Numerical Ability Test - 16

Given:

The given data = 13, 15, 11, 19, 15, 21, 12, 13

Formula:

Range of the data = Upper limit – lower limit

Calculation:

Upper limit = Largest number = 21

Lower limit = Smallest number = 11

Given: 

4, 5, 6, 7, 8, 12, 9, 12, 5, 12

Concept used:

The mode is the most frequently occurring value on the list.

Total = Mean (Average) × number of entities

Calculation:

Given series is

4, 5, 6, 7, 8, 12, 9, 12, 5, 12

Mean of the series = 80 ÷ 10 = 8

Mode of the series = 12

Difference in the mode and the mean of the observations = 12 - 8 = 4

∴ 4 is the difference in the mode and the mean of the observations.

Given

Mean = 9

11, (k-2), 7, (k-1), 11, 16, 12, 15, (k-1), 13

Formula used

Mean = Sum of all observation/Number of observation

Calculation

9 = [11+ (k - 2) + 7+ (k - 1) + 11 + 16 + 12 + 15 + (k - 1) + 13]/10

⇒ 90 = (85 + 3k - 4)

⇒ 90 + 4 = 85 + 3k

⇒ 94 - 85 = 3k

⇒ 9 = 3k

⇒ k = 3

∴ The value of k is 3.

Given:

The arithmetic mean of first n natural number = 100

Formula used:

Arithmetic mean of first n natural number = (n +1)/2

Calculation:

According to the question

(n + 1)/2 = 100

⇒ n + 1 = 200

⇒ n = 199

∴ n is 199

Given:

The following are the weights (in kg) of 25 students:
58, 55, 53, 50, 53, 51, 52, 54, 53, 52, 54, 53, 58, 53, 59, 55, 53, 52, 51, 54, 53, 59, 55, 53, 52

Concept:

Range = It is the difference between the highest and lowest values in a given data. 

Calculation:

Highest weight = 59 kg 

Lowest weight = 50 kg 

Range = 59 - 50 = 9 

∴ The required result is 9.

Given

Mean - Median = 3

mean = 11

Concept used

Mean - Mode = 3(Mean - Median)

Calculation

Mean - Mode = 3(Mean - Median)

⇒ 11 - Mode = 3 × 3

Mode = 11 - 9

Mode = 2

Given:

5.02, 5.18, 5.12, 5.007 and 5.018

Concept used:

1. If the total number of observations (n) is an odd number,

Median = (n + 1)/2th term

If the total number of observations (n) is an even number,

Median = {(n/2)th term + [(n/2) + 1]th term}/2

2. Mean = Sum of all the observations / Number of observations

Calculations:

Arranging the given data in ascending order,

5.007, 5.018, 5.02, 5.12, 5.18

Number of observations = 5

Median = (5 + 1)/2^th term

⇒ Median = 3rd term

⇒ Median = 5.02

Mean = (5.007 + 5.018 + 5.02 + 5.12 + 5.18)/5

⇒ 25.345/5

⇒ 5.069

Sum of mean and median of the numbers = 5.02 + 5.069

⇒ 10.089

∴ The sum of mean and median of the given numbers is 10.089

Given:

Mean of the observation = 7

Formula used:

Mean = Sum of all the terms/Total number of terms

Concept used:

If the total number of observations given is odd, then the formula to calculate the median is: 

Median = {\(\rm \frac{n+1}{2}\)}th term

If the total number of observations is even, then the median formula is:

Median  = [\(\rm \frac{n}{2}\)th term + (\(\rm \frac{n}{2}\)+ 1)th term]/2

Where n is the number of observations.

Calculation:

Mean = (6 + 8 + 5 + 7 + x + 4)/6

⇒ 7 × 6 = 30 + x

⇒ 42 - 30 = x

⇒ x = 12

Arranging the observation in ascending order

4, 5, 6, 7, 8, 12

Total number of term n = 6

Median  = [\(\rm \frac{n}{2}\)th term + (\(\rm \frac{n}{2}\)+ 1)th term]/2

⇒ (6 + 7)/2 = 6.5

∴ The median of these observation is 6.5

Calculation:

First 6 multiples of 5 are 5, 10, 15, 20, 25, 30

⇒ Middle value = 15 and 20

Median = Sum of middle value/2

⇒ Median  = (15 + 20)/2 

⇒ Median = 17.5

∴ The median of first 6 multiples of 5 is 17.5

The correct option is 2 i.e. 17.5

Given:

37, 31, 42, 43, 46, 25, 39, 43, 32

Formula used:

When the data set is an odd number

⇒ (n + 1)/2 th term

n is the number of data set

Calculation:

37, 31, 42, 43, 46, 25, 39, 43, 32

Data set arrange in the ascending order

⇒ 25, 31, 32, 37, 39, 42, 43, 43, 46 

According to the question,

⇒ n = 9

⇒ (n + 1)/2 th term

⇒ (9 + 1)/2 = 5th term

⇒ 5th term = 39

∴ The median is 39.