Question

If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a)4
(b)8
(c)6
(d)10

Answer 1

SOLUTION :  

The correct option is (b) : 8

Given : Mode =  Mean + 12

Mode - Mean = 12 ……………(1)

The Empirical relationship between the three measures of Central tendency mean, mode and median is given by :  

MODE = 3 median - 2 mean

Mode - Mean = 3 median - mean  

Mode - mean = 3 median -  mean

12  = 3 median -  mean

Median -  Mean = 12 /3 = 4

Median -  Mean = 4 …………….(2)

We know that, MODE = 3 median - 2 mean

On multiplying by 2 on both sides

2(MODE) = 2[3 median - 2 mean]

2 Mode = 6 median - 4 mean  

Mode - Mean + Mode - median = 6 median - 4 Mean - mean - median  

[On subtracting 1 mean and 1 median from both sides]

(Mode - Mean) + (Mode - median) = 5 median - 5 Mean  

(Mode - Mean) + (Mode - median) = 5 (median - Mean )

12 + (Mode - median) = 5 (4 )

[From eq 1 & 2 ]

12 + (Mode - median) = 20

(Mode - median) = 20 - 12 = 8  

(Mode - median) = 8

Mode = 8  + median

Hence, the mode exceeds the median by  8 .

HOPE THIS ANSWER WILL HELP YOU……

Answer 2

If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a)4
(b)8
(c)6
(d)10

Ans:
Suppose mean be x
So, mode will be x + 12
Then 3 median will be:
= x + 12+ 2x
= 3x + 12
or 3 (x+4)

So, 1 median or median will be:

(x+4)

Here, Mode - Median 
= (x+12) - (x+4)
= x + 12 - x + 4
= 8
∴ The mode exceeds the median by 8.