In what time will Rs 1000 amount to Rs 1331 at 10% per annum,
Question: In what time will Rs 1000 amount to Rs 1331 at 10% per annum, compound interest? Solution: Let the time be $\mathrm{n}$ years. Then, $\mathrm{A}=\mathrm{P}\left(1+\frac{10}{100}\right)^{\mathrm{n}}$ $1,331=1,000\left(1+\frac{10}{100}\right)^{\mathrm{n}}$ $(1.1)^{\mathrm{n}}=\frac{1,331}{1,000}$ $(1.1)^{\mathrm{n}}=1.331$ $(1.1)^{\mathrm{n}}=(1.1)^{3}$ On comparing both the sides, we get: n = 3 Thus, the required time is three years....
Read More →If A is a 3×3 non-singular matrix such that
Question: If $\mathrm{A}$ is a $3 \times 3$ non-singular matrix such that $A A^{\top}=A^{\top} A$ and $\mathrm{B}=A^{-1} A^{\top}$, then $B B^{\top}=$_____________ Solution: Given: $A$ is a $3 \times 3$ non-singular matrix $A A^{\top}=A^{\top} A$ $B=A^{-1} A^{\top}$ Now, $B B^{\mathrm{T}}=\left(A^{-1} A^{\mathrm{T}}\right)\left(A^{-1} A^{\mathrm{T}}\right)^{\mathrm{T}}$ $=\left(A^{-1} A^{\mathrm{T}}\right)\left(A^{\mathrm{T}}\right)^{\mathrm{T}}\left(A^{-1}\right)^{\mathrm{T}}$ $=\left(A^{-1} A^...
Read More →If one of the zeroes of the quadratic
Question: If one of the zeroes of the quadratic polynomial $(k-1) x^{2}+k x+1$ is $-3$, then the value of $k$ is (a) $\frac{4}{3}$ (b) $\frac{-4}{3}$ (c) $\frac{2}{3}$ (d) $\frac{-2}{3}$ Solution: (a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x2+ kx + 1 is $-3$, then $\quad p(-3)=0$ $\Rightarrow \quad(k-1)(-3)^{2}+k(-3)+1=0$ $\Rightarrow \quad 9(k-1)-3 k+1=0$ $\Rightarrow \quad 9 k-9-3 k+1=0$ $\Rightarrow \quad 6 k-8=0$ $\therefore$$k=4 / 3$...
Read More →A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter
Question: A solid metallic sphere of diameter $28 \mathrm{~cm}$ is melted and recast into a number of smaller cones, each of diameter $4 \frac{2}{3} \mathrm{~cm}$ and height $3 \mathrm{~cm}$. Find the number of cones so formed. Solution: We have, Radius of the metallic sphere, $R=\frac{28}{2}=14 \mathrm{~cm}$, Radius of the smaller cone, $r=\frac{1}{2} \times\left(4 \frac{2}{3}\right)=\frac{1}{2} \times \frac{14}{3}=\frac{7}{3} \mathrm{~cm}$ and Height of the smaller cone, $h=3 \mathrm{~cm}$ Now...
Read More →The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20.
Question: The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum. Solution: Given: $\mathrm{CI}-\mathrm{SI}=\mathrm{Rs} 20$ $\left[\mathrm{P}\left(1+\frac{4}{100}\right)^{2}-\mathrm{P}\right]-\frac{\mathrm{P} \times 4 \times 2}{100}=20$ $\mathrm{P}\left[\left(1.04^{2}-1\right)\right]-0.08 \mathrm{P}=20$ $0.0816 \mathrm{P}-0.08 \mathrm{P}=20$ $0.0016 \mathrm{P}=20$ $\mathrm{P}=\frac{20}{0.0016}$ $=12,500$ Thus, the required sum is Rs ...
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}0 3 \\ 2 0\end{array}\right]$ and $A^{-1}=\lambda(\operatorname{adj} A)$, then $\lambda=$ Solution: Given: $A=\left[\begin{array}{ll}0 3 \\ 2 0\end{array}\right]$ $A^{-1}=\lambda(\operatorname{adj} A)$ Now, $A=\left[\begin{array}{ll}0 3 \\ 2 0\end{array}\right]$ $\Rightarrow|A|=\left|\begin{array}{ll}0 3 \\ 2 0\end{array}\right|$ $=-6$ As we know that, $A^{-1}=\frac{\operatorname{adj} A}{|A|}$ $\Rightarrow \lambda(\operatorname{adj} A)=\frac{1}{|A|}(\operat...
Read More →In what time will Rs 4400 become Rs 4576 at 8%
Question: In what time will Rs 4400 become Rs 4576 at 8% per annum interest compounded half-yearly? Solution: Let the time period be $\mathrm{n}$ years. $\mathrm{R}=8 \%=4 \%$ (Half $-$ yearly $)$ Thus, we have : $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $4,576=4,400\left(1+\frac{4}{100}\right)^{\mathrm{n}}$ $4,576=4,400(1.04)^{\mathrm{n}}$ $(1.04)^{\mathrm{n}}=\frac{4,576}{4,000}$ $(1.04)^{\mathrm{n}}=1.04$ $(1.04)^{\mathrm{n}}=1.04^{1}$ On comparing both the sid...
Read More →A milk container is made of metal sheet in the shape of frustum of a cone whose volume is
Question: A milk container is made of metal sheet in the shape of frustum of a cone whose volume is $10459 \frac{3}{7} \mathrm{~cm}^{3}$. The radii of its lower and upper circular ends are 8 $\mathrm{cm}$ and $20 \mathrm{~cm}$, respectively. Find the cost of metal sheet used in making the container at the rate of ₹ $1.40$ per $\mathrm{cm}^{2}$. Solution: We have, Radius of the upper end, $R=20 \mathrm{~cm}$ and Radius of the lower end, $r=8 \mathrm{~cm}$ Let the height of the container be $h$. A...
Read More →Show that one and only one out of n,
Question: Show that one and only one out of n, n + 4 , n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer. Solution: Given numbers are n, (n+ 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer. Then, let n = 5q, 5g + 1, 5g + 2, 5g + 3, 5q + 4 for q N [by Euclids algorithm] Then, in each case if we put the different values of n in the given numbers. We definitely get one and only one of given numbers is divisible by 5. Hence, one and only one out of n, n+...
Read More →Solve this
Question: If $A=\left[a_{i j}\right] 2 \times 2$, where $a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$, then $A^{-1}=$ Solution: Given: $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$ $a_{i j}=\left\{\begin{array}{l}i+j, \text { if } i \neq j \\ i^{2}-2 j, \text { if } i=j\end{array}\right.$ $a_{11}=(1)^{2}-2(1)=-1$ $a_{12}=1+2=3$...
Read More →In how much time would Rs 5000 amount to Rs 6655 at 10%
Question: In how much time would Rs 5000 amount to Rs 6655 at 10% per annum compound interest? Solution: Let the time period be $n$ years. Thus, we have : $\mathrm{CI}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}$ $6,655=5,000\left(1+\frac{10}{100}\right)^{\mathrm{n}}-5,000$ $11,655=5,000(1 \cdot 10)^{\mathrm{n}}$ $(1.1)^{\mathrm{n}}=\frac{11,655}{5,000}$ $(1.1)^{\mathrm{n}}=2.331$ $(1.1)^{\mathrm{n}}=(1.1)^{3}$ On comparing both the sides, we get: n =3 Thus, the requi...
Read More →A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3 of water.
Question: A bucket is in the form of a frustum of a cone with a capacity of12308.8 cm3of water. The radii of the top and bottom circular ends are20 cm and 12 cm, respectively. Find the height of the bucket.[Use $\pi=3.14$ ] Solution: We have, Radius of the upper end, $R=20 \mathrm{~cm}$ and Radius of the lower end, $r=12 \mathrm{~cm}$ Let the height of the bucket be $h$. As, Volume of the bucket $=12308.8 \mathrm{~cm}^{3}$ $\Rightarrow \frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)=12308.8$ $\Rig...
Read More →For any positive integer n,
Question: For any positive integer n, prove that n3 n is divisible by 6. Solution: Let $a=n^{3}-n \Rightarrow a=n \cdot\left(n^{2}-1\right)$ $\Rightarrow \quad a=n \cdot(n-1) \cdot(n+1)$ $\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]$ $\Rightarrow \quad a=(n-1) \cdot n \cdot(n+1)$ $\ldots$ (i) We know that, If a number is completely divisible by 2 and 3, then it is also divisible by 6. If the sum of digits of any number is divisible by 3, then it is also divisible by 3. If one of the ...
Read More →If A is a non-singular matrix of order 3,
Question: If $A$ is a non-singular matrix of order 3, then $\operatorname{adj}(\operatorname{adj} A)$ is equal to__________ Solution: Given: $A$ is a non-singular matrix of order 3 As we know, $\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A$, where $n$ is the order of $A$ $\Rightarrow \operatorname{adj}(\operatorname{adj} A)=|A|^{3-2} A \quad(\because$ Order of $A$ is 3$)$ $\Rightarrow \operatorname{adj}(\operatorname{adj} A)=|A|^{1} A$ $\Rightarrow \operatorname{adj}(\operatorname{adj} A)...
Read More →The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum.
Question: The interest on a sum of Rs 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20. Solution: Let the time period be $\mathrm{n}$ years. Then, we have: $\mathrm{CI}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}$ 163. $20=2,000\left(1+\frac{4}{100}\right)^{\mathrm{n}}-2,000$ $2,163.20=2,000(1.04)^{\mathrm{n}}$ $(1.04)^{\mathrm{n}}=\frac{2,163.20}{2,000}$ $(1.04)^{\mathrm{n}}=1.0816$ $(1.04)^{\...
Read More →If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm,
Question: If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, then find its capacity and total surface area. Solution: We have, Height, $h=28 \mathrm{~cm}$, Radius of the upper end, $R=28 \mathrm{~cm}$ and Radius of the lower end, $r=7 \mathrm{~cm}$ Also, The slant height, $l=\sqrt{(R-r)^{2}+h^{2}}$ $=\sqrt{(28-7)^{2}+28^{2}}$ $=\sqrt{21^{2}+28^{2}}$ $=\sqrt{441+784}$ $=\sqrt{1225}$ $=35 \mathrm{~cm}$ Now, Capacity of the bucket $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\...
Read More →Prove that one of any three consecutive
Question: Prove that one of any three consecutive positive integers must be divisible by 3. Solution: Any three consecutive positive integers must be of the form n, (n + 1)and (n + 2), where n is any natural number, i.e., n = 1,2, 3 Let, a = n,b = n+ 1 and c = n + 2 Order triplet is (a, b, c) = (n, n + 1, n + 2), where n = 1,2, 3, (i) At n = 1; (a, b, c) = (1,1 + 1,1 + 2) = (1,2, 3) At n = 2; (a,b,c) = (1,2 + 1,2 + 2) = (2, 3, 4) At n = 3; (a, b, c) = (3, 3 + 1, 3 + 2) = (3, 4, 5) At n = 4; (a, ...
Read More →Rachana borrowed a certain sum at the rate of 15% per annum.
Question: Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1290 as interest compounded annually, find the sum she borrowed. Solution: Let the money borrowed by Rachana be Rs $\mathrm{x}$. Then, we have: $\mathrm{CI}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}-\mathrm{P}$ $1,290=\mathrm{x}\left[\left(1+\frac{15}{100}\right)^{2}-1\right]$ $1,290=\mathrm{x}[0.3225]$ $\mathrm{x}=\frac{1,290}{0.3225}$ = 4,000 Thus, Rachana borrowed R...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ and $A(\operatorname{adj} A)=\left[\begin{array}{ll}k 0 \\ 0 k\end{array}\right]$, then $k=$ Solution: Given: $A=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ $A(\operatorname{adj} A)=\left[\begin{array}{ll}k 0 \\ 0 k\end{array}\right]$ Now, $A=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ $\Rightarrow|A|=\left|\begin{array}{cc}\cos x \sin x \\ -\sin x...
Read More →The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50.
Question: The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum. Solution: Given; $\mathrm{CI}-\mathrm{SI}=\mathrm{Rs} 283.50$ $\mathrm{R}=15 \%$ $\mathrm{n}=3$ years Let the sum be Rs $\mathrm{x}$. We know that: $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$ $=\mathrm{x}\left(1+\frac{15}{100}\right)^{3}$ $=\mathrm{x}(1.15)...
Read More →Prove that one and only one out of n,
Question: Prove that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer, Solution: Let a = n, b = n + 2 and c =n + 4 Order triplet is (a, b, c) = (n, n + 2, n + 4) (i) Where, n is any positive integer i.e.,n = 1,2, 3, At n = 1; (a, b, c) = (1,1 + 2,1 + 4) = (1, 3, 5) At n = 2; (a, b, c) = (2,2 + 2,2 + 4)= (2, 4, 6) At n = 3; (a, b,c) = (3, 3 + 2,3 + 4) = (3,5,7) At n =4; (a,b, c) =(4, 4 + 2, 4 +4) = (4, 6, 8) At n = 5; (a, b,c) = (5, 5 + 2, 5 +4)= (...
Read More →A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.
Question: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphereof same radius. The total height of the toy is 15.5 cm. Find the totalsurface area of the toy. Solution: We have, Radius of the hemisphere $=$ Radius of the cone $=r=3.5 \mathrm{~cm}$ and Height of the cone $=15.5-3.5=12 \mathrm{~cm}$ Also, The slant height of the cone, $l=\sqrt{h^{2}+r^{2}}$ $=\sqrt{12^{2}+3.5^{2}}$ $=\sqrt{144+12.25}$ $=\sqrt{156.25}$ $=12.5 \mathrm{~cm}$ Now, Total surface area of the toy $=$ CSA...
Read More →If k is a scalar and I is a unit matrix of order 3,
Question: If $k$ is a scalar and $/$ is a unit matrix of order 3, then adj $(k l)=$_________ Solution: Given: $l$ is a unit matrix of order 3 As we know, $A(\operatorname{adj} A)=|A| I$ $\Rightarrow A^{-1} A(\operatorname{adj} A)=A^{-1}|A| I$ $\Rightarrow\left(A^{-1} A\right)(\operatorname{adj} A)=A^{-1}|A| I$ $\Rightarrow I(\operatorname{adj} A)=|A|\left(A^{-1} I\right)$ $\Rightarrow \operatorname{adj} A=|A| A^{-1}$ $\Rightarrow \operatorname{adj}(k I)=|k I|(k I)^{-1}$ $\Rightarrow \operatornam...
Read More →Show that the cube of a positive integer
Question: Show that the cube of a positive integer of the form 6q + r,q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r. Solution: Let a be an arbitrary positive integer. Then, by Euclids division algorithm, corresponding to the positive integers a and 6, there exist non-negative integersg and r such that a = 6 q + r, where, 0 r 6 $a=6 q+r$, where, $0 \leq r6$ $\Rightarrow \quad a^{3}=(6 q+r)^{3}=216 q^{3}+r^{3}+3 \cdot 6 q \cdot r(6 q+r)$ $\left[\because(a+b)^{3}=a^{3}+b^{3}+3...
Read More →Let A be a square matrix of order 3 and B
Question: Let $A$ be a square matrix of order 3 and $B=|A| A^{-1}$. If $|\mathrm{A}|=-5$, then $|\mathrm{B}|=$_________ Solution: Given: $A$ is a square matrix of order 3 $B=|A| A^{-1}$ $|A|=-5$ Now, $B=|A| A^{-1}$ $\Rightarrow|B|=|| A\left|A^{-1}\right|$ $\Rightarrow|B|=|A|^{3}\left|A^{-1}\right| \quad(\because$ order of $A$ is 3$)$ $\Rightarrow|B|=|A|^{3} \frac{1}{|A|} \quad\left(\because\left|A^{-1}\right|=\frac{1}{|A|}\right)$ $\Rightarrow|B|=|A|^{2}$ $\Rightarrow|B|=(-5)^{2} \quad(\because|...
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