What sum of money will amount to Rs 45582.25 at
Question: What sum of money will amount to Rs $45582.25$ at $6 \frac{3}{4} \%$ per annum in two years, interest being compounded annually? Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $45,582.25=\mathrm{P}\left(1+\frac{27}{400}\right)^{2}$ $\mathrm{P}(1.0675)^{2}=45,582.25$ $\mathrm{P}=\frac{45,582.25}{1.13955625}$ $\mathrm{P}=40,000$ Thus, the required sum is Rs 40,000 ....
Read More →If A, B, C are three non-: square matrices of the same order,
Question: IfA,B,Care three non-: square matrices of the same order, write the condition onAsuch thatAB=AC⇒B=C. Solution: Consider $A B=A C$. On multiplying both sides by $A^{-1}$, we get $A A^{-1} B=A A^{-1} C$ $\Rightarrow I B=I C \quad$ [Because $A A^{-1}=I$ where $I$ is the identity matrix] $\Rightarrow B=C$ Therefore, the required condition is $A$ must be invertible or $|A| \neq 0$....
Read More →The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378.
Question: The compound interest on Rs 1800 at 10% per annum for a certain period of time is Rs 378. Find the time in years. Solution: $\mathrm{CI}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}$ $\Rightarrow 378=1,800\left(1+\frac{10}{100}\right)^{\mathrm{n}}-1,800$ $1,800\left(1+\frac{10}{100}\right)^{\mathrm{n}}=2,178$ $\left(1+\frac{10}{100}\right)^{\mathrm{n}}=\frac{2,178}{1,800}$ $(1.1)^{\mathrm{n}}=1.21$ $(1.1)^{\mathrm{n}}=(1.1)^{2}$ On comparing both the sides, w...
Read More →If the zeroes of the quadratic polynomial
Question: If the zeroes of the quadratic polynomial ax2+ bx+ c, where c0, are equal, then (a) c and a have opposite signs (b) c and b have opposite signs (c) c and a have same signs (d) c and b have the same signs Solution: (c) The zeroes of the given quadratic polynomial ax2+ bx + c, c 0 are equal. If coefficient of x2and constant term have the same sign i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero. e.g., (i) $x^{2}+4 x+4=0$ (ii) $x^{2}-...
Read More →If A is a non-singular square matrix
Question: If $A$ is a non-singular square matrix such that $|A|=10$, find $\left|A^{-1}\right|$. Solution: For any non-singular matrixA, $\left|A^{-1}\right|=\frac{1}{|A|}$ $\therefore\left|A^{-1}\right|=\frac{1}{10} \quad[\because|A|=10]$...
Read More →If A is a non-singular square matrix
Question: If $A$ is a non-singular square matrix such that $|A|=10$, find $\left|A^{-1}\right|$. Solution: For any non-singular matrixA, $\left|A^{-1}\right|=\frac{1}{|A|}$ $\therefore\left|A^{-1}\right|=\frac{1}{10} \quad[\because|A|=10]$...
Read More →At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years
Question: At what rate percent will a sum of Rs 1000 amount to Rs 1102.50 in 2 years at compound interest? Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $1102.50=1000\left(1+\frac{\mathrm{R}}{100}\right)^{2}$ $\frac{1102.50}{1000}=(1+0.01 \mathrm{R})^{2}$ $(1+0.01 \mathrm{R})^{2}=1.1025$ $(1+0.01 \mathrm{R})^{2}=(1.05)^{2}$ On comparing both the sides, we get: $1+0.01 \mathrm{R}=1.05$ $0.01 \mathrm{R}=0.05$ $\mathrm{R}=5$ Thus, the required rate percent is 5...
Read More →The zeroes of the quadratic polynomial x2 + kx + k
Question: The zeroes of the quadratic polynomial x2+ kx + k where k 0, (a) cannot both be positive (b) cannot both be negative (c) are always unequal (d) are always equal Solution: (a)Let p(x) = x2+ kx + k, k0 On comparing p(x) with ax2+ bx + c, we get $a=1, b=k$ and $c=k$ Here, we see that k(k 4) 0 ⇒ k (-, 0) u (4, ) Now, we know that In quadratic polynomial ax2+ bx + c If a 0, b 0, c 0 or a 0, b 0,c 0, then the polynomial has always all negative zeroes. and if a 0, c 0 or a 0, c 0, then the po...
Read More →If A is a square matrix of order 3 such that |adj A| = 64, find |A|.
Question: IfAis a square matrix of order 3 such that |adjA| = 64, find |A|. Solution: For any square matrix of ordern, $|a d j A|=|A|^{n-1}$ $\Rightarrow 64=|A|^{2} \quad[\because|a d j A|=64]$ $\Rightarrow|A|=\pm 8$...
Read More →If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.
Question: If $A$ is a square matrix of order 3 such that $|A|=5$, write the value of $|\operatorname{adj} A|$. Solution: For any square matrix of ordern, $|\operatorname{adj} A|=|A|^{n-1}$ $\Rightarrow|\operatorname{adj} A|=|A|^{2}=5^{2}=25$...
Read More →If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.
Question: If $A$ is a square matrix of order 3 such that $|A|=5$, write the value of $|\operatorname{adj} A|$. Solution: For any square matrix of ordern, $|\operatorname{adj} A|=|A|^{n-1}$ $\Rightarrow|\operatorname{adj} A|=|A|^{2}=5^{2}=25$...
Read More →In how much time will a sum of Rs 1600 amount to Rs 1852.20 at 5%
Question: In how much time will a sum of Rs 1600 amount to Rs 1852.20 at 5% per annum compound interest? Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $1852.20=1600\left(1+\frac{5}{100}\right)^{\mathrm{n}}$ $\frac{1852.20}{1600}=(1.05)^{\mathrm{n}}$ $(1.05)^{\mathrm{n}}=1.157625$ $(1.05)^{\mathrm{n}}=(1.05)^{3}$ On comparing both the sides, we get: n = 3 Thus, the required time is three years....
Read More →The zeroes of the quadratic polynomial
Question: The zeroes of the quadratic polynomial x2+ 99x + 127 are (a) both positive (b) both negative (c) one positive and one negative (d) both equal Solution: (b) Let given quadratic polynomial be p(x) =x2+ 99x + 127. On comparing p(x) with ax2+ bx + c, we get a = 1, b = 99 and c = 127 We know that, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ [by quadratic formula] $=\frac{-99 \pm \sqrt{(99)^{2}-4 \times 1 \times 127}}{2 \times 1}$ $=\frac{-99 \pm \sqrt{9801-508}}{2}$ $=\frac{-99 \pm \sqrt{9293...
Read More →Choose the correct answer of the following question:
Question: Choose the correct answer of the following question:A metallic solid sphere of radius 9 cm is melted to form a solid cylinderof radius 9 cm. The height of the cylinder is(a) 12 cm (b) 18 cm (c) 36 cm (d) 96 cm Solution: We have, Radius of the solid sphere, $R=9 \mathrm{~cm}$ and Radius of the solid cylinder, $r=9 \mathrm{~cm}$ Let the height of the cylinder be $h$. Now, Volume of the cylinder $=$ Volume of the sphere $\Rightarrow \pi r^{2} h=\frac{4}{3} \pi R^{3}$ $\Rightarrow h=\frac{...
Read More →A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years.
Question: A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited. Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $10,404=\mathrm{P}\left(1+\frac{2}{100}\right)^{2}$ $10,404=\mathrm{P}(1.02)^{2}$ $\mathrm{P}=\frac{10,404}{1.0404}$ $\mathrm{P}=10,000$ Thus, the required sum is Rs 10,000 ....
Read More →Choose the correct answer of the following:
Question: Choose the correct answer of the following:The radius (in cm) of the largest right circular cone that can be cut outfrom a cube of edge 4.2 cm is(a) 2.1 (b) 4.2 (c) 8.4 (d) 1.05 Solution: Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm So, the radius of the base of the largest cone $=\frac{4.2}{2}=2.1 \mathrm{~cm}$ Hence, the correct answer is option (a)....
Read More →At what rate percent per annum will a sum of Rs 4000
Question: At what rate percent per annum will a sum of Rs 4000 yield compound interest of Rs 410 in 2 years? Solution: Let the rate percent be $\mathrm{R}$. We know that: $\mathrm{CI}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}$ $410=4,000\left(1+\frac{\mathrm{R}}{100}\right)^{2}-4,000$ $4,410=4,000\left(1+\frac{\mathrm{R}}{100}\right)^{2}$ $\left(1+\frac{\mathrm{R}}{100}\right)^{2}=\frac{4,410}{4,000}$ $\left(1+\frac{\mathrm{R}}{100}\right)^{2}=1.1025$ $\left(1+\frac...
Read More →If one of the zeroes of the cubic polynomial
Question: If one of the zeroes of the cubic polynomial x3+ ax2+ bx + c is -1, then the product of the other two zeroes is (a) b a +1 (b) b a -1 (c) a b +1 (d) a b -1 Solution: (a) Let p(x) = x3+ ax2+ bx + c Let a, p and y be the zeroes of the given cubic polynomial p(x). = -1 [given] and p(1) = 0 ⇒ (-1)3+ a(-1)2+ b(-1) + c = 0 ⇒ -1 + a- b + c = 0 ⇒ c = 1 -a + b (i) We know that, Product of all zeroes $=(-1)^{3} \cdot \frac{\text { Constant term }}{\text { Coefficient of } x^{3}}=-\frac{c}{1}$ $\...
Read More →If A is a square matrix such that A (adj A) 5I,
Question: If $A$ is a square matrix such that $A(\operatorname{adj} A) 5 /$, where $/$ denotes the identity matrix of the same order. Then, find the value of $|A|$. Solution: We know $A(\operatorname{adj} A)=|A| I$ Here, $A(\operatorname{adj} A)=5 I$ $\therefore|A|=5$...
Read More →A solid piece of iron in the form a cuboid of dimensions (49 cm × 33 cm × 24 cm) is moulded into a solid sphere.
Question: A solid piece of iron in the form a cuboid of dimensions (49 cm 33 cm 24 cm) is moulded into a solid sphere. The radius of the sphere is(a) 19 cm(b) 21 cm(c) 23 cm(d) 25 cm Solution: (b) 21 cm Volume of the cuboid $=(l \times b \times h)=49 \times 33 \times 24 \mathrm{~cm}^{3}$ Let the radius of the sphere be $r \mathrm{~cm}$. Volume of the sphere $=\frac{4}{3} \pi r^{3}$ The volume of the sphere and the cuboidare the same.Therefore, $\frac{4}{3} \pi r^{3}=49 \times 33 \times 24$ $\Rig...
Read More →Ishita invested a sum of Rs 12000 at 5% per annum compound interest.
Question: Ishita invested a sum of Rs 12000 at 5% per annum compound interest. She received an amount of Rs 13230 afternyears. Find the value ofn. Solution: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $13,230=12,000\left(1+\frac{5}{100}\right)^{\mathrm{n}}$ $(1.05)^{\mathrm{n}}=\frac{13,230}{12,000}$ $(1.05)^{\mathrm{n}}=1.1025$ $(1.05)^{\mathrm{n}}=(1.05)^{2}$ On comparing both the sides, we get: n = 2 Thus, the value of $\mathrm{n}$ is two years....
Read More →The difference in simple interest and compound interest on a certain sum of money
Question: The difference in simple interest and compound interest on a certain sum of money at $6 \frac{2}{3} \%$ per annum for 3 years is Rs 46 . Determine the sum. Solution: Given: $\mathrm{CI}-\mathrm{SI}=46$ $\mathrm{P}\left[\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-1\right]-\frac{\mathrm{PRT}}{100}=46$ $\mathrm{P}\left[\left(1+\frac{20}{300}\right)^{3}-1\right]-\frac{\mathrm{P} \times 20 \times 3}{3 \times 100}=46$ $\frac{4,096}{3,375} \mathrm{P}-\frac{\mathrm{P}}{5}-\mathrm{P}=46$...
Read More →Write the adjoint of the matrix
Question: Write the adjoint of the matrix $A=\left[\begin{array}{cc}-3 4 \\ 7 -2\end{array}\right]$. Solution: Let $C_{i j}$ be a cofactor of $a_{i j}$ in $A$. Now, $C_{11}=-2$ $C_{12}=-7$ $C_{21}=-4$ $C_{22}=-3$ $\therefore \operatorname{adj} A=\left[\begin{array}{ll}-2 -7 \\ -4 -3\end{array}\right]^{T}=\left[\begin{array}{ll}-2 -4 \\ -7 -3\end{array}\right]$...
Read More →The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360.
Question: The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum. Solution: Let the sum be $P$. Thus, we have: $\mathrm{CI}-\mathrm{SI}=360$ $\left[\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}\right]-\frac{\mathrm{P} \times 7.5 \times 2}{100}=360$ $\mathrm{P}\left[\left(1+\frac{7.5}{100}\right)^{2}-1\right]-\frac{\mathrm{P} \times 7.5 \times 2}{100}=360$ $\mathrm{P}[1.155625-1]-0.15 \mathr...
Read More →Solve this
Question: If $A=\left[\begin{array}{lll}x 5 2 \\ 2 y 3 \\ 1 1 z\end{array}\right], x y z=80,3 x+2 y+10 z=20$ and $A \operatorname{adj} A=k l$, then $k=$__________ Solution: Given: $A=\left[\begin{array}{lll}x 5 2 \\ 2 y 3 \\ 1 1 z\end{array}\right]$ $x y z=80$ $3 x+2 y+10 z=20$ $A(\operatorname{adj} A)=k l$ Now, $A=\left[\begin{array}{lll}x 5 2 \\ 2 y 3 \\ 1 1 z\end{array}\right]$ $\Rightarrow|A|=\left|\begin{array}{lll}x 5 2 \\ 2 y 3 \\ 1 1 z\end{array}\right|$ $\Rightarrow|A|=x(y z-3)-2(5 z-2)...
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