Question

Class 11 Mathematics
Linear Inequalities

Find all pairs of consecutive even positive integers both of which are larger than 8 such that their sum is less than 25.

Additional Information:
I'm getting 4 < x < 5.75 by assuming two consecutive even numbers as 2x and 2x+2. What to do next?

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Linear Inequalities have been used. We see that we need to find all the pairs of consecutive even positive integers both of which are larger than 8 and such that their sum is less than 25 . So first we need to take a pair of even consecutive numbers . The we can apply them in inequalities one by and one and get our answer .

Let's do it !!

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★ What to do further :-

The answer which is got by questionnaire here is so small than the required answer . The numbers should be larger . This might be due to increased assumption . For simple calculations and accuracy its advised to use x and x + 2 as even integers .

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★ Solution :-

Let the smaller even integer be 'x' . The reason because surely both the integers in the pair differ in their value . Then let the other even positive integer be '(x + 2)' .

For first case we need to simply compare x with 8 according to the question.

~ Case I ::

$\\\;\;\bf{\pink{:\rightarrow\;\;x\;\;>\;\;8}}$

And,

$\\\;\;\bf{\pink{:\rightarrow\;\;x\;+\;2\;\;>\;\;8}}$

~ Case II ::

Here its given that the sum of both integers is less than 25 . Then,

$\\\;\;\bf{:\rightarrow\;\;x\;+\;x\;+\;2\;\;<\;\;25}$

$\\\;\;\bf{:\rightarrow\;\;2x\;+\;2\;\;<\;\;25}$

• Subtracting 2 on both sides, we get

$\\\;\;\bf{:\rightarrow\;\;2x\;+\;2\;-\;2\;\;<\;\;25\;-\;2}$

$\\\;\;\bf{:\rightarrow\;\;2x\;\;<\;\;23}$

• Dividing both sides by 2, we get,

$\\\;\;\bf{:\rightarrow\;\;\dfrac{2x}{2}\;\;<\;\;\dfrac{23}{2}}$

$\\\;\;\bf{\pink{:\rightarrow\;\;x\;\;<\;\;\dfrac{23}{2}}}$

Now combining Case I and Case II, we get

$\\\;\;\bf{:\mapsto\;\;8\;\;<\;\;x\;\;<\;\;\dfrac{23}{2}}$

$\\\;\;\bf{:\mapsto\;\;8\;\;<\;\;x\;\;<\;\;11.5}$

Adding 2 to each and every term,

$\;\;\bf{:\mapsto\;\;(8 + 2)\;\;<\;\;(x + 2)\;\;<\;\;(11.5 + 2)}$

Since, 12 < 13.5

Here we see 11.5 is a odd integer . And 9 is also a odd integer . Thus a left value between 8 and 11.5 which is even is 10 . Then,

✒ x = 10

✒ x + 2 = 10 + 2 = 12

This means that there is only one pair as per required.

$\\\;\underline{\boxed{\tt{Hence,\;\;only\;\;required\;\;pair\;=\;\bf{\purple{10\;and\;12}}}}}$

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★ Let's now solve this according to the assumption of Questionnaire ::

We see the two integers are 2x and 2x + 2 .

Then, using cases, we get,

~ Case I ::

$\\\;\;\bf{\gray{:\Longrightarrow\;\;2x\;\;>\;\;2(8)}}$

$\\\;\;\bf{\gray{:\Longrightarrow\;\;2x\;+\;2\;\;>\;\;2(8)}}$

The reason because here the variable is multiplied by two.

~ Case II ::

$\\\;\;\sf{:\Longrightarrow\;\;2x\;+\;2x\;+\;2\;\;<\;\;2(25)\;-\;2}$

Since, x is twice here.

$\\\;\;\sf{:\Longrightarrow\;\;4x\;\;<\;\;2(23)}$

$\\\;\;\sf{:\Longrightarrow\;\;4x\;\;<\;\;46}$

$\\\;\;\sf{:\Longrightarrow\;\;2x\;\;<\;\;23}$

Combining both cases,

$\\\;\;\bf{:\mapsto\;\;2(8)\;\;<\;\;2x\;\;<\;\;23}$

Cancelling each terms by 2, we get

$\\\;\;\bf{:\mapsto\;\;8\;\;<\;\;x\;\;<\;\;11.5}$

Here also we got same answer . But calculation turned out to be complicated . Both methods are applicable and correct .

$\\\;\underline{\boxed{\tt{Hence,\;\;only\;\;required\;\;pair\;=\;\bf{\green{10\;and\;12}}}}}$

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★ Verification ::

Case I :-

→ x, x + 2 > 8

→ 10,12 > 8

Condition Satisfies.

Case II :-

→ x + x + 2 < 25

→ 10 + 12 + 2 < 25

→ 24 < 25

This condition also is satisfied.

Here both conditions are satisfied. So our answer is correct.

Answer 2

Given :-

• Consecutive even positive integers both of which are larger than 8.
• Their sum is less than 25.

To find :-

• All pairs of Consecutive Even positive integers.

Solution :-

Let a Pair of Even consecutive Integers be

→ x and x + 2

But According to the question We should have both number to larger. Here smaller number is 8. According to it x should be greater than 8 and their sum should be less than 25. Conditions necessary for that is x should be positive then obviously x+2 will be positive and x be even.

x should be greater than 8, So,

→ x > 8

Also their sum should be less than 23. So,

→ x + x + 2 < 25

That is

→ x + x + 2 < 25

→ 2x + 2 < 25

→ 2x + 2 < 25

→ 2x < 25 - 2

→ 2x < 23

→ 2x < 23

→ x < 23 / 2

→ x < 11.5

So, Value of x should be in between of

8 < x < 11 . 5

So, all even numbers which are comming in between 8 and 11.5 are

• 10 only.

So,

→ x = 8 &

→ x + 2 = 8 + 2 = 10 also

→ 10 + 2 = 12

Here 10 and 12 should be the pair because 8 is not greater and we have asked the pair .

So, Consecutive even positive integers are :-

→ (10, 12 )