Give reasons for the following observation: <br/><br/>The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food you have to go close.
Solution The smell of hot sizzling food reaches severed meters away, as the particles of hot food have more kinetic energy and hence the rate of diffusion is more than the particles of cold food....
Read More →Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution: Yes. According to Euclid’s 5th postulate, when n line falls on l and m and if $\angle \mathrm{l}+\angle 2180^{\circ}$, then $\angle 3+\angle 4180^{\circ}$, producing line $/$ and $m$ further will meet in the side of $\angle 1$ and $\angle 2$ which is less than $180^{\circ}$. If $\angle \mathrm{l}+\angle 2=180^{\circ}$, then $\angle 3+\angle 4=180^{\circ}$ The lines / and $m$ neither meet at the side of $\angle 1$ and $\angle 2$ nor at the side of $\angle 3$ and $\angle 4$. This means t...
Read More →How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution: Two lines are said to be parallel if they are equidistant from one other and they do not have any point of intersection. In order to understand it easily, let us take any line l and a point P not on l. Then, by Playfair’s axiom (equivalent to the fifth postulate), there is a unique line m through P which is parallel to l. The distance of a point from a line is the length of the perpendicular from the point to the line. Let AB be the distance of any point on m from l and CD be the dista...
Read More →Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Solution: Axiom 5 states that the whole is greater than the part. This axiom is known as a universal truth because it holds true in any field, and not just in the field of mathematics. Let us take two cases − one in the field of mathematics, and one other than that. Case I Let t represent a whole quantity and only a, b, c are parts of it. t = a + b + c Clearly, t will be greater than all its parts a, b, and c. Therefore, it is rightly said that the whole is greater than the part. Case II Let us ...
Read More →Which of the following is matter? <br/>Chair, air, love, smell, hate, almonds, thought, cold, lemon water, the smell of perfume.
Solution Anything which has mass and occupies space is called matter. It is made up of particles. In the above question, chair, air, almonds, and lemon water are matters. And love, smell, hate, thought, cold and smell of perfume are not in the category of matters because they are feeling and emotion of human beings and do not acquires any space....
Read More →In the following figure, if AC = BD, then prove that AB = CD.
Solution: From the figure, it can be observed that $\mathrm{AC}=\mathrm{AB}+\mathrm{BC}$ $B D=B C+C D$ It is given that $A C=B D$ $\mathrm{AB}+\mathrm{BC}=\mathrm{BC}+\mathrm{CD}$(1) According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal. Subtracting BC from equation (1), we obtain $\mathrm{AB}+\mathrm{BC}-\mathrm{BC}=\mathrm{BC}+\mathrm{CD}-\mathrm{BC}$ $\mathrm{AB}=\mathrm{CD}$...
Read More →In the above question, point C is called a mid-point of line segment AB, prove that every line segment has one and only one mid-point.
Solution: Let there be two mid-points, $C$ and $D$. $C$ is the mid-point of $A B$. $\mathrm{AC}=\mathrm{CB}$ $\mathrm{AC}+\mathrm{AC}=\mathrm{BC}+\mathrm{AC}$ (Equals are added on both sides) $\ldots$ (1) Here, $(B C+A C)$ coincides with $A B$. It is known that things which coincide with one another are equal to one another. $\therefore \mathrm{BC}+\mathrm{AC}=\mathrm{AB} \ldots$(2) It is also known that things which are equal to the same thing are equal to one another. Therefore, from equations...
Read More →If a point $C$ lies between two points $A$ and $B$ such that $A C=B C$, then prove that $A C=\frac{1}{2} A B .$ Explain by drawing the figure.
Solution: It is given that, $\mathrm{AC}=\mathrm{BC}$ $\mathrm{AC}+\mathrm{AC}=\mathrm{BC}+\mathrm{AC}$ (Equals are added on both sides) $\ldots$ (1) Here, $(\mathrm{BC}+\mathrm{AC})$ coincides with $\mathrm{AB}$. It is known that things which coincide with one another are equal to one another. $\therefore \mathrm{BC}+\mathrm{AC}=\mathrm{AB} \ldots$(2) It is also known that things which are equal to the same thing are equal to one another. Therefore, from equations (1) and (2), we obtain $\mathr...
Read More →Give the geometric representations of 2x + 9 = 0 as an equation <br/><br/> (i) in one variable <br/><br/>(i) in two variables
Solution: (i) In one variable, $2 x+9=0$ represents a point $x=\frac{-9}{2}=-4.5$ as shown in the following figure. (ii) In two variables, $2 x+9=0$ represents a straight line passing through point $(-4.5,0)$ and parallel to $y$-axis. It is a collection of all points of the plane, having their $x$-coordinate as $4.5$....
Read More →Give the geometric representation of y = 3 as an equation <br/><br/>(i) in one variable <br/><br/>(ii) in two variables
Solution: In one variable, y = 3 represents a point as shown in following figure. In two variables, y = 3 represents a straight line passing through point (0, 3) and parallel to x-axis. It is a collection of all points of the plane, having their y-coordinate as 3....
Read More →From the choices given below, choose the equation whose graphs are given in the given figures.
Solution: Points on the given line are $(-1,1),(0,0)$, and $(1,-1)$. It can be observed that the coordinates of the points of the graph satisfy the equation $x+y=0 .$ Therefore, $x+y=0$ is the equation corresponding to the graph as shown in the first figure. Hence, (ii) is the correct answer. Points on the given line are $(-1,3),(0,2)$, and $(2,0)$. It can be observed that the coordinates of the points of the graph satisfy the equation $y=$ $-x+2$ Therefore, $y=-x+2$ is the equation correspondin...
Read More →If the point $(3,4)$ lies on the graph of the equation $3 y=a x+7$, find the value of $a$.
Solution: Putting $x=3$ and $y=4$ in the given equation, $3 y=a x+7$ $3(4)=a(3)+7$ $5=3 a$ $a=\frac{5}{3}$...
Read More →Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution: It can be observed that point $(2,14)$ satisfies the equation $7 x-y=0$ and $x-y+12=0$ Therefore, $7 x-y=0$ and $x-y+12=0$ are two lines passing through point $(2,14)$. As it is known that through one point, infinite number of lines can pass through, therefore, there are infinite lines of such type passing through the given point....
Read More →Draw the graph of each of the following linear equations in two variables:<br/><br/>(i) $x+y=4$ <br/><br/>(ii) $x-y=2$ <br/><br/>(iii) $y=3 x$ <br/><br/>(iv) $3=2 x+y$
Solution: (i) $x+y=4$ It can be observed that $x=0, y=4$ and $x=4, y=0$ are solutions of the above equation. Therefore, the solution table is as follows. The graph of this equation is constructed as follows. (ii) $x-y=2$ It can be observed that $x=4, y=2$ and $x=2, y=0$ are solutions of the above equation. Therefore, the solution table is as follows. The graph of the above equation is constructed as follows. (iii) $y=3 x$ It can be observed that $x=-1, y=-3$ and $x=1, y=3$ are solutions of the a...
Read More →Find the value of $k$, if $x=2, y=1$ is a solution of the equation $2 x+3 y=k$.
Solution: Putting $x=2$ and $y=1$ in the given equation, $2 x+3 y=k$ $\Rightarrow 2(2)+3(1)=k$ $\Rightarrow 4+3=k$ $\Rightarrow k=7$ Therefore, the value of $k$ is 7 ....
Read More →Check which of the following are solutions of the equation x − 2y = 4 and which are not: <br/><br/>(i) $(0,2$ <br/><br/>(ii) $(2,0)$<br/><br/> (iii) $(4,0)$<br/><br/>(iv) $(\sqrt{2}, 4 \sqrt{2})$<br/><br/> (v) $(1,1)$
Solution: (i) $(0,2)$ Putting $x=0$ and $y=2$ in the L.H.S of the given equation, $x-2 y=0-2 \times 2=-4 \neq 4$ L.H.S ≠ R.H.S Therefore, $(0,2)$ is not a solution of this equation. (ii) $(2,0)$ Putting $x=2$ and $y=0$ in the L.H.S of the given equation, $x-2 y=2-2 \times 0=2 \neq 4$ L.H.S ≠ R.H.S Therefore, $(2,0)$ is not a solution of this equation. (iii) $(4,0)$ Putting $x=4$ and $y=0$ in the L.H.S of the given equation, $x-2 y=4-2(0)$ $=4=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$ Therefo...
Read More →Write four solutions for each of the following equations: <br/><br/>(i) $2 x+y=7$ <br/><br/>(ii) $\pi x+y=9$ <br/><br/>(iii) $x=4 y$
Solution: (i) $2 x+y=7$ For $x=0$, $2(0)+y=7$ $\Rightarrow y=7$ Therefore, $(0,7)$ is a solution of this equation. For $x=1$, $2(1)+y=7$ $\Rightarrow y=5$ Therefore, $(1,5)$ is a solution of this equation. For $x=-1$ $2(-1)+y=7$ $\Rightarrow y=9$ Therefore, $(-1,9)$ is a solution of this equation. For $x=2$, $2(2)+y=7$ $\Rightarrow y=3$ Therefore, $(2,3)$ is a solution of this equation. (ii) $\pi x+y=9$ For $x=0$, $\pi(0)+y=9$ $\Rightarrow y=9$ Therefore, $(0,9)$ is a solution of this equation. ...
Read More →Which one of the following options is true, and why? <br/><br/>$y=3 x+5$ has<br/><br/>(i) a unique solution, <br/><br/>(ii) only two solutions,<br/><br/> (iii) infinitely many solutions
Solution: $y=3 x+5$ is a linear equation in two variables and it has infinite possible solutions. As for every value of $x$, there will be a value of $y$ satisfying the above equation and vice-versa. Hence, the correct answer is (iii)....
Read More →The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. <br/><br/>(Take the cost of a notebook to be Rs x and that of a pen to be Rs y.)
Solution: Let the cost of a notebook and a pen be $x$ and $y$ respectively. Cost of notebook $=2 \times$ Cost of pen $x=2 y$ $x-2 y=0$...
Read More →Plot the point (x, y) given in the following table on the plane, choosing suitable units of distance on the axis.
Solution: The given points can be plotted on the Cartesian plane as follows....
Read More →In which quadrant or on which axis do each of the points $(-2,4),(3,-1),(-1,0),(1,2)$ and $(-3,-5)$ lie? Verify your answer by locating them on the Cartesian plane.
Solution: The point $(-2,4)$ lies in the $I I^{\text {nd }}$ quadrant in the Cartesian plane because for point $(-2,4), x$-coordinate is negative and $y$-coordinate is positive. Again, the point $(3,-1)$ lies in the IV $^{\text {th }}$ quadrant in the Cartesian plane because for point $(3,-1), x$-coordinate is positive and $y$-coordinate is negative. The point $(-1,0)$ lies on negative $x$-axis because for point $(-1,0)$, the value of $y$-coordinate is zero and the value of $x$-coordinate is neg...
Read More →How will you describe the position of a table lamp on your study table to another person?
Solution: Consider that the lamp is placed on the table. Choose two adjacent edges, DC and AD. Then, draw perpendiculars on the edges DC and AD from the position of lamp and measure the lengths of these perpendiculars. Let the length of these perpendiculars be 30 cm and 20 cm respectively. Now, the position of the lamp from the left edge (AD) is 20 cm and from the lower edge (DC) is 30 cm. This can also be written as (20, 30), where 20 represents the perpendicular distance of the lamp from edge ...
Read More →What are the possible expressions for the dimensions of the cuboids whose volumes are given below?<br/><br/>(i) Volume: $3 x^{2}-12 x$ <br/><br/>(ii) Volume: $12 k y^{2}+8 k y-20 k$
Solution: Volume of cuboid $=$ Length $\times$ Breadth $\times$ Height The expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height. (i) $2 x^{2}-12 x=3 x(x-4)$ One of the possible solutions is as follows. Length $=3$, Breadth $=x$, Height $=x-4$ (ii) $12 k y^{2}+8 k y-20 k=4 k\left(3 y^{2}+2 y-5\right)$ $=4 k\left[3 y^{2}+5 y-3 y-5\right]$ $=4 k[y(3 y+5)-1(3 y+5)]$ $=4 k(3 y+5)(y-1)$ One of the...
Read More →Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given: <br/> <br/>(i) Area: $25 \mathrm{a}^{2}-35 a+12$ <br/><br/>(ii) Area: $35 y^{2}+13 y-12$
Solution: Area $=$ Length $\times$ Breadth The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth. (i) $25 a^{2}-35 a+12=25 a^{2}-15 a-20 a+12$ $=5 a(5 a-3)-4(5 a-3)$ $=(5 a-3)(5 a-4)$ Therefore, possible length $=5 a-3$ And, possible breadth $=5 a-4$ (ii) $35 y^{2}+13 y-12=35 y^{2}+28 y-15 y-12$ $=7 y(5 y+4)-3(5 y+4)$ $=(5 y+4)(7 y-3)$ Therefore, possible length $=5 y+4$ And, possible breadth $=7 y-3$...
Read More →Without actually calculating the cubes, find the value of each of the following: <br/><br/> (i) $(-12)^{3}+(7)^{3}+(5)^{3}$<br/><br/>(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
Solution: (i) $(-12)^{3}+(7)^{3}+(5)^{3}$ Let $x=-12, y=7$, and $z=5$ It can be observed that, $x+y+z=-12+7+5=0$ It is known that if $x+y+z=0$, then $x^{3}+y^{3}+z^{3}=3 x y z$ $\therefore(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$ $=-1260$ (ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$ Let $x=28, y=-15$, and $z=-13$ It can be observed that, $x+y+z=28+(-15)+(-13)=28-28=0$ It is known that if $x+y+z=0$, then $x^{3}+y^{3}+z^{3}=3 x y z$ $\therefore(28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)$ $=16380$...
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