ONe half of a certain integer plus one fifth of the next consecutive integer equals 48. Find the two integers.

Question

Please tell me how you found ur answer... THANKS

Answer 1

Let the first integer be x. Then the next integer will be x+1
Given 1/2 x + 1/5 (x+1) = 48
simplifying we get x/2 + x/5 + 1/5 = 48
Let us bring it to a common denominator which we get by multiplying 2 and 5 i.e. 10
or that is 5x + 2x + 2 = 480
Taking the x to left hand side and numbers to right hand side we get
7x = 480-2 = 478
Hence x = 478/7 = 68.3 roundin we get 68

Let us verify the answer
half of 68 plus one fifth of 68+1 i.e. 69 comes to 34+ 14 (rounded) gives 48 so the answer is verified as correct.

Answer: The two integers are 68 and 69

Answer 2

If x is an integer, then x + 1 is the next consecutive integer.
1/2 (x) + 1/5(x + 1) = 48

To get rid of the fractions, multiply through by 10:
5x + 2(x + 1) = 480
7x + 2 = 480
7x = 478. This doesn't have an integer solution. x = 68 2/7

Looking at some numbers that are close to this and also have the property that x is even (divides by 2) and x + 1 ends in 0 or 5 (divides by 5), we get a guess of 64.

Using 64, half of it is 32, and 1/5 of 65 is 13. 32 and 13 add to give 45. Too small.

Try 74, half is 37. 1/5 of 75 is 15. 37 and 15 add to 52. So, I conclude there is no integer solution.

Answer 3

You must have copied the problem wrong. A solvable problem would be 1/2 of a certain integer plus one fifth of the next consecutive integer equals 45.

Let x = the first integer.
Let x+1 = the next consecutive integer.

After that, you set your equation up exactly the way your words explained it:

x/2 + (x+1)/5 = 45

Multiply x/2 times 5/5 and (x+1)/5 times 2/2 to get a common denominator:

5x/10 + (2x + 2)/10 = 45

Multiply both sides by 10

5x + 2x + 2 = 450

7x = 448

x = 64

The consecutive integers are 64 and 65.

(If you use 48 instead of 45, your 'integer' is 68.286, which, obviously, is not an integer.)

Answer 4

Let the first integer = x
Let the next consecutive integer = x + 1
(1/2)x + (1/5)(x + 1) = 48
1/2x + 1/5x + 1/5 = 48
5/10x + 2/10x + 1/5 = 48
7/10x = 47 4/5 Multiply both sides by 10
7x = 478
x = 68 2/7

Answer 5

let X equal the first integer, and so X+1 = the second integer.

we know that
1/2*X + 1/5*(X+1) = 48

which is

0.5X + 0.2(X+1) = 48
0.5X + 0.2X + 0.2 = 48
0.7X + 0.2 = 48
0.7X + 0.2 -0.2 = 48 - 0.2
0.7X = 47.8
X = 68 and 2/7
(and the X+1 is 69 and 2/7)
but X and X+1 are not integers!
so i would say that your problem does not have an integer solution.

but if instead of 48, the sum of the fractions equalled 38, then your two integers are 54 and 55.
or if they were supposed to equal 45, then the integers are 64 and 65.
or if they were supposed to equal 486, then the integers are 694 and 695.

Answer 6

Sounds like the question is messed up.

Let x be the integer, and so x+1 is the next consecutive integer. Then we have

1/2*x+1/5(x+1)=48

And solving for x you get x = 478/7, not an integer.

Edit: if you meant 45 instead of 48, then x = 64

Answer 7

Okay.

Let x = the first integer
Let x + 1 = the next integer. Adding one makes them consecutive.

1/2x + 1/5(x + 1) = 48
.5x + .2(x + 1) = 48
.5x + .2x + .2 = 48
.7x = 47.8
x = 68.29

Hm... are you sure you have your numbers correct?

Answer 8

.5(x) + .2(x+1)=48
1/2x+(x+1)/5=48
5x+2x+2=480
7x=478
x=68 2/7
But we're looking for integers, so there isn't any answer

Answer 9

me too...

0.5X + 0.2*(X+1) = 48

0.7X = 47.8

X = 478/7 = not an integer

so there is no solution

anything else you want to know ?

Answer 10

do your own homework or you'll never learn