Question

Find two consecutive even numbers such that four-fifths of the greater may exceed three-fourths of the smaller number by 23/10.​

Answer 1

Given :-

▪ Two consecutive even numbers such that four-fifths of the greater number may exceed three-fourths of the smaller number by 23/10.

To Find :-

▪ The numbers.

Solution :-

Even numbers are the numbers which leave remainder zero when divided by 2.

Let the first number be 2x then the next even number will be 2x + 2. That way both numbers are even and hence are divisible by 2.

According to the question:

⇒ 4(Greater number) / 5 = 23/10 + 3(smaller number)/4

⇒ 4(2x + 2) / 5 = 23/10 + 3(2x) / 4

⇒ (8x + 8) / 5 = (23 × 2 + 6x × 5)/20

⇒ (8x + 8) / 5 = (46 + 30x) / 20

⇒ 20(8x + 8) = 5(46 + 30x)

⇒ 160x + 160 = 230 + 150x

⇒ 160x - 150x = 230 - 160

⇒ 10x = 70

⇒ x = 7

So, We assumed the numbers to be 2x and 2x + 2

So,

• 2x = 2×7 = 14
• 2x + 2 = 2×7 + 2 = 16

Hence, The two even numbers are 14 and 16.

Answer 2

Answer:

14,16 vvdiccgjgggjffgjdhorrihifih