Question

Solve these all Questions please

⚠️No Spamming ⚠️
– Quality answer required
​

Answer 1

Answer:

A1) Let the breadth of the room be 'x'. According to the question, the length of the room is 50% more than its breadth. This implies:

→ Length = x + 50% of x

→ Length = x + 0.5x = 1.5x

Given the cost of carpeting is Rs. 38.50 per m² & total cost comes out to be Rs. 924. Carpeting covers only the Area of the floor. Hence Area of the floor is given as:

→ Area = length × breadth

→ Area = 1.5x × x

→ Area = 1.5x² m²

Now applying cost we get:

Total Cost = Area × Cost per m²

→ Rs. 924 = 1.5x² × Rs. 38.5

→ 1.5x² = 924 / 38.5

→ 1.5x² = 24

→ x² = 24 / 1.5

→ x² = 16

⇒ x = √16 = 4 units.

Hence the breadth of the room is 4 units.

Length of room = 1.5 × 4 = 6 units.

In the second part of the question, it is given that the cost of painting the walls is Rs. 5.5 per m² and the total cost incurred is 1320. Assuming all the 4 walls are painted we get:

→ Total Area to be painted = 4 × Area of 1 wall

→ Area of wall = Lateral Surface Area of the room

Let us assume height to be 'h'

→ Area of 4 walls = 2h ( l + b )

→ Area of 4 walls = 2h ( 4 + 6 ) = 2h ( 10 )

→ Area of 4 walls = 20h m²

Total Cost = Total Area × Cost per m²

→ Rs. 1320 = 20h × Rs. 5.5

→ 20h = 1320 / 5.5

→ 20h = 240

→ h = 240 / 20 = 12

Hence height of the room is 12 units.

The dimensions of the room are 6m × 4m × 12m.

A2) Cost of ( x - 3 ) articles is x² - 5x + 6

→ [ x² - 2x - 3x + 6 ]

→ [ x ( x - 2 ) -3 ( x - 2 ) ]

→ ( x - 3 ) ( x - 2 ) ⇒ Cost of ( x - 3 ) articles

Applying the law of proportions we get:

$\rightarrow ( x - 3 )\:articles : ( x - 3 ) ( x - 2 ) :: (x + 2 ) \:articles :\: y\\\\\rightarrow ( x - 3 ) \times y = ( x -3 ) ( x - 2 ) \times (x+2)\\\\\rightarrow y = \dfrac{ (x -3)(x-2)(x+2)}{(x-3)}\\\\\rightarrow y = (x-2)(x+2) = (x^2 - 4)\\\\\text{Hence cost of (x+2) articles is}\:(x^2-4).$

A3) Formula for calculating the Mean observation is written as:

$\rightarrow Mean = \dfrac{ \text{Sum of all observations}}{\text{Total number of observations}}$

According to the question, the sum of all observations is ( x² + x - 18 ) and the number of observations is ( x + 2 ). Substituting the values we get:

$\rightarrow Mean = \dfrac{ (x^2 + x - 18) }{ (x + 2)}$

Since ( x + 2 ) is not a factor, we would get a reminder as well. Mean would hence be written in decimals if the value of 'x' is specified.

A4) The given question is of the form ( x + y + z )². Hence we would get:

→ ( x + y + z )² = x² + y² + z² + 2xy + 2yz + 2zx.

On evaluating the LHS with the RHS of above equation, we get:

→ x² = a² ⇔ ( x = a )

→ y² = b²/4 ⇔ ( y = b/2 )

→ z² = c²/9 ⇔ ( z = c/3 )

→ 2xy = 2 ( a ) ( b/2 ) = ab

→ 2yz = 2 ( b/2 ) ( c/3 ) = bc/3

→ 2zx = 2 ( a ) ( c/3) = 2ac/3

Hence on factorizing, we get ( a + b/2 + c/3 )² as the answer.

A5) Area of upper surface ( face ) = 184 cm²

→ πr² = 154 cm²

$\rightarrow \dfrac{22}{7} \times r^2 = 154\\\\\rightarrow r^2 = \dfrac{154 \times 7}{22}\\\\\rightarrow r^2 = 49\;cm^2\\\\\rightarrow r = \sqrt{49} = 7\:cm$

Also it is given that Lateral Surface Area of the cylinder is 880 cm².

$\rightarrow LSA = 880\:cm^2\\\\\rightarrow 2\pi rh = 880\\\\\rightarrow 2 \times \dfrac{22}{7} \times 7 \times h = 880\: cm^2\\\\\rightarrow 44 \times h = 880\: cm^2\\\\\rightarrow h = \dfrac{880}{44} = 20\: cm$

Hence the radius of the cylinder is 7 cm and the height of the cylinder is 20 cm.

Answer 2

Answer:

• 1) The length of a room is 50% more than it's breadth. The cost of carpeting the room at the rate of Rs 38.50 per m² is Rs 924 and the cost of painting the walls at the rate of Rs 5.50 per m² is Rs 1320. Find the dimensions of the room.

Solution:

Let the breadth of the room be x m.

50% of x = x/2

Therefore, Length will be x + x/2 = 3x/2

Area of carpet = 924/38.50

$\sf{\therefore}$ Area of room = 24 m²

Also,

$\boxed{\sf{Area \ of \ rectangle = Length\times \ Breadth}}$

Area of carpet = Length × Breadth

$\sf{\therefore}$ Area of carpet = (3x/2) (x)

$\sf{\therefore}$ 3x²/2 = 24

$\sf{\therefore}$ x² = 16

$\sf{\therefore}$ x = 4 m

[Since, length can't be negative]

=> Breadth = 4 m and Length = 3(4)/2 = 6 m

For painting the walls we need to find the lateral surface area of the cuboid shaped room.

Cost of painting walls is Rs 5.50 per m² and Total cost of painting the walls is Rs 1320

$\sf{\therefore}$ Area of walls = 1320/5.50

$\sf{\therefore}$ Area of walls = 240 m²

$\boxed{\sf{Lateral \ surface \ area \ of \ cuboid=2(l+b)h}}$

$\sf{\therefore}$ 240 = 2 (l + b) h

But, Length = 6 m and Breadth = 4 m

$\sf{\therefore}$ 240 = 2(6 + 4) h

$\sf{\therefore}$ h = 240/20

$\sf{\therefore}$ h = 12 m

Therefore, the dimensions of the room length, breadth and height are 6 m, 4m and 12 m respectively.

__________________________________

• 2) If the cost of (x - 3) articles is x² - 5x + 6 then, find the cost of such (x + 2) articles.

Solution:

Let the price of (x + 2) articles be n.

Cost is directly proportional to the number of articles.

In direct proportion $\sf{\dfrac{x}{y}}$ is constant.

$\sf{\therefore{\dfrac{x-3}{x^{2}-5x+6}=\dfrac{x+2}{n}}}$

$\sf{\therefore{n=\dfrac{(x^{2}-5x+6)(x+2)}{x-3}}}$

$\sf{\therefore{n=\dfrac{x^{3}-5x^{2}+6x+2x^{2}-10x+12}{x-3}}}$

$\sf{\therefore{n=\dfrac{x^{3}-3x^{2}-4x+12}{x-3}}}$

$\sf{\therefore{n=\dfrac{x^{2}(x-3)-4(x-3)}{x-3}}}$

$\sf{\therefore{n=\dfrac{(x-3)(x^{2}-4)}{x-3}}}$

$\sf{\therefore{n=x^{2}-4}}$

Therefore, the cost of (x +2) articles is x² - 4.

_________________________________

• 3) The sum of observations (x + 9) is⠀⠀⠀⠀⠀⠀⠀⠀⠀
• x²+ 7x - 18, find the mean observation.

Solution:

Here,

Number of observations = (x + 9)

Sum of (x + 9) observations = x² + 7x - 18

$\sf{Mean \ observation =\dfrac{Sum \ of \ observations}{Number \ of \ observations}}$

$\sf{\therefore{Mean \ observation=\dfrac{x^{2}+7x-18}{x+9}}}$

$\sf{\therefore{Mean \ observation=\dfrac{x^{2}+9x-2x-18}{x+9}}}$

$\sf{\therefore{Mean \ observation=\dfrac{x(x+9)-2(x+9)}{x+9}}}$

$\sf{\therefore{Mean \ observation=\dfrac{(x+9)(x-2)}{x+9}}}$

$\sf{\therefore{Mean \ observation=x-2}}$

Therefore, the mean observation is x - 2.

_________________________________

• 4) Factorize:

$\sf{a^{2}+\dfrac{b^{2}}{4}+\dfrac{c^{2}}{9}+ab+\dfrac{bc}{3}+\dfrac{2ca}{3}}$

Solution:

$\sf{\leadsto{a^{2}+\dfrac{b^{2}}{4}+\dfrac{c^{2}}{9}+ab+\dfrac{bc}{3}+\dfrac{2ca}{3}}}$

$\sf{Comparing \ with}$

$\boxed{\sf{(a_{1}+b-{1}+c_{1})^{2}=a-{1}^{3}+b_{1}^{3}+c_{1}^{3}+2a_{1}b_{1}+2b_{1}c_{1}+2a_{1}c_{1}}}$

$\sf{we \ get, \ a_{1}=a, \ b_{1}=\dfrac{b}{2} \ and \ c_{1}=\dfrac{c}{3}}$

$\sf{\leadsto{(a+\dfrac{b}{2}+\dfrac{c}{3})(a+\dfrac{b}{2}+\dfrac{c}{3})}}$

Therefore, factorized form of

$\sf{a^{2}+\dfrac{b^{2}}{4}+\dfrac{c^{2}}{9}+ab+\dfrac{bc}{3}+\dfrac{2ca}{3}}$

is $\sf{(a+\dfrac{b}{2}+\dfrac{c}{3})(a+\dfrac{b}{2}+\dfrac{c}{3})}$

_________________________________

• 5) In a cylindrical pipe ( in the figure ), the area of the surfaces are given. Find the radius and the height of the cylinder.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(1.25,-1)(2.5,0)\qbezier(0,0)(1.25,1)(2.5,0)\qbezier(0,3.5)(1.25,4.5)(2.5,3.5)\qbezier(0,3.5)(1.25,2.5)(2.5,3.5)\multiput(0,0)(2.5,0){2}{\line(0,1){3.5}}\put(0.3,3.4){ \bf 154 \ sq.units }\put(0.3,1.7){ \bf 880 \ sq.units }\end{picture}$

Solution:

The area of upper circular surface is 154 unit²

$\boxed{\sf{Area \ of \ circle=\pi \ r^{2}}}$

$\sf{\therefore{154=\dfrac{22}{7}\times \ r^{2}}}$

$\sf{\therefore{r^{2}=\dfrac{154\times7}{22}}}$

$\sf{\therefore{r^{2}=49}}$

$\sf{\therefore{r=7 \ units}}$

[Since, length can't be negative]

=> $\sf{\therefore}$ Radius = 7 units

For cylinder,

$\boxed{\sf{\therefore{Curved \ surface \ area=2\pi\times \ r\times \ h}}}$

$\sf{\therefore{880=2\times\dfrac{22}{7}\times7\times \ h}}$

$\sf{\therefore{h=\dfrac{880}{44}}}$

$\sf{\therefore{h=20 \ units}}$

Therefore, the radius and height of the cylinder are 7 units and 20 units respectively.