Question

46. The median, mode and range of a data 3, 3, 4, 4,
13, 13, 18, 18, x, y, z are y, y and 15 respectively. If
z>y> x and x, y, z are prime numbers, then the
value of z is

Answer 1

Solution :-

→ Median of data = y

→ Mode of data = y

→ Range of data = 15

we know that,

→ Range = Highest observation - Least observation

Case 1) :-

→ x = smallest prime number = 2 = Least observation .

then,

→ Highest will be = 2 + 15 = 17 ≠ Not possible since 18 is also given .

Case 2) :-

→ Least observation = 3,

→ Highest observation = 18

So,

→ Range = 18 - 3 = 15 .

So, data in ascending order will be :- 3, 3, 4, 4,

13, 13, 18, 18 and x ,y and z are in between .

Now, Mode of data is y which is a prime number . As we can see that, 3 and 13 are two prime numbers which comes 2 times in given data .

Taking y = 3,

• if y = 3, x must be smaller than 3, which is 2 . But from range it is not possible .

Then,

→ y = 13

since z > y ,

→ z = 17

So, data in ascending order will be :- 3, 3, 4, 4,

13, 13, 13, 17, 18, 18 and x is in between .

Also,

→ x < y

→ x < 13

→ x = 2, 3, 5, 7 and 11 .

since 2 and 3 are not possible .

Possible values of x will be 5, 7 and 11 .

Now, when x = 5 :-

→ Data = 3, 3, 4, 4, 5, 13, 13, 13, 17, 18, 18

→ Median = (11 + 1)/2 th observation = 6th observation = 13 = y.

when x = 7 :-

→ Data :- 3, 3, 4, 4, 7, 13, 13, 13, 17, 18, 18

→ Median = (11 + 1)/2 th observation = 6th observation = 13 = y .

when x = 11 :-

→ Data :- 3, 3, 4, 4, 11, 13, 13, 13, 17, 18, 18

→ Median = (11 + 1)/2 th observation = 6th observation = 13 = y .

Hence, value of z is equal to 17 .

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