Question

Find two consecutive numbers such that  twice the larger is 40 less than 3 times the smaller ( for class 8 ) ​

Answer 1

Answer:

$42$ and $43$

Step-by-step explanation:

Let, the two consecutive numbers be $x$ and $(x+1)$.

According to given condition, the equation becomes

$2(x+1)=3x-40$

Now, solve this equation to get the value of .

∴ $2x+2=3x-40$

∴ $3x-2x=40+2$

∴ $x= 42$

Now, $(x+1) = (42+1) = 43$

∴ The two consecutive numbers are $42$ and $43$.

Answer 2

Let,

The two numbers be $x$ and $(x+1)$

The given condition is,

Twice the larger is $40$ less than $3$ times the smaller

The suitable condition is

$2(x+1)=3x-40$

Simplify the above condition, we get

$2x+2=3x-40$

$3x-2x=40+2$

$x=42$

The second consecutive number is $(x+1)$ in the place of $x$ substitute the value

$x+1=42+1$

$=43$

Therefore, the two consecutive numbers are $42$ and $43$.