Exercise-13.1, Class 10th, Maths, Chapter 13, NCERT

Q1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants: 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14
Number of houses: 1, 2, 1, 5, 6, 2, 3

Solution (Direct method — numbers are small so direct is simplest).
Class marks $x_i$:

$$1,3,5,7,9,11,13$$

Frequencies $f_i$:

$$1,2,1,5,6,2,3$$

Compute $f_i{x}_i$:

$$1\cdot 1=1,2\cdot 3=6,1\cdot 5=5,5\cdot 7=35,6\cdot 9=54,2\cdot 11=22,3\cdot 13=39$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=1+6+5+35+54+22+39=162}$
${\displaystyle \mathrm{\Sigma }{f}_i=20.}$

Mean $\bar{x}={\displaystyle \frac{\mathrm{\Sigma }{f}_i{x}_i}{\mathrm{\Sigma }{f}_i}}={\displaystyle \frac{162}{20}}=8.1$

Answer: $\boxed{{\displaystyle 8.1\text{ plants per house}}}$

---

Q2.
Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (₹): 500–520, 520–540, 540–560, 560–580, 580–600
Number of workers: 12, 14, 8, 6, 10

Find the mean daily wages.

Solution (Direct / class-mark method).
Class marks $x_i$: $510,530,550,570,590$.
Frequencies $f_i$: $12,14,8,6,10$.

Compute $f_i{x}_i$:

$$12\cdot 510=6120,14\cdot 530=7420,8\cdot 550=4400,6\cdot 570=3420,10\cdot 590=5900$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=6120+7420+4400+3420+5900=27260}$
${\displaystyle \mathrm{\Sigma }{f}_i=50.}$

Mean $\bar{x}={\displaystyle \frac{27260}{50}}=545.2$

Answer: $\boxed{{\displaystyle \text{₹}545.20\text{ per day (mean)}}}$

---

Q3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency $f$.

Daily pocket (₹): 11–13, 13–15, 15–17, 17–19, 19–21, 21–23, 23–25
Number of children: 7, 6, 9, 13, $f$, 5, 4

Solution.
Class marks $x_i$: $12,14,16,18,20,22,24$. Frequencies as above.

Compute known part of $\mathrm{\Sigma }{f}_i{x}_i$:

$$7\cdot 12=84,6\cdot 14=84,9\cdot 16=144,13\cdot 18=234,5\cdot 22=110,4\cdot 24=96$$

Sum of known products $=84+84+144+234+110+96=752$

Term with unknown frequency: $f\cdot 20=20f\mathrm{}$

Total frequency $=7+6+9+13+f+5+4=44+f\mathrm{}$

Given mean $=18$, so

$$\frac{752+20f}{44+f}=18.$$

Solve:

$$752+20f=18(44+f)=792+18f$$
$$20f-18f=792-752\Rightarrow 2f=40\Rightarrow f=20.$$

Answer: $\boxed{{\displaystyle f=20}}\mathrm{}$

---

Q4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women.

Number of heartbeats per minute: 65–68, 68–71, 71–74, 74–77, 77–80, 80–83, 83–86
Number of women: 2, 4, 3, 8, 7, 4, 2

Solution (Direct method with class marks).
Class marks $x_i$: $66.5,69.5,72.5,75.5,78.5,81.5,84.5$
Frequencies $f_i$: $2,4,3,8,7,4,2$

Compute $f_i{x}_i$:

$$\begin{gathered} 2\cdot 66.5=133.0 \\ 4\cdot 69.5=278.0 \\ 3\cdot 72.5=217.5 \\ 8\cdot 75.5 \end{gathered}$$

$$\begin{gathered} =604.0 \\ 7\cdot 78.5=549.5 \\ 4\cdot 81.5=326.0 \\ 2\cdot 84.5=169.0 \end{gathered}$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=133+278+217.5+604+549.5+326+169=2277.0}$
${\displaystyle \mathrm{\Sigma }{f}_i=30.}$

Mean $\bar{x}={\displaystyle \frac{2277.0}{30}}=75.9$

Answer: $\boxed{{\displaystyle 75.9\text{ heartbeats per minute}}}$

---

Q5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes: 50–52, 53–55, 56–58, 59–61, 62–64
Number of boxes: 15, 110, 135, 115, 25

Find the mean number of mangoes kept in a packing box.

Solution (Direct method).
Class marks $x_i$: $51,54,57,60,63$
Frequencies $f_i$: $15,110,135,115,25$

Compute $f_i{x}_i$:

$$\begin{gathered} 15\cdot 51=765 \\ 110\cdot 54=5940 \\ 135\cdot 57=7695 \\ 115\cdot 60=6900 \\ 25\cdot 63=1575 \end{gathered}$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=765+5940+7695+6900+1575=22875}$
${\displaystyle \mathrm{\Sigma }{f}_i=15+110+135+115+25=400}$

Mean $\bar{x}={\displaystyle \frac{22875}{400}}=57.1875$

Answer: $\boxed{{\displaystyle 57.1875\approx 57.19\text{ mangoes per box.}}}$

(Method used: direct class-mark method — frequencies are large but class marks are simple.)

---

Q6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (₹): 100–150, 150–200, 200–250, 250–300, 300–350
Number of households: 4, 5, 12, 2, 2

Find the mean daily expenditure on food.

Solution (Direct method).
Class marks $x_i$: $125,175,225,275,325$.
Frequencies $f_i$: $4,5,12,2,2$

Compute $f_i{x}_i$:

$$4\cdot 125=500,5\cdot 175=875,12\cdot 225=2700,2\cdot 275=550,2\cdot 325=650$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=500+875+2700+550+650=5275}$
${\displaystyle \mathrm{\Sigma }{f}_i=25}$

Mean $\bar{x}={\displaystyle \frac{5275}{25}}=211.0$

Answer: $\boxed{{\displaystyle \text{₹}211.00\text{ per day (mean)}}}$

---

Q7.
To find out the concentration of SO${}_2$ in the air (in ppm), the data was collected for 30 localities:

Concentration (ppm): 0.00–0.04, 0.04–0.08, 0.08–0.12, 0.12–0.16, 0.16–0.20, 0.20–0.24
Frequencies: 4, 9, 9, 2, 4, 2

Find the mean concentration of SO${}_2$.

Solution (Direct method with small decimals).
Class marks $x_i$: $0.02,0.06,0.10,0.14,0.18,0.22$
Frequencies $f_i$: $4,9,9,2,4,2$

Compute $f_i{x}_i$:

$$\begin{gathered} 4\cdot 0.02=0.08 \\ 9\cdot 0.06=0.54 \\ 9\cdot 0.10=0.90 \\ 2\cdot 0.14=0.28 \\ 4\cdot 0.18=0.72 \\ 2\cdot 0.22=0.44 \end{gathered}$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=0.08+0.54+0.90+0.28+0.72+0.44=2.96}$
${\displaystyle \mathrm{\Sigma }{f}_i=30}$

Mean $\bar{x}={\displaystyle \frac{2.96}{30}}=0.098666\dots$

Answer: $\boxed{{\displaystyle 0.09867\text{ ppm (approx)}}}$

---

Q8.
A class teacher has the following absentee record of 40 students for the whole term. Find the mean number of days a student was absent.

Number of days: 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40
Number of students: 11, 10, 7, 4, 4, 3, 1

Solution (Direct method).
Class marks $x_i$: $3,8,12,17,24,33,39$
Frequencies $f_i$: $11,10,7,4,4,3,1$

Compute $f_i{x}_i$:

$$11\cdot 3=33,10\cdot 8=80,7\cdot 12=84,4\cdot 17=68,4\cdot 24=96,3\cdot 33=99,1\cdot 39=39$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=33+80+84+68+96+99+39=499}$
${\displaystyle \mathrm{\Sigma }{f}_i=40}$

Mean $\bar{x}={\displaystyle \frac{499}{40}}=12.475$

Answer: $\boxed{{\displaystyle 12.475\text{ days (mean)}\approx 12.48\text{ days}}}$

---

Q9.
The following table gives the literacy rate (in %) of 35 cities. Find the mean literacy rate.

Literacy rate (%): 45–55, 55–65, 65–75, 75–85, 85–95
Number of cities: 3, 10, 11, 8, 3

Solution (Direct method).
Class marks $x_i$: $50,60,70,80,90$
Frequencies $f_i$: $3,10,11,8,3$

Compute $f_i{x}_i$:

$$3\cdot 50=150,10\cdot 60=600,11\cdot 70=770,8\cdot 80=640,3\cdot 90=270$$

${\displaystyle \mathrm{\Sigma }{f}_i{x}_i=150+600+770+640+270=2430}$
${\displaystyle \mathrm{\Sigma }{f}_i=35.}$

Mean $\bar{x}={\displaystyle \frac{2430}{35}}=69.428571\dots$

Answer: $\boxed{{\displaystyle 69.42857\mathrm{\%}\approx 69.43\mathrm{\%}}}$