A toy is in the form of a cylinder with hemispherical ends.
Question: A toy is in the form of a cylinder with hemispherical ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, then find the cost of painting the toy at the rate of 70 paise per sq cm. Solution: We have, the total height of the toy $=90 \mathrm{~cm}$ and the radius of the toy, $r=\frac{42}{2}=21 \mathrm{~cm}$ Also, the height of the cylinder, $h=90-42=48 \mathrm{~cm}$ Now, the total surface area of the toy $=$ CSA of cylinder $+2 \times$ CSA of a hemisphere $=2 \pi r h+...
Read More →Anil can do a piece of work in 5 days and Ankur in 4 days.
Question: Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together? Solution: Time taken by Anil to do the work $=5$ days Time taken by Ankur to do the work $=4$ days $\therefore$ Work done by Anil in 1 day $=\frac{1}{5}$ Work done by Ankur in 1 day $=\frac{1}{4}$ $\therefore$ Work done by Anil and Ankur in one day $=\frac{1}{5}+\frac{1}{4}$ $=\frac{4+5}{20}=\frac{9}{20}$ Thus, Anil and Ankur can do the work in $\frac{20}{9}$ d...
Read More →A ΔABC can be constructed in
Question: A ΔABC can be constructed in which B = 60, C =45, and AB + BC + CA = 12 cm. Solution: True We know that, sum of angles of a triangle is 180. A + B + C = 180 Here, B + C = 60+ 45 = 105 180, Thus, ΔABC with given conditions can be constructed....
Read More →Rohan can paint
Question: Rohan can paint $\frac{1}{3}$ of a painting in 6 days. How many days will he take to complete the painting? Solution: Rohan can paint $\frac{1}{3} \mathrm{rd}$ of a painting in 6 days. $\therefore$ Time taken by Rohan to complete the painting $=(6 \times 3)$ days $=18$ days....
Read More →A ΔABC can be constructed in
Question: A ΔABC can be constructed in which B =105, C = 90 and AB +BC + CA = 10 cm. Solution: False Here, B = 105, C = 90 and AB + BC + CA = 10cm We know that, sum of angles of a triangle is 180. A + B + C = 180 Here, B + C = 105+90 = 195 180 which is not true. Thus, ΔABC with given conditions cannot be constructed....
Read More →Rohan can paint
Question: Rohan can paint $\frac{1}{3}$ of a painting in 6 days. How many days will he take to complete the painting? Solution: Rohan can paint $\frac{1}{3} \mathrm{rd}$ of a painting in 6 days. $\therefore$ Time taken by Rohan to complete the painting $=(6 \times 3)$ days $=18$ days....
Read More →A vessel is in the form of a hemispherical bowl surmounted
Question: A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity. Solution: Diameter of the hemisphere = 21 cmTherefore, radius of the hemisphere = 10.5 cm Volume of the hemisphere $=\frac{2}{3} \pi \mathrm{r}^{3}=\frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5=2425.5 \mathrm{~cm}^{3}$ Height of the cylinder $=$ Total height of the vessel-Radius o...
Read More →ΔABC can be constructed in
Question: ΔABC can be constructed in which BC = 6 cm, C = 30 and AC AB =4 cm. Solution: True We know that, a triangle can be constructed if sum of its two sides is greater than third side. i.e., in ΔABC, AB + BC AC = BC AC AB = 6 4, which is true, so ΔABC with given conditions can be constructed....
Read More →Rakesh can do a piece of work in 20 days.
Question: Rakesh can do a piece of work in 20 days. How much work can he do in 4 days? Solution: It is given that Rakesh can do a piece of work in 20 days. $\therefore$ Rakesh 's 1 day's work $=\frac{1}{20}$ $\therefore$ Rakesh's work for 4 days $=\frac{4}{20}=\frac{1}{5}$ Thus, he can do $\frac{1}{5}$ th of the work in 4 days....
Read More →ΔABC can be constructed in which AB = 5 cm,
Question: ΔABC can be constructed in which AB = 5 cm, A = 45 and BC + AC = 5 cm. Solution: False We know that, a triangle can be constructed, if sum of its two sides is greater than third side. Here, BC + AC = AB = 5 cm So, ΔABC cannot be constructed....
Read More →A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream.
Question: A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, then find the radius of the ice-cream cone. Solution: We have, the base radius of the cylindrical container, $R=6 \mathrm{~cm}$, the height of the container, $H=15 \mathrm{~cm}$, Let the base radius and the height of the ice - cream cone b...
Read More →An angle of 42.5° can be constructed.
Question: An angle of 42.5 can be constructed. Solution: False We know that, 42.5 = x 85 and an angle of 85 cannot be constructed with the help of ruler and compass....
Read More →An angle of 52.5° can be constructed.
Question: An angle of 52.5 can be constructed. Solution: True To construct an angle of 52.5 firstly construct an angle of 90, then construct an angle of 120 and then plot an angle bisector of 120 and 90 to get an angle 105 (90 + 15). Now, bisect this angle to get an angle of 52.5...
Read More →The construction of a ΔABC,
Question: The construction of a ΔABC, given that BC = 3 cm, C = 60 is possible when difference of AB and AC is equal to (a)3.2 cm (b)3.1 cm (c)3 cm (d)2.8 cm Solution: (d)Given, BC = 3 cm and C=60 We know that, the construction of a triangle is possible, if sum of two sides is greater than the third side of the triangle i.e., AB+ BC AC = BC AC AB = 3 AC AB So, if AC AB = 2.8 cm, then construction of ΔABC with given conditions is possible....
Read More →In 15 days, the earth picks up
Question: In 15 days, the earth picks up 1.2 108kg of dust from the atmosphere. In how many days it will pick up 4.8 108kg of dust? Solution: Let $x$ be the number of days taken by the earth to pick up $4.8 \times 10^{8} \mathrm{~kg}$ of dust. Since the amount of dust picked up by the earth and the number of days are in direct variation, we get: $\frac{15}{x}=\frac{1.2 \times 10^{8}}{4.8 \times 10^{8}}$ $\Rightarrow x=15 \times \frac{4.8}{1.2}$ $\Rightarrow x=60$ Thus,the requirednumberofdayswil...
Read More →The construction of ΔABC,
Question: The construction of ΔABC, given that BC = 6 cm, B = 45 is not possible when difference of AB and AC is equal to (a)6.9 cm (b)5.2 cm (c)5.0 cm (d)4.0 cm Solution: (a)Given, BC = 6 cm and B= 45 We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle. i.e., AB + BC AC = BC AC AB = 6 AC-AB So, if AC AB= 6.9 cm, then construction of ΔABC with given conditions is not possible....
Read More →A toy is in the shape of a cone mounted on a hemisphere of same base radius
Question: A toy is in the shape of a cone mounted on a hemisphere of same baseradius. If the volume of the toy is 231 cm3and its diameter is 7 cm, then find the height of the toy. Solution: We have, Base radius of cone $=$ Base radius of hemisphere $=r=\frac{7}{2}=3.5 \mathrm{~cm}$, As, the volume of the toy $=231 \mathrm{~cm}^{3}$ $\Rightarrow$ Volume of cone $+$ Volume of hemisphere $=231$ $\Rightarrow \frac{1}{3} \pi r^{2} h+\frac{2}{3} \pi r^{3}=231$ $\Rightarrow \frac{1}{3} \pi r^{2}(h+2 r)...
Read More →In 10 days, the earth picks up
Question: In 10 days, the earth picks up 2.6 108pounds of dust from the atmosphere. How much dust will it pick up in 45 days? Solution: Let the amount of dustpicked up by the earth in 45 days bex pounds. Since the amount of dust picked up by the earth and the number of days are in direct variation, we have: Ratio of the dust picked up by the earth in pounds = ratio of the number of days taken $\Rightarrow \frac{10}{45}=\frac{2.6 \times 10^{8}}{x}$ $\Rightarrow x \times 10=45 \times 2.6 \times 10...
Read More →In 10 days, the earth picks up
Question: In 10 days, the earth picks up 2.6 108pounds of dust from the atmosphere. How much dust will it pick up in 45 days? Solution: Let the amount of dustpicked up by the earth in 45 days bex pounds. Since the amount of dust picked up by the earth and the number of days are in direct variation, we have: Ratio of the dust picked up by the earth in pounds = ratio of the number of days taken $\Rightarrow \frac{10}{45}=\frac{2.6 \times 10^{8}}{x}$ $\Rightarrow x \times 10=45 \times 2.6 \times 10...
Read More →With the help of a ruler and a
Question: With the help of a ruler and a compass it is not possible to construct an angle of (a)37.5 (b)40 (c)22.5 (d)67.5 Solution: (b)With the help of a ruler and a compass, we can construct the angels, 90, 60, 45, 22.5, 30 etc., and its bisector of an angle. So, it is not possible to construct an angle of 40....
Read More →The amount of extension in an elastic spring varies directly with the weight hung on it.
Question: The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm. Solution: Letxcm be the extension produced by the weight of 700 gm. Since the amount of extension in an elastic spring varies and the weight hung on it is in direct variation, we have : $\frac{250}{700}=\frac{3.5}{x}$ $\Rightarrow x \times 250=3.5 \times 700$ $\Rightarrow x=\frac{3.5 \times 7...
Read More →A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face.
Question: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy. Solution: The radius of hemisphere as well as that of base of cone isr= 3.5 cm.The height of toy = 15.5 cm.The height of hemisphere,h= 3.5 cm and height of cone = 15.5-3.5 = 12 cm. Using Pythagoras Theorem, slant height of cone is $l=\sqrt{(3.5)^{2}+(12)^{2}}=\sqrt{12.25+144}=\sqrt{156.25}=12.5 \math...
Read More →The amount of extension in an elastic string varies directly as the weight hung on it.
Question: The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm? Solution: Letxgm be the weight that would produce an extension of 17.4 cm. Since the amount of extension in an elastic string and the weight hung on it are in direct variation, we have: $\frac{150}{x}=\frac{2.9}{17.4}$ $\Rightarrow 17.4 \times 150=2.9 \times x$ $\Rightarrow x=\frac{17.4 \tim...
Read More →A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm
Question: A solid is in the shape of a cone surmounted on a hemisphere, theradius of each of them being 3.5 cm and the total height of the solid is9.5 cm. Find the volume of the solid. Solution: We have, Radius of cone $=$ Radius of hemisphere $=r=3.5 \mathrm{~cm}$ or $\mathrm{AD}=\mathrm{BD}=\mathrm{CD}=3.5 \mathrm{~cm}$, Total height of the solid, $O C=9.5 \mathrm{~cm}$ $\Rightarrow \mathrm{OD}+\mathrm{CD}=9.5$ $\Rightarrow \mathrm{OD}+3.5=9.5$ $\Rightarrow \mathrm{OD}=6 \mathrm{~cm}$ $\Righta...
Read More →In figure, O is the centre of the circle,
Question: In figure, O is the centre of the circle, BD = OD and CD AB. Find CAB. Solution: Given, in the figure $B D=O D, C D \perp A B$. $\ln \triangle O B D$ $B D=O D$[given] $O D=O B$ [both are the radius of circle] $\therefore$ $O B=O D=B D$ Thus, $\triangle O D B$ is an equilateral triangle. $\therefore \quad \angle B O D=\angle O B D=\angle O D B=60^{\circ}$ In $\triangle M B C$ and $\triangle M B D$, $M B=M B$ [common side] $\angle C M B=\angle B M D=90^{\circ}$ and $C M=M D$ [in a circle...
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