Given, length of cuboid (l) = 15 cm, breadth of cuboid (b) = 10 cm and height of cuboid (h) = 3.5 cm
\(\therefore\) Volume of cuboid = l \(\times\) b \(\times\) h
= 15 \(\times\) 10 \(\times\)3.5 = 525 cm³
Also, radius of conical depression,r = 0.5 cm
and height of conical depression, h = 1.4 cm
\(\therefore\) Volume of one conical depression \(=\frac{1}{3} \pi \times r^2 \times h\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{3} \times \frac{1}{2} \times \frac{1}{2} \times \frac{2}{10}=\frac{11}{30} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 4 conical depressions = 4 \(\times\)Volume of one conical depression

\(=4 \times \frac{11}{30}=\frac{22}{15} \mathrm{~cm}^3\)
Hence, the volume of wood in the entire stand = Volume of cuboid - Volume of 4 conical depressions
\(=525-\frac{22}{15}=525-1.47=523.53 \mathrm{~cm}^3\)

Given, solid is a combination of a cone and a hemisphere.
Also, radius of the cone, r = radius of the hemisphere
= 1 cm

Height of the cone, h = 1 cm
\(\therefore\) Required volume of the solid = Volume of the cone + Volume of the hemisphere
\(\begin{aligned} & =\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3=\frac{1}{3} \pi(1)^2(1)+\frac{2}{3} \pi(1)^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\pi}{3}+\frac{2}{3} \pi=\frac{(\pi+2 \pi)}{3}=\frac{3 \pi}{3}=\pi \mathrm{cm}^3 \end{aligned}\)

Given, the volume of each cube is 64 cm³.
Let each side of the cube be a cm.

\(\because\) Volume of cube = (Side)³
\(\therefore\)   64 = a³      [given]
\(\Rightarrow\) a = 4 cm
When two cubes are joined end-to-end, then we get a cuboid whose length, l = (4 + 4) = 8 cm, breadth, b = 4 cm and height, h = 4 cm.
Now, surface area of the resulting cuboid
= 2(lb + bh + hl) = 2(8 \(\times\) 4 + 4 \(\times\)4 + 4 \(\times\) 8)
= 2(32 + 16 + 32) = 2(80) = 160 cm²

Given, radius of a circle, AO = 10 cm and \(\angle\)AOC = 90°
Area of \(\triangle A O C=\frac{1}{2} \times O A \times O C=\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2\)
\(\begin{aligned} \text { Area of sector } O A E C O & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)

(i) Area of minor segment AECDA
= Area of sector OAECO - Area of \(\Delta\)AOC
= 78.5 - 50 = 28.5 cm²
(ii) Area of major sector OAFGCO
= Area of circle - Area of sector OAECO
= 3.14 \(\times\)(10)² - 78.5
= 314 - 78.5 = 235.5 cm²

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Given, side of the cube = Diameter of the hemisphere = l units
\(\therefore\) Radius of the hemisphere, \(r=\frac{l}{2}\) units

Now, required surface area of the remaining solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere
\(\begin{aligned} & =6 \times(\text { Edgc })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times l^2+2 \pi \times\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 l^2+2 \pi \times \frac{l^2}{4}-\pi \frac{l^2}{4}=6 l^2+\pi \frac{l^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{l^2}{4}(\pi+24) \text { sq units } \end{aligned}\)

Given, a cubical block is surmounted by a hemisphere. Therefore, diameter of hemisphere must be equal to the side of cubical block and it is the greatest diameter of hemisphere.

For cubical portion,
Edge = 7 cm
For hemispherical portion,
Diameter = 7 cm
\(\therefore\)   Radius, r = \(\frac{7}{2}\)cm
Now, required surface area of solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere.
\(\begin{aligned} & =6 \times(\text { Edge })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times(7)^2+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}-\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+11 \times 7-\frac{11 \times 7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+77-\frac{77}{2} \end{aligned}\)
= 371 - 38.5 = 332.5 cm²

Given, diameter of circle, d = 35 mm
\(\therefore\) Circumference of circle = \(\pi\)d      [\(\because\) d = 2r]
\(=\frac{22}{7} \times 35=110 \mathrm{~mm}^2\)
Now, length of 5 diameters = 5 \(\times\) 35 = 175 mm
(i) Total length of the silver wire = \(\pi\)d + 5d
= 110 + 175 = 285 mm²
(ii) Here, we see that total circle is divided into 10 sectors.
\(\therefore\) Angle of each sector = \(\frac{360^{\circ}}{10}=36^{\circ}\)
Then, area of each sector ofthe brooch = \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(\begin{aligned} & =\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2 \quad\left[\because r=\frac{d}{2}=\frac{35}{2} \mathrm{~mm}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}=\frac{11 \times 35}{2 \times 2}=\frac{385}{4} \mathrm{~mm}^2 \end{aligned}\)

Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Figure). The radius BO of the hemisphere (as well as of the cone) \(=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm} .\)
So, volume of the toy = \(\begin{aligned} & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \end{aligned}\)
\(\begin{aligned} & =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3 \end{aligned}\)
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm
So, the colume required = volume of the right circular cylinder - volume of the toy
= 3.14 \(\times\)2² \(\times\)4 - 25.12) cm³
= 25.12 cm³
hence, the required difference of the two volumes = 25.12 cm³.

(i) The cumulative frequency table of given frequency distribution is

Here, n = 68
\(\therefore \quad \frac{n}{2}=34\)
Since, the cumulative frequency just greater than 34 is 42 and the corresponding class is 125- 145. So, the median class is 125-145.
Now, l = 125, f = 20, cf = 22 and h = 20
\(\begin{aligned} & \therefore \text { Median }=l+\left\{\frac{\frac{n}{2}-c f}{f}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} \quad=125+\left\{\frac{34-22}{20}\right\} \times 20=125+12=137 \text { units } \end{aligned}\)
(ii) Let the assumed mean, a = 135
and width of the class, h = 20
Table for the given data is

We have, N = 68, and \(\sum\)f_iu_i = 7
By step deviation method,
Mean = \(\begin{aligned} \text { Mean } & =a+b \times \frac{1}{N} \times \Sigma f_i u_i=135+20 \times \frac{1}{68} \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =135+\frac{35}{17}=135+2.05=137.05 \text { units } \end{aligned}\)
(iii) Here, the modal class is 125-145 having maximum frequency, f₁ = 20.
Now, f₀ = 13, f₂ = 14, l = 125 and h = 20
\(\begin{aligned} \therefore \text { Mode } & =l+\left\{\frac{f_1-f_0}{2 f_1-f_0-f_2}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} & =125+\left\{\frac{20-13}{40-13-14}\right\} \times 20 \\ \end{aligned}\)
\(\begin{aligned} & =125+\frac{7 \times 20}{13}=125+\frac{140}{13} \end{aligned}\)
= 125 + 10.77 = 135.77 units
Hence, median = 137 units, mean = 137.05 units and mode = 135.77 units
So, we conclude that three measures are approximately the same.
(iv) Electricity consumption can be reduced by
(a) switching off electric appliances when not in use.
(b) using star rated AC's, refrigerators.

Recall that to find the mean marks, we require the product of each x_i with the corresponding frequency f_i. So, let us put them in a column as shown in Table 1.
Table - 1

Now, \(\bar{x}=\frac{\sum f_{i} x_{i}}{\Sigma f_{i}}=\frac{1779}{30}=59.3\)
Therefore, the mean marks obtained is 59.3.
In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean.
Let us convert the ungrouped data example into grouped data by forming class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table. (see Table - 2)
Table - 2

Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is,
\(\text { Class mark }=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\)
With reference to Table - 2, for the class 10-25, the class mark is \(\frac{10+25}{2}\) i.e.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table - 3. These class marks serve as our x_i’s. Now, in general, for the ith class interval, we have the frequency f_i corresponding to the class mark x_i. We can now proceed to compute the mean in the same manner as in Example.
Table - 3

The sum of the values in the last column gives us \(\Sigma \)f_i x_i. So, the mean \(\bar {x}\) of the given data is given by
\(\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{1860.0}{30}=62\)
This new method of finding the mean is known as the Direct Method.
We observe that Table -1 and Table - 3 are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table - 3, 59.3 being the exact mean, while 62 an approximate mean.
Sometimes when the numerical values of x_i and f_i are large, finding the product of x_i and f_i becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations.
We can do nothing with the f_i’s, but we can change each x_i to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these x_i’s? Let us try this method. The first step is to choose one among the x_i’s as the assumed mean, and denote it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that x_i which lies in the centre of x₁, x₂, . . ., x_n. So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5
The next step is to find the difference d_i between a and each of the x_i’s, that is,the deviation of ‘a’ from each of the x_i’s.
\(\text { i.e., } \quad d_{i}=x_{i}-a=x_{i}-47.5\)
The third step is to find the product of d_i with the corresponding f_i, and take the sum of all the f_i d_i’s. The calculations are shown in Table - 4.
Table - 4

So, from Table - 4, the mean of the deviations \(\bar{d}=\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
Now, let us find the relation between \(\bar {d}\) and \(\bar {x}\) .
Since in obtaining d_i, we subtracted ‘a’ from each x_i, so, in order to get the mean \(\bar {x}\) , we need to add ‘a’ to \(\bar {d}\) . This can be explained mathematically as:
Mean of deviations \(\bar{d}=\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
So, \(\bar{d}=\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}} \)
\(\bar{d}=\frac{\Sigma f_{i}\left(x_{i}-a\right)}{\Sigma f_{i}}\)
\(=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}-\frac{\sum f_{i} a}{\Sigma f_{i}}\)
\(=\bar{x}-a \frac{\Sigma f_{i}}{\Sigma f_{i}} \)
\(=\bar{x}-a \) 
So,
\(\bar{x} =a+\bar{d} \)
i.,e
\(\bar{x} =a+\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
Substituting the values of a, \(\Sigma\)f_id_i and \(\Sigma\)f_i from Table - 4, we get
\(\bar{x}=47.5+\frac{435}{30}=47.5+14.5=62\)
Therefore, the mean of the marks obtained by the students is 62.
The method discussed above is called the Assumed Mean Method.
Activity 1 : From the Table - 3 find the mean by taking each of x_i (i.e., 17.5, 32.5, and so on) as ‘a’. What do you observe? You will find that the mean determined in each case is the same, i.e., 62. (Why ?)
So, we can say that the value of the mean obtained does not depend on the choice of ‘a’.
Observe that in Table - 4, the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with f_i. (Here, 15 is the class size of each class interval.)
So, let \(u_{i}=\frac{x_{i}-a}{h}\) where a is the assumed mean and h is the class size.
Now, we calculate ui in this way and continue as before (i.e., find f_iu_i and then \(\Sigma\)f_i ​​​​​​u_i). Taking h = 15, let us form Table - 5.
Table - 5

Let \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)
Here, again let us find the relation between \(\bar {u}\)​​​​​ and \(\bar {x}\).
We have, \(u_{i} =\frac{x_{i}-a}{h} \)
Therefore, \(\bar{u} =\frac{\sum f_{i} \frac{\left(x_{i}-a\right)}{h}}{\Sigma f_{i}}=\frac{1}{h}\left[\frac{\Sigma f_{i} x_{i}-a \Sigma f_{i}}{\Sigma f_{i}}\right] \)
\(=\frac{1}{h}\left[\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}-a \frac{\Sigma f_{i}}{\Sigma f_{i}}\right]\)
\(=\frac{1}{h}[\bar{x}-a] \)
So, \(h \bar{u} =\bar{x}-a \)
i.e., \(\bar{x} =a+h \bar{u} \)
So, \(\bar{x} =a+h\left(\frac{\sum f_{i} u_{i}}{\Sigma f_{i}}\right)\)
Now, substituting the values of a, h, \(\Sigma\)f_iu_i and \(\Sigma\)f_i from Table - 5, we get
\(\bar{x} =47.5+15 \times\left(\frac{29}{30}\right) \)
\(=47.5+14.5=62\)
So, the mean marks obtained by a student is 62.
The method discussed above is called the Step-deviation method.

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

The cumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than  n/2 (i.e 400/2 = 200) is 216
belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
\(\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \)
\(=3000+\left(\frac{200-130}{86}\right) \times 500 \)
\(=3000+\frac{70 \times 500}{86}\)
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.

To find the class mark (x_i) for each interval, the following relation is used.
Class size = 50
Taking 225 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows.

From the table, we obtain
\(\sum f_{i}=25 \)
\(\sum f_{i} u_{i}=-7 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{1} u_{i}}{\sum f_{i}}\right) x h \)
\(=225+\left(\frac{-7}{25}\right) \times(50)\)
= 225 - 14
= 211
Therefore, mean daily expenditure on food is Rs 211.

Given, circumference of a circle = 22 cm
\(\begin{array}{lll} \Rightarrow & 2 \pi r=22 \quad \Rightarrow \quad 2 \times \frac{22}{7} \times r=22 \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & \quad r=\frac{22 \times 7}{2 \times 22}=\frac{7}{2} \mathrm{~cm} \end{array}\)
Now, area of a quadrant of a circle \(=\frac{\pi r^2}{4}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{\left(\frac{7}{2}\right)^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{11}{14} \times \frac{49}{4}=\frac{77}{8} \mathrm{~cm}^2 \end{aligned}\)

Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass \(=\pi r^{2} h\)
\(=3.14 \times 2.5 \times 2.5 \times 10 \mathrm{~cm}^{3}=196.25 \mathrm{~cm}^{3}\)
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e., it is less by \(\frac{2}{3} \pi r^{3}=\frac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \mathrm{~cm}^{3}=32.71 \mathrm{~cm}^{3}\)
So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= (196.25 – 32.71) cm³
=163.54 cm³