Advertisements
Advertisements
प्रश्न
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Advertisements
उत्तर
Let the two numbers are a and b.
∵ 28 is the mean proportional
∵ a : 28 : : 28 : b
∴ ab = (28)2 = 784
⇒ a = `(784)/b` ...(i)
∵ 224 is the third proportional
∴ a : b : : b : 224
⇒ b2 = 224a ...(ii)
Substituting the value of a in (ii)
b2 = `24 xx (784)/b`
⇒ b3 = 224 x 784
⇒ b2 = 175616 = (56)3
∴ b = 56
Now substituting the value of b in (i)
a = `(784)/(56)` = 14
Hence numbers are 14, 56.
APPEARS IN
संबंधित प्रश्न
Using the properties of proportion, solve for x, given `(x^4 + 1)/(2x^2) = 17/8`.
Find the mean proportion of the following :
0.09 and 0.25
If y is the mean proportional between x and z, show that :
xyz (x+y+z)3 =(xy+yz+xz)3
If ( a+c) : b = 5 : 1 and (bc + cd) : bd = 5 : 1, then prove that a : b = c : d
Find the third proportional to Rs. 3, Rs. 12
Determine if the following numbers are in proportion:
4, 6, 8, 12
Determine if the following numbers are in proportion:
150, 200, 250, 300
If `x/a = y/b = z/c`, prove that `x^3/a^2 + y^3/b^2 + z^3/c^2 = (x+ y+ z)^3/(a + b+ c)^2`
A gets double of what B gets and B gets double of what C gets. Find A : B and B : C and verify whether the result is in proportion or not
Unequal masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down.
Unequal masses will balance when the following proportion is true:
`("mass"1)/("length"2) = ("mass"2)/("length"1)`
Two children can be balanced on a seesaw when
`("mass"1)/("length"2) = ("mass"2)/("length"1)`. The child on the left and child on the right are balanced. What is the mass of the child on the right?
