Latest Hexaware Aptitude Question SOLUTION: the ratio of the length : breadth :height of cuboid is 5:4:3 the volumeof 7500 what will be the surface area of the bar

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the ratio of the length : breadth :height of cuboid is 5:4:3 the volumeof 7500 what will be the surface area of the bar

Asked In Hexaware diviyapriyanka (8 years ago)

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GopalaRao  (8 years ago)

volume= 5x X 4x X 3x=7500
so, x=5
5x=25, 4x= 20, 3x=15
suface area= 2(lb + bh + hl)= 2(25X20 + 20X15 + 15X25) = 2350

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Sayed Imran Hossain  (8 years ago)

volume= 5x X 4x X 3x=7500
so, x=5
5x=25, 4x= 20, 3x=15
suface area= 2(lb + bh + hl)= 2(25X20 + 20X15 + 15X25) = 2350

Is this solution Helpfull? Yes (5) | No   

Devakumar  (6 years ago)

Let the Length, Breadth and Height of Cuboid is 5X, 4X, 3X.
Volume = 7500
5X * 4X * 3X = 7500
60 X^3 = 7500
X^3 = 750/6
X^3 = 125
X = 5
Therefore Length = 25, Breadth = 20, Height = 15
Surface Area = 2(Length + Breadth + Height)
Surface Area = 2(60)
Surface Area = 120

Is this solution Helpfull? Yes (2) | No (7)  

Priyanka ChandraSekar  (5 years ago)

1^3+1=2
2^3+2=10
3^3+3=30
4^3+4=68
so 5 = 5^3+5=130 is the ans

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Haripriya  (4 years ago)

5x*4x*3x=60x^3
60x^3=7500
x^3=7500/60
x^3=125
x=5
surface area =2(l+b+h)
volume=2(5*5+4*5+3*5)
2(60)
120

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