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The average of two numbers is 90. If the difference between the number

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Bunuel
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The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   17 Jul 2020, 04:22
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The average of two numbers is 90. If the difference between the numbers is 12, what is the product of the two numbers?

A. 8064
B. 8100
C. 8400
D. 8640
E. 9600
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A
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Re: The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   17 Jul 2020, 04:54
Bunuel wrote:
The average of two numbers is 90. If the difference between the numbers is 12, what is the product of the two numbers?

A. 8064
B. 8100
C. 8400
D. 8640
E. 9600


a-b=12
(a+b)/2 = 90

=>a=96 & b=84

OA: A
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Re: The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   17 Jul 2020, 09:48
Let one number be x. Then the other number would be x+12.
x+x+12/2 = 90
=> 2x + 12 = 180
=> x = 84
Product of two numbers = 84*96 = 8064

Hence, OA is (A).
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Re: The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   17 Jul 2020, 14:27
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Bunuel wrote:
The average of two numbers is 90. If the difference between the numbers is 12, what is the product of the two numbers?

A. 8064
B. 8100
C. 8400
D. 8640
E. 9600


Solution


    • Let us assume that the two numbers are a and b.
    • \(\frac{a+b}{2} = 90 ⟹ a + b = 180……Eq.(i)\)
    • And, \(a – b = 12 ……….Eq.(ii)\)
    • Adding Eq.(i) and Eq.(ii) we get,
      o \( 2a = 192 ⟹ a = 96\)
      o Substituting the value of a in Eq.(ii), we get,
         \(b = a – 12 = 96 – 12 = 84\)
    • Hence, \(a*b = 96*84 = 8064\)
Thus, the correct answer is Option A.

Alternately,


    • Squaring both sides of Eq.(i), we get,
      o \((a+b)^2 = (180)^2 ……..Eq.(iii)\)
    • Squaring both sides of Eq.(ii), we get,
      o \((a – b)^2 = (12)^2 ………Eq.(iv)\)
    • We know that \( (a+b)^2 – (a-b)^2 = 4*a*b\), therefore from Eq.(iii) and Eq.(iv) we get,
      o \( 4*a*b = 180^2 – 12^2 ⟹ a*b = 90^2 – 6^2 ⟹ a*b = 8100 – 36 ⟹ a*b = 8064\)
Thus, the correct answer is Option A.
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The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   17 Jul 2020, 22:04
Average = 90 = (a+b)/2 ---- (A)
Also. a-b=12 (given)

Alternate Approach:

a+b=180 ---- From (A)
(a+b)^2 = 180^2
Rewriting (a+b)^2, we get:
(a-b)^2 + 4ab = 180^2
ab = (1/4){(180)^2-(12)^2} ---- (Given a-b=12)

Solving, we get:
ab = 8064

Hence, the answer is 8064. Option (D).
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Re: The average of two numbers is 90. If the difference between the number [#permalink] The average of two numbers is 90. If the difference between the number   20 Jul 2020, 01:53
As mentioned,
Average = 90 = (a+b)/2 ..... so a+ b = 180 ....a = 180 -b
a-b=12 .....180 -b-b = 12 ....b= 168

Substituting b=168 in a-b=12
a-b=12 ......a= 84

Product of the two numbers = 84 *168 = 8064

OA : A
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Re: The average of two numbers is 90. If the difference between the number [#permalink]
20 Jul 2020, 01:53
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Bunuel
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