Question

4. Find a rational number between 1/4 and 1/3.​

Answer 1

Answer ♡

Given

Perimeter of Rectangle is 60m

length is 4m more than four times it's breadth

To findi

length and breadth of the Rectangle

Slove

Let us assume the breadth of Rectangle be 'x'

according to the question

length would be => 4m+4x

Perimeter is 60m

Perimeter of Rectangle=2(length+Breadth)

=>60= 2( 4+4x +x)

=>60=2(4+5x)

=>2(4+5x)=60

=>4+5x=30

=>5x= 30-4

=>5x= 26

=>x= 26÷ 5

=>x=5.2

Now , breadth is 5.2m

length = 4+4x=4+4×5.2

length= 4+20.8=24.8m

Check

Perimeter of Rectangle=2( 24.8+5.2)

Perimeter of Rectangle=2(30)

Perimeter of Rectangle= 60m

Extra information=>

Perimeter is the total distance occupy by a solid 2D figure around its edge.

Area of Rectangle= length × breadth

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Answer 2

Given:

• Rational numbers : $\sf \dfrac{1}{4} and \dfrac{1}{3}$

To Find:

• A rational numbers between them

Solution:

➪ Now, we have been given 2 rational numbers and said that we have to find few more rational numbers in between them

How do you find them?

Nothing much to worry about though. it's just a piece of cake So, firstly what we do here is that we find the common multiple of both the denominator for the rational numbers and then we make both the denominators equal and them list out the rational numbers in between the given rational numbers.

Let's start now :

→ ln the first case :

$\tt \longrightarrow \dfrac{1}{4}$

• 1 is the numerator and 4 is the denominator

→ln the second case :

$\tt \longrightarrow \dfrac{1}{3}$

• 1 is the numerator and 3 is the denominator

✪ As I have mentioned above now let's find the l.c.m of 3 and 4

$\begin{gathered}\begin{gathered}\:\:\:\:\:\:\:\:\:\:\:\begin{gathered} \begin{array}{c|c} \underline{\sf{3}}& {\sf{ \underline{ \red{3,4} \: \: \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\ \underline{\sf{4}}&{\sf{ \underline{1 ,4\: \: \: }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\\underline{\sf{1}}&{\sf{ \underline{1,1\: \: \: \: \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\ \sf{} & \sf{1 \: \: \: \: \: } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{array}\end{gathered}\end{gathered}\end{gathered}$

➼ Now the least common multiple is

• 3 × 4 × 1 × 1

• 12

➠ Hence their L.C.M is 12

✪ we know that,

• 4 × 3 = 12
• 3 × 4 =12

So, we are gonna multiply the first denominator with 3 and the second one with 4 so that the denominators are equal

$\tt \longrightarrow \frac{1 \times 4}{3 \times 4} \\ \\ \\ \tt \longrightarrow \frac{4}{12} \: \: \: \: \:$

• Hence, the new number = 4 / 12

let's do the same with the second one too

$\tt \longrightarrow \frac{1 \times 3}{4 \times 3} \\ \\ \\ \tt \longrightarrow \: \frac{3}{12} \: \: \: \: \:$

Now,

As there is no possiblity of a rational number in between let's multiply the numerator and the denominator with a greater number i.e 10

$\tt \longrightarrow \frac{4 \times 10}{12 \times10 } , \frac{3 \times 10}{12 \times 10} \\ \\ \\ \tt \longrightarrow \frac{40}{120} , \frac{30}{120} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence:

• The numbers in between are :$\\ \tt \dfrac{31}{120} , \dfrac{32}{120} , \dfrac{33}{120} , \dfrac{37}{120} , \dfrac{39}{120}$

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