Advertisements
Advertisements
Question
Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96.
Advertisements
Solution
Let a and b be the two numbers, whose mean proportional is 12.
∴ ab = 122
`=>` ab = 144
`=> b = 144/a` ...(i)
Now, third proportional is 96
∴ a : b :: b : 96
`=>` b2 = 96a
`=> (144/a)^2 = 96a`
`=> (144)^2/a^2 = 96a`
`=> a^3 = (144 xx 144)/96`
`=>` a3 = 216
`=>` a = 6
`b = 144/6 = 24`
Therefore, the numbers are 6 and 24.
RELATED QUESTIONS
If x + 5 is the mean proportional between x + 2 and x + 9; find the value of x.
If `x/a = y/b = z/c` prove that `(2x^3 - 3y^3 + 4z^3)/(2a^3 - 3b^3 + 4c^3) = ((2x - 3y + 4z)/(2a - 3b + 4c))^3`
If a, b, c and dare in continued proportion, then prove that
(a+ d)(b+ c)-(a+ c)(b+ d)= (b-c)2
What quantity must be added to each term of the ratio a + b: a - b to make it equal to (a + b)2 : (a - b)2 ?
If a, b, c, d are in continued proportion, prove that:
`sqrt(ab) - sqrt(bc) + sqrt(cd) = sqrt((a - b + c) (b - c + d)`
Write (T) for true and (F) for false in case of the following:
32 kg : Rs 36 : : 8 kg : Rs 9
40 men can finish a piece of work in 26 days. How many men will be needed to finish it in 16 days?
In covering 111 km, a car consumes 6 L of petrol. How many kilometers will it go to 10 L of petrol?
If `a/c = c/d = c/f` prove that : `(a^2)/(b^2) + (c^2)/(d^2) + (e^2)/(f^2) = "ac"/"bd" + "ce"/"df" + "ae"/"df"`
What is the term "d" called in the expression a : b :: c : d?
