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a , b , c and d are different positive numbers. The average (arithmeti
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21 Sep 2023, 23:36
21 Sep 2023, 23:36
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a , b , c and d are different positive numbers.
The average (arithmetic mean) of a and b is 30.
The average of a , b , c and d is 40.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
The average (arithmetic mean) of a and b is 30.
The average of a , b , c and d is 40.
Quantity A |
Quantity B |
The greatest possible value of d |
99 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: a , b , c and d are different positive numbers. The average (arithmeti
[#permalink]
23 Sep 2023, 10:46
23 Sep 2023, 10:46
2
Given that a , b , c and d are different positive numbers.
The average (arithmetic mean) of a and b is 30.
=> \(\frac{a + b}{2}\) = 30
=> a + b = 60
The average of a , b , c and d is 40.
=> \(\frac{a + b + c + d}{4}\) = 40
=> a + b + c + d = 160
=> 60 + c + d = 160
=> c + d = 100
Now, c and d are distinct positive numbers (NOTE numbers is mentioned and not INTEGERS, numbers can be decimal also)
=> c can be 0.1 or smaller also also
Making d = 100 = 0.1 = 99.9
=> Quantity A(Maximum Value of d) > Quantity B(99)
So, Answer will be A
Hope it helps!
Watch the following video to Learn the Basics of Statistics
The average (arithmetic mean) of a and b is 30.
=> \(\frac{a + b}{2}\) = 30
=> a + b = 60
The average of a , b , c and d is 40.
=> \(\frac{a + b + c + d}{4}\) = 40
=> a + b + c + d = 160
=> 60 + c + d = 160
=> c + d = 100
Now, c and d are distinct positive numbers (NOTE numbers is mentioned and not INTEGERS, numbers can be decimal also)
=> c can be 0.1 or smaller also also
Making d = 100 = 0.1 = 99.9
=> Quantity A(Maximum Value of d) > Quantity B(99)
So, Answer will be A
Hope it helps!
Watch the following video to Learn the Basics of Statistics
a , b , c and d are different positive numbers. The average (arithmeti
[#permalink]
22 Jan 2024, 06:12
22 Jan 2024, 06:12
Carcass sir, i have very silly doubt , in case of mean or avg there is no need to arrange series in ascending or descending order ?
as if we see here a+b=60 thus c+d=100 and we take the series, same as question then it will be a , b , c, d = 30 , 30 , 30, 70 (here i took a, b , c as 30 in order to make d greatest) >> here d is less than 99
and if the series is like this c , a , b , d = 1 , 30, 30 , 99 >> here d is equal to 99
but if we directly take without arranging them in an order than we can solve it in the same way as BrushMyQuant did , so in case of mean or avg there is no need to arrange series in ascending or descending order ?
as if we see here a+b=60 thus c+d=100 and we take the series, same as question then it will be a , b , c, d = 30 , 30 , 30, 70 (here i took a, b , c as 30 in order to make d greatest) >> here d is less than 99
and if the series is like this c , a , b , d = 1 , 30, 30 , 99 >> here d is equal to 99
but if we directly take without arranging them in an order than we can solve it in the same way as BrushMyQuant did , so in case of mean or avg there is no need to arrange series in ascending or descending order ?
