The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively.
Question: The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively. Find the slant height of the frustum. Solution: We have, Height of the frustum, $h=6 \mathrm{~cm}$, Radii of the circular ends, $R=14 \mathrm{~cm}$ and $r=6 \mathrm{~cm}$ Let the slant height of the frustum be $l$. Now, $l=\sqrt{(R-r)^{2}+h^{2}}$ $=\sqrt{(14-6)^{2}+6^{2}}$ $=\sqrt{8^{2}+6^{2}}$ $=\sqrt{64+36}$ $=\sqrt{100}$ $=10 \mathrm{~cm}$ So, the slant height of the frustum is 10 cm....
Read More →Find the amount that David would receive if he invests Rs 8192 for 18 months at
Question: Find the amount that David would receive if he invests Rs 8192 for 18 months at $12 \frac{1}{2} \%$ per annum, the interest being compounded half-yearly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 8,192$ $\mathrm{R}=12.5 \%$ p. a. $\mathrm{n}=1.5$ years When the interest is compounded half - yearly, we have : $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=\operatorname{Rs} 8,192\left(1+\frac{12.5}{200}\right)^{3}$ $=\operatorname{Rs} 8,192(1.0625)^{3}$ $=\o...
Read More →If A is a unit matrix of order n,
Question: If $A$ is a unit matrix of order $n$, then $A(\operatorname{adj} A)=$____________ Solution: As we know that, $A(\operatorname{adj} A)=|A| I$. But it is given that $A$ is a unit matrix of order $n$ Therefore, $A(\operatorname{adj} A)=|I| I=(1) \mid=1$ Hence, if $A$ is a unit matrix of order $n$, then $A(\operatorname{adj} A)=\underline{l}$....
Read More →A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm.
Question: A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire. Solution: We have, Radius of the sphere, $R=\frac{18}{2}=9 \mathrm{~cm}$ and Radius of the wire, $r=\frac{4}{2}=2 \mathrm{~mm}=0.2 \mathrm{~cm}$ Let the length of the wire be $l$. Now, Volume of the wire $=$ Volume of the copper sphere $\Rightarrow \pi r^{2} l=\frac{4}{3} \pi R^{3}$ $\Rightarrow l=\frac{4 R^{3}}{3 r^{2}}$ $\Rightarrow l=\frac{4 \times 9 \times 9 \times 9}{3 \times 0.2 ...
Read More →Show that the square of any positive
Question: Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. Solution: Let a be an arbitrary positive integer. Then, by Euclids divisions Algorithm, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that $a=5 m+r$, where $0 \leq r5$ $\Rightarrow \quad a^{2}=(5 m+r)^{2}=25 m^{2}+r^{2}+10 m r \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $\Rightarrow \quad a^{2}=5\left(5 m^{2}+2 m r\right)+r^{...
Read More →Romesh borrowed a sum of Rs 245760 at 12.5% per annum, compounded annually.
Question: Romesh borrowed a sum of Rs 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years. Solution: Given: $\mathrm{P}=\mathrm{Rs} 245,760$ $\mathrm{R}=12.5 \%$ p. a. $\mathrm{n}=2$ years When compounded annually, we have : $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=$ Rs $245,760\left(1+\frac{12.5}{100}\right)^{2}$ $=$ Rs 311,040 Whe...
Read More →A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm.
Question: A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone. Solution: We have, Radius of the hemisphere, $R=6 \mathrm{~cm}$ and Height of the cone, $h=75 \mathrm{~cm}$ Let the radius of the base of the cone be $r$. Now, Volume of the cone = Volume of the hemisphere $\Rightarrow \frac{1}{3} \pi r^{2} h=\frac{2}{3} \pi R^{3}$ $\Rightarrow r^{2}=\frac{2 R^{3}}{h}$ $\Rightarrow r^{2}=\frac{2 \times 6 \times 6 \times 6}{75...
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}2 \lambda -3 \\ 0 2 5 \\ 1 1 3\end{array}\right]$, then $A^{-1}$ exists if (a) $\lambda=2$ (b) $\lambda \neq 2$ (c) $\lambda \neq-2$ (d) none of these Solution: Given: $A=\left[\begin{array}{ccc}2 \lambda -3 \\ 0 2 5 \\ 1 1 3\end{array}\right]$ $A^{-1}$ exists only if $|A| \neq 0$ $\left|\begin{array}{ccc}2 \lambda -3 \\ 0 2 5 \\ 1 1 3\end{array}\right| \neq 0$ $\Rightarrow 2(6-5)+1(5 \lambda+6) \neq 0$ $\Rightarrow 2(1)+5 \lambda+6 \neq 0$ $\Rightarrow 2+...
Read More →A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each.
Question: A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are formed? Solution: We have, Radius of the metallic cone, $r=12 \mathrm{~cm}$, Height of the metallic cone, $h=24 \mathrm{~cm}$ and Radius of the sphere, $R=2 \mathrm{~cm}$ Now, The number of spheres so formed $=\frac{\text { Volume of the metallic cone }}{\text { Volume of a sphere }}$ $=\frac{\left(\frac{1}{3} \pi r^{2} h\right)}{\left(\frac{4}{3} \pi R^{3}\right)}...
Read More →Rakesh lent out Rs 10000 for 2 years at 20% per annum, compounded annually.
Question: Rakesh lent out Rs 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly? Solution: Given: $\mathrm{P}=\mathrm{Rs} 10,000$ $\mathrm{R}=20 \%$ p. $\mathrm{a} .$ $\mathrm{n}=2$ years $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=\operatorname{Rs} 10,000\left(1+\frac{20}{100}\right)^{2}$ $=\operatorname{Rs} 10,000(1.2)^{2}$ $=\operatorname{Rs} 14,400$ When the interest is compounded half...
Read More →If A and B are invertible matrices, then which one of the following is not correct?
Question: IfAandBare invertible matrices, then which one of the following is not correct? (a) $\operatorname{adj} A=|A| A^{-1}$ (b) $\operatorname{det}\left(A^{-1}\right)=[\operatorname{det}(A)]^{-1}$ (c) $(A B)^{-1}=B^{-1} A^{-1}$ (d) $(A+B)^{-1}=B^{-1}+A^{-1}$ Solution: (a) adj $A=|A| A^{-1}$ As we know, $A^{-1}=\frac{\operatorname{adj} A}{|A|}$ $\Rightarrow A^{-1}|A|=\operatorname{adi} A$ Thus, $\operatorname{adj} A=|A| A^{-1}$ is correct. (b) $\operatorname{det}\left(A^{-1}\right)=[\operator...
Read More →If A and B are invertible matrices, then which one of the following is not correct?
Question: IfAandBare invertible matrices, then which one of the following is not correct? (a) $\operatorname{adj} A=|A| A^{-1}$ (b) $\operatorname{det}\left(A^{-1}\right)=[\operatorname{det}(A)]^{-1}$ (c) $(A B)^{-1}=B^{-1} A^{-1}$ (d) $(A+B)^{-1}=B^{-1}+A^{-1}$ Solution: (a) adj $A=|A| A^{-1}$ As we know, $A^{-1}=\frac{\operatorname{adj} A}{|A|}$ $\Rightarrow A^{-1}|A|=\operatorname{adi} A$ Thus, $\operatorname{adj} A=|A| A^{-1}$ is correct. (b) $\operatorname{det}\left(A^{-1}\right)=[\operator...
Read More →Show that cube of any positive integer
Question: Show that cube of any positive integer is of the form 4m, 4m + 1 or Am + 3, for some integer m. Solution: Let a be an arbitrary positive integer. Then, by Euclids division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that a = 4q + r, where 0 r 4 $a=4 q+r$, where $0 \leq r4$ $\Rightarrow \quad a^{3}=(4 q+r)^{3}=64 q^{3}+r^{3}+12 q r^{2}+48 q^{2} r$ $\left[\because(a+b)^{3}=a^{3}+b^{3}+3 a b^{2}+3 a^{2} b\right]$ $\Rightarrow \...
Read More →How many lead shots each 3 mm
Question: How many lead shots each 3 mm in diameter can be made from acuboid of dimensions 9 cm11 cm12 cm? Solution: We have, Radius of a lead shot, $r=\frac{3}{2}=1.5 \mathrm{~mm}=0.15 \mathrm{~cm}$ and Dimensions of the cuboid are $9 \mathrm{~cm} \times 11 \mathrm{~cm} \times 12 \mathrm{~cm}$ Now, The number of the lead shots $=\frac{\text { Volume of the cuboid }}{\text { Volume of a lead shot }}$ $=\frac{9 \times 11 \times 12}{\left(\frac{4}{3} \pi r^{3}\right)}$ $=\frac{9 \times 11 \times 1...
Read More →Abha purchased a house from Avas Parishad on credit.
Question: Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half. Solution: Given: $\mathrm{P}=\mathrm{Rs} 64,000$ $\mathrm{R}=5 \%$ p. a. $\mathrm{n}=1.5$ years When the interest is compounded half - yearly, we have : $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=$ Rs $64,000\left(1+\frac{5}{200}\right)^{3}$ ...
Read More →A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm.
Question: A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls obtained. Solution: We have, Radius of the solid metallic sphere, $R=8 \mathrm{~cm}$ and Radius of the spherical ball, $r=2 \mathrm{~cm}$ Now, The number spherical balls obtained $=\frac{\text { Volume of the solid metallic sphere }}{\text { Volume of a spherical ball }}$ $=\frac{\left(\frac{4}{3} \pi R^{3}\right)}{\left(\frac{4}{3} \pi r^{3}\right)}$...
Read More →Kamal borrowed Rs 57600 from LIC against her policy at
Question: Kamal borrowed Rs 57600 from LIC against her policy at $12 \frac{1}{2} \%$ per annum to build a house. Find the amount that she pays to the LIC after $1 \frac{1}{2}$ years if the interest is calculated half-yearly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 57,600$ $\mathrm{R}=12.5 \%$ p. a. $\mathrm{n}=1.5$ years When the interest is compounded half-yearly, we have: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=$ Rs $57,600\left(1+\frac{12.5}{200}\right)^...
Read More →Show that the square of any positive integer
Question: Show that the square of any positive integer is either of the form 4qor 4g + 1 for some integerq. Solution: Let a be an arbitrary positive integer. Then, by, Euclid's division algorithm, corresponding to the positive integers a and 4 , there exist non-negative integers $m$ and $r$, such that $a=4 m+r$, where $0 \leq r4$ $\Rightarrow \quad a^{2}=16 m^{2}+r^{2}+8 m r$ $\ldots$ (i) where, $0 \leq r4$ $\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ Case I When $r=0$, then putting $r=0$ ...
Read More →If x, y, z are non-zero real numbers,
Question: If $x, y, z$ are non-zero real numbers, then the inverse of the matrix $A=\left[\begin{array}{lll}x 0 0 \\ 0 y 0 \\ 0 0 z\end{array}\right]$, is (a) $\left[\begin{array}{ccc}x^{-1} 0 0 \\ 0 y^{-1} 0 \\ 0 0 z^{-1}\end{array}\right]$ (b) $x y z\left[\begin{array}{ccc}x^{-1} 0 0 \\ 0 y^{-1} 0 \\ 0 0 z^{-1}\end{array}\right]$ (c) $\frac{1}{x y z}\left[\begin{array}{lll}x 0 0 \\ 0 y 0 \\ 0 0 z\end{array}\right]$ (d) $\frac{1}{x y z}\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\...
Read More →Find the amount and the compound interest on Rs 8000 for
Question: Find the amount and the compound interest on Rs 8000 for $1 \frac{1}{2}$ years at $10 \%$ per annum, compounded half-yearly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 8,000$ $\mathrm{R}=10 \%$ p. a. $\mathrm{n}=1.5$ years When compounded half - yearly, we have: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=\operatorname{Rs} 8,000\left(1+\frac{10}{200}\right)^{3}$ $=\operatorname{Rs} 8,000(1.05)^{3}$ $=\operatorname{Rs} 9,261$ Also, $\mathrm{CI}=\mathrm{A}...
Read More →The surface areas of two spheres are in the ratio of 4 : 25.
Question: The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes. Solution: Let the radii of the two spheres be $r$ and $R$. As, $\frac{\text { Surface area of the first sphere }}{\text { Surface area of the second sphere }}=\frac{4}{25}$ $\Rightarrow \frac{4 \pi r^{2}}{4 \pi R^{2}}=\frac{4}{25}$ $\Rightarrow\left(\frac{r}{R}\right)^{2}=\frac{4}{25}$ $\Rightarrow \frac{r}{R}=\sqrt{\frac{4}{25}}$ $\Rightarrow \frac{r}{R}=\frac{2}{5} \quad \ldots \ldots(\mathr...
Read More →Find the amount of Rs 4096 for 18 months at
Question: Find the amount of Rs 4096 for 18 months at $12 \frac{1}{2} \%$ per annum, the interest being compounded semi-annually. Solution: Given: $\mathrm{P}=\mathrm{Rs} 4,096$ $\mathrm{R}=12.5 \%$ p. a. $\mathrm{n}=18$ months $=1.5$ years We have : $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ When the interest is compounded semi-annually, we have: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=\operatorname{Rs} 4,096\left(1+\frac{12.5...
Read More →A rational number in its decimal expansion is 327.7081.
Question: A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form p/q ? Give reasons. Solution: 327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form 2mx5n. Thus, $327.7081=\frac{3277081}{10000}=\frac{p}{q}$ $\therefore$$q=10^{4}=2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5$ $=2^{4} \times 5^{4}=(2 \times 5)^{4}$ Hence, the...
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}2 -1 \\ 3 -2\end{array}\right]$, then $A^{n}=$ (a) $A=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$, if $n$ is an even natural number (b) $A=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]$, if $n$ is an odd natural number (c) $A=\left[\begin{array}{cc}-1 0 \\ 0 1\end{array}\right]$, if $n \in N$ (d) none of these Solution: Disclaimer: In all option, the power of $A$ (i.e. $n$ is missing) (a) $A^{n}=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\ri...
Read More →Without actually performing the long division,
Question: Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer. Solution: Yes, after simplification denominator has factor 53-22and which is of the type 2m. 5″. So, this is terminating decimal. $\because \quad \frac{987}{10500}=\frac{47}{500}=\frac{47}{5^{3} \cdot 2^{2}} \times \frac{2}{2}$ $=\frac{94}{5^{3} \times 2^{3}}=\frac{94}{(10)^{3}}=\frac{94}{1000}=0.094$...
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