Sum of three positive even integers is 30 and their product is 960. Sum of the two smaller integers

M4maths Previous Puzzles

31 Oct 2012 even integers

Sum of three positive even integers is 30 and their product is 960. Sum of the two smaller integers will be?

OPtion

1) 6
2) 8
3) 10
4) 12
5) 14
6) 16
7) 18
8) 20
9) 22
10) none of these

Solution

let the integers are 2a, 2b and 2c
according to the given condition
2a+2b+2c = 30 so a+b+c = 15 (1)
and 2a*2b*2c = 960
so abc = 120 (2)
on putting the value of a from (1) in (2)
we get (15-b-c)bc = 120
now check b = 1,2,3,4,5,6,7,8,9,10,11,12,13,14 after that 15-b-c will be negative
out of these 14 numbers 14,13,11,9 and 7 are not multiple of 120 so rejected. besides it 1,2 and 3 will give imaginary roots so rejected.
For b = 8,10 and 12 roots are not integers. Finally for b = 4,5 and 6 we get a and c as integer
values of a,b and c are 4,5 and 6
so integers are 8,10 and 12

Correct Option: 7 Best Solution (6)

Pranaya   11 years AGO

consider 3 even integer are 2x,2x+2,2x+4
then 2x+2x+2+2x+4=30
x=4
then 2x+2x+2=18

Like? Yes (10) | No (3)  

Piyush Jain   11 years AGO

x+y+x=30
xyz=960
xyz=96*10=48*2*10=24*4*10=12*8*10
x=12,y=10,z=8
y+z=18

Like? Yes (5) | No (3)  

susmit   11 years AGO

8+10+12=30
and 8*10*12=960
therefore three integers are 8,10,12
hence some of two small integers is 8+10=18
option no.7

Like? Yes (5) | No (4)  

Devendra Marghade   11 years AGO

Integers are 8,10 & 12
Sum of two smaller integers=8+10=18

Like? Yes (3) | No (3)  

sagar arora   11 years AGO

let the 3 nos be x,x+2,x+4
so,3x+6=30
x=8
thoughwe cant suppose thse nos as x,x+2,x+4 as they are not mentioned to be consecutive
but after calculating like this we get nos as 8,10,12 and their sum is 30 and product is 960 only so the are consecutive and the ans is option 18(8+10)

Like? Yes (3) | No (3)  

bhargavi   11 years AGO

the answer is 18
take x-2,x,x+2 as even integers
nw sum is 30 and product is 960
den num are 8,10,12
sum of 8 and 10

Like? Yes (3) | No (3)