Hello students, here the **solution of Class 10 math paper** is given below.

Class 10 math paper solution 2023 |

*Note:- some answers are not given because we are solving them and will be updated in some time*

### Class 10 math paper solution 2023

#### 1. In what ratio, does the x-axis divide the line segment joining the points A(3,6) and B(-12,-3)?

#### 2. In the given figure,PQ is tangent to the circle centered at O. If angle AOB=90°, then the measure of angle ABQ will be

#### 3. If 2tanA=3, then the value of 4sinA+3cosA/4sinA-3cosA is

#### 4. In a group of 20 people, 5 cannot swim. If one person is selected at random ,then the probability that he / she can swim is.

#### 5. The distribution below gives the marks obtained by 80 students on a test:

Class 10 cbse board maths paper solution 2023 |

#### 6. The curved surface area of a cone having height 24 cm and radius 7 cm, is

#### 7. The end-points of a diameter of a circle are (2,4) and (-3,-1). The radius of the circle is

#### 8. Which of the following is a quadratic polynomial with zeroes 5/3 and 0?

#### 9. The graph of y=p(x) is given , for a polynomial p(x). The number of zeroes of p(x) from the graph is

#### 10. The value of k for which the pair of equations kx=y+2 and 6x=2y+3 has infinite many solutions,

#### 11. form A.P. with common difference then the value a-2b-c is equal to

#### 12. If the value of each observation of a statistical data is increased by 3, then the mean of the data

#### 13. probability of happening of an event is denoted by p and probability of non happening of the event is denoted by q. Relation between P and q is.

#### 14. a girl calculates that the probability of her winning the first prize in a lottery is 0.08. if 6000 tickets are sold, how many tickets has see bought?.

#### 15. If alpha, beta are the zeroes of a polynomial p(x)= x2+x-1, then 1/a+1/b equals to.

#### 16. The least positive value of k, for which the quadratic equations 2x2+kx-4=0 has rational roots, is.

#### 17. [5/8 sec260°-tan260°+cos245°] is equal to.

#### 18. Curved surface area of a cylinder of height 5 cm is 94.2cm2. radius of the cylinder is (take π=3.14)

#### 19. Assertion (A) : The perimeter of Triangle ABC is a rational number.

Reason(R): The sum of the squares of two rational numbers is always rational.

#### 20. Assertion (A) : Point P(0,2) is the point of intersection of y-axis with the line 3x+2y=4.Reason(R): The distance of point P(0,2) from x-axis is 2 units.

#### 21. Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case.

#### 22. A bag contains 4 red, 3 blue and 2 yellow balls. One ball is drawn at random from the bag . Find the probability that drawn ball is (a) red (b) Yellow

#### 23. (A) solve the pair of Equations x=5 and y=7 graphically.

(B) using graphically method, find whether pair of equations x=0 and y= -3 is consistent or not.

#### 24. (A) if sin 0 + cos 0= √3 , then find the value of sin0.cos0

(B) if sin a = 1/√2 and cot b = √3 , then find the value of cosec a + cosec b.

#### 25. In the given figure, XZis parallel to BC. AZ= 3cm, ZC =2 cm, BM= 3 cm and MC =5cm. Find the length XY.

#### 26. The centre of a circle is (2a, a-7). Find the values of 'a' if the circle passes through the point (11, -9). Radius of the circle is 5√2 cm.

#### 27. (a) Two tangents TP and TQ are drawn to circle with centre O from an external point T. Prove that angle PTQ= 2 Angle OPQ.

#### (B) In the given figure, a circle is inscribed in a quadrilateral ABCD in which angle B= 90°. If AD = 17 cm, AB = 20 cm and DS 3 cm, then find the radius of the circle.

AP + BS = BQ + CR (Sum of opposite sides of a cyclic quadrilateral)

Substituting AP = AS = x, BS = BP = y, and CR = CQ = z, we get:

x + y = z + DS ...(1)

Also, since triangle ABP is a right triangle, we have:

AP^2 + BP^2 = AB^2

Substituting AP = x, BP = y, and AB = 20 cm, we get:

x^2 + y^2 = 400 ...(2)

Similarly, since triangle CQR is a right triangle, we have:

CQ^2 + QR^2 = CR^2

Substituting CQ = z, QR = r, and CR = z + r, we get:

z^2 + r^2 = (z + r)^2

Simplifying and rearranging, we get:

r = (z^2)/(2z + 2DS) ...(3)

Now, we need to find x, y, and z in terms of the given quantities. From equation (1), we have:

x + y - DS = z

Adding and subtracting equation (2), we get:

2x = 400 - 2y, and

2y = 400 - 2x

Substituting these into the above equation, we get:

2x - DS = 400 - 2x

Simplifying and solving for x, we obtain:

x = (400 - DS)/4

Similarly, we can solve for y:

y = (400 - DS)/4

Substituting these values of x and y into equation (1), we get:

(400 - DS)/2 = z

Substituting this into equation (3), we obtain:

r = [(400 - DS)^2]/[8(400 - DS)] = (400 - DS)/8 = 3/8

Therefore, the radius of the inscribed circle is 3/8 cm.

#### 28. Half of the difference between two numbers is 2. The sum of the greater number and twice the smaller number is 13. Find the numbers.

From the first condition, we have:

(x - y)/2 = 2

Multiplying both sides by 2, we get:

x - y = 4

From the second condition, we have:

x + 2y = 13

Now we have two equations with two unknowns:

x - y = 4 ...(1) x + 2y = 13 ...(2)

We can solve for x and y by eliminating one variable. Multiplying equation (1) by 2, we get:

2x - 2y = 8

Adding this equation to equation (2), we get:

3x = 21

Dividing both sides by 3, we get:

x = 7

Substituting this value of x into equation (1), we get:

7 - y = 4

Solving for y, we get:

y = 3

Therefore, the two numbers are 7 and 3.

#### 29. A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains [ 1408/21] m3 of air. ( Take π= 22/7)

Let the height of the cylindrical part be 'h' and the base radius of the hemisphere be 'r'. Then, the total height of the room will be:

H = h + 2r ...(1)

The volume of the room can be calculated as the sum of the volume of the cylindrical part and half the volume of the hemispherical part, which is given as:

V = (1/2)πr^2h + (2/3)πr^3 ...(2)

Substituting the value of r in terms of h from the given condition, we get:

r = h/2

Substituting this value of r in equation (2), we get:

V = (1/2)π(h^2/4)h + (2/3)π(h^3/8)

Simplifying, we get:

V = πh^3/12 + πh^3/12

V = πh^3/6

Equating this to the given volume, we get:

πh^3/6 = 1408/21

Solving for h, we get:

h^3 = (1408/21) x 6/π

h^3 = 128

h = 4

Substituting this value of h in equation (1), we get:

H = h + 2r

H = 4 + 2(2)

H = 8

Therefore, the total height of the room is 8 m.