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Q:

Consider the following frequency distribution:

x     f

8     6

5     4

6     5

10    8

9     9

4     6

7     4

What is the median for the distribution?

A) 6 B) 7
C) 8 D) 9

Answer:   C) 8



Explanation:
Summation of frequencies = 6+4+5+8+9+6+4 = 42
Median = mid value = average of 21st and 22nd value
Arranging data in increasing order we get

x     f

4     6

5     4

6     5

7     4

8     6

9     9

10    8

 
So mid value i.e 21st and 22nd value = 8
Subject: Average
Exam Prep: Bank Exams
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The average age of a husband and his wife was 23 years at the time of their marriage. After  five years they have a one-year old child. The average age of the family now is : 

A) 25 B) 23
C) 19 D) 18
 
Answer & Explanation Answer: C) 19

Explanation:

Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) + 1 = 57 years.

Required average = (57/3) = 19 years.

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Q:

In Arun's opinion, his weight is greater than 65 kg but leas than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all of them are correct in their estimation, what is the average of diferent probable weights of Arun ?

A) 55.5 kg B) 66.5 kg
C) 77.5 kg D) 88.5 kg
 
Answer & Explanation Answer: B) 66.5 kg

Explanation:

Let Arun's weight be X kg.

According to Arun, 65 < X < 72.

According to Arun's brother, 60 < X < 70.

According to Arun's mother, X < 68.

The values satisfying all the above conditions are 66 and 67.

Required average = (66 + 67) / 2 = 66.5 kg

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Q:

The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is :

A) 25 B) 35
C) 45 D) 55
 
Answer & Explanation Answer: B) 35

Explanation:

Excluded number = (27 x 5)  -  (25 x 4) = 135 - 100 = 35.

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31 18906
Q:

The average of 7 consecutive numbers is 20. The largest of these numbers is :

A) 21 B) 22
C) 23 D) 24
 
Answer & Explanation Answer: C) 23

Explanation:

Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,

Then   (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or  7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

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Q:

In an examination, a pupil's average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination ?

A) 8 B) 9
C) 10 D) 11
 
Answer & Explanation Answer: D) 11

Explanation:

Let the number of papers be x. Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11.

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Q:

David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics,Physics, Chemistry and Biology What are his average marks ?

A) 75 B) 68
C) 65 D) 60
 
Answer & Explanation Answer: A) 75

Explanation:

Average =   (76 + 65 + 82 + 67 + 85 )/ 5 = 375/5   =  75.

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