Question

10. The dimensions of a rectangular box are in
the ratio 4 :: 2 : 3. The difference between
cost of covering it with paper at 12 per m2
and with paper at the rate of 13.50 per m² is
1,248. Find the dimensions of the box.​

Answer 1

AnswEr :

• L : B : H = 4 : 2 : 3
• Difference between cost of covering it with paper at 12 @m² and with paper at the rate of 13.50 @m² is Rs. 1,248
• Find the Dimensions of Box

Let the Dimensions of Cuboidal box be 4n, 2n and, 3n respectively.

• According to the Question Now :

$\longrightarrow \tt (TSA \times High \:Rate) - (TSA \times Low \:Rate) = Rs. 1248 \\ \\\longrightarrow \tt TSA( High \:Rate - Low \:Rate) = Rs. 1248 \\ \\\longrightarrow \tt TSA( 13.50 - 12) = Rs. 1248 \\ \\\longrightarrow \tt TSA \times 1.5= 1248 \\ \\\longrightarrow \tt TSA = \cancel\dfrac{1248 \times 10}{15} \\ \\\longrightarrow \tt TSA = 832\:{m}^{2} \\ \\\longrightarrow \tt\cancel{2}(LB + BH + HL) =\cancel{832 \:{m}^{2}} \\ \\\longrightarrow \tt(LB + BH + HL) =416\:{m}^{2} \\\\\longrightarrow \tt(4n \times 2n) + (2n \times 3n) + (3n \times 4n) = 416 \\ \\\longrightarrow \tt8{n}^{2} + 6 {n}^{2} + 12 {n}^{2} = 416 \\ \\\longrightarrow \tt26 {n}^{2} = 416 \\ \\\longrightarrow \tt {n}^{2} = \cancel\dfrac{416}{26} \\ \\\longrightarrow \tt {n}^{2} = 16 \\ \\\longrightarrow \tt n =\sqrt{16} \\ \\\longrightarrow \blue{\tt n = 4}$

$\rule{300}{1}$

• D I M E N S I O N S :

◗ Length = 4n = 4(4) = 16 m

◗ Breadth = 2n = 2(4) = 8 m

◗ Height = 3n = 3(4) = 12 m

⠀

∴ Dimensions of Cuboidal box will be 16 m, 8 m and 12 m respectively.

$\rule{300}{2}$

• I M P O R T A N T ⠀F O R M U L A E :

1 ) Cuboid :

↠ Volume = lbh

↠ Surface Area = 2(l + b) × h

↠ TSA = 2(lb + bh + hl)

↠ Diagonal = √(l² + b² + h²)

$\rule{100}{2}$

2 ) Cube :

↠ Volume = (Side)³

↠ Surface Area = 4 × (Side)²

↠ TSA = 6 × (Side)²

↠ Diagonal = √3 Side

$\rule{100}{2}$

3 ) Cylinder :

↠ Volume = πr²h

↠ CSA = 2πrh

↠ TSA = 2πr(r + h)

$\rule{100}{2}$

4 ) Cone :

↠ Volume = 1 /3 × πr²h

↠ CSA = πrl

↠ TSA = πr(r + l)

↠ Slant Height ( l ) = √(r² + h²)

$\rule{100}{2}$

5 ) Sphere :

↠ Volume = 4 /3 × πr³

↠ Surface Area = 4πr²

$\rule{100}{2}$

6 ) Hemisphere :

↠ Volume = 2 /3 × πr³

↠ CSA = 2πr²

↠ TSA = 3πr²

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\red{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

The dimensions of a rectangular box are in the ratio 4:2:3. The difference between cost of covering it with paper at 12/m² & with paper at the rate of Rs.13.50/m².

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The dimensions of the box.

$\bf{\Large{\underline{\boxed{\rm{\purple{Explanation\::}}}}}}$

Let the ratio of the dimensions of rectangular box be R

$\bf{We\:have\begin{cases}\rm{The\:length\:of\:rectangular\:box\:=\:4R}\\ \rm{The\:breadth\:of\:rectangular\:box\:=\:2R}\\ \rm{The\:height\:of\:rectangular\:box\:=\:3R}\\ \sf{Cost\:of\:covering\:with\:paper\:=12/m^{2} }\\ \rm{The\:rate\:of\:paper\:=\:Rs.13.50/m^{2}\:is\:1248 }\end{cases}}$

Therefore,

$\rm{Area\:of\:sheet\:paper\:required\:for\:covering\:=\:Total\:surface\:area\:of\:cuboid}$

We know that formula of the Total surface area of cuboid:

$\leadsto\rm{\orange{2(Length*breadth+breadth+height*length*height)}}$

$\longmapsto\rm{2(4R*2R+2R*3R+4R*3R)}$

$\longmapsto\rm{2(8R^{2}+6R^{2}+12R^{2})}$

$\longmapsto\rm{2*26R^{2}}$

$\longmapsto\rm{52R^{2}m^{2}}$

$\bf{\large{\boxed{\rm{Cost\:of \:covering\: it\: with\: sheet \:of \:paper\: at \:Rs.9.50/m^{2} }}}}}$

$\longmapsto\rm{(9.50*52R^{2} )}$

$\longmapsto\rm{\orange{Rs.(494R^{2})}}$

&

$\bf{\large{\boxed{\rm{Cost\:of\: covering \:it\: with\: sheet\: of\: paper \:at\: rate\: Rs.8/m2}}}}}$

$\longmapsto\rm{(8*52R^{2} )}$

$\longmapsto\rm{\orange{Rs.(416R^{2})}}$

• According to the question:

→ 494R² - 416R² = 1248

→ 78R² = 1248

→ R² = $\rm{\cancel{\frac{1248}{78} }}$

→ R² = 16

→ R = √16

→ R = 4m

Thus,

• $\rm{The\:length\:of\:rectangular\:box\:=\:4(4)=16m}$

• $\rm{The\:breadth\:of\:rectangular\:box\:=\:2(4)=8m}$

• $\rm{The\:height\:of\:rectangular\:box\:=\:3(4)=12m}$