Asha got 86.875% marks in the annual examination.
Question: Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination. Solution: Let $x$ be the total number of marks of the examination. Then, $86.875 \%$ of $x=695$ $\Rightarrow \frac{86.875}{100} x=695$ $\Rightarrow x=\frac{695 \times 100}{86.875}$ $=800$ $\therefore$ The total number of marks of the examination is 800 ....
Read More →Find the cost of laying grass in
Question: Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2.Thinking Process (i) First, determine the semi-perimeter s by using formula $s=\frac{a+b+c}{2}$ and then determine the area of a triangle by using Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$ (ii) Cost of laying grass = Area of field $\left(\right.$ in $\left.m^{2}\right) \times\left(\right.$ Cost of laying grass in $\left.1 m^{2}\right)$ Solution: Since, $A B C$ is a triangular fi...
Read More →A coolie deposits Rs 150 per month in his post office Savings Bank account.
Question: A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income. Solution: Let his monthly income be $\mathrm{Rs} \mathrm{x}$. Then, $15 \%$ of $x=150$ $\Rightarrow \frac{15}{100} x=150$ $\Rightarrow x=\frac{150 \times 100}{15}$ $=1000$ $\therefore$ His monthly income is Rs 1000 ....
Read More →Water is flowing through a cylindrical pipe of internal diameter 2 cm,
Question: Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour. Solution: We have, the internal radius of the cylindrical pipe, $r=\frac{2}{2}=1 \mathrm{~cm}$ and the base radius of cylindrical tank, $R=40 \mathrm{~cm}$, Also, the rate of water flow, $h=0.4 \mathrm{~m} / \mathrm{s}=40 \mathrm{~cm} / \mathrm{s}$ Let the rise in level of ...
Read More →A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm.
Question: A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres. Solution: We have, height of cone, $h=60 \mathrm{~cm}$, the base radius of cone, $r=30 \mathrm{~cm}$, the height of cylinder, $H=180 \mathrm{~cm}$ and the base radius of the cylinder, $R=60 \mathrm{~cm}$ Now, Volume of water left in the cylinder = Volume of cylinder-Volume o...
Read More →In a triangle, the sides are given as 11 cm,
Question: In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having 12 cm. Solution: True Since the sides of a triangle are $a=11 \mathrm{~cm}, b=12 \mathrm{~cm}$ and $c=13 \mathrm{~cm}$. Now, semi-perimeter, $\quad s=\frac{a+b+c}{2}$ $=\frac{11+12+13}{2}=\frac{36}{2}=18 \mathrm{~cm}$ Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula] $=\sqrt{18(18-11)(18-12)(18-13)}$ $=\sqrt{18 \times 7 \times 6 \times 5...
Read More →x is 5% of y, y is 24% of z.
Question: xis 5% ofy,yis 24% ofz. Ifx= 480, find the values ofyandz. Solution: Given: $x=\frac{5}{100} y$ $\Rightarrow y=\frac{100}{5} x$ $\therefore y=20 x$ Also, $y=\frac{24}{100} z$ $\therefore z=\frac{100}{24} y$ Putting $x=480$ in the above equations, we get: $y=20(480)=9600$ $z=\frac{100}{24} \times 9600=40000$...
Read More →The rain water from a 22 m × 20 m roof drains into a cylindrical vessel of diameter 2 m and height 3.5 m.
Question: The rain water from a $22 \mathrm{~m} \times 20 \mathrm{~m}$ roof drains into a cylindrical vessel of diameter $2 \mathrm{~m}$ and height $3.5 \mathrm{~m}$. If the rain water collected from the roof fills $\frac{4}{5}$ th ofthe cylindrical vessel, then find the rainfall in centimetre. Solution: We have, the length of the roof, $l=22 \mathrm{~m}$, the width of the roof, $b=20 \mathrm{~m}$, the base radius of the cylindrical vessel, $R=\frac{2}{2}=1 \mathrm{~m}$ and the height of the cyl...
Read More →The cost of levelling the ground in the
Question: The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs. 3 per m2 is Rs. 918. Solution: True Let sides of a triangle be $a=51 \mathrm{~m} b=37$ mand $c=20 \mathrm{~m}$ Now, semi-perimeter of triangle, $s=\frac{a+b+c}{2}=\frac{51+37+20}{2}=\frac{108}{2}=54 \mathrm{~m}$ $\therefore \quad$ Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula] $=\sqrt{54(54-51)(54-37)(54-20)}$ $=\sqrt{54 \times 3 \times 17 \times 34}$ $...
Read More →The rain water from a roof of 44 m × 20 m drains into a cylindrical tank having diameter of base 4 m and height 3.5 m.
Question: The rain water from a roof of 44 m20 m drains into a cylindrical tankhaving diameter of base 4 m and height 3.5 m. If the tank is just full, then findthe rainfall in cm. Solution: We have, the length of the roof, $l=44 \mathrm{~m}$, the width of the roof, $b=20 \mathrm{~m}$, the height of the cylindrical tank, $H=3.5 \mathrm{~m}$ and the base radius of the cylindrical tank, $R=\frac{4}{2}=2 \mathrm{~m}$ Let the height of the rainfall be $h$. Now, Volume of rainfall $=$ Volume of cylind...
Read More →Find the number a, if
Question: Find the numbera, if (i) 8.4% ofais 42 (ii) 0.5% ofa is 3 (iii) $\frac{1}{2} \%$ of $a$ is 50 (iv) $100 \%$ of $a$ is 100 Solution: (i) $8.4 \%$ of $a$ is 42 . i. e., $\frac{8.4}{100} \times a=42$ $\Rightarrow a=\frac{42 \times 100}{8.4}$ $\therefore a=500$ (ii) $0.5 \%$ of $a$ is 3 . i. e., $\frac{0.5}{100} \times a=3$ $\Rightarrow a=\frac{3 \times 100}{0.5}$ $\therefore a=600$ (iii) $\frac{1}{2} \%$ of $a$ is 50 . i. e., $\frac{1}{2} \times \frac{1}{100} \times a=50$ $\Rightarrow a=5...
Read More →The area of regular hexagon of side
Question: The area of regular hexagon of side a is the sum of the areas of the five equilateral triangles with side a. Solution: FalseWe know that regular hexagon is divided into six equilateral triangles. Area of regular hexagon of side a = Sum of area of the six equilateral triangles....
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: $\left[\begin{array}{cc}3 10 \\ 2 7\end{array}\right]$ Solution: $A=\left[\begin{array}{ll}3 10\end{array}\right.$ $\left.\begin{array}{ll}2 7\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ll}3 10\end{array}\right.$ $2 \quad 7]=\left[\begin{array}{ll}1 0\end{array}\right.$ $\left.\begin{array}{ll}0 1\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ll}3-2 10-7\end{array}...
Read More →The base and the corresponding altitude
Question: The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm2. Solution: False Given, parallelogram in which base = 10 cm and altitude = 3.5 cm Area of a parallelogram = Base x Altitude = 10 x 3.5 = 35 cm2....
Read More →A hemispherical tank, full of water, is emptied by a pipe at the rate of
Question: A hemispherical tank, full of water, is emptied by a pipe at the rate of $\frac{25}{7}$ litres per second. How much time will it take to empty half the tank if the diameter ofthe base of the tank is 3 m? Solution: We have, the radius of the hemispherical $\operatorname{tank}, r=\frac{3}{2} \mathrm{~m}$ Volume of the hemispherical $\operatorname{tank}=\frac{2}{3} \pi r^{3}$ $=\frac{2}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}$ $=\frac{99}{14} \mathr...
Read More →If the side of a rhombus is 10 cm and
Question: If the side of a rhombus is 10 cm and one diagonal is 16 cm, then area of the rhombus is 96 cm2. Solution: True Given, side of a rhombus PQRS is 10 cm and one of the diagonal is 16 cm. i.e., PQ = QR = RS =SP = 10 cm and PR = 16 cm In ΔPOQ, PQ2= OP2+ OQ2[by Pythagoras theorem] [ since, the diagonal of rhombus bisects each other at 90] = OQ2= PQ2 OP2= (10)2 (8)2 = OQ2=100 64 = 36 = OQ = 6 cm [taking positive square root because length is always positive] SQ =2 x OP = 2 x 6= 12 cm Area of...
Read More →Find: (i) 22% of 120
Question: Find: (i) 22% of 120 (ii) 25% of Rs 1000 (iii) 25% of 10 kg (iv) 16.5% of 5000 metre (v) 135% of 80 cm (vi) 2.5% of 10000 ml Solution: (i) $22 \%$ of $120=\frac{22}{100} \times 120$ $=26.4$ (ii) $25 \%$ of $1000=\frac{25}{100} \times 1000$ $=250$ (iii) $25 \%$ of $10 \mathrm{~kg}=\frac{25}{100} \times 10 \mathrm{~kg}$ $=\frac{25}{10} \mathrm{~kg}=\frac{5}{2} \mathrm{~kg}$ $=2.5 \mathrm{~kg}$ (iv) $16.5 \%$ of $5000 \mathrm{~m}=\frac{16.5}{100} \times 5000 \mathrm{~m}$ $=16.5 \times 50 ...
Read More →The area of the equilateral triangle
Question: The area of the equilateral triangle is 20 3 cm2whose each side is 8 cm. Solution: False Given, side of an equilateral triangle be 8 cm. Area of the equilateral triangle = 3/4 (Side)2 = (3/4) x (8)2= (64/4) 3 [ side = 8 cm] = 16 3cm2...
Read More →The area of the isosceles triangle is
Question: The area of the isosceles triangle is (5/4) 11 cm2if the perimeter is 11 cm and the base is 5 cm. Solution: True Let equal sides of an isosceles triangle be $b$. $\therefore$ Perimeter of a triangie, $\quad 2 s=b+b+5 \quad[\because 2 s=a+b+c]$ $\therefore$$11=2 b+5$ $\Rightarrow \quad 2 b=11-5 \Rightarrow 2 b=6$ $\Rightarrow \quad b=\frac{6}{2}=3 \mathrm{~cm}$ We know that, area of an isosceles triangle $=\frac{a}{4} \sqrt{4 b^{2}-a^{2}}$ Here, sides of triangle are $a=5 \mathrm{~cm}$ ...
Read More →A hemispherical bowl of internal radius 9 cm is full of water.
Question: A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel. Solution: We have, the radius of the hemispherical bowl, $R=9 \mathrm{~cm}$ and the internal base radius of the cylindrical vessel, $r=6 \mathrm{~cm}$ Let the height of the water in the cylindrical vessel be $h$. As, Volume of water in the cylindrical vessel $=$ Volume of hemispherical bowl $\Right...
Read More →Express the following as decimal fractions:
Question: Express the following as decimal fractions: (i) 27% (ii) 6.3% (iii) 32% (iv) 0.25% (v) 7.5% (vi) $\frac{1}{8} \%$ Solution: (i) $27 \%=\frac{27}{100}$ $=0.27$ (ii) $6.3 \%=\frac{6.3}{100}$ $=0.063$ (iii) $32 \%=\frac{32}{100}$ $=0.32$ (iv) $0.25 \%=\frac{0.25}{100}$ $=0.0025$ $(\mathrm{v}) 7.5 \%=\frac{7.5}{100}$ $=0.075$ $(\mathrm{vi}) \frac{1}{8} \%=\frac{1}{8} \times \frac{1}{100}$ $=\frac{1}{800}$ $=0.00125$...
Read More →The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross-section.
Question: The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross-section. If the length of the wire is 108 m, find its diameter. Solution: Diameter of sphere $=18 \mathrm{~cm}$ Radius of the sphere $=9 \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times 9 \times 9 \times 9 \mathrm{~cm}^{3}$ Length of wire $=108 \mathrm{~m}=10800 \mathrm{~cm}$ Let radius of the wire be $r \mathrm{~cm} .$ Volume of the wire $=\pi r^{2} ...
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: Solution: $A=\left[\begin{array}{ll}2 5\end{array}\right.$ $\left.\begin{array}{ll}1 3\end{array}\right]$ We know $A=I A$ $\Rightarrow\left[\begin{array}{ll}2-1 5-3\end{array}\right.$ $\left.\begin{array}{ll}1 3\end{array}\right]=\left[\begin{array}{ll}1-0 0-1\end{array}\right.$ $0 \quad 1] A \quad\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}-R_{2}\right]$ $\Rightarrow\left[\begin{array...
Read More →Convert the following percentages to fractions and ratios:
Question: Convert the following percentages to fractions and ratios: (i) 25% (ii) 2.5% (iii) 0.25% (iv) 0.3% (v) 125% Solution: (i) $25 \%=\frac{25}{100}$ $=\frac{5}{20}=\frac{1}{4}$ $\therefore$ Ratio $=1: 4$ (ii) $2.5 \%=\frac{2.5}{100}$ $=\frac{25}{1000}=\frac{5}{200}$ $=\frac{1}{40}$ $\therefore$ Ratio $=1: 40$ (iii) $0.25 \%=\frac{0.25}{100}$ $=\frac{25}{10000}=\frac{5}{2000}$ $=\frac{1}{400}$ $\therefore$ Ratio $=1: 400$ (iv) $0.3 \%=\frac{0.3}{100}$ $=\frac{3}{1000}$ $\therefore$ Ratio $=...
Read More →The diameter of a sphere is 42 cm.
Question: The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire. Solution: Diameter of sphere $=42 \mathrm{~cm}$ Radius of sphere $=21 \mathrm{~cm}$ Volume of sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times 21 \times 21 \times 21 \mathrm{~cm}^{3}$ Diameter of wire $=2.8 \mathrm{~cm}$ Radius of wire $=1.4 \mathrm{~cm}$ Let the length of the wire be $/ \mathrm{cm}$. Volume of the wire $=\pi r^{2} l=\pi \times 1.4 \ti...
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