Answer:

Let the first number, the second number and the third number be f, s and t respectively.

f=s+3313100s=43

f=t+50100t=32t

f=43s=32t

⇒s98t=98(100)%oft

i.e., 112.5% of t

Verified answer

Answer

[tex] \rm \: The \: percentage \: is \: \bf112.5 \%[/tex]

Step-by-step-Explaination

Here are three numbers , upon which first no. is given and said that it is 33⅓% more than the second number .

And again Second no. is 50% more than the third number.

Here we need to point out two thing that 3rd no. is largest among three and First no. is smallest.

• First no. is 33⅓% more than the second number .
• Second no. is 50% more than the third number.

Thus , we can let the third no. now

Let :-

• Let the Third number be X.
• Let the Second number be Y.
• Let the First number be Z.
• Let the no. be K.

Then :-

Relationship of First no. with Second no.

[tex] \rm \: First \: number \: = Second \: no. + 33 \dfrac{1}{300} \times Second \: no. [/tex]

[tex] \rm \: z \: = y+ 33 \dfrac{1}{300} \times y[/tex]

[tex] \rm \: z \: = y+ \dfrac{100}{300} \times y[/tex]

[tex] \rm \: z \: = y+ \dfrac{1}{3} \times y[/tex]

[tex] \rm \: z \: = \dfrac{4}{3} \times y[/tex]

[tex] \boxed{ \rm \bf\: z \: = \dfrac{4}{3} y}[/tex]

Relationship of First no. with Third no.

[tex] \rm \: z \: = x + \dfrac{50}{100} x[/tex]

[tex] \rm \: z \: = x + \dfrac{x}{2} [/tex]

[tex] \boxed{\rm \bf \: z \: = \dfrac{3}{2} x}[/tex]

Now , According to the Problem,

[tex] \rm \bf \: \: \dfrac{k}{100} \times x = y[/tex]

[tex] \rm \bf \: \: \dfrac{y}{x} \times 100 = k[/tex]

Value of y in terms of z

[tex] \rm \: y = \dfrac{3}{4} z[/tex]

Value of x in terms of z

[tex] \rm \: x= \dfrac{2}{3} z[/tex]

Now , putting the values of relationships,

[tex] \rm \: k= \dfrac{ \frac{3}{4} z}{ \frac{2}{3} z} \times 100[/tex]

[tex] \rm \: k = \dfrac{3 \times 3}{4 \times 2} \times 100[/tex]

[tex] \rm \: k = \dfrac{9 \times 25}{ 2} [/tex]

[tex] \rm \: k = 112.5 \%[/tex]

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