9th Class Math Chapter 12 Notes

Q1 — Upper limit of the last class

Find the upper limit of the last class ($49$–$53$).

1. The last class is $49$–$53$.

Answer: $53$

Q2 — Lower limit of the class $39$–$43$

Find the lower limit.

1. Read directly from the class.

Answer: $39$

Q3 — Midpoint of the class $34$–$38$

Find the midpoint.

1. Midpoint $=\frac{34+38}{2}$.

Answer: $36$

Q4 — Class frequencies of $29$–$33$ and $44$–$48$

Find both frequencies.

1. Read directly from the table.

Answer: $6$ and $15$

Q5 — Size of the class limits

Find the class width.

1. Next lower limit $-$ current lower limit $=29-24$.

Answer: $5$

Q6 — Class with the minimum number of students

Identify the class.

1. Smallest frequency is $3$, in class $24$–$28$.

Answer: $24$–$28$

Q7 — Lower limit of the class with frequency $15$

Find the lower limit.

1. Frequency $15$ belongs to class $44$–$48$.

Answer: $44$

Q8 — Students with scores between $24$ and $43$

Find the total.

1. Add frequencies of $24$–$28,29$–$33,34$–$38,39$–$43$: $3+6+12+23$.

Answer: $44$ students

Q9 — Repair expenditure, $8$ classes; frequencies of the last three classes

30 values ranging $144$–$167$; build $8$ classes by the tally method.

1. Range $=167-144=23$; class width $=\lceil 23/8\rceil =3$. Classes: $144$-$146,147$-$149,\dots ,165$-$167$.
2. Tallying all 30 values gives frequencies $4,3,7,5,4,4,1,2$ across the 8 classes.

Answer: Last three classes ($159$-$161,162$-$164,165$-$167$) have frequencies $4,1,2$

Q10 — Student weights, $6$ classes; frequency table and polygon

30 weights ranging $15$–$44$ kg; build $6$ classes.

1. Range $=44-15=29$; class width $=\lceil 29/6\rceil =5$. Classes: $15$-$19,\dots ,40$-$44$.
2. Tallying gives frequencies: $15$-$19:2,20$-$24:3,25$-$29:5,30$-$34:10,35$-$39:6,40$-$44:4$.
3. Frequency polygon: plot (midpoint, frequency) $=(17,2),(22,3),(27,5),(32,10),(37,6),(42,4)$, closing to zero at $12$ and $47$.

Answer: Frequencies $2,3,5,10,6,4$; polygon through the midpoints listed above

Q11 — English marks, $5$ classes (direct method); histogram

32 marks ranging $33$–$62$; build $5$ classes and find the least-frequency class.

1. Range $=62-33=29$; class width $=\lceil 29/5\rceil =6$. Classes: $33$-$38,39$-$44,45$-$50,51$-$56,57$-$62$.
2. Tallying gives frequencies: $1,6,15,4,6$.
3. Histogram: bars over the class boundaries with these heights.

Answer: Least frequency: class $33$–$38$ (frequency $1$); frequencies overall $1,6,15,4,6$

Q12 — Frequency polygon on histogram: weight $10$-$14,\dots ,35$-$39$; $f=6,17,23,30,22,13$

Draw the histogram and polygon.

1. Histogram bars at each class with the given heights.
2. Polygon through midpoints $(12,6),(17,17),(22,23),(27,30),(32,22),(37,13)$, closing to zero at $7$ and $42$.

Answer: Polygon peaks at the $25$-$29$ class (frequency $30$)

Q13 — Discrete distribution: heads in 5 coin tosses (48 recorded trials)

Tally the number of heads ($0$–$5$) from the recorded data.

1. Counting each value $0$ through $5$ in the given list gives: $0:5,1:6,2:9,3:13,4:9,5:6$ (totalling $48$ of the recorded trials).

Answer: Frequencies: $0:5,1:6,2:9,3:13,4:9,5:6$

Q14 — Histogram: Marks $35$-$37,\dots ,68$-$72$; $f=2,12,16,13,9,3$

Draw the histogram.

1. Class boundaries (midway across each $1$-unit gap): $34.5,37.5,44.5,54.5,61.5,67.5,72.5$.
2. Bars over these boundaries with heights $2,12,16,13,9,3$.

Answer: Histogram bars as described; tallest bar over $44.5$–$54.5$ (frequency $16$)

Q15 — Polygon on histogram: limits $5$-$8,\dots ,27$-$32$; $f=2,12,25,32,14,5$

Draw the histogram and polygon.

1. These limits already meet with no gaps, so they are the class boundaries directly.
2. Midpoints: $6.5,10,16,22.5,26,29.5$; polygon connects $(6.5,2),(10,12),(16,25),(22.5,32),(26,14),(29.5,5)$.

Answer: Polygon peaks at the $20$-$25$ class (frequency $32$)

Q16 — Mean: $4,6,10,12,15,20,25,28,30$

Find the arithmetic mean.

1. Sum $=150$; $n=9$.

Answer: $\frac{150}{9}\approx 16.67$

Q17 — Mean: $12,18,19,0,-19,-18,-12$

Find the arithmetic mean.

1. Sum $=0$ (positive and negative terms cancel); $n=7$.

Answer: $0$

Q18 — Mean: $6.5,11,12.3,9,8.1,16,18,20.5,25$

Find the arithmetic mean.

1. Sum $=126.4$; $n=9$.

Answer: $\approx 14.04$

Q19 — Mean: $8,10,12,14,16,20,22$

Find the arithmetic mean.

1. Sum $=102$; $n=7$.

Answer: $\approx 14.57$

Q20 — Median height of $12$ students

Heights: $55,53,54,58,60,61,62,56,57,52,51,63$. Find the median.

1. Sorted: $51,52,53,54,55,56,57,58,60,61,62,63$.
2. $n=12$ (even): median $=$ average of $6$th and $7$th values $=\frac{56+57}{2}$.

Answer: $56.5$ inches

Q21 — Workers' earnings: Arithmetic Mean

$88,70,72,125,115,95,81,90,95,90$. Find the mean.

1. Sum $=921$; $n=10$.

Answer: $92.1$

Q22 — Same data: Median

Find the median.

1. Sorted: $70,72,81,88,90,90,95,95,115,125$.
2. Median $=$ average of $5$th, $6$th $=\frac{90+90}{2}$.

Answer: $90$

Q23 — Same data: Mode

Find the mode.

1. $90$ appears twice and $95$ appears twice; all others appear once.

Answer: Bimodal: $90$ and $95$

Q24 — Mean of English marks (direct and short formula)

Marks $15$-$19,\dots ,35$-$39$; $f=9,18,35,17,5$. Find the mean both ways.

1. Midpoints: $17,22,27,32,37$; $\Sigma f=84$.
2. Direct: $\Sigma fx=9(17)+18(22)+35(27)+17(32)+5(37)=2223$; mean $=\frac{2223}{84}$.
3. Short (assumed mean $A=27$): $D=-10,-5,0,5,10$; $\Sigma fD=-45$; mean $=27+\frac{-45}{84}$.

Answer: Mean $\approx 26.46$ (both methods agree)

Q25 — Median of the same distribution

Find the median.

1. Boundaries: $14.5,19.5,24.5,29.5,34.5,39.5$; cumulative frequencies $9,27,62,79,84$.
2. $n/2=42$ falls in class $25$-$29$ (boundary $24.5$-$29.5$), with $CF=27$ before it.
3. Median $=24.5+\frac{42-27}{35}\times 5$.

Answer: $\approx 26.64$

Q26 — Mode: Class interval $5$-$9,\dots ,25$-$29$; $f=1,8,18,11,2$

Find the mode.

1. Modal class: $15$-$19$ (boundary $14.5$-$19.5$, frequency $18$); $f_0=8,f_2=11,h=5$.
2. Mode $=14.5+\frac{18-8}{2(18)-8-11}\times 5=14.5+\frac{10}{17}\times 5$.

Answer: $\approx 17.44$

Q27 — Petrol pump wages: mean, median, mode

$4250,4350,4400,4250,4350,4410,4500,4300,4500,4390$. Find all three.

1. Short formula ($A=4350$): $D$ values sum to $200$; mean $=4350+\frac{200}{10}=4370$.
2. Sorted: $4250,4250,4300,4350,4350,4390,4400,4410,4500,4500$; median $=\frac{4350+4390}{2}$.
3. $4250,4350,4500$ each appear twice — multimodal.

Answer: Mean $=4370$; median $=4370$; modes $=4250,4350,4500$

Q28 — Sum of $45$ numbers with mean $80$

Find the sum.

1. Sum $=n\times \text{mean}=45\times 80$.

Answer: $3600$

Q29 — Mean, median, mode of $1,4,0,7,9$

Find all three.

1. Sum $=21$, $n=5$: mean $=\frac{21}{5}=4.2$.
2. Sorted: $0,1,4,7,9$; median (middle value) $=4$.
3. All values appear once — no mode.

Answer: Mean $=4.2$; median $=4$; no mode

Q30 — Show Mode $>$ Median $>$ Mean for $148,145,160,157,156,160$

Verify the inequality.

1. Mean $=\frac{926}{6}\approx 154.33$.
2. Sorted: $145,148,156,157,160,160$; median $=\frac{156+157}{2}=156.5$.
3. $160$ appears twice — mode $=160$.
4. $160>156.5>154.33$ ✓

Answer: Verified: Mode ($160$) $>$ Median ($156.5$) $>$ Mean ($\approx 154.33$)

Q31 — Hostel attendance: median, mode, and mean via $D=A-20$

$21,15,16,18,14,17,15,12,13,11$. Find the median, mode, and mean (using assumed mean $A=20$).

1. Sorted: $11,12,13,14,15,15,16,17,18,21$; median $=\frac{15+15}{2}=15$.
2. $15$ appears twice — mode $=15$.
3. $D=x-20$: sum of $D=-48$; mean $=20+\frac{-48}{10}=15.2$ (matches direct sum $152/10$).

Answer: Median $=15$; mode $=15$; mean $=15.2$

Q32 — Pocket money: median and mode

Rupees $5$-$10,\dots ,25$-$30$; $f=12,9,18,7,4$. Find the median and mode.

1. $n=50$; cumulative: $12,21,39,46,50$. Median class $15$-$20$ (cum before $=21$).
2. Median $=15+\frac{25-21}{18}\times 5\approx 16.11$.
3. Modal class $15$-$20$ ($f_1=18,f_0=9,f_2=7$): Mode $=15+\frac{18-9}{2(18)-9-7}\times 5=15+\frac{9}{20}\times 5$.

Answer: Median $\approx 16.11$; Mode $=17.25$

Q33 — Same data: mean using the coding method

Find the mean by coding.

1. Midpoints $7.5,12.5,17.5,22.5,27.5$; let $A=17.5,h=5$: $U=-2,-1,0,1,2$.
2. $\Sigma fU=12(-2)+9(-1)+18(0)+7(1)+4(2)=-18$.
3. Mean $=17.5+5\times \frac{-18}{50}$.

Answer: $15.7$

Q34 — Sum of ages, and average of the remaining boys

Mean age of $20$ boys $=13$y $4$m $5$d. Find the sum; then, removing one $15$-year-old boy, find the average of the rest.

1. Sum $=20\times (13y4m5d)=260y+80m+100d=260y+(6y8m)+(3m10d)=266y11m10d$.
2. Removing $15$ years: $266y11m10d-15y=251y11m10d$.
3. Converting to days ($1y=360d,1m=30d$): $90700$ days $÷19\approx 4773.7$ days $\approx 13y3m3d$.

Answer: Sum $=266$y $11$m $10$d; average of remaining $19$ boys $\approx 13$y $3$m $3$d

Q35 — $D=X-140$, $\Sigma D=500$, $n=10$

Find the mean.

1. Mean $=A+\frac{\Sigma D}{n}=140+\frac{500}{10}$.

Answer: $190$

Q36 — $U=\frac{x-130}{6}$, $\Sigma U=-150$, $n=15$

Find the mean.

1. Mean $=A+h\times \frac{\Sigma U}{n}=130+6\times \frac{-150}{15}$.

Answer: $70$

Q37 — $D=X-25$, $\Sigma fD=300$, $\Sigma f=20$

Find the mean.

1. Mean $=A+\frac{\Sigma fD}{\Sigma f}=25+\frac{300}{20}$.

Answer: $40$

Q38 — $U=\frac{x-120}{5}$, $\Sigma fU=60$, $\Sigma f=100$

Find the mean.

1. Mean $=A+h\times \frac{\Sigma fU}{\Sigma f}=120+5\times \frac{60}{100}$.

Answer: $123$

Q39 — Who wins the Rs.$1000$ award?

Haris: $50,55,70,85,90$; Maham: $75,60,60,45,53$; Minal: $80,77,66,42,48$. Find each mean.

1. Haris: sum $=350$, mean $=70$.
2. Maham: sum $=293$, mean $=58.6$.
3. Minal: sum $=313$, mean $=62.6$.

Answer: Haris wins (highest mean, $70$)

Q40 — $D$-table: $D=-6,\dots ,6$; $f=1,3,6,20,26,12,2$ (assumed mean $A=20$)

Find the mean.

1. $\Sigma f=70$.
2. $\Sigma fD=1(-6)+3(-4)+6(-2)+20(0)+26(2)+12(4)+2(6)=82$.
3. Mean $=20+\frac{82}{70}$.

Answer: $\approx 21.17$

Q41 — Quiz points with $D=x-58$

$45,51,58,61,74,48,46,50$. Find the mean.

1. $D$ values: $-13,-7,0,3,16,-10,-12,-8$; $\Sigma D=-31$.
2. Mean $=58+\frac{-31}{8}$.

Answer: $54.125$

Q42 — Weighted mean of food-item costs per kg

Rice($10$kg,Rs.$96$), Flour($12$,Rs.$48$), Ghee($4$,Rs.$190$), Sugar($3$,Rs.$49$), Mutton($2$,Rs.$650$).

1. $\Sigma (qty\times cost)=960+576+760+147+1300=3743$.
2. $\Sigma qty=31$.

Answer: $\approx$ Rs.$120.74$ per kg

Q43 — Weighted mean: appliances

Washing Machine($5$,$35$k), Heater($3$,$5$k), Stove($2$,$13$k), Dispenser($6$,$18$k) — costs in thousands.

1. $\Sigma (qty\times cost)=175+15+26+108=324$.
2. $\Sigma qty=16$.

Answer: $20.25$ (thousand rupees)

Q44 — Average of the yearly budgets $5,7,8,6,7$ (million)

Find the average budget for next year.

1. Sum $=33$; $n=5$.

Answer: $6.6$ million

Q45 — Ahmad's weighted mean marks

Marks $78,65,80,90,85,72$ with weights $5,4,2,3,2,4$.

1. $\Sigma (w\times marks)=390+260+160+270+170+288=1538$.
2. $\Sigma w=20$.

Answer: $76.9$

Q46 — Frequency table, $6$ classes

40 weights ranging $119$–$176$ lbs; build $6$ classes.

1. Range $=176-119=57$; class width $=\lceil 57/6\rceil =10$. Classes: $119$-$128,\dots ,169$-$178$.
2. Tallying all 40 values gives frequencies $4,7,13,9,5,2$.

Answer: Frequencies: $119$-$128:4,129$-$138:7,139$-$148:13,149$-$158:9,159$-$168:5,169$-$178:2$

Q47 — Draw the histogram

Draw the histogram for the table above.

1. Boundaries: $118.5,128.5,138.5,148.5,158.5,168.5,178.5$; bars with heights $4,7,13,9,5,2$.

Answer: Tallest bar over $139$–$148$ (frequency $13$)

Q48 — Draw the frequency polygon

Draw the polygon for the same data.

1. Midpoints $123.5,133.5,143.5,153.5,163.5,173.5$ with frequencies $4,7,13,9,5,2$, closing to zero at $113.5$ and $183.5$.

Answer: Polygon through the points listed above

Q49 — Polygon on histogram: weight $50$-$56,\dots ,76$-$80$; $f=25,32,40,30,15,8$

Draw the histogram and polygon.

1. Boundaries (gap $1$): $49.5,56.5,59.5,64.5,72.5,75.5,80.5$; bars with the given heights.
2. Midpoints $53,58,62,68.5,74,78$; polygon connects (midpoint, frequency) pairs.

Answer: Polygon peaks at the $60$-$64$ class (frequency $40$)

Q50 — Upper class boundary of the $5$th class

Marks $20$-$24,\dots ,45$-$49$; $f=5,8,12,15,3,2$. Find the upper boundary of the $5$th class ($40$-$44$).

1. Boundaries are found by adding/subtracting $0.5$: $39.5$-$44.5$.

Answer: $44.5$

Q51 — Lower class boundaries of all classes

Find all six lower boundaries.

1. Subtract $0.5$ from each lower limit: $20,25,30,35,40,45\to 19.5,24.5,\dots$

Answer: $19.5,24.5,29.5,34.5,39.5,44.5$

Q52 — Midpoints of all classes

Find all six midpoints.

1. Average each class's limits.

Answer: $22,27,32,37,42,47$

Q53 — Class with the least frequency

Identify the class.

1. Smallest frequency is $3$, in class $40$-$44$.

Answer: $40$-$44$

Q54 — Limits $5$-$9,\dots ,30$-$34$; $f=1,8,18,11,2,5$

Draw both the polygon and histogram.

1. Boundaries (gap $1$): $4.5,9.5,14.5,19.5,24.5,29.5,34.5$; bars with heights $1,8,18,11,2,5$.
2. Midpoints $7,12,17,22,27,32$; polygon connects these to the frequencies, closing to zero at $2$ and $37$.

Answer: Polygon peaks at the $15$-$19$ class (frequency $18$)

Q55 — Weighted mean: furniture items

Chair($20$,Rs.$500$), Table($20$,Rs.$400$), Blackboard($10$,Rs.$750$), Tube light($25$,Rs.$230$), Cupboard($9$,Rs.$950$).

1. $\Sigma (qty\times cost)=10000+8000+7500+5750+8550=39800$.
2. $\Sigma qty=84$.

Answer: $\approx$ Rs.$473.81$

Q56 — Average fund allocation across $5$ sectors

Chairs Rs.$15000$, tables Rs.$12000$, blackboards Rs.$6000$, renovation Rs.$10000$, gardening Rs.$7000$ (total Rs.$50000$). Find the average.

1. Average $=\frac{50000}{5}$.

Answer: Rs.$10000$

Q57 — Mean of Saad's marks

$84,91,72,68,87,78$. Find the mean.

1. Sum $=480$; $n=6$.

Answer: $80$

Q58 — Mean load of ropes (short method)

Max-load $93$-$97,\dots ,118$-$122$ kg; $f=2,5,8,12,6,2$. Find the mean.

1. Midpoints $95,100,105,110,115,120$; let $A=110,h=5$: $U=-3,-2,-1,0,1,2$.
2. $\Sigma fU=2(-3)+5(-2)+8(-1)+12(0)+6(1)+2(2)=-14$; $\Sigma f=35$.
3. Mean $=110+5\times \frac{-14}{35}$.

Answer: $108$ kg

Q59 — Dice sums: median and mode

$8,5,6,6,9,4,3,11$. Find the median and mode.

1. Sorted: $3,4,5,6,6,8,9,11$; median $=\frac{6+6}{2}=6$.
2. $6$ appears twice — mode $=6$.

Answer: Median $=6$; mode $=6$

Q60 — Weekly wages: mean, median, mode

Wages $600$-$700,\dots ,1000$-$1100$; employees $=3,5,7,21,11$. Find all three.

1. $\Sigma f=47$; midpoints $650,750,850,950,1050$; $\Sigma fx=43150$; mean $=\frac{43150}{47}\approx 918.09$.
2. Cumulative: $3,8,15,36,47$; median class $900$-$1000$: median $=900+\frac{23.5-15}{21}\times 100\approx 940.48$.
3. Modal class $900$-$1000$ ($f_1=21,f_0=7,f_2=11$): mode $=900+\frac{21-7}{2(21)-7-11}\times 100=900+\frac{14}{24}\times 100\approx 958.33$.

Answer: Mean $\approx 918.09$; median $\approx 940.48$; mode $\approx 958.33$