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Numerical Ability
Arithmetic
If p is greater than q, then which of the following is true?
1) 0.9^p=0.9^q
2) 0.9^p>0.9^q
3) 9^p>9^q
Read Solution (Total 16)
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- Since p>q so (9)^p>(9)^q
- 10 years agoHelpfull: Yes(13) No(0)
- option 3 is right.
say p=2 , q=1 then 0.9^2=0.81 while, 0.9^1=0.9.
and 9^2=81 whereas 9^1=9. - 10 years agoHelpfull: Yes(7) No(0)
- clearly 3rd option is true!!!
- 10 years agoHelpfull: Yes(1) No(0)
- 3 is true
- 10 years agoHelpfull: Yes(1) No(0)
- 0.9^p>0.9^q beause p is greater than q
- 10 years agoHelpfull: Yes(0) No(5)
- 9^p>9^q
explanation:p is greater than q.so 0.9^p will be not equal to 0.9^q.if you assign values for p as 2 and q as 1 then it is clear dat option 3 is correct. - 10 years agoHelpfull: Yes(0) No(0)
- 9^p>9^q, as given that p>q ,assume p=2 and q=1 ie. 9^2>9^1 = 81>9
- 10 years agoHelpfull: Yes(0) No(0)
- option 3
let us assume p to be 3 and q to be 2 (p is greater than q)
then 729>81 - 10 years agoHelpfull: Yes(0) No(0)
- p>q so (9)^p>(9)^q
- 10 years agoHelpfull: Yes(0) No(0)
- 3) since 9>1
- 10 years agoHelpfull: Yes(0) No(0)
- 3rd option is true
- 10 years agoHelpfull: Yes(0) No(0)
- Option 3 is true
- 10 years agoHelpfull: Yes(0) No(0)
- (753*753+247*247-753*247)/(753*753*753+247*247*247)
here 753=a;247=b
so,
(a^2+b^2-ab)/(a^3+b^3)=1/(a+b) [a^3+b^3=(a^2+b^2-a*b)(a+b)]
=1/1000 ans..
- 10 years agoHelpfull: Yes(0) No(0)
- It can also be : 0.9^p < 0.9^q
- 10 years agoHelpfull: Yes(0) No(1)
- option 3 is correct
9^p>9^q - 10 years agoHelpfull: Yes(0) No(0)
- 3 assume p and q
- 10 years agoHelpfull: Yes(0) No(0)
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