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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
Related Questions
Q:

On a school’s Annual day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?

A) 14 B) 16
C) 12 D) 15
 
Answer & Explanation Answer: D) 15

Explanation:

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

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10 9561
Q:

A batsman scores 26 runs and increases his average from 14 to 15. Find the runs to be made if he wants top increasing the average to 19 in the same match ?

A) 74 B) 79
C) 72 D) 60
 
Answer & Explanation Answer: A) 74

Explanation:

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

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9 7876
Q:

There are 44 students in a hostel, due to the administration,  15 new students has joined. The expense of the mess increase by Rs. 33 per day. While the average expenditure per head diminished by Rs. 3, what was the original expenditure of the mess ?

A) Rs. 404 B) Rs. 514
C) Rs. 340 D) Rs. 616
 
Answer & Explanation Answer: D) Rs. 616

Explanation:

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

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11 8200
Q:

Sripad has scored average of 65 marks in three objects. In no subjects has he secured less than 58 marks. He has secured more marks in Maths than other two subjects. What could be his maximum score in Maths ?

A) 79 B) 77
C) 76 D) 73
 
Answer & Explanation Answer: A) 79

Explanation:

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

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5 8102
Q:

Three maths classes: A, B and C take an algebra test. The average score of class A is 83. The average score of class B is 76. The average score of class C is 85. The average score of class A and B is 79 and average score of class B and C is 81. What is the average score of classes A, B, C ?

A) 81 B) 78
C) 80.5 D) 81.5
 
Answer & Explanation Answer: D) 81.5

Explanation:

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5

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12 9006
Q:

Without any stoppage, a person travels a certain distance at an average speed of 42 km/h, and with stoppages he covers the same distance at an average speed of 28 km/h. How many minutes per hour does he stop?

A) 15 minutes B) 20 minutes
C) 18 minutes D) 22 minutes
 
Answer & Explanation Answer: B) 20 minutes

Explanation:

Let the total distance to be covered is 84 kms.
Time taken to cover the distance without stoppage = 84/42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 84/28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.

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27 23244
Q:

The average monthly salary of the employees of a firm is Rs.60 out of the employees 12 officers get an average of Rs.400 per month and the rest an average of Rs. 56 per month , Find the number of emp

Answer

Actual average salary of 12 officers = 400


\inline \fn_jvn \therefore Total salary of 12 officers = 400 x 12 = 4800


Average of these 12  officers as a whole =60


Total salary of 12 officers as a whole = 60 x 12 = 720


The total salary of rest  = 4800 - 720 = 4080


Difference in average of the rest of officers  = 60 - 56 = 4


Rest of the officers = \inline \fn_jvn \frac{4080}{4}=1020


Total number of employees = 1020 + 12 = 1032

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51 10869
Q:

The mean temperature of Monday to Wednesday was 37C and of Tuesday to Thursday was 34C . If the temperature on Thursday was (4/5) th that of Monday, the temperature on Thursday was?

Answer

\inline \fn_jvn M+T+W =(37\times 3)^{0}=111^{0}C   -----------(1)


\inline \fn_jvn T+W+Th=(34\times 3)^{0}=102^{0}C


\inline \fn_jvn \Rightarrow T+W+\frac{4}{5}M = 102^{0}C                --------------(2)


(1) - (2) gives


\inline \fn_jvn \frac{1}{5} th of temperature on Monday = 


\inline \fn_jvn \Rightarrow Temperature on Monday = \inline \fn_jvn 45^{0}C


\inline \fn_jvn \therefore Temperature on Thursday = \inline \fn_jvn \frac{4}{5}\times 45^{0}=36^{0}

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