The area of a ΔABC is 8 cm2 in
Question: The area of a ΔABC is 8 cm2in which AB = AC = 4 cm and A = 90. Solution: True We have, ABC is a right angled triangle at A, where sides are given AB = AC = 4 cm Area of a triangle ABC = (Base x Height) = x AC x AB= x 4 x 4=8 cm2...
Read More →The area of a triangle with base 4 cm
Question: The area of a triangle with base 4 cm and height 6 cm is 24 cm2. Solution: False We know that, area of a triangle= (Base x Height) Here Base = 4 cm and Height = 6 cm Area of a triangle = x 4 x 6= 12 cm2...
Read More →A solid sphere of radius 3 cm is melted and then cast into small spherical balls, each of diameter 0.6 cm.
Question: A solid sphere of radius 3 cm is melted and then cast into small spherical balls, each of diameter 0.6 cm. Find the number of balls obtained. Solution: Radius of solid sphere $=3 \mathrm{~cm}$ Volume of the sphere $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times 3 \times 3 \times 3 \mathrm{~cm}^{3}$ Radius of each new ball $=0.3 \mathrm{~cm}$ Volume of each new ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} \math...
Read More →Write each of the following as percent:
Question: Write each of the following as percent: (i) $\frac{7}{25}$ (ii) $\frac{14}{625}$ (iii) $\frac{5}{8}$ (iv) 0.8 (v) 0.005 (vi) 3 : 25 (vii) 11 : 80 (viii) 111 : 125 (ix) 13 : 75 (x) 15 : 16 (xi) 0.18 (xii) $\frac{7}{125}$ Solution: (i) $\frac{7}{25}=\frac{7}{25} \times 100$ $=\frac{700}{25}=28 \%$ (ii) $\frac{14}{625}=\frac{14}{25 \times 25} \times 100$ $=\frac{1400}{625}$ $=2.24 \%$ (iii) $\frac{5}{8}=0.625 \times 100$ $=62.5 \%$ (iv) $0.8=\frac{8}{10} \times 100$ $=80 \%$ (v) $0.005=0....
Read More →The edges of a triangular board are 6 cm,
Question: The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm2is (a)Rs. 2.00 (b)Rs. 2.16 (c)Rs. 2.48 (d)Rs. 3.00 Solution: (b) Since,the edges of a triangular board are $a=6 \mathrm{~cm} b=8 \mathrm{~cm}$ and $c=10 \mathrm{~cm}$. Now, semi-perimeter of a triangular board, $s=\frac{a+b+c}{2}$ $=\frac{6+8+10}{2}=\frac{24}{2}=12 \mathrm{~cm}$ Now, area of a triangular board $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula] $=\sqrt{12(12-6)(1...
Read More →How many lead balls, each of radius 1 cm,
Question: How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm? Solution: Radius of the sphere $=R=8 \mathrm{~cm}$ Volume of the sphere $=\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi \times 8 \times 8 \times 8=\frac{4}{3} \pi \times 512 \mathrm{~cm}^{3}$ Radius of each new ball $=r=1 \mathrm{~cm}$ Volume of each ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times 1 \times 1 \times 1=\frac{4}{3} \pi \times 1 \mathrm{~cm}^{3}$ Total number of new balls that can be m...
Read More →A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm.
Question: A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm. Find the number of cubes so formed. Solution: Diameter of the spherical ball= 21 cm Radius of the ball $=\frac{21}{2} \mathrm{~cm}$ Volume of spherical ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}=11 \times 21 \times 21=4851 \mathrm{~cm}^{3}$ Volume of each cube $=1^{3}=1 \mathrm{~cm}^{3}$ Number of cubes $=\frac{\te...
Read More →A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm.
Question: A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm. Find the number of cubes so formed. Solution: Diameter of the spherical ball= 21 cm Radius of the ball $=\frac{21}{2} \mathrm{~cm}$ Volume of spherical ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}=11 \times 21 \times 21=4851 \mathrm{~cm}^{3}$ Volume of each cube $=1^{3}=1 \mathrm{~cm}^{3}$ Number of cubes $=\frac{\te...
Read More →The area of an isosceles triangle having
Question: The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is (a) $\sqrt{15} \mathrm{~cm}^{2}$ (b) $\sqrt{\frac{15}{2}} \mathrm{~cm}^{2}$ (c) $2 \sqrt{15} \mathrm{~cm}^{2}$ (d) $4 \sqrt{15} \mathrm{~cm}^{2}$ Solution: (a) Let $A B C$ be an isosceles triangle in which $A B=A C=4 \mathrm{~cm}$ and $B C=2 \mathrm{~cm}$. In right angled $\triangle A D B$, $A B^{2}=A D^{2}+B D^{2}$ [by Pythagoras theorem] $\Rightarrow \quad(4)^{2}=A D^{2}+1$ $\Rightarr...
Read More →Find the inverse of each of the following matrices by using elementary row transformations:
Question: Find the inverse of each of the following matrices by using elementary row transformations: $\left[\begin{array}{cc}1 6 \\ -3 5\end{array}\right]$ Solution: $A=\left[\begin{array}{ll}1 6\end{array}\right.$ $\left.\begin{array}{ll}-3 5\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ll}1 6\end{array}\right.$ $-3 \quad 5]=\left[\begin{array}{ll}1 0\end{array}\right.$ $0 \quad$ 1] $A$ $\Rightarrow\left[\begin{array}{ll}1 6\end{array}\right.$ $-3+3 \quad 5+18]=\left[\beg...
Read More →After a 20% hike, the cost of Chinese Vase is Rs 2000.
Question: After a 20% hike, the cost of Chinese Vase is Rs 2000. What was the original price of the object? Solution: Let the original price of the object be $x$. According to the question, we have: $20 \%$ of $x+x=2000$ $\Rightarrow \frac{20 x}{100}+x=2000$ $\Rightarrow \frac{120}{100} x=2000$ $\Rightarrow x=\frac{200000}{120}$ $\Rightarrow x=1666.67$ Thus, the original price of the object was Rs $1,666.67$....
Read More →A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm.
Question: A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone. Solution: Radius of hemisphere = 9 cm Volume of hemisphere $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi \times 9 \times 9 \times 9 \mathrm{~cm}^{3}$ Height of cone = 72 cmLet the radius of the cone be r cm. Volume of the cone $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \pi \mathrm{r}^{2} \times 72 \mathrm{~cm}^{3}$ The volumes of the hemisphere and co...
Read More →Rs 3500 is to be shared among three people so that the first person gets 50% of the second,
Question: Rs 3500 is to be shared among three people so that the first person gets 50% of the second, who in turn gets 50% of the third. How much will each of them get? Solution: Let $x, y$ and $z$ be the amounts received by the first, second and the third person, respectively. We have : $x=50 \%$ of $y$ $=\frac{50}{100} y=\frac{1}{2} y$ $\therefore y=2 x$ Again, $y=50 \%$ of $z$ $=\frac{1}{2} z$ $\therefore z=2 y$ $=2(2 x)$ $=4 x$ Also, $x+y+z=3500$ Substituting the values of $z$ and $y$, we ge...
Read More →A spherical shell of lead, whose external and internal diameters are 24 cm and 18 cm,
Question: A spherical shell of lead, whose external and internal diameters are 24 cm and 18 cm, is melted and recast into a right circular cylinder 37 cm high. Find the diameter of the base of the cylinder. Solution: External diameter of the shell = 24 cmExternal radius of the shell = 12 cmInternaldiameter of the shell = 18 cmInternal radius of the shell = 9 cm Volume of the shell $=\frac{4}{3} \pi\left(12^{3}-9^{3}\right)=\frac{4}{3} \pi(1728-729)=\frac{4}{3} \pi \times(999)=4 \pi \times(333) \...
Read More →Shikha's income is 60% more than that of Shalu.
Question: Shikha's income is 60% more than that of Shalu. What percent is Shalu's income less than Shikha's? Solution: Let Shalu's income be Rs $x$. $\therefore$ Shikha's income $=\operatorname{Rs}\left(\mathrm{x}+\frac{60 x}{100}\right)=\mathrm{Rs} \frac{160 x}{100}=\mathrm{Rs} \frac{16 x}{10}$ Difference in the incomes of Shikha and Shalu $=\frac{16 x}{10}-x=\frac{16 x-10 x}{10}=$ Rs $\frac{6 x}{10}$ Percentage of the difference in the incomes of Shikha and Shalu to that of Shikha's income $=\...
Read More →A spherical ball of radius 3 cm is melted and recast into three spherical balls.
Question: A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball. Solution: Radius of sphere = 3 cmRadius of first ball = 1.5 cmRadius of second ball = 2 cmLet radius of the third ball bercm.Volume of third ball = Volume of original sphere - Sum of volumes of other two balls $=\frac{4}{3} \pi \times 3^{3}-\left(\frac{4}{3} \pi \times \frac{3}{2}^{3}+\frac{4}{3} \pi \times 2^{3}\right)...
Read More →A spherical ball of radius 3 cm is melted and recast into three spherical balls.
Question: A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball. Solution: Radius of sphere = 3 cmRadius of first ball = 1.5 cmRadius of second ball = 2 cmLet radius of the third ball bercm.Volume of third ball = Volume of original sphere - Sum of volumes of other two balls $=\frac{4}{3} \pi \times 3^{3}-\left(\frac{4}{3} \pi \times \frac{3}{2}^{3}+\frac{4}{3} \pi \times 2^{3}\right)...
Read More →The sides of a triangle are 35 cm,
Question: The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude (a)1675 cm (b)1o75 cm (c)2475 cm (d)28 cm Thinking Process (i) First, determine the semi-perimeter, s and then determine the area of triangle by using Herons formula. (ii) For the longest altitude, take base as the smallest side. Apply the formula, Area = x Base x Altitude (iii) Equate the area obtained using the two formulas and obtain the required height. Solution: (c) Let $A B C$ be ...
Read More →A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm.
Question: A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm. Find the height of the cone. Solution: Diameter of cannon ball $=28 \mathrm{~cm}$ Radius of the cannon ball $=14 \mathrm{~cm}$ Volume of ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times(14)^{3} \mathrm{~cm}^{3}$ Diameter of base of cone $=35 \mathrm{~cm}$ Radius of base of cone $=\frac{35}{2} \mathrm{~cm}$ Let the height of the cone be $\mathrm{h...
Read More →Mohan's income is Rs 15500 per month.
Question: Mohan's income is Rs 15500 per month. He saves 11% of his income. If his income increases by 10%, then he reduces his saving by 1%, how much does he save now? Solution: Mohan's saving $=11 \%$ of 15500 $=\frac{11}{100} \times 15500$ $=$ Rs. 1705 It is given that Mohan's income increases by $10 \%$. $\therefore$ Increase in income $=\frac{10}{100} \times 15500=\mathrm{Rs} 1550$ Increased income $=$ Rs $(15500+1550)=$ Rs 17050 Now, percentage of saving $=(11-1)=10 \%$ $\therefore$ Saving...
Read More →In the new budget, the price of petrol rose by 10%.
Question: In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase? Solution: We have to reduce the consumption such that the expenditure does not increase. For this, we use the following formula: $\left(\frac{r}{r+100}\right) \times 100$, where $\mathrm{r}=$ percentage rise in the price of the commodity $\therefore$ Percentage reduction in the consumption $=\left(\frac{10}{10+100}\right) \times 100=\left(\f...
Read More →A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones,
Question: A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed. Solution: Diameter of sphere $=21 \mathrm{~cm}$ Radius of sphere $=\frac{21}{2} \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4 \times 21 \times 21 \times 21 \pi}{3 \times 2 \times 2 \times 2}=\frac{21 \times 21 \times 21 \pi}{3 \times 2} \mathrm{~cm}^{3}$ Diameter of the cone $=3.5 \mathrm{~c...
Read More →If the area of an equilateral triangle is
Question: If the area of an equilateral triangle is 163 cm2, then the perimeter of the triangle is (a)48 cm (b)24 cm (c)12 cm (d)36 cm Solution: (b)Given, area of an equilateral triangle = 163 cm2 Area of an equilateral triangle = 3/4 (side)2 3/4(Side)2= 163 = (Side)2= 64 = Side =8 cm [taking positive square root because side is always positive] Perimeter of an equilateral triangle = 3 x Side= 3 x 8 = 24 cm Hence, the perimeter of an equilateral triangle is 24 cm....
Read More →Ankit was given an increment of 10% on his salary.
Question: Ankit was given an increment of 10% on his salary. His new salary is Rs 3575. What was his salary before increment? Solution: Let the initial salary be Rs $x$. We know that: Salary before increment $+$ increment given on salary $=$ new salary $\therefore x+10 \%$ of $x=3575$ $\Rightarrow x+\frac{10}{100} x=3575$ $\Rightarrow 100 x+10 x=357500$ $\Rightarrow 110 x=357500$ $\Rightarrow x=\frac{357500}{110}$ $\Rightarrow x=3250$ $\therefore$ Salary before increment $=$ Rs 3,250...
Read More →A hemispherical bowl of internal diameter 30 cm contains some liquid.
Question: A hemispherical bowl of internal diameter 30 cm contains some liquid. This liquid is to be poured into cylindrical bottles of diameter 5 cm and height 6 cm each. Find the number of bottles required. Solution: Inner diameter of the bowl $=30 \mathrm{~cm}$ Inner radius of the bowl $=\frac{30 \mathrm{~cm}}{2}=15 \mathrm{~cm}$ Inner volume of the bowl $=$ Volume of liquid $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \times \pi \times 15^{3} \mathrm{~cm}^{3}$ Radius of each bottle $=2.5 \mathrm{~cm}...
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