Question

there are 3 whole , positive number. the product of ist ans second number is 24. product of second and third number is 48, and that of ist and third is 32, let's calculate to find the three numbers.​

Answer 1

Given :-

♦ There are 3 positive whole numbers.

♦ The product of the first and second numbers is 24.

♦ The product of second and third numbes is 48

♦ The product of first and third numbers is 32

To find :-

♦ The the positive whole numbers .

Solution :-

Let the three positive whole numbers be X,Y and Z

Given that

The product of the first and second numbers=24

=> XY = 24 --------------(1)

=> X = 24/Y --------------(2)

The product of second and third numbes=48

=> YZ = 48 -----------------(3)

=> Z = 48/Y ------------------(4)

The product of first and third numbers=32

=> ZX = 32 -----------------------(5)

=> (48/Y)(24/Y) = 32

From (1) & (4)

=> (48×24)/(Y×Y) = 32

=> 1152/Y² = 32

=> 1152 = 32×Y²

=> 1152/32 = Y²

=> Y² = 1152/32

=> Y² = 36

=> Y = ±√36

=> Y = ±6

=> Y = 6

Since, Given numbers are positive whole numbers.

On substituting the value of Y in (2) then

X = 24/Y

=> X = 24/6

=> X = 4

On substituting the value of Y in (4) then

Z = 48/Y

=> Z = 48/6

=> Z = 8

Therefore, X = 4 , Y = 6 , Z = 8

Answer :-

The required three positive whole numbers are

4 , 6 and 8

Check :-

The three positive whole numbers are 4 , 6 and 8

The product of the first and second numbers

= 4×6

= 24

The product of second and third numbes

= 6×8

= 48

The product of first and third numbers

= 4×8

= 32

Verified the given relations in the given problem.

Answer 2

Answer:

Appropriate Question :-

• There are 3 whole, positive number. The product of first and second number is 24. Product of second and third number is 48, and that of first and third is 32. Let's calculate to find the three numbers.

Given :-

• There are 3 whole, positive number.
• The product of first and second number is 24.
• Product of second and third number is 48, and that of first and third is 32.

To Find :-

• What is the three numbers.

Solution :-

Let,

$\mapsto \bf First\: Number =\: a$

$\mapsto \bf Second\: Number =\: b$

$\mapsto \bf Third\: Number =\: c$

According to the question :

$\bigstar$ The product of first and second number is 24.

So,

$\implies \sf a \times b =\: 24$

$\implies \sf\bold{ab =\: 24\: ------\: (Equation\: No\: 1)}$

Again,

$\bigstar$ The product of second and third number is 48.

So,

$\implies \sf b \times c =\: 48$

$\implies \sf\bold{bc =\: 48\: ------\: (Equation\: No\: 2)}$

Again,

$\bigstar$ The product of first and third is 32.

So,

$\implies \sf a \times c =\: 32$

$\implies \sf\bold{ac =\: 32\: ------\: (Equation\: No\: 3)}$

From, the equation no 1 and equation no 2 we get :

$\implies \sf \dfrac{a\cancel{b}}{\cancel{b}c} =\: \dfrac{24}{48}$

$\implies \sf \dfrac{a}{c} =\: \dfrac{1}{2}$

By doing cross multiplication we get,

$\implies \sf\bold{c =\: 2a\: ------\: (Equation\: No\: 4)}$

By putting the equation no 4 into the equation no 3 we get :

$\implies \sf ac =\: 32$

$\implies \sf a(2a) =\: 32$

$\implies \sf 2a^2 =\: 32$

$\implies \sf a^2 =\: \dfrac{32}{3}$

$\implies \sf a^2 =\: 16$

$\implies \sf a =\: \sqrt{16}$

$\implies \sf\boxed{\bold{a =\: 4}}$

By putting the value of a in the equation no 3 we get :

$\implies \sf ac =\: 32$

$\implies \sf 4c =\: 32$

$\implies \sf c =\: \dfrac{32}{4}$

$\implies \sf\boxed{\bold{c =\: 8}}$

Again, by putting the value of c in the equation no 2 we get :

$\implies \sf bc =\: 48$

$\implies \sf b(8) =\: 48$

$\implies \sf 8b =\: 48$

$\implies \sf b =\: \dfrac{48}{8}$

$\implies \sf\boxed{\bold{b =\: 6}}$

Hence, the required numbers are :

$\dashrightarrow \sf First\: Number =\: a =\: \sf\bold{\underline{4}}$

$\dashrightarrow \sf Second\: Number =\: b =\: \sf\bold{\underline{6}}$

$\dashrightarrow \sf Third\: Number =\: c =\: \sf\bold{\underline{8}}$

$\therefore$ The three whole, positive numbers are 4 , 6 and 8 respectively.