Question

The product of two successive multiples of 10 is 1200 .Then find these multiples​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{Product of two successive multiples of 10 is 1200}$

$\underline{\textsf{To find:}}$

$\textsf{The those multiples of 10}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the two successive multiples of 10 be 10x and 10(x+1)}$

$\textsf{As per given data,}$

$\mathsf{10x\;{\times}\;10(x+1)=1200}$

$\mathsf{100\,x(x+1)=1200}$

$\mathsf{x(x+1)=12}$

$\mathsf{x^2+x-12=0}$

$\mathsf{(x+4)(x-3)=0}$

$\implies\mathsf{x=-4,3}$

$\mathsf{When\;x=3}$

$\mathsf{10x=10{\times}3=30}$

$\mathsf{10(x+1)=10(3+1)=40}$

$\mathsf{When\;x=-4}$

$\mathsf{10x=10{\times}(-4)=-40}$

$\mathsf{10(x+1)=10(-4+1)=-30}$

$\underline{\textsf{Answer:}}$

$\textsf{The required multiples of 10 are 30,40 (or) -40,-30}$

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Answer 2

Given :- The product of two successive multiples of 10 is 1200 . Then find these multiples ?

Solution :-

Let us assume that, first multiple of 10 is 10x and next multiple of 10 is 10y .

then,

→ First multiple * next multiple = 1200

→ 10x * 10y = 1200

→ 100xy = 1200

dividing both sides by 100,

→ xy = 12

Now,

• if x = 1, y = 12 => both are not consecutive numbers .
• if x = 2, y = 6 => both are not consecutive numbers .
• if x = 3, y = 4 => both are consecutive numbers .

Therefore,

→ First Multiple = 10x = 10 * 3 = 30

→ Next Multiple = 10y = 10 * 4 = 40 .

Hence, Required multiples are 30 and 40 respectively .

Note :-

As we know , if we multiply two negative numbers we get a positive number, since in question natural number is not mentioned , we can conclude that, given two successive multiples of 10 can be negative numbers also , which are (-30) and (-40) .

As,

• (-30) * (-40) = 1200 .

Therefore, Required multiples are also (-30) and (-40) respectively .

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