Question

Find three consecutive odd integers such that the sum of the last two is 15 less than 5 times the first. What is the product?

Answer 1

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✤ Required Answer:

✒ GiveN:

• There are 3 consecutive odd integers.
• Sum of last two is 15 less than 5 times the 1st.

✒ To FinD:

• Product of the numbers......?

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✤ How to solve?

First of we have to frame equations by taking a single variable, because we have to solve it.

• Consecutive odd integers have difference of 2.
• Then, Taking one variable and framing the second condition, we will get our 1st number. Accordingly, the second and third one.

☔ So, Let's solve this question..$.$

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✤ Solution:

Let,

• 1st number be x
• Then, 2nd number will be x + 2
• And, 3rd number will be x + 4

According to question,

• 5 × 1st number - 15 = sum of 2nd and 3rd.

Then,

➝ 5 × x - 15 = (x + 2) + (x + 4)

➝ 5x - 15 = x + 2 + x + 4

➝ 5x - 15 = 2x + 6

➝ 5x - 2x = 15 + 6

➝ 3x = 21

➝ x = 7

As per our consideration,

• First number x = 7
• Second number x + 2 = 9
• Third number x + 4 = 11

Then, product of numbers,

➝ 7 × 9 × 11

➝ 693

✒ Our Required product = 693

☀️ Hence, solved !!

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Answer 2

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Let, X is the first number,

The others consecutives odd integers must be

X+2

And

X+4

That said, the resulting equation

Is pretty simple:

x + (x+2) + (x+4) + 15= 5x

And so :

2x + 6 + 15 = 5x

21=3x

• x = 7

If seven is our first odd number ,the others two must be :

• x+2 = 7+2 = 9

• x+4 = 7+4 = 11

Hence, Numbers Are = 7, 9, 11

So, Product = 7 × 9 × 11

= 693

$\bf\</strong>p<strong>i</strong><strong>n</strong><strong>k</strong><strong>{Hope\ it\ helps.}$

$\</strong><strong>s</strong><strong>f</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{Plz\ Mark\ As\ Brainliest.}$