MATHEMATICS : CLASS VI

OPEN TEXT – BASED ASSESSMENT

SUMMATIVE EXAMINATION - IIMARCH 2015

THEMES

1.2011 CRICKET WORLD CUP 2.PERIMETER AND AREA OF VARIOUS

SHAPES

OPEN

TEXT MATERIAL

OPEN TEXT MATERIAL1. Theme – 2011 Cricket World Cup

Abstract:

The 2011 ICC Cricket World Cup was the tenth Cricket World Cup. It was played in India, Sri Lanka, and (for the first time) Bangladesh. Pakistan was also scheduled to be a co-host, but after the 2009 attack on the Sri Lanka national cricket team in Lahore, the International Cricket Council (ICC) cancelled that, and the headquarters of the organising committee, originally in Lahore, was transferred to Mumbai. Pakistan was to have held 14 matches, including one semi-final. Eight of the games (including the semi-final) were awarded to India, four to Sri Lanka, and two to Bangladesh.

India won the tournament, defeating Sri Lanka by 6 wickets in the final in Mumbai, thus becoming the first country to win the Cricket World Cup final on home soil. India's Yuvraj Singh was declared the man of the tournament. This was the first time in World Cup history that two Asian teams had appeared in the final. It was also the first time since the 1992 World Cup that the final match did not feature Australia.List of qualified teams:The following 14 teams qualified for the final tournament.

Group A: 1. Australia 2. Pakistan 3. New Zealand

4. Sri Lanka 5. Zimbabwe 6. Kanada 7. Kenya

Group B: 1. India 2. South Africa 3. England

4. West Indies 5. Bangladesh 6. Ireland 7. Netherlands

Prize money: The prize money was distributed as follows:

US$ 0.75 million – To each team exiting after the quarter-finals (4 teams) US$1.5 million – Fourth placed team US$1.5 million – Third placed team US$4.5 million – Runner up US$10.5 million – Winner

Capacity (No. of persons) of some stadium were as follows:

Stadium No. of persons Capacity (Approx.)

Eden Gardens, Kolkata 65,000

M. A. Chidambaram Stadium, Chennai 35,000

Feroz Shah Kotla, New Delhi 40,000

Vidarbha CricketAssociation Stadium, Nagpur

45,000

Sardar Patel Stadium, Ahemdabad 55,000

Wankhede Stadium, Mumbai 45,000

Punjab CricketAssociation Stadium, Mohali

40,000

M. Chinnaswamy Stadium, Banglore 35,000

Group stage:The top four teams from the two groups qualified for the quarter-finals.

Points table of Group A :

Points table of Group B :

Run Rate of Group B Teams :India : 0.9 South Africa :2.026 England: 0.072West Indies: 1.066 Bangla Desh : (- )1.361 Ireland : (-) 0.096 Netherlands: (–) 2.045

Knockout stage:

After Quarter Finals four team selected for Semi Final.

Semi-finals:Semi Final Date and Venue Teams and Score Result1 29 March 2011

R Premadasa Stadium, Colombo

New Zealand , 17 (48.5 overs) Sri Lanka won by 5 wickets

Sri Lanka, 220/5 (47.5 overs)

2 30 March 2011Punjab Cricket Association Stadium, Punjab

India, 260/9 (50 overs) India won by 29 runsPakistan, 231(49.5 overs)

Final:Date and Venue Teams and Score Result2 April 2011 Wankhede Stadium, Mumbai

Sri Lanka. 274/6 (50 overs) India won by 6 wicketsIndia, 277/4 (48.2 overs)

The 2011 ICC Cricket World Cup Final was played between India and Sri Lanka at the Wankhede Stadium in Mumbai on 2 April

2011. This was the 10th ICC Cricket World Cup. India won the match by six wickets, its second World Cup victory, having previously won in 1983. India became the third nation to have won the World Cup more than once, along with Australia and the West Indies. This was the first World Cup where both the finalists

were Asian teams. This was the second time in World Cup history that a host nation won the World Cup Final and the first time to win on their home ground.

Sample Questions :Q.1 (a) Represent the data of capacity (No. of persons ) of given stadium

of India by Bar graph using suitable scale.

(3)(b) Which stadium(s) have minimum capacity? (2)

Q.2 (a) By studying the Bar graph of point table of Group B, Which two teams have equal points? Name them. (1)(b) In which years India won the Cricket World Cup (1)(c) How many balls are left of India Inning in final match? (3)

Marking Scheme:Q.1 (a) Scale (1/2 mark)

Correct Values on axis (1/2 mark)Equal width of bars and equal gap between bars(1/2 mark)Correct lengths of bars (2 marks)

(b) M. A. Chidambaram Stadium, Chennai (1 mark)

M. Chinnaswamy Stadium, Banglore (1 mark)Q.2 (a) West Indies and Bangladesh (1/2 +1/2 mark)

(b) 1983 and 2011 (1/2 +1/2 mark)(c) Balls left = 50 overs - 48.2 overs

= 50 overs – (48 overs + 2 balls)= 1 over + 4 balls= 10 balls (3 marks)

2. THEME- PERIMETER AND AREA OF VARIOUS SHAPES

All the Kendriya Vidyalayas across the country are famous for their huge play grounds. One such Vidyalaya has a playground as shown below. The management of the school developed a beautiful garden for their primary children at one corner of the playground. The children of the Vidyalaya always spend their free time in that garden.

In this year the school management has decided to lay path of width 5m around and inside the park and some modifications in the garden. The supervision of that work is given to the mathematics teacher. She decided to discuss the matter with the students of class VI.

She informed the class about the proposal of the V.M.C. She invited the suggestions from the students for the good looking of the garden. She drew the plan on the black board, and indicated the measurements as shown in the figure. The students were curious and are happy to see map of their beloved garden on the black board.

She announced in the class that the work will be supervised by the students only. She divided the students into 3 teams. Team A, Team B, and Team C.

Team Team leader Work allotted.A Ayushi To find the perimeters and

areas of the given shapes that are there in the garden

B Manish To look after the required purchasings

C Namrata To look after the work force and their payments.

D Rahul To give any suggestions for beautification of the garden.

Immediately the teams entered into their allotted assignments. At first Rahul with his team mates went around the garden. He found that everything is fine. But one of his team mates, Yash suggested that, if some plants are planted around the inner boarder to separate the path from the ground, then it looks more beautiful. Everybody in the team liked that idea and got approval from their teacher. Then they decided to plant the saplings with 1 metre gap between any two plants around inner edge of the path. They also decided that the path should be filled with the square tiles of ½ m each. Their teacher added that four flower beds are also

to be put around the four corners. The outer boundary is going to be fenced with a fencing wire whose cost is about Rs. 50 per metre.

After the necessary modifications the plan of the school garden is as shown in the figure. The shaded region indicates the path. This plan is given to all teams to have coordination.

On the next day Ayushi selected her team members and went around in the garden to take the measurements. She took a tape and some papers and pen to check the exactness in the plan and in the actual things. She found that the plan is as per scale and it is correct. She also came to know about the proposal of Rahuls team.

Then Manish and his teammates of Team B started to list out the things to be purchased. They calculated the no. of saplings to be purchased. When they enquired in the nursery about the saplings that each one costs Rs. 5.

Namrata was informed by her teacher, that 2 female workers and one male worker is available to do the necessary changes in the garden on each day. The labour charges are as follows. The male worker will be paid Rs. 300 per day and the female worker will be paid Rs. 200 per day. The team C estimated that the work will be done in two days.

Teacher: what are the outer length and breadth of the rectangular garden?

Nishi: length 1oo m and breadth 70 m.

Teacher: what is the width of the path?

Ayushi: 5 m

Teacher: The width of the path around the garden is uniform and is same at any point.

Ashish: Madam I can find out the length of inner rectangle.

Teacher: Good! How do you find out?

Ashish: Just by subtracting the width of the path on both sides. That is 100 – (5+5)= 90 m

Teacher: very good. What about the breadth.

Ashish: Similarly we can calculate. It is 60m.

Teacher: Sadiq, if we subtract area of inner rectangle from area of outer rectangle what do we get?

Sadiq: Madam, we get area of path.

Teacher: can you see the circular shape? What is it?

Mary: Madam it is the water fountain. I regularly sit near by it only.

Teacher: This time we are going to develop four flower beds at the four corners of the garden. The side of each is 6 m. The outer boundary is going to be fenced with a fencing wire whose cost is about Rs. 50 per metre. Various other shapes are seen around the fountain. They are the seats for children to sit and enjoy the fountain.

Now answer the following questions.

I. (i) What are perimeters of the shapes that you see in the map except circle and the four flower beds 2 marks (ii) How many saplings may be planted around the inner boarder and what is the total cost of the saplings? 2 marks (iii) Name the two shapes whose side is 5 m 1 mark

II. (i) What is the total cost of the fencing wire to fence the outer boundary?

2 marks(ii) What will be the labour charges if all the work force work for two days

2 marks. (iii) What is the area of each tile.? 1 mark.

6 m

6 m6 m

6 m 6 m

6 m

6 m

70m

5m

5m.

5m 7m

6 m

10 m8 m

5 m

7 m7 m5 m 5 m

5 m

6 m

5 m

5 m

5 m

5 m

6 m

100 m

100 m

6 m

6 m. 6m 6 m.

70 mMarking Scheme.

1. (i) Perimeters of all shapes except circle and four flower beds = (3 X 7) + (5 X5 ) + ( 5 X 6) 1 mark = 21 + 25 + 30

= 76 m 1 mark(ii) Inner perimeter = 2 ( 60+90) = 2 X 150 = 300 m No. of saplings = 300 1 mark Cost of saplings = 300 X 5 = Rs. 1500 1 mark(iii) regular pentagon, regular hexagon. 1 mark.

2. (i) Total length of outer boundary = 2 ( 100+70) = 2 X 340 = 680 m 1 markCost of fencing wire = 680 X 50 = Rs. 52 000 1 mark(ii) labour charges for one day = 300+ (2x200) = 300+400 = 700 1mark Labour charges for two days = 700 X 2 = Rs. 1400 1 mark(iii) area of each tile = ½ X ½ = ¼ sq. metres. 1 mark.

6 m 6 m6 m