Tick (✓) the correct answer:
Question: Tick (✓) the correct answer: The edges of a cuboid are in the ratio 1 : 2 : 3 and its surface area is 88 cm2. The volume of the cuboid is (a) 48 cm3 (b) 64 cm3 (c) 96 cm3 (d) 120 cm3 Solution: (a) $48 \mathrm{~cm}^{3}$ Let $a$ be the length of the smallest edge. Then the edges are in the proportion $a: 2 a: 3 a$. Now, surface area $=2(a \times 2 a+a \times 3 a+2 a \times 3 a)=2\left(2 a^{2}+3 a^{2}+6 a^{2}\right)=22 a^{2}=88 \mathrm{~cm}^{2}$ $\Rightarrow a=\sqrt{\frac{88}{22}}=\sqrt{4...
Read More →At a birthday party,
Question: At a birthday party, the children spin a wheel to get a gift. Find the probability of (a) getting a ball (b) getting a toy car (c) getting any toy except a chocolate. Solution: (a) The probability of getting a ball $=\frac{\text { Number of events of getting a ball }}{\text { Total number of events }}=\frac{2}{8}=\frac{1}{4}$ (b) The probability of getting a toy car $=\frac{\text { Number of events of getting a toy car }}{\text { Total number of events }}=\frac{3}{8}$ (c) The probabili...
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Question: Tick (✓) the correct answer: How many cubes of 10 cm edge can be put in a cubical box of 1 m edge? (a) 10 (b) 100 (c) 1000 (d) 10000 Solution: (c) 1000 Volume of the smaller cube $=(10 \mathrm{~cm})^{3}=1000 \mathrm{~cm}^{3}$ Volume of box $=(100 \mathrm{~cm})^{3}=1000000 \mathrm{~cm}^{3} \quad[1 \mathrm{~m}=100 \mathrm{~cm}]$ $\therefore$ Total no. of cubes $=\frac{100 \times 100 \times 100}{10 \times 10 \times 10}=1000$...
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Question: Tick (✓) the correct answer: How many bricks, each measuring 25 cm 11.25 cm 6 cm, will be needed to build a wall 8 m long, 6 m high and 22.5 cm thick? (a) 5600 (b) 6000 (c) 6400 (d) 7200 Solution: (c) 6400 Volume of each brick $=25 \times 11.25 \times 6=1687.5 \mathrm{~cm}^{3}$ Volume of the wall= $800 \times 600 \times 22.5=10800000 \mathrm{~cm}^{3}$ $\therefore$ No. of bricks $=\frac{10800000}{1687.5}=6400$...
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Question: Tick (✓) the correct answer: The cost of painting the whole surface area of a cube at the rate of 10 paise per cm2is Rs 264.60. Then, the volume of the cube is (a) 6859 cm3 (b) 9261 cm3 (c) 8000 cm3 (d) 10648 cm3 Solution: (b) $9261 \mathrm{~cm}^{3}$ Rate of painting $=10$ paise per sq $\mathrm{cm}=\mathrm{Rs} 0.1 / \mathrm{cm}^{2}$ Total cost $=$ Rs $264.60$ Now, total surface area $=\frac{264.6}{0.1}=2646 \mathrm{~cm}^{2}$ Also, length of edge, $a=\sqrt{\frac{2646}{6}}=\sqrt{441}=21 ...
Read More →Find the modulus of each of the following complex numbers and hence
Question: Find the modulus of each of the following complex numbers and hence express each of them in polar form: $\left(\sin 120^{\circ}-i \cos 120^{\circ}\right)$ Solution: = sin(90 + 30 ) - icos(90 + 30) $=\cos 30^{\circ}+i \sin 30^{\circ}$ Since, sin(90+ ) = cos And cos(90 + ) = -sin $=\frac{\sqrt{3}}{2}+i \frac{1}{2}$ Hence it is of the form $\mathrm{Z}=\frac{\sqrt{3}}{2}+i \frac{1}{2}=\mathrm{r}(\cos \theta+i \sin \theta)$ Therefore r = 1 Hence its modulus is 1 and argument is $\frac{\pi}{...
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Question: Tick (✓) the correct answer: The volume of a cube is 343 cm3. Its total surface area is (a) 196 cm2 (b) 49 cm2 (c) 294 cm2 (d) 147 cm2 Solution: (c) $294 \mathrm{~cm}^{2}$ Volume $=a^{3}=343 \mathrm{~cm}^{3}$ $\Rightarrow a=\sqrt[3]{343}=7 \mathrm{~cm}$ $\therefore$ Total surface area $=6 a^{2}=6 \times 7 \times 7=294 \mathrm{~cm}^{2}$...
Read More →The following data represents the approximate
Question: The following data represents the approximate percentage of water in various oceans. Prepare a pie chart of the given data. Solution: The pie chart is as follows:...
Read More →Find the modulus of each of the following complex numbers and hence
Question: Find the modulus of each of the following complex numbers and hence express each of them in polar form: $\frac{(1-i)}{\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)}$ Solution: $=\frac{1-i}{\frac{1}{2}+i \frac{\sqrt{3}}{2}}$ $=\frac{2-2 i}{1+i \sqrt{3}}$ $=\frac{2-2 i}{1+\sqrt{3} i} \times \frac{1-\sqrt{3} i}{1-\sqrt{3} i}$ $=\frac{2-2 \sqrt{3} i-2 i+2 \sqrt{3} i^{2}}{1-3 i^{2}}$ $=\frac{(2-2 \sqrt{3})+i(2 \sqrt{3}+2)}{4}$ $=\frac{(1-\sqrt{3})+i(\sqrt{3}+1)}{2}$ Let $Z=\frac{(1-\...
Read More →A survey was carried out to
Question: A survey was carried out to find the favourite beverage preferred by a certain group of young people. The following pie chart shows the findings of this survey. From this pie chart, answer the following: (i) Which type of beverage is liked by the maximum number of people? (ii) If 45 people like tea, how many people were surveyed? Solution: (i) The percentage of people preferring cold drinks is maximum. So, cold drinks is liked by the maximum number of people. (ii) From the pie chart, n...
Read More →In the time table of a school,
Question: In the time table of a school, periods allotted per week to different teaching subjects are given below: Draw a pie chart for this data. Solution: From the question, Total number periods allotted = 7 + 8 + 8 + 8 + 7 + 4 + 3 = 45 periods Central angle of pie chart for each subject, Hindi = (7/45) 360 English = (8/45) 360o = 64o Maths = (8/45) 360o = 64o Science = (8/45) 360o = 64o Social Science = (7/45) 360o = 56o Computer = (4/45) 360o = 32o Sanskrit = (3/45) 360o = 24o...
Read More →Find the modulus of each of the following complex numbers and hence
Question: Find the modulus of each of the following complex numbers and hence express each of them in polar form: $\left(\mathrm{i}^{25}\right)^{3}$ Solution: $=\mathrm{i}^{75}$ $=i^{4 n+3}$ where $n=18$ Since $i^{4 n+3}=-i$ $\mathrm{i}^{75}=-\mathrm{i}$ Let $Z=-i=r(\cos \theta+i \sin \theta)$ Now , separating real and complex part , we get 0 = rcos .eq.1 -1 = rsin eq.2 Squaring and adding eq.1 and eq.2, we get $1=r^{2}$ Since r is always a positive no., therefore, r = 1, Hence its modulus is 1....
Read More →For the development of basic infrastructure
Question: For the development of basic infrastructure in a district, a project of Rs 108 crore approved by Development Bank is as follows: Draw a pie chart for this data. Solution: From the question, Total number approved = 108 crore Central angle of pie chart for each infrastructure, Road = (43.2/108) 360o = 144o Electricity = (16.2/108) 360o Drinking water = (27/108) 360o = 90o Sewerage = (21.6/108) 360o = 72o...
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Question: Tick (✓) the correct answer: The total surface area of a cube is 150 cm2. Its volume is (a) 216 cm3 (b) 125 cm3 (c) 64 cm3 (d) 1000 cm3 Solution: (b) $125 \mathrm{~cm}^{3}$ Total surface area $=6 a^{2}=150 \mathrm{~cm}^{2}$, where $a$ is the length of the edge of the cube. $\Rightarrow 6 a^{2}=150$ $\Rightarrow a=\sqrt{\frac{150}{6}}=\sqrt{25}=5 \mathrm{~cm}$ $\therefore$ Volume $=a^{3}=5^{3}=125 \mathrm{~cm}^{3}$...
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Question: Tick (✓) the correct answer: The maximum length of a pencil that can be kept in a rectangular box of dimensions 12 cm 9 cm 8 cm, is (a) 13 cm (b) 17 cm (c) 18 cm (d) 19 cm Solution: (b) 17 Length of the diagonal of a cuboid $=\sqrt{l^{2}+b^{2}+h^{2}}$ $\therefore \sqrt{l^{2}+b^{2}+h^{2}}=\sqrt{12^{2}+9^{2}+8^{2}}=\sqrt{144+81+64}=\sqrt{289}=17 \mathrm{~cm}$...
Read More →In a district, the number of branches
Question: In a district, the number of branches of different banks is given below: Draw a pie chart for this data. Solution: Total number of branches $=30+17+15+10=72$...
Read More →The length of a metallic tube is 1 metre, its thickness is 1 cm and its inner diameter is 12 cm.
Question: The length of a metallic tube is 1 metre, its thickness is 1 cm and its inner diameter is 12 cm. Find the weight of the tube if the density of the metal is 7.7 grams per cubic centimetre. Solution: Length $=1 \mathrm{~m}=100 \mathrm{~cm}$ Inner diameter $=12 \mathrm{~cm}$ Radius $=6 \mathrm{~cm}$ Now, inner volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 6 \times 6 \times 100=11314.286 \mathrm{~cm}^{3}$ Thickness $=1 \mathrm{~cm}$ Total radius $=7 \mathrm{~cm}$ Now, we have...
Read More →Find the modulus of each of the following complex numbers and hence
Question: Find the modulus of each of the following complex numbers and hence express each of them in polar form: $-\sqrt{3}-\mathrm{i}$ Solution: Let $Z=-i-\sqrt{3}=r(\cos \theta+i \sin \theta)$ Now, separating real and complex part, we get $-\sqrt{3}=r \cos \theta \ldots \ldots \ldots . . q .1$ $-1=r \sin \theta \ldots \ldots \ldots \ldots$ eq.2 Squaring and adding eq.1 and eq.2, we get $4=r^{2}$ Since r is always a positive no., therefore, r = 2 Hence its modulus is 2. Now, dividing eq.2 by e...
Read More →A cylinder is open at both ends and is made of 1.5-cm-thick metal.
Question: A cylinder is open at both ends and is made of 1.5-cm-thick metal. Its external diameter is 12 cm and height is 84 cm. What is the volume of metal used in making the cylinder? Also, find the weight of the cylinder if 1 cm3of the metal weighs 7.5 g. Solution: Thickness of the cylinder $=1.5 \mathrm{~cm}$ External diameter $=12 \mathrm{~cm}$ i.e., radius $=6 \mathrm{~cm}$ also, internal radius $=4.5 \mathrm{~cm}$ Height $=84 \mathrm{~cm}$ Now, we have the following: Total volume $=\pi \m...
Read More →The following histogram shows the frequency
Question: The following histogram shows the frequency distribution of teaching experiences of 30 teachers in various schools: (a) What is the class width? (b) How many teachers are having the maximum teaching experience and how many have the least teaching experience? (c) How many teachers have teaching experience of 10 to 20 years? Solution: (a) In the histogram, we see that the class width Is 5. (b) By the histogram, it is clear that two teachers have the maximum teaching experience,i.e. 15-20...
Read More →A road roller takes 750 complete revolutions to move once over to level a road.
Question: A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of the road roller is 84 cm and its length is 1 m. Solution: Diameter $=84 \mathrm{~cm}$ i.e., radius $=42 \mathrm{~cm}$ Length $=1 \mathrm{~m}=100 \mathrm{~cm}$ Now, lateral surface area $=2 \pi \mathrm{rh}=2 \times \frac{22}{7} \times 42 \times 100=26400 \mathrm{~cm}^{2}$ $\therefore$ Area of the road = lateral surface area $\times$ no. of rotations $=26400 \times...
Read More →A well of inner diameter 14 m is dug to a depth of 12 m.
Question: A well of inner diameter 14 m is dug to a depth of 12 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of the embankment so formed. Solution: Inner diameter $=14 \mathrm{~m}$ i.e., radius $=7 \mathrm{~m}$ Depth $=12 \mathrm{~m}$ Volume of the earth dug out $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 7 \times 7 \times 12=1848 \mathrm{~m}^{3}$ Width of embankment $=7 \mathrm{~m}$ Now, total radius $=7+7=14 m$ ...
Read More →In a hypothetical sample of 20 people,
Question: In a hypothetical sample of 20 people, the amount of money (in t thousands) with each was found to be as follows: 114, 108, 100, 98, 101, 109, 117, 119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118 Draw a histogram of the frequency distribution, taking one of the class intervals as 50-100. Solution: Before preparing histogram of the given data, we will prepare the frequency distribution table....
Read More →Find the modulus of each of the following complex numbers and hence
Question: Find the modulus of each of the following complex numbers and hence express each of them in polar form: $\sqrt{\frac{1+\mathrm{i}}{1-\mathrm{i}}}$ Solution: $=\sqrt{\frac{1+i}{1-i}} \times \sqrt{\frac{1+i}{1+i}}$ $=\sqrt{\frac{(1+i)^{2}}{1-i^{2}}}$ $=\frac{1+i}{\sqrt{2}}$ $=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$ Let $Z=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=r(\cos \theta+i \sin \theta)$ Now, separating real and complex part, we get $\frac{1}{\sqrt{2}}=r \cos \theta$ .eq.1 $\frac{1}{\sqr...
Read More →How many cubic metres of earth must be dug out to sink a well which is 20 m deep and has a diameter of 7 metres?
Question: How many cubic metres of earth must be dug out to sink a well which is 20 m deep and has a diameter of 7 metres? If the earth so dug out is spread over a rectangular plot 28 m by 11 m, what is the height of the platform so formed? Solution: Diameter $=7 \mathrm{~m}$ Radius $=3.5 \mathrm{~m}$ Depth $=20 \mathrm{~m}$ Volume of the earth dug out $=\pi r^{2} \mathrm{~h}=\frac{22}{7} \times 3.5 \times 3.5 \times 20=770 \mathrm{~m}^{3}$ Volume of the earth piled upon the given plot $=28 \tim...
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