Question

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

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Answer 1

$\rm\Huge\underline{\text{Question}}$

Three consecutive integers that when taken in increasing order and multiplied by 2, 3, and 4 respectively, add up to 74. Find these numbers.

$\rm\Huge\underline{\text{Idea}}$

Consecutive integers:-

Let's assume the three numbers are $x-1,x,x+1$.

$\rm\Huge\underline{\text{Solution}}$

$\rm\large\underline{\text{Finding the numbers:-}}$

The least number gets multiplied by 2:-

$\rm\rightarrow2\times\text{(least number)}=2(x-1)$

The middle number gets multiplied by 3:-

$\rm\rightarrow3\times\text{(middle number)}=3x$

The largest number gets multiplied by 4:-

$\rightarrow4\times\text{(largest number)}=4(x+1)$

$\rm\large\underline{\text{Establishing an equation:-}}$

Thus, $2(x-1),3x,4(x+1)$ sums to 74.

$\rm\rightarrow\underline{2(x-1)+3x+4(x+1)=74}$

Let's solve for $x$.

$\rm\large\underline{\text{Solving the equation:-}}$

$\rm\rightarrow2(x-1)+3x+4(x+1)=74$

BODMAS- brackets first.

$\rm\rightarrow2x-2+3x+4x+4=74$

$\rm\rightarrow(2x+3x+4x)+(-2+4)=74$

$\rm\rightarrow9x+2=74$

$\rm\rightarrow9x=72$

$\rm\rightarrow\red{\underline{x=8}\tiny\text{//}}$

The three consecutive numbers are 7, 8, and 9.

($7\times2+8\times3+9\times4=14+24+36=74$, hence verified.)

$\rm\Huge\underline{\text{Advanced examples}}$

$\rm\large\underline{\text{Question:-}}$

The nine consecutive numbers are arranged in increasing order. If each number gets multiplied by numbers from 2 to 10 respectively in increasing order, find the numbers if the sum is $438$.

$\rm\large\underline{\text{Hint:-}}$

Suppose the middle number is $x$, when there are odd numbers.

$\rm\large\underline{\text{Solution:-}}$

In this case, we can assume the numbers are consecutive integers from $x-4$ to $x+4$.

$\rm\large\underline{\text{Extablishing an equation:-}}$

$\rm\rightarrow\underline{2(x-4)+3(x-3)+4(x-2)+5(x-1)+6x+7(x+1)+8(x+2)+9(x+3)+10(x+4)=438}$

BODMAS- brackets first, then let's group the numbers.

$\rm\iff(2+3+4+5+6+7+8+9+10)x+(-8-9-8-5+7+16+27+40)=438$

$\rm\iff6\times9x+60=438$

$\rm\iff\underline{54x+60=48}$

$\rm\large\underline{\text{Solving the equation:-}}$

$\rm\rightarrow54x+60=438$

$\rm\iff54x=378$

$\rm\iff \red{\underline{x=7}\tiny\text{//}}$

The nine integers in increasing order are 3, 4, 5, 6, 7, 8, 9, 10, 11.

Answer 2

Step-by-step explanation:

given :

• Three consecutive integers are such that when they are taken in increasing order

• and multiplied by 2, 3 and 4 respectively, they add up to 74.

to find :

• Find these numbers = ?

• Find these numbers = ?

knowledge :

we will assume we have = 1, x, x + 1

solution :

we will do it with the help of BODMAS :

• 2 + 3(x+1) + 4(x+2) = 74

• 2+ 3 +3 + 4 + 8 = 74

• 9 + 11 = 74

• 9 = 74 - 11

• value = 63/9

• value =7

integer number is 7, 8 and 9

verify :

• integer number = 2 × 7+3 8+4 × 9

• integer number = 14+ 24 +36

• integer number = 74

learn more :

three integer number we will multiplied

by 2, 3, and 4 then we will add with. 74