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) |
Answer: Option
Explanation:
| Given Exp. = |
|
| = | xa | + | xb | + | xc |
| (xa + xb + xc) | (xa + xb + xc) | (xa + xb + xc) |
| = | (xa + xb + xc) |
| (xa + xb + xc) |
= 1.
Discussion:
24 comments Page 3 of 3.
SenthilRaja said:
6 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).
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).
Shantanu Gupta said:
5 years ago
Here,
a, b, and c in cyclic order = 1.
a, b, and c in cyclic order = 1.
John said:
3 years ago
Thanks all for explaining the answer.
Rutuja said:
3 years ago
It's great. Thanks all.
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