Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 8)
8.
1  + 1 + 1 = ?
1 + x(b - a) + x(c - a) 1 + x(a - b) + x(c - b) 1 + x(b - c) + x(a - c)
0
1
xa - b - c
None of these
Answer: Option
Explanation:

Given Exp. =
1  +  1  +  1
1 + xb + xc
xa xa
1 + xa + xc
xb xb
1 + xb + xa
xc xc

   = xa + xb + xc
(xa + xb + xc) (xa + xb + xc) (xa + xb + xc)

   = (xa + xb + xc)
(xa + xb + xc)

   = 1.

Discussion:
24 comments Page 1 of 3.

Rutuja said:   2 years ago
It's great. Thanks all.

John said:   2 years ago
Thanks all for explaining the answer.

Shantanu Gupta said:   3 years ago
Here,

a, b, and c in cyclic order = 1.

SenthilRaja said:   5 years ago
Step 1: (1/(1+(x^b/x^a)+(x^c+x^a)))+(1/(1+(x^a/x^b)+(x^c+x^b)))+(1/(1+(x^b/x^c)+(x^a+x^c)))

Step 2: (1/((x^a+x^b+x^c)/x^a))+(1/((x^b+x^a+x^c)/x^b))+(1/((x^c+x^b+x^a)/x^c))

Step 3: (x^a/(x^a+x^b+x^c)) +(x^b/(x^a+x^b+x^c))+(x^c/(x^a+x^b+x^c)) Note: Take LCM in Denominator which is in (step 2 )

Step 4: (x^a+x^b+x^c)/(x^a+x^b+x^c) = 1 (Answer).

Ragav said:   6 years ago
Here, assign x=1.

Aarambh said:   6 years ago
I cannot understand the second step. Please explain me.

Ayush said:   6 years ago
@Garima

How? Please explain in detail.

Garima said:   7 years ago
Short cut method.

Let a=b=c=1.
We get 1/3 +1/3 +1/3 =3/ 3 =1.

Garima said:   7 years ago
Short cut method:
Let a=b=c,
We get , 1/3 +1/3 +1/3 = 3/3 =1.

Ekta said:   7 years ago
In the second line,the denominator is solved as:-

(1+x^b/x^a+x^c/x^a) (solving 1st denominator of the first fraction)
Which comes as( x^a+x^b+x^c)/x^a.

Now, the first fraction is : 1/((x^a+x^b+x^c)/x^a) which can be also written as x^a/(x^a+x^b+x^c).

Similarly other two fractions are also solved and finally x^a/(x^a+x^b+x^c) + x^b/(x^a+x^b+x^c) + x^c/(x^a+x^b+x^c) = 1.


Post your comments here:

Your comments will be displayed after verification.