Advertisements
Advertisements
प्रश्न
6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.
Advertisements
उत्तर
6 is the mean proportion between two numbers x and y,
`i.e 6 = sqrt(xy)`
So, 36 = xy ...(1)
It is given that 48 is the third proportional to x and y
So, y2 = 48x … (2)
From (1) and (2), we get
`y^2 = 48 (36/y) => y^3 = 1728`
Hence, y = 12
`:. x = 36/y = 36/12 = 3`
Thus, the required numbers are 3 and 12.
APPEARS IN
संबंधित प्रश्न
Using properties of proportion, solve for x. Given that x is positive:
`(2x + sqrt(4x^2 -1))/(2x - sqrt(4x^2 - 1)) = 4`
If a/b = c/d prove that each of the given ratio is equal to `sqrt((3a^2 - 10c^2)/(3b^2 - 10d^2))`
Find the value of the unknown in the following proportion :
5 : 12 :: 15 : x
If ( a+c) : b = 5 : 1 and (bc + cd) : bd = 5 : 1, then prove that a : b = c : d
Find the smallest number that must be subtracted from each of the numbers 20, 29, 84 and 129 so that they are in proportion.
If a : b : : c : d, then prove that
2a+7b : 2a-7b = 2c+7d : 2c-7d
If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.
If 24 workers can build a wall in 15 days, how many days will 8 workers take to build a similar wall?
If a, b, c and d are in proportion, prove that: (ma + nb) : b = (mc + nd) : d
7 Persons is to 49 Persons as 11 kg is to 88 kg
