Class 10

[CBSE 2025] The students of a class are made to stand equally in rows. If 3 students are extra in . . .

Question : The students of a class are made to stand equally in rows. If 3 students are extra in each row, there would be 1 row less. If 3 students are less in a row, there would be 2 more rows. Find the number of students in the class.

CBSE 2025

Solution :
Let the number of students in each row = x
and the number of rows = y

Total number of students = xy

ATQ

(x+3)(y-1)=xy
x(y-1)+3(y-1)=xy
xy-x+3y-3=xy
-x+3y-3=0
-x+3y=3 — (1)

Also

(x-3)(y+2)=xy
x(y+2)-3(y+2)=xy
xy+2x-3y-6=xy
2x-3y-6=0
2x-3y=6 — (2)

Solving equation (1) and (2)

-x+3y=3
2x-3y=6
--------------
x+0y=9
--------------

x=9

putting in equation (1)
-9+3y=3
3y=3+9
3y=12
y=12/3
y=4

Number of students in each row = 9
and the number of rows = 4

Total number of students
=xy
=9×4
=36

Hence, the total number of students in the class is 36.

[CBSE 2025] The perimeter of a rectangle is 70 cm. The length of the rectangle is 5 cm . . .

Question : The perimeter of a rectangle is 70 cm. The length of the rectangle is 5 cm more than twice its breadth. Express the given situation as a system of linear equations in two variables and hence solve it.

CBSE 2025

Solution :

Let the length and breath of a rectangle be x cm and y cm respectively.

ATQ

Perimeter of rectangle = 2(l+b)
70 = 2(x+y)
70/2=x+y
35=x+y
x+y=35 — (1)

Also,

x=5+2y
x-2y=5 — (2)

Solving equation (1) and (2)

x+y=35

x-2y=5
- + -
---------------
0 +3y=30
---------------

3y=30

y=30/3
y=10

putting in equation (1)

x+10=35
x=35-10
x=25

Hence, length = 25 cm and breadth = 10 cm

The two angles of a right angled triangle other than . . .

Question : The two angles of a right angled triangle other than 90° are in the ratio 2:3. Express the given situation algebraically as a system of linear equations in two variables and hence solve it.

CBSE 2025

Solution :

Let the measures of the two angles be x and y.

ATQ
x+y+90°=180° (ASP)
x+y=180°-90°
x+y=90° — (1)

Also
x:y=2:3
x/y=2/3
3x=2y
3x-2y=0 — (2)

Solving equation (1) and (2)

[x+y=90°]×2

[3x-2y=0]×1

2x+2y=180°
3x-2y=0
----------------
5x+0y=180
----------------

5x=180
x=180/5
x=36°

putting in equation (1)

36°+y=90°
y=90°-36°
y=54°

Hence, the required values of two angles is 36° and 54°.

The area of a rectangle gets reduced by 9 square units . . .

Question : The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Doubt by Noor

Solution :

Let length of the rectangle be x units

and breadth of the rectangle be y units

Area of rectangle = l×b
= xy square units.

ATQ

(x-5)×(y+3)=(xy-9)
x(y+3)-5(y+3)=xy-9
xy+3x-5y-15=xy-9

3x-5y-15=-9

3x-5y=-9+15
3x-5y=6 — (1)

Also

(x+3)×(y+2)=(xy+67)
x(y+2)+3(y+2)=xy+67
xy+2x+3y+6=xy+67
2x+3y+6=67
2x+3y=67-6
2x+3y=61 — (2)

Solving equations (1) and (2)

3x-5y=6

2x+3y=61

[3x-5y=6 ]×2

[2x+3y=61]×3

6x-10y=12

6x+9y=183
- - -
-----------------
0x-19y=-171
-----------------

-19y=-128
19y=128
y=128/19
y=9

putting y=9 in equation (1)
3x-5(9)=6
3x-45=6
3x=6+45
3x=51
x=51/3
x=17

Hence, length of rectangle is 17 units and breadth of rectangle is 9 units.

[CBSE 2025] √2x+√3y = 5, √3x-√8y =-√6 Solve for x and y . . .

Question : Solve for x and y :

√2x+√3y = 5

√3x-√8y =-√6

CBSE 2025

Solution :

√2x+√3y = 5 — (1)

√3x-√8y =-√6 — (2)

[√2x+√3y = 5]×√3

[√3x-√8y =-√6]×√2

√6x+3y=5√3

√6x-4y=-√12
- + +
---------------------------
0x + 7y =5√3+√12
---------------------------

7y=5√3+2√3
7y=7√3
y=7√3/7
y=√3

Putting in equation (1)

√2x+√3(√3) = 5
√2x+3=5
√2x=5-3
√2x=2
x=2/√2
x=√2

Hence, x=√2 and y=√3

[CBSE 2025] 30x+44y=10, 40x+55y=13 Solve the following system of equations algebraically . . .

Question : Solve the following system of equations algebraically :

30x+44y=10

40x+55y=13

CBSE 2025

Solution :

30x+44y=10 — (1)

40x+55y=13 — (2)

[30x+44y=10]×4

[40x+55y=13]×3

120x+176y=40
120x+165y=39

- - -
----------------------
0x + 11y = 1
----------------------

11y=1
y=1/11

Putting in equation (1)
30x+44y=10
30x+44(1/11)=10

30x+4=10
30x=10-4
30x=6
x=6/30

x=1/5
Hence, x=1/5 and y=1/11.

[CBSE 2025] Solve the following system of equations algebraically . . .

Question : Solve the following system of equations algebraically :

37x+63y=137

63x+37y=163

CBSE 2025

Solution :

37x+63y=137 — (1)

63x+37y=163 — (2)

Adding equation (1) and (2)

37x+63y=137

63x+37y=163
-----------------------
100x+100y=300

-----------------------

100(x+y)=300
x+y=300/100

x+y=3 — (3)

Subtracting equation (2) from (1)

37x+63y=137

63x+37y=163
- - -
-----------------------
-26x+26y=-26

-----------------------

26(-x+y)=-26
-x+y=-26/26
-x+y=-1 — (4)

Now Solving equation (3) and (4) by elimination method

x+y=3

-x+y=-1
-------------
0 +2y=2

-------------

2y=2

y=2/2

y=1

Putting in equation (3)
x+1=3
x=3-1
x=2

Hence, x=2 and y=1

[CBSE 2025] A school is organizing a grand cultural event to show . . .

Case Study Based Question on Chapter - 3
(Pair of Linear Equations in Two Variables)

A school is organizing a grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹ 50 and for each table is ₹ 200. The school spends ₹ 30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are 300.

(i) Write down the pair of linear equations representing the given information. [1 Marks]

(ii) (a) Find the number of chairs and number of tables rented by the school. [2 Marks]
OR

(b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. [2 Marks]

(iii) What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? [1 Marks]

CBSE 2025

Detailed Solution :

(i) Write down the pair of linear equations representing the given information. [1 Marks]

Ans :
Let number of chairs = x

and number of tables = y

ATQ

50x+200y=30000

50[x+4y]=30000

x+4y=30000/50

x+4y=600 — (1)

Also

x+y=300 — (2)

(ii) (a) Find the number of chairs and number of tables rented by the school. [2 Marks]

Ans :
Solving equation (1) and (2)

x + 4y=600

x + y=300
- - -
----------------
0+3y=300
----------------

3y=300

y=300/3

y=100

putting in equation (2)

x+y=300
x+100=300

x=300-100

x=200

Number of chairs = x = 200

Number of tables = y = 100

(b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. [2 Marks]

Ans :

50x+200y=27000

50[x+4y]=27000

x+4y=540

x+y=30

Solving the above equations

x+4y=540

x+ y=300
- - -
---------------
0+3y=240
---------------

y=240/3

y=80

putting in equation (2)

x+80=300

x=300-80

x=220

Number of chairs = 220

Number of tables = 80

(iii) What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? [1 Marks]

Ans :

Using Equation (1)

x + 4y=600

x=0
y=?

0+4y=600

4y=600

y=600/4
y=150

Maximum number of tables which can be rented in 30,000 is 150, if no chairs are rented.

[CBSE 2025] A bag contains some red and blue balls . . .

Question : A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of 24. If three times the number of red balls exceeds the number of blue balls by 20, find the number of red and blue balls.

CBSE 2025

Solution :

Let the number of red balls = x

and the number of blue balls = y

ATQ

10% of x + 20% of y = 24
(10/100)x + (20/100)y = 24
x/10 + 2y/10 = 24
(x+2y)/10 = 24
x+2y = 240 — (1)

3x-y = 20 — (2)

Solving equation (1) and (2)

x + 2y = 240

[3x - y = 20]×2

x + 2y = 240

6x - 2y = 40

---------------------
7x + 0 = 280
---------------------
7x=280
x=280/7
x=40

putting in equation (2)

3(40)-y = 20
120-y=20
120-20 = y
100 = y
y = 100

Hence, number of red balls = 40 and number of blue balls = 100.

[CBSE 2025] Vijay invested certain amounts of money in two schemes A and . . .

Question : Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received 20 more as annual interest. How much money did he invest in each scheme ?

CBSE 2025

Solution :

Let amount invested by Vijay at 8% interest rate be ₹ x

and the amount invested by Vijay at 9% interest rate be ₹ y

We know,
Simple Interest (SI) = [Principle×Rate×Time]/100

Here Time is 1 Year in both the case.

SI on Amount of ₹ x which is invested at 8% interest rate

= [Principle×Rate×Time]/100
= [x×8×1]/100
= 8x/100

SI on Amount of ₹ y which is invested at 9% interest rate

= [Principle×Rate×Time]/100
= [y×9×1]/100
= 9y/100

ATQ
8x/100 + 9y/100 = 1860
(8x+9y)/100 = 1860
8x+9y=186000 — (1)

Similarly
9x+8y= (1860+20)×100
9x+8y=188000 — (2)

Solving equation (1) and (2)

[8x+9y=186000] ×9

[9x+8y=188000] ×8

72x+81y=1674000
72x+64y=1504000
- - -
----------------------------
0 + 17y = 170000
----------------------------

17y=170000
y=170000/17
y=10000

putting in equation (1)

8x+9[10000]=186000
8x+90000=186000
8x=186000-90000
8x=96000
x=96000/8
x=12000

Hence, money invested in scheme A is ₹12000 and scheme B is ₹10000.

Pages