4 Marks Questions - Maths STD 9 Questions - Vidyadip

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4 Marks Questions

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Question 14 Marks

Find the capacity of a closed rectangular cistern whose length is 8m, breadth 6m and depth 2.5m. Also, find the area of the iron sheet required to make the cistern.

Answer

Length of the cistern, l = 8m
Breadth of the cistern, b = 6m
Height (or depth) of the cistern, h = 2.5m
$\therefore$ Capacity of the cistern
= Volume of the cistern
= l × b × h
= 8 × 6 × 2.5
= 120m³
Also,
Area of the iron sheet required to make the cistern
= Total surface area of the cistern
= 2(lb + bh + hl)
= 2(8 × 6 + 6 × 2.5 + 2.5 × 8)
= 2 × 83
= 166m²

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Question 24 Marks

Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length = 24m, breadth = 25cm and height = 6m

Answer

Here, l = 24m, b = 25cm = 0.25m, h = 6m
Volume of the cuboid = l × b × h
= (24 × 0.25 × 6)m³
= 36m³
Total surface area = 2(lb × lh × bh)
= 2(24 × 0.25 + 24 × 6 + 0.25 × 6)m²
= 2(6 + 144 + 1.5)m²
= 2 × 151.5m²
= 303m²
Lateral surface area = 2(l + b) × h
= [2(24 + 0.25) × 6]m²
= [2 × 24 × 6]m²
= 291m²

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Question 34 Marks

It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer

Diameter of a cylinder = 140cm
⇒ Radius, r = 70cm
Height (h) of a cylinder = 1m = 100cm
Now,
Area of sheet required = Total surface area of cylinder
$=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times70(100+70)\Big]\text{cm}^2$
$=\big[2\times22\times10\times170\big]\text{cm}^2$
$=74800\text{cm}^2$
$=7.48\text{m}^2$

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Question 44 Marks

A cylindrical water tank of diameter 1.4m and height 2.1m is being fed by a pipe of diameter 3.5cm through which water flows at the rate of 2m per second. In how much time will the tank be filled?

Answer

Suppose the tank is filled in x minutes. Then,
Volume of the water that flows out through the pipe in x minutes
= Volume of the tank
$\Rightarrow\pi\times\Big(\frac{3.5}{2\times100}\Big)^2\times(2\times60\text{x})=\pi\times(0.7)\times2.1$
$\Rightarrow\Big(\frac{35}{2000}\Big)^2\times120\text{x}=\Big(\frac{7}{10}\Big)^2\times\Big(\frac{21}{10}\Big)$
$\Rightarrow\frac{35}{2000}\times\frac{35}{2000}\times120\text{x}=\frac{7}{10}\times\frac{7}{10}\times\frac{21}{10}$
$\Rightarrow\text{x}=\frac{7\times7\times21\times2000\times2000}{10\times10\times10\times35\times35\times120}$
$\Rightarrow\text{x}=28$
Hence, the tank will be flled in 28 minutes.

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Question 54 Marks

A man uses a piece of canvas having an area of 551m², to make a conical tent of base radius 7m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately 1m², find the volume of the tent that can be made with it. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Radius of a conical tent, r = 7m
Area of 49 canvas used in making conical tent = (551 - 1)m² = 550m²
⇒ Curved surface areaof a conical tent = 550m²
$\Rightarrow\pi\text{rl}=550$
$\Rightarrow\frac{22}{7}\times7\times\text{l}$
$\Rightarrow\text{l}=\frac{550}{22}=25\text{m}=\text{slant}\ \text{height}$
Now, l² = r² + h²
⇒ 25² = 7² + h²
⇒ h² = 25² - 7² = 625 - 49 = 576
⇒ h = 24m = height
$\therefore$ Volume of conical tent $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times\frac{22}{7}\times7\times7\times24\Big)\text{m}^3$
$=1232\text{m}^3$

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Question 64 Marks

The sum of length, breadth and depth of a cuboid is 19cm and the length of its diagonal is 11cm. The surface area of the cuboid.

Answer

Let the length and height (or depth) of the cuboid be l cm, b cm and h cm, respectively.
$\therefore$ l + b + h = 19 ....(1)
Also,
Length of the diagonal = 11cm
$\Rightarrow\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}=11$
$\Rightarrow\text{l}^2+\text{b}^2+\text{h}^2=121\ ...(2)$
Squaring (1) we get
(l + b + h)² = 19²
⇒ l² + b² + h² + 2(lb + bh + hl) = 361
⇒ 121 + 2(lb + bh + hl) = 361 [Using (2)]
⇒ 2(lb + bh + hl) = 361 - 121 = 240cm²
Thus, the surface area of the cuboid is 240cm².

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Question 74 Marks

A wall 15m long, 30cm wide and 4m high is made of bricks, each measuring (22cm × 12.5cm × 7.5cm). If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer

Length of the wall = 15m = 1500cm
Breadth of the wall = 30cm
Height of the wall = 4m = 400cm
Volume of wall = 1500 × 30 × 400cm³= 18000000cm³
Now, volume of each brick = 22 × 12.5 × 7.5cm³
= 2062.5cm³
Also, volume of the morter $=\frac{1}{12}\times\text{Volume}\ \text{of}\ \text{the}\ \text{wall}$
$=\frac{18000000}{12}=1500000\text{cm}^3$
Total volume of the bricks in the wall = Volume of the wall - Volume of the mortar
= (18000000 - 1500000)cm³ = 16500000cm³
$\therefore$ Number of briks $=\frac{\text{Volume}\ \text{of}\ \text{bricks}}{\text{Volume}\ \text{of}\ \text{one}\ \text{brick}}$
$=\frac{16500000}{2062.5}=8000\ \text{bricks}$

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Question 84 Marks

The curved surface area of a cone is 308cm² and its slant height is 14cm. Find the radius of the base and total surface area of the cone. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Let r be the radius of a cone.
Slant height of a cone, l = 14cm
Curved surface area of a cone = 308cm²
$\Rightarrow\pi\text{rl}=308$
$\Rightarrow\frac{22}{7}\times\text{r}\times14=308$
$\Rightarrow22\times\text{r}\times2=308$
$\Rightarrow\text{r}=\frac{308}{22\times2}=7\text{cm}$
Total surface area of a cone $=\pi\text{r}(\text{l}+\text{r})$
$=\Big[\frac{22}{7}\times7(14+7)\Big]\text{cm}^2$
$=(22\times21)\text{cm}^2$
$=462\text{cm}^2$

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Question 94 Marks

The dimension of a room are (9m × 8m × 6.5m). It has one door of dimension (2m × 1.5m) and two window, each of dimension (1.5 × 1m). Find the cost of whitewashing the walls at ₹ 25 per square metre.

Answer

Length of the room,l = 9m
Breadth of the room, b = 8m
Height of the room, h = 6.5m
Now,
Area of the walls to be whitewashed
= Curved surface area of the room - Area of the door - 2 × Area of each window
= 2h(l + b) - 2m × 1.5m - 2 × 1.5m × 1m
= 2 × 6.5 × (9 + 8) - 3 - 3
= 221 - 6
= 215m²
$\therefore$ Cost of whitewashing the walls at ₹ 25 per square metre
= Area of the walls to be whitewashed × ₹ 25 per square metre
= 215 × 25
= ₹ 5,375

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Question 104 Marks

How many metres of cloth, 2.5m wide, will be required to make a conical tent whose base radius is 7m and height 24m? $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Here, radius = 7m and height (h) = 24m
$\therefore$ slant height $\text{l}=\sqrt{\text{h}^2+\text{r}^2}$
$=\sqrt{(24)^2+(7)^2}$
$\text{l}=\sqrt{576+49}$
$\text{l}=\sqrt{625}$
$\text{l}=25\text{m}$
Now, area of cloth $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times7\times25\Big)\text{m}^2$
$=550\text{m}^2$
$\therefore$ length of cloth $=\frac{\text{Area}\ \text{of}\ \text{cloth}}{\text{Width}\ \text{of}\ \text{cloth}}$
$=\Big(\frac{550}{2.5}\Big)\text{m}$
$=220\text{m}$
$\therefore$ Length of cloth required to make conical tent = 220m.

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Question 114 Marks

Water in a canal, 30dm wide and 12dm deep, is flowing with a velocity of 20km per hour. How much area will it irrigate, if 9cm of standing water is desired?

Answer

Width of the canal = 30dm = 3m (1m = 10dm)
Depth of the canal = 12dm = 1.2m
Speed of the water flow = 20km/h = 20000m/h
$\therefore$ Volume of water flowing out of the canal in 1h = 3 × 1.2 × 20000 = 72000m³
Height of standing water on field - 9cm - 0.09m (1m - 100cm)
Assume that water flows of the canal for 1h. Then,
Area of the field irrigated
$=\frac{\text{Volume}\ \text{of}\ \text{water}\ \text{flowing}\ \text{out}\ \text{of}\ \text{the}\ \text{canal}}{\text{Height}\ \text{of}\ \text{standing}\ \text{water}\ \text{on}\ \text{the}\ \text{field}}$
$=\frac{72000}{0.09}$
$=800000\text{m}^2$
$=\frac{800000}{10000}$ $\big(1\ \text{hectare}=1000\text{m}^2\big)$
$=80\ \text{hectare}$
Thus, the area of the field irrigated is 80hectares.
Disclaimer: In this question time is not given, so the question is solved assuming that the water flows out of the canal for i hour.

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Question 124 Marks

The curved surface area of a cylinder is 1210cm² and its diameter is 20cm. Find its height and volume.

Answer

Here, curved surface area = 1210cm²
Diameter = 20cm $\Rightarrow\text{radius}=\frac{20}{2}=10\text{cm}$
$\therefore$ Curved surface area of the cylinder $=2\pi\text{rh}$
$\Rightarrow1210=2\times\frac{22}{7}\times10\times\text{h}$
$\Rightarrow\text{h}=\Big(\frac{1210\times7}{2\times22\times10}\Big)\text{cm}=19.25\text{cm}$
$\therefore\text{Height}=19.25\text{cm}$
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10^2\times19.25\Big)\text{cm}^3$
$=\Big(\frac{22}{7}\times10\times10\times19.25\Big)\text{cm}^3$
$=6050\text{cm}^3$
$\therefore$ Volume of the cylinder = 6050cm³.

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Question 134 Marks

Water flows at the rate of 10 metres per minute through a cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40cm and depth 24cm?

Answer

Diameter of the pipe = 5mm = 0.5cm
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe = 10 metres = 1000cm
Volume that flows in 1 min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$
$=51\ \text{min}\ 12\ \text{sec}$

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Question 144 Marks

Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length = 26m, breadth = 14m and height = 6.5m

Answer

Here, l = 26m, b = 14m, h = 6.5m
Volume of the cuboid = l × b × h
= (26 × 14 × 6.5)m³
= 2366m³
Total surface area = 2(lb × lh × bh)
= 2(26 × 14 + 26 × 6.5 + 14 × 6.5)m²
= 2(364 + 169 + 91)m²
= 2 × 624m²
= 1248m²
Lateral surface area = 2(l + b) × h
= [2(26 + 14) × 6.5]m²
= [2 × 40 × 6.5]m²
= 520m²

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Question 154 Marks

Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length = 15m, breadth = 6m and height = 5dm

Answer

Here, l = 15m, b = 64m, h = 5dm = 0.5
Volume of the cuboid = l × b × h
= (15 × 6 × 0.5)m³
= 45m³
Total surface area = 2(lb × lh × bh)
= 2(15 × 6 + 15 × 0.5 + 6 × 0.5)m²
= 2(90 + 7.5 + 3)m²
= 2 × 100m²
= 201m²
Lateral surface area = 2(l + b) × h
= [2(15 + 6) × 0.5]m²
= [2 × 21 × 0.5]m²
= 21m²

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Question 164 Marks

Find the volume, curved surface area and the total surface area of a cone whose height and slant height are 6cm and 10cm respectively. $\big(\text{Take}\ \pi=3.14\big)$

Answer

Here, height (h) = 6cm and slant height (l) = 10cm
$\therefore$ radius (r) $=\sqrt{\text{l}^2-\text{h}^2}$
$\text{r}=\sqrt{10^2-6^2}$
$\text{r}=\sqrt{100-36}$
$\text{r}=\sqrt{64}$
$\text{r}=8\text{cm}$
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\Big(\frac{1}{3}\times3.14\times8\times8\times8\Big)\text{cm}^3$
$=401.92\text{cm}^3$
$\therefore$ Curved surface area $=\pi\text{rl}$
$=\big(3.14\times8\times10\big)\text{cm}^2$
$=251.2\text{cm}^2$
$\therefore$ Total surface area $=\pi\text{r}(\text{l}+\text{r})$
$=\pi\text{r}(10+8)$
$=(3.14\times8\times18\big)\text{cm})^2$
$=452.16\text{cm}^2$

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Question 174 Marks

The barrel of a fountain pen, cylindrical in shape, is 7cm long and 5mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a litre?

Answer

Length = 7cm = (height)
Diameter = 5mm $\Rightarrow\text{radius}=\Big(\frac{5}{2}\Big)\text{mm}=2.5\text{mm}$
$=0.25\text{cm}$
$\therefore$ Volume of the barrel $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times0.25\times0.25\times7\Big)\text{cm}^3$
$=\frac{11}{8}\text{cm}^3$
$\frac{11}{8}\text{cm}^3$ isused for writing 330 words.
So, $\Big(\frac{1}{5}\times1000\Big)\text{cm}^3$ will be used for writing
$\Big(330\times\frac{8}{11}\times\frac{1}{5}\times1000\Big)\text{words}$
$=48000\ \text{words}$

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Question 184 Marks

A bus stop is barricated from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 25 per m², what will be the cost of painting all these cones? $\big(\text{Use}\ \pi=3.14\ \text{and}\ \sqrt{1.04}=1.02\big)$

Answer

Radius of a cone, r = 20cm $=\frac{20}{100}\text{m}=\frac{1}{5}\text{m}$
Height of a cone, h = 1m
$\therefore$ Slant height of a cone, $\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{\Big(\frac{1}{5}\Big)^2+1}$
$\text{l}=\sqrt{\frac{26}{25}}$
$\text{l}=\sqrt{1.04}$
$\text{l}=1.02$
Curved surface area of a cone $=\pi\text{rl}$
$=\Big(3.14\times\frac{1}{5}\times1.02\Big)\text{m}^2$
$=\frac{3.2028}{5}\text{m}^2$
⇒ Curved surface area of 50 cones $=\Big(50\times\frac{3.2028}{5}\Big)\text{m}^2$
$=32.028\text{m}^2$
Cost of painting = ₹ 25 per m²
⇒ Cost of painting 32.028m² area = ₹ (25 × 32.028) = ₹ 800.70

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Question 194 Marks

The volume of a sphere is 38808cm³. Find its radius and hence its surface area. $\big(\text{Take}\ \pi=\frac{22}{7}\big).$

Answer

Volume of the sphere = 38808cm³
Suppose that r cm is the radius of the given sphere.
$\therefore\frac{4}{3}\pi\text{r}^3=38808$
$\Rightarrow\text{r}^3=\frac{38808\times3\times7}{4\times22}=9261$
$\Rightarrow\text{r}=\sqrt[3]{9261}=21\text{cm}$
$\therefore$ Surface area of the sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=5544\text{cm}^2$

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Question 204 Marks

The surface area of a cuboid is 758cm². Its length and breadth are 14cm and 11cm respectively. Find its height.

Answer

Length of the cuboid = 14cm
Breadth of the cuboid = 11cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758cm²
Then 758 = 2(14 × 11 + 14 × x + 11 × x)
⇒ 758 = 2(154 + 14x + 11x)
⇒ 758 = 2(154 + 25x)
⇒ 758 = 308 + 50x
⇒ 50x = 758 - 308 = 450
$\Rightarrow\text{x}=\frac{450}{50}=9$
$\therefore$ The height of the cuboid is 9cm.

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Question 214 Marks

The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28cm. The length of the pipe is 35cm. Find the mass of the pipe, if 1cm³ of wood has a mass of 0.6g.

Answer

Internal diameter of a cylinder = 24cm
⇒ Internal radius of a cylinder, r = 12cm
External diameter of a cylinder = 28cm
⇒ External radius of a cylinder, R = 14cm
Length of the pipe, i.e. height, h = 35cm
Now,
Volume of pipe = Volume of cylinder $=\pi(\text{R}^2-\text{r}^2)\text{h}$
$=\Big[\frac{22}{7}(14^2-12^2)\times35\Big]\text{cm}^3$
$=\big[22\times(196-144)\times5\big]\text{cm}^3$
$=(22\times52\times5)\text{cm}^3$
$=5720\text{cm}^3$
Given, 1cm3 = 0.6g
$\therefore$ Mass of pipe = (5720 × 0.6)g = 3432g = 3.432kg

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Question 224 Marks

Find the volume, the lateral surface area and the total surface area of the cuboid whose dimensions are:
Length = 12cm, breadth = 8cm and height = 4.5cm

Answer

Here, l = 12cm, b = 8cm, h = 4.5cm
Volume of the cuboid = l × b × h
= (12 × 8 × 4.5)cm³
= 432cm³
Total surface area = 2(lb × lh × bh)
= 2(12 × 8 + 12 × 4.5 + 8 × 4.5)cm²
= 2(96 + 54 + 36)cm²
= 2 × 186cm²
= 372cm²
Lateral surface area = 2(l + b) × h
= [2(12 + 8) × 4.5]cm²
= [2(20) × 4.5]cm²
= 40 × 4.5cm²
= 180cm²

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Question 234 Marks

A conical tent is 10m high and the radius of its base is 24m. Find the slant height of the tent. If the cost of 1m² canvas is ₹ 70, find the cost of canvas required to make the tent. $\Big(\text{Use}\ \pi=\frac{22}{7}\Big).$

Answer

Radius of a conical tent, r = 24m
Height of a conical tent, h = 10m
$\therefore$ Slant height of a conical tent,
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{24^2+10^2}$
$\text{l}=\sqrt{576+100}$
$\text{l}=\sqrt{676}$
$\text{l}=26\text{cm}$
Curved surface area of a conical tent $=\pi\text{rl}$
$=\Big(\frac{22}{7}\times24\times26\Big)\text{m}^2$
$=\frac{13728}{7}\text{m}^2$
Cost of 1m² canvas = ₹ 70
$\Rightarrow\text{Cost}\ \text{of}\ \frac{13728}{7}\text{m}^2\ \text{canvas}$
$=₹\ \Big(70\times\frac{13728}{7}\Big)=₹\ 137280$

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Question 244 Marks

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4cm and its length is 25cm. The thickness of the metal is 8mm everywhere. Calculate the volume of the metal.

Answer

Internal diameter of the tube = 10.4cm
Internal radius $=\Big(\frac{10.4}{2}\Big)\text{cm}=5.2\text{cm}$
and length = 25cm
and external radius $= (5.2 + 0.8)\text{cm}= 6\text{cm}$
Required volume $=\big[\pi\times(6)^2\times25-\pi\times(5.2)^2\times25\big]\text{cm}^2$
$=\pi\times25\big[(6)^2-(5.2)^2\big]\text{cm}^3$
$=\frac{22}{7}\times25\big[36-27.04\big]\text{cm}^3$
$=\Big(\frac{22}{7}\times25\times8.96\Big)\text{cm}^3$
$=704\text{cm}^3$
$\therefore$ the volume of the metal = 704cm³.

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Question 254 Marks

A cylindrical bucket, 28cm in diameter and 72cm and high, is full of water. The water is emptied into a rectangular tank, 66cm long and 28cm wide. Find the height of the water level in the tank.

Answer

Here, cylindrical bucket has diameter = 28cm.
$\therefore$ radius $=\Big(\frac{28}{2}\Big)\text{cm}=14\text{cm}$ and height = 72cm.
Length of the tank = 66cm
Breadth of the tank = 28cm
$\therefore$ Volume of the tank = Volume of cylindrical bucket
$\Rightarrow\text{l}\times\text{b}\times\text{h}=\pi\text{r}^2\text{h}$
$\Rightarrow66\times28\times\text{h}=\frac{22}{7}\times14\times14\times72$
$\Rightarrow\text{h}=\Big(\frac{22\times2\times14\times72}{66\times28}\Big)\text{cm}$
$\Rightarrow\text{h}=24\text{cm}.$
$\therefore$ The height of the water level in the tank = 24cm.

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Question 264 Marks

Find the weight of a solid cylinder of radius 10.5cm and height 60cm if the material of the cylinder weighs 5g per cm³.

Answer

Here, radius (r) = 10.5cm and height = 60cm.
$\therefore$ Volume of the cylinder $=(\pi\text{r}^2\text{h})$
$=\Big(\frac{22}{7}\times10.5\times10.5\times60\Big)\text{cm}^3$
$=20790\text{cm}^3$
$\therefore$ Weight of the solid cylinder if the material of the cylinder.
Weighs 5g per cm3 = (20790 × 5) = 103950g
$=\frac{103950}{1000}$ $\big[\therefore1000\text{g}=1\text{kg}\big]$
$=103.95\text{kg}$

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Question 274 Marks

A classroom is 10m long, 6.4m wide and 5m high. If each student be given 1.6m² of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?

Answer

Length of the classroom = 10m
Breadth of the classroom = 6.4m
Height of the classroom = 5m
Area of the floor = length × breadth
= 10 × 6.4m²
No. of students $=\frac{\text{Area}\ \text{of}\ \text{the}\ \text{floor}}{\text{Area}\ \text{given}\ \text{to}\ \text{student}\ \text{on}\ \text{the}\ \text{floor}}$
$=\frac{640}{16}=40\ \text{students}$
Now, volume of the classroom = 10 × 6.4 × 5m³
$\therefore$ Air required by each student $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{room}}{\text{Number}\ \text{of}\ \text{students}}$
$=\frac{10\times6.4\times5}{40}\text{m}^3$
$=8\text{m}^3$

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Question 284 Marks

The curved surface area of a cylinder is 4400cm² and the circumference of its base is 110cm. Find the height and the volume of the cylinder.

Answer

Let base radius be r and height be h
Then, $2\pi\text{rh}=4400\text{cm}^2$
And, $2\pi\text{r}=110\text{cm}$
$\Rightarrow\frac{2\pi\text{rh}}{2\pi\text{r}}=\frac{4400}{110}$
$\Rightarrow\text{h}=40\text{cm}$
$\therefore2\times\frac{22}{7}\times\text{r}\times\text{h}\times40=4400\text{cm}.$
$\Rightarrow\text{r}=\Big(\frac{4400\times7}{44\times40}\Big)\text{cm}=\frac{35}{2}\text{cm}.$
$\therefore$ Volume of the cylinder $=\pi\text{r}^2\text{h}$
$=\Big(\frac{22}{7}\times\frac{35}{2}\times\frac{35}{2}\times40\Big)\text{cm}^3$
$=38500\text{cm}^3.$

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Question 294 Marks

How many planks of dimensions (5m × 25cm × 10cm) can be stored in a pit which is 20m long, 6m wide and 80cm deep?

Answer

Number of planks $=\frac{\text{Volume}\ \text{of}\ \text{the}\ \text{pit}\ \text{in}\ \text{cm}^3}{\text{Volume}\ \text{of}\ 1\ \text{plank}\ \text{in}\ \text{cm}^3}$
Volume of one plank = (l × b × h)cm³
= 500 × 25 × 10cm³
= 125000cm³
Volume of the pit = (l × b × h)cm³
Here, l = 20m = 2000cm, b = 6m = 600cm, h = 80cm
i.e., volume of the pit = 2000 × 600 × 80cm³
= 96000000cm³
$\therefore$ Number of planks $=\frac{96000000}{125000}$
$=\frac{96000}{125}=768$

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Question 304 Marks

The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. Find the area of the metal sheet needed to make it.

Answer

Volume of a cylinder = 15.4 litres = 15400cm³
Height (h) of a cylinder = 1m = 100cm
Volume of a cylinder $=\pi\text{r}^2\text{h}$
$\Rightarrow15400=\Big(\frac{22}{7}\times\text{r}^2\times100\Big)$
$\Rightarrow\text{r}^2=\frac{15400\times7}{22\times100}=49$
$\Rightarrow\text{r}=7\text{cm}$
Now,
Area of metal sheet needed = Total surface area of cylinder
$=2\pi\text{r}(\text{h}+\text{r})$
$=\Big[2\times\frac{22}{7}\times7(100+7)\Big]\text{cm}^2$
$=\big[2\times22\times107\big]\text{cm}^2$
$=4708\text{cm}^2$

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Question 314 Marks

A box made of sheet metal costs ₹ 6480 at ₹ 120 per square metre. If the box is 5m long and 3m wide, find its height.

Answer

Length of the box = 5m
Breadth of the box = 3m
Area of the sheet required $=\frac{\text{Total}\ \text{cost}}{\text{Cost}\ \text{per}\ \text{metre}\ \text{square}}$
Let h m be the height of the box.
Then area of the sheet = total surface area of the box
= 2(lb + lh + bh)m²
=2(5 × 3 + 5 × h + 3 × h)m²
= 2(15 + 8h) = (30 + 16h )m²
Now, $30 + 16\text{h}=\frac{6480}{120}$
⇒ 30 + 16h = 54
⇒ 16h = 24
⇒ h = 1.5m
$\therefore$ The height of the box is 1.5m.

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Question 324 Marks

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

Answer

Let the initial edge of the cube be a units.
$\therefore$ Initial surface area of the cube = 6a² square units
New edge of the cube = a + 50% of a $=\text{a}+\frac{50}{100}\text{a}=1.5\text{a}\ \text{units}$
$\therefore$ New surface of the cube = 6(1.5a)² = 13.5a² square units
Increase in surface area of the cube = 13.5a² - 6a² = 7.5a² square units
$\therefore$ Percentage increase in the surface area of the cube
$=\frac{\text{Increase}\ \text{in}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}{\text{Initial}\ \text{surface}\ \text{area}\ \text{of}\ \text{the}\ \text{cube}}\times100\%$
$=\frac{7.5\text{a}^2}{6\text{a}^2}\times100\%$
$=125\%$

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Question 334 Marks

There are 20 cylindrical pillars in a building, each having a diameter of 50cm and height 4m. Find the cost of cleaning them at ₹ 14 per m².

Answer

Radius (r) of 1 pillar $=\frac{50}{100\times2}=\frac{1}{4}\text{m}$
Height (h) of 1 pillar = 4m
$\therefore$ Lateral surface area of 1 pillar $=2\pi\text{rh}$
$=\Big(2\times\frac{22}{7}\times\frac{1}{4}\times4\Big)\text{m}^2$
$=\frac{44}{7}\text{m}^2$
⇒ Lateral surface area of 20 such pillars $=20\times\frac{44}{7}=\frac{880}{7}\text{m}^2$
Cost of cleaning = ₹ 14/m²
⇒ Cost of cleaning $\frac{880}{7}\text{m}^2=₹\ \Big(14\times\frac{880}{7}\Big)$
$=₹\ 1760$

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Question 344 Marks

Water flows at the rate of 10 metres per minute through a cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter art the surface 40cm and depth 24cm?

Answer

Diameter of the pipe = 5mm = 0.5cm
Radius of the pipe $=\frac{0.5}{2}=0.25\text{cm}$
Length of the pipe = 10 metres = 1000cm
Volume that flows in 1 min $=\big[\pi\times(0.25)^2\times1000\big]\text{cm}^3$
$\therefore$ Volume of the conical vessel $=\Big[\frac{1}{3}\pi\times(20)^2\times24\Big]\text{cm}^3$
$\therefore$ Required time $=\Bigg[\frac{\frac{1}{3}\pi\times(20)^2\times24}{\pi\times(0.25)^2\times1000}\Bigg]\text{min}$
$=\Bigg[\frac{\frac{1}{3}\times400\times24}{\pi\times0.0625\times1000}\Bigg]\text{min}$
$=51.2\ \text{min}$
$=51\ \text{min}\ 12\ \text{sec}$

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Question 354 Marks

It costs ₹ 3300 to paint the inner curved surface of a cylindrical vessel 10m deep at the rate of ₹ 30 per m². Find the:

Inner curved surface area of the vessel.
Inner radius of the base.
Capacity of the vessel.

Answer

Cost of painting inner curved surface area of vassel

= Cost of painting per m² × Inner curved surface of vessel

⇒ ₹ 3300 = ₹ 30 × Inner curved surface of vessel

⇒ Inner curved surface of vessel = 110m²

Let inner radius of the base = r

Depth, h = 10m

Inner curved surface of vessel $=2\pi\text{rh}$

$\Rightarrow10=2\times\frac{22}{7}\times\text{r}\times10$

$\Rightarrow\text{r}=\frac{110\times7}{2\times22\times10}=1.75\text{m}$

Capacity of the vessel $=\pi\text{r}^2\text{h}$

$=\Big(\frac{22}{7}\times1.75\times1.75\times10\Big)\text{m}^3$

$=96.25\text{m}^3$

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Question 364 Marks

1cm³ of gold is drawn into a wire 0.1mm in diameter. Find the length of the wire.

Answer

1cm³ = 1cm × 1cm × 1cm = 0.01m
Therefore,
Volume of the
gold = 0.01m × 0.01m × 0.01m = 0.000001m³ ...(1)
Diameter of the wire drawn = 0.1mm
Radius of the wire drawn $=\frac{0.1}{2}\text{mm}=0.05\text{mm}$
r = 0.00005m ...(2)
Length of the wire = h m ...(3)
Volume pf the wire drawn = Volume of the gold
$\Rightarrow\pi\text{r}^2\text{h}=0.000001$
$\Rightarrow\pi\times0.00005\times0.00005\times\text{h}=0.000001$ [from equations (1), (2) and (3)]
$\text{h}=\frac{0.00000\times7}{0.00005\times0.00005\times22}$
$\therefore\text{h}=127.27\text{m}$
$\therefore$ the length of the wire is 127.27m.

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