find two concecutive numbers such that one-forth of the smaller number is three more than one-fifth of the big

Question

find two concecutive numbers such that one-forth of the smaller number is three more than one-fifth of the bigger number

Answer 1

the numbers are 64 and 65.........
let numbers be x and x+1 as they are consecutive.

one-forth of the smaller number=x/4

one-fifth of the bigger number=(x+1)/5

now, x/4-(x+1)/5=3
=>5x-4x-4=60
=>x-4=60
=>x=64
therfore x+1=65

the numbers are 64 and 65

Answer 2

Consecutive numbers are construct algebraically easily because consecutive numbers are just 4 numbers in numerical order that differ by 1 (i.e. 3,4,5,6 and 100, 101, 102, 103). Since we don't know what the consecutive numbers are, then

Let x = the first number. That would mean
x + 1 = the second number

Since it is given "one-fourth of the smaller number ... " this means
(1/4)(x)

" ... is three more than 1/5 of the big" means
(1/5)(x + 1) + 3

All you have to do now is equate them, because they use the word "is".

(1/4)(x) = (1/5)(x + 1) + 3

Your first step would be to eliminate all fractions, by multiplying both sides by 20. This will leave us with

5x = 4(x + 1) + 60

Then, solve as normal. Expand, bring all x terms to one side and everything else on the other, isolate the x ...

5x = 4x + 4 + 60
x = 64

So we've just solved for our smaller number. The question is asking to find both numbers, but we know the next number in the sequence is 65.

So our two numbers are 64 and 65.

Answer 3

64, 65

Here's how:

x = smaller number
y = larger number

Eq 1:
x + 1 = y

Eq 2:
x / 4 = 3 + y / 5

multiple Eq 2 by 20 remove fractions

5x = 60 + 4y

replace y with x + 1

5x = 60 + 4 (x + 1)
5x = 60 + 4x + 4
5x = 64 + 4x

x = 64

therefore y = 65

Check:

Eq 1:
64 + 1 = 65

Eq 2:
64 / 4 = 3 + 65 / 5
16 = 16

Answer 4

the numbers are 64 and 65.

consider x
consecutive will be x+1
given x/4 = 3+((x+1)/5)
= (15+x+1)/5 = (16+x)/5
==> 5x = 64+4x
==> x = 64


hence the consecutive numbers are 64 and 65.

Answer 5

let x, x+1 be the numbers

given

x/4 = (x + 1) / 5 + 3

x/4 = ( x + 1 + 15 ) / 5

5x = 4x + 4 + 60

x = 64

Numbers 64, 65

Answer 6

1/4(n+1) = 3 + 1/5(n + 2)
1/4n + 1/4 = 3 + 1/5n + 2/5
lcd(4, 5) = 20
5n + 5 = 60 + 4n + 8
5n - 4n = 60 + 8 -5
n = 63
n+ 1 = 63 + 1 = 64
n + 2 = 65
Answer: The consecutive numbers are 64 e 65.
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Answer 7

Let x and x+1 are the two consecutive numbers.

x/4 = 3 + (x+1)/5

Multiply by 20,

5x = 60 + 4x + 4

x = 64
x+1 = 65

Answer 8

1/4 n = 1/5(n+1) + 3
5n = 4n + 4 + 60
n = 64
n+1 = 65
1/4(64) = 16 = 1/5(65) + 3 = 13 + 3

Answer 9

x/4 = (x+1)/5 + 3
Multiply both sides by 20 to get rid of fractions
5x = 4x + 4 + 60
5x -4x = 4x + 64 - 4x
x = 64, and x+1 = 65

Answer 10

(1/4)x = (1/5)(x + 1) + 3
multiply everything by 20
5x = 4(x + 1) + 60
5x = 4x + 4 + 60
5x = 4x + 64
x = 64

ANS : 64 and 65