# CBSE Class 10 Maths Question Paper 2020

Question papers with the appropriate solutions help students develop an in-depth insight into the subject. They get an understanding of the paper pattern, mark distribution, time management and the different types of questions that they may come across. It is very important for any student in Class 10 to practice these CBSE question papers in order to prepare well for their upcoming exams.

Regularly practising these question papers is quite crucial for subjects like Mathematics. In the case of mathematics, students must practice different types of questions to develop a strong base in the subject. Exposing themselves to various types of questions on a specific topic can help them develop their analytical thinking and problem-solving skills.

Students can find the Class 10 CBSE question papers for the past 10 years on Extramarks. Extramarks, India’s leading e-learning portal, help students access important study material for free. They can download the question papers on any device of their choice and even get a printout.  In addition to the question papers, students can also find the solved answer sheets that will help them clarify their doubts and cross-check their answers. Going through the solutions will also help them understand the necessary steps that they need to show while solving a problem, as each step carries marks.

Students who desire to pursue JEE or other competitive examinations can also try solving these Class 10 Maths Question Papers to brush up on their concepts. In addition to that, students applying for national-level exams for government jobs and higher studies should also go through the NCERT material including the Class 10 Maths Question Papers.

## CBSE Class 10 Maths Question Paper 2020

Please check that the question paper contains a total of 25 questions.

Students are requested to write down the question number properly before attempting to answer the questions.

To read the question paper, 15 minutes will be allotted to read the question paper.

Section A

1. K’s value for the quadratic equation 2X2 + kx + 2 = 0. The value of equal roots is
1. 0
2. -4
3. +-4
4. 4

Sol. The quadratic equation stated here is 2X2 + kx + 2 = 0. For an equal amount of roots, D = 0, which is = b2 – 4ac = 0. Therefore, a = 2, c = 2, and b = k, so k2 – 16 = 0, k2 = 16. Therefore, k = +- 4. So, option no. c will be correct.

1. One sphere’s radius in centimeters who has a volume of 12πcm3 will be
1. 3 2/3
2. 3√3
3. 3
4. 3 1/3

Sol. The sphere’s volume will be = 4/3 πr3. Then after the equation, 4/3 πr3 = 12πr cm3, r3 = 12π * 3/ 4π cm3, r3 = 9 cm3, r3 = (3)2, r = (3) 2/3 cm. So the actual radius of the sphere will be (3) 2/3 cm. So the correct option will be C. The sphere’s volume will be = 4/3 πr3. Then after the equation, 4/3 πr3 = 12πr cm3, r3 = 12π * 3/ 4π cm3, r3 = 9 cm3, r3 = (3)2, r = (3) 2/3 cm. So the actual radius of the sphere will be (3) 2/3 cm. So the correct option will be C.

1. The actual distance between the different points (-m, n) and (m, -n) will be
1. m + n
2. √ 2m2 + 2n2
3. √ m2  + n2
4. 2 √ m2 + n2

Sol. Let us consider that B (-m, n) and A (m, – n).

From the formula of distance, it can be figured out that

AB = √(-m -m)2 +(n- (-n))2

AB = √ (-2m)2 + (2n)2

AB = √ 4 m2 + 4 n2

AB = 2 √ m2 + n2.

Henceforth, the actual distance between the two different points (–m, n) and (m, n) will be 2 √ m2 + n2. So, option D will be the correct option. This is the correct answer.

1. From the P external point, two tangents can be drawn, which are PR and PQ. A circle with a radius of 4 cm along with center O. So, if ∟PQR = 90 degrees, then what will be the length of QP?
1. 2 cm
2. 3 cm
3. 2 √ 2 cm
4. 4 cm

Sol. The angle is divided or bisected from the circle’s centre by the line among the tangents.

PRQ = 45 degrees, then in Δ QOP, tan 45 degrees = OQ / PQ. QP = QO, QP = 4 cm.

Therefore the length of QP will be 4 cm. So, option number D will be the correct option. This is the correct answer.

1. If one zero of the polynomial quadratic, x² + 3x + k is 2. What will be the value of K?
1. -2
2. -9
3. -10
4. +10

Sol. The answer will be B = -10. The above situation states that the value of B = -10.

1. What will the LCM and HCF of 15, 21, 12, respectively, be?
1. 420, 3
2. 420, 3
3. 420, 12
4. 3, 140

Sol. The LCM and HCF of 15, 21, 12 will be 3, 420. The HCF of 15,21,12 is 420 and LCM of 15,21,12 is 3.

1. The polynomial quadratic and the summation of the zeroes is -5, and the value of the product will be 6. The quadratic polynomial will be?
1. x² +5x + 6
2. x² – 5x – 6
3. x² -5x + 6
4. x² + 5x +6

sol. The correct answer will be A= x² +5x +6. This is the correct answer.

1. A die is thrown one time. Then what will be the probability of getting a value less than 3?

Sol. When one die is thrown one time, the total probable results will be six. The results will be 5, 4, 3, 2, 1 and 6. So the outcome, which will be less than 3, will be 2 and 1. So the total number of favorable outcomes will be 2. Then the probability of the outcome will be a favorable outcome / total number of the possible outcome. Therefore, the probability of the number less than 3 = 2 / 6 = 1 / 3.

1. Suppose a letter from the English alphabet is randomly chosen. Then what will be the probability of being one consonant for the chosen letter?

Sol. In the English alphabet, the total number of letters is 26. So, the total possible outcome will be 26. The total number of consonants in the English alphabet will be 21. So, the total number of favorable outcomes will be 21. So the probability = favorable outcome / total number of possible results. Probability of consonant = 21 / 26. So the actual probability that the chosen letter will be consonant will be 21 / 26.

1. In a 600 km flight, an aircraft was slowed down since the weather was very bad. So the average speed of the aircraft was reduced by 200 km and the time of the flight amplified by 30 minutes. So what will be the actual duration needed for the flight?

Sol. If the actual speed of the flight can become x km per hour. Then the total distance the aircraft needs to cover is 600 km. But the original duration of the particular flight will be total distance/time = 600 / x. (Here, x is considered as the amount of time)

The modified speed will be x – 200. Therefore the total amount of time taken by the reduced speed will be 600 / x – 200. Since 30 minutes equals to 0.5 hrs. (30 / 60)

As per the question, 600 / x – 200 – 600 / x = 0.5.

600 x – 120000 + 600 x = 0.5 (x² – 200 x)

0.5 x² – 100 x – 240000 = 0,

x² -600 x + 400 x – 240000 = 0

x (x – 600) + 400 (x – 600) = 0.

(x – 600) (x + 400) = 0

x = – 400, 600. Since speed cannot be measured in negative value, the actual speed of the aircraft will be 600 km per hour.

Therefore, the actual duration of time taken by the aircraft will be total distance / actual speed = 600 / 600 = 1. So it can be said that the actual time duration taken by the aircraft will be only 1 hour. So it is concluded, the time duration could be taken for only 60 minutes.

1. Draw one line segment with 7 cm of length. If A is taken as the center, with a radius of 3 cm, draw one circle. After that, take B as the center and draw another circle with a radius of 2 cm. Then build tangents to all the circles from the center point of the other circles.

Sol.

The steps needed in the construction of the tangents.

1. Drawing one line segment with a length of 7 cm which will be AB.
2. Considering B and A as centres for drawing two circles with radii 2 and 3 cm, respectively.
3. Drawing on a perpendicular bisector with the line segments AB. Then a midpoint will come out as C from AB. Then C will be taken as the centre for drawing the circle with the diameter of AB’s length. Then this circle will intersect the circle with # cm radii at the points R and T, and it will also intersect the circle with 2 cm radii at the points Q and P.
4. Then R to B and T to B should be joined. After joining them, two tangents, such as on the circle BR and BT, will be got. They will be drawn from the circle with a radius of 2 cm and then joined P to A and Q to A. Then we can get two tangents named AQ and AP. They will be drawn from the centre point of A circle along with a radius of 3 cm.  They will be drawn from the circle with a radius of 2 cm and then joined P to A and Q to A. Then we can get two tangents named AQ and AP. They will be drawn from the centre point of A circle along with a radius of 3 cm.

1. If the radii of two concentric circles are 4 cm and 5 cm, then each chord of a circle that is tangent to another circle has a length that is

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 2 cm

Solution:

Let, the centre of two concentric circles C1 and C2 be “O”. The radii of the circles are r1=4 cm and r2=5 cm. Now we create a chord named AC of the circle C2 that is tangent to the circle C1 at B. Also, join OB that is perpendicular to AC.

Therefore,    OBC is a right-angled triangle

Therefore, (OB) 2 + (BC) 2 = (OC) 2

(4)2+ (BC) = (OC)

16 + BC² = 25

(AB) ² = 25 – 16

(AB) ² = 9

Taking square root,

AB = 3 cm

We know, BC = 2AB

So, BC = 2(3)

BC = 6 cm

Answer: Therefore, the length of the chord is evaluated as 6 cm.

1. In the figure 1 given below, i of <AOB = 125º, then < COD is equal to

Figure 1

Answer: ABCD is a quadrilateral circumscribing the circle as found from the problem.

We are aware that the opposing sides of a quadrilateral enclosing a circle subtend additional angles at its Centre. So, from the figure we obtain

AOB + COD = 180°

125° + COD = 180°

COD = 55°

Hence,  the value of COD is 55°

1. In the figure given below, AB is a chord of the circle and AOC is the diameter of the circle such that <ACB =50º. If AT is the tangent to the circle at point A, then <BAT =?

(A) 70° (B) 60° (C) 30° (D) 50°

A circle with centre O, diameter AC and ACB = 50°

AT is a tangent to the circle at point A as portrayed in the above figure.

Since the angle in a semicircle is a right angle, Hence the value of angle CBA is 90°. ,

CBA = 90°

By sum angle property of triangle,

ACB + CAB + CBA = 180°

50° + CAB + 90° = 180°

CAB = 40° … (i)

Since the tangent at any point on the circle is perpendicular to the radius through the point of contact,

We get, OA  AT

OAT = 90°

OAT + BAT = 90°

CAT + BAT = 90°

40° + BAT = 90° [from equation (i)]

BAT = 50°

Answer: Hence it has evaluated that the value of BAT is 50°

1. Use Euclid’s division algorithm to find the HCF of:
2. a) 135 and 2254
3. b) 196 and 38220
4. c) 867 and 255
5. a) Answer: Given numbers: 135 and 225

Here, 225>135.

So, we will divide the greater number by the smaller number.

Divide 225 by 135.

The quotient is 1 and the remainder is 90.

225=135×1+90

Divide 135 by 90

The quotient is 1 and the remainder is 45.

135=90×1+45

Divide 90 by 45.

The quotient is 2 and the remainder is 0.

90=2×45+0

Thus, the HCL is 45

1. b) 38220> 196

By Euclid’s division algorithm,

We get,

38220= 196*195+0

In this case,

Remainder = 0

So, the division i.e. 196 is the HCF of 38220 and 196

Therefore, HCF (38220, 196) = 196

1. c) 867 = 255 × 3 + 102.

255 = 102 × 2 + 51.

102 = 51 × 2 + 0.

H.C.F of (867, 255 ) = 51

By evaluating the problem the value of the HCF has been found as 51.

1. Show that any positive odd integer is of form 6q+1, or 6q+ 3, or 6q + 5, where q is some integer.

Answer: 6q + 1, 6q + 3 and 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, and 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form 6q + 1 or 6q + 3, or 6q + 5.

It portrays that any integer value can be incorporated in the form 6q + 1 or 6q + 3, or 6q + 5.

1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2: 3.

Solution: Let the coordinates of point C be (x, y).

X-coordinate of C = mx2+nx1/ m+ n

= 2*4+3*(-1) / 2+3

=8-3/5

=1

Y- Coordinate of C = my2+ny1/ m+ n

= 2*(-3)+3*(7)/2+3

=-6+21/5

=3

Hence, the coordinates of C are (1, 3).

The coordinates of C are found as (1,3)

1. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

(i) First find out the probability of drawing a red ball by using the formula

Probability of an event= Number of possible outcomes/ total no of favorable outcomes

(ii) Then by using the formula of the sum of complementary events you have to, find out the probability of not getting a red ball.

P (E) + P (not E) =1

Solution: Number. of red balls in a bag = 3

Number. of black balls in a bag = 5

Total number. of balls = 3 + 5 =8 N

Probability of drawing red ball = Number of possible outcomes /Total no. of favorable outcomes

Probability of drawing red ball= 3/8

(ii) Probability of not getting red ball P(R) =1-P (not E) = 1-3/8-⅝

1. This question is straightforward – the value of the sum of roots and the product of roots is given. You have to form a quadratic polynomial. Put the values in the general equation of the quadratic polynomial i.e.

K (x2– (sum of roots product of roots)x+ product of roots)

1. ¼, -1

Solution: We know that the general equation of a quadratic polynomial is:

K (x2– (sum of roots product of roots)x+ product of roots)

=K{x2-1/4x+1/4*(-1)}

= K{x2-1/4x-1/4}

Hence, the quadratic polynomial is K{x2-1/4x-1/4}

1. 19. Mangoes were being sold by fruit vendors in a retail market; they were being packed in boxes. The number of mangoes in these cartons varied. The allocation of mangoes based on the number of boxes was as follows.

 Number of mangoes 50 – 52 53 – 55 56 – 58 59 – 61 62 – 64 Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

We know that xi = Upper-class limit + lower-class limit/ 2

Class size, h = 3

Taking assumed mean, a = 57

 Number of mangoes Number of boxes fi xi di = xi  – a u = di i h fiui 49.5 – 52.5 15 51 –6 –2 –30 52.5 – 55.5 110 54 –3 –1 –110 55.5 – 58.5 135 57 (a) 0 0 0 58.5 – 61.5 115 60 3 1 115 61.5 – 64.5 25 63 6 2 50 fi = 400 fiui = 25

From the table of mangoes and boxes given above, we obtain

fi= 400

fi u i = 25

Mean, x = a + (∑ fi u i)/ (∑ fi) * h

x= 57+ (25/ 400)*3

x= 57+1/16*3

x=57+3/16

x=57+0.19

x=57.19

The mean numbers of mangoes kept in a packing box are 57.19.

It portrays the mean value of mangoes that are kept in one single box.

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution: Let, the volume of a cube = a3

Length of the edge is a

As the cubes are joined end to end, they will appear as follows

Volume of the cube,

a3 = 64cm3

a3 = 64cm3

a = 3 64cm3

a = 3 (4cm)3

a = 4cm

Therefore, Length of the resulting cuboid, l = a = 4cm

Breadth of the resulting cuboid, b = a = 4cm

Height of the resulting cuboid, h = 2a = 2´ 4cm = 8cm

Surface area of the resulting cuboid, 2 (lb+ bh+ lh)

= 2 (4 cm ‘ 4 cm + 4cm ‘8 cm)

= 2 (16 cm2 + 32 cm2 + 32 cm2

=2 * 80 cm2

=160

Hence, the value of surface area is 160.

#### 21. Raja tells his son, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.

Solution: Present age of Raja =x years and his son = y

Therefore, 7 years ago, the age of Raja = (x-7) years and his son = (y- 7) years

Using this information and applying the known condition, 7 years ago, Raja was 7 times as old as his son then:

x – 7 =7(y – 7)

x – 7 = 7 y – 49

x – 7 y – 7 = 49 = 0

x – 7 y + 42  0

3 years from now, the age of Raja = (x + 3) years and his son = (y + 3) years, and also Raja will be 3 times as old as his son. Then mathematically,

x+3 =3(y + 3)

x + 3 = 3y + 9

x – 3y +3 – 9 = 0

x – 3y – 6 = 0

Algebraic representations, where x and y are present ages of Raja and his son respectively

x – 7 y + 42 = 0 (1)

x – 3y – 6 = 0 (2)

Therefore, the algebraic representation for equation 1 is:

x – 7 y + 42 = 0

-7 y = -x – 42

7 y = x + 42

y = x +42/ 7

And, algebraic representation for equation (2) is:

x – 3y – 6 = 0

-3y = -x + 6

3y = x – 6

y = x -6/ 3

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in the table shown below.

For equation (1)

 x 21 28 V= x=42/7 9 9

For equation (2)

 x 30 21 V= x-6/3 8 5

The representation of the graph is

Unit: 1 cm = 5 years.

CBSE extra questions

1. Consider the figure below
2. The figure given above represents a shaded region. Evaluate the perimeter of the shaded region where ADC, AEB, and BFC are semi-circles over the diameters AC, AB and BC, respectively.
3. Suppose we consider OABC a quadrant of a circle whose radius is 7 cm. Calculate the area of the shaded portion of the figure. [Provided: π = 22/7, OD=4 cm]
4. The vertices of a triangle are M (-5,7), N (-3,-5), and O (6,7). Find out the area of the triangle with the given measurements of the vertices.
5. Consider points (3,2), (a,b), and (4,6) to be linear to each other. Evaluate the relationship between a and b.
6. If points X (0,2) are equidistant from respective points Y (4,p) and Z (5,p). Calculate the value of P.
7. Find the value of X in the equations below
• √3X2-2√2X-2√3=0
• X25√5X-70=0
1. The roots provided in the below equation are real.

Equations are ax2+2bx+c=0 and bx2-2√ac x+b=0.

Then show that b2=4ac

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