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BLE Maths Past Questions
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Past Years BLE Questions on Mean
Basic Level Examination (Grade 8) – Mathematics Solved Question Bank
Question 1
Kathmandu 2082
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Here, cricket player Rohit Poudel’s run scores in eight matches against America are given:
Runs: 62, 2, 62, 16, 16, 96, 1, 2
- Write the formula to find the mean of individual series.
- What is the average (mean) run score of Rohit Poudel in the eight matches played against America?
- If he wants to make his average (mean) run score 40 including the ninth match, how many runs does he need to score in the ninth match?
Solution:
(a) Formula:
Mean (\(\bar{x}\)) = \(\frac{\sum x}{N}\)
(b) Given data,
Runs (\(x\)) = 62, 2, 62, 16, 16, 96, 1, 2
\(N\) = 8
We know,
Mean (\(\bar{x}\)) = \(\frac{\sum x}{N}\)
or, \(\bar{x}\) = \(\frac{62 + 2 + 62 + 16 + 16 + 96 + 1 + 2}{8}\)
or, \(\bar{x}\) = \(\frac{257}{8}\) = 32.125
\(\therefore\) Average run score = 32.125 runs
Runs (\(x\)) = 62, 2, 62, 16, 16, 96, 1, 2
\(N\) = 8
We know,
Mean (\(\bar{x}\)) = \(\frac{\sum x}{N}\)
or, \(\bar{x}\) = \(\frac{62 + 2 + 62 + 16 + 16 + 96 + 1 + 2}{8}\)
or, \(\bar{x}\) = \(\frac{257}{8}\) = 32.125
\(\therefore\) Average run score = 32.125 runs
(c) Let runs in 9th match = \(x_9\)
Given, Mean (\(\bar{x}\)) = 40, \(N\) = 9
40 = \(\frac{257 + x_9}{9}\)
or, \(40 \times 9\) = 257 + \(x_9\)
or, 360 = 257 + \(x_9\) ⇒ \(x_9\) = 360 – 257 = 103
\(\therefore\) Required runs in 9th match = 103 runs
Given, Mean (\(\bar{x}\)) = 40, \(N\) = 9
40 = \(\frac{257 + x_9}{9}\)
or, \(40 \times 9\) = 257 + \(x_9\)
or, 360 = 257 + \(x_9\) ⇒ \(x_9\) = 360 – 257 = 103
\(\therefore\) Required runs in 9th match = 103 runs
Question 2
Nepalgunj 2080, 2082
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The heights (in cm) of eight students of grade eight are presented below. Find the mean height of the given data.
Heights (cm): 140, 144, 148, 149, 140, 150, 145, 140
Solution:
Given data,Heights (\(x\)) = 140, 144, 148, 149, 140, 150, 145, 140
\(N\) = 8
Mean (\(\bar{x}\)) = \(\frac{\sum x}{N}\)
or, \(\bar{x}\) = \(\frac{140 + 144 + 148 + 149 + 140 + 150 + 145 + 140}{8}\)
or, \(\bar{x}\) = \(\frac{1156}{8}\) = 144.5
\(\therefore\) Mean height = 144.5 cm
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Question 3
Bhaktapur 2081, Butwal 2080
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In a data set, if \(\sum x = m + 77\), \(\sum f = 10\) and mean \(\bar{x} = 8\), find the value of \(m\).
Solution:
Given,\(\sum x\) = \(m + 77\), \(\sum f\) = 10, Mean (\(\bar{x}\)) = 8
Mean (\(\bar{x}\)) = \(\frac{\sum x}{\sum f}\)
or, 8 = \(\frac{m + 77}{10}\)
or, 80 = \(m + 77\) ⇒ \(m\) = 80 – 77 = 3
\(\therefore m\) = 3
Question 4
Kathmandu 2081, Bhaktapur 2082
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The following are the marks obtained by 7 students of grade eight in the First Terminal Examination in Mathematics:
Marks: 23, 30, 25, 26, 23, 28, 27
- What is the average mark obtained by the seven students in the First Terminal Examination?
- If there were 8 subjects and the overall mean mark of all 8 subjects was 26, what mark was obtained in the 8th subject?
Solution:
(a) Average Mark:
Mean (\(\bar{x}\)) = \(\frac{182}{7}\) = 26\(\therefore\) Average mark = 26
(b) Mark in 8th Subject (\(x\)):
26 = \(\frac{182 + x}{8}\)
or, 208 = 182 + \(x\) ⇒ \(x\) = 208 – 182 = 26
\(\therefore\) Mark in 8th subject = 26
26 = \(\frac{182 + x}{8}\)
or, 208 = 182 + \(x\) ⇒ \(x\) = 208 – 182 = 26
\(\therefore\) Mark in 8th subject = 26
Question 5
Lalitpur 2082, Biratnagar 2081
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The number of students playing different games is presented in the table below. Find the average value of the given data.
| Games | Number of Students |
|---|---|
| Volleyball | 55 |
| Football | 20 |
| Cricket | 60 |
| Chess | 45 |
Solution:
Average = \(\frac{55 + 20 + 60 + 45}{4}\) = \(\frac{180}{4}\) = 45\(\therefore\) Average number of students = 45
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Question 6
Pokhara 2080, 2082
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The monthly expenditure of a family is given in the table below. What is the monthly average expenditure of the family?
| Months | Expenditure (in Rs.) |
|---|---|
| Poush | 30,000 |
| Magh | 32,000 |
| Falgun | 35,000 |
| Chaitra | 28,000 |
Solution:
Average Expenditure = \(\frac{30000 + 32000 + 35000 + 28000}{4}\) = \(\frac{125000}{4}\) = Rs. 31,250\(\therefore\) Average monthly expenditure = Rs. 31,250
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