text: Solution:
  
  Now,
  Arranging in ascending order
  42, 52, 62, 72, 82, 90, 100
  Total number (N) = 7
  We know that,
  or, Q3 = 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q3 = 3$$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q3 = 3$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q3 = 3 $$\times$$ 2thitem
  or, Q3 = 6thitem
  $$\therefore$$ The value of Q3is 90.
  
  Solution:
  
  Arrange the values in ascending order of magnitude:
  
  18 20 20 23 23 23 24 27 29
  
  Here, n =7
  
  Lower quartile = value of$$\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thterm
  
  = value of 2nditem
  
  $$\therefore$$ Lower Quartile = 20
  
  Upper Quartile = value of$$\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{24}{4} \end{pmatrix}$$thterm
  
  = value of 6thterm
  
  $$\therefore$$ Upper Quartile = 27
  
  Solution:
  
  Here,
  6, x+5, 12, 14, 17, 20, 21
  Q1= 8
  numbers = 7
  We know that,
  or, Q1 = $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$th item
  or, Q1 = 2th item
  or, Q1 =x + 5
  Now,
  or,Q1= x + 5
  or, 8= x + 5
  or, x = 8 - 5
  $$\therefore$$ x = 5
  
  Solution:
  
  Here,
  numbers = 7
  Q3= 60
  Now,
  or, Q3= 3($$\frac{N + 1}{4}$$)th item
  or, Q3= 3($$\frac{7 + 1}{4}$$)th item
  or, Q3= 3($$\frac{8}{4}$$)th item
  or, Q3= 3 $$\times$$ 2th item
  or, Q3= 6th item
  i.e.Q3= x + 25
  Then,
  or,Q3= x + 25
  or, 60 = x + 25
  or, x + 60 - 25
  $$\therefore$$ x = 35
  
  Solution:
  
  Now,
  Arranging in ascending order
  50, 52, 56, 61, 64, 68, 72
  numbers = 7
  We know that,
  or, Q1 = $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q1 = $$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q1 = $$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q1 = 2thitem
  or,Q1 = 52
  Again,
  or, Q2= $$\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}$$thitem
  or, Q2= $$\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}$$thitem
  or, Q2= $$\begin{pmatrix}\frac{8}{2} \end{pmatrix}$$thitem
  or, Q2= 4thitem
  or, Q2= 61
  Similarly,
  or, Q3= 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q3= 3$$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q3= 3$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q3= 3 $$\times$$2thitem
  or, Q3= 6thitem
  or, Q3 = 68
  
  $$\therefore$$ The value ofQ1, Q2 and Q3is 52, 61 and 68 respectively.
  
  Solution:
  
   Marks No. of students(f) cumulative frequency (c.f.) 10 2 2 20 4 6 30 6 12 40 12 24 50 8 32 60 6 38 70 1 39 N = 39
  
  We have,
  
  Q1= value of $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$th term
  
  = value of $$\begin{pmatrix}\frac{39 + 1}{4} \end{pmatrix}$$th term
  
  = value of 10th term
  
  In c.f. column just greater than 10 is 12 and its coressponding value is 30.
  
  $$\therefore$$ Q1 = 30
  
  Q2 =value of $$\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}$$th term
  
  = value of $$\begin{pmatrix}\frac{39 + 1}{2} \end{pmatrix}$$th term
  
  = value of 20th term
  
  In c.f. column just greater than 20 is 24and its coressponding value is 40.
  
  $$\therefore$$ Q2=40
  
  Q3 =value of $$\begin{pmatrix}\frac{3(n + 1)}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3 * 40}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}$$th term
  
  =value of 30th term
  
  In c.f. column just greater than 30 is 32 and its coressponding value is 50.
  
  $$\therefore$$ Q3= 50
  
  Solution:
  
   Marks Frequency (f) c.f 15 5 5 25 10 15 35 16 31 45 7 38 55 5 43 N = 43
  
  Now,
  or,Q1= $$\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}$$th item
  or,Q1= $$\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}$$th item
  or,Q1= $$\begin{pmatrix}\frac{44}{4} \end{pmatrix}$$th item
  or,Q1= 11th item
  
  In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q1 = 25
  Again,
  or, Q3 = 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$
  or, Q3 = 3$$\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}$$
  or, Q3 = 3$$\begin{pmatrix}\frac{44}{4} \end{pmatrix}$$
  or, Q3 = 3 $$\times$$ 11thitem
  or, Q3 = 33th item
  
  In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.
  So, Q3 is 45.
  
  Solution:
  
   x f c.f 0 -10 6 6 10 - 20 8 14 20 - 30 10 24 30 - 40 7 31 40 - 50 5 36 N = 36
  
  Now,
  or, Q1 = $$\begin{pmatrix}\frac{N}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{36}{4} \end{pmatrix}$$th item
  or, Q1 = 9th item
  In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10 - 20.
  So, Q1 lies in class 10 - 20
  Then,
  l = 10, c.f = 6, f = 8, and i = 10
  
  or, Q1 =L + $$\frac{\frac{N}{4}-C.F}{f}$$ × i
  or,Q1 = 10 + $$\frac{\frac{36}{4}-6}{8}$$ × 10
  or,Q1 = 10 + $$\frac{9 - 6}{8}$$ × 10
  or,Q1 = 10 + $$\frac{3}{8}$$ × 10
  or,Q1 = 10 + $$\frac{3}{4}$$ × 5
  or,Q1 = 10 + 0.75× 5
  or,Q1 = 10 +3.75
  or,Q1 = 13.75
  $$\therefore$$ The value of Q1is 13.75.
  
  Solution:
  
   Class Frequency(f) c.f. 0-5 6 6 5-10 8 14 10-15 10 24 15-20 7 31 20-25 5 36 25-30 4 40 N =40
  
  Now,
  
  Q1 Class = value of ($$\frac{N}{4}$$)th item
  
  = value of ( $$\frac{40}{4}$$)th item
  
  = value of 10th item
  
  In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (5-10).
  
  $$\therefore$$ Q1 lies in the class 5-10
  
  Here,
  
  $$\frac{N}{4}$$ = 10,l= 5, c.f. = 6, f = 8 and i = 5
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{4}− c.f}{f}$$ × i
  
  = 5 + $$\frac{10− 6}{8}$$ ×5
  
  =5 + $$\frac{4}{8}$$ ×5
  
  = 5 + 2.5
  
  = 7.5
  
  Again,
  
  Q3 Class = value of($$\frac{3N}{4}$$)th item
  
  = value of($$\frac{3 * 40}{4}$$)th item
  
  =value of30th item
  
  In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 15-20.
  
  Here,
  
  $$\frac{3N}{4}$$ = 30,l= 5, c.f. = 24, f = 7and i = 5
  
  $$\therefore$$Q3= l + $$\frac{\frac{3N}{4}− c.f}{f}$$ × i
  
  = 5 + $$\frac{30− 24}{7}$$ ×5
  
  = 5 + 4.28
  
  = 9.28
  
  Solution:
  
   Class frequency c.f 5 - 10 4 4 10 - 15 6 10 15 - 20 10 20 20 - 25 15 35 25 - 30 10 45 30 - 35 6 51 35 - 40 5 56 N = 56
  
  Now,
  or, Q3= 3($$\frac{N}{4}$$)th item
  or, Q3= 3($$\frac{56}{4}$$)th item
  or, Q3= 3 $$\times$$ 14th item
  or, Q3= 42th item
  In c.f column, c.f is just greater than 42 is 45 and its corresponding value is 25 - 30.
  Now,
  l = 25, c.f = 35, f = 10, and i = 5
  Then,
  or, Q3= L + $$\frac{\frac{3N}{4}-c.f}{f}$$× i
  or, Q3= 25 + $$\frac{42 - 35}{10}$$× 5
  or, Q3= 25 + $$\frac{7}{2}$$
  or, Q3= 25 +3.5
  or, Q3= 28.5
  $$\therefore$$ The value of Q3 is 28.5.
  
  Solution:
  
   Marks Students (f) c.f. 30-40 5 5 40-50 12 17 50-60 25 42 60-70 15 57 70-80 8 65 80-90 3 68 N =68
  
  Now,
  
  Q1 Class = value of ($$\frac{N}{4}$$)th item
  
  = value of ( $$\frac{68}{4}$$)th item
  
  = value of 17th item
  
  In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (50-60).
  
  $$\therefore$$ Q1 lies in the class 50-60
  
  Here,
  
  $$\frac{N}{4}$$ = 17, l= 50, c.f. = 17, f = 25 and i =10
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{4}− c.f}{f}$$ × i
  
  = 50 + $$\frac{17− 17}{25}$$ ×10
  
  =50 + $$\frac{0}{25}$$ ×10
  
  = 50 +0
  
  = 50
  
  Again,
  
  Q2Class = value of ($$\frac{N}{2}$$)th item
  
  = value of ( $$\frac{68}{2}$$)th item
  
  = value of 34th item
  
  In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (50-60).
  
  $$\therefore$$ Q2lies in the class 50-60
  
  Here,
  
  $$\frac{N}{2}$$ = 34, l = 50, c.f. = 17, f = 25 and i =10
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{2}− c.f}{f}$$ × i
  
  = 50 + $$\frac{34− 17}{25}$$ ×10
  
  =50 + $$\frac{17}{25}$$ ×10
  
  = 50 +6.8
  
  = 56.8
  
  Q3 Class = value of($$\frac{3N}{4}$$)th item
  
  = value of($$\frac{3 * 68}{4}$$)th item
  
  =value of 51th item
  
  In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 60-70.
  
  Here,
  
  $$\frac{3N}{4}$$ = 51, l= 60, c.f. = 42, f = 15 and i = 10
  
  $$\therefore$$Q3= l + $$\frac{\frac{3N}{4}− c.f}{f}$$ × i
  
  = 60 + $$\frac{51− 42}{15}$$ × 10
  
  = 60 + 6
  
  = 66
  
  Solution:
  
   x f c.f 50 8 8 60 16 24 70 12 36 80 4 40 90 3 43 N = 43
  
  Now,
  or, Q1 =($$\frac{N+1}{4}$$)th item
  or, Q1 =($$\frac{43+1}{4}$$)th item
  or, Q1 =($$\frac{44}{4}$$)th item
  or, Q1 =11th item
  In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.
  So,Q1 = 60
  Again,
  or, Q3 =3($$\frac{N+1}{4}$$)th item
  or, Q3 =3($$\frac{43+1}{4}$$)th item
  or, Q3 =3($$\frac{44}{4}$$)th item
  or, Q3 =3 $$\times$$ 11th item
  or, Q3 =33th item
  In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.
  i.e.Q3 = 70
explanation: The provided text extract contains multiple solutions to math problems, specifically dealing with the calculation of quartiles (Q1, Q2, Q3) from given data sets. Each solution includes a step-by-step calculation and a final answer. The text does not contain any explicit questions, but rather presents the solutions directly.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Solution:
  
  Now,
  Arranging in ascending order
  42, 52, 62, 72, 82, 90, 100
  Total number (N) = 7
  We know that,
  or, Q3 = 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q3 = 3$$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q3 = 3$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q3 = 3 $$\times$$ 2thitem
  or, Q3 = 6thitem
  $$\therefore$$ The value of Q3is 90.
  
  Solution:
  
  Arrange the values in ascending order of magnitude:
  
  18 20 20 23 23 23 24 27 29
  
  Here, n =7
  
  Lower quartile = value of$$\begin{pmatrix}\frac{n+1}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{7+1}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thterm
  
  = value of 2nditem
  
  $$\therefore$$ Lower Quartile = 20
  
  Upper Quartile = value of$$\begin{pmatrix}\frac{3(n+1)}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{3(7+1)}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{3 \times 8}{4} \end{pmatrix}$$thterm
  
  = value of$$\begin{pmatrix}\frac{24}{4} \end{pmatrix}$$thterm
  
  = value of 6thterm
  
  $$\therefore$$ Upper Quartile = 27
  
  Solution:
  
  Here,
  6, x+5, 12, 14, 17, 20, 21
  Q1= 8
  numbers = 7
  We know that,
  or, Q1 = $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$th item
  or, Q1 = 2th item
  or, Q1 =x + 5
  Now,
  or,Q1= x + 5
  or, 8= x + 5
  or, x = 8 - 5
  $$\therefore$$ x = 5
  
  Solution:
  
  Here,
  numbers = 7
  Q3= 60
  Now,
  or, Q3= 3($$\frac{N + 1}{4}$$)th item
  or, Q3= 3($$\frac{7 + 1}{4}$$)th item
  or, Q3= 3($$\frac{8}{4}$$)th item
  or, Q3= 3 $$\times$$ 2th item
  or, Q3= 6th item
  i.e.Q3= x + 25
  Then,
  or,Q3= x + 25
  or, 60 = x + 25
  or, x + 60 - 25
  $$\therefore$$ x = 35
  
  Solution:
  
  Now,
  Arranging in ascending order
  50, 52, 56, 61, 64, 68, 72
  numbers = 7
  We know that,
  or, Q1 = $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q1 = $$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q1 = $$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q1 = 2thitem
  or,Q1 = 52
  Again,
  or, Q2= $$\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}$$thitem
  or, Q2= $$\begin{pmatrix}\frac{7 + 1}{2} \end{pmatrix}$$thitem
  or, Q2= $$\begin{pmatrix}\frac{8}{2} \end{pmatrix}$$thitem
  or, Q2= 4thitem
  or, Q2= 61
  Similarly,
  or, Q3= 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$thitem
  or, Q3= 3$$\begin{pmatrix}\frac{7 + 1}{4} \end{pmatrix}$$thitem
  or, Q3= 3$$\begin{pmatrix}\frac{8}{4} \end{pmatrix}$$thitem
  or, Q3= 3 $$\times$$2thitem
  or, Q3= 6thitem
  or, Q3 = 68
  
  $$\therefore$$ The value ofQ1, Q2 and Q3is 52, 61 and 68 respectively.
  
  Solution:
  
   Marks No. of students(f) cumulative frequency (c.f.) 10 2 2 20 4 6 30 6 12 40 12 24 50 8 32 60 6 38 70 1 39 N = 39
  
  We have,
  
  Q1= value of $$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$th term
  
  = value of $$\begin{pmatrix}\frac{39 + 1}{4} \end{pmatrix}$$th term
  
  = value of 10th term
  
  In c.f. column just greater than 10 is 12 and its coressponding value is 30.
  
  $$\therefore$$ Q1 = 30
  
  Q2 =value of $$\begin{pmatrix}\frac{n + 1}{2} \end{pmatrix}$$th term
  
  = value of $$\begin{pmatrix}\frac{39 + 1}{2} \end{pmatrix}$$th term
  
  = value of 20th term
  
  In c.f. column just greater than 20 is 24and its coressponding value is 40.
  
  $$\therefore$$ Q2=40
  
  Q3 =value of $$\begin{pmatrix}\frac{3(n + 1)}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3 * 40}{4} \end{pmatrix}$$th term
  
  =value of $$\begin{pmatrix}\frac{3(39 + 1)}{4} \end{pmatrix}$$th term
  
  =value of 30th term
  
  In c.f. column just greater than 30 is 32 and its coressponding value is 50.
  
  $$\therefore$$ Q3= 50
  
  Solution:
  
   Marks Frequency (f) c.f 15 5 5 25 10 15 35 16 31 45 7 38 55 5 43 N = 43
  
  Now,
  or,Q1= $$\begin{pmatrix}\frac{N + 1}{4} \end{pmatrix}$$th item
  or,Q1= $$\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}$$th item
  or,Q1= $$\begin{pmatrix}\frac{44}{4} \end{pmatrix}$$th item
  or,Q1= 11th item
  
  In c.f column, c.f is just greater than 11 is 15 and its corresponding value is 25. So, Q1 = 25
  Again,
  or, Q3 = 3$$\begin{pmatrix}\frac{n + 1}{4} \end{pmatrix}$$
  or, Q3 = 3$$\begin{pmatrix}\frac{43 + 1}{4} \end{pmatrix}$$
  or, Q3 = 3$$\begin{pmatrix}\frac{44}{4} \end{pmatrix}$$
  or, Q3 = 3 $$\times$$ 11thitem
  or, Q3 = 33th item
  
  In c.f column, c.f is just greater than 33 is 38 and its corresponding value is 45.
  So, Q3 is 45.
  
  Solution:
  
   x f c.f 0 -10 6 6 10 - 20 8 14 20 - 30 10 24 30 - 40 7 31 40 - 50 5 36 N = 36
  
  Now,
  or, Q1 = $$\begin{pmatrix}\frac{N}{4} \end{pmatrix}$$th item
  or, Q1 = $$\begin{pmatrix}\frac{36}{4} \end{pmatrix}$$th item
  or, Q1 = 9th item
  In c.f column, c.f is just greater than 9 is 14 and its corresponding value is 10 - 20.
  So, Q1 lies in class 10 - 20
  Then,
  l = 10, c.f = 6, f = 8, and i = 10
  
  or, Q1 =L + $$\frac{\frac{N}{4}-C.F}{f}$$ × i
  or,Q1 = 10 + $$\frac{\frac{36}{4}-6}{8}$$ × 10
  or,Q1 = 10 + $$\frac{9 - 6}{8}$$ × 10
  or,Q1 = 10 + $$\frac{3}{8}$$ × 10
  or,Q1 = 10 + $$\frac{3}{4}$$ × 5
  or,Q1 = 10 + 0.75× 5
  or,Q1 = 10 +3.75
  or,Q1 = 13.75
  $$\therefore$$ The value of Q1is 13.75.
  
  Solution:
  
   Class Frequency(f) c.f. 0-5 6 6 5-10 8 14 10-15 10 24 15-20 7 31 20-25 5 36 25-30 4 40 N =40
  
  Now,
  
  Q1 Class = value of ($$\frac{N}{4}$$)th item
  
  = value of ( $$\frac{40}{4}$$)th item
  
  = value of 10th item
  
  In c.f. column, c.f. just greater than 10 is 14 and its corresponding class is (5-10).
  
  $$\therefore$$ Q1 lies in the class 5-10
  
  Here,
  
  $$\frac{N}{4}$$ = 10,l= 5, c.f. = 6, f = 8 and i = 5
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{4}− c.f}{f}$$ × i
  
  = 5 + $$\frac{10− 6}{8}$$ ×5
  
  =5 + $$\frac{4}{8}$$ ×5
  
  = 5 + 2.5
  
  = 7.5
  
  Again,
  
  Q3 Class = value of($$\frac{3N}{4}$$)th item
  
  = value of($$\frac{3 * 40}{4}$$)th item
  
  =value of30th item
  
  In c.f. column, c.f.just greater than 30 is 31 and its corresponding class is 15-20.
  
  Here,
  
  $$\frac{3N}{4}$$ = 30,l= 5, c.f. = 24, f = 7and i = 5
  
  $$\therefore$$Q3= l + $$\frac{\frac{3N}{4}− c.f}{f}$$ × i
  
  = 5 + $$\frac{30− 24}{7}$$ ×5
  
  = 5 + 4.28
  
  = 9.28
  
  Solution:
  
   Class frequency c.f 5 - 10 4 4 10 - 15 6 10 15 - 20 10 20 20 - 25 15 35 25 - 30 10 45 30 - 35 6 51 35 - 40 5 56 N = 56
  
  Now,
  or, Q3= 3($$\frac{N}{4}$$)th item
  or, Q3= 3($$\frac{56}{4}$$)th item
  or, Q3= 3 $$\times$$ 14th item
  or, Q3= 42th item
  In c.f column, c.f is just greater than 42 is 45 and its corresponding value is 25 - 30.
  Now,
  l = 25, c.f = 35, f = 10, and i = 5
  Then,
  or, Q3= L + $$\frac{\frac{3N}{4}-c.f}{f}$$× i
  or, Q3= 25 + $$\frac{42 - 35}{10}$$× 5
  or, Q3= 25 + $$\frac{7}{2}$$
  or, Q3= 25 +3.5
  or, Q3= 28.5
  $$\therefore$$ The value of Q3 is 28.5.
  
  Solution:
  
   Marks Students (f) c.f. 30-40 5 5 40-50 12 17 50-60 25 42 60-70 15 57 70-80 8 65 80-90 3 68 N =68
  
  Now,
  
  Q1 Class = value of ($$\frac{N}{4}$$)th item
  
  = value of ( $$\frac{68}{4}$$)th item
  
  = value of 17th item
  
  In c.f. column, c.f. just greater than 17 is 42 and its corresponding class is (50-60).
  
  $$\therefore$$ Q1 lies in the class 50-60
  
  Here,
  
  $$\frac{N}{4}$$ = 17, l= 50, c.f. = 17, f = 25 and i =10
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{4}− c.f}{f}$$ × i
  
  = 50 + $$\frac{17− 17}{25}$$ ×10
  
  =50 + $$\frac{0}{25}$$ ×10
  
  = 50 +0
  
  = 50
  
  Again,
  
  Q2Class = value of ($$\frac{N}{2}$$)th item
  
  = value of ( $$\frac{68}{2}$$)th item
  
  = value of 34th item
  
  In c.f. column, c.f. just greater than 34 is 42 and its corresponding class is (50-60).
  
  $$\therefore$$ Q2lies in the class 50-60
  
  Here,
  
  $$\frac{N}{2}$$ = 34, l = 50, c.f. = 17, f = 25 and i =10
  
  $$\therefore$$Q1= l + $$\frac{\frac{N}{2}− c.f}{f}$$ × i
  
  = 50 + $$\frac{34− 17}{25}$$ ×10
  
  =50 + $$\frac{17}{25}$$ ×10
  
  = 50 +6.8
  
  = 56.8
  
  Q3 Class = value of($$\frac{3N}{4}$$)th item
  
  = value of($$\frac{3 * 68}{4}$$)th item
  
  =value of 51th item
  
  In c.f. column, c.f.just greater than 51 is 57 and its corresponding class is 60-70.
  
  Here,
  
  $$\frac{3N}{4}$$ = 51, l= 60, c.f. = 42, f = 15 and i = 10
  
  $$\therefore$$Q3= l + $$\frac{\frac{3N}{4}− c.f}{f}$$ × i
  
  = 60 + $$\frac{51− 42}{15}$$ × 10
  
  = 60 + 6
  
  = 66
  
  Solution:
  
   x f c.f 50 8 8 60 16 24 70 12 36 80 4 40 90 3 43 N = 43
  
  Now,
  or, Q1 =($$\frac{N+1}{4}$$)th item
  or, Q1 =($$\frac{43+1}{4}$$)th item
  or, Q1 =($$\frac{44}{4}$$)th item
  or, Q1 =11th item
  In c.f column, c.f is just greater than 11 is 24 and its corresponding value is 60.
  So,Q1 = 60
  Again,
  or, Q3 =3($$\frac{N+1}{4}$$)th item
  or, Q3 =3($$\frac{43+1}{4}$$)th item
  or, Q3 =3($$\frac{44}{4}$$)th item
  or, Q3 =3 $$\times$$ 11th item
  or, Q3 =33th item
  In c.f column, c.f is just greater than 33 is 36 and its corresponding value is 70.
  i.e.Q3 = 70
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text extract contains multiple solutions to math problems, specifically dealing with the calculation of quartiles (Q1, Q2, Q3) from given data sets. Each solution includes a step-by-step calculation and a final answer. The text does not contain any explicit questions, but rather presents the solutions directly.
  
  Label: multiple-question-and-answer