CLASS-8 SUBJECT MATHS (FINALS.E) CHAPTER-11 (EXERCISE11.3) MENSURATION

  EVENTS CONVENT HIGH SCHOOL

25/01/2022      CLASS- 8   SESSION 2021-22
SUBJECT :MATHS

CHAPTER-11 (EXERCISE 11.3)

MENSURATION

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 Question 1.There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?

Solution:
(a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 120000 cm3
(b) Volume of cube = (Side)3 = (50)3 = 50 × 50 × 50 = 125000 cm3
Cuboidal box (a) requires lesser amount of material.

Question 2.A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of the suitcase = 80 cm × 48 cm × 24 cm
l = 80 cm, b = 48 cm and h = 24 cm
Total surface area of the suitcase = 2[lb + bh + hl]
= 2 [80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]
= 2 × 6912
= 13824 cm2
Area of tarpaulin = length × breadth = l × 96 = 96l cm2
Area of tarpaulin = Area of 100 suitcase
96l = 100 × 13824
l = 100 × 144 = 14400 cm = 144 m
Hence, the required length of the cloth = 144 m.

Question 3.Find the side of a cube whose surface area is 600 cm2?
Solution:
Total surface area of a cube = 6l2
6l2 = 600
l2 = 100
l = √100 = 10 cm
Hence, the required length of side = 10 cm.

Question 4.Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution:
l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box
= 2 [lb + bh + hl] – lb
= 2[2 × 1.5 + 1.5 × 1 + 1 × 2] – 2 × 1
= 2[3 + 1.5 + 2] – 2
= 2[6.5] – 2
= 13 – 2
= 11 m2
Hence, the required area = 11 m2.

Question 5.Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of the area is painted. How many cans of paint will she need to paint the room?
Solution:
Surface area of a cuboidal hall without bottom = Total surface area – Area of base
= 2 [lb + bh + hl] – lb
= 2 [15 × 10 + 10 × 7 + 7 × 15] – 15 × 10
= 2[150 + 70 + 105] – 150
= 2 [325] – 150
= 650 – 150
= 500 m2
Area of the paint in one can = 100 m2
Number of cans required = 500100 = 5 cans.

CLASS-6 SUBJECT MATHS (FINALS.E) CHAPTER-12 (EXERCISE12.2) RATIOS AND PROPORTIONS

 EVENTS CONVENT HIGH SCHOOL

25/01/2022      CLASS- 6   SESSION 2021-22
SUBJECT :MATHS

CHAPTER-12

RATIOS AND PROPORTION'S

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Exercise 12.2

Question 1.Determine if the following are in proportion,
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solution:

∴ 15 : 45 :: 40 : 120
∴ 15, 45, 40 and 120 are in proportion.

∴ 33, 121, 9 and 96 are in proportion.

∴ 24, 28, 36 and 48 are not in proportion.

Question 2 Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ₹ 40 : ₹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg : 80 kg and 25 g : 625 g

Solution:
(a) 25 cm : 1 m = 25 cm : 100 cm [∵ 1 m = 100 cm]

∴ The given ratios are in proportion.
Extreme terms are 25 cm and ₹ 160.
Middle terms are 1 m and ₹40.

∴ The given ratios are in proportion.
Extreme terms are 39 litres and 10 bottles.
Middle terms are 65 litres and 6 bottles.

∴ The given ratios are not in proportion.

Question 3. Are the following statements true?
(a) 40 persons : 200 persons = ₹15 : ₹75
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
(c) 99 kg : 45 kg = ₹ 44 : ₹ 20

Solution:
(a) 40 persons : 200 persons

∴ Statement (a) is true.

(b) 7.5 litres : 15 litres

∴ Statement (b) is true.

∴ Statement (c) is true.

CLASS-6 SUBJECT SCIENCE (FINALS.E) CHAPTER-5 SEPRATION OF SUBSTANCES

  EVENTS CONVENT HIGH SCHOOL

25/01/2022      CLASS- 6   SESSION 2021-22
SUBJECT :SCIENCE

CHAPTER-5

SEPRATION OF SUBSTANCES

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1. Fill in the blanks:
(a) The method of separating seeds of paddy from its stalks is called .
(b) When milk, cooled after boiling, is poured onto a piece of cloth the cream (malai) is left behind on it. This process of separating cream from milk is an example of ______. 
(a) Salt is obtained from sea water by the process of ____________ .
(b) Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called ____________.
Ans.
(a) threshing
(b)filtration
(b) evaporation
(d) sedimentation and decantation

2. True or false?
(a) A mixture of milk and water can be separated by filtration.
(b) A mixture of powdered salt and sugar can be separated by the process of winnowing.
(c) Separation of sugar from tea can be done with filtration.
(d) Grain and husk can be separated with the process of decantation.
Ans.
(a) False
(b) False
(c) False
(d) False

VERY SHORT ANSWER TYPE QUESTIONS

1. What is strainer?
Ans. Strainer is a kind of sieve which is used to separate a liquid from solid.

2. Name the method used to separate cream from curd.
Ans. Centrifugation.

3. How will you separate mango from a mixture of mango and apple?
Ans. By picking.

4. You are given a mixture of salt and sand. Can you separate them by picking?
Ans. No, we cannot separate them by picking.

5. Name the method used to separate the pieces of stone from grain.
Ans. Handpicking.

SHORT ANSWER TYPE QUESTIONS

l. What is mixture?
Ans. When two or more than two substances are mixed together in any ratio then it is called a mixture.

2. Write various methods of separation of components from their mixture.
Ans.

1. Handpicking
2. Threshing
3. Winnowing
4. Sedimentation
5. Decantation
6. Filtration
7. Evaporation
8. Condensation

3. Define the term handpicking.
Ans. The process used to separate slightly larger particles from a mixture by hand is called handpicking. For example: Stone pieces can be separated from wheat or rice by handpicking.

4. What do you mean by threshing? Where is it used?
Ans. Threshing is a process in which we separate grain from stalks. This process is used by farmer to separate gram, wheat, rice, mustard seeds in his field.

5. Write three methods of separation.
Ans. Handpicking, threshing and winnowing.

6. How will you separate oil and water from their mixture?
Ans. Oil, being lighter than water, will float on it. Two distinct layers are formed and slowly oil is allowed to flow into another container and is separated from water. Separating funnel can also be used to separate the two.