Algebra for elitmus
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Algebra for elitmus |
The importance of algebra is high in elitmus you can expect around 2 questions from algebra.
Important formulae of algebra specific to elitmus.
1. a2 – b2 = (a – b)(a + b)
2. (a+b)2 = a2 + 2ab + b2
3. a2 + b2 = (a – b)2 + 2ab
4. (a – b)2 = a2 – 2ab + b2
5. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
6. (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
7. (a + b)3 = a3 + 3a2 b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
8. (a – b)3 = a3– b3 – 3ab (a - b)
9. a3 – b3 = (a – b)(a2 + ab + b2 )
10. a3 + b3 = (a + b)(a2 – ab + b2 )
Important shortcuts of algebra specific to elitmus
1. if (x+1/x) = a, then (x3+1/x3) = a3-3a and (x4+1/x4 )=a4-4a2+2
2. geometric mean of two number a and b is sqrt(ab)
3. arithmetic mean of two number a and b is (a+b)/2
4. if 1/a+1/b+1/c=1/(a+b+c) where (a+b+c)=0 then (a+b)(b+c)(c+a)=0
Sample questions with the solution of algebra specific to elitmus
1. The sum of a two digit number ‘n’ and the number obtained by interchanging digits of n is 99. The difference of the digits of ‘n’ is 3, with the tens place being larger than the units place. Find the number ‘n’.
Solution:
Let the number be ‘xy’, where x and y are single digits.
=> The number is 10x + y
=> Reverse of the number = yx = 10y + x
=> Sum = 11 x + 11 y = 11 (x + y) =99 (given)
=> x + y = 9
Also, we are given that the difference of the digits is 3 and x > y.
=> x – y = 3
Therefore, x = 6 and y = 3
Thus, the number is 63.
2. If x3 + y3 = 9 and x + y = 3, then the value of x4+y4 is,
solution:squaring both side (x+y)3=33
=> x3+y3+3*x*y*(x+y)=27
=> 9*x*y= 27-9
=> x*y=18/9
=> xy=2
now, (x2+y2)2=x4+y4+2*x2*y2
=> x4+y4 = (x2+y2)2- 2*(xy)2
=> x4+y4 = ((x+y)2 - 2*xy)2 - 8
=> x4+y4 = (9- 4)2 - 8
=> x4+y4 = 17
3. If x+1/2x = 2, find the value of 8x3+1/x3
Solution:
(a+b)3 = a3+b3+3*a*b(a+b)
=> x+1/2x =2
=> 2x+1/x =4
=> cube of both side 8x3+1/x3 + 3 * 2(4) = 64
=> 8x3+1/x3= 64-24 = 40
4. If 5√x +12√x =13√x then value of x is,
Solution:
Solution:
always check with options like trial and error method for these type of question here value of x will be 4.
(5)2+(12)2 = (13)2
=> 25+144=169.
=> 169 = 169
5. If b2 + 1⁄b2 = 1, then value of b3 + 1⁄b3 is ,
Solution:
let (b+1/b) = c
=> b2+1/b2+2=c2
=> 1 = c2 -2
=> c2 = 3
=> c= sqroot(3)
=> (b+1/b)3 = c3
=> b3+ 1/b3 + 3sqroot(3) = 3sqroot(3)
=> b3 + 1⁄b3 = 0
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