx - 5y = 2$

$6x + 2y = 7$

Answer

Given
$kx - 5y = 2$
$6x + 2y = 7$
$$To find: To determine for what value of k the system of equation has no solution
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$_
Where, $a_1 = k, b_1 = -5, c_1 = -2$
and,$ a_2 = 6, b_2 = 2, c_2 = -7$
For solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{\text{k}}{6}=\frac{-5}{2}\neq\frac{2}{7}$
$\frac{\text{k}}{6}=\frac{-5}{2}$
$2k = -30$
$k = -15$
Hence for k = -15 the system of equation have infinitely many solutions.

View full question & answer→

Question 513 Marks

Find the value of k for which the system has (i) a unique solution, and (ii) no solution:

$kx + 2y = 5$

$3x + y = 1$

Answer

Given,
$kx + 2y = 5$
$3x + y = 1$
To find: To determine for what value of k the system of equation has:
Unique solution.
No solution.
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For Unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Here,
$\frac{\text{k}}{3}\neq\frac{2}{1}$
$\text{k}\neq6$
Hence for $\text{k}\neq6$ the system of equation has unique solution.
For no solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_1}$
$\frac{\text{k}}{3}=\frac{2}{1}\neq\frac{5}{1}$
$\text{k}=6$
Hence for k = 6 the system of equation has no solution.

View full question & answer→

Question 523 Marks

For what value of k, the following pair of linear equation has infinitely many solutions?

$10x + 5y - (k - 5) = 0$

$20x + 10y - k = 0$

Answer

The given equation are
$10x + 5y - (k - 5) = 0 .....(i)$
$20x + 10y - k = 0 ......(ii)$
The given equations are of the from
$a_1x + b_1y + c_1 = 0 .....(iii)$
$a_2x + b_2y + c_2 = 0 ......(iv)$
Where, $a_1 = 10, b_1 = 5, c_1 = -(k - 2), $
$a_2= 20, b_2= 10$ and $c_2= -k$
For infinitely solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{10}{20}=\frac{5}{10}=\frac{-(\text{k}-5)}{-\text{k}}$
$\Rightarrow\frac{10}{20}=\frac{-(\text{k}-5)}{-\text{k}}$
$\Rightarrow\frac{1}{2}=\frac{\text{k}-5}{\text{k}}$
$\Rightarrow\text{k}=2\text{k}-10$
$\Rightarrow\text{k}=10$

View full question & answer→

Question 533 Marks

In a competitive examination, one mark is aw awarded for each correct answer while $\frac{1}{2}$ mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly.

Answer

Let x be the number of correct answer of the questions in a competitive examination, then (120 - x) be the number of wrong answers of the questions.
Then, by given condition,
$\text{x}\times1-(120-\text{x})\times\frac{1}{2}=90$
$\Rightarrow\text{x}-60+\frac{\text{x}}{2}=90$
$\Rightarrow\frac{3\text{x}}{2}=150$
$\therefore\text{x}=\frac{150\times2}{3}$
$=50\times2=100$
Hence, Jayanti answered correctly 100 questions.

View full question & answer→

Question 543 Marks

A part of monthly hostel charges in a college are fixed and the remaining depend on the number of days one has taken food in the mess. When a student A takes foods for 20 days, he has to pay Rs. 1000 as hostel charges whereas a students B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day.

Answer

Let the fixed charges of hostel be Rs. x and the cvost of food charges be Rs. y per day.
According to the question
⇒ x + 20y = 1000
⇒ x + 20y - 1000 = 0 .....(i)
⇒ x + 26y = 1180
⇒ x + 26y - 1180 .....(ii)
Now, subtracting equation (i) from eq. (ii)
⇒ x + 26y - 1180 - (x + 20y - 1000) = 0
⇒ x + 26y - 1180 - x - 20y + 1000 = 0
⇒ 6y - 180 = 0
⇒ 6y = 180
$\Rightarrow\text{y}=\frac{180}{6}$
⇒ y = 30
Now, putting the value y in eq. (i)
⇒ x + 20y - 1000 = 0
⇒ x + 20 × 30 - 1000 = 0
⇒ x + 600 - 1000 = 0
⇒ x - 400 = 0
⇒ x = 400
Hence, the fixed chagres of hostel is Rs. 400. The cost of food per day is Rs. 30

View full question & answer→

Question 553 Marks

Find the value of k for which the following system of equations has no solution:

$3x - 4y + 7 = 0$

$kx + 3y - 5 = 0$

Answer

The given equations are,
$3x - 4y + 7 = 0 .....(i)$
$kx + 3y - 5 = 0 ........(ii)$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 3, b_1 = -4, c_1 = 7$
and, $a_2 = k, b_2 = 3, c_2 = -5$
For a solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{3}{\text{k}}=\frac{-4}{3}\neq\frac{7}{-5}$
$\Rightarrow\frac{3}{\text{k}}=\frac{-4}{3}$
$\Rightarrow\text{k}=\frac{-9}{4}$
Thus, the given of equations will have no solution for $\text{k}=\frac{-9}{4}$

View full question & answer→

Question 563 Marks

Write the number of solution of the following pair of linear equations:

$x + 2y - 8 = 0$

$2x + 4y = 16$

Answer

The given equation are
$x + 2y - 8 = 0$
$2x + 4y - 16 = 0$
Where, $a_1 = 1, a_2 = 2, b_1 = 2, b_2= 4, c_1= 8, c_2= 16$
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{2}{4},\frac{\text{c}_1}{\text{c}_2}=\frac{8}{16}$
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{1}{2},\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Every solution of the second equation is also a solution of the first equation.
Hence, there are infinite-solution, the system equation is consistent.

View full question & answer→

Question 573 Marks

Find the value of k for which the following system of equations has a unique solution:

$4x - 5y = k$

$2x - 3y = 12$

Answer

The given system of equation is
$4x - 5y - k = 0$
$2x - 3y - 12 = 0$
The given system of equation is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1= 4, b_1 = -5, c_1 = -k$
And, $a_2 = 2, b_2 = -3, c_2 = -12$
For unique solution we must have,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\therefore\frac{4}{2}\neq\frac{-5}{-3}$
$\therefore\text{k}$ is any real number.
So, the given system of equations will have a unique solution for all real values of k.

View full question & answer→

Question 583 Marks

Find the value of k for which the following system of equations has a unique solution:

$4x + ky + 8 = 0$

$2x + 2y + 2 = 0$

Answer

The given equation are
$4x + ky + 8 = 0 .....(i)$
$2x + 2y + 2 = 0 .......(ii)$
The given equation are of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1= 4, b_1 = k, c_1 = 8$
And, $a_2 = 2, b_2 = 2, c_2 = 2$
For unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\Rightarrow\frac{4}{2}\neq\frac{\text{k}}{2}$
$\Rightarrow\text{k}\neq4$
Thus, the given system of equations has a unique solution if $\text{k}\neq4$

View full question & answer→

Question 593 Marks

Find the value of k for which the following system of equations has a unique solution:

$x + 2y = 3$

$5x + ky + 7 = 0$

Answer

Given
$x + 2y = 3$
$5x + ky + 7 = 0$
To find: To determine to value of k for which the system has a unique solution.
We know that the system of equations,
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$
For unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Here,
$\frac{1}{5}\neq\frac{2}{\text{k}}$
$\text{k}\neq5\times2$
$\text{k}\neq10$
Hence for $\text{k}\neq10$ the system of equation has unique solution.

View full question & answer→

Question 603 Marks

Solve the following systems of equations:

0.4x + 0.3y = 1.7,

0.7x - 0.2y = 0.8.

Answer

The given equations are
0.4x + 0.3y = 1.7 ......(i)
0.7x - 0.2y = 0.8 ......(ii)
Multiplying both sides of (i) and (ii), by 10, we get
4x + 3y = 17 .....(iii)
7x - 2y = 8 ......(iv)
From (iv), we get
7x = 8 + 2y
$\Rightarrow\text{x}=\frac{8+2\text{y}}{7}$
Substituting $\text{x}=\frac{8+2\text{y}}{7}$ in (iii), we get
$4\Big(\frac{8+2\text{y}}{7}\Big)+3\text{y}=17$
$\Rightarrow\frac{32+8\text{y}}{7}+3\text{y}=17$
$\Rightarrow32+29\text{y}=17\times7$
$\Rightarrow29\text{y}=119-32$
$\Rightarrow29\text{y}=87$
$\Rightarrow\text{y}=\frac{87}{29}=3$
Putting y = 3 in $\text{x}=\frac{8+2\text{y}}{7},$ we get
$\text{x}=\frac{8+2\times3}{7}$
$=\frac{8+6}{7}$
$=\frac{14}{7}$
$=2$
Hence, the solution of the given system of equation is x = 2, y = 3.

View full question & answer→

Question 613 Marks

Six years hence a man's age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Answer

Let the present age of a man and his son be x and y years respectively.
6 years leave (later)
Age of man = (x + 6) years
Age of his son = (y + 6) years
3 years ago
Age of man = (x - 3) years
Age of his son = (y - 3) years
According to question
(x + 6) = 3(y + 6)
⇒ x + 6 = 3y + 18
⇒ x - 3y - 12 = 0 .....(i)
and, (x - 3) = 9(y - 3)
⇒ x - 3 = 9y - 27
⇒ x - 9y + 24 = 0 .....(ii)
Subtracting (ii) from (i) we get
⇒ 6y - 36 = 0
⇒ 6y = 36
$\Rightarrow\text{y}=\frac{36}{6}$
⇒ y = 6 years
Putting y = 6 in (i) we get
⇒ x - 3 × 6 - 12 = 0
⇒ x - 18 - 12 = 0
⇒ x - 30 = 0
⇒ x = 30 years
Thus, present ages of man and his son are 30 years and 6 years respectively.

View full question & answer→

Question 623 Marks

3 bags and 4 pens together cost Rs. 257 whereas 4 bags and 3 pens together cost Rs. 324. Find the total cost of 1 bag and 10 pens.

Answer

Let the cost of a be Rs. x and that of a pen be Rs. y. Then,
3x + 4y = 257 .....(i)
and, 4x + 3y = 324 .....(ii)
Multiplying equation (i) by 3 and equation (ii) by 4 we get
9x + 12y = 770 ......(iii)
16x + 12y = 1296 ......(iv)
Subtracting equation (iii) by equation (iv) we get
16x - 9x = 1296 - 771
⇒ 7x = 525
$\Rightarrow\text{x}=\frac{525}{7}=75$
Cost of a bag = Rs. 75
Putting x = 75 in equation (i) we get
3 × 75 + 4y = 257
⇒ 225 + 4y = 257
⇒ 4y = 257 - 225
⇒ 4y = 32
$\Rightarrow\text{y}=\frac{32}{4}=8$
$\therefore$ Cost of a pen = Rs. 8
$\therefore$ Cost of 10 pens = 8 × 10 = Rs 80
Hence, the total cost of 1 bag and 10 pens = 75 + 80 = Rs. 155.

View full question & answer→

Question 633 Marks

Solve the following systems of equations:

2(3u - ν) = 5uν,

2(u + 3ν) = 5uν.

Answer

The given equation are
2(3u − ν) = 5uν ⇒ 6u - 2v = 5uv ......(i)
2(u + 3ν) = 5uν ⇒ 2u + 6v = 5uv .......(ii)
On dividing both sides of (i) and (ii) by uv we get
$\Rightarrow\frac{6}{\text{v}}-\frac{2}{\text{u}}=5\ ......(\text{iii})$
$\Rightarrow\frac{2}{\text{v}}+\frac{6}{\text{u}}=5\ ......(\text{iv})$
Let $\frac{1}{\text{v}}=\text{x}$ and $\frac{1}{\text{u}}=\text{y}$
$\Rightarrow6\text{x}-2\text{y}=5\ ......(\text{v})$
$\Rightarrow2\text{x}+6\text{y}=5\ ......(\text{vi})$
Multiplying (v) by 3 and adding it with (vi) we get
$18\text{x}-6\text{y}=15\\\underline{2\text{x}\ \ +6\text{y}=\ \ 5}\\20\text{x}\ \ \ \ \ \ \ \ \ =20$
$\text{x}=1$
So, that $\text{y}=\frac{1}{2}$
$\because\frac{1}{\text{v}}=\text{x}\Rightarrow\text{v}=\frac{1}{\text{x}}=1$
and $\frac{1}{\text{u}}=\text{y}\Rightarrow\text{u}=\frac{1}{\text{y}}=2$
Thus $\text{u}=0$ and $\text{v}=1$

View full question & answer→

Question 643 Marks

Write the value of k for which the system of equations 3x - 2y = 0 and kx + 5y = 0 has infinitely may solutions.

Answer

The given equation are
3x - 2y = 0
kx + 5y = 0
For the equations to have infinitely number of solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{3}{\text{k}}=\frac{-2}{5}$
By cross multiplication we have
$3\times5=-2\times\text{k}$
$15=-2\text{k}$
$\frac{15}{-2}=\text{k}$
Hence, the value of k for the system of equation 3 × -2y = 0 and kx + 5y = 0 is $\frac{15}{-2}$

View full question & answer→

Question 653 Marks

A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Answer

Let the two-digit number = 10x + y
Case I: Multiplying the sum of the digits by 8 and then subtracting 5 = two-digit number
⇒ 8 × (x + y) - 5 = 10x + y
⇒ 8x + 8y - 5 = 10x + y
⇒ 2x - 7y = -5 ......(i)
Case II: Multiplying the difference of the digits by 16 and then adding 3 = two-digit number
⇒16 × (x - y) + 3 = 10x + y
⇒ 16x - 16y + 3 = 10x + y
⇒ 6x - 17y = -3 ......(ii)
Now, multiplying in eq. (i) by 3 and then subtracting from eq. (ii) we get
(image)
⇒ y = 3
Now, Put the value of y in eq. (i) we get
2x - 7 × 3 = -5
⇒ 2x = 21 - 5 = 16
⇒ x = 8
Hence, the required two-digit number
= 10x + y
= 10 × 8 + 3 = 80 + 3 = 83

View full question & answer→

Question 663 Marks

There are two examination rooms A and B. If 10 candidates are sent from A to B, the number of students in each room is same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in B. Find the number of students in each room.

Answer

Let us take the A examination room will be x and the B examination room will be y.
If 10 candidates are sent from A to B, the number of students in each room is same. According to the above condition equation will be,
y + 10 = x - 10
0 = x - y - 10 - 10
x - y - 20 = 0 ....(i)
If 20 candidates are sent from B to A, the number of students in A is double the number of students in B, then equation will be,
x + 20 = 2(y - 20)
x + 20 = 2y - 40
x + 20 - 2y + 40 = 0
x - 2y + 20 + 40 = 0
x - 2y + 60 = 0 ......(ii)
By subtracting the equation (i) from (ii) we get y = 80 Substituting y = 80 in equation (i) we get,
Hence 100 candidates are in A examination Room, 80 candidates are in B examination Room.

View full question & answer→

Question 673 Marks

Solve the following system of equations by the method of cross-multiplication:

ax + by = a - b,

bx - ay = a + b.

Answer

The given system of equations is
ax + by = a - b ....(i)
bx - ay = a + b .....(ii)
By cross-multiplication we get
$\Rightarrow\frac{\text{x}}{\text{b}\big\{-(\text{a}+\text{b})\big\}-\big\{-(\text{a}-\text{b})\big\}(-\text{a})}=\frac{\text{y}}{-(\text{a}-\text{b})(\text{b})-\big\{-(\text{a}+\text{b})\big\}\text{a}}\\=\frac{1}{\text{a}\times(-\text{a})-\text{b}\times\text{b}}$
$\Rightarrow\frac{\text{x}}{-\text{ab}-\text{b}^2-\text{a}^2+\text{ab}}=\frac{\text{y}}{-\text{ab}+\text{b}^2+\text{a}^2+\text{ab}}=\frac{1}{-\text{a}^2-\text{b}^2}$
$\Rightarrow\frac{\text{x}}{-(\text{a}^2+\text{b}^2)}=\frac{\text{y}}{(\text{a}^2+\text{b}^2)}=\frac{1}{-(\text{a}^2+\text{b}^2)}$
$\text{x}=\frac{-(\text{a}^2+\text{b}^2)}{-(\text{a}^2+\text{b}^2)}=1$ and $\text{y}=\frac{(\text{a}^2+\text{b}^2)}{-(\text{a}^2+\text{b}^2)}=-1$
Thus, x = 1 and y = -1.

View full question & answer→

Question 683 Marks

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?

Answer

Let the present age of Nuri be x years and the present age of sonu be y years.
After 10 years, Nuri's age will be(x + 10) years and the age of sonu will be(y + 10) years. Thus using the given information, we have
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y - 10 = 0
Before 5 years, the age of Nuri was (x - 5) years and the age of sonu was (y - 5) years. Thus using the given information, we have
x - 5 = 3(y - 5)
⇒ x - 5 = 3y - 15
⇒ x - 3y + 10 = 0
So, we have two equations
x - 2y - 10 = 0
x - 3y + 10 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
$\Rightarrow\frac{\text{x}}{(-2)\times10-(-3)\times(-10)}=\frac{-\text{y}}{1\times10-1\times(-10)}\\ =\frac{1}{1\times(-3)-1\times(-2)}$
$\Rightarrow\frac{\text{x}}{-20-30}=\frac{-\text{y}}{10+10}=\frac{1}{-3+2}$
$\Rightarrow\frac{\text{x}}{-50}=\frac{-\text{y}}{20}=\frac{1}{-1}$
$\Rightarrow\frac{\text{x}}{50}=\frac{\text{y}}{20}=1$
$\Rightarrow\text{x}=50,\ \text{y}=20$
Hence, the present age of nuri is 50 years and the present age of sonu is 20 years.

View full question & answer→

Question 693 Marks

Write an equation of a line passing through the point representing solution of the pair of linear equations $x + y = 2$ and $2x - y = 1$. How many such lines can we find?

Answer

Given pair of linear equation is
$x + y - 2 = 0 ......(i)$
and $2x - y - 1 = 0 ......(ii)$
On Comparing with ax + by + c = 0 we get
$a_1 = 1, b_1 = 1$ and $c_1 = -2 $[from eq. (i)]
$a_2 = 2, b_2 = -1$ and $c_2 = -1$ [from eq. (ii)]
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{1}{-1}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{-2}{-1}=\frac{2}{1}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
So, both lines intersect at a point. Therefore, the pair of equations has a unique solution.
Hence, these equations are consistent.
Now$, x + y = 2 $
$\Rightarrow y = 2 - x$
if x = 0, then y = 2 and if x = 2, then y = 0

x	0	2
y	2	0
Points	A	B

and $2x - y - 1 = 0 $
$\Rightarrow y = 2x - 1$
If x = 0, then $y = -1,$
If $\text{x}=\frac{1}{2},$ then y = 0
and if $x = 1$, then$ y = 1$

x	0	$\frac{1}{2}$	1
y	-1	0	1
Points	C	D	E

Plotting the points A(2, 0) and B(0, 2), we get the straight line AB. Plotting the points C(0, -1) and $\text{D}\Big(\frac{1}{2},0\Big)$ we get the straight line CD. The lines AB and CD intersect at E(1, 1).
Hence, infinite lines can pass through the intersection point of linear equations.
$x + y = 2 $and $2x - y = 1 $i.e.,$ E(1, 1)$ like as $y = x, 2x + y = 3, x + 2y = 3$, so no.

View full question & answer→

Question 703 Marks

Solve the following systems of equations:
$\frac{\text{x}+\text{y}}{\text{xy}}=2,$
$\frac{\text{x}-\text{y}}{\text{xy}}=6.$

Answer

The given equations are
$\frac{\text{x}+\text{y}}{\text{xy}}=2$
$\text{x}+\text{y}=2{\text{xy}}\ ......(\text{i})$
$\frac{\text{x}-\text{y}}{\text{xy}}=6$
$\text{x}-\text{y}=6{\text{xy}}\ ......(\text{ii})$
Adding both equations we get,
$\frac{\text{x}\ +\ \text{y}\ =\ 2\text{xy}\\\text{x}\ -\ \text{y}\ =\ 6\text{xy}}{2\text{x}\ \ \ \ \ \ \ =\ 8\text{xy}}$
$\Rightarrow\text{y}=\frac{1}{4}$
Put the value of y in equation (i) we get,
$\text{x}+\frac{1}{4}=2\text{x}\times\frac{1}{4}$
$\Rightarrow\frac{-\text{x}}{2}=\frac{1}{4}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence the values of $\text{x}=-\frac{1}{2}$ and $\text{y}=\frac{1}{4}$

View full question & answer→

Question 713 Marks

5 books and 7 pens together cost Rs. 79 whereas 7 books and 5 pens together cost Rs. 77. Find the total cost of 1 book and 2 pens.

Answer

Let cost of a book and a pen be Rs. x and y respectively.
According to question,
5x + 7y = 79 .....(i)
7x + 5y = 77 .......(ii)
Adding (i) and (ii) we get
⇒ 12x + 12y = 156
⇒ x + y = 13 .....(iii)
Subtracting (i) from (ii) we get
⇒ 2x - 2y = -2
⇒ x - y = -1 ......(iv)
Adding x = 6 in (iii) we get
⇒ 6 + y = 13
⇒ y = 7
Thus total cost of 1 book and 2 pens = (6 + 2 × 7)
= (6 + 14)
= Rs. 20

View full question & answer→

Question 723 Marks

The sum of digits of a two digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number?

Answer

Let the digit in the units place be x and digit in the ten's place be y. Then
x + y = 13 [given] ......(i)
and, Number = 10y + x
Number obtained by reversing the digits = 10x + y
It is given that the number is subtracted from the one obtained by interchanging the digits, the result is 45.
i.e., (Number obtained by interchanging the digits) - Number = 45
$\therefore$ 10x + y - (10y + x) = 45
⇒ 9x - 9y = 45
⇒ 9(x - y) = 45
⇒ x - y = 5 .....(ii)
Adding equation (i) and equation (ii), we get
2x = 13 + 5
⇒ 2x = 18
$\Rightarrow\text{x}=\frac{18}{2}$
⇒ x = 9
Putting x = 9 in equation (i) we get
9 + y = 13
⇒ y = 13 - 9
⇒ y = 4
Hence, the number is 10y + x = 10 × 4 + 9 = 49

View full question & answer→

Question 733 Marks

Solve the following systems of equations:

0.5x + 07y = 0.74

0.3x + 0.5y = 0.5

Answer

The given equations are
0.5x + 0.7y = 0.74 ...(i)
0.3x + 0.5y = 0.5 .....(ii)
Multiply equation (i) by 0.5 and (ii) by 2 and subtract equation (ii) from (i) we get
0.25x + 0.35y = 0.37
(image)
⇒ x = 0.5
Put the value of x in equation (i), we get
0.5 × 0.5 + 0.7y = 0.74
⇒ y = 0.7
Hence the value of x = 0.5 and y = 0.7

View full question & answer→

Question 743 Marks

For what value of k, the following system of equations will represent the coincident lines?

$x + 2y + 7 =0$

$2x + ky + 14 = 0$

Answer

The given system of equations may be written as
$x + 2y + 7 =0$
$2x + ky + 14 = 0$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 1, b_1 = 2, c_1 = 7$
And, $a_2 = 2, b_2 = k$, and $c_2 = 14$
The givenequations will represent coincident lines if they have infinitely many solutions.
The condition for which is,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_1}$
$\Rightarrow\frac{1}{2}=\frac{2}{\text{k}}$
$\Rightarrow\text{k}=4$
Hence, the given system of equations will represent coincident lines, if k = 4.

View full question & answer→

Question 753 Marks

A shopkeeper gives books pn rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹ 22 for a book kept for 6 days, while Anand paid ₹ 16 for the book kept for four days. Find the fixed charges and charge for each extra.

Answer

Let Latika takes a fixed charge for the first two day is ₹ x and additional charge for each day thereafter is ₹ y.
Now, by first condition,
Latika paid ₹ 22 for a book kept for six days
i.e., x + 4y = 22 .....(i)
and by second condition,
Anand paid ₹ 16 for a book kept for four days
i.e., x + 2y = 16 ....(ii)
Now, subtracting Eq. (ii) from Eq. (i), we get
2y = 6
⇒ y = 3
On putting the value of y in Eq. (ii), we get
x + 2 × 3 = 16
x = 16 - 6 = 10
Hence, the fixed charge = ₹ 10 and the charge for each extra day = ₹ 3

View full question & answer→

Question 763 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$kx + 3y = 2k + 1$

$2(k + 1)x + 9y = 7k + 1$

Answer

Given,
$kx + 3y = 2k + 1$
$2(k + 1)x + 9y = 7k + 1$
To find: To determine for what value of k the system of equation has infinitely many solutions
We know that the system of equations
$a_1x + b_1y = c_1$
$a_2x + b_2y = c_2$_
Where, $a_1 = k, b_1 = 3, c_1 = -(2k + 1)$
and, $a_2 = 2(k + 1), b_2 = 9,$ and $c_2 = -(7k + 1)$
For infinitely many solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{\text{k}}{2(\text{k}+1)}=\frac{3}{9}=\frac{2\text{k}+1}{7\text{k}+1}$
Consider the following relation to find k,
$\frac{\text{k}}{2(\text{k}+1)}=\frac{3}{9}$
$\Rightarrow 9k = 6(k + 1)$
$\Rightarrow 9k - 6k - 6 = 0$
$\Rightarrow 3k = 6$
$\Rightarrow k = 2$
Now consider the following
$\frac{3}{9}=\frac{2\text{k}+1}{7\text{k}+1}$
$\Rightarrow 3(7k + 1) = 9(2k + 1)$
$\Rightarrow 21k + 3 = 18k + 9$
$\Rightarrow 21k - 18k = 9 - 3$
$\Rightarrow 3k = 6$
$\Rightarrow k = 2$
Hence for $k = 2$ the system of equation have infinitely many solutions.

View full question & answer→

Question 773 Marks

Write the number of solutions of the following pair of linear equations:

$x + 3y - 4 = 0$

$2x + 6y = 7$

Answer

The given equation are
$x + 3y - 4 = 0 .....(i)$
$2x + 6y - 7 = 0 ......(ii)$
The given equations are of the form
$a_1x + b_1y + c_1 = 0 .....(iii)$
$a_2x + b_2y + c_2 = 0 ......($iv)
Where, $a_1 = 1, b_1 = 3, c_1 = -4, a_2= 2, b_2= 6$ and $c_2= -7$
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{-4}{-7}=\frac{4}{7}$
Thus, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
Thus, given pair of linear equation has no solution.

View full question & answer→

Question 783 Marks

Write the value of k for which the system of equations $x + ky = 0, 2x - y = 0$ has unique solution.

Answer

The given equations are,
$x + ky = 0 .....(i)$
$2x - y = 0 ......(ii)$
The given equations are of the form,
$a_1x + b_1y + c_1 = 0 .....(iii)$
$a_2x + b_2y + c_2 = 0 ......(iv)$
Where, $a_1 = 1, b_1 = k, c_1 = 0, a_2= 2, b_2= -1$ and $c_2= 0$
For unique solution,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\Rightarrow\frac{1}{2}\neq\frac{\text{k}}{-1}$
$\Rightarrow\text{k}\neq\frac{-1}{2}$
Thus, given system of equations has unique solution if $\text{k}\neq\frac{-1}{2}$

View full question & answer→

Question 793 Marks

In the following system of equation determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

$2x + y = 5$

$4x + 2y = 10$

Answer

The given equations are,
$2x + y = 5 .....(i)$
$4x + 2y = 10 .......(ii)$
The given equations are of the from,
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where $a_1 = 2, b_1 = 1, c_1= -5$
$a_2 = 4, b_2 = 2$ and $c_2 = -10$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2},$
$\frac{\text{b}_1}{\text{b}_2}=\frac{1}{2},$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-5}{-10}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given system of equations has infinitely many solution.

View full question & answer→

Question 803 Marks

The car hire charges in a city comprise of a fixed charges together with the charge for the distance covered. For a journey of 12km, the charge paid is Rs. 89 and the journey of 20km, the charge paid is Rs. 145. What will a person have to pay for travelling a distance of 30km?

Answer

Let the fixed charges of car be Rs. x per km and the running charges be Rs. y km/hr. According to the given condition we have x + 12y = 89 .....(i) x + 20y = 145 ......(ii)

$\text{y}=\frac{-56}{-8}$ y = 7 Putting y = 7 in equation (i) we get x + 12y = 89 x + 12 × 7 = 89 x + 84 = 89 x = 89 - 84 x = 5 Therefore, Total charges for travelling distance of 30km = x + 30y = 5 + 30 × 7 = 5 + 210 = Rs. 215 Hence, a person have to pay Rs. 215 for travelling a distance of 30km.

View full question & answer→

Question 813 Marks

Write the value of k for which the system of equations has infinitely many solutions.

$2x - y = 5$

$6x + ky = 15$

Answer

The given equation are
$2x - y = 5 ....(i)$
$6x + ky = 15 .....(ii)$
The given equations are of the from
$a_1x + b_1y + c_1 = 0 .....(iii)$
$a_2x + b_2y + c_2 = 0 ......(iv)$
Where, $a_1 = 2, b_1 = -1, c_1 = -5, a_2= 6, b_2= k, c_2= -15$
For infinitely solution,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{6}=\frac{-1}{\text{k}}=\frac{-5}{15}$
$\Rightarrow\frac{2}{6}=\frac{-1}{\text{k}}$
$\Rightarrow\frac{1}{3}=\frac{-1}{\text{k}}$
$\Rightarrow\text{k}=-3$
Thus, $k = -3$

View full question & answer→

Question 823 Marks

Find the value of k for which the following system of equations has no solution:

$2x - ky + 3 = 0$

$3x + 2y - 1 = 0$

Answer

The given system of equations is
$2x - ky + 3 = 0$
$3x + 2y - 1 = 0$
The system of equations is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where, $a_1 = 2, b_1 = -k, c_1 = 3$
and,$ a_2 = 3, b_2 = 2, c_2 = -1$
For no solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{3}$
and, $\frac{\text{c}_1}{\text{c}_2}=\frac{3}{-1}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given system of equations will have no solution, if
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
i.e., $\frac{2}{3}=\frac{-\text{k}}{2}$
$\Rightarrow\text{k}=\frac{-4}{3}$
Hence, the given system of equations will have no solution, if $\text{k}=\frac{-4}{3}$

View full question & answer→

Question 833 Marks

A two-digit number is $4$ times the sum of its digits and twice the product of the digits. Find the number.

Answer

Let the digit in the unit's place be x and the digit at the ten's place be y. Then,
Number $= 10y + x$
The number obtained by reversing the order of the digits is $10x + y$
According to the given conditions we have
$10y + x = 4(x + y)$
$\Rightarrow 10y + x = 4x + 4y$
$\Rightarrow 10y - 4y + x - 4x = 0$
$\Rightarrow 6y - 3x = 0$
$\Rightarrow -3x + 6y = 0$
$\Rightarrow 3x - 6y = 0$
$\Rightarrow 3(x- 2y) = 0$
$\Rightarrow x - 2y = 0 ......(i)$
and, $10y + x = 2(xy)$
$\Rightarrow 10y + x = 2xy .......(ii)$
Substituting x = 2y in equation (ii) we get
$10y + 2y = 2 \times (2y) \times y$
$\Rightarrow 12y = 4y^2$
$\Rightarrow 3y = y^2$
$\Rightarrow y^2 - 3y = 0$
$\Rightarrow y(y - 3) = 0$
$\Rightarrow\text{y}\neq0$ or y = 3 [ $\because$ Ten's digit can not be 0]
Putting y = 3 in equation (i) we get
$x = 2 × 3 = 6$
Hence, the required number is 10y + x
$= 10 × 3 + 6 = 36$

View full question & answer→

Question 843 Marks

Solve the following systems of equations:
$\frac{2}{\text{x}}+\frac{5}{\text{y}}=1,$
$\frac{60}{\text{x}}+\frac{40}{\text{y}}=19,\text{x}\neq0,\text{y}\neq0.$

Answer

The given equations are
$\frac{2}{\text{x}}+\frac{5}{\text{y}}=1\ ......(\text{i})$
$\frac{60}{\text{x}}+\frac{40}{\text{y}}=19\ ......(\text{ii})$
Multiply equation (i) by 8 and subtract (ii) from equation (i), we get
$-\frac{44}{\text{x}}=-11$
$\Rightarrow\text{x}=4$
Put the value of x in equation (i) we get
$\Rightarrow\frac{2}{4}+\frac{5}{\text{y}}=1$
$\Rightarrow\frac{5}{\text{y}}=1-\frac{2}{4}$
$\Rightarrow\text{y}=10$
Hence the value of x = 4 and y = 10.

View full question & answer→

Question 853 Marks

Write the value of k for which the system of equations $x + y - 4 = 0$ and $2x + ky - 3 = 0$ has no solution.

Answer

The given system of equation is
$x + y - 4 = 0$
$2x + ky - 3 = 0$
$a_1 = 1, a_2= 2, b_1 = 1, b_2 = k, c_1 = 4, c_2 = 3$
For the equations to have no solutions $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\frac{1}{2}=\frac{1}{\text{k}}$
By cross-multiplication we get,
$1 \times k = 1 \times 2$
$k = 2$
Hence, the value of k is 2 when system equations has no solution.

View full question & answer→

Question 863 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$4x + 5y = 3$

$kx + 15y = 9$

Answer

The given system of equation is
$4x + 5y - 3 = 0$
$kx + 15y - 9 = 0$
The given system of equation is of the form
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
Where,$ a_1 = 4, b_1 = 5, c_1 = -3$
$a_2 = k, b_2 = 15, c_2 = -9$
For a unique solution we must have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{4}{\text{k}}=\frac{5}{15}=\frac{-3}{-9}$
Now,
$\frac{4}{\text{k}}=\frac{5}{15}$
$\Rightarrow\frac{4}{\text{k}}=\frac{1}{3}$
$\Rightarrow\text{k}=12$
Hence, the given system of equations will have infinitely many solutions if k = 12.

View full question & answer→

Generate Paper Free All Linear Equations in Two Variables topics