3
Q:

If a, b, c be the terms of a GP then the value of (q-r) log a + (r-p) log b + (p-q) log c is :

A) 0 B) 1
C) -1 D) pqr

Answer:   A) 0



Explanation:

Let m be the first term and k be the common ratio of GP, then 

                             \inline a=mk^{p-1}

                             \inline b=mk^{q-1}

                             \inline c=mk^{r-1}

\therefore \inline (q-r)\log a + (r-p) \log b + (p-q) \log c = \log \left [ mk^{(p-1)} \right ]^{q-r}+\log \left [ mk^{(q-1)} \right ]^{r-p}+\log \left [ mk^{(r-1)} \right ]^{p-q}

       = \inline \log (m^{q-r+r-p+p-q})(k^{p-1})^{q-r}(k^{q-1})^{r-p}(k^{r-1})^{p-q}

       = \inline \log m^{0}k^{0}=\log 1=0

Q:

Solve the equation   ?

A) -1/2 B) 1/2
C) 1 D) -1
 
Answer & Explanation Answer: A) -1/2

Explanation:

Rewrite equation as

Leads to 2x + 1 = 0 

Solve for x : x = -1/2

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6 804
Q:

If  = m, then  is equal to ?

A) 1/(1+2m) B) (1+2m)/2
C) 2m/(2m+1) D) (2m+1)/2m
 
Answer & Explanation Answer: B) (1+2m)/2

Explanation:

=
=

.

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11 1067
Q:

If  , then 

A) 1 B) 2
C) 4 D) 8
 
Answer & Explanation Answer: B) 2

Explanation:

Given  

Now  = 

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9 1056
Q:

If log 64 = 1.8061, then the value of log 16 will be (approx)?

A) 1.9048 B) 1.2040
C) 0.9840 D) 1.4521
 
Answer & Explanation Answer: B) 1.2040

Explanation:

Given that, 

 

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17 2784
Q:

A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.

A) 100 m B) 125 m
C) 150 m D) 175 m
 
Answer & Explanation Answer: C) 150 m

Explanation:

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19 1967