Question

2. A number consists of two digits whose sum is 8. If 36 is added to the
number, the digits are reversed. Find the number.
3. The numerator of a rational number is less than its denominator by 3. If
the numerator becomes 3 times and the denominator is increased by 20,
the new number becomes . Find the original number.
4. The length of a rectangular room exceeds its breadth by 4m. If the length
is increased by 6 m and breadth is diminished by 4 m, there would be no
the area Find the length and breadth of the room.

Answer 1

Solution 2:

Let the ten's place digit be r & one's place digit be m respectively;

$\boxed{\bf{Original\:number=10r+m}}\\\boxed{\bf{Reversed\:number=10m+r}}$

A/q

$\mapsto\sf{r+m=8}\\\\\mapsto\sf{r=8-m.....................(1)}$

&

$\mapsto\sf{10r+m+36 = 10m+r}\\\\\mapsto\sf{10r-r + m-10m = -36}\\\\\mapsto\sf{9r -9m = -36}\\\\\mapsto\sf{9(r-m) = -36}\\\\\mapsto\sf{r-m=\cancel{-36/9}}\\\\\mapsto\sf{r-m=-4}\\\\\mapsto\sf{(8-m) -m=-4\:\:[from(1)]}\\\\\mapsto\sf{8-m-m=-4}\\\\\mapsto\sf{8-2m=-4}\\\\\mapsto\sf{-2m=-4-8}\\\\\mapsto\sf{-2m=-12}\\\\\mapsto\sf{m=\cancel{-12/-2}}\\\\\mapsto\bf{m=6}$

∴ Putting the value of m in equation (1),we get;

$\mapsto\sf{r=8-6}\\\\\mapsto\bf{r=2}$

Thus;

The number is 10r + m = 10(2) + 6 = 20 + 6 = 26 .

Solution 3:

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

The numerator of a rational number is less than it's denominator by 3. If the numerator becomes 3 times & the denominator is increased by 20, the new number becomes 1/8 .

$\underline{\bf{Explanation\::}}$

Let the denominator place be (r) & the numerator place be (r-3) respectively;

$\boxed{\bf{The \:original\:number=\frac{r-3}{r} }}$

A/q

• New numerator formed = 3(r-3) .
• New denominator formed = r + 20 .

Now;

$\mapsto\sf{\dfrac{3(r-3)}{r+20} =\dfrac{1}{8} }\\\\\mapsto\sf{\dfrac{3r -9}{r+20} =\dfrac{1}{8} }\\\\\mapsto\sf{8(3r - 9) = 1(r+20) \:\:\underbrace{\sf{Cross-multiplication}}}\\\\\mapsto\sf{24r -72 = r + 20}\\\\\mapsto\sf{24r - r = 20 + 72}\\\\\mapsto\sf{23r = 92}\\\\\mapsto\sf{r=\cancel{92/23}}\\\\\mapsto\bf{r=4}$

Thus;

• Numerator is r - 3 = (4 - 3) = 1
• Denominator is r = 4

$\boxed{\bf{The \:original\:number=\frac{r-3}{r} =\frac{4-3}{4} =\boxed{\bf{\frac{1}{4} }}}}$

Solution 4:

Let the breadth of a rectangular room be r m & the length of the rectangular room be (r+4) m .

As we know that formula of the area of rectangle;

$\boxed{\bf{Area\:_{(rectangle)} = Length \times breadth \:\:(sq,unit)}}$

We have rectangular dimensions :

• Length = (r+4) m
• Breadth = r m

$\longrightarrow\sf{Area \:_{(rectangle)} = Length \times breadth }\\\\\longrightarrow\sf{Area \:_{(rectangle)} = (r+4)\times r}\\\\\longrightarrow\sf{Area \:_{(rectangle)} = r^{2} + 4r}$

&

If the length is increased by 6 m & breadth is diminished by 4 m.

• New length of rectangular room formed = (r+4) + 6 = (r+10) m
• New breadth of rectangular room formed = (r-4) m

Now;

$\longrightarrow\sf{Area \:_{(rectangle)} = new\:length \times new\:breadth }\\\\\longrightarrow\sf{Area \:_{(rectangle)} = (r+10) (r-4)}\\\\\longrightarrow\sf{Area \:_{(rectangle)} = r^{2} - 4r + 10r -40}\\\\\longrightarrow\sf{Area \:_{(rectangle)} = r^{2} +6r -40}$

$\underline{\boldsymbol{According\:to\:the\:question\::}}$

$\longrightarrow\sf{\cancel{r^{2}} + 4r = \cancel{r^{2}} + 6r-40}\\\\\longrightarrow\sf{4r = 6r -40}\\\\\longrightarrow\sf{4r - 6r = -40}\\\\\longrightarrow\sf{-2r = -40}\\\\\longrightarrow\sf{r=\cancel{-40/-2}}\\\\\longrightarrow\bf{r=20\:m}$

Thus;

The original dimensions of the of the rectangular room;

• Length, (L) = r = 20 m .
• Breadth, (B) = (r+4) = (20 + 4) = 24 m .

Answer 2

Step-by-step explanation:

Q.2)

Given :-

• A number consists of two digits whose sum is 8.
• If 36 is added to the number, the digits are reversed.

To find :-

• The number.

Solution :-

Let the tens digit of the number be x and unit digit of the number be y.

Then, the number = 10x+y

According to the 1st condition,

• A number consists of two digits whose sum is 8.

x+y = 8

→ x = 8-y...............(i)

According to the 2nd condition,

• If 36 is added to the number, the digits are reversed.

10x+y+36 = 10y+x

→10(8-y)+y+36= 10y+8-y

→ 80-10y+y+36= 9y+8

→ -9y-9y = -108

→ -18y = -108

→ y = 6

Now put y = 6 in eq(i)

x = 8-6

→ x = 2

Therefore,

The number = 2×10+6 = 26

__________________

Q.3)

Given :-

• The numerator of a rational number is less than its denominator by 3.
• If the numerator becomes 3 times and the denominator is increased by 20, the new number becomes 1/8.

To find :-

• The original number.

Solution :-

Let the denominator of the rational number be x.

★ The numerator of a rational number is less than its denominator by 3.

• Numerator = (x-3)

★ If the numerator becomes 3 times and the denominator is increased by 20, the new number becomes 1/8.

• Numerator = 3(x-3)
• Denominator = x+20

According to the question,

$\to\sf{\dfrac{3(x-3)}{x+20}=\dfrac{1}{8}}$

$\to\sf{\dfrac{3x-9}{x+20}=\dfrac{1}{8}}$

→ 24x-72=x+20

→ 24x-x = 92

→ 23x = 92

→ x = 4

★ Denominator = 4

★ Numerator = (4-3) = 1

Therefore,

${\boxed{\sf{Original\: number=\dfrac{1}{4}}}}$

_____________________

Q.4)

Given :-

• The length of a rectangular room exceeds its breadth by 4m.
• If the length is increased by 6 m and breadth is diminished by 4 m, there would be no change in the area.

To find :-

• Length and breadth of the room.

Solution :-

Let the breadth of the rectangular room be x m.

★ The length of a rectangular room exceeds its breadth by 4m.

Then,

Length of the rectangular room = (x+4) m

Area of the rectangular room,

= [x(x+4) ] m²

=( x² + 4x ) m²

★ If the length is increased by 6 m and breadth is diminished by 4 m, there would be no change in the area.

• Length = (x+4+6) = (x+10) m
• Breadth = (x-4) m

Now, area of the rectangular room,

=[ (x+10)(x-4) ] m²

=[ x²-4x+10x-40] m²

= (x²+6x-40) m²

According to the question,

x²+4x = x²+6x-40

→ 4x = 6x-40

→ 4x-6x = -40

→ -2x = -40

→ x = 20

Therefore,

★Breadth of the rectangular room = 20 m.

★ Length of the rectangular room = (20+4) = 24 m.

__________________