| If \[A+C=B,\]then \[tan\,A\text{ }tan\,B\text{ }tan\,C\]is equal to |
A) \[\tan \,B+\tan \,A+\tan \,C\]
B) \[\tan \,B-\tan \,A+\tan \,C\]
C) \[\tan \,B-\tan \,A-\tan \,C\]
D) \[\tan \,B+\tan \,A-\tan \,C\]
Correct Answer: C
Solution :
| Let the average age of 8 men be x yr. |
| \[\therefore \] Total age = 8x |
| According to the question, |
| \[\frac{8x-(21+23)+(Age\,of\,two\,\text{new}\,\text{men})}{8}=x+2\] |
| \[\Rightarrow \] \[8x-44+(\text{Age of two new men})=8x+16\] |
| \[\Rightarrow \]\[Age\text{ }of\text{ }two\text{ }new\text{ }men=16+44=60\text{ }yr\] |
| \[\therefore \] Average age \[=\frac{60}{2}=30\,yr\] |
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