Question
$\text{m}-\frac{\text{m}-1}{2}=1-\frac{\text{m}-2}{3}$
✓
Answer
$\text{m}-\frac{\text{m}-1}{2}=1-\frac{\text{m}-2}{3}$
$=\frac{2\text{m}-\text{m}+1}{2}=\frac{3-\text{m}+2}{3}$
$=\frac{\text{m}+1}{2}=\frac{5-\text{m}}{3}$
$=\frac{\text{m}+1}{2}=\frac{5}{3}-\frac{\text{m}}{3}$
$=\frac{\text{m}}{2}+\frac12=\frac53-\frac{\text{m}}{3}$
Transposing $\frac{\text{m}}{3}$ to $L.H.S$. and $\frac12$ to $R.H.S.,$
we get $=\frac{\text{m}}{2}+-\frac{\text{m}}{3}=\frac{5}{3}-\frac12$
$=\frac{3\text{m}+2\text{m}}{6}=\frac{10-3}{6}$ Multiplying both sides by $6,$
we get $=\frac{5\text{m}}{6}\times6=\frac{7}{6}\times6$
$=5\text{m}=7$ Dividing both sides by $5$,
we get $=\frac{5\text{m}}{5}=\frac75$
$=\text{m}=\frac75$ Verification: Substituting $\text{m}=\frac75$ on both sides,
we get $\frac75-\frac{7-5}{10}=1-\frac{7-10}{15}$
$\frac75-\frac{2}{10}=\frac{15+3}{15}$
$\frac{14-2}{10}=\frac{15+3}{15}$
$\frac{12}{10}=\frac{18}{15}$
$\frac{6}{5}=\frac{6}{5}$
$\text{L.H.S.} =\text{R.H.S.}$
Hence, verified.
$=\frac{2\text{m}-\text{m}+1}{2}=\frac{3-\text{m}+2}{3}$
$=\frac{\text{m}+1}{2}=\frac{5-\text{m}}{3}$
$=\frac{\text{m}+1}{2}=\frac{5}{3}-\frac{\text{m}}{3}$
$=\frac{\text{m}}{2}+\frac12=\frac53-\frac{\text{m}}{3}$
Transposing $\frac{\text{m}}{3}$ to $L.H.S$. and $\frac12$ to $R.H.S.,$
we get $=\frac{\text{m}}{2}+-\frac{\text{m}}{3}=\frac{5}{3}-\frac12$
$=\frac{3\text{m}+2\text{m}}{6}=\frac{10-3}{6}$ Multiplying both sides by $6,$
we get $=\frac{5\text{m}}{6}\times6=\frac{7}{6}\times6$
$=5\text{m}=7$ Dividing both sides by $5$,
we get $=\frac{5\text{m}}{5}=\frac75$
$=\text{m}=\frac75$ Verification: Substituting $\text{m}=\frac75$ on both sides,
we get $\frac75-\frac{7-5}{10}=1-\frac{7-10}{15}$
$\frac75-\frac{2}{10}=\frac{15+3}{15}$
$\frac{14-2}{10}=\frac{15+3}{15}$
$\frac{12}{10}=\frac{18}{15}$
$\frac{6}{5}=\frac{6}{5}$
$\text{L.H.S.} =\text{R.H.S.}$
Hence, verified.
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Observe the following data:
$a.$ Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
$b.$ Which class has the maximum number of students?
$c.$ In which class, the difference of total students and number of students present is minimum?
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Government School, Chandpur
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Daily Attendance
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Date $: 15.4.2009$
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Class
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Total Students
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Number of Students Present on that Day
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$VI$
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$90$
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$81$
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$VII$
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$82$
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$76$
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$VIII$
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$95$
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$91$
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$IX$
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$70$
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$65$
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$X$
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$63$
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$62$
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$b.$ Which class has the maximum number of students?
$c.$ In which class, the difference of total students and number of students present is minimum?
$d.$ Find the ratio of numts of Clasber of students present to the total number of studens $IX$.
$e.$ What per cent of Class $VI$ students were absent?