The average (arithmetic mean) of a and b is 90; the average (arithmeti

given
a+b= 180 and
a+c= 300
subtract
b-c= 120
so b-c/2= 60
IMO D

The average (arithmetic mean) of a and b is 90; the average (arithmetic mean) of a and c is 150; what is the value of (b-c)/2?

A) 60
B) 120
C) 30
D) -60
E) -40

a+b=90*2=180
a+c=150*2=300
b-c=-120
b-c/2=-120/2=-60

Ans (D)

The average (arithmetic mean) of a and b is 90; the average (arithmetic mean) of a and c is 150; what is the value of (b-c)/2?

A) 60
B) 120
C) 30
D) -60
E) -40

a + b = 180 - Eqn. 1
a + c + 300 - Eqn. 2
Eqn. 1 - Eqn. 2
a + b - a - c = 180 - 300
b - c = - 120
\(\frac{b-c}{2} = - 60\)

Answer D.

a+b = 180
a+c = 300
------------ (Subtracting both equations)
b - c = -120
\(\frac{(b-c)}{2}\) = -60

IMO, D

From the above,(a+b)/2=90, implies a+b=180.................(1)
(a+c)/2=150, implies a+c=300..................(2)
(1)-(2)=> b-c=-120
Hence (b-c)/2=-60.

The answer is therefore D.

So, \(a + b = 180\) ; \(a + c = 300\)

Now, \((a + b ) - ( a + c ) = a - c\)

Or, \(\frac{a - c }{2}= \frac{180 - 300}{2}\)

Or, \(\frac{a - c }{2} = -60\), Hence, Answer must be (D) -60

The average (arithmetic mean) of a and b is 90; the average (arithmetic mean) of a and c is 150; what is the value of \(\frac{(b-c)}{2}\) ?

a+b =180
a+c =300
--> a+b-a -c= 180-300 --> b-c =-120
--> \(\frac{(b-c)}{2} = -\frac{120}{2}= -60\)

The answer is D.

The average (arithmetic mean) of a and b is 90; the average (arithmetic mean) of a and c is 150; what is the value of (b-c)/2?

A) 60
B) 120
C) 30
D) -60
E) -40

Solution : a+b/2= 90 there fore a+b= 180
a+c/2=150 Therefore a+c=300

a+b=180
- a+c=300
=b-c=-120

Therefor b-c/2= -60

IMO D

(a+b)/2 = 90.....(i)
(a+c)/2 = 150.....(ii)

(i) - (ii) gives us:
(a+b-a-c)/2 = 90-150
=> (b-c)/2 = -60

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OFFICIAL EXPLANATION

Translate the question into algebra, using the formula that the average equals the sum of the terms divided by the number of terms.
90 = (a + b)/2
Multiply by 2: 180 = a + b

150 = (a + c)/2
Multiply by 2: 300 = a + c

At this point, there are two ways to manipulate the algebra.

Method 1
Line up the two equations so that you can subtract them.
Equation 1: a+b = 180
Equation 2: a+c = 300
Notice that the equations are lined up such that if you subtract them, you end up with b-c on the left and a number on the right. Since you are trying to find (b-c)/2, this setup will get you most of the way to the answer.
Subtract the two equations:
Equation 1: a+b = 180
Equation 2: a+c = 300
Equation 1 - 2: a - a + b - c = 180 - 300 = -120
Equation 1 - 2 (cont.): b-c = -120
With b-c = -120, if you divide by two, you arrive at the equation that you are being asked for:
(b-c)/2 = -120/2 = -60

Method 2
Since the question asks for the value of (b-c)/2, solve the two equations for the variable a so that they can be combined.
a = 180 – b
a = 300 – c
Combine the two equations by setting them equal to each other (since a = a) and manipulate them such that (b-c)/2 is on one side of the equation and a number is on the other.
180 – b = 300 – c
180 – 300 = b – c
-120 = b-c
-60 = (b – c)/2

We can create the equations:

a + b = 180

and

a + c = 300

Subtracting the second equation from the first, we have:

(b - c) = -120

Therefore, (b - c)/2 = -120/2 = -60.

Answer: D

The average (arithmetic mean) of a and b is 90, this should translate to:

(a + b)/2 = 90

=> a + b = 180 ------ (1)

Similarly,

a + c = 300 ------ (2)


Now the question requires us to find (b - c), thus, subtract (2) from (1) & we will get:

b - c = -120

divide both the sides by 2 to get the requisite value;

(b - c)/2 = -60

Thus, the correct answer is D.

\(\frac{(a+b)}{2}=90\) and \(\frac{(a+c)}{2}=150\)

(a+b)=180 and (a+c)=300

\(a=180-b\) and \(a=300-c\)

\(180-b=300-c\)

\(b=180\)
\(c=300\)

\(\frac{(180-300)}{2}=-60\)


Answer: D