Math class 10 Answer key 2023

 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

Below the answers of class 10 Math paper solution is given.

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

Answer:- 2:1
= K×-3+1×6/k+1=0
= -3k=-6
= k=-6/-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

And:- 47.5°

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

And:- 3

=4×3/√3 + 3×2/√3/4×3/√3 - 3×2/√3
= 18/√3/6/√3

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.

Answer:- 3/4

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

Class 10 cbse board maths paper solution 2023
Answer:- 50-60

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

Answer:- 550cm2

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

Answer:- 5√2

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

Answer:- 3x(3x-5)

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

Answer:- 1

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

Answer:- k=3

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

Answer:- -2a-4d

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

Answer:- increase by 3

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.

Answer:- p+q =1

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?.

Answer:- 480

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

Answer:- 1

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

Answer:- 2

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

Answer:- 0

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

Answer:- 2.9 cm

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.

Answer:- Both assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of assertion (A)

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.

Answer:- Assertion (A) is false but Reason (R) is true.

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

Answer:- 5375

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.

Answer:- However, the equation x = 5 represents a vertical line passing through the point (5,0) on the x-axis. The equation y = 7 represents a horizontal line passing through the point (0,7) on the y-axis. These lines do not intersect, as they are parallel to each other.

Therefore, there is no solution to this pair of linear equations.

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.

Answer:- 6cm

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.

Let the tangents TP and TQ intersect at point X, and let the radius of the circle be r. Since OP and OQ are radii of the circle, we have OP = OQ = r.

Since TP and TQ are tangents to the circle, we have:

∠OTP = ∠OTQ = 90° (tangents are perpendicular to the radius at the point of contact)

Now, we can apply the angle sum property of triangle TPX to obtain:

∠PTX + ∠PTQ + ∠XTQ = 180° (sum of angles in a triangle)

Since TP = TQ (both are tangents from point T), we have ∠PTX = ∠QTX, so:

2∠PTQ + ∠XTQ = 180°

Also, we know that ∠OTP = ∠OTQ, so:

∠OPQ = 1/2 (∠OTP + ∠OTQ) = 1/2 (2∠OTP) = ∠OTP

Therefore, we have:

∠XTQ = 2∠OPQ

Substituting this into the earlier equation, we get:

2∠PTQ + 2∠OPQ = 180°

Dividing both sides by 2, we obtain:

∠PTQ + ∠OPQ = 90°

Now, since ∠PTQ and ∠OPQ are both angles in triangle TPQ, we have:

∠PTQ + ∠OPQ + ∠PQT = 180°

Substituting ∠PTQ + ∠OPQ = 90°, we get:

90° + ∠PQT = 180°

Simplifying gives:

∠PQT = 90°

Therefore, in triangle OPQ, we have:

∠OPQ + ∠PQO + ∠OQP = 180° (sum of angles in a triangle)

Substituting ∠OPQ = ∠PTQ/2 and ∠PQT = 90°, we get:

∠PTQ/2 + ∠PQO + ∠OQP = 180°

Simplifying and rearranging gives:

∠PTQ/2 = ∠PQO + ∠OQP

Therefore, we have:

∠PTQ/2 = 2∠OPQ

Or equivalently,

∠PTQ = 2∠OPQ

Hence, we have proved that angle PTQ is twice 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.

Answer:- Since the circle is inscribed in quadrilateral ABCD, we have:

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.

Answer:- Let the two numbers be x and y, where x is the greater number and y is the smaller number.

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.

(b) An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone which is 1/6 of the volume of  the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take л= 3.14)

Answer:- The volume of the empty cone can be calculated using the formula:

Volume of cone = (1/3)πr^2h

Substituting the given values, we get:

Volume of cone = (1/3) x 3.14 x 3^2 x 12

Volume of cone = 113.04 cm^3

The lower part of the cone which is unfilled is 1/6 of the volume of the cone, so the volume of the unfilled part is:

Volume of unfilled = (1/6) x Vcone

Volume of unfilled = (1/6) x 113.04

Volume of unfilled = 18.84 cm^3

The volume of the ice-cream can be calculated as the difference between the volume of the hemisphere and the volume of the unfilled part of the cone. The radius of the hemisphere can be calculated as the radius of the cone, which is 3 cm. The volume of the hemisphere can be calculated using the formula:

Volume of hemisphere = (2/3)πr^3

Substituting the value of r, we get:

Volume of hemisphere = (2/3) x 3.14 x 3^3

Volume of hemisphere = 56.52 cm^3

Therefore, the volume of the ice-cream is:

Volume of ice-cream = Volume of hemisphere - Volume of unfilled

Volume of ice-cream = 56.52 - 18.84

Volume of ice-cream = 37.68 cm^3

Hence, the volume of the ice-cream is 37.68 cm^3.

30. Prove that √5 is irrational number.

Answer:- To prove that √5 is an irrational number, we need to assume the opposite, i.e., we assume that √5 is a rational number. By definition, a rational number can be expressed as the ratio of two integers, say p and q, where q is not equal to zero and p and q have no common factors other than 1.

Assuming that √5 is rational, we can write:

√5 = p/q

Squaring both sides, we get:

5 = p^2/q^2

Multiplying both sides by q^2, we get:

5q^2 = p^2

This means that p^2 is divisible by 5, which implies that p must be a multiple of 5. Let p = 5k, where k is an integer. Substituting this value in the above equation, we get:

5q^2 = (5k)^2

5q^2 = 25k^2

Dividing both sides by 5, we get:

q^2 = 5k^2

This means that q^2 is divisible by 5, which implies that q must also be a multiple of 5.

However, we had assumed that p and q have no common factors other than 1, which contradicts the fact that p and q are both multiples of 5. Therefore, our assumption that √5 is a rational number leads to a contradiction, and we can conclude that √5 is an irrational number.

31. Prove that (cosecA-sinA) (secA-cisA)= 1/cotA+tanA


32. A ladder set against a wall at an angle 45° to the ground. If the foot of the ladder is pulled away from the wall through a distance of 4 m, its top slides a distance of 3 m down the wall making an angle 30° with the ground. Find the final height of the top of the ladder from the ground and length of the ladder.


33. The ratio of the 11th term to 17th term of an A.P. is 3; 4. Find the ratio of 5th term to 21st term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.


34. 250 logs are stacked in the following manner: 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?


35. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle. 


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