Std. 7 _Perimeter & Area

Question 1:

If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?

ANSWER:

Let the length of old rectangle = l
Let the breadth of old rectangle = b
Perimeter of the old rectangle = 2(length + breadth) = 2(l + b)
When the length and the breadth of the rectangle are doubled, then
length of the new rectangle = 2l
breadth of the new rectangle = 2b
∴ Perimeter of new rectangle = 2(length + breadth)
= 2(2l + 2b)
= 2 × 2(l + b)
= 2 × perimeter of the old rectangle    [∵ perimeter of old rectangle = 2(l + b)]
Hence, the perimeter of the new rectangle will become two times of the perimeter of the old rectangle.

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Question 2:

If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?

ANSWER:

Let the length of each side of the old square = s
Then, perimeter of the old square = 4 × side = 4 × s = 4s
When the length of each side of the square is tripled, then
length of each side of the new square = 3s
∴ perimeter of the new square = 4 × side 
= 4 × 3s
= 3 × 4s
= 3 × perimeter of the old square    [∵ perimeter of the old square = 4s]
Hence, the perimeter of the new square will become three times of the perimeter of the old square. 

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Question 3:

Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.

ANSWER:

It is known that, perimeter of a polygon is equal to sum of the lengths of all the sides of the polygon.
Let us mark the vertices of the given polygon as A, B, C, D, E and F.
The given figure can be broken into rectangle and square by drawing a line CG || BA.

Now, ABCG is a rectangle and GDEF is a square.
Since, opposite sides of rectangle are equal, then
GC = AB = 10 m and AG = BC = 15 m
Since length of each sides of square is equal, then
GD = DE = EF = FG = 15 m
Perimeter of ABCDEFG = AB + BC + CD + DE + EF + FG + GA
= 10 + 15 + 5 + 15 + 15 + 15 + 15
= 90 m

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Question 4:

As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins ?

ANSWER:

Length of each side of a square piece of cloth = 1 m
Now, four napkins all of same size were made from this square piece of cloth.
∴ Length of each side of a napkin, s = Length of each side of a square piece of cloth ÷ 2
= 1 ÷ 2
= 0.5 m
Length of lace needed to trim all the four sides of a napkin = perimeter of a napkin
= 4 × s
= 4 × 0.5
= 2 m
∴ Length of lace needed to trim all the four sides of four napkins = 4 × length of lace needed for a napkin
= 4 × 2
= 8 m

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Question 1:

If the side of a square is 12 cm, find its area.

ANSWER:

Length of each side of square, s = 12 cm
∴ Area of square = s²
= (12)²
= 12 × 12
= 144 cm²
Thus, the area of the square is 144 cm².

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Question 2:

If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.

ANSWER:

Length of the rectangle, l = 15 cm
Breadth of the rectangle, b = 5 cm
∴ Area of rectangle = l × b
= 15 × 5
= 75 cm²
Hence, the area of rectangle is 75 cm².

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Question 3:

The area of a rectangle is 102 sqcm. If its length is 17 cm, what is its perimeter ?

ANSWER:

Area of rectangle = 102 cm²
Length of the rectangle, l = 17 cm
∴ Breadth of the rectangle, b = area of rectangle÷ length of rectangle  (∵ area of rectangle = length × breadth)

⇒b = 102÷17 = 6 cm
Perimeter of rectangle = 2(length + breadth)
= 2(17 + 6)
= 2 × 23
= 46 cm

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Question 4:

If the side of a square is tripled, how many times will its area be as compared to the area of the original square ?

ANSWER:

Let the length of each side of original square = s
Area of original square = side × side = s²
When the side of the square is tripled, then
length of each side of new square = 3s
Area of new square = side × side
= 3s × 3s
= 9s²
= 9 × s²
= 9 × area of original square    [∵ area of original square = s²]
Hence, the area of the new square will become nine times the area of the original square.

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Question 1:

A page of a calendar is 45 cm long and 26 cm wide. What is its area ?

ANSWER:

Length of the calendar, l = 45 cm
Breadth of the calendar, b = 26 cm
∴ Area of calendar = l × b
= 45 × 26
= 1170 cm²

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Question 2:

What is the area of a triangle with base 4.8 cm and height 3.6 cm ?

ANSWER:

We have, base of a triangle = 4.8 cm
Height of a triangle = 3.6 cm
∴ Area of a triangle = 12$\frac{1}{2}$ × base × height
= 12$\frac{1}{2}$ × 4.8 × 3.6
= 8.64 cm²

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Question 3:

What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of 1000 rupees per square metre ?

ANSWER:

Length of the rectangular plot of land, l = 75.5 m
Breadth of the rectangular plot of land, b = 30.5 m
∴ Area of rectangular plot of land = l × b
= 75.5 × 30.5
= 2302.75 m²
Cost of 1 m² of rectangular plot of land = 1000 rupees
∴ Cost of 2302.75 m² of rectangular plot of land = 1000 × 2302.75
= 2302750 rupees
Hence, the value of the rectangular plot of land is Rs 23,02,750.

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Question 4:

A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall ? How many would be required if tiles of side 15 cm were used?

ANSWER:

Length of the rectangular hall, l = 12 m
Breadth of the rectangular hall, b = 6 m
Area of the floor of the rectangular hall = l × b
= 12 × 6
= 72 m²
Length of each side of square tile used in the first case = 30 cm = 30÷ 100= 0.3 m    (∵ 100 cm = 1 m)
Area of each tile = side × side
= 0.3 × 0.3
= 0.09 m²
∴ Required number of square tiles each of side 30 cm = area of floor ÷ area of each tile 
= 72÷ 0.09
= 800
Length of each side of square tile used in the second case = 15 cm = 15÷ 100= 0.15 m   
 Area of each tile = side × side
= 0.15 × 0.15
= 0.0225 m²
∴ Required number of square tiles each of side 15 cm = area of floor÷ area of each tile
= 72÷ 0.0225
= 3200

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Question 5:

Find the perimeter and area of a garden with measures as shown in the figure alongside.

ANSWER:

Let us mark the vertices of the given polygon as A, B, C, D, E, F, G, H, I, J, K and L.
Perimeter of garden ABCDEFGHIJKL = sum of all the sides of polygon = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL
= 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13
= 143 m

The given figure of the garden can be broken into five squares as follows.



Now, length of each side of each of the five squares is equal to 13 m.
Area of each square = side × side
= 13 × 13
= 169 m²
∴ Area of the five squares = 5 × area of each square
= 5 × 169
= 845 m²
Hence, the area of the given garden is 845 m².

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Question 1:

Find the total surface area of cubes having the following sides.

(i) 3 cm (ii) 5 cm  (iii) 7.2 m (iv) 6.8 m (v) 5.5 m

ANSWER:

(i) Length of each side of cube, l = 3 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (3)²
= 6 × 9
= 54 cm²

(ii) Length of each side of cube, l = 5 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (5)²
= 6 × 25
= 150 cm²

(iii) Length of each side of cube, l = 7.2 m
∴ Total surface area of the cube = 6 × l²
= 6 × (7.2)²
= 6 × 51.84
= 311.04 m²

(iv) Length of each side of cube, l = 6.8 m
∴ Total surface area of the cube = 6 × l²
= 6 × (6.8)²
= 6 × 46.24
= 277.44 m²

(v) Length of each side of cube, l = 5.5 m
∴ Total surface area of the cube = 6 × l²
= 6 × (5.5)²
= 6 × 30.25
= 181.50 m²

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Question 2:

Find the total surface area of the cuboids of length, breadth and height as given below:

(i) 12 cm, 10 cm, 5 cm  (ii) 5 cm, 3.5 cm, 1.4 cm

(iii) 2.5 cm, 2 m, 2.4 m  (iv) 8 m, 5 m, 3.5 m

ANSWER:

(i) Length of the cuboid, l = 12 cm
Breadth of the cuboid, b = 10 cm
Height of the cuboid, h = 5 cm
∴ Total surface area of the cuboid = 2(l × b + b × h + h × l)
= 2(12 × 10 + 10 × 5 + 5 × 12)
= 2(120 + 50 + 60)
= 2 × 230
= 460 cm²

(ii) Length of the cuboid, l = 5 cm
Breadth of the cuboid, b = 3.5 cm
Height of the cuboid, h = 1.4 cm
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(5 × 3.5 + 3.5 × 1.4 + 1.4 × 5)
= 2(17.5 + 4.9 + 7.0)
= 2 × 29.4
= 58.8 cm²

(iii) Length of the cuboid, l = 2.5 cm = 2.5100$\frac{2.5}{100}$ m = 0.025 m    [∵ 100 cm = 1 m]
Breadth of the cuboid, b = 2 m
Height of the cuboid, h = 2.4 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(0.025 × 2 + 2 × 2.4 + 2.4 × 0.025)
= 2(0.05 + 4.8 + 0.06)
= 2 × 4.91
= 9.82 m²

(iv)  Length of the cuboid, l = 8 m
Breadth of the cuboid, b = 5 m
Height of the cuboid, h = 3.5 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(8 × 5 + 5 × 3.5 + 3.5 × 8)
= 2(40 + 17.5 + 28)
= 2 × 85.5
= 171 m²

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Question 3:

A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so ?

ANSWER:

Length of the matchbox, l = 4 cm
Breadth of the matchbox, b = 2.5 cm
Height of the matchbox, h = 1.5 cm
∴ Surface area of the box = 2(l × b + b × h + h × l)
= 2(4 × 2.5 + 2.5 × 1.5 + 1.5 × 4)
= 2(10 + 3.75 + 6)
= 2 × 19.75
= 39.5 cm² 
Hence, 39.5 cm² of the craft paper will be needed to cover the matchbox.

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Question 4:

An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box ? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?

ANSWER:

Length of the open box, l = 1.5 m
Breadth of the open box, b = 1 m
Height of the open box, h = 1 m
Surface area of open box = Total surface area of the box − Area of the top
= 2(l × b + b × h + h × l) − l × b
= 2(1.5 × 1 + 1 × 1 + 1 × 1.5) − 1.5 × 1
= 2(1.5 + 1 + 1.5) − 1.5
= 2 × 4 − 1.5
= 8 − 1.5
= 6.5 m²
Hence, 6.5 m² of the sheet will be needed to make the open box.
Now, it is given that the inside and the outside surface of the open box is to be painted with rust proof paint.
∴ Total area of the box to be painted = 2 × Surface area of open box
= 2 × 6.5
= 13 m²
Now, cost of painting of 1 m² of area = 150 rupees
∴ Cost of painting of 13 m² of area = 13 × 150 = 1950 rupees
Hence, it will cost 1950 rupees to paint the open box from inside and outside with rust proof paint.

LIGHT (Std.6)

Question 1:

Choose an appropriate word and fill in the blanks.
(a) A ........... is a natural source of light.
(b) A ......... is an artificial source of light.
(c) When light passes through a prism, it gets separated into ....... colours.
(d) The image obtained in the pinhole camera is ......... .
(e) A shadow is formed when an ........... object comes in the way of light.
(f) When a ........... object comes in the way of light, light passes ......... it.
(options : seven, star, through, transparent, opaque, colors, shape, erect, inverted, luminous, candle)

ANSWER:

(a) A star is a natural source of light.
(b) A candle is an artificial source of light.
(c) When light passes through a prism, it gets separated into seven colours.
(d) The image obtained in the pinhole camera is inverted.
(e) A shadow is formed when an opaque object comes in the way of light.
(f) When a transparent object comes in the way of light, light passes through it.

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Question 2:

Write whether of the following objects are luminous or non-luminous.

Object	Luminous/Non-luminous
A book
A burning candle
A wax cloth
A pencil
A pen
A light bulb
A tyre
A torch

ANSWER:

Object	Luminous/Non-luminous
A book	Non-luminous
A burning candle	Luminous
A wax cloth	Non-luminous
A pencil	Non-luminous
A pen	Non-luminous
A light bulb	Luminous
A tyre	Non-luminous
A torch	Luminous

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Question 3:

Match the following.

Group A	Group B
(a) Mirror	(1) Non-luminous
(b) Firefly	(2) Inverted image
(c) Pinhole camera	(3) Reflection
(d) Moon	(4) Luminous

ANSWER:

Group A	Group B
(a) Mirror	(3) Reflection
(b) Firefly	(4) Luminous
(c) Pinhole camera	(2) Inverted image
(d) Moon	(1) Non-luminous

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Question 4:

Write the answers to the following.
(a) What things are necessary for the formation of a shadow ?
(b) When can an object be seen?
(c) What is a shadow?

ANSWER:

(a) The things necessary for the formation of a shadow are:

• A source of light
• An opaque object whose shadow has to be formed
• A screen on which shadow is to be formed

(b) An object is seen when the light rays reflected from its surface reaches our eyes.

(c) Shadow is a dark patch obtained on a screen where the light from a light source is blocked by an opaque object.

SOUND

Question 1: Fill in the blanks with the proper words. 

(a) The propagation of sound does not occur through a ........... . 

(b) Noise pollution is a .......... issue. 

(c) The sound which is disagreeable to the ears is called..... .

 (d) Noise has adverse effects on our .......... .

ANSWER: (a) The propagation of sound does not occur through a vacuum.

(b) Noise pollution is a social issue.

(c) The sound which is disagreeable to the ears is called noise.

 (d) Noise has adverse effects on our health.

 Page No 96: 

Question 2: 

What should we do ?

 (a) The silencer of a motorcycle is broken. (b) A factory in the surroundings is producing continuous loud noise.

 ANSWER: 

a) If the silencer of a motorcycle is broken, then take the motorcycle to a mechanic and get it reinstalled or repaired.

 (b) If a factory in the surroundings is producing continuous loud noise, then we can lodge a complaint against the factory to a local government body citing the troubles faced by the residents around and the danger it is causing to the environment.

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 Question 3:

 Write the answers in your own words.

 (a) What is meant by vibration ?

 (b) Explain with the help of practical examples how sound is propagated through solids.

 (c) What is meant by noise pollution? 

(d) What measures will you take to control noise pollution ? 

 ANSWER: (a) The to and fro motion of a body about its mean position is known as vibration.

 (b) Sound is propagated in the form of waves in solids. It can be explained clearly from the below examples: Place your ear on one end of a long table. Ask your friend to tap the table from the other end. You can hear the produced sound. Sound reaches your ear after travelling through the table. This indicates that sound can travel through solids and in the form of waves. ​Stethoscope is another example of sound travelling through solids. Doctors use stethoscopes to listen to our heartbeat.

 (c) Noise pollution is the excess of unwanted sound in the environment. It can adversely affect human and other living beings health.

 (d) Following measures can be taken to control noise pollution: To control noise pollution, we must control its source. Hence, silencers must be installed in vehicles such as motorcycles, cars, trucks, buses, and other noise producing machines. We should watch television and listen to music at a low volume. Also, use of loudspeakers as well as horns of buses and trucks should be minimized. Regular maintenance of automobiles should be done so that noise produced by them can be kept under check. All industrial work should be done away from residential areas. More trees should be planted in residential areas as they help in reducing noise.

Direct Proportion

Question 1:

If 7 kg onions cost 140 rupees, how much must we pay for 12 kg onions?

ANSWER:

Let us suppose the cost of 12 kg onions is x rupees.
The number of onions and their cost vary in direct proportion.
∴7140=12x⇒x=12×1407$$\begin{gathered} \therefore \frac{7}{140}=\frac{12}{x} \\ \Rightarrow x=\frac{12\times 140}{7} \end{gathered}$$
= 240 rupees
Hence, the cost of 12 kg onions is 240 rupees.

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Question 2:

If 600 rupees buy 15 bunches of feed, how many will 1280 rupees buy?

ANSWER:

Let us suppose x bunches of feed can be bought in 1280 rupees.
The number of bunches of feed and their cost vary in direct proportion.
∴15600=x1280⇒x=15×1280600$$\begin{gathered} \therefore \frac{15}{600}=\frac{x}{1280} \\ \Rightarrow x=\frac{15\times 1280}{600} \end{gathered}$$
= 32
Hence, 32 bunches of feed can be bought in 1280 rupees.

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Question 3:

For 9 cows, 13 kg 500 g of food supplement are required every day. In the same proportion, how much will be needed for 12 cows?

ANSWER:

Let us suppose x kg of food supplement required for 12 cows.
The quantity of food supplement and the number of cows vary in direct proportion.
∴913.5=12x⇒x=12×13.59$$\begin{gathered} \therefore \frac{9}{13.5}=\frac{12}{x} \\ \Rightarrow x=\frac{12\times 13.5}{9} \end{gathered}$$
= 18 kg
Hence, 18 kg of food supplement required for 12 cows.

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Question 4:

The cost of 12 quintals of soyabean is 36,000 rupees. How much will 8 quintals cost?

ANSWER:

Let us suppose the cost of 8 quintals of soyabean is x rupees.
The quantity of soyabeans and their cost vary in direct proportion. 
∴123600=8x⇒x=8×360012$$\begin{gathered} \therefore \frac{12}{3600}=\frac{8}{x} \\ \Rightarrow x=\frac{8\times 3600}{12} \end{gathered}$$
= 24000 rupees
Hence, the cost of 8 quintals of soyabean is 24000 rupees.

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Question 5:

Two mobiles cost 16,000 rupees. How much money will be required to buy 13 such mobiles ?

ANSWER:

Let us suppose the cost of 13 mobiles is x rupees.
The number of mobiles and their cost vary in direct proportion.
∴216000=13x⇒x=13×160002$$\begin{gathered} \therefore \frac{2}{16000}=\frac{13}{x} \\ \Rightarrow x=\frac{13\times 16000}{2} \end{gathered}$$
= 104000 rupees
Hence, the cost of 13 mobiles is 104000 rupees.

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Question 1:

Five workers take 12 days to weed a field. How many days would 6 workers take ? How many would 15 take ?

ANSWER:

Let us suppose 6 workers will take x days to weed a field. 
As the number of workers increases, the number of days decreases.
So, the number of workers and number of days are in inverse proportion. 
∴ 5 × 12 = 6 × x
⇒x=606$\Rightarrow x=\frac{60}{6}$
⇒ x = 10 days
Let us suppose 15 workers will take y days to weed a field. 
∴ 5 × 12 = 15 × y
⇒y=6015$\Rightarrow y=\frac{60}{15}$
⇒ y = 4 days
Hence, 6 workers will take 10 days, while 15 workers will take 4 days to weed a field.

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Question 2:

Mohanrao took 10 days to finish a book, reading 40 pages every day. How many pages must he read in a day to finish it in 8 days?

ANSWER:

Let us suppose Mohanrao will have to read x pages every day to finish the book in 8 days.
As the number of days decreases, the number of pages increases.
So, the number of days and number of pages are in inverse proportion. 
∴ 10 × 40 = 8 × x
⇒x=400/8$\Rightarrow x=\frac{\mathrm{400/}}{8}$
⇒ x = 50 pages
Hence, Mohanrao will have to read 50 pages every day to finish the book in 8 days

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Question 3:

Mary cycles at 6 km per hour. How long will she take to reach her Aunt’s house which is 12 km away ? If she cycles at a speed of 4 km/hr, how long would she take ?

ANSWER:

Given:
Case -1:
Speed = 6 km/hr
Distance = 12 km
∴Time=DistanceSpeed=126$$\begin{gathered} \therefore \mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ =\frac{12}{6} \end{gathered}$$
= 2 hours
Case -2:
​Speed = 4 km/hr
Distance = 12 km
∴Time=DistanceSpeed=124$$\begin{gathered} \therefore \mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ =\frac{12}{4} \end{gathered}$$
= 3 hours
Hence, if the speed of cycle is 6 km/hr then, Marry will take 2 hours and if the speed of cycle is 4 km/hr then, she will take 3 hours to reach her Aunt’s house.

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Question 4:

The stock of grain in a government warehouse lasts 30 days for 4000 people. How many days will it last for 6000 people ?

ANSWER:

Let us suppose the stock of grain in a government warehouse lasts x days for 6000 people.
As the number of people increases, the number of days decreases.
So, the number of days and number of people are in inverse proportion. 
∴ 30 × 4000 = 6000 × x
⇒x=1200006000$\Rightarrow x=\frac{120000}{6000}$
⇒ x = 20 days
Hence, the stock of grain in a government warehouse lasts 20 days for 6000 people.

PARTNERSHIP 

Page No 68:

Question 1:

Suresh and Ramesh together invested 144000 rupees in the ratio 4:5 and bought a plot of land. After some years they sold it at a profit of 20%. What is the profit each of them got?

ANSWER:

The proportion of Suresh's and Ramesh’s investment is 4:5.
The profit is shared in the same proportion as the investment, hence, proportion of profit is 4:5.
Now, profit = 20100×144000$\frac{20}{100}\times 144000$
= 28800 rupees
Therefore, the profit of Suresh and Ramesh  is given by
Suresh's profit = 49×28800$\frac{4}{9}\times 28800$
= 12800 rupees
Ramesh's profit = 59×28800$\frac{5}{9}\times 28800$
= 16000 rupees
Hence, Suresh and Ramesh got the profit of 12800 and 16000 rupees respectively.

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Question 2:

Virat and Samrat together invested 50000 and 120000 rupees to start a business. They suffered a loss of 20%. How much loss did each of them incur ?

ANSWER:

The proportion of Virat's and Samrat’s investment is given by
50000:120000 = 5:12
The loss is shared in the same proportion as the investment, hence, proportion of profit is 5:12.
Now, Loss = 20100×(50000+120000)$\frac{20}{100}\times \left(50000+120000\right)$
20100×(170000)$\frac{20}{100}\times \left(170000\right)$
= 34000 rupees
Therefore, the loss incurred by Virat and Samrat is given by
Virat's loss = 517×34000$\frac{5}{17}\times 34000$
= 10000 rupees
Samrat's loss = 1217×34000$\frac{12}{17}\times 34000$
= 24000 rupees
Hence, Virat and Samrat incurred the loss of ​10000 and 24000 rupees respectively.

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Question 3:

Shweta, Piyush and Nachiket together invested 80000 rupees and started a business of selling sheets and towels from Solapur. Shweta’s share of the capital was 30000 rupees and Piyush’s 12000. At the end of the year they had made a profit of 24%. What was Nachiket’s investment and what was his share of the profit?

ANSWER:

Nachiket’s investement = Total investment − (Shweta’s investment + Piyush’s investment)
= 80000 − (30000 + 12000)
= 80000 − 42000
= 38000 rupees
The proportion of Shweta's, Piyush's and Nachiket's investment is given by
30000:12000:38000 = 15:6:19
The profit is shared in the same proportion as the investment, hence, proportion of profit is 15:6:19.
Now, Profit = 24100×(80000)$\frac{24}{100}\times \left(80000\right)$
= 19200 rupees
Therefore, Nachiket’s share of the profit is given by
= 1940×19200$\frac{19}{40}\times 19200$
= 9120 rupees
Hence, Nachiket’s investment and his share of the profit are ​38000 and 9120 rupees respectively.

Page No 68:

Question 4:

A and B shared a profit of 24500 rupees in the proportion 3:7. Each of them gave 2% of his share of the profit to the Soldiers’ Welfare Fund. What was the actual amount given to the Fund by each of them?

ANSWER:

Amount of share to the Soldiers’ Welfare Fund = 2% of 24500
= 490 rupees
The profit is shared in the proportion 3:7.
Therefore, A’s share to the Fund is given by
= 310×490$\frac{3}{10}\times 490$
= 147 rupees
Therefore, B’s share to the Fund is given by
= ​710×490$\frac{7}{10}\times 490$
= 343 rupees
Hence, A's and B's share to the fund are 147 and 343 rupees respectively.

Page No 68:

Question 5:

Jaya, Seema, Nikhil and Neelesh put in altogether 360000 rupees to form a partnership, with their investments being in the proportion 3 : 4 : 7 : 6. What was Jaya’s actual share in the capital ? They made a profit of 12%. How much profit did Nikhil make ?

ANSWER:

Total investment = 360000 rupees
Total profit = 12100×360000$\frac{12}{100}\times 360000$
= 43200 rupees
The profit is shared in the same proportion as the investment, hence, proportion of profit is 3:4:7:6.
Jaya's share is given by
320×360000$\frac{3}{20}\times 360000$
= 54000 rupees
Nikhil's share in the profit is given by
720×43200$\frac{7}{20}\times 43200$
= 15120 rupees
Hence, Jaya's share and Nikhil's profit are 54000 and 15120 rupees respectively.

Circle

Question 1:

Complete the table below

Sr. No.	Radius (r)	Diameter (d)	Circumference (c)
(i)	7 cm	....................	.................
(ii)	............	28 cm	..............
(iii)	..............	...............	616 cm
(iv)	...........	.............	72.6 cm

ANSWER:

(i) Radius, r = 7 cm
 Diameter, d = 2r = 2 × 7 = 14 cm
∴ Circumference, c = πd 
= 22/7 × 14
= 22 × 2
= 44 cm

(ii) Diameter, d = 28 cm
Radius, r = d2$\frac{d}{2}$ = 282$\frac{28}{2}$ = 14 cm
∴ Circumference, c = 2πr
= 2 × 22/7 × 14
= 88 cm

(iii) Circumference, c = 616 cm
Now, c = 2πr     (where 'r' is the radius)
⇒616 = 2 × 22/7× r
⇒r = 616 x 7/44
⇒r = 98
So, radius = 98 cm
Diameter, d = 2r = 2 × 98 = 196 cm

(iv) Circumference, c = 72.6 cm
Now, c = 2πr     (where 'r' is the radius)
⇒72.6 = 2 × 22/7 × r
⇒r = 72.6 × 7/44
⇒r = 11.55
So, radius = 11.55 cm
Diameter, d = 2r = 2 × 11.55 = 23.1 cm

The complete table is shown below.
 

Sr. No.	Radius (r)	Diameter (d)	Circumference (c)
(i)	7 cm	14 cm	44 cm
(ii)	14 cm	28 cm	88 cm
(iii)	98 cm	196 cm	616 cm
(iv)	11.55 cm	23.1 cm	72.6 cm

Page No 77:

Question 2:

If the circumference of a circle is 176 cm, find its radius.

ANSWER:

Circumference, c = 176 cm
Now, c = 2πr    (where 'r' is the radius of circle)
⇒176 = 2 × 22/7 × r
⇒r = 176 / 2 x 3.14
⇒r = 28
∴ Radius of the circle = 28 cm

Page No 77:

Question 3:

The radius of a circular garden is 56 m. What would it cost to put a 4-round fence around this garden at a rate of 40 rupees per metre ?

ANSWER:

Radius of the circular garden, r = 56 m
Circumference of the circular garden, c = 2πr
= 2 × 22/7 × 56
= 352 m
∴ Length of the wire needed for one round of fencing = c = 352 m
Cost of one round of fencing = length of wire × cost per metre
= 352 × 40
= 14080 rupees
Cost of four round of fencing = 4 × 14080 = 56320 rupees

Page No 77:

Question 4:

The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km ?

ANSWER:

Diameter of the wheel, d = 1.4 m
Circumference, c = πd
= 22/7 × 1.4
= 4.4 m
When the wheel completes 1 rotation, it covers a distance that is equal to its circumference.
So, number of rotations taken by the wheel to cover 4.4 m = 1
Now, the wheel covered a total distance of 1.1 km.
We know that, 1 km = 1000 m
∴ 1.1 km = 1.1 × 1000 m = 1100 m
∴ Total number of rotations taken by wheel = total distancecircumference$\frac{\mathrm{total} \mathrm{distance}}{\mathrm{circumference}}$ 
= 11004.4$\frac{1100}{4.4}$
= 1100044$\frac{11000}{44}$
= 250
Hence, the wheel completes 250 rotations to cover a distance of 1.1 km.

Page No 79:

Question 1:

Choose the correct option.

If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120^° then m(arc AYB) = 12 eryyt4$\boxed{12 eryyt4}$
 

(i) 140^°   (ii) 60^°   (iii) 240^°  (iv) 160^°

 

ANSWER:

Consider that arc AXB is the minor arc and arc AYB is the corresponding major arc.
It is known that, measure of major arc = 360^∘ − measure of the corresponding minor arc.
We have, m(arc AXB) = 120^°.
So, m(arc AYB) = 360^∘ − m(arc AXB) = 360^∘ − 120^∘ = 240^∘
Hence, the correct answer is option (iii).

Page No 79:

Question 2:

Some arcs are shown in the circle with centre ‘O’. Write the names of the minor arcs, major arcs and semicircular arcs from among them.

ANSWER:

Minor arc : An arc of a circle having measure less than 180^∘.
Major arc : An arc of a circle having measure greater than 180^∘.
Semicircular arc : An arc of a circle having measure equal to 180^∘.

Names of minor arcs : 
(i) arc PXQ
(ii) arc PR
(iii) arc RY
(iv) arc XP
(v) arc XQ
(vi) arc QY

Names of major arcs :
(i) arc PYQ
(ii) arc PQR
(iii) arc RQY
(iv) arc XQP
(v) arc QRX

Names of semicircular arcs :
(i) arc QPR
(ii) arc QYR

Page No 79:

Question 3:

In a circle with centre O, the measure of a minor arc is 110^°. What is the measure of the major arc PYQ?

ANSWER:

Suppose PQ is the minor arc and then m(arc PQ) = 110^∘.
We know that, measure of major arc = 360^∘ − measure of corresponding minor arc.
∴ m(arc PYQ) = 360^∘ − m(arc PQ) 
=  360^∘ − 110^∘
= 250^∘
Hence, the measure of major arc PYQ is 250^∘.

Simple Machines

Question 1:

Classify the following as a lever, a pulley and an inclined plane :
A wedge, a needle, a staircase, a slide, the wheel of a flagpole, nutcrackers, scissors, an opener, an axe, a crane, a knife.

ANSWER:

A wedge: An inclined plane
A needle: An inclined plane
A staircase: An inclined plane
A slide: An inclined plane
The wheel of a flagpole: A pulley
Nutcrackers: A lever
Scissors: A lever
An opener: A lever
An axe: An inclined plane
A crane: A pulley
A knife: An inclined plane

Page No 90:

Question 2:

Fill in the blanks using the proper word and complete the statements.
(a) The ....... in the centre, the .......... on one side and the ...... on the other side make a lever of the first order.
(b) The ..... in the centre, the ........ on one side and the ......... on the other side make a lever of the second order.
(c) The .......... in the centre, the ....... on one side and the........on the other side make a lever of the third order.

ANSWER:

(a) The fulcrum in the centre, the load on one side and the effort on the other side make a lever of the first order.
(b) The load in the centre, the fulcrum on one side and the effort on the other side make a lever of the second order.
(c) The effort in the centre, the load on one side and the fulcrum on the other side make a lever of the third order.

Page No 90:

Question 3:

Which machines will you use to do the following work? Write their types.
(a) To remove the lid of a tin.
(b) To lift bricks to the top of a tall building.
(c) To cut vegetables.
(d) To draw water from a well.
(e) To hold a papad for roasting it.

ANSWER:

(a) To remove the lid of a tin, we should use an opener which is a second order lever.

(b) To lift bricks to the top of a tall building, we should use a crane which is a pulley.

(c) To cut vegetables, we should use a knife which is a wedge.

(d) To draw water from a well, we should use a pulley system or wheel-axle.

(e) To hold a papad for roasting it, we should use a tong which is a third order lever.

Page No 90:

Question 4:

Write the answers to the following questions in your own words.
(a) What is meant by simple machines?
(b) Mention the advantages of using a machine.
(c) What is meant by complex machines?
(d) What is a lever ? How are the orders of the lever determined?

ANSWER:

(a) Simple machines are the tools that help people work faster and better. These help us lift heavy loads, change the speed of the motion or the direction of force. Examples are simple machines are knife, nutcrackers etc.

(b) The advantages of using a machine are:

• More work can be done in lesser time as well as with great accuracy
• Less effort has to be applied to accomplish the task

(c) A machine made up of two or more simple machines is called a complex machine. It consists of different parts that carry out different task and together contribute to the working of the machine. The most common example is that of a bicycle.

(d) A lever is a simple machine consisting of a rigid rod which is capable of turning around a pivot called a fulcrum. It has three parts, namely, effort, load and fulcrum.

• Fulcrum: The rod of the lever rests on it and the lever rotates about it.
• Load: The weight lifted by the liver is called the load.
• Effort: The force applied on the other end of the rod to lift the load is called the effort.

The orders of the lever are determined depending on the position of the effort, the fulcrum and the load.
Lever of first order: When the fulcrum is situated between load and effort, we call it a lever of first order. For example, beam balance, a crowbar, a see-saw.

Lever of second order: When the fulcrum and effort are situated at the two opposite ends of the lever and a load is placed in between them, we call it a lever of second order. For example, a nutcracker, a wheel-barrow, etc.

Lever of third order: When the fulcrum and load is situated at the opposite ends of the lever and an effort is applied somewhere between them, we call it a lever of third order. For example, a pair of tongs, a fishing rod, etc.

Page No 90:

Question 5:

Why is this so?
(a) Traveller’s bags have wheels.
(b) Machines have to be maintained.
(c) A bicyle is said to be a complex machine.

ANSWER:

(a) Traveller's bag has wheels so that it can be dragged easily because of reduced friction between the ground and the wheel. Thus, the wheel, a simple machine, helps to decrease the effort to be applied on the load which is a bag here.

(b) A machine is composed of many parts. These parts rub against one another when in use. Due to this, these parts wear out with time. Also, soil and dust creates more friction between the parts which further deteriorates their condition. Even, some parts get affected by weather and thus rust and erode. Because of these factors, the machine can become useless if proper care is not given to it. So, it is very necessary that we have proper equipments and methods for the maintenance of machines. Thus, machines should be sent for maintenance check at fixed interval of time to ensure their proper working.

(c) A bicycle is composed of many simple machines, such as a screw, wheel and axle, lever, pulley, which carry out different task and together contribute to its working. Thus, the bicycle is said to be a complex machine.

Page No 90:

Question 6:

Name the levers mentioned in the following passage. Identify the fulcrum, load and effort of each and say which type of lever it is.
Ravi and Savita sit on a sea-saw in a garden. In the mean time, a gardener is trimming trees in the garden. He puts the leaves and other garbage in the wheelbarrow. Later, Ravi gets thirsty and he buys lemon sherbet. The sherbet seller cuts the lemon and squeezes it using a lemon squeezer. He puts small pieces of ice in the glass with the help of the tongs

ANSWER:

Levers mentioned in the passage are:
(a) Sea-saw: It is a first order lever.

(b) Wheel barrow: It is a second order lever.

(c) Lemon squeezer: It is a second order lever.

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