The length of the longest pole that can be kept in a room (12 m × 9 m ×8 m) is
Question: The length of the longest pole that can be kept in a room (12 m 9 m 8 m) is(a) 29 m(b) 21 m(c) 19 m(d) 17 m Solution: (d) 17 mLength of the longest pole that can be kept in a room =Length of the diagonal of the room $=\sqrt{l^{2+} b^{2}+h^{2}} \mathrm{~m}$ $=\sqrt{(12)^{2}+(9)^{2}+(8)^{2}} \mathrm{~m}$ $=\sqrt{289} \mathrm{~m}$ $=17 \mathrm{~m}$...
Read More →The length of the longest pole that can be kept in a room (12 m × 9 m ×8 m) is
Question: The length of the longest pole that can be kept in a room (12 m 9 m 8 m) is(a) 29 m(b) 21 m(c) 19 m(d) 17 m Solution: (d) 17 mLength of the longest pole that can be kept in a room =Length of the diagonal of the room $=\sqrt{l^{2+} b^{2}+h^{2}} \mathrm{~m}$ $=\sqrt{(12)^{2}+(9)^{2}+(8)^{2}} \mathrm{~m}$ $=\sqrt{289} \mathrm{~m}$ $=17 \mathrm{~m}$...
Read More →Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly.
Question: Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months. Solution: Given: $\mathrm{P}=$ Rs 7,500 $\mathrm{R}=12 \%$ p. a. $=3 \%$ quarterly $\mathrm{T}=9$ months $=3$ quarters We know that: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $\mathrm{A}=7,500\left(1+\frac{3}{100}\right)^{3}$ $=7,500(1.03)^{3}$ $=8,195.45$ Thus, the required amount is Rs $8,195.45$....
Read More →A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere tucked at each end.
Question: A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere tucked at each end. The length of the entire capsule is 2 cm. The capacity of the capsule is(a) 0.33 cm2(b) 0.34 cm2(c) 0.35 cm2(d) 0.36 cm2 Solution: (d) 0.36 cm3Radius of the capsule $=\frac{0.5}{2} \mathrm{~cm}$ = 0.25 cmLet the length of the cylindrical part of the capsule bexcm.Then, $0.25+x+0.25=2$ $\Rightarrow 0.5+x=2$ $\Rightarrow x=1.5$ Hence, the capacity of the capsule $=2 \times$ (Volume o...
Read More →Find the compound interest on Rs 64000 for 1 year at the rate of 10%
Question: Find the compound interest on Rs 64000 for 1 year at the rate of 10% per annum compounded quarterly. Solution: To calculate the interest compounded quarterly, we have: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{400}\right)^{4 \mathrm{n}}$ $=64,000\left(1+\frac{10}{400}\right)^{4 \times 1}$ $=64,000(1.025)^{4}$ $=70,644.03$ Thus, the required amount is Rs $70,644.03$. Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 70,644.025-\mathrm{Rs} 64,000$ $=\mathrm{Rs} 6,644.03$...
Read More →A metallic spherical shell of internal and external diameters 4 cm and 8 cm,
Question: A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast in the form of a cone of base diameter 8 cm. The height of the cone is(a) 12 cm(b) 14 cm(c) 15 cm(d) 8 cm Solution: (b) 14 cmLet the internal and external radii of the spherical shell berandR, respectively. Volume of the spherical shell= $\frac{4}{3} \pi\left(R^{3}-r^{3}\right)$ $=\frac{4}{3} \pi\left[(4)^{3}-(2)^{3}\right] \mathrm{cm}^{3} \quad\left[\right.$ Since $\left.\ma...
Read More →Find the compound interest at the rate of 10%
Question: Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs 200 as simple interest. Solution: $\mathrm{SI}=\frac{\mathrm{PRT}}{100}$ $\therefore \mathrm{P}=\frac{\mathrm{SI} \times 100}{\mathrm{RT}}$ $=\frac{200 \times 100}{10 \times 2}$ $=\mathrm{RS} 1,000$ $\mathrm{~A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=1,000\left(1+\frac{10}{100}\right)^{2}$ $=1,000(1.10)^{2}$ $=\mathrm{...
Read More →Find the compound interest on Rs 8000 for 9 months
Question: Find the compound interest on Rs 8000 for 9 months at 20% per annum compounded quarterly. Solution: $\mathrm{P}=$ Rs 8,000 $\mathrm{~T}=9$ months $=3$ quarters $\mathrm{R}=20 \%$ per annum $=5 \%$ per quarter $\mathrm{A}=8,000\left(1+\frac{5}{100}\right)^{3}$ $=8,000(1.05)^{3}$ $=9,261$ The required amount is Rs 9,261 . Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 9,261-\mathrm{Rs} 8,000$ $=\mathrm{Rs} 1,261$...
Read More →A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and
Question: A hollow cube of internal edge $22 \mathrm{~cm}$ is filled with spherical marbles of diameter $0.5 \mathrm{~cm}$ and $\frac{1}{8}$ space of the cube remains unfilled. Number of marbles required is(a) 142296(b) 142396(c) 142496(d) 142596 Solution: (a) 142296 Since $\frac{1}{8}$ th of the cube remains unfulfilled, volume of the cube $=22 \times 22 \times 22 \mathrm{~cm}^{3}$ Space filled in the cube $=\left(\frac{7}{8} \times 22 \times 22 \times 22\right) \mathrm{cm}^{3}$ $=(7 \times 133...
Read More →Mewa Lal borrowed Rs 20000 from his friend Rooplal at 18% per annum simple interest.
Question: Mewa Lal borrowed Rs 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years. Solution: SI for Mewa Lal $=\frac{\text { PRT }}{100}$ $=\frac{20,000 \times 18 \times 2}{100}$ $=\operatorname{Rs} 7,200$ Thus, he has to pay Rs 7,200 as interest after borrowing. CI for Mewa Lal $=\mathrm{A}-\mathrm{P}$ $=20,000\left(1+\frac{18}{100}\right)^{2}-20,000$ $=20,000(1.18)^{2}-20,000$ $=27,848-20,00...
Read More →For which values of a and b,
Question: For which values of a and b, the zeroes of q(x) = x3+ 2x2+ a are also the zeroes of the polynomial p(x) = x5 x4 4x3+ 3x2+ 3x + b? Which zeroes of p(x) are not the zeroes of p(x)? Solution: Given that the zeroes of q(x) = x3+ 2x2+ a are also the zeroes of the polynomial p(x) = x5 x4 4x3+ 3xz+ 3x + b i.e., q(x) is a factor of p(x). Then, we use a division algorithm. If (x3+ 2x2+ a) is a factor of (x5 x4 4x3+ 3x2+ 3x + b), then remainder should be zero. i.e., (1 + a) x2+ (3 + 3a) x + (b 2...
Read More →Roma borrowed Rs 64000 from a bank for
Question: Roma borrowed Rs 64000 from a bank for $1 \frac{1}{2}$ years at the rate of $10 \%$ per annum. Compute the total compound interest payable by Roma after $1 \frac{1}{2}$ years, if the interest is compounded half-yearly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 64,000$ $\mathrm{R}=10 \%$ p. a. $\mathrm{n}=1.5$ years $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=64,000\left(1+\frac{10}{200}\right)^{3}$ $=64,000(1.05)^{3}$ $=\mathrm{Rs} 74,088$ Now, $\mathrm...
Read More →If x – √5 is a factor of the cubic polynomial
Question: If x 5 is a factor of the cubic polynomial x3 35x2+ 13x 35, then find all the zeroes of the polynomial. Solution: Let f(x) = x3 35x2+ 13x 35 and given that, (x 5) is a one of the factor of f(x). Now, using division algorithm, $\therefore \quad x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}=\left(x^{2}-2 \sqrt{5} x+3\right) \times(x-\sqrt{5})$ $[\because$ dividend $=$ divisor $\times$ quotient $+$ remainder $]$ $=(x-\sqrt{5})\left[x^{2}-\{(\sqrt{5}+\sqrt{2})+(\sqrt{5}-\sqrt{2})\} x+3\right] \qu...
Read More →If A is a square matrix with |A| = 4 then find the value
Question: If $A$ is a square matrix with $|A|=4$ then find the value of $|A .(a d j A)|$. Solution: Given: $A$ is a square matrix $|A|=4$ Now, $A(\operatorname{adj} A)=|A| I$ $\Rightarrow|A(\operatorname{adj} A)|=|| A|I|$ $\Rightarrow|A(\operatorname{adj} A)|=|A|^{n}|I|$, where $n$ is the order of $I$ $\Rightarrow|A(\operatorname{adj} A)|=|A|^{n} \times 1$ $\Rightarrow|A(\operatorname{adj} A)|=4^{n}$ $(\because|A|=4)$ Hence, $|A(\operatorname{adj} A)|=\underline{4^{n}}$....
Read More →Swati took a loan of Rs 16000 against her insurance policy at the rate of
Question: Swati took a loan of Rs 16000 against her insurance policy at the rate of $12 \frac{1}{2} \%$ per annum. Calculate the total compound interest payable by Swati after 3 years. Solution: Given: $\mathrm{P}=$ Rs 16,000 $\mathrm{R}=12.5 \%$ p. a. $\mathrm{n}=3$ years We know that: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=16,000\left(1+\frac{12.5}{100}\right)^{3}$ $=16,000(1.125)^{3}$ $=\operatorname{Rs} 22,781.25$ Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ ...
Read More →Find the compound interest on Rs 160000 for one year at the rate of 20% per annum,
Question: Find the compound interest on Rs 160000 for one year at the rate of 20% per annum, if the interest is compounded quarterly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 16,000$ $\mathrm{R}=20 \%$ p. a. $\mathrm{n}=1$ year We know that: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{400}\right)^{4 \mathrm{n}}$ $=16,000\left(1+\frac{20}{400}\right)^{4}$ $=16,000(1.05)^{4}$ $=\mathrm{Rs} 19,448.1$ Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 19,448.1-\mathrm{Rs} 16,000$ $=\mathrm{R...
Read More →Find k, so that x2 + 2x + k is a factor of
Question: Find k, so that x2+ 2x + k is a factor of 2x4+ x3 14x2+ 5x + 6. Also, find all the zeroes of the two polynomials. Solution: Given that, x2+ 2x+ k is a factor of 2x4+ x3-14x2+ 5x+ 6, then we apply division algorithm, Since, $\left(x^{2}+2 x+k\right)$ is a factor of $2 x^{4}+x^{3}-14 x^{2}+5 x+6$ So, when we apply division algorithm remainder should be zero. $\therefore \quad(7 k+21) x+\left(2 k^{2}+8 k+6\right)=0 \cdot x+0$ $\Rightarrow \quad 7 k+21=0$ and $2 k^{2}+8 k+6=0$ $\Rightarrow...
Read More →In the following matrix equation use elementary operation
Question: In the following matrix equation use elementary operation $R_{2} \rightarrow R_{2}+R_{1}$ and the equation thus obtained: $\left[\begin{array}{ll}2 3 \\ 1 4\end{array}\right]\left[\begin{array}{rr}1 0 \\ 2 -1\end{array}\right]=\left[\begin{array}{ll}8 -3 \\ 9 -4\end{array}\right]$ Solution: $\left[\begin{array}{ll}2 3 \\ 1 4\end{array}\right]\left[\begin{array}{rr}1 0 \\ 2 -1\end{array}\right]=\left[\begin{array}{rr}8 -3 \\ 9 -4\end{array}\right]$ By applying elementary operation $R_{2...
Read More →Use elementary column operation
Question: Use elementary column operation $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+2 \mathrm{C}_{1}$ in the following matrix equation : $\left(\begin{array}{ll}2 1 \\ 2 0\end{array}\right)=\left(\begin{array}{ll}3 1 \\ 2 0\end{array}\right)\left(\begin{array}{rr}1 0 \\ -1 1\end{array}\right)$ Solution: $\left(\begin{array}{ll}2 1 \\ 2 0\end{array}\right)=\left(\begin{array}{ll}3 1 \\ 2 0\end{array}\right)\left(\begin{array}{rr}1 0 \\ -1 1\end{array}\right)$ Applying $\mathrm{C}_{2} \rightarrow...
Read More →Find the compound interest on Rs 1000 at the rate of
Question: Find the compound interest on Rs 1000 at the rate of $8 \%$ per annum for $1 \frac{1}{2}$ years when interest is compounded half-yearly. Solution: Given: $\mathrm{P}=\mathrm{Rs} 1,000$ $\mathrm{R}=8 \%$ p. a. $\mathrm{n}=1.5$ years We know that: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{200}\right)^{2 \mathrm{n}}$ $=1,000\left(1+\frac{8}{200}\right)^{3}$ $=1,000(1.04)^{3}$ $=\mathrm{Rs} 1,124.86$ Now, $\mathrm{CI}=\mathrm{A}-\mathrm{P}$ $=\mathrm{Rs} 1,124.86-\mathrm{Rs} 1,000$ $...
Read More →The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
Question: The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is(a) 9 : 16(b) 16 : 9(c) 3 : 4(d) 4 : 3 Solution: (b) 16 : 9Let the radii of the spheres beRandr.Then, ratio of their volumes $=\frac{\frac{4}{3} \pi R^{3}}{\frac{4}{3} \pi r^{3}}$ Therefore, $\frac{\frac{4}{3} \pi R^{3}}{\frac{4}{3} \pi r^{3}}=\frac{64}{27}$ $\Rightarrow \frac{R^{3}}{r^{3}}=\frac{64}{27}$ $\Rightarrow\left(\frac{R}{r}\right)^{3}=\left(\frac{4}{3}\right)^{3}$ $\Rightarrow \frac{R}{r}...
Read More →If √2 is a zero of the cubic polynomial
Question: If 2 is a zero of the cubic polynomial 6x3+ 2x2 10x 42, the find its other two zeroes. Solution: Let f(x) = 6x3+ 2x2-10x 42 and given that. 2 is one of the zeroes of f(x) i.e.,(x 2) is one of the factor of given cubic polynomial. Now, using division algorithm, $\therefore \quad 6 x^{3}+\sqrt{2} x^{2}-10 x-4 \sqrt{2}=\left(6 x^{2}+7 \sqrt{2} x+4\right) \times(x-\sqrt{2})+0$ $[\because$ dividend $=$ divisor $x$ quotient $+$ remainder $]$ $=(x-\sqrt{2})\left(6 x^{2}+4 \sqrt{2} x+3 \sqrt{2...
Read More →Rohit deposited Rs 8000 with a finance company for 3 years at an interest of 15% per annum.
Question: Rohit deposited Rs 8000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years? Solution: We know that amount $A$ at the end of n years at the rate of $R \%$ per annum is given by $A=P\left(1+\frac{R}{100}\right)^{n}$. Given: $\mathrm{P}=\mathrm{Rs} 8,000$ $\mathrm{R}=15 \%$ p. a $\mathrm{n}=3$ years Now, $\mathrm{A}=8,000\left(1+\frac{15}{100}\right)^{3}$ $=8,000(1.15)^{3}$ $=\mathrm{Rs} 12,167$ And, CI $=\mathrm...
Read More →The slant height of a bucket is 45 cm and the radii of its top and bottom are 28 cm and 7 cm, respectively.
Question: The slant height of a bucket is 45 cm and the radii of its top and bottom are 28 cm and 7 cm, respectively. The curved surface area of the bucket is(a) 4953 cm2(b) 4952 cm2(c) 4951 cm2(d) 4950 cm2 Solution: (d) 4950 cm2Let the radii of the top and bottom of the bucket beRandrand let its slant height bel. Then, $R=28 \mathrm{~cm}, r=7 \mathrm{~cm}, l=45 \mathrm{~cm}$ Curved surface area of the bucket $=\pi l(R+r)$ $=\frac{22}{7} \times 45 \times(28+7) \mathrm{cm}^{2}$ $=4950 \mathrm{~cm...
Read More →Write the value
Question: Write $A^{-1}$ for $A=\left[\begin{array}{ll}2 5 \\ 1 3\end{array}\right]$ Solution: $|A|=\left|\begin{array}{ll}2 5 \\ 1 3\end{array}\right|=1 \neq 0$ Let $C_{i j}$ be the cofactor of $a_{i j}$ in $A$. The cofactors of element $A$ are given by $C_{11}=3$ $C_{12}=-1$ $C_{21}=-5$ $C_{22}=2$ $\operatorname{adj} A=\left[\begin{array}{cc}3 -1 \\ -5 2\end{array}\right]^{T}=\left[\begin{array}{cc}3 -5 \\ -1 2\end{array}\right]$ $|A|=6-5=1$ $\therefore A^{-1}=\frac{1}{|A|}$ adj $A=\left[\begi...
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