Expand the following
Question: Expand the following (i) $(3 a-2 b)^{3}$ (ii) $\left(\frac{1}{x}+\frac{y}{3}\right)^{3}$ (iii) $\left(4-\frac{1}{3 x}\right)^{3}$ Solution: (i) $(3 a-2 b)^{3}=(3 a)^{3}+(-2 b)^{3}+3(3 a)(-2 b)(3 a-2 b)$ [using identity, $(a-b)^{3}=a^{3}-b^{3}+3 a(-b)(a-b)$ ] $=27 a^{3}-8 b^{3}-18 a b(3 a-2 b)$ $=27 a^{3}-8 b^{3}-54 a^{2} b+36 a b^{2}$ $=27 a^{3}-54 a^{2} b+36 a b^{2}-8 b^{3}$ (ii) $\left(\frac{1}{x}+\frac{y}{3}\right)^{3}=\left(\frac{1}{x}\right)^{3}+\left(\frac{y}{3}\right)^{3}+3\left...
Read More →Solve this
Question: If $\left|\begin{array}{ccc}a b-y c-z \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$, then using properties of determinants, find the value of $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$, where $x, y, z \neq 0$. Solution: $\left|\begin{array}{ccc}a b-y c-z \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$ $R_{1} \rightarrow R_{1}-R_{2}$ $\left|\begin{array}{ccc}x -y 0 \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$ $R_{2} \rightarrow R_{2}-R_{3}$ $\Rightarrow\left|\begin{array}{ccc}x -y 0 \\ 0 y...
Read More →Solve this
Question: If $\left|\begin{array}{ccc}a b-y c-z \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$, then using properties of determinants, find the value of $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$, where $x, y, z \neq 0$. Solution: $\left|\begin{array}{ccc}a b-y c-z \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$ $R_{1} \rightarrow R_{1}-R_{2}$ $\left|\begin{array}{ccc}x -y 0 \\ a-x b c-z \\ a-x b-y c\end{array}\right|=0$ $R_{2} \rightarrow R_{2}-R_{3}$ $\Rightarrow\left|\begin{array}{ccc}x -y 0 \\ 0 y...
Read More →Find the volume of a cube,
Question: Find the volume of a cube, one face of which has an area of 64 m2. Solution: Area of a face of cube is given by: $A=s^{2}$, where $s=$ Side of the cube Further, volume of a cube is given by: $V=s^{3}$, where $s=$ Side of the cube It is given that the area of one face of the cube is64m2. Therefore we have: $s^{2}=64 \Rightarrow s=\sqrt{64}=8 \mathrm{~m}$ Now, volume is given by: $V=s^{3}=8^{3} \Rightarrow V=8 \times 8 \times 8=512 \mathrm{~m}^{3}$ Thus, the volume of the cube is 512 m3....
Read More →If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2.
Question: If a + b + c = 9 and ab + bc + ca = 26, find a2+ b2+c2. Solution: Given. $a+b+c=9$ and $a b+b c+c a=26$ ...(i) Now, $\quad a+b+c=9$ On squaring both sides, we get $(a+b+c)^{2}=(9)^{2}$ $\Rightarrow \quad a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=81$ [using identity, $\left.(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$ $\Rightarrow \quad a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=81$ $\Rightarrow \quad a^{2}+b^{2}+c^{2}+2(26)=81$ [from Eq. (i)] $\Rightarrow \quad a^{2}+b^{2}+c^{2}=81-52=29$...
Read More →If a,b and c are all non-zero
Question: If $a, b$ and $c$ are all non-zero and $\left|\begin{array}{ccc}1+a 1 1 \\ 1 1+b 1 \\ 1 1 1+c\end{array}\right|=0$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0$ Solution: We have, $\left|\begin{array}{ccc}1+a 1 1 \\ 1 1+b 1 \\ 1 1 1+c\end{array}\right|=0$ $C_{1} \rightarrow C_{1}-C_{2}$ $\left|\begin{array}{ccc}a 1 1 \\ -b 1+b 1 \\ 0 1 1+c\end{array}\right|=0$ $C_{2} \rightarrow C_{2}-C_{3}$ $\left|\begin{array}{ccc}a 0 1 \\ -b b 1 \\ 0 -c 1+c\end{array}\right|=0$ Expandin...
Read More →What happens to the cube of a number if the number is multiplied by
Question: What happens to the cube of a number if the number is multiplied by (i) 3? (ii) 4? (iii) 5? Solution: (i) Let us consider a number $n$. Its cube would be $n^{3}$. If $n$ is multiplied by 3 , it becomes $3 n$. Let us now find the cube of 3n,we get: $(3 n)^{3}=3^{3} \times n^{3}=27 n^{3}$ Therefore, the cube of 3nis 27 times of the cube ofn.Thus, if a number is multiplied by 3, its cube is 27 times of the cube of that number. (ii)Let us consider a number $n$. Its cube would be $n^{3}$. I...
Read More →A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°.
Question: A man on the deck of a ship, 16 m above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 60 and 30. Calculate the distance of the cliff from the ship and height of the cliff. Solution: Let $A B$ be the deck of the ship above the water level and $D E$ be the cliff. Now, $A B=16 \mathrm{~m}$ such that $C D=16 \mathrm{~m}$ and $\angle B D A=30^{\circ}$ and $\angle E B C=60^{\circ}$. If $A D=x \mathrm{~m}$ and $D E=h \mathrm...
Read More →Factorise the following
Question: Factorise the following (i) 9x2+4y2+ 16z2+12xy-16yz -24xz (ii) 25x2+ 16y2+ 4z2 40xy + 16yz 20xz (iii) 16x2+ 4y2+ 9z2 16xy 12yz + 24xz Solution: (i) $9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z$ $=(3 x)^{2}+(2 y)^{2}+(-4 z)^{2}+2(3 x)(2 y)+2(2 y)(-4 z)+2(-4 z)(3 x)$ [using identity, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$ ] $=(3 x+2 y-4 z)^{2}=(3 x+2 y-4 z)(3 x+2 y-4 z)$ (ii) $25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z$ $=(-5 x)^{2}+(4 y)^{2}+(2 z)^{2}+2(-5 x)(4 y)+2(4 y...
Read More →Solve the following determinant equations:
Question: Solve the following determinant equations: (i) $\left|\begin{array}{ccc}x+a b c \\ a x+b c \\ a b x+c\end{array}\right|=0$ (ii) $\left|\begin{array}{ccc}x+a x x \\ x x+a x \\ x x x+a\end{array}\right|=0, a \neq 0$ (iii) $\left|\begin{array}{ccc}3 x-8 3 3 \\ 3 3 x-8 3 \\ 3 3 3 x-8\end{array}\right|=0$ (iv) $\left|\begin{array}{lll}1 x x^{2} \\ 1 a a^{2} \\ 1 b b^{2}\end{array}\right|=0, a \neq b$ (v) $\left|\begin{array}{ccc}x+1 3 5 \\ 2 x+2 5 \\ 2 3 x+4\end{array}\right|=0$ (vi) $\lef...
Read More →Prove that if a number is trebled then its cube is 27
Question: Prove that if a number is trebled then its cube is 27 times the cube of the given number. Solution: Let us consider a number $n$. Then its cube would be $n^{3}$. If the number $n$ is trebled, i.e., $3 n$, we get: $(3 n)^{3}=3^{3} \times n^{3}=27 n^{3}$ It is evident that the cube of 3nis 27 times of the cube ofn. Hence, the statement is proved....
Read More →Expand the following
Question: Expand the following (i) (4a-b + 2c)2 (ii) (3a 5b c)2 (iii) (-x + 2y-3z)2 Solution: (i) $(4 a-b+2 c)^{2}=(4 a)^{2}+(-b)^{2}+(2 c)^{2}+2(4 a)(-b)+2(-b)(2 c)+2(2 c)(4 a)$ [using identity, $\left.(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$ $=16 a^{2}+b^{2}+4 c^{2}-8 a b-4 b c+16 a c$ (ii) $(3 a-5 b-c)^{2}=(3 a)^{2}+(-5 b)^{2}+(-c)^{2}+2(3 a)(-5 b)+2(-5 b)(-c)+2(-c)(3 a)$ [using identity, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$ ] $=9 a^{2}+25 b^{2}+c^{2}-30 a b+10 b c...
Read More →By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
Question: By which smallest number must the following numbers be divided so that the quotient is a perfect cube? (i) 675 (ii) 8640 (iii) 1600 (iv) 8788 (v) 7803 (vi) 107811 (vii) 35721 (viii) 243000 Solution: (i)On factorising 675 into prime factors, we get: $675=3 \times 3 \times 3 \times 5 \times 5$ On grouping the factors in triples of equal factors, we get: $675=\{3 \times 3 \times 3\} \times 5 \times 5$ It is evident that the prime factors of 675 cannot be grouped into triples of equal fact...
Read More →Factorise the following
Question: Factorise the following (i) 9x2-12x+ 3 (ii) 9x2-12xy + 4 Solution: (i) $9 x^{2}-12 x+3=3\left(3 x^{2}-4 x+1\right)$ [taking 3 common factor] $=3\left(3 x^{2}-3 x-x+1\right)$[by splitting middle term] $=3[3 x(x-1)-1(x-1)]=3(3 x-1)(x-1)$ (ii) $9 x^{2}-12 x+4=(3 x)^{2}-2 \times 3 x \times 2+(2)^{2}$ $=(3 x-2)^{2}$[using identity, $(a-b)^{2}=a^{2}-2 a b+b^{2}$ ] $=(3 x-2)(3 x-2)$...
Read More →The angles of depression of the top and bottom of a tower as seen from the top of a 60
Question: The angles of depression of the top and bottom of a tower as seen from the top of a $60 \sqrt{3}-\mathrm{m}-\mathrm{high}$ cliff are $45^{\circ}$ and $60^{\circ}$, respectively. Find the height of the tower. Solution: Let AD be the tower and BC be the cliff.We have, $\mathrm{BC}=60 \sqrt{3} \mathrm{~m}, \angle \mathrm{CDE}=45^{\circ}$ and $\angle \mathrm{BAC}=60^{\circ}$ Let $\mathrm{AD}=h$ $\Rightarrow \mathrm{BE}=\mathrm{AD}=h$ $\Rightarrow \mathrm{CE}=\mathrm{BC}-\mathrm{BE}=60 \sqr...
Read More →Factorise the following
Question: Factorise the following (i) $4 x^{2}+20 x+25$ (ii) $9 y^{2}-66 y z+121 z^{2}$ (iii) $\left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}$ Solution: (i) $4 x^{2}+20 x+25=(2 x)^{2}+2 \times 2 x \times 5+(5)^{2}$ $=(2 x+5)^{2}$[using identity, $a^{2}+2 a b+b^{2}=(a+b)^{2}$ ] (ii) $9 y^{2}-66 y z+121 z^{2}=(3 y)^{2}-2 \times 3 y \times 11 z+(11 z)^{2}$ $=(3 y-11 z)^{2} \quad$ [using identity, $a^{2}-2 a b+b^{2}=(a-b)^{2}$ ] (iii) $\left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{...
Read More →Show that x = 2 is a root of the equation
Question: Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x x+2\end{array}\right|=0$ and solve it completely. Solution: Let $\Delta=\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x x+2\end{array}\right|$ $=\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3-x 2 x+6 x+3\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$ $=(x+3)\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -1 2 1\end{array}\right|$ ...
Read More →What is the smallest number by which the following numbers must be multiplied,
Question: What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes? (i) 675 (ii) 1323 (iii) 2560 (iv) 7803 (v) 107811 (vi) 35721 Solution: (i)On factorising 675 into prime factors, we get: $675=3 \times 3 \times 3 \times 5 \times 5$ On grouping the factors in triples of equal factors, we get: $675=\{3 \times 3 \times 3\} \times 5 \times 5$ It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such ...
Read More →Using suitable identity, evaluate the following
Question: Using suitable identity, evaluate the following (i) 1033 (ii) 101 x 102 (iii) 9992 Thinking Process Firstlyadjust the given number into two number such that one is a multiple of 10 and use the proper identity. Solution: (i) $103^{3}=(100+3)^{3}$ $=(100)^{3}+(3)^{3}+3 \times 100 \times 3 \times(100+3)$ [using identity, $\left.(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$ $=1000000+27+900(103)$ $=1000027+92700=1092727$ (ii) $101 \times 102=(100+1)(100+2)$ $=(100)^{2}+100(1+2)+1 \times 2 \quad...
Read More →Show that x = 2 is a root of the equation
Question: Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x x+2\end{array}\right|=0$ and solve it completely. Solution: Let $\Delta=\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x x+2\end{array}\right|$ $=\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3-x 2 x+6 x+3\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$ $=(x+3)\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -1 2 1\end{array}\right|$ ...
Read More →The angle of elevation of the top of a vertical tower from a point on the ground is 60°.
Question: The angle of elevation of the top of a vertical tower from a point on the ground is 60. From another point 10 m vertically above the first, its angle of elevation is 30. Find the height of the tower. Solution: Let PQ be the tower.We have, $\mathrm{AB}=10 \mathrm{~m}, \angle \mathrm{MAP}=30^{\circ}$ and $\angle \mathrm{PBQ}=60^{\circ}$ Also, $\mathrm{MQ}=\mathrm{AB}=10 \mathrm{~m}$ Let $\mathrm{BQ}=x$ and $\mathrm{PQ}=h$ So, $\mathrm{AM}=\mathrm{BQ}=x$ and $\mathrm{PM}=\mathrm{PQ}-\math...
Read More →Prove the following
Question: Factorise (i) 2x3-3x2-17x + 30 (ii) x3-6x2+11 x-6 (iii) x3+ x2 4x 4 (iv) 3x3 x2 3x +1 Thinking Process (i) Firstly, find the prime factors of constant term in given polynomial. (ii) Substitute these factors in place of x in given polynomial and find the minimum factor which satisfies given polynomial and write it in the form of a linear polynomial. (iii) Now, adjust the given polynomial in such a way that it becomes the product of two factors, one of them is a linear polynomial and oth...
Read More →Show that
Question: If $\left|\begin{array}{lll}p b c \\ a q c \\ a b r\end{array}\right|=0$, find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}, p \neq a, q \neq b, r \neq c$ Solution: Let $\Delta=\left|\begin{array}{lll}p b c \\ a q c \\ a b r\end{array}\right|$. Now, $\Delta=\left|\begin{array}{lll}p b c \\ a q c \\ a b r\end{array}\right|$ $=\left|\begin{array}{ccc}p b c \\ 0 q-b c-r \\ a b r\end{array}\right|$ $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$ $=p[r(q-b)-...
Read More →Which of the following are cubes of odd natural numbers?
Question: Which of the following are cubes of odd natural numbers? 125, 343, 1728, 4096, 32768, 6859 Solution: We know that the cubes of all odd natural numbers are odd. The numbers 125, 343, and 6859 are cubes of odd natural numbers.Any natural numbers could be either even or odd. Therefore, if a natural number is not even, it is odd. Now, the numbers 125, 343 and 6859 are odd (It could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8). Non...
Read More →Which of the following are cubes of even natural numbers?
Question: Which of the following are cubes of even natural numbers? 216, 512, 729, 1000, 3375, 13824 Solution: We know that the cubes of all even natural numbers are even.The numbers 216, 512, 1000 and 13824 are cubes of even natural numbers.The numbers 216, 512, 1000 and 13824 are even and it could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.Thus, the cubes of even natural numbers are 216, 512, 1000 and 13824....
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