Answer
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Hint: Take the proportionate as ‘x’. thus get the length, breadth and height of the original cuboid. Then find l,b & h, new, increased side. Then find the volume of the original cuboid and new increased volume of the cuboid. Subtract them to find the increases in volume of cuboid.
Complete step-by-step answer:
We have been given a cuboid whose length, breadth and height in the ratio 1:2:3. Now let us take ‘x’ as the side.
$\therefore $ The length of cuboid $=1.x=x$ .
The breadth of cuboid $=2.x=2x$ .
The height of cuboid $=3.x=3x$ .
Now their dimensions of the cuboid are increased by 100%, 200% and 200%. i.e. the length of cuboid is increased by 100%, the breadth is increased by 200% and the height is increased by 200%.
Hence, the new dimensions of the cuboid are
Length $=x+\dfrac{100}{100}x=x+x=2x$ .
Breadth $=2x+\dfrac{200}{100}\times 2x=2x+4x=6x$ .
Height $=3x+\dfrac{200}{100}\times 3x=3x+6x=9x$ .
Hence, the original dimensions of cuboid $=x,2x,3x$ .
New dimensions of the cuboid $=2x,6x,9x$ .
We know that volume of a cuboid $=length\times breadth\times height$ .
$\therefore $ Original volume of cuboid $=x\times 2x\times 3x=6{{x}^{3}}$
Now increased volume of cuboid $=2x\times 6x\times 9x=108{{x}^{3}}$
Thus the increase in volume = New increased volume – original volume of cuboid
$=108{{x}^{3}}-6{{x}^{3}}=102{{x}^{3}}$
Thus we can write the increase in volume as,
$102{{x}^{3}}=17\times 6{{x}^{3}}$
We know that $6{{x}^{3}}$ is the original value.
$\therefore $ Increase in volume $=17\times \text{ Original volume}$
Thus, when comparing the original volume the increase in the volume of cuboid is 17 times.
$\therefore $ Option (D) is the correct answer.
Note: It is said length, breadth and height are in ratio. Don’t take different variables for the 3 quantities as l,b,h etc. take as they are in proportionate as ‘x’. putting 3 variables is complicated and you won't get the answer.
Complete step-by-step answer:
We have been given a cuboid whose length, breadth and height in the ratio 1:2:3. Now let us take ‘x’ as the side.
$\therefore $ The length of cuboid $=1.x=x$ .
The breadth of cuboid $=2.x=2x$ .
The height of cuboid $=3.x=3x$ .
Now their dimensions of the cuboid are increased by 100%, 200% and 200%. i.e. the length of cuboid is increased by 100%, the breadth is increased by 200% and the height is increased by 200%.
Hence, the new dimensions of the cuboid are
Length $=x+\dfrac{100}{100}x=x+x=2x$ .
Breadth $=2x+\dfrac{200}{100}\times 2x=2x+4x=6x$ .
Height $=3x+\dfrac{200}{100}\times 3x=3x+6x=9x$ .
Hence, the original dimensions of cuboid $=x,2x,3x$ .
New dimensions of the cuboid $=2x,6x,9x$ .
We know that volume of a cuboid $=length\times breadth\times height$ .
$\therefore $ Original volume of cuboid $=x\times 2x\times 3x=6{{x}^{3}}$
Now increased volume of cuboid $=2x\times 6x\times 9x=108{{x}^{3}}$
Thus the increase in volume = New increased volume – original volume of cuboid
$=108{{x}^{3}}-6{{x}^{3}}=102{{x}^{3}}$
Thus we can write the increase in volume as,
$102{{x}^{3}}=17\times 6{{x}^{3}}$
We know that $6{{x}^{3}}$ is the original value.
$\therefore $ Increase in volume $=17\times \text{ Original volume}$
Thus, when comparing the original volume the increase in the volume of cuboid is 17 times.
$\therefore $ Option (D) is the correct answer.
Note: It is said length, breadth and height are in ratio. Don’t take different variables for the 3 quantities as l,b,h etc. take as they are in proportionate as ‘x’. putting 3 variables is complicated and you won't get the answer.
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