The average (arithmetic mean) of 20, 40, and 60 is 5 more than the ave

Question

The average (arithmetic mean) of 20, 40, and 60 is 5 more than the average of 10, 70, and what number?

(A) 15
(B) 25
(C) 35
(D) 45
(E) 55

CounterSniper · Answer

a1 = 120/3 = 40 
a2 = a1 -5 = 35 
sum of second list = 35*3 = 105
therefore the number = 105 -80 = 25

stonecold · Answer

As per Question =>

Mean(1) => 40 (Notice that it is an Ap series)
Mean(2) = 80+x/3

As per Question => 40=80+x/3 +5 => x=25

Hence B

Sirakri · Answer

(20+40+60)/3=5+ 
=> 40= 
=> 105=80+x
=> x=25

IdiomSavant · Answer

You can do this one in your head forget calculating a mean.

20+40+60=120

5 more than the average with 3 numbers is 15 more than the sum so you are looking for 10+70+x to equal 105

Abhishek009 · Answer

\frac{( 20 + 40 + 60 )}{3} - \frac{( 10 + 70 + x )}{3} = 5

120 - 80 - x = 15

Or, x  = 25 
 
Hence, answer will be (B)

kanusha · Answer

a1 = \frac{120}{3} = 40 
a2 = a1 -5 = 35 
sum of second list = 35*3 = 105
therefore the number = 105 -80 = 25

BrushMyQuant · Answer

Theory

➡ Average = Sum of all the Values / Total Number of Values 
 ➡ Sum of All the values = Average * Total Number of Values

The average (arithmetic mean) of 20, 40, and 60 is 5 more than the average of 10, 70, and what number

Let the unknown number be x

Average of 20, 40, and 60 =  \frac{Sum}{3}  =  \frac{20+40+60}{3}  =  \frac{120}{3}

Average of 10, 70 and x =  \frac{Sum}{3}  =  \frac{10+70+x}{3}  =  \frac{80+x}{3}

Average of 20, 40 and 60 = 5 + Average of 10, 70 and x (given)

=>  \frac{120}{3}  = 5 +  \frac{80+x}{3} 
=> 120 = 5*3 + 80 + x = 15 + 80 + x = 95 + x
=> x = 120 - 95 = 25

So,  Answer will be B .
Hope it helps!

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