CRE 4 - Means

2-digit numbers which are divisible by 7 are: 14, 21, 28. …, 98.

We know the Arithmetic mean of an A.P. is same as the average of first and last term.

⇒ AM = (14 + 98)/2 = 56.

Hence, 56.

The given terms are in geometric progression.

We know the geometric mean of a G.P. is same as the geometric mean of first and last term.

∴ Geometric mean = √(4×1024) = 2 × 32 = 64.

Hence, 64.

Geometric mean = (40 × 54 × 100)^1/3

= (2³ × 5 × 3³ × 2 × 2² × 5²)^1/3

= (2⁶ × 3³ × 5³)^1/3

= 2² × 3 × 5 = 60.

Hence, option (b).

We know for any two positive numbers 

When numbers are equal: AM = GM

When numbers are distinct: AM > GM.

Hence, option (c).

Here we need to insert three arithmetic means between 3 and 23.

Let the AM’s be a, b and c

∴ 3, a, b, c and 23 are in AP.

Let the common different of this AP be ‘d’.

⇒ 23 = 3 + (5 - 1) × d

⇒ d = 5.

First AM = 3 + d = 8
Second AM = 8 + d = 13
Third AM = 13 + d = 18

⇒ AMs are 8, 13, 18.

∴ Sum of the three AMs = 8 + 13 + 18 = 39.

Hence, option (d).

Here we need to insert three geometric means between 2 and 162.

Let the geometric means be a, b and c.

∴ 2, a, b, c and 162 are in G.P.

Let r be the common ratio of this G.P.

⇒ 162 = 2 × r⁴

⇒ r⁴ = 81

⇒ r = 3.

∴ G₃ = 2 × r³ = 2 × 27 = 54.

Hence, option (c).

Since the numbers are in the ratio of 9 : 1, let the numbers be 9x and x.

Now, 

A.M. = (a + b)/2 = 5x and 

G.M. = √ab = 3x

Given, A.M. = G.M. + 6 

⇒ 5x = 3x + 6

⇒ x = 3

∴ The numbers are 27 and 3.

Hence, option (c).

We know for two positive numbers: AM ≥ GM ≥ HM.

∴ AM = 15 (highest of the three numbers)

GM = 12

HM = 144/15 (lowest of the three numbers)

Hence, option (c).

We know for two positive numbers, GM² = AM × HM.

∴ 18² = 27 × HM

⇒ HM = 324/27 = 12

Hence, option (c).

Let the two no’s be a & b.

We know, G.M. = √ab and H.M. = 2ab/(a+b)

∴ $\frac{{\displaystyle \frac{2ab}{a+b}}}{\sqrt{ab}}=\frac{12}{13}$

⇒ $\frac{2ab}{\sqrt{ab}}=\frac{12}{13}(a+b)$

⇒ 13√ab = 6a + 6b

Dividing the whole equation by √ab, we get

⇒ 13 = $6\sqrt{\frac{a}{b}}+6\sqrt{\frac{b}{a}}$

Substituting $\sqrt{\frac{a}{b}}$ = x

⇒ 13 = 6x + 6/x

⇒ 6x² – 13x + 6 = 0

⇒ (3x - 2)(2x - 3) = 0

⇒ x = 2/3 or 3/2

∴ a/b = x² = 4/9 or 9/4.

Hence, option (a).