OP Malhotra Class 10 Maths Solutions Chapter 18 Arithmetic Mean Median Mode and Quartiles Exercise 18 (A)

Question 1. Find Q1, Q3 for the following sets data :
(i) 38, 7, 43, 25, 20, 15, 12, 18, 11
(ii) 5, 12, 17, 23, 28, 31, 37, 41, 42, 49, 54, 58, 65, 68, 73, 77
(iii) 22, 21, 12, 15, 17, 18, 18, 20, 19, 1, 6, 25
Answer:
(i) First, arrange the given data in ascending order:
7, 11, 12, 15, 18, 20, 25, 38, 43.
The number of terms, \( n \), is 9, which is an odd number.

Lower Quartile (Q1):
\( \text{Q1} = \text{Size of } \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 9 + 1 }{ 4 }\text{th term} = \frac { 10 }{ 4 }\text{th term} = 2.5\text{th term} \)
The 2.5th term lies between the 2nd term (11) and the 3rd term (12).
\( \text{Q1} = \text{2nd term} + 0.5 \times (\text{3rd term} - \text{2nd term}) \)
\( = 11 + 0.5 \times (12 - 11) \)
\( = 11 + 0.5 \times 1 \)
\( = 11 + 0.5 = 11.5 \)

Upper Quartile (Q3):
\( \text{Q3} = \text{Size of } \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(9+1) }{ 4 }\text{th term} = \frac { 3 \times 10 }{ 4 }\text{th term} = \frac { 30 }{ 4 }\text{th term} = 7.5\text{th term} \)
The 7.5th term lies between the 7th term (25) and the 8th term (38).
\( \text{Q3} = \text{7th term} + 0.5 \times (\text{8th term} - \text{7th term}) \)
\( = 25 + 0.5 \times (38 - 25) \)
\( = 25 + 0.5 \times 13 \)
\( = 25 + 6.5 = 31.5 \)

(ii) First, arrange the given data in ascending order:
5, 12, 17, 23, 28, 31, 37, 41, 42, 49, 54, 58, 65, 68, 73, 77.
The number of terms, \( n \), is 16, which is an even number.

Lower Quartile (Q1):
\( \text{Q1} = \text{Size of } \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 16 + 1 }{ 4 }\text{th term} = \frac { 17 }{ 4 }\text{th term} = 4.25\text{th term} \)
The 4.25th term lies between the 4th term (23) and the 5th term (28).
\( \text{Q1} = \text{4th term} + 0.25 \times (\text{5th term} - \text{4th term}) \)
\( = 23 + 0.25 \times (28 - 23) \)
\( = 23 + 0.25 \times 5 \)
\( = 23 + 1.25 = 24.25 \)

Upper Quartile (Q3):
\( \text{Q3} = \text{Size of } \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(16+1) }{ 4 }\text{th term} = \frac { 3 \times 17 }{ 4 }\text{th term} = \frac { 51 }{ 4 }\text{th term} = 12.75\text{th term} \)
The 12.75th term lies between the 12th term (58) and the 13th term (65).
\( \text{Q3} = \text{12th term} + 0.75 \times (\text{13th term} - \text{12th term}) \)
\( = 58 + 0.75 \times (65 - 58) \)
\( = 58 + 0.75 \times 7 \)
\( = 58 + 5.25 = 63.25 \)

(iii) First, arrange the given data in ascending order:
1, 6, 12, 15, 17, 18, 18, 19, 20, 21, 22, 25.
The number of terms, \( n \), is 12, which is an even number.

Lower Quartile (Q1):
\( \text{Q1} = \text{Size of } \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 12 + 1 }{ 4 }\text{th term} = \frac { 13 }{ 4 }\text{th term} = 3.25\text{th term} \)
The 3.25th term lies between the 3rd term (12) and the 4th term (15).
\( \text{Q1} = \text{3rd term} + 0.25 \times (\text{4th term} - \text{3rd term}) \)
\( = 12 + 0.25 \times (15 - 12) \)
\( = 12 + 0.25 \times 3 \)
\( = 12 + 0.75 = 12.75 \)

Upper Quartile (Q3):
\( \text{Q3} = \text{Size of } \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(12+1) }{ 4 }\text{th term} = \frac { 3 \times 13 }{ 4 }\text{th term} = \frac { 39 }{ 4 }\text{th term} = 9.75\text{th term} \)
The 9.75th term lies between the 9th term (20) and the 10th term (21).
\( \text{Q3} = \text{9th term} + 0.75 \times (\text{10th term} - \text{9th term}) \)
\( = 20 + 0.75 \times (21 - 20) \)
\( = 20 + 0.75 \times 1 \)
\( = 20 + 0.75 = 20.75 \)
In simple words: To find quartiles, first put all the numbers in order from smallest to largest. Then, use formulas to find where Q1 (the first quarter mark) and Q3 (the third quarter mark) fall in the list. If the position is a decimal, you need to calculate a value between the two terms it falls between.

🎯 Exam Tip: Always arrange the data in ascending order before calculating quartiles. For fractional positions, use interpolation: \( \text{term_before} + \text{decimal_part} \times (\text{term_after} - \text{term_before}) \).

 

Question 2. In a class of ten students, the marks obtained by each student are shown against their Roll No's given below:

Calculate the lower and the upper quartiles.
Answer: First, arrange the marks obtained by the students in ascending order:
2, 4, 8, 10, 12, 15, 20, 25, 30, 40.
The number of terms, \( n \), is 10.

Lower Quartile (Q1):
\( \text{Position of Q1} = \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 10 + 1 }{ 4 }\text{th term} = \frac { 11 }{ 4 }\text{th term} = 2.75\text{th term} \)
The 2.75th term lies between the 2nd term (4) and the 3rd term (8).
\( \text{Q1} = \text{2nd term} + 0.75 \times (\text{3rd term} - \text{2nd term}) \)
\( = 4 + 0.75 \times (8 - 4) \)
\( = 4 + 0.75 \times 4 \)
\( = 4 + 3 = 7 \text{ marks} \)

Upper Quartile (Q3):
\( \text{Position of Q3} = \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(10+1) }{ 4 }\text{th term} = \frac { 3 \times 11 }{ 4 }\text{th term} = \frac { 33 }{ 4 }\text{th term} = 8.25\text{th term} \)
The 8.25th term lies between the 8th term (25) and the 9th term (30).
\( \text{Q3} = \text{8th term} + 0.25 \times (\text{9th term} - \text{8th term}) \)
\( = 25 + 0.25 \times (30 - 25) \)
\( = 25 + 0.25 \times 5 \)
\( = 25 + 1.25 = 26.25 \text{ marks} \)
In simple words: First, list all the marks from smallest to largest. Then, use formulas to find the position for the lower quartile (Q1) and the upper quartile (Q3). If the position is a decimal, you calculate the value between the two closest marks.

🎯 Exam Tip: Always sort raw data before finding quartiles. Remember that quartiles split data into four equal parts, helping understand the spread of values.

 

Question 3. From the data given below calculate the value of 3rd quartile.

Answer: To find the 3rd quartile (Q3), we first need to calculate the cumulative frequency (c.f.).
(See table above for c.f. column)

The total number of students, \( n \), is 76.

Third Quartile (Q3):
\( \text{Position of Q3} = \frac { 3n }{ 4 }\text{th term} \)
\( = \frac { 3 \times 76 }{ 4 }\text{th term} = \frac { 228 }{ 4 }\text{th term} = 57\text{th term} \)
Now, we look for the 57th term in the cumulative frequency (c.f.) column.
The c.f. value of 57 falls within the range where the cumulative frequency changes from 56 to 65.
The mark corresponding to a c.f. of 65 is 17.
Therefore, the 57th term is 17.
\( \text{Q3} = 17 \text{ marks} \)
In simple words: For data shown in groups, add up the frequencies to find the cumulative frequency. Then, use the formula for Q3 to find its position. Look at the cumulative frequency column to find which 'marks obtained' value matches this position.

🎯 Exam Tip: For discrete frequency distributions, calculate the cumulative frequency (c.f.) first. The quartile value is the data point corresponding to the c.f. that is just greater than or equal to the quartile position.

 

Question 4. Find both quartiles for the following distribution.
(i)
Answer: To find the quartiles, we first create a cumulative frequency (c.f.) table for the given data.

The total number of students, \( n \), is 100.

Lower Quartile (Q1):
\( \text{Position of Q1} = \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 100 + 1 }{ 4 }\text{th term} = \frac { 101 }{ 4 }\text{th term} = 25.25\text{th term} \)
The 25.25th term lies within the c.f. of 40 (since 21 < 25.25 <= 40). The mark corresponding to this c.f. is 60.
\( \text{Q1} = 60 \)

Upper Quartile (Q3):
\( \text{Position of Q3} = \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(100+1) }{ 4 }\text{th term} = \frac { 3 \times 101 }{ 4 }\text{th term} = \frac { 303 }{ 4 }\text{th term} = 75.75\text{th term} \)
The 75.75th term lies within the c.f. of 86 (since 60 < 75.75 <= 86). The mark corresponding to this c.f. is 80.
\( \text{Q3} = 80 \)

(ii)

The total frequency, \( n \), is 30.

Lower Quartile (Q1):
\( \text{Position of Q1} = \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 30 + 1 }{ 4 }\text{th term} = \frac { 31 }{ 4 }\text{th term} = 7.75\text{th term} \)
The 7.75th term lies within the c.f. of 15 (since 7 < 7.75 <= 15). The size corresponding to this c.f. is 6.
\( \text{Q1} = 6 \)

Upper Quartile (Q3):
\( \text{Position of Q3} = \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(30+1) }{ 4 }\text{th term} = \frac { 3 \times 31 }{ 4 }\text{th term} = \frac { 93 }{ 4 }\text{th term} = 23.25\text{th term} \)
The 23.25th term lies within the c.f. of 24 (since 15 < 23.25 <= 24). The size corresponding to this c.f. is 7.
\( \text{Q3} = 7 \)
In simple words: For distributions with frequencies, first create a running total called cumulative frequency. Then use the formulas with \( n+1 \) to find the quartile positions. Match these positions to the data values in the cumulative frequency table.

🎯 Exam Tip: Always make sure your cumulative frequency column is correctly calculated. This is a common point of error that can lead to incorrect quartile values.

 

Question 5. Find the median size of the shoes from figures given below :

Also calculate the Quartiles.
Answer: We need to calculate the cumulative frequency (c.f.) first, which is shown in the table above.
The total frequency, \( n \), is 457.

Median:
\( \text{Position of Median} = \frac { n + 1 }{ 2 }\text{th term} \)
\( = \frac { 457 + 1 }{ 2 }\text{th term} = \frac { 458 }{ 2 }\text{th term} = 229\text{th term} \)
The 229th term lies within the c.f. of 294 (since 212 < 229 <= 294). The shoe size corresponding to this c.f. is 8.5.
\( \text{Median} = 8.5 \)

Lower Quartile (Q1):
\( \text{Position of Q1} = \frac { n + 1 }{ 4 }\text{th term} \)
\( = \frac { 457 + 1 }{ 4 }\text{th term} = \frac { 458 }{ 4 }\text{th term} = 114.5\text{th term} \)
The 114.5th term lies within the c.f. of 117 (since 57 < 114.5 <= 117). The shoe size corresponding to this c.f. is 7.5.
\( \text{Q1} = 7.5 \)

Upper Quartile (Q3):
\( \text{Position of Q3} = \frac { 3(n+1) }{ 4 }\text{th term} \)
\( = \frac { 3(457+1) }{ 4 }\text{th term} = \frac { 3 \times 458 }{ 4 }\text{th term} = \frac { 1374 }{ 4 }\text{th term} = 343.5\text{th term} \)
The 343.5th term lies within the c.f. of 413 (since 309 < 343.5 <= 413). The shoe size corresponding to this c.f. is 9.5.
\( \text{Q3} = 9.5 \)
In simple words: To find the median and quartiles from grouped data, first add up frequencies to get cumulative frequencies. Then, use formulas to find the position of the median, Q1, and Q3. Finally, find the data value that matches each position in the cumulative frequency table.

🎯 Exam Tip: When dealing with discrete series (data with frequencies), always construct a cumulative frequency column. The median and quartile values are found by locating their position in this cumulative frequency column and then identifying the corresponding data value.