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#### Question 1:

By using variables x and y form any five linear equations in two variables.

Linear equations in two variables x and y will be
$x+y=1\phantom{\rule{0ex}{0ex}}2x+3y=8\phantom{\rule{0ex}{0ex}}x-y=6\phantom{\rule{0ex}{0ex}}5x-2y=3\phantom{\rule{0ex}{0ex}}x+2y=3$

#### Question 2:

Write five solutions of the equation x + y = 7.

The given equation is + y = 7.
The possible solution are

 x 0 7 1 6 4 y 7 0 6 1 3

#### Question 3:

Solve the following sets of simultaneous equations.

(i) x + y = 4 ; 2x $-$ 5y = 1

(ii) 2x + y = 5; 3x $-$  y = 5

(iii) 3x$-$ 5y =16;  x $-$ 3y = 8

(iv) 2y$-$x =0; 10x + 15y = 105

(v) 2x + 3y + 4 = 0;   x$-$ 5y = 11

(vi)  2x $-$7y = 7; 3x + y = 22

(i) x + y = 4            .....(I)
2$-$ 5y = 1            .....(II)
From (I) we have
$x=4-y$
Putting this value of in (II)
$2\left(4-y\right)-5y=1\phantom{\rule{0ex}{0ex}}⇒8-2y-5y=1\phantom{\rule{0ex}{0ex}}⇒8-7y=1\phantom{\rule{0ex}{0ex}}⇒8-1=7y\phantom{\rule{0ex}{0ex}}⇒7=7y\phantom{\rule{0ex}{0ex}}⇒y=1$
Putting the value of in (I) we have
$x+1=4\phantom{\rule{0ex}{0ex}}⇒x=3$
Thus, $\left(x,y\right)=\left(3,1\right)$

(ii)  2x + y = 5              .....(I)
3x $-$  y = 5                   .....(II)
$5x=10\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{5}=2$
Putting this value in (I) we have
$2×2+y=5\phantom{\rule{0ex}{0ex}}⇒4+y=5\phantom{\rule{0ex}{0ex}}⇒y=1\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,1\right)$

(iii) 3x$-$ 5=16          .....(I)
$-$ 3= 8 .....(II)
Multiply (II) with 3
$3x-9y=24$ .....(III)
(I) $-$ (III)

(iv) 2$-$ = 0 $⇒x=2y$              .....(I)
10+ 15= 105                          .....(II)
Taking out 5 common from (II)
$2x+3y=21$                                 .....(III)
From (I) and (III)
$2\left(2y\right)+3y=21\phantom{\rule{0ex}{0ex}}⇒4y+3y=21\phantom{\rule{0ex}{0ex}}⇒7y=21\phantom{\rule{0ex}{0ex}}⇒y=3\phantom{\rule{0ex}{0ex}}x=2×3=6$

(v) 2+ 3y + 4 = 0                  .....(I)
x$-$ 5y = 11                              .....(II)
From (I) we have
2+ 3y = $-$4                          .....(III)
Multiply (II) with 2
$2x-10y=22$                           .....(IV)
From (III) and (IV) we have
2+ 3y = $-$4
$2x-10y=22$
(III) $-$ (IV)
$13y=-26\phantom{\rule{0ex}{0ex}}⇒y=-2$
x = 11 + 5y
x = 1

(vi)  2x $-$7= 7                     .....(I)
3+ y = 22                            .....(II)
Multiply (II) with 7
21x + 7y = 154                    .....(III)
(I) + (III)
23x = 161
x = 7, y = 1

#### Question 1:

In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.

Disclaimer: There is an error in the given question. Instead of "Number of 5 rupee notes are less by 10 than number of 10 rupee notes"
the sentence should have been "Number of 5 rupee notes are greater by 10 than number of 10 rupee notes".

Let the number of Rs 5 notes be x and the number of Rs 10 notes be y.

Number of 5 rupee notes are greater by 10 than number of 10 rupee notes
So, $x-y=10$
$⇒x=10+y$                       .....(II)
From (I) and (II) we have
$10+y+2y=70\phantom{\rule{0ex}{0ex}}⇒10+3y=70\phantom{\rule{0ex}{0ex}}⇒3y=60\phantom{\rule{0ex}{0ex}}⇒y=20$
Thus, = 10 + 20 = 30
Number of Rs 5 notes = 30, number of Rs 10 notes = 20

#### Question 2:

The denominator of a fraction is 1 more than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 1 : 2. Find the fraction.

Disclaimer: There is an error in the given question. With the given statements, the two linear equations formed will be the same.

#### Question 3:

The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today's ages.

Let the age of Priyanka be x years and that of Deepika be y years.
The sum of ages of Priyanka and Deepika is 34 years.
$x+y=34$                  .....(I)
Priyanka is elder to Deepika by 6 years.
$x-y=6$                  .....(II)
Adding (I) and (II) we have
$2x=40\phantom{\rule{0ex}{0ex}}⇒x=20$
Putting the value of in (I) we have
= 14
Thus, Priyanka's age = 20 years and Deepika's age = 14 years.

#### Question 4:

The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is140. Then find the number of lions and peacocks in the zoo.

Let the number of lions be x and the number of peacocks be y
The total number of lions and peacocks in a certain zoo is 50.
So, $x+y=50$                     .....(I)
The total number of their legs is140.
Lion has 4 legs while peacock has 2.

Subtracting (II) from (I) we have
$x=20\phantom{\rule{0ex}{0ex}}y=30$
Thus, the number of lions = 20 and the number of peacocks = 30.

#### Question 5:

Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.

Let the fixed monthly income be Rs x
Annual increament be Rs y
After 4 years, his monthly salary was Rs. 4500
Monthly salary + annual increament of 4 years = 4500
$x+4y=4500$                        .....(I)
After 10 years his monthly salary became 5400 rupees
Monthly salary + annual increament of 10 years = 5400
$x+10y=5400$                        .....(II)
(II) $-$ (I)
6= 900
y = 150
x = 4500 $-$ 150 $×$ 4 = 3900
Thus, the monthly salary = Rs 3900
Annual increament = Rs 150

#### Question 6:

The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the price of 2 chairs and 2 tables.

Let the price of 1 chair be Rs and the price of 1 table be Rs y.
The price of 3 chairs and 2 tables is 4500 rupees

Price of 5 chairs and 3 tables is 7000 rupees

Multiply (I) with 3 and (II) with 2

Price of 2 chairs and 2 tables will be
$=2x+2y\phantom{\rule{0ex}{0ex}}=2\left(x+y\right)\phantom{\rule{0ex}{0ex}}=2\left(500+1500\right)\phantom{\rule{0ex}{0ex}}=2×2000\phantom{\rule{0ex}{0ex}}=4000$
Thus, the price of 2 chairs and 2 tables will be Rs 4000

#### Question 7:

The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.

Let the number at the ten's place be x and that at the one's place be y
Number obtained = 10x + y
The sum of the digits in a two-digits number is 9.
$x+y=9$                      .....(I)
The number obtained by interchanging the digits exceeds the original number by 27.

$2y=12\phantom{\rule{0ex}{0ex}}⇒y=6$
x = 3
Thus, the number is 36.

#### Question 8:

In $∆$ABC, the measure of angle A is equal to the sum of the measures of $\angle$ B and $\angle$ C. Also the ratio of measures of $\angle$ B and $\angle$C is 4 : 5. Then find the measures of angles of the triangle.

$\angle$A = $\angle$B + $\angle$C               .....(I)

We know by angle sum property,
$\angle$A + $\angle$B + $\angle$C = 180$°$
From (I)
$\angle$B + $\angle$C +$\angle$B + $\angle$C = 180$°$
$⇒2\left(\angle \mathrm{B}+\angle \mathrm{C}\right)=180°\phantom{\rule{0ex}{0ex}}⇒2\left(\angle \mathrm{B}+\frac{5\angle \mathrm{B}}{4}\right)=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{B}+\frac{5\angle \mathrm{B}}{4}=90°\phantom{\rule{0ex}{0ex}}⇒4\angle \mathrm{B}+5\angle \mathrm{B}=360°\phantom{\rule{0ex}{0ex}}⇒9\angle \mathrm{B}=360°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{B}=40°$
$5×40°=4\angle \mathrm{C}\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{C}=50°$
And $\angle$A = $\angle$B + $\angle$C = $40°+50°=90°$.

#### Question 9:

Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to $\frac{1}{3}$ of the larger part. Then find the length of the larger part.

Let the smaller part be cm and larger part be cm.
$x+y=560$                    .....(I)
Twice the length of the smaller part is equal to $\frac{1}{3}$ of the larger part.
$x=\frac{1}{3}y\phantom{\rule{0ex}{0ex}}⇒3x=y$
Putting the value of y in (I) we have
$x+3x=560\phantom{\rule{0ex}{0ex}}⇒4x=560\phantom{\rule{0ex}{0ex}}⇒x=140$
And thus, $y=3×140=420$
Thus, the larger part = 420 cm and smaller part is 140 cm.

#### Question 10:

In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong ?

Let the number of correct answers be and the number of incorrect answers be y.
Total answers = Number of correct + number of incorrect = 60
$x+y=60$                      .....(I)
Total marks obtained = 90
$2x-y=90$                     .....(II)
Adding (I) and (II) we have
$3x=150\phantom{\rule{0ex}{0ex}}⇒x=50\phantom{\rule{0ex}{0ex}}y=10$
Number of questions he got wrong is 10.

#### Question 1:

Choose the correct alternative answers for the following questions.

(i) If 3x + 5y = 9 and 5x + 3y= 7 then What is the value of  x  y ?
(A) 2 (B) 16 (C) 9 (D) 7

(ii) 'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.' What is the mathematical form of the statement ?
(A)  x $-$ = 8        (B)  x  +  y = 8    (C)  x +  y = 23      (D) 2x +  y = 21

(iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age ?
(A) 20 (B) 15 (C) 10 (D) 5

(i) 3x + 5y = 9             .....(I)
5x + 3y= 7                   .....(II)
Adding (I) and (II) we have
$8x+8y=16\phantom{\rule{0ex}{0ex}}⇒x+y=2$
Hence, the correct answer is option (A).

(ii) Let the length be and the breadth be y
Perimeter of the rectangle = 2(l + b) = 26
2(l + b) = 26
$⇒2\left[\left(x-5\right)+\left(y-5\right)\right]=26\phantom{\rule{0ex}{0ex}}⇒2\left[x+y-10\right]=26\phantom{\rule{0ex}{0ex}}⇒x+y=23\phantom{\rule{0ex}{0ex}}$
Hence, the correct answer is option (C).

(iii) Let Ajay's age be years and Vijay's age be years.
Ajay is younger than Vijay by 5 years.

Sum of their ages is 25 years.

Adding (I) and (II) we have
$2y=30\phantom{\rule{0ex}{0ex}}⇒y=15\phantom{\rule{0ex}{0ex}}x=10$
Thus, Ajay's age is 10 years.
Hence, the correct answer is option (C).

#### Question 2:

Solve the following simultaneous equations.

(i) 2x + y = 5 ; 3$-$  y  = 5

(ii) $-$2  y  = $-$1 ; 2$-$  y  = 7

(iii) +  y = 11 ; 2 $-$ 3  y  = 7

(iv) 2 +  y  = $-$2 ; 3$-$  y  = 7

(v) 2 $-$ = 5 ; 3 + 2 = 11

(vi) x$-$ 2 = $-$2 ; + 2 = 10

(i) 2+ y = 5              .....(I)
3$-$  y  = 5 .....(II)
(I) + (II)
$5x=10\phantom{\rule{0ex}{0ex}}⇒x=2$
Putting the value of x in (I)
$2×2+y=5\phantom{\rule{0ex}{0ex}}⇒4+y=5\phantom{\rule{0ex}{0ex}}⇒y=1$

(ii)  $-$2 = $-$1           .....(I)
2$-$  y  = 7                   .....(II)
Multiply (I) with 2
$2x-4y=-2$                  .....(III)
Subtracting (III) from (II)
$-3y=-9\phantom{\rule{0ex}{0ex}}⇒y=3$
Putting the value of in (I) we get
$x-2×3=-1\phantom{\rule{0ex}{0ex}}⇒x-6=-1\phantom{\rule{0ex}{0ex}}⇒x=5$

(iii)  +  y = 11             .....(I)
2 $-$ 3 = 7 .....(II)
Multiply (I) with 3
$3x+3y=33$ .....(III)
$5x=40\phantom{\rule{0ex}{0ex}}⇒x=8$

Putting the value of in (I)
$8+y=11\phantom{\rule{0ex}{0ex}}⇒y=3$

(iv) 2 +  y  = $-$2               .....(I)
3$-$  y  = 7                         .....(II)
$5x=5\phantom{\rule{0ex}{0ex}}⇒x=1$
Putting the value of in (I)
$2×1+y=-2\phantom{\rule{0ex}{0ex}}⇒2+y=-2\phantom{\rule{0ex}{0ex}}⇒y=-4$

(v) 2 $-$ = 5                     .....(I)
3 + 2 = 11 .....(II)
Multiplying (I) with 2 we get
$4x-2y=10$ .....(III)
$7x=21\phantom{\rule{0ex}{0ex}}⇒x=3$
Putting the value of x in (I) we get
$2×3-y=5\phantom{\rule{0ex}{0ex}}⇒6-y=5\phantom{\rule{0ex}{0ex}}⇒y=1$

(vi) $-$ 2 = $-$2               .....(I)
+ 2 = 10                      .....(II)
$2x=8\phantom{\rule{0ex}{0ex}}⇒x=4$
Putting the value of in (I) we get
$4-2y=-2\phantom{\rule{0ex}{0ex}}⇒-2y=-6\phantom{\rule{0ex}{0ex}}⇒y=3$

#### Question 3:

By equating coefficients of variables, solve the following equations.

(i) 3x $-$ 4 y = 7; 5 + 2 = 3

(ii) 5 + 7 y  = 17 ; 3$-$2 = 4

(iii) $-$2 = $-$10; 3 $-$ 5= $-$12

(iv) 4+ y  = 34 ;   + 4 = 16

(i) 3$-$ 4 y = 7                     .....(I)
5 + 2 = 3 .....(II)
Multiply (II) with 2

10 + 4 = 6                        .....(III)
$13x=13\phantom{\rule{0ex}{0ex}}⇒x=1$
Putting the value of x in (I) we get
$3×1-4y=7\phantom{\rule{0ex}{0ex}}⇒3-4y=7\phantom{\rule{0ex}{0ex}}⇒y=-1$

Thus, x = 1 and $y=-1$.

(ii) 5 + 7 y  = 17            .....(I)
3$-$2 = 4                      .....(II)
Multiply (I) with 3 and (II) with 5
15 + 21 y  = 51              .....(III)
15$-$10 = 20                .....(IV)
Subtracting (IV) from (III) we get
$31y=31\phantom{\rule{0ex}{0ex}}⇒y=1$
Putting this value of y in (I) we get
$5x+7×1=17\phantom{\rule{0ex}{0ex}}⇒5x=10\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, .

(iii) $-$2 = $-$10                     .....(I)
3 $-$ 5$-$12 .....(II)
Multiply (I) with 3
$3x-6y=-30\phantom{\rule{0ex}{0ex}}$ .....(III)
Subtracting (II) from (III) we get
$-y=-18\phantom{\rule{0ex}{0ex}}⇒y=18$
Putting the value of y in (I) we get
$x-2×18=-10\phantom{\rule{0ex}{0ex}}⇒x=-10+36=26$
Thus, $x=26,y=18$

(iv) 4+ y  = 34                   .....(I)
+ 4 = 16                         .....(II)
Adding (I) and (II) we get

Subtracting (II) from (I) we get

(III) + (IV)
$2x=16\phantom{\rule{0ex}{0ex}}⇒x=8$
Putting the value of x in (I) we get
$4\left(8\right)+y=34\phantom{\rule{0ex}{0ex}}⇒y=34-32=2\phantom{\rule{0ex}{0ex}}⇒y=2$
Thus, $x=8,y=2$

#### Question 4:

Solve the following simultaneous equations.

(i)

(ii)

(iii)

(i)              .....(I)
.....(II)
Multiplying (I) with LCM of 3 and 4 which is 12 we get
$4x+3y=48$                     .....(III)
Multiplying (II) with LCM of 2 and 4 which is 4 we get
$2x-y=4$                     .....(IV)
Multiplying (IV) with 2
$4x-2y=8$                    .....(V)
Subtracting (V) from (III)
$5y=40\phantom{\rule{0ex}{0ex}}⇒y=8$
Putting this value of in (IV) we get
$2x-8=4\phantom{\rule{0ex}{0ex}}⇒2x=12\phantom{\rule{0ex}{0ex}}⇒x=6$

(ii)

Multiplying (III) with 4 we get
$4x+60y=156$            .....(V)
Subtracting (IV) from (V)
$59y=118\phantom{\rule{0ex}{0ex}}⇒y=2$
Putting the value of y in (IV) we get
$4x+2=38\phantom{\rule{0ex}{0ex}}⇒4x=36\phantom{\rule{0ex}{0ex}}⇒x=9$

(iii)
Let
So, the equations obtained are

#### Question 5:

A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this  number, the sum is equal to the number obtained by interchanging the digits. Find the  number.

Let the two digit number be 10x + y.
A two digit number is 3 more than 4 times the sum of its digits.

If 18 is added to this  number, the sum is equal to the number obtained by interchanging the digits.

(I) $-$ (II)
$x=3,y=5$
Thus, the number obtained is 35.

#### Question 6:

The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5 pens is 77 rupeess. Find the cost of 1 book and 2 pens.

Disclaimer: There is an error in the given question. Instead of 7 books and 5 pens there should be 7 books and 6 pens.

Let the cost of 1 book be and the cost of 1 pen be y
The total cost of 6 books and 7 pens is 79 rupees.

Total cost of 7 books and 6 pens is 77 rupeess.

Subtracting (II) from (I)

$2x=10\phantom{\rule{0ex}{0ex}}⇒x=5$
= 7
Cost of 1 book and 2 pens is 5 + 14 = Rs 19.

#### Question 7:

The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each.

Let the incomes of the two persons be x and y.

Savings of the two persons = Rs 200
Expenses = Income $-$ Savings

Multiply (I) with 3 and (II) with 7.

Putting the value y in (I) we get
= 1800
Thus, the income of the two persons is Rs 1800 and Rs 1400.

#### Question 8:

If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.

Let the length be and the breadth be b
Area of rectangle = lb
Length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units.

If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units.

Multiply (I) with 2 and (II) with 3

Putting this value in (I) we get
$3l-5\left(205\right)=7\phantom{\rule{0ex}{0ex}}⇒3l=7+1025=1032\phantom{\rule{0ex}{0ex}}⇒l=344$
Thus, length = 344 units and breadth = 205 units.

#### Question 9:

The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.

Let the speed of the faster car be x km/h and that of the slower car be km/h.
Total distance between them is 70 km.
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}$
When both cars travel in the same direction, they meet in 7 hours.
Total speed = x − y

When both cars move towards each other,
Total speed = x + y

$2x=80\phantom{\rule{0ex}{0ex}}⇒x=40$
Putting the value of in (I)
$40-y=10\phantom{\rule{0ex}{0ex}}⇒y=40-10=30$
Thus, speed of the cars are 40 km/h and 30 km/h.

#### Question 10:

The sum of a two digit number and the number obtained by interchanging its digits  is 99. Find the number.

$⇒11x+11y=99\phantom{\rule{0ex}{0ex}}⇒x+y=9$