Aptitude - Surds and Indices - Discussion
Discussion Forum : Surds and Indices - General Questions (Q.No. 14)
14.
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xb |  |
(b + c - a) | . | |
xc | |
(c + a - b) | . | |
xa | |
(a + b - c) | = ? |
| xc | xa | xb |
Answer: Option
Explanation:
| Given Exp. |
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Discussion:
22 comments Page 2 of 3.
Anand kumar said:
8 years ago
I should not understand this.
Someone's can explain this properly?
Someone's can explain this properly?
Jegan said:
8 years ago
Next step they are taking (a+b)(b-C) = A*2-B*2 & adding in common X.
(1)
Shivam Raj said:
8 years ago
Can anyone explain the 2nd step in easy solution?
Ankit singhai said:
9 years ago
x^(b - c)(b + c - a) How it is possible?
Susav karki said:
9 years ago
Can anyone explain the third step?
Gaurav joshi said:
9 years ago
X^(b-c)((b+c)-a).
We get, X^(b-c)(b+c)-a(b-c).
I hope we all get understand by this.
We get, X^(b-c)(b+c)-a(b-c).
I hope we all get understand by this.
Karan said:
9 years ago
Dot (.) is multiplication sigh you can use x or (.).
Ammar said:
9 years ago
What does the dot (.) means between those numbers, help anybody?
Ganesh said:
1 decade ago
Second step is the simple representation through indices and hence in the third step the exponents are multiplied and again with the help of indices the sum is solved it is a very simple but tricky question.
Hope you understand my explanation. All the best.
Hope you understand my explanation. All the best.
Mansi said:
1 decade ago
In the second step:
From the first term: x^(b-c)(b+c-a).
There are two brackets term, from the second bracket -a is taken out. As it comes out, it has to be multiplied by the first bracket.
So it gets multiplied and we get x^(b-c)(b+c)-a(b-c).
Similar thing is done for the rest.
From the first term: x^(b-c)(b+c-a).
There are two brackets term, from the second bracket -a is taken out. As it comes out, it has to be multiplied by the first bracket.
So it gets multiplied and we get x^(b-c)(b+c)-a(b-c).
Similar thing is done for the rest.
(1)
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