Pritam bought a plot of land for Rs 640000.
Question: Pritam bought a plot of land for Rs 640000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years? Solution: Given: $\mathrm{P}=\operatorname{Rs} 64,000$ $\mathrm{R}=5 \%$ for every six months Value of the plot after two years $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $\Rightarrow 64,000\left(1+\frac{5}{200}\right)^{4}$ $=64,000(1.025)^{4}$ $=706,440.25$ Thus, the value of the plot after two ye...
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Question: If $A=\left[\begin{array}{cc}2 3 \\ 5 -2\end{array}\right]$ be such that $A^{-1}=k A$, then find the value of $k$. Solution: $A=\left[\begin{array}{ll}2 3\end{array}\right.$ $5-2]$ $\therefore|A|=\mid 2 \quad 3$ $5-2 \mid=-14-15=-19$ The value is non-zero, so $A^{-1}$ exists. By definition, we have $A^{-1} A=I$ $[I$ is the identity matrix $]$ $\Rightarrow k A \cdot A=I$ [Substituting $\left.A^{-1}=k A\right]$ $\Rightarrow k\left[\begin{array}{ll}2 3\end{array}\right.$ $5-2]\left[\begin...
Read More →The value of a machine depreciates at the rate of 10% per annum.
Question: The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000? Also, find the total depreciation during this period. Solution: Value of the machine after two years $=\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $\Rightarrow 100,000\left(1-\frac{10}{100}\right)^{2}$ $=100,000(0.90)^{2}$ $=81,000$ Thus, the value of the machine after two years will be Rs 81,000 . Depreciation $=$ Rs $100,000-$ Rs ...
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Question: Choose the correct answer of the following question:A solid is hemispherical at the bottom and conical (of same radius)above it. If the surface areas of the two parts are equal, then the ratio ofits radius and the slant height of the conical part is(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Solution: Let the radius of the hemisphere or the radius of the cone be $r$ and the slant height of the cone be $l$. Now, Surface area of the hemisphere = Surface area of the cone $\Rightarrow 2 \pi r^...
Read More →Ms. Cherian purchased a boat for Rs 16000.
Question: Ms. Cherian purchased a boat for Rs 16000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years. Solution: Value of the boat after two years $=\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $\Rightarrow 16,000\left(1-\frac{5}{100}\right)^{2}$ $=16,000(0.95)^{2}$ $=14,440$ Thus, the value of the boat after two years will be Rs 14,440 ....
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Question: Choose the correct answer of the following question:The radii of the circular ends of a bucket of height 40 cm are 24 cm and15 cm. The slant height (in cm) of the bucket is(a) 41 (b) 43 (c) 49 (d) 51 Solution: We have, Height of the bucket, $h=40 \mathrm{~cm}$, Radius of the upper end, $R=24 \mathrm{~cm}$ and Radius of the lower end, $r=15 \mathrm{~cm}$ Now, The slant height, $l=\sqrt{(R-r)^{2}+h^{2}}$ $=\sqrt{(24-15)^{2}+40^{2}}$ $=\sqrt{9^{2}+40^{2}}$ $=\sqrt{81+1600}$ $=\sqrt{1681}$...
Read More →Jitendra set up a factory by investing Rs 2500000.
Question: Jitendra set up a factory by investing Rs 2500000. During the first two successive years his profits were 5% and 10% respectively. If each year the profit was on previous year's capital, compute his total profit. Solution: Profit at the end of the first year $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)$ $=2,500,000\left(1+\frac{5}{100}\right)$ $=2,500,000(1.05)$ $=2,625,000$ Profit at the end of the second year $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)$ $=2,625,000\left(1+...
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Question: Choose the correct answer of the following question:A solid right circular cone is cut into two parts at the middle of itsheight by a plane parallel to its base. The ratio of the volume of thesmaller cone to the whole cone is(a) 1 : 2 (b) 1 : 4 (c) 1 : 6 (d) 1 : 8 Solution: Let the radii of the smaller and given cones be $r$ and $R$, respectively; and their heights be $h$ and $H$, respectively. We have, $H=2 h \quad \ldots \ldots$ (i) In $\Delta \mathrm{AQD}$ and $\Delta \mathrm{APC}$,...
Read More →The population of a city increases each year by 4% of what it had been at the beginning of each year.
Question: The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997. Solution: (i) Population of the city in $2001=P\left(1+\frac{R}{100}\right)^{2}$ $=6760000\left(1+\frac{4}{100}\right)^{2}$ $=6760000(1.04)^{2}$ $=7311616$ Thus, Population of the city in 2001 is 7311616 . (ii) Population of the city in $1997=P\left(1+\frac{R}{100}\right)^{-2}$ $=6760000\...
Read More →If A is a square matrix such that
Question: If $A$ is a square matrix such that $A(\operatorname{adj} A)=\left[\begin{array}{lll}5 0 0 \\ 0 5 0 \\ 0 0 5\end{array}\right]$, then write the value of $|\operatorname{adj} A|$. Solution: Given : $A(\operatorname{adj} A)=\left[\begin{array}{lll}5 0 0\end{array}\right.$ $\begin{array}{lll}0 5 0\end{array}$ $\left.\begin{array}{lll}0 0 5\end{array}\right]$ $\Rightarrow|A| I_{n}=5\left[\begin{array}{lll}1 0 0\end{array}\right.$ $\begin{array}{lll}0 1 0\end{array}$ $\left.\begin{array}{ll...
Read More →The production of a mixi company in 1996 was 8000 mixies.
Question: The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years? Solution: Production after three years $=\mathrm{P}\left(1+\frac{\mathrm{R}_{1}}{100}\right)^{2}\left(1-\frac{\mathrm{R}_{2}}{100}\right)$ $=8,000\left(1+\frac{15}{1,000}\right)^{2}\left(1-\frac{5}{100}\right)$ $=8,000(1.15)^{2}(0.95)$ $=10,051$ Thus, the pr...
Read More →If A is an invertible matrix
Question: If $A$ is an invertible matrix such that $\left|A^{-1}\right|=2$, find the value of $|A|$. Solution: We know, $|A|^{-1}=\frac{1}{|A|}$ $\Rightarrow 2=\frac{1}{|A|} \quad\left[\because|A|^{-1}=2\right]$ $\Rightarrow|A|=\frac{1}{2}$...
Read More →The population of a town increases at the rate of 40 per thousand annually.
Question: The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago. Solution: Population after three years $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{3}$ $175,760=\mathrm{P}\left(1+\frac{40}{1000}\right)^{3}$ $175,760=\mathrm{P}(1.04)^{3}$ $\mathrm{P}=\frac{175,760}{1.124864}$ $=156,250$ Thus, the population three years ago was 156,250 ....
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Question: Choose the correct answer of the following question:A sphere of diameter 18 cm is dropped into a cylindrical vessel ofdiameter 36 cm, partly filled with water. If the sphere is completelysubmerged, then the water level rises by(a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm Solution: We have, Radius of the sphere, $r=\frac{18}{2}=9 \mathrm{~cm}$ and Radius of the cylindrical vessel, $R=\frac{36}{2}=18 \mathrm{~cm}$ Let the rise of water level be $H$. Now, Volume of the water rised $=$ Volume of th...
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Question: If $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$ and $A(\operatorname{adj} A=)\left[\begin{array}{cc}k 0 \\ 0 k\end{array}\right]$, then find the value of $k$. Solution: $A=\left[\begin{array}{ll}\cos \theta \sin \theta\end{array}\right.$ $-\sin \theta \quad \cos \theta]$ $\therefore|A|=\mid \cos \theta \quad \sin \theta$ $-\sin \theta \quad \cos \theta \mid=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0$ Thus, $A^{-1}$ exists. Now, $A^{-1...
Read More →Aman started a factory with an initial investment of Rs 100000.
Question: Aman started a factory with an initial investment of Rs 100000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years. Solution: Aman's profit for three years $=\mathrm{P}\left(1-\frac{\mathrm{R}_{1}}{100}\right)\left(1+\frac{\mathrm{R}_{2}}{100}\right)\left(1+\frac{\mathrm{R}_{3}}{100}\right)$ $=100,000\left(1-\frac{5}{100}\right)\left(1+...
Read More →6400 workers were employed to construct a river bridge in four years.
Question: 6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year? Solution: Number of workers $=6,400$ At the end of the first year, $25 \%$ of the workers were retrenched. $\t...
Read More →Choose the correct answer of the following question:
Question: Choose the correct answer of the following question:If the height of a bucket in the shape of frustum of a cone is 16 cm and the diameters of its two circular ends are 40 cm and 16 cm, then its slant height is (a) $20 \mathrm{~cm}$ (b) $12 \sqrt{5} \mathrm{~cm}$ (c) $8 \sqrt{13} \mathrm{~cm}$ (d) $16 \mathrm{~cm}$ Solution: We have, Height of the frustum, $h=16 \mathrm{~cm}$, Radii of the circular ends, $R=\frac{40}{2}=20 \mathrm{~cm}$ and $r=\frac{16}{2}=8 \mathrm{~cm}$ Now, The slant...
Read More →The population of a certain city was 72000 on the last day of the year 1998.
Question: The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000? Solution: Population at the end of the year $2000=\mathrm{P}\left(1+\frac{\mathrm{R}_{1}}{100}\right)\left(1-\frac{\mathrm{R}_{2}}{100}\right)$ $=72,000\left(1+\frac{7}{100}\right)\left(1-\frac{10}{100}\right)$ $=72,000(1.07)(0.9)$ $=69,336$ Thus, the popula...
Read More →Choose the correct answer of the following question:
Question: Choose the correct answer of the following question:If the radius of a sphere becomes 3 times, then its volume will become(a) 3 times (b) 6 times (c) 9 times (d) 27 times Solution: Let the radius of the given sphere and that of the new sphere be $r$ and $R$, respectively; Also, the volume of the given sphere and that of the new sphere be $v$ and $V$, respectively. We have, $R=3 r \quad \ldots \ldots(\mathrm{i})$ Now, $\frac{\text { Volume of the new sphere }}{\text { Volume of the give...
Read More →The count of bacteria in a culture grows by 10% in the first hour,
Question: The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours? Solution: Given: $\mathrm{R}_{1}=10 \%$ $\mathrm{R}_{2}=-8 \%$ $\mathrm{R}_{3}=12 \%$ $\mathrm{P}=$ Original count of bacteria $=13,125,000$ We know that: $\mathrm{P}\left(1+\frac{\mathrm{R}_{1}}{100}\right)\left(1-\frac{\mathrm{R}_{2}}{100...
Read More →If A is a non-singular symmetric matrix,
Question: If $A$ is a non-singular symmetric matrix, write whether $A^{-1}$ is symmetric or skew-symmetric. Solution: Let $A$ be an invertible symmetric matrix. Then, $|A| \neq 0$ and $A^{T}=A$ Now, $\left(A^{-1}\right)^{T}=\left(A^{T}\right)^{-1}$ $\Rightarrow\left(A^{-1}\right)^{T}=A^{-1} \quad\left[\because A^{T}=A\right]$ Thus, $A^{-1}$ is symmetric matrix....
Read More →Choose the correct answer of the following question:
Question: Choose the correct answer of the following question:If the surface area of a sphere is 616 cm2, its diameter (in cm) is(a) 7 (b) 14 (c) 28 (d) 56 Solution: Let the radius of the sphere be $r$. As, Surface area of the sphere $=616 \mathrm{~cm}^{2}$ $\Rightarrow 4 \pi r^{2}=616$ $\Rightarrow 4 \times \frac{22}{7} \times r^{2}=616$ $\Rightarrow r^{2}=\frac{616 \times 7}{4 \times 22}$ $\Rightarrow r^{2}=49$ $\Rightarrow r=\sqrt{49}$ $\Rightarrow r=7 \mathrm{~cm}$ $\therefore$ Diameter of t...
Read More →Let A be a 3 × 3 square matrix,
Question: Let $A$ be a $3 \times 3$ square matrix, such that $A(\operatorname{adj} A)=2 I$, where $I$ is the identity matrix. Write the value of $|\operatorname{adj} A|$. Solution: We know that for a matix $A$ of order $n, A \cdot(\operatorname{adj} A)=|A| I_{n}$, where $I$ is the identity matrix. Given: $A \cdot(\operatorname{adj} A)=2 I$ $\Rightarrow|A| I=2 I$ $\Rightarrow|A|=2$ Now, $|a d j A|=|A|^{n-1}$ $\Rightarrow|a d j A|=2^{2}=4$...
Read More →Let A be a 3 × 3 square matrix, such that A (adj A) = 2 I,
Question: Let $A$ be a $3 \times 3$ square matrix, such that $A(\operatorname{adj} A)=2 I$, where $I$ is the identity matrix. Write the value of $|\operatorname{adj} A|$. Solution: We know that for a matix $A$ of order $n, A \cdot(\operatorname{adj} A)=|A| I_{n}$, where $I$ is the identity matrix. Given: $A \cdot(\operatorname{adj} A)=2 I$ $\Rightarrow|A| I=2 I$ $\Rightarrow|A|=2$ Now, $|a d j A|=|A|^{n-1}$ $\Rightarrow|a d j A|=2^{2}=4$...
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