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Maths 4 Marks Questions

Linear Equations In Two Variables · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 209 questions.

Questions · Page 2 of 5

4 Marks Questions

Question 514 Marks
If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.
Answer
Let the length and breadth of the rectangle be x and y units respectively.
Then area of rectangle = xy square units
It is given that if length is increared and breadth reduced each by 2 units. then the area is reduced by 28 square units.
⇒ (x + 2) (y - 2) = xy - 28
⇒ xy - 2x + 2y - 4 = xy - 28
⇒ -2x + 2y - 4 + 28 = 0
⇒ -2x + 2y + 24 = 0
⇒ 2x - 2y - 24 = 0 ......(i)
It is also given that the length is reduced by 1 unit and breath is increcred by 2 units then the area is increared by 33 square units.
⇒ (x - 1) (y + 2) = xy + 33
⇒ xy + 2x - y - 2 = xy + 33
⇒ xy + 2x - y - 2 - xy - 33 = 0
⇒ 2x - y - 35 = 0 ......(ii)
Now, subtracting eq. (ii) from eq. (i) and we get
⇒ 2x - 2y - 24 - (2x - y - 35) = 0
⇒ 2x - 2y - 24 - 2x + y + 35 = 0
⇒ -y + 11 = 0
⇒ y = 11
Putting the value of y in eq. (i)
⇒ 2x - 2y - 24 = 0
⇒ 2x - 2 × 11 - 24 = 0
⇒ 2x - 22 - 24 = 0
⇒ 2x - 46 = 0
⇒ x = 23
Hence, the length of the rectangle is 23 and breadth of the rectangle is 11
Area os rectangle = length × breadth
= 23 × 11
= 253 square units.
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Question 524 Marks
Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:
3x + 2y -4 = 0
2x - 3y -7 = 0.
Answer
The system of given equations is,
3x + 2y -4 = 0
2x - 3y -7 = 0
Now, 3x + 2y -4 = 0
⇒ 3x = 4 - 2y
$\Rightarrow\text{x}=\frac{4-2\text{y}}{3}$
When y = 5, we have,
$\text{x}=\frac{4-2\times5}{3}=-2$
When y = 8, we have,
$\text{x}=\frac{4-2\times8}{3}=-4$
Thus, we have the following table,
x
 -2
-4
y
5
8
We have, 2x - 3y -7 = 0
⇒ 2x = 3y + 7
$\Rightarrow\text{x}=\frac{3\text{y}+7}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1+7}{2}=5$
When y = -1, we have,
$\text{x}=\frac{3\times(-1)+7}{2}=2$
Thus, we have the following table,
x
 5
2
y
 1
-1
Graph of the given system of equations.

Clearly, the two lines intersect at P(2, -1).
Hence, x = 2, y = -1 is the solution of the given system of equations.
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Question 534 Marks
Jamila spld a table and chair for ₹ 1050, there by making a profit of 10% on a table and 25% on the chair. If she had taken profit of 25% on the table and 10% on the chair she would have got ₹ 1065. Find the cost price of each.
Answer
Let the cost price of the table be ₹ x
and the cost price of the chair by ₹ y
The selling price of the table, when it is sold at a profit of 10%
$=₹\Big(\text{x}+\frac{10}{100}\text{x}\Big)$
$=₹\frac{110}{100}\text{x}$
The selling price of the chir when it is sold at a profit of 25%
$=₹\Big(\text{y}+\frac{25}{100}\text{y}\Big)$
$=₹\frac{125}{100}\text{y}$
So, $\frac{110}{100}\text{x}+\frac{125}{100}\text{y}=1050\ .....(\text{i})$
When the table is sold at a profit of 25%, its
Selling price $=₹\Big(\text{x}+\frac{25}{100}\text{x}\Big)$
$=₹\frac{125}{100}\text{x}$
When the chair is sold at a profit of 10%, its
Selling price $=₹\Big(\text{y}+\frac{10}{100}\text{y}\Big)$
$=₹\frac{110}{100}\text{y}$
So, $\frac{125}{100}\text{x}+\frac{110}{100}\text{y}=1065\ ....(\text{ii})$
From equations (i) and (ii) we get
110x + 125y = 105000
and 125x + 110y = 106500
On adding and subtracting these equations, we get
235x + 235y = 211500
and 15x – 15y= 1500
i.e., x + y = 900 ......(iii)
and x – y = 100 ......(iv)
Solving equation (iii) and (iv), we get
2x = 1000
x = 500
500 + y = 900
y = 900 - 500
y = 400
x = 500, y = 400
So, the cost price of the table is ₹ 500 and the cost price of the chair is ₹ 400
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Question 544 Marks
On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs. 2000. But if he sells the T.V. at 10% gain the fridge at 5% loss. He gains Rs. 1500 on the transaction. Find the actual prices of T.V. and fridge.
Answer
Let the price of a T.V. be Rs. x and that of a fridge be Rs. y. Then we have
$\frac{5\text{x}}{100}+\frac{10\text{y}}{100}=2000$
⇒ 5x + 10y = 200000
⇒ 5(x + 2y) = 200000
⇒ x + 2y = 40000 ......(i)
And, $\frac{10\text{x}}{100}-\frac{5\text{y}}{100}=1500$
⇒ 10x - 5y = 150000
⇒ 5(2x - y) = 150000
⇒ 2x - y = 30000 ....(ii)
Multiplying equation (ii) by 2 we get
4x - 2y = 60000 .....(iii)
Adding equation (i) and equation (iii) we get
x + 4x = 40000 + 60000
⇒ 5x = 100000
⇒ x = 20000
Putting x = 20000 in equation (i) we get
20000 + 2y = 40000
⇒ 2y = 40000 - 20000
$\Rightarrow\text{y}=\frac{20000}{2}=10000$
Hence, the actual price of T.V = Rs. 20,000 and, the actual price of fridge = Rs. 10, 000
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Question 554 Marks
A wizard having powers of mystic in candations and magical medicines seeing a cock, fight going on, spoke privately to both the owners of cocks. To one he said, if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that'. Going to the other, he promised in the same way to give three fourths. From both of them his gain would be only 12 gold coins. Find the the stake of money each of the cock-owners have.
Answer
Let the stake money of first and second cock-owners be Rs. x and Rs. y respectively. Then,
$\text{y}-\frac{2}{3}\text{x}=12$
⇒ 3y - 2x = 36 .....(i)
And, $\text{x}-\frac{3}{4}\text{y}=12$
⇒ 4x - 3y = 48 .....(ii)
Multiplying equation (i) by 2 we get
6y - 4x = 72 .....(iii)
Adding equation (ii) and (iii) we get
-3y + 6y = 48 + 72
⇒ 3y = 120
$\Rightarrow\text{y}=\frac{120}{3}$
⇒ y = 40
Putting y = 40 in equation (ii) we get
4x - 3 × 40 = 48
⇒ 4x - 120 = 48
⇒ 4x = 48 + 120
⇒ 4x = 168
$\Rightarrow\text{x}=\frac{168}{4}$
⇒ x = 42
Hence, the stake of money of its cock-owner = 42 gold and, the stake of money of II nd cock-owner = 40 gold coins.
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Question 564 Marks
Solve the following systems of equations:
21x + 47y = 110,
47x + 21y = 162.
Answer
21x + 47y = 110 ....(i)
47x + 21y = 162 .....(ii)
Adding (i) and (ii) we get
68x + 68y = 272
⇒ x + y = 4 ....(iii)
Subtracting (i) from (ii), we get
26x - 26y = 52
⇒ x - y = 2 ......(iv)
Adding (iii) and (iv), we get
2x = 6
⇒ x = 3
Putting x = 3 in (iv) we get
3 - y = 2
⇒ y = 1.
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Question 574 Marks
A is elder to B by 2 years. A's father F is twice as old as A and B is twice as old as his sister S. If the ages of the father and sister differ by 40 years, find the age of A.
Answer
Let age of A = x years
and age of B = y years
According to the conditions,
x = y + 2
⇒ y = x – 2 ….(i)
Age of A’s’ father = 2x
Age of B’s sisters $=\frac{\text{y}}{2}$
$2\text{x}-\frac{\text{y}}{2}=40$
4x – y = 80 ….(ii)
4x – (x – 2) = 80
⇒ 4x – x + 2 = 80
3x = 80 – 2 = 78
= x = 26
A’s age = 26 years
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Question 584 Marks
Solve the following system of equations by the method of cross-multiplication:
2(ax - by) + a + 4b = 0
2(bx + ay) + b - 4b = 0
Answer
The given equations are,
2(ax - by) + a + 4b = 0 .....(i)
2(bx + ay) + b - 4b = 0 ........(ii)
By cross-multiplication method we get,
$\Rightarrow\frac{\text{x}}{(-2\text{b})(\text{b}-4\text{a})-(\text{a}+4\text{b})2\text{a}}=\frac{\text{y}}{(\text{a}+4\text{b})2\text{b}-2\text{a}(\text{b}-4\text{a})}\\=\frac{1}{2\text{a}\times2\text{a}-2\text{b}(-2\text{b})}$
$\Rightarrow\frac{\text{x}}{-2\text{b}^2+8\text{ab}-2\text{a}^2-8\text{ab}}=\frac{\text{y}}{2\text{ab}+8\text{b}^2-2\text{ab}+8\text{a}^2}\\=\frac{1}{4\text{a}^2+4\text{b}^2}$
$ \Rightarrow\frac{\text{x}}{-2(\text{b}^2+\text{a}^2)}=\frac{\text{y}}{8(\text{a}^2+\text{b}^2)}=\frac{1}{4(\text{a}^2+\text{b}^2)}$
$\text{x}=\frac{-2(\text{b}^2+\text{a}^2)}{4(\text{a}^2+\text{b}^2)}=-\frac{1}{2}$
$\text{y}=\frac{8(\text{a}^2+\text{b}^2)}{4(\text{a}^2+\text{b}^2)}=2$
Thus, $\text{x}=-\frac{1}{2}$ and $\text{y}=2$
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Question 594 Marks
Show graphically that the following system of equation is in-consistent (i.e. has no solution):
3x - 5y = 20
6x - 10y = -40
Answer
We have,
3x - 5y = 20
6x - 10y = -40
Now, 3x - 5y = 20
$\Rightarrow\text{x}=\frac{5\text{y}+20}{3}$
When y = -1, we have,
$\text{x}=\frac{5(-1)+20}{3}=5$
When y = -4, we have,
$\text{x}=\frac{5(-4)+20}{3}=0$
Thus we have the following table giving points on the line 3x - 5y = 20.
x
5
0
y
-1
-4
Now, 6x - 10y = -40
⇒ 6x = -40 + 10y
$\Rightarrow\text{x}=\frac{-40+10\text{y}}{6}$
When y = 4, we have,
$\text{x}=\frac{-40+10\times4}{6}=0$
When y = 1, we have,
$\text{x}=\frac{-40+10\times1}{6}=-5$
Thus we have the following table giving points on the line 6x - 10y = -40
x
0
-5
y
4
1
Graph of the given equations:

Clearly, there is no common points between these two lines.
Hence, given system of equations is in-consistent.
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Question 604 Marks
Solve the following systems of equations:
$\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2,$
$\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1.$
Answer
The given equations are
$\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2$
$\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1$
Let $\frac{1}{\text{x}+\text{y}}=\text{u}$ and $\frac{1}{\text{x}+\text{y}}=\text{v}$ then equations are
$3{\text{u}+2\text{v}}=2\ ......(\text{i})$
$9{\text{u}-4\text{v}}=1\ ......(\text{ii})$
Multiply equation (i) by 2 and add both equation we get
$\frac{6{\text{u}\ +\ 4\text{v}}\ \ \ =\ 4\\9{\text{u}\ -\ 4\text{v}}\ \ \ =\ 1}{15\text{u}\ \ \ \ \ \ \ \ \ =\ 5}$
$\Rightarrow\text{u}=\frac{1}{3}$
Put the value of u in equation (i) we get
$3\times\frac{1}{3}+2\text{v}=2$
$\Rightarrow2\text{v}=1$
$\Rightarrow\text{v}=\frac{1}{2}$
Then $\frac{1}{\text{x}+\text{y}}=\frac{1}{3}$
$\Rightarrow\text{x}+\text{y}=3$
$$$\frac{1}{\text{x}-\text{y}}=\frac{1}{2}$
$\Rightarrow\text{x}-\text{y}=2$
Add both equations we get
$\frac{\text{x}\ +\ \text{y}\ =\ 3\\ \text{x}\ +\ \text{y}\ =\ 2}{2\text{x}\ \ \ \ \ \ \ =\ 5}$
$\Rightarrow\text{x}=\frac{5}{2}$
Put the value of x in first equation we get
$\frac{5}{2}+\text{y}=3$
$\Rightarrow\text{y}=\frac{1}{2}$
Hence the value of $\text{x}=\frac{5}{2}$ and $\text{y}=\frac{1}{2}$
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Question 614 Marks
Point A and B are 70km. a part on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if the travel towards each other, the meet in one hour. Find the speed of the two cars.
Answer
Let x km/hr and y km/hr be the speed of cars A and B respectvely.
When they move in same direction,
Supposr they meet at C. Then
Distance cover by A = AC
Distance cover by B = BC
Distance = Speed × time
Distance cover by A = 7x km
Distance cover by B = 7y km
Cleary, AC - BC = AB
7x - 7y = 70
x - y = 10 ......(i)
When they move in opposite direction
Suppose they meet at 0
Distance cover by A = AD
Distance cover by B = BD
Distance cover by A = x km
Distance cover by B = y km
Clearly, x + y = 70 ......(ii)
Adding (i) and (ii) we get
⇒ 2x = 80
$\Rightarrow\text{x}=\frac{80}{2}$
⇒ x = 40km/hr
Putting x = 40 in (ii) we get
⇒ 40 + y = 70
⇒ y = 30km/hr
Thus, speed of car A and B is 40km/hr and 30km/hr.
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Question 624 Marks
A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.
Answer
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x
The sum of the digits of the number is 5. Thus, we have x + y = 5
After interchanging the digits, the number becomes 10x + y
The number obtained by interchanging the digits is greater by 9 from the original number.
Thus, we have
10x + y = 10y + x + 9
⇒ 10x + y - 10y - x = 9
⇒ 9x - 9y = 9
⇒ 9(x - y) = 9
$\Rightarrow\text{x}-\text{y}=\frac{9}{9}$
⇒ x - y = 1
So, we have two equations
x + y = 5
x - y = 1
Here x and y are unknowns. we have to solve the above equations for x and y.
Adding the two equations, we have
(x + y) + (x - y) = 5 + 1
⇒ x + y + x - y = 6
⇒ 2x = 6
$\Rightarrow\text{x}=\frac{6}{2}$
⇒ x = 3
Substituting the value of x in the first equation, we have
3 + y = 5
⇒ y = 5 - 3
⇒ y = 2
Hence, the number is 10 × 2 + 3 = 23
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Question 634 Marks
Solve the following system of equations by the method of cross-multiplication:
$2x + y = 35,$
$3x + 4y = 65.$
Answer
The given system of equations may be written as
$2x + y - 35 = 0$
$3x + 4y - 65 = 0$
Here, $a_1 = 2, b_1 = 1, c_1 = -35$
$a_2 = 3, b_2 = 4,$ and $c_2 = -65$
By cross-multiplication, we have
$\Rightarrow\ \frac{\text{x}}{1\times(-65)-(-35)\times4}=\frac{-\text{y}}{2\times(-65)-(-35)\times3}\\=\frac{1}{2\times4-1\times3}$
$\Rightarrow\ \frac{\text{x}}{-65+140}=\frac{-\text{y}}{-130+105}=\frac{1}{8-3}$
$\Rightarrow\ \frac{\text{x}}{75}=\frac{-\text{y}}{-25}=\frac{1}{5}$
$\Rightarrow\ \frac{\text{x}}{75}=\frac{\text{y}}{25}=\frac{1}{5}$
Now, $\frac{\text{x}}{75}=\frac{1}{5}$
$\Rightarrow\ \text{x}=\frac{75}{5}=15$
and, $\frac{\text{y}}{25}=\frac{1}{5}$
$\Rightarrow\ \text{y}=\frac{25}{5}=5$
Hence, $x = 15, y = 5$ is the solution of the given system of equations.
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Question 644 Marks
Solve the following system of equations by the method of cross-multiplication:
$\frac{\text{a}^2}{\text{x}}-\frac{\text{b}^2}{\text{y}}=0$
$\frac{\text{a}^2\text{b}}{\text{x}}+\frac{\text{b}^2\text{a}}{\text{y}}=\text{a}+\text{b}$ $\text{x},\text{y}\neq0$
Answer
Taking $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}.$ Then, the given system of equations become
$a^2u - b^2v = 0$
$a^2bu + b^2av - (a + b) = 0$
Here,
$a_1= a^2, b_1 = -b^2, c_1 = 0$
$a_2^= a^2b, b_2 = b^2a$, and $c_2 = -(a + b)$
By cross multiplication we have,
$\Rightarrow\frac{\text{u}}{\text{b}^2(\text{a}+\text{b})-0\times\text{b}^2\text{a}}=\frac{-\text{v}}{-\text{a}^2(\text{a}+\text{b})-0\times\text{a}^2\text{b}}=\frac{1}{\text{a}^3\text{b}^2+\text{a}^2 \text{b}^3}$
$\Rightarrow\frac{\text{u}}{\text{b}^2(\text{a}+\text{b})}=\frac{\text{v}}{\text{a}^2(\text{a}+\text{b})}=\frac{1}{\text{a}^2\text{b}^2(\text{a}+ \text{b})}$
Now, $\frac{\text{u}}{\text{b}^2(\text{a}+\text{b})}=\frac{1}{\text{a}^2\text{b}^2(\text{a}+ \text{b})}$
$\Rightarrow\text{u}=\frac{\text{b}^2(\text{a}+\text{b})}{\text{a}^2\text{b}^2(\text{a}+ \text{b})}$
$\Rightarrow\text{u}=\frac{1}{\text{a}^2}$
And, $\frac{\text{v}}{\text{a}^2(\text{a}+\text{b})}=\frac{1}{\text{a}^2\text{b}^2(\text{a}+\text{b})}$
$ \Rightarrow{\text{v}}=\frac{\text{a}^2(\text{a}+\text{b})}{\text{a}^2\text{b}^2(\text{a}+\text{b})}$
$ \Rightarrow{\text{v}}=\frac{1}{\text{b}^2}$
Now, ${\text{x}}=\frac{1}{\text{u}}=\text{a}^2$
And, ${\text{y}}=\frac{1}{\text{v}}=\text{b}^2$
Hence, $x = a^2, y = b^2$​​​​​​​ is the solution of the given system of equations.
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Question 654 Marks
Find the value of k for which each of the following system of equations have infinitely many solutions:
$2x - 3y = 7$
$(k + 2)x - (2k + 1)y = 3(2k - 1)$
Answer
The given system of equations may be written as
$2x - 3y - 7 = 0$
$(k + 2)x - (2k + 1)y - 3(2k - 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 = -3, c_1 = -7$
and, $a_2 = k + 2, b_2 = -(2k + 1), c_2 = -3(2k - 1)$
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{2}{\text{k}+2}=\frac{-3}{-(2\text{k}+1)}=\frac{-7}{-3(2\text{k}-1)}$
$\Rightarrow\frac{2}{\text{k}+2}=\frac{-3}{-(2\text{k}+1)}$ and $\frac{-3}{-(2\text{k}+1)}=\frac{-7}{-3(2\text{k}-1)}$
$\Rightarrow 2(2k + 1) = 3(k + 2)$ and $3 \times 3(2k - 1) = 7(2k + 1)$
$\Rightarrow 4k + 2 = 3k + 6$ and $18k - 9 = 14k + 7$
$\Rightarrow 4k - 3k = 6 - 2$ and $18k - 14k = 7 + 9$
$\Rightarrow k = 4$ and $4k = 16 \Rightarrow k = 4$
$\Rightarrow k = 4$ and $k = 4$
Hence the given system of equations will have infinitely many solutions, if $k = 4$
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Question 664 Marks
A man travels 600km partly by train and partly by car. If the covers 400km by train and the rest by car, it takes him 6 hours 30 minutes. But, if the travels 200km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Answer
Let the speed of the train be x km/ hr. and that of the car be y km/ hr. we have following cases:
Case I: When Ramesh traveis 400km by train and the rest by car:
In this case we have
Time taken by the man to travel 400km by train $=\frac{400}{\text{x}}$
Times taken by the man to travel (600-400) = 200km by car $=\frac{200}{\text{y}}$
In this case, total times of the journey is 6 hrs 30 minutes.
$\therefore\frac{400}{\text{x}}+\frac{200}{\text{y}}=6\text{hrs. }30\text{ minutes}$
$\Rightarrow\frac{400}{\text{x}}+\frac{200}{\text{y}}=6\frac{1}{2}$
$\Rightarrow\frac{400}{\text{x}}+\frac{200}{\text{y}}=\frac{13}{2}\ .....(\text{i})$
Case II: when he travels 200km by train and the rest by car:
In this case we have
Times taken by the man to travel 200km by train $=\frac{200}{\text{x}}\text{hrs.}$
Times taken by the man to travel (600 - 200) = 400 km by car $=\frac{400}{\text{y}}\text{hrs}$
In this case, total time of journey is $\Big(\frac{13}{2}+\frac{1}{2}\Big)=7\text{hrs.}$
$\therefore\frac{200}{\text{x}}+\frac{400}{\text{y}}=7\ ......(\text{ii})$
Putting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v},$ in equation (i) and (ii) we get
$400\text{u}+200\text{v}=\frac{13}{2}\ .....(\text{iii})$
$200\text{u}+400\text{v}=7\ .....(\text{iv})$
Multiplying equation (iii) by 2 we get
$800\text{u}+400\text{v}=13\ .....(\text{v})$
Subtracting equation (iv) by equation (v) we get
$$$800\text{u}-200\text{u}=13-7$
$\Rightarrow600\text{u}=6$
$\Rightarrow\text{u}=\frac{6}{600}=\frac{1}{100}$
Putting $\text{u}=\frac{1}{100}$ in equation (iv) we get
$200\times\frac{1}{100}+400\text{v}=7$
$\Rightarrow2+400\text{v}=7$
$\Rightarrow400\text{v}=7-2$
$\Rightarrow400\text{v}=5$
$\text{v}=\frac{5}{400}=\frac{1}{80}$
Now, $\text{u}=\frac{1}{100}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{100}$
$\Rightarrow\text{x}=100$
And, $\text{v}=\frac{1}{80}$
$\Rightarrow\frac{1}{\text{y}}=\frac{1}{80}$
$\Rightarrow\text{y}=80$
Hence, speed of the train = 100km/ hr. Speed of the car = 80 km/ hr.
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Question 674 Marks
In a $\triangle\text{ABC},\angle\text{A}=\text{x}^\circ,\angle\text{B}=3\text{x}^\circ$ and $\angle\text{C}=\text{y}^\circ$ if 3y - 5x = 30, , prove that the triangle is right angled.
Answer
We have,
$\angle\text{A}=\text{x}^\circ\ ....(\text{i})$
$\angle\text{B}=3\text{x}^\circ\ .....(\text{ii})$
And, $\angle\text{C}=\text{y}^\circ\ ......(\text{iii})$
We know that, the sum of angles of a triangle is 180°
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
x + 3x + y = 180 [using (i), (ii) and (iii)]
⇒ 4x + y = 180 .....(iv)
Now, 3y - 5x = 30 .....(v) [given]
Multiplying equation (iv) by 3 we get
12x + 3y = 540 .....(vi)
Subtracting equation (v) from equation (vi) we get
12x + 5x = 540 - 30
⇒ 17x = 510
$\Rightarrow\text{x}=\frac{510}{17}$
⇒ x = 30
Putting x = 30 in equation (iv) we get
4 × 30 + y = 180
⇒ 120 + y = 180
⇒ y = 180 - 120
⇒ y = 60
Now, $\angle\text{B}=3\text{x}^\circ$
$\Rightarrow\angle\text{B}=3\times30^\circ$
$\Rightarrow\angle\text{B}=90^\circ$
$\therefore\triangle\text{ABC}$ is the right angle triangle.
Hence prived.
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Question 684 Marks
Find the value of k for which each of the following system of equations have infinitely many solutions:
$x + (k + 1)y = 4$
$(k + 1)x + 9y = 5k + 2$
Answer
The given system of equations may be written as
$x+(k+1) y-4=0$
$(k+1) x+9 y-(5 k+2)=0$
The system of equations is of the form
$a_1 x+b_1 y+c_1=0$
$a_2 x+b_2 y+c_2=0$
Where, $a_1=1, b_1=k+1, c_1=-4$
$\text { and, } a_2=k+1, b_2=9 \text {, and } c_2=-(5 k+2)$
For infinitely many solutions, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{1}{ k +1}=\frac{ k +1}{9}=\frac{-4}{-(5 k+2)}$
$\Rightarrow \frac{1}{ k +1}=\frac{ k +1}{9} \text { and } \frac{ k +1}{9}=\frac{- 4 }{-(5 k +2)}$
$\Rightarrow 9=(k+1)^2 \text { and }(k+1)(5 k+2)=36$
$\Rightarrow 9=k^2+1+2 k \text { and } 5 k^2+2 k+5 k+2=36$
$\Rightarrow k^2+2 k+1-9=0 \text { and } 5 k^2+7 k+2-36=0$
$\Rightarrow k^2+2 k-8=0 \text { and } 5 k^2+7 k-34=0$
$\Rightarrow k^2+4 k-2 k-8=0 \text { and } 5 k^2+17 k-10 k-34=0$
$\Rightarrow k(k+4)-2(k+4)=0 \text { and } k(5 k+17)-2(5 k+7)=0$
$\Rightarrow(k+4)(k-2)=0 \text { and }(5 k+17)(k-2)=0$
$\Rightarrow(k=-4 \text { or } k=2) \text { and }\left(k=\frac{-17}{5} \text { or } k=2\right)$
$\Rightarrow k=2 \text { satisfies both the conditions }$
Hence, the given system of equations will have infinitely many solutions, if $k=2$
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Question 694 Marks
Solve graphically that the following system of equation has infinitely many solutions:
x - 2y + 11 = 0
3x - 6y + 33 = 0
Answer
The given equations are
x - 2y + 11 = 0 .......(i)
3x - 6y + 33 = 0 ..........(ii)
Putting x = 0 in equation (i), we get,
⇒ 0 - 2y = -11
$\Rightarrow\text{y}=\frac{11}{2}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{11}{2}$
Putting y = 0 in equation (i), we get,
⇒ x - 2 × 0 = -11
⇒ x = -11
⇒ x = -11, y = 0
Use the following table to draw the graph,
x
0
-11
y
$\frac{11}{2}$
0
Draw the graph by plotting the two points $\text{A}\Big(0,\frac{11}{2}\Big),$ B(-11, 0) from table.

Graph of the equation,
3x - 6y = -33 .......(ii)
Putting x = 0 in equation (ii), we get,
⇒ 3 × 0 - 6y = -33
$\Rightarrow\text{y}=\frac{11}{2}$
$\Rightarrow\text{x}=0,\text{y}=\frac{11}{2}$
Putting y = 0 in equation (ii), we get,
⇒ 3x - 6 × 0 = -33
⇒ x = -11
⇒ x = -11, y = 0
Use the following table to draw the graph.
x
0
-11
y
$\frac{11}{2}$
0
Draw the graph by plotting the two points $\text{C}\Big(0,\frac{11}{2}\Big),$ (11, 0) from table. Thus the graph of the two equations are coincide Consequently, every solution of one equation is a solution of the other.
Hence the equations have infinitely many solutions.
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Question 704 Marks
Vijay had some bananas, and he divided them into two lots A and B. He sold first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of ₹ 1 per banana and got a total of ₹ 400. If he had sold the first lot at the rate of ₹ 1 per banana and the second lot at the rate of ₹ 4 per five bananas, his total collection would have been ₹ 460. Find the total number of bananas he had.
Answer
Let the number of bananas in lots A and B be x and y, respectively. Case I: Cost of the first lot at the rate of ₹ 2 for 3 bananas + Cost of the second lot at the rate of ₹ 1 per banana = Amount received $\Rightarrow\frac{2}{3}\text{x}+\text{y}=400$ ⇒ 2x + 3y = 1200 .....(i)Case II:
Cost of the first lot at the rate of ₹ 1 per banana + Cost of the second lot at the rate of ₹ 4 for 5 bananas = Amount received $\Rightarrow\text{x}+\frac{4}{5}\text{y}=460$ ⇒ 5x + 4y = 2300 .....(ii) On multiplying in the Eq. (i) by 4 and Eq. (ii) by 3 and then subtracting them, we get, (image) Now, putting the value of x in Eq. (i), we get 2 x 300 + 3y = 1200 ⇒ 600 + 3y = 1200 ⇒ 3y = 1200 - 600 ⇒ 3y = 600 ⇒ y = 200 Total number of bananas = Number of bananas in lot A + Number of bananas in lot B = x + y = 300 + 200 = 500 Hence, he had 500 bananas.
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Question 714 Marks
Solve the following systems of equations graphically:
x - 2y = 6
3x - 6y = 0
Answer
We have
x - 2y = 6
3x - 6y = 0
Now, x - 2y = 6
⇒ x = 6 + 2y
When y = -2, we have,
x = 6 + 2 × -2 = 2
When y = -3, we have,
x = 6 + 2 × -3 = 0
Thus, we have the following table giving points on the line x - 2y = 6
x
2
0
y
-2
-3
Now, 3x - 6y = 0
3x = 6y
x = 2y
When y = 0, we have,
x = 0
When y = 1, we have,
x = 2
Thus, we have following table giving points on the line 3x - 6y = 0.
x
0
2
y
0
1
Graph of the given equation are,

Clearly, two lines are parallel to each other. So, the two lines have no common point.
Hence, the given system of equations has no solution.
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Question 724 Marks
Find the values of a and b for which the following system of equations has infinitely many solutions:
$(a - 1)x + 3y = 2$
$6x + (1 - 2b)y = 6$
Answer
The given equations are
$(a-1) x+3 y=2 \ldots$
$6 x+(1-2 b) y=6$
The given equations are of the form
$a_1 x+b_1 y+c_1=0$
$a_2 x+b_2 y+c_2=0$
Where $a_1=(a-1), b_1=3, c_1=-2$
$a_2=6, b_2=(1-2 b) \text {, and } c_2=-6$
For infinite many solution
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{(a-1)}{6}=\frac{3}{(1-2 b)}=\frac{-2}{-6}$
$\Rightarrow \frac{(a-1)}{6}=\frac{-2}{-6} \text { and } \frac{3}{(1-2 b)}=\frac{-2}{-6}$
$\Rightarrow 3(a-1)=6 \text { and } 3 \times 3=1-2 b$
$\Rightarrow 3 a=6+3 \text { and } 9=1-2 b$
$\Rightarrow a=\frac{9}{3} \text { and } 2 b=1-9$
$\Rightarrow a=3 \text { and } b=-4$
Thus, $a=3$ and $b=-4$
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Question 734 Marks
Draw the graphs of the following equations on the same graph paper.
2x + 3y = 12,
x - y = 1.
Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis.
Answer
The given equations are
2x + 3y = 12 .......(i)
x - y = 1 .........(ii)
Putting x = 0 in equation (i), we get,
⇒ 2 × 0 + 3y = 12
⇒ y = 4
⇒ x = 0, y = 4
Putting y = 0 in equation (i), we get,
⇒ x + 3 × 0 = 6
⇒ x = 6
⇒ x = 6, y = 0
Use the following table to draw the graph
x
0
6
y
4
0
Draw the graph by plotting the two points A(0, 2) and B(6, 0) from table.

Putting x = 0 in equation (ii) we get,
⇒ 0 - y = 1
⇒ y = -1
⇒ x = 0, y = -1
Putting y = 0 in equation (ii) we get,
⇒ x - 0 = 1
⇒ x = 1
⇒ x = 1, y = 0
Use the following table to draw the graph.
x
0
1
y
-1
0
Draw the graph by plotting the two points C(0, -1), D(1, 0) from table. Draw the graph by plotting the two points from table. The intersection point is P(3, 2). Three points of the triangle are A(0, 4), C(0, -1) and P(3, 2).
Hence the value of x = 3 and y = 2.
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Question 744 Marks
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the number of student in the class.
Answer
Let the number of students be x and the number of row be y. Then,
Number of students in each row $=\frac{\text{x}}{\text{y}}$
Where three students is extra in each row, there are one row less that is when each row has $\Big(\frac{\text{x}}{\text{y}}+3\Big)$
Students the number of rows is (y - 1)
Total number of students = no. of rows x no. of students in each row
$\text{x}=​​\Big(\frac{\text{x}}{\text{y}}+3\Big)(\text{y}-1)$
$\text{x}=\Big(\text{x}+3\text{y}-\frac{\text{x}}{\text{y}}-3\Big)$
$0=\frac{-\text{x}}{\text{y}}+\text{x}-\text{x}+3\text{y}-3$
$0=\frac{-\text{x}}{\text{y}}+\text{x}-\text{x}+3\text{y}-3$
$0=\frac{-\text{x}}{\text{y}}+3\text{y}-3$
If three students are less in each row then there are 2 rows more that is when each row has
$​​\Big(\frac{\text{x}}{\text{y}}-3\Big)(\text{y}+2)$
Therefore, total number of students = Number of rows × Number of students in each row
$\text{x}=​​\Big(\frac{\text{x}}{\text{y}}-3\Big)(\text{y}+2)$
$\text{x}=\text{x}-3\text{y}+\frac{2\text{x}}{\text{y}}-6$
$0=\frac{2\text{x}}{\text{y}}+\text{x}-\text{x}-3\text{y}-6$
$0=\frac{2\text{x}}{\text{y}}-3\text{y}-6.......(\text{ii})$
Putting $\frac{\text{x}}{\text{y}}=\text{u}$ in (i) and (ii) equation we get
$\text{u}+3\text{y}-3=0\ .....(\text{iii})$
$2\text{u}-3\text{y}-6=0\ .....(\text{iv})$
Adding (iii) and (iv) equation we get
$-\text{u}\ +\ 3\text{y}\ -\ 3\ =\ 0\\2\text{u}\ -\ 3\text{y}\ -\ 6\ =\ 0\over\text{u} \ \ \ \ \ \ \ \ \ \ \ \ 9\ \ \ \ \ \ \ \ \ \ \ =\ 0$
u = 9
Putting u = 9 in equation (iii) we get
$\text{u}+3\text{y}-3=0$
$-9+3\text{y}-3=0$
$3\text{y}-12=0$
$3\text{y}=12$
$\text{y}=\frac{12}{3}$
$\text{y}=4$
$\text{u}=9$
$\frac{\text{x}}{\text{y}}=9$
$\frac{\text{x}}{4}=9$
$\text{x}=9\times4$
$\text{x}=36$
Hence, the number of students in the class is 36
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Question 754 Marks
Find the values of a and b for which the following system of linear equations has infinite number of solutions:
$2x - 3y = 7$
$(a + b)x - (a + b - 3)y = 4a + b$
Answer
The given equations are,
$2x - 3y = 7 ....(i)$
$(a + b)x - (a + b - 3)y = 4a + b ....(ii)$
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 $a_2 = (a + b), b_2 = -(a + b - 3)$, and $c_2 = -(4a + b)$
For infinitely many solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{(\text{a}+\text{b})}=\frac{-3}{-(\text{a}+\text{b}-3)}=\frac{-7}{-(4\text{a}+\text{b})}$
$\Rightarrow\frac{2}{(\text{a}+\text{b})}=\frac{-3}{-(\text{a}+\text{b}-3)}$ and $\frac{-3}{-(\text{a}+\text{b}-3)}=\frac{-7}{-(4\text{a}+\text{b})}$
$\Rightarrow 2(a + b - 3) = 3(a + b)$ and $3(4a + b) = 7(a + b - 3)$
$\Rightarrow 2a + 2b - 6 = 3a + 3b$ and $12a + 3b = 7a + 7b - 21$
$\Rightarrow a + b + 6 = 0 .....(iii)$ and $5a - 4b = -21 ....(iv)$
Multiplying eq. (iii) by $5$ ther subtract from eq. (iv), we get
$-9b = 9$ and $b = -1$
Put $b = -1$ eq. (iii) we get $a = -5$
Thus $a = 5$ and $b = -1$
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Question 764 Marks
A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.
Answer
Let the digits at units and tens place of the given number
be x and y respectively. Thus, the number is 10y + x
The number is 3 more than 4 times the sum of the two digits. Thus, we have
10y + x = 4(x + y) + 3
⇒ 10y + x = 4x + 4y + 3
⇒ 4x + 4y - 10y - x = -3
⇒ 3x - 6y = -3
$\Rightarrow\text{x}-2\text{y}=-\frac{3}{3}$
⇒ x - 2y = -1
After interchanging the digits, the number becomes 10x + y
If 18 is added to the number, the digits are reversed. Thus, we have
(10y + x) + 18 = 10x + y
⇒ 10x + y - 10y - x = 18
⇒ 9x - 9y = 18
⇒ 9(x - y) = 18
$\Rightarrow\text{x}-\text{y}=\frac{18}{9}$
⇒ x - y = 2
So, we have the systems of equations
x - 2y = -1
x - y = 2
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
Subtracting the first equation from the second, we have
(x - y) - (x - 2y) = 2 - (-1)
⇒ x - y - x + 2y = 3
⇒ y = 3
Substituting the value of y in the first equation, we have
x - 2 × 3 = -1
⇒ x - 6 = -1
⇒ x = -1 + 6
⇒ x = 5
Hence, the number is 10 × 3 + 5 = 35
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Question 774 Marks
Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of y.
x + 2y - 7 = 0,
2x - y - 4 = 0.
Answer
We have,
x + 2y - 7 = 0
Now, 2x - y - 4 = 0
x + 2y - 7 = 0
x = 7 - 2y
when, y = 1, x = 5
y = 2, x = 3
x
5
3
y
1
2
Also, 2x - y - 4 = 0
y = 2x - 4
x
2
0
y
0
-4

From the graph, the solution is A(3, 2).
Also, the coordinates of the points where the lines meet the y-axis are B(0, 3, 5) and C(0, -4).
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Question 784 Marks
Solve the following system of equations by the method of cross-multiplication:
$(a - b)x + (a + b)y = 2a^2 - 2b^2$,
$(a + b)(x + y) = 4ab$.
Answer
Given
$(a - b)x + (a + b)y = 2a^2 - 2b^2$
$(a + b)(x + y) = 4ab$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation.
$(a - b)x + (a + b)y - 2a^2 + 2b^2 = 0$
$(a + b)(x + y) - 4ab = 0$
By cross multiplication method we get,
$\Rightarrow\frac{\text{x}}{(-4\text{ab})(\text{a}+\text{b})-(\text{a}+\text{b})(-2\text{a}^2+2\text{b}^2)}=\frac{-\text{y}}{(-4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2\text{a}^2+2\text{b}^2)}\\=\frac{1}{(\text{a}+\text{b})(\text{a}-\text{b})-(\text{a}+\text{b})^2}$
$\Rightarrow\frac{\text{x}}{(\text{a}+\text{b})\big((-4\text{ab})-(-2\text{a}^2+2\text{b}^2)\big)}=\frac{-\text{y}}{(-4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2\text{a}^2+2\text{b}^2)}\\=\frac{1}{(\text{a}+\text{b})\big((\text{a}-\text{b})-(\text{a}+\text{b})\big)}$
Consider the following for x
$\Rightarrow\frac{\text{x}}{(\text{a}+\text{b})\big((-4\text{ab})-(-2\text{a}^2+2\text{b}^2)\big)}=\frac{1}{(\text{a}+\text{b})\big((\text{a}-\text{b})-(\text{a}+\text{b})\big)}$
$\Rightarrow\frac{\text{x}}{(-4\text{ab})-(-2\text{a}^2+2\text{b}^2)}=\frac{1}{(\text{a}-\text{b})-(\text{a}+\text{b})}$
$\text{x}=\frac{4\text{ab}+2\text{a}^2-2\text{b}^2}{2\text{b}}$
$\text{x}=\frac{2\text{ab}+\text{a}^2-\text{b}^2}{\text{b}}$
Now consider the following for y
$\Rightarrow\frac{-\text{y}}{(-4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2\text{a}^2+2\text{b}^2)}=\frac{1}{(\text{a}+\text{b})\big((\text{a}-\text{b})-(\text{a}+\text{b})\big)}$
$\Rightarrow\frac{\text{y}}{(4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2\text{a}^2+2\text{b}^2)}=\frac{1}{(\text{a}+\text{b})(-2\text{b})}$
$\Rightarrow\frac{\text{y}}{(4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2)(\text{a}^2-\text{b}^2)}=\frac{1}{(\text{a}+\text{b})(-2\text{b})}$
$\Rightarrow\frac{\text{y}}{(4\text{ab})(\text{a}-\text{b})-(\text{a}+\text{b})(-2)(\text{a}-\text{b})(\text{a}+\text{b})}=\frac{1}{(\text{a}+\text{b})(-2\text{b})}$
$\Rightarrow\frac{\text{y}}{(\text{a}-\text{b})(\text{a}^2-\text{b}^2)}=\frac{1}{(\text{a}+\text{b})\text{b}}$
$\text{y}=\frac{(\text{a}-\text{b})(\text{a}^2+\text{b}^2)}{(\text{a}+\text{b})\text{b}}$
Hence we get the value of $\text{x}=\frac{2\text{ab}+\text{a}^2-\text{b}^2}{\text{b}}$ and $\text{y}=\frac{(\text{a}-\text{b})(\text{a}^2+\text{b}^2)}{(\text{a}+\text{b})\text{b}}$
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Question 794 Marks
Find the solution of the pair of equations $\frac{\text{x}}{10}+\frac{\text{y}}{5}-1=0$ and $\frac{\text{x}}{8}+\frac{\text{y}}{6}=15.$ Hence, find $\lambda,$ if $\text{y}=\lambda\text{x}+5.$
Answer
Given pair of equation is
$\frac{\text{x}}{10}+\frac{\text{y}}{5}-1=0\ ......(\text{i})$
and $\frac{\text{x}}{8}+\frac{\text{y}}{6}=15\ .......(\text{ii})$
Now, multiplying both sides of eq. (i) by
L.C.M (10, 5) = 10, we get
x + 2y - 10 = 0
⇒ x + 2y = 10 .....(iii)
Again multiplying both sides of eq. (ii) by
L.C.M (8, 6) = 24, we get
3x + 4y = 360 ......(iv)
On, multiplying eq. (iii) by 2 and then subtracting from eq. (iv) we get

Put the value of x in eq (iii) we get
340 + 2y = 10
⇒ y = - 165
Given that the linear relation between x, y and $\lambda$ is $\text{y}=\lambda\text{x}+5$
Now Put the value of x and y in above relation we get
$-165=\lambda(340)+5$
$\Rightarrow340\lambda=-170$
$\Rightarrow\lambda=\frac{-1}{2}$
Hence the solution of the pair of equations is x = 340, y = -165 and the requations is x = 340, y = -165 and the required values of $\lambda$ is $\frac{-1}{2}.$
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Question 804 Marks
Solve the following systems of equations graphically:
x + y = 3
2x + 5y = 12
Answer
The given equations are:
x + y = 3 ......(i)
2x + 5y = 12 .....(ii)
From (i), y = 3 - x ......(iii)
Puting x = 0 in (iii), we get y = 3
Puting x = 1 in (iii), we get y = 2
Puting x = 2 in (iii), we get y = 1
x
0
1
2
y
3
2
1
From (ii), $\text{y}=\frac{12-2\text{x}}{5}\ .......(\text{iv})$
Puting x = -4 in (iv), we get y = 4
Puting x = 1 in (iv), we get y = 2
Puting x = 6 in (iv), we get y = 0
x
-4
1
6
y
4
2
0
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Question 814 Marks
Solve the following system of equations by the method of cross-multiplication:
$x + 2y + 1 = 0,$
$2x - 3y - 12 = 0.$
Answer
The given system of equation is
$x + 2y + 1 = 0$
$2x - 3y - 12 = 0$
Here,
$a_1 = 1, b_1 = 2, c_1 = 1$
$a_2 = 2, b_2 = -3$ and $c_2 = -12$
By cross-multiplication, we get
$\Rightarrow\frac{\text{x}}{2\times(-12)-1\times(-3)}=\frac{-\text{y}}{1\times(-12)-1\times2}\\=\frac{1}{1\times(-3)-2\times2 }$
$\Rightarrow\frac{\text{x}}{-24+3}=\frac{-\text{y}}{-12-2}=\frac{1}{-3-4}$
$\Rightarrow\frac{\text{x}}{-21}=\frac{-\text{y}}{-14}=\frac{1}{-7}$
Now,
$\frac{\text{x}}{-21}=\frac{1}{-7}$
$\Rightarrow\text{x}=\frac{-21}{-7}=3$
And,
$\frac{-\text{y}}{-14}=\frac{1}{-7}$
$ \Rightarrow\ \frac{\text{y}}{14}=\frac{-1}{7}$
$\Rightarrow\ \text{y}=\frac{-14}{7}=-2$
Hence, the solution of the given system of equations is $x = 3, y = -2.$
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Question 824 Marks
If 2 is added to the numerator of a fraction, it reduces to $\frac{1}{2}$ and if 1 is subtracted from the denominator, it reduces to $\frac{1}{3}.$ 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 2 is added to the numerator of the fraction, it reduces to $\frac{1}{2}$
Thus we have
$\frac{\text{x}+2}{\text{y}}=\frac{1}{2}$
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x - y + 4 = 0
If 1 is subtracted from the denominator, the fraction reduces to $\frac{1}{3}$
Thus we have
$\frac{\text{x}}{\text{y}-1}=\frac{1}{3}$
⇒ 3x = y - 1
⇒ 3x - y + 1 = 0
So, we have two equations
2x - y + 4 = 0
3x - 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)\times4}=\frac{-\text{y}}{2\times1-3\times4}\\=\frac{1}{2\times(-1)-3\times(-1)}$
$ \Rightarrow\frac{\text{x}}{-1+4}=\frac{-\text{y}}{2-12}=\frac{1}{-2+3}$
$\Rightarrow\frac{\text{x}}{3}=\frac{-\text{y}}{-10}=1$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{y}}{10}=1$
$\Rightarrow\text{x}=3,\ \text{y}=10$
Hence, the fraction is $\frac{3}{10}$
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Question 834 Marks
Solve the following system of equations by the method of cross-multiplication:
$(a + 2b)x + (2a − b)y = 2,$
$(a - 2b)x + (2a + b)y = 3.$
Answer
The given system of equations may be written as,
$(a + 2b)x + (2a − b)y - 2 = 0$
$(a - 2b)x + (2a + b)y - 3 = 0$
Here, $a_1 = a + 2b, b_1 = 2a - b, c_1 = -2$
$a_2 = a - 2b, b_2 = 2a + b$ and $c_2 = -3$
By cross multiplication, we have
$\Rightarrow\frac{\text{x}}{-3(2\text{a}-\text{b})-(-2)(2\text{a}+\text{b})}=\frac{-\text{y}}{3(\text{a}+2\text{b})-(-2)(\text{a}-2\text{b})}\\=\frac{1}{(\text{a}+2\text{b})(2\text{a}+\text{b})-(2\text{a}-\text{b})(\text{a}-2\text{b})}$
$\Rightarrow\frac{\text{x}}{-6\text{a}+3\text{b}+4\text{a}+2\text{b}}=\frac{-\text{y}}{-3\text{a}-6\text{b}+2\text{a}-4\text{b}}\\=\frac{1}{2\text{a}^2+\text{a}\text{b}+4\text{a}\text{b}+2\text{b}^2-(2\text{a}^2-4\text{a}\text{b}-\text{a}\text{b}+2\text{b}^2)}$
$\Rightarrow\frac{\text{x}}{-2\text{a}+5\text{b}}=\frac{-\text{y}}{-\text{a}-10\text{b}}\\=\frac{1}{2\text{a}^2+\text{a}\text{b}+4\text{a}\text{b}+2\text{b}^2-(2\text{a}^2-4\text{a}\text{b}-\text{a}\text{b}+2\text{b}^2)}$
$\Rightarrow\frac{\text{x}}{-2\text{a}+5\text{b}}=\frac{-\text{y}}{-(\text{a}+10\text{b})}=\frac{1}{10\text{ab}}$
$\Rightarrow\frac{\text{x}}{-2\text{a}+5\text{b}}=\frac{-\text{y}}{\text{a}+10\text{b}}=\frac{1}{10\text{ab}}$
Now,
$\Rightarrow\frac{\text{x}}{-2\text{a}+5\text{b}}=\frac{1}{10\text{ab}}$
$\Rightarrow\text{y}=\frac {\text{a}+10\text{b}}{10\text{ab}}$
and,
$\Rightarrow\frac{\text{y}}{\text{a}+10\text{b}}=\frac{1}{10\text{ab}}$
$\Rightarrow\text{y}=\frac{\text{a}+10\text{b}}{10\text{ab}}$
Hence, $ \text{x}=\frac{5\text{b}-2\text{a}}{10\text{ab}},\text{y}=\frac{\text{a}+10\text{b}}{10\text{ab}}$ is the solution of the given system of equations.
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Question 844 Marks
Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?
Answer
Let Salim and his daughter’s age be x and y year respectively.
Now, by first condition.
Two years ago, Salim was thrice as old as his daughter.
i.e., x - 2 = 3(y - 2)
⇒ x - 2 = 3y - 6
⇒ x - 3y = -4 .....(i)
and by second condition,
six years later, Salim will be four years older than twice her age,
x + 6 = 2(y + 6) + 4
⇒ x - 2y = 16 - 6
⇒ x - 2y = 10 .....(ii)
On subtracting eq. (i) from eq. (ii), we get
y = 14
Put the value of y in eq. (ii), we get
x - 2 × 14 = 10
⇒ x = 10 + 28
⇒ x = 38
Hence, Salim and his daughter’s age are 38 years and 14 years, respectively.
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Question 854 Marks
Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of y.
2x - 5y + 4 = 0,
2x + y - 8 = 0.
Answer
We have,
2x - 5y + 4 = 0
2x + y - 8 = 0
Now, 2x - 5y + 4 = 0
⇒ 2x = 5y - 4
$\Rightarrow\text{x}=\frac{5\text{y}-4}{2}$
When y = 2, we have,
$\text{x}=\frac{5\times2-4}{2}=3$
When y = 4, we have,
$\text{x}=\frac{5\times4-4}{2}=8$
Thus, we have the following table giving points on the line 2x - 5y + 4 = 0
x
3
8
y
2
4
Now, 2x + y - 8 = 0
⇒ 2x = 8 - y
$\Rightarrow\text{x}=\frac{8-\text{y}}{2}$
When y = 4, we have,
$\text{x}=\frac{8-4}{2}=2$
When y = 2, we have,
$\text{x}=\frac{8-2}{2}=3$
Thus, we have the following table points on the line 2x + y - 8 = 0
x
2
3
y
4
2
Graph of the given equations,

Clearly, two intersect at P(3, 2).
Hence, x = 3, y = 2 is the solution of the given system of equations.
We also observe that lines represented by 2x - 5y + 4 and 2x + y - 8 = 0 meet y-axis at $\text{A}\Big(0,\frac{4}{5}\Big)$ and B(0, 8) respectively.
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Question 864 Marks
Solve the following systems of equations graphically:
2x + y + 3 = 0
2x - 3y - 7 = 0
Answer
The given equations are:
2x + y + 3 = 0 ......(i)
2x - 3y - 7 = 0 .........(ii)
From (i), y = 3 - 2x ........(iii)
Putting x = 0 in (3), we get y = 3
Putting x = 1 in (3), we get y = 1
Putting x = 2 in (3), we get y = -1
x
0
1
2
y
3
1
-1
From (ii), $\text{y}=\frac{2\text{x}-7}{3}\ ......(\text{iv})$
Putting x = 2 in (iv), we get y = -1
Putting x = 5 in (iv), we get y = 1
Putting x = 8 in (iv), we get y = 3
x
2
5
8
y
-1
1
3

Clearly, from the above graph solution of above systam of equations is x = 2 and y = -1.
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Question 874 Marks
Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.
Answer
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is 10y + x
The difference between the two digits of the number is 3. Thus, we have
After interchanging the digits, the number becomes 10x + y
Seven times the number is equal to four times the number obtained by reversing the order of the digits.
Thus, we have
7(10y + x) = 4(10x + y)
⇒ 70y + 7x = 40x + 4y
⇒ 40x + 4y - 70y - 7x = 0
⇒ 33x - 66y = 0
⇒ 33(x - 2y) = 0
⇒ x - 2y = 0
So, we have two systems of simultaneous equations
x - y = 3
⇒ x - 2y = 0
⇒ x - y = -3
⇒ x - 2y = 0
Here x and y are unknowns. We have to solve the above systems of equations,
  1. First, we solve the system
x - y = 3

x - 2y = 0

Multiplying the first equation by 2 and then subtracting from the second equation, we have

(x - 2y) - 2(x - y) = 0 - 2 × 3

⇒ x - 2y - 2x + 2y = -6

⇒ -x = -6

⇒ x = 6

Substituting the value of x in the first equation, we have

6 - y = 3

⇒ y = 6 - 3

⇒ y = 3

Hence, the number is 10 × 3 + 6 = 36
  1. Now, we solve the system
x - y = -3

x - 2y = 0

Multiplying the first equation by 2 and then subtracting from the second equation, we have

(x - 2y) - 2(x - y) = 0 - (-3 × 2)

⇒ x - 2y - 2x + 2y = 6

⇒ -x = 6

⇒ x = -6

Substituting the value of x in the first equation, we have

-6 - y = -3

⇒ y = -6 + 3

⇒ y = -3

But, the digits of the number can’t be negative. Hence, the second case must be removed.
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Question 884 Marks
While covering a distance of 30km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.
Answer
Let the speed of Ajeet and Amit be x km/ hr and y km/ hr respectively. Then,
Time taken by Ajeet to cover $30\text{km}=\frac{30}{\text{x}}\text{hrs}$
Time taken by Amit to cover $30\text{km}=\frac{30}{\text{y}}\text{hrs}$
By the given conditions, we have
$\frac{30}{\text{x}}-\frac{30}{\text{y}}=2\ ....(\text{i})$
If Ajeet doubles his speed then speed of Ajeet is 2x km/ hr
Time taken by Ajeet to cover $30\text{km}=\frac{30}{2\text{x}}\text{hrs}$
Time taken by Amit to cover $30\text{km}=\frac{30}{\text{y}}\text{hrs}$
According to the given condition we have
$\frac{-15\text{x}}{\text{x}}+\frac{30}{\text{y}}=1\ ...(\text{ii})$
Putting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ in equation (i) and (ii) we get
$30\text{u}-30\text{v}=2\ ....(\text{iii})$
$-15\text{u}+30\text{v}=1\ .....(\text{iv})$
Adding equations (iii) and (iv) we get
$\ \ \ \ 30\text{u}-30\text{v}=2\\\underline{-15\text{u}+30\text{v}\ =1}\\\ \ \ 15\text{u}\ \ \ \ \ \ \ \ \ \ \ \ =3$
$\text{u}=\frac{3}{15}$
$\text{u}=\frac{1}{5}$
Putting $\text{u}=\frac{1}{5}$ in equation (iii) we get
$30\text{u}-30\text{v}=2$
$30\times\frac{1}{5}-30\text{v}=2$
$6-30\text{v}=2$
$-30\text{v}=2-6$
$-30\text{v}=-4$
$\text{v}=\frac{-4}{-30}$
$\text{v}=\frac{2}{15}$
Now, $\text{u}=\frac{1}{5}$
$\frac{1}{\text{x}}=\frac{1}{5}$
$\text{x}=5$
And $\text{v}=\frac{2}{15}$
$\frac{1}{\text{y}}=\frac{2}{15}$
$\text{y}=7.5$
Hence, the speed of Ajeet is 5km/ hr. The speed of Amit is 7.5 km/ hr.
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Question 894 Marks
Determine, graphically whether the system of equation x - 2y = 2, 4x - 2y = 5 is consistent or in-consistent.
Answer
We have
x - 2y = 2
4x - 2y = 5
Now x - 2y = 2
⇒ x = 2 + 2y
When y = 0, we have,
x = 2 + 2 × 0 = 2
when y = -1, we have,
x = 2 + 2 × (-1) = 0
Thus, we have the following table giving points on the line x - 2y = 2
x
2
0
y
0
-1
Now, 4x - 2y = 5
⇒ 4x = 5 + 2y
$\Rightarrow\text{x}=\frac{5+2\text{y}}{4}$
When y = 0, we have
$\text{x}=\frac{5+2\times0}{4}=\frac{5}{4}$
When y = 1, we have
$\text{x}=\frac{5+2\times1}{4}=\frac{7}{4}$
Thus, we have the following table giving points on the line 4x - 2y = 5
x $\frac{5}{4}$ $\frac{7}{4}$
y 0 1
Graph of the given equations,

Clearly, the two lines intersect at (i!).
Hence, the system of equations is consistent.
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Question 904 Marks
Solve the following system of linear equations graphically.
4x - 5y - 20 = 0
3x + 5y - 15 = 0
Determine the vertices of the triangle formed by the lines representing the above equation and the y-axis.
Answer
The given equations are 4x - 5y - 20 = 0 .......(i) 3x + 5y - 15 = 0 ........(ii) From (i), $\text{y}=\frac{4\text{x}-20}{5}\ ......(\text{iii})$ Putting x = 0 in equations (i), we get y = -4 Putting x = 5 in equations (i), we get y = 0
x
0
5
y
-4
0
From (ii), $\text{y}=\frac{15-3\text{x}}{5}\ ......(\text{iv})$ Putting x = 0 in equations (i), we get y = 3 Putting x = 5 in equations (i), we get y = 0
x
0
5
y
3
0

When we solve these equations we get x = 5 and y = 0. Thus the vertices of $\triangle$ are (0, 3), (0, -4) and (5, 0).
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Question 914 Marks
Solve the following systems of equations graphically:
2x - 3y + 13 = 0
3x - 2y + 12 = 0
Answer
We have,
2x - 3y + 13 = 0
3x - 2y + 12 = 0
Now, 2x - 3y + 13 = 0
⇒ 2x = 3y - 13
$\Rightarrow\text{x}=\frac{3\text{y}-13}{2}$
When y = 1, we have,
$\text{x}=\frac{3\times1-13}{2}=-5$
When y = 3, we have,
$\text{x}=\frac{3\times3-13}{2}=-2$
Thus, we have the following table giving points on the line 2x - 3y + 13 = 0
x
-5
-2
y
1
3
Now, 3x - 2y + 12 = 0
⇒ 3x = 2y - 12
$\Rightarrow\text{x}=\frac{2\text{y}-12}{3}$
When y = 0, we have,
$\Rightarrow\text{x}=\frac{2\times0-12}{3}=-4$
When y = 3, we have,
$\text{x}=\frac{2\times3-12}{3}=-2$
Thus, we have the following table giving points on the line 3y - 2y + 12 = 0
x
-4
-2
y
0
3
Graph of the given equations are,

Clearly, two lines intersect at (-2, 3)
Hence, x = -2, y = 3 is the solution of the given system of equations.
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Question 924 Marks
Solve the following system of equations by the method of cross-multiplication:
$ax + by = a^2,$
$bx + ay = b^2.$
Answer
Given,
$ax + by = a^2$
$bx + ay = b^2$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation,
$ax + by - a^2 = 0$
$bx + ay - b^2 = 0$
By cross multiplication method we get,
$\Rightarrow\frac{\text{x}}{\text{b}\times(-\text{b}^2)-\text{a}\times(-\text{a}^2)}=\frac{-\text{y}}{\big(\text{a}\times(-\text{b}^2)-\big(\text{b}\times(-\text{a}^2)\big)}\\=\frac{1}{\big(\text{a}\times(\text{a})\big)-\big((\text{b}\times\text{b})\big)}$
$\frac{\text{x}}{(\text{a}^3-\text{b}^3)}=\frac{-\text{y}}{(-\text{ab}^2+\text{a}^2\text{b})}=\frac{1}{(\text{a}^2-\text{b}^2)}$
$\frac{\text{x}}{(\text{a}^3-\text{b}^3)}=\frac{1}{(\text{a}^2-\text{b}^2)}$
$\text{x}=\frac{(\text{a}^3-\text{b}^3)}{(\text{a}^2-\text{b}^2)}$
$\text{x}=\frac{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{(\text{a}^2-\text{b}^2)}$ $\big\{\text{since}(\text{a}^3-\text{b}^3)=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big\}$
$\text{x}=\frac{(\text{a}^2+\text{ab}+\text{b}^2)}{(\text{a}+\text{b})}$ $\{\text{since}(\text{a}^2-\text{b}^2)=(\text{a}+\text{b})(\text{a}-\text{b})\big\}$
And, $\frac{-\text{y}}{(-\text{ab}^2+\text{a}^2\text{b})}=\frac{1}{(\text{a}^2-\text{b}^2)}$
$ \text{y}=\frac{(-\text{ab}^2+\text{a}^2\text{b})}{(\text{a}^2-\text{b}^2)}$
$\text{y}=\frac{\big(-\text{ab}(\text{a}-\text{b})\big)}{(\text{a}^2-\text{b}^2)}$
$\text{y}=\frac{(-\text{ab}(\text{a}-\text{b}))}{(\text{a}-\text{b})(\text{a}+\text{b})}$ $\big\{\text{since}(\text{a}^2-\text{b}^2)=(\text{a}+\text{b})(\text{a}-\text{b})\big\}$
$\text{y}=\frac{-\text{ab}}{(\text{a}+\text{b})}$
Hence we get the value of $ \text{x}=\frac{(\text{a}^2+\text{ab}+\text{b}^2)}{(\text{a}+\text{b})}$ and $\text{y}=\frac{-\text{ab}}{(\text{a}+\text{b})}.$
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Question 934 Marks
Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.
2x + y = 6,
x - 2y = -2.
Answer
The given system of equation is, 2x + y = 6 .......(i) x - 2y = -2 ..........(ii) 2x + y = 6 .......(i) y = 6 - 2x Substituting some different values of x, we get their corresponding values of y as shown below;
x
1
2
3
y
4
2
0
Now plot the points and join them similarly in the equation. x - 2y = -2 ⇒ x = 2y - 2
x
-2
0
2
y
0
1
2
Now plot the points and join them we see that these two lines intersect each other at (2, 2) x = 2, y = 2. Here two lines also meet x-axis at (3, 0) and (-2, 0) respectively as shown in the.
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Question 944 Marks
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs. 216 and one full and one half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservation charge?
Answer
Let take first class full of fare is Rs. x and reservation charge is Rs. y per ticket.
Then, half of the ticket as on full ticket $=\frac{\text{x}}{2}$
According to the question,
$\Rightarrow\text{x}+\text{y}=216\ ....(\text{i})$
$\Rightarrow\text{x}+\text{y}+\frac{\text{x}}{2}+\text{y}=327$
$\Rightarrow\frac{3\text{x}}{2}+2\text{y}-327=0\ ....(\text{ii})$
Now, multiplying eq. (i) by 2
$\Rightarrow2\text{x}+2\text{y}-432=0\ .....(\text{iii})$
Now, subtracting eq. (iii) from eq. (ii)
$\Rightarrow\frac{3\text{x}}{2}+2\text{y}-327-(2\text{x}+2\text{y}-432)$
$\Rightarrow \frac{3\text{x}}{2}+2\text{y}-327-2\text{x}-2\text{y}+432$
$\Rightarrow \frac{3}{2}\text{x}-2\text{x}+105$
$\Rightarrow \frac{\text{-x}}{2}+105=0$
$\Rightarrow \frac{\text{-x}}{2}=-105$
$\Rightarrow\text{x}=2\times105$
$\Rightarrow\text{x}=210$
Now, putting the value of x in eq. (i)
$\Rightarrow210+\text{y}=216$
$\Rightarrow\text{y}=216-210$
$\Rightarrow\text{x}=6$
Hence, the basic first class full fare is Rs. 210, The reservation charqe is Rs. 6.
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Question 954 Marks
Solve the following systems of equations:
$3\text{x}-\frac{\text{y}+7}{11}+2=10,$
$3\text{y}-\frac{\text{x}+11}{7}=10.$
Answer
The given systems of equations is,
$3\text{x}-\frac{\text{y}+7}{11}+2=10\ ......(\text{i})$
$3\text{y}-\frac{\text{x}+11}{7}=10\ ......(\text{ii})$
From (i), we get
$\frac{33\text{x}-\text{y}-7+22}{11}=10$
$\Rightarrow 33\text{x} -\text{y} + 15 = 10 × 11$
$\Rightarrow 33\text{x} + 15 - 110 = \text{y}$
$\Rightarrow \text{y} = 33\text{x} - 95$
From (ii), we get
$\frac{21\text{y}-\text{x}-11}{7}=10$
$21\text{y}-\text{x}-11=70$
$21\text{y}-\text{x}=70+11$
$\text{x}-21\text{y}=81$
$\text{x}-21\text{y}=-81\ ...(\text{iii})$
Substitution value of y in equation (iii)
$\text{x}-21(33\text{x}-95)=-81$
$\text{x}-693\text{x}+1995=-81$
$-692\text{x}=-81-1995$
$-692\text{x}=-2076$
$\text{x}=\frac{2076}{692}=3$
So that $\text{y}=33\times3-95$
$\text{y}=99-95$
$\text{y}=4$
Hence solution of the given system of equation x = 3, y = 4.
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Question 964 Marks
Solve the following system of equations by the method of cross-multiplication:
$\frac{\text{x}}{\text{b}}=\frac{\text{y}}{\text{b}},$
$\text{ax}+\text{by}=\text{a}^2+\text{b}^2.$
Answer
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0$
$\text{a}\text{x}+\text{b}\text{y}=\text{a}^2+\text{b}^2$
The given pair of linear equetions is
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0\dots(\text{i})$
$\text{a}\text{x}+\text{b}\text{y}=\text{a}^2+\text{b}^2\dots(\text{ii})$
From equetion (i), we have
$\frac{\text{y}}{\text{b}}=\frac{\text{x}}{\text{a}}$
$\Rightarrow\text{y}=\frac{\text{b}}{\text{a}}\text{x}\dots(\text{iii})$
Substituting the value of y in equetion (ii), we get
$\text{a}\text{x}+\text{b}\Big(\frac{\text{b}}{\text{a}}\text{x}\Big)=\text{a}^2+\text{b}^2$
$\Rightarrow\text{a}\text{x}+\frac{\text{b}^2}{\text{a}}\text{x}=\text{a}^2+\text{b}^2$
$\Rightarrow\text{a}^2\text{x}+\text{b}^2\text{x}=\text{a}(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{x}=\frac{\text{a}(\text{a}^2+\text{b}^2)}{\text{a}^2+\text{b}^2}=\text{a}$
Substituting the value of x in equetion (iii), we get
$\text{y}=\frac{\text{b}}{\text{a}}\text{x}$
$\Rightarrow\text{y}=\frac{\text{b}}{\text{a}}\times\text{a}=\text{b}$
Hence, $\text{x}=\text{a},\text{y}=\text{b}$
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Question 974 Marks
Solve the following system of equations by the method of cross-multiplication:
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=13,$
$\frac{5}{\text{x}}-\frac{4}{\text{y}}=-2,$ where $\text{x}\neq0$ and $\text{y}\neq0.$
Answer
The given equations are,
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=13\ ...(\text{i})$
$\frac{5}{\text{x}}-\frac{4}{\text{y}}=-2\ ...(\text{ii})$
Let $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}.$
$2\text{u}+3\text{v}-13=0\ ...(\text{iii})$
$5\text{u}-4\text{v}+2=0\ ....(\text{iv})$
By cross-multiplication we get,
$\Rightarrow\frac{\text{u}}{3\times2-(-13)(-4)}=\frac{\text{v}}{(-13)(5)-2\times2}\\=\frac{1}{2(-4)-3(5)}$
$\Rightarrow\frac{\text{u}}{6-52}=\frac{\text{v}}{-65-4}=\frac{1}{-8-15}$
$\Rightarrow\frac{\text{u}}{-46}=\frac{\text{v}}{-69}=\frac{1}{-23}$
$\text{u}=\frac{-46}{-23}=2$ and $\text{v}=\frac{-69}{-23}=3$
$\because\frac{1}{\text{x}}=\text{u}\ \Rightarrow\text{x}=\frac{1}{\text{u}}\ \Rightarrow\text{x}=\frac{1}{2}$
and $\frac{1}{\text{y}}=\text{v}\ \Rightarrow\text{y}=\frac{1}{\text{v}}\ \Rightarrow\text{y}=\frac{1}{3}$
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Question 984 Marks
Solve the following systems of equations graphically:
x + y = 6
x - y = 2
Answer
The given equations are
x + y = 6 .......(i)
x - y = 2 ..........(ii)
Putting x = 0 in equation (i), we get,
⇒ 0 + y = 6
⇒ y = 6
x = 0, y = 6
Putting y = 0 in equation (i), we get,
⇒ x + 0 = 6
⇒ x = 6
x = 6, y= 0
Use the following table to draw the graph.
x
0
6
y
6
0
Draw the graph by plotting the two points A(0, 6) and B(6, 0) from table.

Graph of the equation,
x - y = 2 .......(ii)
Putting x = 0 in equation (ii) we get,
⇒ 0 - y = 2
⇒ y = -2
⇒ x = 0, y = -2
Putting y = 0 in equation (ii), we get,
⇒ x - 0 = 2
⇒ x = 2
⇒ x = 2, y = 0
Use the following table to draw the graph.
x
0
2
y
-2
0
Draw the graph by plotting the two points C(0, -2) and D(2, 0) from table.
The two lines intersect at points P(4, 2).
Hence x = 4, y = 2 is the solution.
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Question 994 Marks
A two-digit number is such that the product of its digits is $20$. If $9$ is added to the number, the digits interchange their places. Find the number.
Answer
Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10y + x$
The product of the two digits of the number is $20$. Thus, we have $xy = 20$
After interchanging the digits, the number becomes $10x + y$
If 9 is added to the number, the digits interchange their places. Thus, we have
$(10y + x) + 9 = 10x + y$
$\Rightarrow 10y + x + 9 = 10x + y$
$\Rightarrow 10x + y - 10y - x = 9$
$\Rightarrow 9x - 9y = 9$
$\Rightarrow 9(x - y) = 9$
$\Rightarrow\text{x}-\text{y}=\frac{9}{9}$
$\Rightarrow x - y = 1$
So, we have the systems of equations
$xy = 20$
$x - y = 1$
Here $x$ and $y$ are unknowns. We have to solve the above systems of equations for $x$ and $y$
Substituting $x = 1 + y$ from the second equation to the first equation, we get
$(1 + y)y = 20$
$\Rightarrow y + y^2 = 20$
$\Rightarrow y + y^2 - 20 = 0$
$\Rightarrow y^2 + 5y - 4y - 20 = 0$
$\Rightarrow y(y + 5) - 4(y + 5) = 0$
$\Rightarrow (y + 5)(y - 4) = 0$
$\Rightarrow y = -5$ or $y = 4$
Substituting the value of y in the second equation, we have
y
$-5$
$4$
x
$-4$
$5$
Hence, the number is $10 \times 4 + 5 = 45$
Note: That in the first pair of solution the values of x and y are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.
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Question 1004 Marks
A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you." B replies, "if you give me 10, I will have thrice as many as left with you." How many mangoes does each have?
Answer
To find:
Total mangoes of A.
Total mangoes of B.
Suppose A has x mangoes and B has y mangoes,
According to the given conditions,
x + 30 = 2(y - 30)
x + 30 = 2y - 60
x - 2y + 30 + 60 = 0
x - 2y + 90 = 0 ......(i)
y + 10 = 3(x - 10)
y + 10 = 3x - 30
y - 3x + 10 + 30 = 0
y - 3x + 40 = 0 .......(ii)
Multiplying eq. (i) by (3)
3x - 6y + 270 = 0 ......(iii)
and now adding eq. (ii) and eq (iii)
5y = 310
$\text{y}=\frac{310}{5}$
y = 62
x - 2 × 62 + 90 = 0
x - 124 + 90 = 0
x - 34 = 0
x = 34
Hence A has 34 mangoes and B has 62 mangoes.
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