What number must be added to each of the numbers 10/22 11 and 24 so that they are in proportion?

Given:

The numbers 104, 34, 110 and 36

Concept used:

If a, b, c and d are four numbers and they are in continued proportion

Then a ∶ b ∶∶ c ∶ d 

⇒ ad = cb 

⇒ a/b = c/d

Calculation

Let the number x added

Now, according to question,

(104 + x) ∶ (34 + x) ∶∶ (110 + x) ∶ (36 + x)

⇒ (104 + x) × (36 + x) = (110 + x) × (34 + x) 

⇒ 3744 + 36x + 104x + x2 = 3740 + 34x + 110x + x2

⇒ 3744 + 140x = 3740 + 144x

⇒ 4x = 4

Then x = 1

∴ The required number is 1.

Alternate Method

Calculations:

Let the number x added

Now, according to question,

(104 + x) ∶ (34 + x) ∶∶ (110 + x) ∶ (36 + x)

⇒ (104 + x)/(34 + x) = (110 + x)/(36 + x)

Let’s check options one by one,

For 1 option,

After adding 9 to each number, numbers will be 113, 43, 119 and 45

For them to be in continued proportion this ratio must be equal

⇒ 113/43 ≠ 119/45

Hence option 1 is not correct

For 2 option,

After adding 3 to each number, numbers will be 107, 37, 113 and 39

For them to be in continued proportion this ratio must be equal

⇒ 107/37 ≠ 113/39

Hence option 2 is not correct

For 3 option,

After adding 1 to each number, numbers will be 105, 35, 111 and 37

For them to be in continued proportion this ratio must be equal

⇒ 105/35 = 111/37 = 3

Hence option 3 is correct

For 4 option,

After adding 4 to each number, numbers will be 108, 38, 114 and 40

For them to be in continued proportion this ratio must be equal

⇒ 108/38 ≠ 114/40

Hence option 4 is not correct

∴ The correct answer is option 3.


Define x:

Let x be the number that we need to add to each number

Solve x:

Sequence = 10, 18, 22, 38

(term 1 + x) /(term 2 + x) = (term 3 + x) / term 4 + x)

(10 + x) / (18 + x) = (22 + x) / (38 + x)

(10 + x)(38 + x) = (18 + x) (22 + x)

380 + 10x + 38x + x² = 396 + 18x + 22x + x²

380 + 48x + x² = 396 + 40x + x²

48x - 40x = 396 - 380

8x = 16

x = 2

Answer: 2 must be added the numbers

Toplist

Latest post

TAGs