The sum of 5 positive numbers is 45. If average of lower 3 numbers is 7. Then maximum value of the p

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18 Jan 2011 5 Positive number

The sum of 5 positive numbers is 45. If average of lower 3 numbers is 7. Then maximum value of the product of higher 2 numbers is

OPtion

1) 150
2) 156
3) 164
4) 144
5) 143
6) 140
7) 135
8) 128
9) 119
10) 188

Solution

sum of 5 positive numbers = 45
sum of lower 3 numbers = 7*3 = 21
so sum of higher 2 numbers = 24

maximum product can be obtained only by splitting the number in 2 equal parts, so product = 12*12 = 144

Correct Option: 4 Best Solution (14)

Ravi B Mali   13 years AGO

a+b+c+d+e=45
given (a+b+c)/3=7
therefore a+b+c=21
thus, d+e=45-21=24

for the product to be maximum d = e = 24/2 = 12
thus product=12*12=144

Like? Yes (6) | No   

dharmaveer   13 years AGO

first 3 numbers avg=7. means sum of first lower 3 numbers is 21.
then, higher 2 values sum becomes 24.
and the maximum value we will get by doing 12 X 12=144.

Like? Yes (3) | No   

Dr.Dipin Singh   13 years AGO

The sum of 5 positive numbers is 45. If average of lower 3 numbers is 7. sum of 3 smaller numbers = 7*3=21.
sum of 2 larger numbers = 45-21=24.
Numbers can be 12,12 or 13,11 or 14,10.
For max value of product, numbers should be 12,12.
Then maximum value of the product of higher 2 numbers is 12*12=144

Like? Yes (3) | No (1)  

dhiresh   13 years AGO

let x1+x2+x3+x4+x5=45
x1+x2+x3=7*3=21
so....x4+x5=45-21=24
now max(x4*x5)=?
for x4=x5....x4*x5 wioll be max so
x4=x5=12
and X4*x5=144.

Like? Yes (2) | No   

goutam Choudhury   13 years AGO

7*3=21.45-21=24. maximum product possible only 12*12=144.

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Suresh Varma P   13 years AGO

as the average of lower three numbers (x,y,z) = 7
(x+y+z)/3 = 7
(x+y+z) = 21
x+y+z+a+b = 45
Assuming higher 2 numbers as (a, b)
then a+b = 45 - 21 = 24
For getting the maximum value of the product of higher 2 numbers is 12 * 12 = 144 Ans: Option - 4

Like? Yes (2) | No (1)  

Sneha Latha M   13 years AGO

The average of 3 lower numbers is 7 that means sum of 3 lower numbers is 3*7=21
The sum of 5 positive numbers is=45.
Then the sum of two higher numbers is 45-21=24
That means, the higher numbers sum must be 24
The numbers that satisfies the above condition are as follows
{12,12},{11,13},{10,14}........ and so on
Here we have to pickup the maximum product of higher numbers. so,the maximum product of higher numbers is obviously 12*12=144

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kapil    13 years AGO

from give data sum of highest two no = 24
let 1st no = x other is 24-x
and their product is max when x=12 by d(prod)/dx=0
hence max product = 144

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vivek mittal   13 years AGO

as avg of lower 3 is 7 so
7+7+7=21
45-21=24
so the sum of higher 2 numbers is 24
the maximum value is coming for 12*12=144

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Nishant singh   13 years AGO

45-3*7=24
12*12=144

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karthik   13 years AGO

avg of 3 num is 7, so sum is 21
then frm given 45-21=24
hence the max value of product of remaining 2 num
is 12*12=144

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SUDESHNA SEN   13 years AGO

Sum of 5 positive numbers is 45.Average of lower 3 numbers is 7, so their sum is 3*7= 21 and sum of the higher 2 numbers is= 45-21= 24.
The possible pairs for the higher 2 numbers are: (1,23),(2,22),(3,21),(4,20),(5,19),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13),(12,12) of which maximum value of the product is generated by the pair (12,12) which is 12*12= 144..... Option 4)

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mohd anas   13 years AGO

a+b+c+d+e=45
avg of a,b,c=7
sum a+b+c=3*7=21
therefore d+e=45-21=24
avg of d,e=12
max pproduct=12*12=144

Like? Yes (1) | No   

Parbati Sen   13 years AGO

Sum of 5 positive numbers is 45.Average of lower 3 numbers is 7, so their sum is 3*7= 21 and sum of the higher 2 numbers is= 45-21= 24.
The possible pairs for the higher 2 numbers are: (1,23),(2,22),(3,21),(4,20),(5,19),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13),(12,12) of which maximum value of the product is generated by the pair (12,12) which is 12*12= 144..... Option 4)

Like? Yes (1) | No