The length of each side of an equilateral
Question: The length of each side of an equilateral triangle having an area of 93 cm2is (a)8 cm (b)36 cm (c)4 cm (d)6 cm Solution: (d)Given, area of an equilateral triangle = 93 cm2 Area of an equilateral triangle = 3/4(Side)2 = 3/4 (Side)2= 93 = (Side)2= 36 Side = 6 cm[taking positive square root because side is always positive] Hence, the length of an equilateral triangle is 6 cm....
Read More →The population of a town increases by 10% annually.
Question: The population of a town increases by 10% annually. If the present population is 22000, find its population a year ago. Solution: Let the population of the town one year ago $=x$ Now, it is given that population of the town increases by $10 \%$ annually. Present population $=x+10 \%$ of $x$ $=x+\frac{10}{100} \times x$ $=x+\frac{x}{10}$ $=\frac{11 x}{10}$ But present population of the town $=22000$ So, $\frac{11 x}{10}=22000$ $\Rightarrow 11 x=220000$ $\Rightarrow x=\frac{220000}{11}$ ...
Read More →The area of an equilateral triangle
Question: The area of an equilateral triangle with side 23 cm is (a)5.196 cm2 (b)0.866 cm2 (c)3.496 cm2 (d)1.732 cm2 Solution: (a)Given, side of an equilateral triangle is 23 cm. Area of an equilateral triangle = 3/4 (Side)2 = 3/4 (23)2= (3/4) x 4 x 3 = 33 = 3 x 1.732 = 5.196 cm2 Hence, the area of an equilateral triangle is 5.196 cm2....
Read More →A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of uniform thickness and length 10 m.
Question: A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of uniform thickness and length 10 m. Find the thickness of the wire. Solution: We have, the radius of the copper rod, $R=\frac{2}{2}=1 \mathrm{~cm}$, the height of the copper rod, $H=10 \mathrm{~cm}$ and the height of the wire, $h=10 \mathrm{~m}=1000 \mathrm{~cm}$ Let the radius of the wire be $r$. As, Volume of the wire $=$ Volume of the rod $\Rightarrow \pi r^{2} h=\pi R^{2} H$ $\Rightarrow r^{2} h=R^{2} H$ $\Righta...
Read More →The sides of a triangle are 56 cm,
Question: The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is (a)1322 cm2 (b)1311 cm2 (c)1344 cm2 (d)1392cm2 Thinking Process (i) First, determine the semi-perimeter of a triangle by using the formula, s = (a + b +c)/2 (ii) Further, determine the area of triangle by using the formula, area of triangle (Fierons formula) = $\sqrt{s(s-a)(s-b)(s-c)}$ Solution: (c) Since, the three sides of a triangle are $a=56 \mathrm{~cm}, b=60 \mathrm{~cm}$ and $c=52 \mathrm{...
Read More →The population of a town increases by 10% annually.
Question: The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years? Solution: Present population $=60000$ It increases by $10 \%$ annually. $\therefore$ Increase in the population in the first year $=10 \%$ of $60000=\frac{10}{100} \times 60000$ $=6000$ $\therefore$ Population after 1 year $=60000+6000=66000$ Increase in the population in the second year, $10 \%$ of $66000=\frac{10}{100} \times 66000$ $=6600$ Thus, populati...
Read More →The value of a machine depreciates every year by 5%.
Question: The value of a machine depreciates every year by 5%. If the present value of the machine be Rs 100000, what will be its value after 2 years? Solution: It is given that the value of the machine depreciates by $5 \%$ every year. Present value of the machine $=$ Rs 100000 $\therefore$ For the first year, $5 \%$ of $100000=\frac{5}{100} \times 100000$ $=$ Rs 5000 $\therefore$ Value of the machine after one year $=100000-5000=$ Rs 95000 Value of the machine in the second year $=$ Rs 95000 $...
Read More →The perimeter of an equilateral triangle is 60 m.
Question: The perimeter of an equilateral triangle is 60 m. The area is (a)103 m2 (b)153 m2 (c)203 m2 (d)1003m2 Thinking Process (i) First, determine the side of an equilateral by usingformula, perimeter=3x. (ii) Further, substitute the value of x in the formula, area of an equilateral triangle = 3/4 (a)2 and simplify it. Solution: (d)Let each side of an equilateral be x. Then, perimeter of an equilateral triangle = 60 m x + x + x = 60 = 3x = 60 = x = 60/3 = 20 m Area of an equilateral triangle ...
Read More →Aman's income is 20% less than that of Anil.
Question: Aman's income is 20% less than that of Anil. How much percent is Anil's income more than Aman's income? Solution: Let Anil's income be $x$. Then, Aman's income $=x-\frac{20 x}{100}$ $=\frac{8 x}{10}$ Difference in the incomes of Anil and Aman $=x-\frac{8 x}{10}$ $=\frac{2 x}{10}$ $\therefore$ Percentage of the difference in the incomes of Anil and Aman to that of Aman's incom $e=\left(\frac{2 x}{10}\right) /\left(\frac{8 x}{10}\right)$$\times 100$ $=\frac{1}{4} \times 100$ $=25 \%$...
Read More →A certain school has 300 students, 142 of whom are boys.
Question: A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female? Solution: Total number of female student $s=300-142$ $=158$ Total number of female teachers $=30-12$ $=18$ Total number of females $=158+18$ $=176$ Total population of the school $=300+30$ =330 Percentage of females $=\frac{176}{330} \times 100$ $=\frac{160}{3} \%$ or $53.33 \%$...
Read More →A certain school has 300 students, 142 of whom are boys.
Question: A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female? Solution: Total number of female student $s=300-142$ $=158$ Total number of female teachers $=30-12$ $=18$ Total number of females $=158+18$ $=176$ Total population of the school $=300+30$ $=330 Percentage of females $=\frac{176}{330} \times 100$ $=\frac{160}{3} \%$ or $53.33 \%$...
Read More →A motorist travelled 122 kilometres before his first stop.
Question: A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride? Solution: Let the length of total ride be $x \mathrm{~km}$. Then, $10 \%$ of $x=122$ $\Rightarrow \frac{10}{100} x=122$ $\Rightarrow x=1220$ $\therefore$ The length of total ride is $1,220 \mathrm{~km}$....
Read More →An alloy contains 32% copper, 40% nickel and rest zinc.
Question: An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy. Solution: Percentage of copper in the alloy $=32$ Percentage of nickel in the alloy $=40$ Percentage of zinc in the alloy $=100-32-40=28$ $\therefore$ Amount of zinc in $1 \mathrm{~kg}$ of the alloy $=(0.28 \times 1) \mathrm{kg}$ $=280 \mathrm{gm}$...
Read More →An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper.
Question: An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy? Solution: Composition of the alloy $=15$ parts of tin $+105$ parts of copper $=120$ parts $\therefore$ Percentage of tin $=\frac{15}{120} \times 100$ $=12.5 \%$ Also, percentage of copper $=\frac{105}{120} \times 100$ $=87.5 \%$...
Read More →Gunpowder contains 75% nitre and 10% sulphur.
Question: Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur? Solution: Let the amount of gunpowder that contains $9 \mathrm{~kg}$ nitre be $\mathrm{x} \mathrm{kg}$. i.e., $\frac{75}{100} \times \mathrm{x}=9$ $\Rightarrow \frac{100}{75}=\frac{x}{9}$ $\Rightarrow x=\frac{900}{75}$ $\Rightarrow x=12$ $\therefore$ The amount of gunpowder containing $9 \mathrm{~kg}$ nitre is $12 \mathrm{~kg}$....
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: $\left[\begin{array}{ll}5 2 \\ 2 1\end{array}\right]$ Solution: $A=\left[\begin{array}{ll}5 2\end{array}\right.$ $\left.\begin{array}{ll}2 1\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ll}5 2\end{array}\right.$ $2 \quad 1]=\left[\begin{array}{ll}1 0\end{array}\right.$ $0 \quad 1] A \quad\left[\right.$ Applying $\left.\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-2 \mathrm{R}_{...
Read More →The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively.
Question: The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed. Solution: Internal diameter of the hemispherical shell = 6 cmTherefore, internal radius of the hemispherical shell = 3 cmExternal diameter of the hemispherical shell = 10 cmExternal radius of the hemispherical shell = 5 cm Volume of hemispherical shell $=\frac{2}{3} \pi\left(5^{3}-3...
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: $\left[\begin{array}{cc}7 1 \\ 4 -3\end{array}\right]$ Solution: $A=\left[\begin{array}{ll}7 1\end{array}\right.$ $4-3]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ll}7 1\end{array}\right.$ $4-3]=\left[\begin{array}{ll}1 0\end{array}\right.$ $\begin{array}{ll}0 1] A\end{array}$ $4-3]=\left[\begin{array}{ll}\frac{1}{7} 0\end{array}\right.$ $0 \quad 1] A$ (Applying $R_{1} \rightarrow \frac{1}...
Read More →Rohit deposits 12% of his income in a bank.
Question: Rohit deposits 12% of his income in a bank. He deposited Rs 1440 in the bank during 1997. What was his total income for the year 1997? Solution: Let the total income for the year 1997 be Rs $x$. i.e., $12 \%$ of $x=1440$ $\Rightarrow \frac{12}{100} x=1440$ $\Rightarrow x=\frac{144000}{12}$ $\Rightarrow x=12000$ $\therefore$ Rohit's total income was Rs 12,000 ....
Read More →Radha earns 22% of her investment.
Question: Radha earns 22% of her investment. If she earns Rs 187, then how much did she invest? Solution: Let the investment be Rs $x$. i. e., $22 \%$ of $x=187$ $\Rightarrow \frac{22}{100} x=187$ $\Rightarrow x=\frac{18700}{22}$ $=850$ $\therefore$ Radha invest $s$ Rs $850 .$...
Read More →An isosceles right triangle has area 8 cm2.
Question: An isosceles right triangle has area 8 cm2. The length of its hypotenuse is (a)32 cm (b)16 cm (c)48 cm (d)24 Solution: (a) Given, area of an isosceles right triangle $=8 \mathrm{~cm}^{2}$ Area of an isosceles triangle $=1 / 2$ (Base $x$ Height) $\Rightarrow 8=1 / 2$ (Base $\times$ Base $)$ $[\therefore$ base $=$ height, as triangle is an isosceles triangle $]$ $\Rightarrow(\text { Base })^{2}=16 \Rightarrow$ Base $=4 \mathrm{~cm}$ In ΔABC, using Pythagoras theore AC2= AB2+ BC2= 42+ 42=...
Read More →A cricketer hit 120 runs in 150 balls during a test match.
Question: A cricketer hit 120 runs in 150 balls during a test match. 20% of the runs came in 6's, 30% in 4's, 25% in 2's and the rest in 1's. How many runs did he score in (i) 6's (ii) 4's (iii) 2's (iv) singles What % of his shots were scoring ones? Solution: (i) Let the cricketer score w runs in 6's. $\therefore 20 \%$ of $120=w$ $\Rightarrow \frac{20}{100} \times 120=w$ $\Rightarrow w=24$ (ii) Let the cricketer score $x$ runs in 4 's. $\therefore 30 \%$ of $120=x$ $\Rightarrow \frac{30}{100} ...
Read More →The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively.
Question: The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder. Solution: We have, the internal base radius of spherical shell, $r_{1}=3 \mathrm{~cm}$, the external base radius of spherical shell, $r_{2}=5 \mathrm{~cm}$ and the base radius of solid cylinder, $r=\frac{14}{2}=7 \mathrm{~cm}$ Let the height of the cylinder be $h$. As, Volume of solid c...
Read More →A cricketer scored a total of 62 runs in 96 balls.
Question: A cricketer scored a total of 62 runs in 96 balls. He hit 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in (i) Sixes (ii) fours (iii) twos (iv) singles Solution: (i) The cricketer hits 3 sixes. i.e., $3 \times 6=18$ $\therefore \frac{18}{62} \times 100=29.03 \%$ (ii) The cricketer hits 8 fours. i. e., $8 \times 4=32$ $\therefore \frac{32}{62} \times 100=51.61 \%$ (iii) The cricketer hits 2 twos. i. e., $2 \times 2=4$ $\therefore \frac{4}{62} \times 100=...
Read More →A solid metal cone with base radius of 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each.
Question: A solid metal cone with base radius of 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls thus formed. Solution: Radius of the cone = 12 cmHeight of the cone = 24 cm Volume $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \pi \times 12 \times 12 \times 24=48 \times 24 \times \pi \mathrm{cm}^{3}$ Radius of each ball $=3 \mathrm{~cm}$ Volume of each ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times 3 \times 3 ...
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