IBM
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Numerical Ability
Area and Volume
A room has width half of its length. When 6 is decreased from both length and width then its area is differed by 108 so find the length.
Read Solution (Total 8)
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- let length = x n width = x/2
so as per question
{(x*x/2)-(x-6)(x/2-6)}=108
so after solving 9x =144
x=16
so length is 16.. - 14 years agoHelpfull: Yes(95) No(7)
- Answer is legth=16; Because if we consider length as x then width is (x/2).Area will be (x^2/2).after that the length and width are decreased by 6 so new length and width are (x-6),(x/2-6) and the difference between these two areas is 108 so we get the equation (x^2/2)-((x-6)*(x^2/2-6))=108.On solving this we get the value of x(length)=16.
- 14 years agoHelpfull: Yes(15) No(5)
- answer will be 8
- 10 years agoHelpfull: Yes(10) No(11)
- answer is :8
{(x*x/2)-(x-6)(x/2-6)}=108
18X+72=108*2
X+4=12
X=8 - 9 years agoHelpfull: Yes(2) No(8)
- l^2-(l-6)^2=108*2 will be the equation.
so the ans will be 21 - 11 years agoHelpfull: Yes(1) No(14)
- 24
A1-A2=108
L^2/2-(L^2/2-6L-3L+36)=108
3L=108-36
L=72/3=24
L=24 Answer.....
- 10 years agoHelpfull: Yes(1) No(9)
- x^2-18x+72=2a-216
where 2a=x^2
length =16 - 8 years agoHelpfull: Yes(0) No(0)
- l and l/2
a=l^2/2
a'=(l-6)(l/2-6)=(l-6)(l-12)/2=l^2-18l+72/2
a-a'=108
solving this equation,
l=16 - 6 years agoHelpfull: Yes(0) No(0)
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