The product of the ages of 4 elementary school students is 5040. a is older than b by a year. b is older than c by a year. c is older than d by a year. what are the ages of the students?

The age of student A= 10 year.

The age of student B=9 years.

The age of student C=8 years.

The age of student D=7 years.

Explanation:

Given:

The product of the ages of 4 elementary school students is 5040.  

A is older than B by a year.  

B is older than C by a year.  

C is older than D by a year.  

To find:

The ages of the students?  

Solution:

Now, let the age of D be x,

Then, age of C will be x + 1.

Consequently, age of B and A will be x + 2 and x + 3

Now, we know that, product of ages = 5040  

x(x + 1)(x + 2)(x + 3) = 5040  

x(x + 1)(x + 2)(x + 3) = 504 x 10  [ factorizing the number 5040 ]

x(x + 1)(x + 2)(x + 3) = 84 x 6 x 10  

x(x + 1)(x + 2)(x + 3) = 21 x 4 x 6 x 10  

x(x + 1)(x + 2)(x + 3) = 7 x 3 x 4 x 3 x 2 x 10  

x(x + 1)(x + 2)(x + 3) = 7 x (4 x 2) x (3 x 3) x 10  

x(x + 1)(x + 2)(x + 3) =7 x 8 x 9 x 10  

x(x + 1)(x + 2)(x + 3) = 7(7 + 1)(7 + 2)(7 + 3)  

So, by comparison x = 7 .

Then, ages of A, B, C, D are 7 + 3 = 10, 7 +2 = 9, 7 + 1= 8 and 7 respectively.

Hence, the ages of students are 10 years, 9 years, 8 years, and 7 years for A, B, C, D respectively.

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.