CBSE Board 2020 Solution for Math for Class 10 Section D

# CBSE Board 2020

## Mathematics

### Question Paper Solution

#### Section D

##### 4 Marks Questions

Question 35: The mean of the following frequency distribution is 18. The frequency f in the class interval 19-21 is missing. Determine f.

 Class Interval Frequency 11-13 13-15 15-17 17-19 19-21 21-23 23-25 3 6 9 13 f 5 4

Class IntervalFrequency (fi)Class Mark (xi)fixi
11-1331236
13-1561484
15-17916144
17-191318234
19-21f2020f
21-23522110
23-2542496
TotalΣfi = 40 + fΣfixi = 704 + 20f

x=(Σf_ix_i)/(Σf_i)

Or, (700+20f)/(40+f)=18

Or, 700+20f=720+18f

Or, 700+2f=720

Or, 2f=7=20

Or, f=10

OR

The following table gives production yield per hectare of wheat of 100 farms of a village:

 Production Yield No. of farms 40-45 45-50 50-55 55-60 60-65 65-70 4 6 16 20 30 24

Change the distribution to a ‘more than’ type distribution and draw its ogive.

Production YieldCumulative Frequency
More than 40100
More than 4596
More than 5090
More than 5574
More than 6054
More than 6524

Question 36: From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer: Let us assume the height of building = p = 20 m

Total height of building and transmission tower = p’

The angle of elevation to the top of the building = 45° and that to the top of transmission tower = 60°

Height of building subtracted from total height will give the height of transmission tower.

1st case:
text(tan) 45°=p/b

Or, (20)/(b)=1

Or, b=20 m

2nd case:
text(tan) 60°=(p’)/(b)

Or, sqrt3=(p’)/(20)

Or, p’=20sqrt3

Height of transmission tower can be calculated as follows:

p'- p = 20sqrt3 - 20

= 20(sqrt3 - 1)

= 20(1.732 - 1)

= 20 x× 0.732 = 14.64  m

Question 37: It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Answer: 12A + 12B = 1 (Case 1)

Case 2: 4A + 9B = 1/2

Or, 8A + 18B = 1

Or, 8A + 18B = 12A + 12B

Or, 12A – 8A = 18B – 12B

Or, 4A = 6B

Or, A = 3/2B

Substituting the value of A in first equation, we get:

12×3/2B+12B=1

Or, 18B + 12B = 1

Or, 30B = 1

Or, B = 1/(30)

B will fill the pool in 30 hours

Substituting the value of B in second equation, we get:

8A+18×1/(30)=1

Or, 8A+3/5=1

Or, (40A+3)/5=1

Or, 40A + 3 = 5

Or, 40A = 3 – 5 = 2

Or, A=2/(40)=1/(20)

So, A will fill the pool in 20 hours

A takes 20 hours and B takes 30 hours to fill the swimming pool

Question 38: Prove that sqrt5 is an irrational number.

Answer: Let us assume the contrary, i.e. sqrt5 is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;

sqrt5=a/b

Or, bsqrt5=a

Squaring on both sides, we get;

5b^2=a^2

This means that a2 is divisible by 5 and hence a is also divisible by 5.

This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.

This also contradicts our earlier assumption that sqrt5 is irrational.

Hence, sqrt5 is irrational.

Question 39: Draw a circle of radius 3.5 cm. From a point P, 6 cm from its center, draw two tangents to the circle.

Draw a line segment OP = 6 cm

Make perpendicular bisector of OP which intersects OP at point O’.

Take O’P as radius and draw another circle.

From point P, draw tangents to points of intersection between the two circles.

OR

Construct a Δ ABC with AB = 6 cm, BC = 5 cm and ∠B = 60°. Now construct another triangle whose sides are 2/3 times the corresponding sides of Δ ABC.

Draw a line segment BC = 5 cm.

Make a 60° angle from point B and draw AB = 6 cm.

Join A and C to get the triangle ABC.

Dividing the base:

Draw a ray at an acute angle from BC.

Plot 3 points on BA so that BB1 = B1B2 = B2B3

Join B3 to point A.

Draw a line from B2 parallel B3C so that it meets BC at point C’.

Draw A’C’ || AC.

Triangle A’C’B is the required triangle.

Question 40: A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. (Take π=(22)/7)

Answer: Volume of cone =1/3πr^2h

=1/3×(22)/7×7^2×3.5

=1/3×22×7×3.5=(539)/3

Volume of hemisphere =2/3πr^3

=2/3×(22)/7×7^3

=(2156)/3

Volume of solid =(539+2156)/3=(2695)/3=898.33 sq cm