Mathematics 211 Solved Assignment 2021 - 22 | NIOS Secondary Solved Assignment (2021 - 22)

 NIOS Solved Assignment 2021 - 22
Mathematics 211 Solved Assignment 2021 - 22
Tutor Marked Assignment
Max Marks: 20

Numbers with a negative sign are less than zero

Right of zero represented by ‘+’ sign and to the left of zero represented by ‘–’ sign.

+ Sign numbers can be denoted as simply numbers  

On a number line the number increases as we move to the right and decreases as we move to the left.  

1: Draw a line and mark some points at equal distance on it

Mark a point as zero on it. Points to the right of zero are positive integers and are marked simply 1, 2, 3  and so on , while points to the left of zero are negative integers and are marked – 1, – 2, – 3 and so on

The temperature in Leh was 9°C below zero

Go 9 step left to zero and Mark Leh

In Amritsar it was 1°C above zero.  

Go 1 step right to zero and Mark Amritsar

In Mount Abu was 2 °C.

Go 2 steps right to zero and Mark Mount Abu

(b) A rectangular sheet of paper 44cm × 18cm is rolled along its length and a cylinder is formed. Find the volume of the cylinder. (See Lesson 21)

Let the radius be r

We know that the length of the sheet = circumference of the circle and height of the cylinder = 18cm

Now, 

Circumference of the base = 2πr=44

=> 2 * 22/7 * r = 44

=> r = 7

Again, 

Height of the cylinder=breadth of the rectangle = 18cm

Now, 

Volume of cylinder = πr2h

 = 22/7 * 72 *18

= 22*7*18

= 2772 Cm3

2. Answer any one of the following questions.

गणित (211) | Mathematics 211 NIOS Free Solved Assignment 2021 – 22 (Hindi Medium)

(a) Diagonals of a quadrilaterals PQRS bisect each other. If ∠P = 400. Determine ∠Q. (Lesson-13)

Ans: If the diagonals bisect each other than it is a parallelogram.

Now, Sum of adjacent angles of a parallelogram is 180°  

So, Angle P + Angle Q = 180°

Given, Angle P = 400 

Then, Angle Q will be =1800 – 400 = 140°

(b) Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7) (See Lesson-19)

Let the line divides the points in k: 1 ratio according to section formula 

(2k+1/k+1, 7k+3/k+1) = (x, y)
It must satisfy the given equation so
=> 3(2k+1/k+1) + (7k+3/k+1) = 9
=> 6k+3+7k+3/k+1 = 9
=> 13k+6=9k+9
=> 13k-9k=9-6
=> 4k=3
=> k=3/4

Hence, required ratio will be 3:4.

3. Answer any one of the following questions.

(a) The length of a rectangle is 12 m more than twice the width. The area of the rectangle is 320 Square meter. Write an equation that can be used to find the length and width of the rectangle. Also find the dimensions of the rectangle. (See Lesson-6)

Solution:

Let the width of the rectangle be X mtr

Then Length of the rectangle will be = (2X+12) mtrs

Area of the rectangle = 320 square mtr

Now,

Area of Rectangle = length × width = 320

=> (2X+12) × X = 320 

=> (X + 6) x X = 160 (dividing both side by 2)

=> X2 + 6X 160 = 0

=> X2 – 10X + 16X - 160 = 0

=> X(X-10) + 16(X-10) = 0 

=> (X-10) (X+16) = 0

=> Either X-10=0 or X+16=0

=> X = 10 or X = -16

=> X should be positive value, X = 10

Therefore
Width of the rectangle = X = 10 mtr

And Length of the rectangle = 2X + 12 = 2×10+12 = 20 + 12 = 32 mtr

(b) There are two rectangular paths in the middle of a park as shown in the adjoining figure.

Find the area of the path, it is give that (See Lesson-20)

AB = CD = 100m

AD = BC = 70 m

HE = SR = 5 m

Now, 

Width of 1st road (SPQR) =5 m

Length of first parallel road 70 m

Area of first road 70×5=350 sq m

Again, 

Width of 2nd road (HEFG) = 5 m

Length of second road 100 m

Area of second road =100×5=500 sq m

Again, 

Area of the common part of cross road with 5 m width that lies at the center of the park 5×5 = 25 sq m

Now, 

Area of Path = Area of first road + Area of second cross road−common area

= (350+500) − 25

= 825 sq m

4. Answer any one of the following questions.

(a) Kamla invested Rs. 8000 in a business. She would be paid interest at 5% per annum compounded annually. Find (i) The amount credited against her name at the end of the second year. (ii) The interest for the third year. (See lesson 8 & 9)

Ans: Answer of any one question is necessary

(b) If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. (See lesson 14)

Solution: 

Consider the ∆ABC, let AD be the bisector of ∠A and BD = CD. It is required to prove ∆ABC is an isosceles triangle i.e. AB = AC. For this draw a line from C parallel AD and extend BA. Let they meet at E.

It is given that

∠BAD = ∠CAD    ... (1)

CE||AD

∴ ∠BAD = ∠AEC  (Corresponding angles)  ... (2)

And ∠CAD = ∠ACE (Alternate interior angles)  ... (3)

From (1), (2) and (3)

∠ACE = ∠AEC

In ∆ACE, ∠ACE = ∠AEC

∴ AE = OAC (Sides opposite to equal angles)  ... (4)

In ∆BEC, AD||CE and D is the mid-point of BC using converse of mid-point theorem A is the mid-point of BE.

∴ AB = AE

⇒ AB = AC  [Using (4)]

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

Hence, proved

5. Answer any one of the following questions.

(a) If 6 times the 6th term of an A.P is equal to 9 times its 9th term, then show that its 15th term is zero. (See Lesson -7)

Solution: 

(b) The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the Distance between the two towers and also the height of the other tower. (See Lesson-23)

6. Prepare any one project out of the given below:

(a) Conduct a survey of at least 50 households from your locality/village, regarding population and family income.

(i) Present the data related to family members in tabular form mentioning frequencies.

(ii) Calculate the average family size. How many families are above the average family size?

(iii) Draw the Bar graph for top 10 earning families. (See Lesson-24 and 25)

Ans: (i) Data of 50 households from your locality/village, regarding population and family income.

(ii) The data related to family members in tabular form mentioning frequencies:

(iii) the average family size = (2 + 3 + 7 + 5 + 5 + 2 + 7 + 2 + 4 + 3 + 2 + 4 + 3 + 3 + 4 + 3 + 3 + 6 + 6 + 4 + 3 + 4 + 3 + 5 + 6 + 2 + 3 + 4 + 2 + 4 + 6 + 2 + 4 + 5 + 3 + 6 + 2 + 6 + 6 + 5 + 7 + 2 + 5 + 7 + 5 + 2 + 3 + 4 + 6 + 2) / 50

= 4.04 = 4

Number of families are above the average family size = 19

(iv) The Bar graph for top 10 earning families is given below in the attachment:

(b) Observe a cricket match to be held in the current year. Prepare a project report indicating comparison of performances of the two teams with respect to:

(i) data of individual scores.

(ii) team-wise average score per over

(iii) run rate of individual batsman who did batting

(iv) run rate of individual bowler who did bowling

(v) draw a bar graph displaying the 5 top scoring batsman of the winning team. (See Lesson-24 & 25)

Ans: Answer of any one project is necessary.