A solid cube of metal each of whose sides measures 2.2 cm
Question: A solid cube of metal each of whose sides measures 2.2 cm is melted to form a cylindrical wire of radius 1 mm. Find the length of the wire so obtained. Solution: Length of the edge, $a=2.2 \mathrm{~cm}$ Volume of the cube $=a^{3}=(2.2)^{3}=10.648 \mathrm{~cm}^{3}$ Volume of the wire $=\pi \mathrm{r}^{2} \mathrm{~h}$ Radius $=1 \mathrm{~mm}=0.1 \mathrm{~cm}$ As volume of cube = volume of wire, we have: $h=\frac{\text { volume }}{\pi \mathrm{r}^{2}}=\frac{10.648 \times 7}{22 \times 0.1 \...
Read More →Draw a histogram for the following data.
Question: Draw a histogram for the following data. Solution:...
Read More →A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm.
Question: A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It is drawn out into a wire of diameter 1 mm. What will be the length of the wire so obtained? Solution: Diameter of the given wire $=1 \mathrm{~cm}$ Radius $=0.5 \mathrm{~cm}$ Length $=11 \mathrm{~cm}$ Now, volume $=\pi r^{2} \mathrm{~h}=\frac{22}{7} \times 0.5 \times 0.5 \times 11=8.643 \mathrm{~cm}^{3}$ The volumes of the two cylinders would be the same. Now, diameter of the new wire $=1 \mathrm...
Read More →A rectangular vessel 22 cm by 16 cm by 14 cm is full of water.
Question: A rectangular vessel 22 cm by 16 cm by 14 cm is full of water. If the total water is poured into an empty cylindrical vessel of radius 8 cm, find the height of water in the cylindrical vessel Solution: Volume of the rectangular vessel $=22 \times 16 \times 14=4928 \mathrm{~cm}^{3}$ Radius of the cylindrical vessel $=8 \mathrm{~cm}$ Volume $=\pi \mathrm{r}^{2} \mathrm{~h}$ As the water is poured from the rectangular vessel to the cylindrical vessel, we have: Volume of the rectangular ve...
Read More →Find the cost of painting 15 cylindrical pillars of a building at Rs 2.50
Question: Find the cost of painting 15 cylindrical pillars of a building at Rs 2.50 per square metre if the diameter and height of each pillar are 48 cm and 7 metres respectively. Solution: Diameter $=48 \mathrm{~cm}$ Radius $=24 \mathrm{~cm}=0.24 \mathrm{~m}$ Height $=7 \mathrm{~m}$ Now, we have: Lateral surface area of one pillar $=\pi \mathrm{dh}=\frac{22}{7} \times 0.48 \times 7=10.56 \mathrm{~m}^{2}$ Surface area to be painted $=$ total surface area of 15 pillars $=10.56 \times 15=158.4 \ma...
Read More →A particular brand of talcum powder is available in two packs,
Question: A particular brand of talcum powder is available in two packs, a plastic can with a square base of side 5 cm and of height 14 cm, or one with a circular base of radius 3.5 cm and of height 12 cm. Which of them has greater capacity and by how much? Solution: For the cubic pack: Length of the side, $a=5 \mathrm{~cm}$ Height $=14 \mathrm{~cm}$ Volume $=a^{2} h=5 \times 5 \times 14=350 \mathrm{~cm}^{3}$ For the cylindrical pack: Base radius $=r=3.5 \mathrm{~cm}$ Height $=12 \mathrm{~cm}$ V...
Read More →The following pie chart depicts the expenditure
Question: The following pie chart depicts the expenditure of a state government under different heads: (i) If the total spending is 10 crore, how much money was spent on roads? (ii) How many times is the amount of money spents on education compared to the amount spent on roads? (iii) What fraction of the total expenditure is spents on both roads and public welfare together? Solution: (i) Money spent on roads $=10 \%$ of total spending $=\frac{10}{100} \times 10$ crore $=1$ crore (ii) Money spent...
Read More →Find the modulus of each of the following
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}{\sq...
Read More →The curved surface area of a cylinder is 4400 cm
Question: The curved surface area of a cylinder is 4400 cm2and the circumference of its base is 110 cm. Find the volume of the cylinder. Solution: Curved surface area $=2 \pi r h=4400 \mathrm{~cm}^{2}$ Circumference $=2 \pi \mathrm{r}=110 \mathrm{~cm}$ Now, height $=h=\frac{\text { curved surface area }}{\text { circumference }}=\frac{4400}{110}=40 \mathrm{~cm}$ Also, radius, $r=\frac{4400}{2 \pi \mathrm{h}}=\frac{4400 \times 7}{2 \times 22 \times 40}=\frac{35}{2}$ $\therefore$ Volume $=\pi \mat...
Read More →The radius and height of a cylinder are in the ratio 7 : 2.
Question: The radius and height of a cylinder are in the ratio 7 : 2. If the volume of the cylinder is 8316 cm3, find the total surface area of the cylinder. Solution: We have: $\frac{\text { radius }}{\text { height }}=\frac{7}{2}$ i.e., $r=\frac{7}{2} h$ Now, volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\pi\left(\frac{7}{2} \mathrm{~h}\right)^{2} \mathrm{~h}=8316 \mathrm{~cm}^{3}$ $\Rightarrow \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h^{3}=8316$ $\Rightarrow h^{3}=\frac{8316 \times...
Read More →Shoes of the following brands are sold
Question: Shoes of the following brands are sold in November 2007 at a shoe store. Construct a pie chart for the given data. Solution: Total number of pairs of shoes sold $=(130+120+90+40+20)=400$ $\therefore$ Central angle of pie chart representing the brand (i) $A=\frac{130}{400} \times 360^{\circ}=117^{\circ}$ (ii) $B=\frac{120}{400} \times 360^{\circ}=108^{\circ}$ (iii) $C=\frac{90}{40} \times 360^{\circ}=81^{\circ}$ (iv) $D=\frac{40}{400} \times 360^{\circ}=36^{\circ}$ (v) $E=\frac{20}{400}...
Read More →The volume of a cylinder of height 8 cm is 1232 cm
Question: The volume of a cylinder of height 8 cm is 1232 cm3. Find its curved surface area and the total surface area. Solution: Height $=8 \mathrm{~cm}$ Volume $=\pi r^{2} \mathrm{~h}=1232 \mathrm{~cm}^{3}$ Now, radius $=r=\sqrt{\frac{1232}{\pi \mathrm{h}}}=\sqrt{\frac{1232 \times 7}{22 \times 8}}=\sqrt{49}=7 \mathrm{~cm}$ Also, curved surface area $=2 \pi \mathrm{rh}=2 \times \frac{22}{7} \times 7 \times 8=352 \mathrm{~cm}^{2}$ $\therefore$ Total surface area $=2 \pi \mathrm{r}(\mathrm{h}+\ma...
Read More →The lateral surface area of a cylinder of length 14 m is 220 m
Question: The lateral surface area of a cylinder of length 14 m is 220 m2. Find the volume of the cylinder. Solution: Length $=$ height $=14 \mathrm{~m}$ Lateral surface area $=2 \pi \mathrm{rh}=220 \mathrm{~m}^{2}$ Radius $=r=\frac{220}{2 \pi \mathrm{h}}=\frac{220 \times 7}{2 \times 22 \times 14}=\frac{10}{4}=2.5 \mathrm{~m}$ $\therefore$ Volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 2.5 \times 2.5 \times 14=275 \mathrm{~m}^{3}$...
Read More →The circumference of the base of a cylinder is 88 cm and its height is 60 cm.
Question: The circumference of the base of a cylinder is 88 cm and its height is 60 cm. Find the volume of the cylinder and its curved surface area. Solution: Circumference of the base $=88 \mathrm{~cm}$ Height $=60 \mathrm{~cm}$ Area of the curved surface $=$ circumference $\times$ height $=88 \times 60=5280 \mathrm{~cm}^{2}$ Circumference $=2 \pi r=88 \mathrm{~cm}$ Then radius $=r=\frac{88}{2 \pi}=\frac{88 \times 7}{2 \times 22}=14 \mathrm{~cm}$ $\therefore$ Volume $=\pi \mathrm{r}^{2} \mathrm...
Read More →The weights (in kg) of 30 students of a class are as follows:
Question: The weights (in kg) of 30 students of a class are as follows: 39, 38, 36, 38, 40, 42, 43, 44, 33, 33, 31, 45, 46, 38, 37, 31, 30, 39, 41, 41, 46, 36, 35, 34, 39, 43, 32, 37, 29, 26 Prepare a frequency distribution table using one class interval as (30-35), 35 not included. (i) Which class has the least frequency? (ii) Which class has the maximum frequency? Solution: (i) The class interval $25-30$ has the least frequency, i.e. $2 .$ (ii) The class interval $35-40$ has the maximum freque...
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: $\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}$ Solution: $=\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i} \times \frac{5-\sqrt{3} i}{5-\sqrt{3} i}$ $=\frac{10+28 \sqrt{3} i-18 i^{2}}{25-3 i^{2}}$ $=\frac{28 \sqrt{3} i+28}{28}$ $=\sqrt{3} \mathrm{i}+1$ Let $Z=\sqrt{3} \mathrm{i}+1=\mathrm{r}(\cos \theta+\mathrm{i} \sin \theta)$ Now , separating real and complex part , we get 1 = rcos .eq.1 $\sqrt{3}=r \s...
Read More →Prepare a histogram from the frequency
Question: Prepare a histogram from the frequency distribution table obtained in question 93. Solution:...
Read More →The marks obtained (out of 20) by 30 students
Question: The marks obtained (out of 20) by 30 students of a class in a test are as follows: 14, 16, 15, 11, 15, 14, 13, 16, 8, 10, 7, 11, 18, 15, 14, 19, 20, 7, 10, 13, 12, 14, 15, 13, 16, 17, 14, 11, 10, 20. Prepare a frequency distribution table for the above data using class intervals of equal width in which one class interval is 4-8 (excluding 8 and including 4). Solution:...
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: $\frac{-16}{1+\sqrt{3} i}$ Solution: $=\frac{-16}{1+\sqrt{3} i} \times \frac{1-\sqrt{3} i}{1-\sqrt{3} i}$ $=\frac{-16+16 \sqrt{3} i}{1-3 i^{2}}$ $=\frac{16 \sqrt{3} i-16}{4}$ $=4^{\sqrt{3}} \mathrm{i}-4$ Let $Z=4^{\sqrt{3}} i-4=r(\cos \theta+i \sin \theta)$ Now , separating real and complex part , we get $-4=r \cos \theta \ldots \ldots \ldots .$ eq. 1 $4 \sqrt{3}=r \sin \theta \ldot...
Read More →Following are the numbers of members in 25 families of a village:
Question: Following are the numbers of members in 25 families of a village: 6, .8, 7, 7, 6, 5, 3, 2, 5, 6, 8, 7, 7, 4, 3, 6, 6,6, 7, 5, 4, 3, 3, 2, 5 Prepare a frequency distribution table for the data using class intervals 0-2, 2-4 etc. Solution:...
Read More →A closed cylindrical tank of diameter 14 m and height 5 m is made from a sheet of metal.
Question: A closed cylindrical tank of diameter 14 m and height 5 m is made from a sheet of metal. How much sheet of metal will be required? Solution: Diameter $=14 \mathrm{~m}$ Radius $=\frac{14}{2}=7 \mathrm{~m}$ Height $=5 \mathrm{~m}$ $\therefore$ Area of the metal sheet required $=$ total surface area $=2 \pi \mathrm{r}(\mathrm{h}+\mathrm{r})$ $=2 \times \frac{22}{7} \times 7(5+7) \mathrm{m}^{2}$ $=44 \times 12 \mathrm{~m}^{2}$ $=528 \mathrm{~m}^{2}$...
Read More →The volume of a circular iron rod of length 1 m is 3850 cm
Question: The volume of a circular iron rod of length 1 m is 3850 cm3. Find its diameter. Solution: Volume $=\pi \mathrm{r}^{2} \mathrm{~h}=3850 \mathrm{~cm}^{3}$ Height $=1 \mathrm{~m}=100 \mathrm{~cm}$ Now, radius, $r=\sqrt{\frac{3850}{\pi \times \mathrm{h}}}=\sqrt{\frac{3850 \times 7}{22 \times 100}}=1.75 \times 7=3.5 \mathrm{~cm}$ Volume $=\pi \mathrm{r}^{2} \mathrm{~h}=3850 \mathrm{~cm}^{3}$ Height $=1 \mathrm{~m}=100 \mathrm{~cm}$ Now, radius, $r=\sqrt{\frac{3850}{\pi \times \mathrm{h}}}=\...
Read More →Look at the histogram below and
Question: Look at the histogram below and answer the questions that follow. (a) How many students have height more than or equal to 135 cm, but less than 150 cm ? (b) Which class interval has the least number of students? (c) What is class size? (d) How many students have height less than 140 cm? Solution: (a) Number of students who have height more than or equal to 135 cm, but less than 150 cm = 14+ 18+ 10 = 42 (b) The class interval 150-155 has the least number of students, i.e. 4. (c) We-know...
Read More →Find the height of the cylinder whose volume is 1.54 m
Question: Find the height of the cylinder whose volume is 1.54 m3and diameter of the base is 140 cm? Solution: Diameter $=2 r=140 \mathrm{~cm}$ i.e., radius, $r=70 \mathrm{~cm}=0.7 \mathrm{~m}$ Now, volume $=\pi \mathrm{r}^{2} \mathrm{~h}=1.54 \mathrm{~m}^{3}$ $\Rightarrow \frac{22}{7} \times 0.7 \times 0.7 \times h=1.54$ $\therefore h=\frac{1.54 \times 7}{0.7 \times 0.7 \times 22}=\frac{154 \times 7}{154 \times 7}=1 \mathrm{~m}$...
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: $\frac{5-1}{2-3 \mathrm{i}}$ Solution: $=\frac{5-i}{2-3 i} \times \frac{2+3 i}{2+3 i}$ $=\frac{10-3 i^{2}+13 i}{4-9 i^{2}}$ $=\frac{+13 i+13}{13}$ = i + 1 Let Z = 1 + i = r(cos + isin) Now , separating real and complex part , we get 1 = rcos .eq.1 1 = rsin eq.2 Squaring and adding eq.1 and eq.2, we get $2=r^{2}$ Since r is always a positive no., therefore, $r=\sqrt{2}$ Hence its modu...
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