Question

The sum of the successors of two
numbers is 40 and the difference of
their predecessors is 8. Find the
numbers.
​

Answer 1

Answer:

The two numbers are:26&14

Step-by-step explanation:

Let the two numbers be x and y.

(x+1)+(y+1)=42

x+y=40

Also,

(x−1)−(y−1)=12

x−y=12

Solving these equations,

2x=52

x=26

Hence, the two numbers are 26 and 14.

Answer 2

The numbers are 23 and 15

Step-by-step explanation:

Given:

the sum of successors of numbers= 40

difference of their predecessor= 8

To find:

The numbers

Solution:

Let's take the numbers a & b

Now, according to given

The sum of the successors of the two numbers is 40

(a+1)+ (b+1)=40

∴ a+b = 38..................(i)

The difference between their predecessors is 8

(a-1) - (b-1)= 8

∴ a-b = 8.......................(ii)

Adding the equations (i) & (ii)

a+b = 38

a-b =   8

2a   = 46

∴ 2a = 46

∴ a = 46/2

∴ a = 23

Putting the value of a in equation (i)

a+b = 38

23+b = 38

b= 38-23

∴ b = 15

Thus, the two numbers are 23 & 15

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