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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.
Which number should be subtracted from 12,16 and 21 so that resultant numbers are in continued proportion?
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Suggest Corrections
1
Let the number x be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.∴ (12 − x), (16 − x), (21 − x) are in continued proportion.⇒ `(12 - x)/( 16 - x) = ( 16 - x)/( 21 - x )`⇒ `( 16 - x)^2 = ( 12 -x ) xx ( 21 - x)`⇒ `256 + x^2 - 32x = 252 - 12x - 21x + x^2`⇒ `256 - 32x = 252 - 33x`⇒ `33x - 32x = 252 - 256`⇒ `x =- 4`
Thus, the number −4 must be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion.