If the areas of three adjacent faces of a cuboid are x, y and z, respectively, the volume of the cuboid is
Question: If the areas of three adjacent faces of a cuboid arex,yandz, respectively, the volume of the cuboid is(a)xyz(b) 2xyz (c) $\sqrt{x y z}$ (d) $3 \sqrt{x y z}$ Solution: (c) $\sqrt{x y z}$ Let the length of the cuboid =lbreadth of the cuboid =band height of the cuboid =hSince, the areas of the three adjacent faces arex, yandz, we have: $l b=x$ $b h=y$ $l h=z$ Therefore, $l b \times b h \times l h=x y z$ $\Rightarrow l^{2} b^{2} h^{2}=x y z$ $\Rightarrow l b h=\sqrt{x y z}$ Hence, the volu...
Read More →How many diagonals does each of the following have?
Question: How many diagonals does each of the following have? (i) A convex quadrilateral (ii) A regular hexagon (iii) A triangle Solution: An $\mathrm{n}$-sided convex polygon has $\frac{n(n-3)}{2}$ diagonals. (i) A quadrilateral has $\frac{4(4-3)}{2}=2$ diagonals. There are 2 diagonals in the convex quadrilateral. (ii) A regular hexagon has $\frac{6(6-3)}{2}=9$ diagonals. There are 9 diagonals in a regular hexagon.(iii) A triangle does not have any diagonal in it....
Read More →If the lines given by 3x+ 2ky = 2
Question: If the lines given by 3x+ 2ky = 2 and 2x + 5y = 1 are parallel, then the value ofkis (a) $-\frac{5}{4}$ (b) $\frac{2}{5}$ (c) $\frac{15}{4}$ (d) $\frac{3}{2}$ Solution: (c) Condition for parallel lines is $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ Given lines, $3 x+2 k y-2=0$ and $2 x+5 y-1=0$ Here, $a_{1}=3, b_{1}=2 k, c_{1}=-2$ and $a_{2}=2, b_{2}=5, c_{2}=-1$ From Eq. (i), $\frac{3}{2}=\frac{2 k}{5}$ $\therefore$ $k=\frac{15}{4}$...
Read More →Following are some figures: Classify each of these fugures on the basis of the following:
Question: Following are some figures: Classify each of these fugures on the basis of the following: (i) Simple curve (ii) Simple closed curve (iii) Polygon (iv) Convex polygon (v) Concave polygon (vi) Not a curve Solution: (i) It is a simple closed curve and a concave polygon. (ii) It is a simple closed curve and a convex polygon. (iii) It is not a curve; hence, it is not a polygon. (iv) It is not a curve; hence, it is not a polygon. (v) It is a simple closed curve but not a polygon. (vi) It is ...
Read More →The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3.
Question: The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3. The breadth of the wall is(a) 30 cm(b) 40 cm(c) 22.5 cm(d) 25 cm Solution: Note : It should be 128 m3 instead of 12.8 m3 (b) 40 cmLet the breadth of the wall bexcm.Then, its height = 5xcm and its length $=(8 \times 5 x) \mathrm{cm}$ = 40xcm Hence, the volume of the wall $=(40 x \times x \times 5 x) \mathrm{cm}^{3}$ It is given that the volume of the wall = 128 m3.Therefore, $\Rightarrow 4...
Read More →Illustrate, if posible, each one of the following with a rough diagram:
Question: Illustrate, if posible, each one of the following with a rough diagram: (i) A colsed curve that is not a polygon. (ii) An open curve made up entirely of line segments. (iii) A polygon with two sides. Solution: (i) Polygons are made up of straight lines, not curves.(ii) An open curve made up entirely of line segments:(iii) Not possible because polygons are closed figures....
Read More →The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3.
Question: The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3. The breadth of the wall is(a) 30 cm(b) 40 cm(c) 22.5 cm(d) 25 cm Solution: Note : It should be 128 m3 instead of 12.8 m3 (b) 40 cmLet the breadth of the wall bexcm.Then, its height = 5xcm and its length $=(8 \times 5 x) \mathrm{cm}$ = 40xcm Hence, the volume of the wall $=(40 x \times x \times 5 x) \mathrm{cm}^{3}$ It is given that the volume of the wall = 128 m3.Therefore, $\Rightarrow 4...
Read More →Draw a polygon and shade its interior.
Question: Draw a polygon and shade its interior. Also draw its diagonals, if any. Solution: In polygon ABCD, AC and BD are the diagonals....
Read More →For what value of k,
Question: For what value of $k$, do the equations $3 x-y+8=0$ and $6 x-k y=-16$ (a) $\frac{1}{2}$ (b) $-\frac{1}{2}$ (c) 2 (d) $-2$ Solution: (c)Condition for coincident lines is $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Given lines.$3 x-y+8=0$ and $6 x-k y+16=0$ Here,$a_{1}=3, b_{1}=-1, c_{1}=8$ and $a_{2}=6, b_{2}=-k, c_{2}=16$ From Eq. (i), $\frac{3}{6}=\frac{-1}{-k}=\frac{8}{16}$ $\Rightarrow$ $\frac{1}{k}=\frac{1}{2}$ $\therefore$ $k=2$...
Read More →Classify the following curves as open or closed:
Question: Classify the following curves as open or closed: Solution: Open curve: A curve in which the beginning and end points are different or do not cut each other. Closed curve: A curve in which the beginning and end points are the same and cut each other. By the above definitions, we can classify the given figures as follows: (i) Open curve (ii) Closed curve (iii) Closed curve (iv) Open curve (v) Open curve (vi) Closed curve...
Read More →Draw rough diagrams to illustrate the following:
Question: Draw rough diagrams to illustrate the following: (i) Open curve (ii) Closed curve Solution: (i) Open curve: A curve in which the beginning and the end points does not cut each other or are different.(ii) Closed curve: A curve in which the beginning and the end points are the same and cuts each other....
Read More →The area of the base of a rectangular tank is 6500 cm2 and the volume of water contained in it is 2.6 m3.
Question: The area of the base of a rectangular tank is 6500 cm2and the volume of water contained in it is 2.6 m3. The depth of water in the tank is(a) 3.5 m(b) 4 m(c) 5 m(d) 8 m Solution: (b) 4 m Area of the base of a rectangular tank $=6500 \mathrm{~cm}^{2}$ $=\left(\frac{6500}{100 \times 100}\right) \mathrm{m}^{2}$ $=\frac{13}{20} \mathrm{~m}^{2}$ Let the depth of the water bedmetres.Then, $\frac{13}{20} \times d=2.6$ $\Rightarrow d=\left(\frac{26}{10} \times \frac{20}{13}\right) \mathrm{m}$ ...
Read More →The pair of equations x = a and
Question: The pair of equations x = a and y = b graphically represents lines which are (a) parallel (b) intersecting at (b, a) (c) coincident (d) intersecting at (a, b) Solution: (d)By graphically in every condition, if a, b0; a, b 0, a0, b 0; a0, b0 but a = b 0. The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b). If a, b 0 Similarly, in all cases two lines intersect at (a, b)....
Read More →How many bricks, each measuring (25 cm × 11.25 cm × 6 cm),
Question: How many bricks, each measuring (25 cm 11.25 cm 6 cm), will be required to construct a wall (8 m 6 m 22.5 cm)?(a) 8000(b) 6400(c) 4800(d) 7200 Solution: (b) 6400 Volume of the wall $=(800 \times 600 \times 22.5) \mathrm{cm}^{3}$ Volume of each brick $=(25 \times 11.25 \times 6) \mathrm{cm}^{3}$ Number of bricks $=\frac{\text { Volume of the wall }}{\text { Volume of each brick }}$ $=\frac{800 \times 600 \times 22.5}{25 \times 11.25 \times 6}$ $=6400$...
Read More →The pair of equations y = 0 and
Question: The pair of equations y = 0 and y = 7 has (a) one solution (b) two solutions (c) infinitely many solutions (d) no solution Solution: (d) The given pair of equations are $y=0$ and $y=-7$. By graphically, both lines are parallel and having no solution...
Read More →The total surface area of a cube is 864 cm2. Its volume is
Question: The total surface area of a cube is 864 cm2. Its volume is(a) 3456 cm3(b) 432 cm3(c) 1728 cm3(d) 3456 cm3 Solution: (c) 1728 cm3Let the edge of the cube bea.Total surface area of the cube = 6a2Therefore, $6 a^{2}=864$ $a^{2}=144$ $a=12 \mathrm{~cm}$ Therefore, volume of the cube =a3 $=(12 \times 12 \times 12) \mathrm{cm}^{3}$ $=1728 \mathrm{~cm}^{3}$...
Read More →The total surface area of a cube is 864 cm2. Its volume is
Question: The total surface area of a cube is 864 cm2. Its volume is(a) 3456 cm3(b) 432 cm3(c) 1728 cm3(d) 3456 cm3 Solution: (c) 1728 cm3Let the edge of the cube bea.Total surface area of the cube = 6a2Therefore, $6 a^{2}=864$ $a^{2}=144$ $a=12 \mathrm{~cm}$ Therefore, volume of the cube =a3 $=(12 \times 12 \times 12) \mathrm{cm}^{3}$ $=1728 \mathrm{~cm}^{3}$...
Read More →If a pair of linear equations is consistent,
Question: If a pair of linear equations is consistent, then the lines will be (a) parallel (b) always coincident (c) intersecting or coincident (d) always intersecting Solution: (c) Condition for a consistent pair of linear equations $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ [intersecting lines having unique solution] and $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ [coincident or dependent]...
Read More →The volume of a cube is 2744 cm2. Its surface area is
Question: The volume of a cube is 2744 cm2. Its surface area is(a) 196 cm2(b) 1176 cm2(c) 784 cm2(d) 588 cm2 Solution: (b) 1176 cm2Let the edge of the cube beacm.Then, volume of the cube =a3Or, $a^{3}=2744$ $a^{3}=\left(2^{3} \times 7^{3}\right)$ $a=2 \times 7$ $a=14 \mathrm{~cm}$ Therefore, surface area of the cube $=6 a^{2}$ $=(6 \times 14 \times 14) \mathrm{cm}^{2}$ $=1176 \mathrm{~cm}^{2}$...
Read More →The pair of equations x + 2y + 5 = 0 and – 3x – 6y +1 = 0 has
Question: The pair of equations x + 2y + 5 = 0 and 3x 6y +1 = 0 has (a) a unique solution (b) exactly two solutions (c) infinitely many solutions (d) no solution Solution: (d) Given, equations are $x+2 y+5=0$ and $-3 x-6 y+1=0$ Here, $\quad a_{1}=1, b_{1}=2, c_{1}=5$ and $a_{2}=-3, b_{2}=-6, c_{2}=1$ $\therefore \quad \frac{a_{1}}{a_{2}}=-\frac{1}{3}, \frac{b_{1}}{b_{2}}=-\frac{2}{6}=-\frac{1}{3}$, $\frac{c_{1}}{c_{2}}=\frac{5}{1}$ $\therefore$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac...
Read More →Daljit received a sum of Rs. 40000 as a loan from a finance company.
Question: Daljit received a sum of Rs. 40000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years. Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=40,000\left(1+\frac{7}{100}\right)^{2}$ $=40,000(1.07)^{2}$ $=45,796$ Thus, the required amount is Rs 45,796 . Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 45,796-\mathrm{Rs} 40,000$ $=\mathrm{Rs} 5,7...
Read More →The length of the diagonal of a cube is
Question: The length of the diagonal of a cube is $6 \sqrt{3} \mathrm{~cm}$. Its total surface area is (a) 144 cm2(b) 216 cm2(c) 180 cm2(d) 108 cm2 Solution: (b) 216 cm2Let the edge of the cube beacm. Then, length of the diagonal $=\sqrt{3} a$ Or, $\sqrt{3} a=6 \sqrt{3}$ $\Rightarrow a=6 \mathrm{~cm}$ Therefore, the total surface area of the cube = 6a2 $=(6 \times 6 \times 6) \mathrm{cm}^{3}$ $=216 \mathrm{~cm}^{3}$...
Read More →Surabhi borrowed a sum of Rs 12000 from a finance company to purchase a refrigerator.
Question: Surabhi borrowed a sum of Rs 12000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years. Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=12,000\left(1+\frac{5}{100}\right)^{3}$ $=12,000(1.05)^{3}$ $=13,891.50$ Thus, the required amount is Rs $13,891.50$. Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs...
Read More →Graphically, the pair of equations
Question: Graphically, the pair of equations 6x 3y + 10 = 0 2x y + 9 = 0 represents two lines which are (a) intersecting at exactly one point (b) intersecting exactly two points (c) coincident (d) parallel Solution: The given equations are $6 x-3 y+10=0$ $\Rightarrow 2 x-y+\frac{10}{3}=0$ [dividing by 3 ]... (i) and $\quad 2 x-y+9=0$ .......(ii) Now, table for $2 x-y+\frac{10}{3}=0$, and table for $2 x-y+9=0$, Hence, the pair of equations represents two parallel lines....
Read More →Anil borrowed a sum of Rs 9600 to install a handpump in his dairy.
Question: Anil borrowed a sum of Rs 9600 to install a handpump in his dairy. If the rate of interest is $5 \frac{1}{2} \%$ per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years. Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=9,600\left(1+\frac{5.5}{100}\right)^{3}$ $=9,600(1.055)^{3}$ $=\mathrm{Rs} 11,272.72$ Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 11,272.72-\mathrm{Rs} 9,600$ $=\mathrm{Rs} 1...
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