Find the value of
Question: Find the value of (i) x3+y3-12xy + 64,when x+y = -4. (ii) x3-8y3-36xy-216,when x = 2y + 6. Solution: (i) Here, we see that, $x+y+4=0$ $\therefore \quad x^{3}+y^{3}+(4)^{3}=3 x y(4)$ [using identily, $a+b+c=0$, then $a^{3}+b^{3}+c^{3}=3 a b c$ ] $=12 x y$ ...(i) Now, $\quad x^{3}+y^{3}-12 x y+64=x^{3}+y^{3}+64-12 x y$ $=12 x y-12 x y=0$ [from Eq. (i)] (ii) Here, we see that, $x-2 y-6=0$ $\therefore$$x^{3}+(-2 y)^{3}+(-6)^{3}=3 x(-2 y)(-6)$ [using identity, $a+b+c=0$, then $a^{3}+b^{3}+c...
Read More →By taking three different values of n verify the truth of the following statements:
Question: By taking three different values ofnverify the truth of the following statements: (i) Ifnis even , thenn3is also even. (ii) Ifnis odd, thenn3is also odd. (iii) Ifnleaves remainder 1 when divided by 3, thenn3also leaves 1 as remainder when divided by 3. (iv) If a natural numbernis of the form 3p+ 2 thenn3also a number of the same type. Solution: (i)Let the three even natural numbers be 2, 4 and 8. Cubes of these numbers are: $2^{3}=8,4^{3}=64,8^{3}=512$ By divisibility test, it is evide...
Read More →Using determinants show that the following points are collinear:
Question: Using determinants show that the following points are collinear: (i) $(5,5),(-5,1)$ and $(10,7)$ (ii) $(1,-1),(2,1)$ and $(4,5)$ (iii) $(3,-2),(8,8)$ and $(5,2)$ (iv) $(2,3),(-1,-2)$ and $(5,8)$ Solution: (i) If the points $(5,5),(-5,1)$ and $(10,7)$ are collinear, then $\Delta=\left|\begin{array}{ccc}5 5 1 \\ -5 1 1 \\ 10 7 1\end{array}\right|=0$ $=\left|\begin{array}{ccc}5 5 1 \\ -10 -4 0 \\ 10 7 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\ri...
Read More →As observed form the top of a lighthouse, 100 m above sea level,
Question: As observed form the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30 to 60. Determine the distance travelled by the ship during the period of observation. Solution: Let $O A$ be the lighthouse and $B$ and $C$ be the two positions of the ship. Thus, we have: $O A=100 \mathrm{~m}, \angle O B A=30^{\circ}$ and $\angle O C A=60^{\circ}$ Let: $O C=x \mathrm{~m}$ and $B C=y \mathrm{~m}$ In the right $\triangle O A C$...
Read More →Using determinants show that the following points are collinear:
Question: Using determinants show that the following points are collinear: (i) $(5,5),(-5,1)$ and $(10,7)$ (ii) $(1,-1),(2,1)$ and $(4,5)$ (iii) $(3,-2),(8,8)$ and $(5,2)$ (iv) $(2,3),(-1,-2)$ and $(5,8)$ Solution: (i) If the points $(5,5),(-5,1)$ and $(10,7)$ are collinear, then $\Delta=\left|\begin{array}{ccc}5 5 1 \\ -5 1 1 \\ 10 7 1\end{array}\right|=0$ $=\left|\begin{array}{ccc}5 5 1 \\ -10 -4 0 \\ 10 7 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\ri...
Read More →Without finding the cubes,
Question: Without finding the cubes, factorise (x- 2y)3+ (2y 3z)3+ (3z x)3. Solution: We know that, a3+ b3+ c3 3 abc = (a + b + c)(a2+ b2+ c2-ab-bc-ca) Also, if a + b + c = 0, then a3+ b3+ c3= 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x)....
Read More →Without actually calculating the cubes,
Question: Without actually calculating the cubes, find the value of 36xy-36xy = 0 (i) $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$ (ii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$ Thinking Process In this method firstly check the values of a + b+ c, then . if a + b+c = Q, now use the identity a3+ b3+ c3 = 3abc. Solution: (i) Given, $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$ or $\left(\frac{1}{2}\right)^{3}+\left(\frac{...
Read More →For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be
Question: For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be (a) multiplied so that the product is a perfect cube. (b) divided so that the quotient is a perfect cube. Solution: The only non-perfect cube in question number 20 is 243. (a)On factorising 243 into prime factors, we get: $243=3 \times 3 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors, we get: $243=\{3 \times 3 \times 3\} \times 3 \times 3$ It is evident that the ...
Read More →The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30°
Question: The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30 On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60. Show that the height of the tower is 129.9 metres. Solution: Let $A B$ be the tower. We have: $C D=150 \mathrm{~m} . \angle A C B=30^{\circ}$ and $\angle A D B=60^{\circ}$ Let: $A B=h \mathrm{~m}$ and $B D=x \mathrm{~m}$ In the right $\triangle A B D$, we have: $\frac{A B}{A D}=\tan 60^{\circ}=\sq...
Read More →Prove the following
Question: Factorise (i) a3-8b3-64c3-2Aabc (ii) 22a3+8b3-27c3+182abc Solution: (i) $a^{3}-8 b^{3}-64 c^{3}-24 a b c=(a)^{3}+(-2 b)^{3}+(-4 c)^{3}-3 \times(a) \times(-2 b) \times(-4 c)$ $=(a-2 b-4 c)\left[(a)^{2}+(-2 b)^{2}+(-4 c)^{2}-a(-2 b)-(-2 b)(-4 c)-(-4 c)(a)\right]$ [using identity, $\left.a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$ $=(a-2 b-4 c)\left(a^{2}+4 b^{2}+16 c^{2}+2 a b-8 b c+4 a c\right)$ (ii) $2 \sqrt{2} a^{3}+8 b^{3}-27 c^{3}+18 \sqrt{2} ...
Read More →Which of the following numbers are not perfect cubes?
Question: Which of the following numbers are not perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1728 Solution: (i)On factorising 64 into prime factors, we get: $64=2 \times 2 \times 2 \times 2 \times 2 \times 2$ On grouping the factors in triples of equal factors, we get: $64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$ It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube. (ii)On factorising 2...
Read More →Find the area of the triangle with vertices at the points:
Question: Find the area of the triangle with vertices at the points: (i) $(3,8),(-4,2)$ and $(5,-1)$ (ii) $(2,7),(1,1)$ and $(10,8)$ (iii) $(-1,-8),(-2,-3)$ and $(3,2)$ (iv) $(0,0),(6,0)$ and $(4,3)$. Solution: (i) $\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 8 1 \\ -4 2 1 \\ 5 -1 1\end{array}\right|$ $\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 8 1 \\ -7 -6 0 \\ 5 -1 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$ $\Delta=\frac{1}{2}\left|\begin{arr...
Read More →Prove the following
Question: Fi nd (2x y + 3z) (4x2+ y2+ 9z2+ 2xy + 3yz 6xz). Solution: $(2 x-y+3 z)\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$ $=2 x\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)-y\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$ $+3 z\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$ $=8 x^{3}+2 x y^{2}+18 x z^{2}+4 x^{2} y+6 x y z-12 x^{2} z-4 x^{2} y-y^{3}-9 y z^{2}-2 x y^{2}$ $-3 y^{2} z+6 x y z+12 x^{2} z+3 y^{2} z+27 z^{3}+6 x y z+9 y z^{2}-18 x z^{2}$ $=8 x^{3}+\left(...
Read More →Find the area of the triangle with vertices at the points:
Question: Find the area of the triangle with vertices at the points: (i) $(3,8),(-4,2)$ and $(5,-1)$ (ii) $(2,7),(1,1)$ and $(10,8)$ (iii) $(-1,-8),(-2,-3)$ and $(3,2)$ (iv) $(0,0),(6,0)$ and $(4,3)$. Solution: (i) $\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 8 1 \\ -4 2 1 \\ 5 -1 1\end{array}\right|$ $\Delta=\frac{1}{2}\left|\begin{array}{ccc}3 8 1 \\ -7 -6 0 \\ 5 -1 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$ $\Delta=\frac{1}{2}\left|\begin{arr...
Read More →Find the cubes of the following numbers by column method:
Question: Find the cubes of the following numbers by column method: (i) 35 (ii) 56 (iii) 72 Solution: (i) We have to find the cube of 35 using column method. We have: $a=3$ and $b=5$ Thus, cube of 35 is 42875. (ii)We have to find the cube of 56 using column method. We have: $a=5$ and $b=6$ Thus, cube of 56 is 175616. (iii)We have to find the cube of 72 using column method. We have: $a=7$ and $b=2$ Thus, cube of 72 is 373248....
Read More →The angle of elevation of an aeroplane from a point on the ground is 45°.
Question: The angle of elevation of an aeroplane from a point on the ground is 45. After flying for 15 seconds, the elevation changes to 30. If the aeroplane is flying at a height of 2500 metres, find the speed of the aeroplane. Solution: Let the height of flying of the aeroplane be PQ = BCand point A be the point of observation.We have, $\mathrm{PQ}=\mathrm{BC}=2500 \mathrm{~m}, \angle \mathrm{PAQ}=45^{\circ}$ and $\angle \mathrm{BAC}=30^{\circ}$ In $\Delta \mathrm{PAQ}$, $\tan 45^{\circ}=\frac...
Read More →Prove the following
Question: Factorise (i) 1 + 64x3 (ii) a3-22b3 Thinking Process (i) Firstly, adjust the given number either in the farm 0f a3+ b3or a3-b3 (ii) Further, use any of the identities i.e., a3+ b3=(a + b)(a2+b2-ab) and a3-b3=(a-b)(a2+ b2+ ab), then simplify it, to get the factor. Solution: (i) $1+64 x^{3}=(1)^{3}+(4 x)^{3}$ $=(1+4 x)\left[(1)^{2}-(1)(4 x)+(4 x)^{2}\right]$ [using identity, $\left.a^{3}-b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$ $=(1+4 x)\left(1-4 x+16 x^{2}\right)$ (ii) $a^{3}-2 \...
Read More →Find the cubes of the following numbers by column method:
Question: Find the cubes of the following numbers by column method: (i) 35 (ii) 56 (iii) 72 Solution: (i) We have to find the cube of 35 using column method. We have: $a=3$ and $b=5$...
Read More →Find the following products
Question: Find the following products (i) $\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$ (ii) $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$ Solution: (i) $\left(\frac{x}{2}+2 y\right)\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)=\frac{x}{2}\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)+2 y\left(\frac{x^{2}}{4}-x y+4 y^{2}\right)$ $=\frac{x}{2}\left(\frac{x^{2}}{4}\right)+\frac{x}{2}(-x y)+\frac{x}{2}\left(4 y^{2}\right)+2 y\left(\frac{x^{2}}{4}\right)+2 y(-x y)+2 y\left(4 y^{2}\r...
Read More →Write the units digit of the cube of each of the following numbers:
Question: Write the units digit of the cube of each of the following numbers: 31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125 Solution: Properties: If a numbers ends with digits 1, 4, 5, 6 or 9, its cube will have the same ending digit. If a number ends with 2, its cube will end with 8. If a number ends with 8, its cube will end with 2. If a number ends with 3, its cube will end with 7. If a number ends with 7, its cube will end with 3. From the above properties, we get: Cube of the numbe...
Read More →Evaluate the following:
Question: Evaluate the following: (i) $\left\{\left(5^{2}+12^{2}\right)^{1 / 2}\right\}^{3}$ (ii) $\left\{\left(6^{2}+8^{2}\right)^{1 / 2}\right\}^{3}$ Solution: (i)To evaluate the value of the given expression, we can proceed as follows: $\left\{\left(5^{2}+12^{2}\right)^{1 / 2}\right\}^{3}$ $=\left\{(25+144)^{1 / 2}\right\}^{3}$ $=\left\{(169)^{1 / 2}\right\}^{3}$ $=\{\sqrt{(169)}\}^{3}$ $=\{\sqrt{13 \times 13}\}^{3}$ $=\{13\}^{3}$ $=13 \times 13 \times 13=2197$ (ii)To evaluate the value of th...
Read More →Using properties of determinants,
Question: Using properties of determinants, prove that $\left|\begin{array}{ccc}1 1 1+3 x \\ 1+3 y 1 1 \\ 1 1+3 z 1\end{array}\right|=9(3 x y z+x y+y z+z x)$. Solution: Let $\Delta=\left|\begin{array}{ccc}1 1 1+3 x \\ 1+3 y 1 1 \\ 1 1+3 z 1\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$ $\Rightarrow \Delta=\left|\begin{array}{ccc}1 1 1+3 x \\ 3 y 0 -3 x \\ 0 3 z -3 x\end{array}\right|$ Expanding along $R_{1}$, we get $\Delta=1(0+9 x z)-1(-9 x y-0)+(1+3...
Read More →Factorise the following
Question: Factorise the following (i) $1-64 a^{3}-12 a+48 a^{2}$ (ii) $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$ Solution: (i) $1-64 a^{3}-12 a+48 a^{2}=(1)^{3}-(4 a)^{3}-3 \times 1^{2} \times 4 a+3 \times 1 \times(4 a)^{2}$ [using identity, $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$ ] $=(1-4 a)^{3}=(1-4 a)(1-4 a)(1-4 a)$ (ii) $8 p^{3}+\frac{12}{5} p^{2}+\frac{6}{25} p+\frac{1}{125}$ $=(2 p)^{3}+3 \times(2 p)^{2} \times \frac{1}{5}+3 \times(2 p) \times\left(\frac{1}{5}\right)^{2}...
Read More →The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°.
Question: The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60. At a point Y, 40 m vertically above X, the angle of elevation is 45. Find the height of tower PQ. [Take $\sqrt{3}=1.73$ ] Solution: We have, $\mathrm{XY}=40 \mathrm{~m}, \angle \mathrm{PXQ}=60^{\circ}$ and $\angle \mathrm{MYQ}=45^{\circ}$ Let $\mathrm{PQ}=h$ Also, $\mathrm{MP}=\mathrm{XY}=40 \mathrm{~m}, \mathrm{MQ}=\mathrm{PQ}-\mathrm{MP}=h-40$ In $\Delta \mathrm{MYQ}$ $\tan 45^{\circ}=\frac...
Read More →Find the volume of a cube whose surface area is
Question: Find the volume of a cube whose surface area is 384 m2. Solution: Surface area of a cube is given by: $S A=6 s^{2}$, 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 surface area of the cube is 384 m2. Therefore, we have: $6 s^{2}=384 \Rightarrow s=\sqrt{\frac{384}{6}}=\sqrt{64}=8 \mathrm{~m}$ Now, volume is given by: $V=s^{3}=8^{3} \Rightarrow V=8 \times 8 \times 8=512 \mathrm{~m}^{3}$ Thus, the required vol...
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