Sum of Algebra
The collection of Sum of Algebra of different competitive exams is given here.
| 1) If {a}^x = b, {b}^x =c and {c}^x = a then find the value of xyz. { a }^{ xyz }\quad =\quad { ({ a }^{ x }) }^{ yz } { a }^{ xyz }\quad =\quad {b }^{ yz } { a }^{ xyz }\quad =\quad {{b}^y}^z { a }^{ xyz }\quad =\quad { a }^z { a }^{ xyz }\quad = a on comparison we get xyz = 1 |
| 2) If a+b+c = 9 and ab+bc+ca = 40 then find the value of {a}^2+{b}^2+{c}^2. By formula {(a+b+c)}^2 = {a}^2 + {b}^2+{c}^2 + 2(ab+bc+ca) {9}^2 = {a}^2 + {b}^2+{c}^2 + 2\times 40 81 = {a}^2 + {b}^2+{c}^2 + 80 {a}^2 + {b}^2+{c}^2 =81 - 80 {a}^2 + {b}^2+{c}^2 = 1 |
| 3) If a+b=7 then find the value of {(a-b)}^2 +4ab+ 1 . {(a-b)}^2 +4ab+ 1 .= {a}^2 +{b}^2 - 2ab +4ab+ 1 . = {a}^2 +{b}^2 + 2ab + 1 . = {(a+b)}^2 + 1 . = 49 + 1 = 50. |
| 4) If 2x+y = 3 and xy = 1 then find the value of {(x+y)}^{x-y}. {(2x-y)}^2 = {(2x+y)}^2 - 4\times 2x \times y = {3}^2 - 4\times 2\times1 =9 – 8 = 1 2x – y = 1 ————(1) 2x + y = 3 ———–(2) solving 1 and 2 we get x = 1 and y = 1 Now, {(x+y)}^{x-y}. = {2}^0 =1 {Something to the power Zero is 1} |
| 5) If x = (b-c) (a-d), y = (c-a) (b-d) and z = (a-b) (c-d) then find the value of {x}^3 + {y}^3 +{z}^3 . x+y+z = (b-c)(a-d)+ (c-a)(b-d) + (a-b)(c-d) x+y+z = 0 cubing both side we get {x}^3 + {y}^3 + {z}^3 = 3xyz |
| 6) If {(a-2)}^2 + {(b+3)}^2 + {(c-4)}^2 = 0 then find the value of a-b+c By given a = 2 b = -3 c = 4 a-b+c = 2-(-3)+4 =2+3+4 9. |
| 7) If {(x)}^2+{(y)}^2+4x+4y+8 =0 then find the value of x+y . {(x)}^2+{(y)}^2+4x+4y+8 =0 ({x}^2+4x+4)+({y}^2+4y+4) = 0 {(x+2)}^2 + {(y+2)}^2 = 0 x+2 = 0 and y+2 = 0 so x=-2 and y = -2, therefore x+y = (-2) + (-2) = -4 |
| 8) If a+b+c = 0,then find the value of {a}^2(b+c) + {b}^2 (c+a) + {c}^2 (a+b) . {a}^2(b+c) + {b}^2 (c+a) + {c}^2 (a+b) . = {a}^2(a+b+c-a) + {b}^2 (c+a+b-b) + {c}^2 (a+b+c-c) . = - ({a}^3+{b}^3+{c}^3) = – 3abc {a+b+c = 0, then {a}^3+{b}^3+{c}^3 = 3abc |
| 9) If a+b+c = 13, and ab+bc+ca = 50 then find the value {a}^2 +{b}^2 +{c}^2. By formula, {a}^2 +{b}^2 +{c}^2. = {(a+b+c)}^2 - 2(ab+bc+ca). = {13}^2 – 2 × 50 = 169 – 100 = 69 |
| 10) If p+q = r and pqr = 30 then find the value of {p}^3 + {q}^3 - {r}^3. Given that , P+q = r p + q – r = 0 {p}^3 +{q}^3 +{(-r)}^3 = 3pq(-r) = 3 × (-30) =-90 |
| 11) If {x}^2+8{y}^2+9{z}^2 - 4xy - 12xz = 0, then which option is correct. a) 3x = 2y = z b) x=y=z c) x=2y=z d) x+2y+3z=0 {x}^2+8{y}^2+9{z}^2 - 4xy - 12xz = 0 {x}^2 - 4xy+4{y}^2+4{y}^2 - 12xz+9{z}^2 = 0 {(x-2y)}^2 + {(2y-3z)}^2 = 0 Now, x-2y = 0 and 2y-3z = 0. so, x= 2y = 3z. |
| 12) If {3}^{(x-y)} = 27, and {3}^{(x+y)} = 243, then find the value of x. {3}^{(x-y)} = 27 {3}^{(x-y)} = {3}^3 therefore x- y = 3 ——————— (1) {3}^{(x+y)} = 243 {3}^{(x+y)} = {3}^5 therefore x+y = 5 ————————(2) on addition of equation 1 and 2 we get 2x = 8 so, x = 4 |
| 13) If x = 12 and y = 4 then find the value of {(x+y)} ^{x/y} . {(x+y)} ^{x/y} ={(12+4)}^ 3 ={16}^3 = 4096. |
| 14) If {(a-1)}^2 + {(b+2)}^2 + {(c+1)}^2 = 0, then find the value of 2a+3b+7c. {(a-1)}^2 + {(b+2)}^2 + {(c+1)}^2 = 0 therefore, a-1 = 0 , b+2 = 0, c+1 = 0. so, a = 1, b = -2, c = -1. Now, 2a -3b +7c = 2+6-7 = 1 |
| 15) If p = 124 then \sqrt [ 3 ]{ p({ p }^{ 2 }+3p+3)+1 } = ?. \sqrt [ 3 ]{ p({ p }^{ 2 }+3p+3)+1 } = {[{(p+1)}^3]}^{1/3} = (p+1) =(124+1) =125 |
| 16) {x}^a.{x}^b.{x}^c = 1 then find the value of {a}^3 +{b}^3+{c}^3. {x}^a.{x}^b.{x}^c = 1 {x}^{(a+b+c)} = {x}^0 so, a+b+c = 0 Now, {a}^3 +{b}^3+{c}^3 - 3abc = (a+b+c)({a}^2+{b}^2)+{c}^2-ab-bc-ca) {a}^3 +{b}^3+{c}^3 - 3abc = 0 {a}^3 +{b}^3+{c}^3 = 3abc. |
| 17) If (x+y+z) = 16 and (xy+yz+zx) = 78 then find the value of {x}^3+{y}^3+{z}^3 -3 xyz. (x+y+z) = 16 On squaring both side {(x+y+z)}^2 =256 {x}^2+{y}^2+{z}^2 + 2(xy+yz+zx) = 256 {x}^2+{y}^2+{z}^2 =100 we know that, {x}^3+{y}^3+{z}^3 -3 xyz = (x+y+z)({x}^2+{y}^2+{z}^2 -xy-yz-zx) = 16 × (100-78) = 16 × 22 = 352. |
| 18) If {a}^2 + {b}^2 + 4{c}^2 = 2(a+b-2c) – 3 and a,b,c are real numbers then find the value of ({a}^2+{b}^2+{c}^2). {a}^2 + {b}^2 + 4{c}^2 = 2(a+b-2c) – 3 {a}^2 + {b}^2 + 4{c}^2 – 2(a+b-2c) + 3 = 0 {a}^2-2a+1+{b}^2-2b+1+{(2c)}^2 +4c+1 = 0 {(a-1)}^2 +{(b-1)}^2 +{(2c+1)}^2 =0 therefore, {(a-1)}^2 = 0, {(b-1)}^2 =0, {(2c+1)}^2 = 0 so, a=1,b=1, c= 1/2, Now, ({a}^2+{b}^2+{c}^2) = 1+1+1/4 = 2\frac { 1 }{ 4 } |
| 19) If {a}^2 +1 =a, then find the value of {a}^{12} + {a}^6+1. {a}^2 +1 =a {a}^2 +1 -a=0 Multiplying both side by (a+1), (a+1)({a}^2 +1 -a)=0 {a}^3+1 = 0 so, a = -1. Now, {a}^{12} + {a}^6+1 = {{a}^3}^4 + {{a}^3}^2 + 1 = {(-1)}^4 + {(-1)}2 +1 = 1+1+1 =3 |
| 20) If x = a-b, y = b-c, z = c-a, then find the value of {x}^3+{y}^3 + {z}^3 -3xyz . x+y+z = (a-b)+(b-c)+(c-a) x+y+z=0 cubing both side we get, {x}^3+{y}^3+{z}^3-3xyz = 0 |
| 21) If x=y=333 and z = 334 then find the value of {x}^3+{y}^3+(z)^3-3xyz. {x}^3+{y}^3+(z)^3-3xyz. = \frac { 1 }{ 2 }(x+y+z)[{(x-y)}^2 + {(y-z)}^2 + {(z-x)}^2] =\frac { 1 }{ 2 } <strong><em>×</em></strong> 1000[0+1+1] = 1000 |
| 22) {a}^2 + {b}^2 + {c}^2 = ab +bc + ca, then find the value of \frac{a+c}{b} {a}^2 + {b}^2 + {c}^2 = ab +bc + ca multiplying both sides by 2 we get, 2 {a}^2 +2 {b}^2 + 2{c}^2 = 2ab +2bc +2 ca ({a}^2 - 2ab+{b}^2) + ({b}^2 - 2bc+{c}^2)+({c}^2 - 2ca+{c}^2) = 0 So, {(a-b)}^2 +{(b-c)}^2 + {(c-a)}^2 = 0 Therefore a = b = c Now, \frac{(a+c)}{b} = 2 |
| 23) If a+b+c = 4 then find the value of {a}^3 + {b}^3 + {c}^3 - 12{c}^2 +48c-64. a+b+c = 4 a+b+c-4 = 0 We Know that If a+b+c= 0 then, {a}^3 + {b}^3 + {c}^3 = 3abc So, {a}^3 + {b}^3 + {(c-4)}^3 = 3ab(c-4) {a}^3 + {b}^3 +[ {c}^3-64-12c(c-4)] = 3abc-12ab {a}^3 + {b}^3 + {c}^3-64-12{c}^2+48c = 3abc-12ab {a}^3 + {b}^3 + {c}^3-12{c}^2+48c-64 = 3abc-12ab |
| 24) 4x -5z = 16 and xz = 12 then find the value of 64{x}^3 - 125{z}^3. 4x -5z = 16 On squaring both side we get, 16{x}^2+25{z}^2 - 40 zx = 256 16{x}^2+25{z}^2 = 256 + 480 16{x}^2+25{z}^2 = 736 16{x}^3-25{z}^3 ={(4x)}^3 - {(5z)}^3 = (4x – 5z)(16{x}^2 + 25{z}^2 + 20xz) = 16(736+240) = 15616. |
Click Here to buy this book for SSC advanced maths.
For the collection of Arithmetic, sums click Here
Please Share if you care for others