## Maharashtra State Board Class 7 Maths Solutions Chapter 7 Joint Bar Graph Practice Set 31

Question 1.
The number of saplings planted by schools on World Tree Day is given in the table below. Draw a joint bar graph to show these figures.

 School Name\Name of Sapling Almond Karanj Neem Ashok Gulmohar Nutan Vidyalaya 40 60 72 15 42 Bharat Vidyalaya 42 38 60 25 40

Solution: Question 2.
The table below shows the number of people who had the different juices at a juice bar on a Saturday and a Sunday. Draw a joint bar graph for this data.

 Days\Fruits Sweet Lime Orange Apple Pineapple Saturday 43 30 56 40 Sunday 59 65 78 67

Solution: Question 3.
The following numbers of votes were cast at 5 polling booths during the Gram Panchayat elections. Draw a joint bar graph for this data.

 Persons\Booth No. 1 2 3 4 5 Men 200 270 560 820 850 Women 700 240 340 640 470

Solution: Question 4.
The maximum and minimum temperatures of five Indian cities are given in °C. Draw a joint bar graph for this data.

 City\Temperature Delhi Mumbai Kolkata Nagpur Kapurthala Maximum temperature 35 32 37 41 37 Minimum temperature 26 25 26 29 26

Solution: Question 5.
The numbers of children vaccinated in one day at the government hospitals in Solapur and Pune are given in the table. Draw a joint bar graph for this data:

 City\Vaccine D.P.T. (Booster) Polio (Booster) Measles Hepatitis Solapur 65 60 65 63 Pune 89 87 88 86

Solution: Question 6.
The percentage of literate people in the states of Maharashtra and Gujarat are given below. Draw a joint bar graph for this data.

 State\Year 1971 1981 1991 2001 2011 Maharashtra 46 57 65 77 83 Gujarat 40 45 61 69 79

Solution: Maharashtra Board Class 7 Maths Chapter 7 Joint Bar Graph Practice Set 31 Intext Questions and Activities

Question 1.
Observe the graph shown below and answer the following questions. (Textbook pg. no. 51)

1. In which year did Ajay and Vijay both produce equal quantities of wheat?
2. In year 2014, who produced more wheat?
3. In year 2013, how much wheat did Ajay and Vijay each produce? Solution:

1. Both produced equal quantities of wheat in the year 2011.
2. Ajay produced more wheat in the year 2014.
3. Ajay’s wheat production in 2013 = 40 quintal.
Vijay’s wheat production in 2013 = 30 quintal.

Question 2.
The minimum and maximum temperature in Pune for five days is given. Read the joint bar graph and answer the questions below: (Textbook pg. no. 52) 1. What data is shown on X- axis?
2. What data is shown on Y- axis?
3. Which day had the highest temperature?
4. On which day is the minimum temperature the highest?
5. On Thursday, what is the difference between the minimum and maximum temperature?
6. On which day is the difference between the minimum and maximum temperature the greatest?

Solution:

1. Five days of a week are shown on X – axis.
2. Temperature in the city of Pune is shown on Y – axis.
3. Monday had the highest temperature.
4. The minimum temperature was highest on Wednesday.
5. Maximum temperature = 29.5° C
Minimum temperature = 15° C
∴ Difference in temperature = 29.5° C – 15° C = 14.5 ° C
6. The difference in minimum and maximum temperature is greatest on Thursday.

Question 3.
Collect various kinds of graphs from newspapers and discuss them. (Textbook pg. no. 53)
i. Histogram ii. Line graph iii. Pie chart Solution:
(Students should attempt the above activities on their own.)

## Maharashtra State Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1

Practice Set 14.1 Class 8 Question 1.
Find the amount and the compound interest.

 No Principal (Rs) Rate (p.c.p.a.) Duration (years) i. 2000 5 2 ii. 5000 8 3 iii. 4000 7.5 2

Solution:
i. Here P = Rs 2000, R = 5 p.c.p.a. and N = 2 years = 5 × 441
∴ A = Rs 2205
I = Amount (A) – Principal (P)
= 2205 – 2000
= Rs 205
∴ The amount is Rs 2205 and the compound interest is Rs 205.

ii. Here, P = Rs 5000, R = 8 p.c.p.a. and N = 3 years ∴ A = Rs 6298.56
I = Amount (A) – Principal (P)
= 6298.56 – 5000
= Rs 1298.56
∴ The amount is Rs 6298.56 and the compound interest is Rs 1298.56.

iii. Here, P = Rs 4000, R = 7.5 p.c.p.a. and N = 2 years ∴A = Rs 4622.50
I = Amount (A) – Principal (P)
= 4622.50 – 4000
= Rs 622.50
∴The amount is Rs 4622.50 and the compound interest is Rs 622.50.

Compound Interest Practice Set 14.1 Question 2.
Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Solution:
Here, P = Rs 12,500, R = 12 p.c.p.a. and
N = 3 years = 0.8 × 28 × 28 × 28
= Rs 17,561.60
Sameerrao should pay Rs 17,561.60 to clear his loan.

8th Standard Maths Practice Set 14.1 Question 3.
To start a business Shalaka has taken a loan of Rs 8000 at a rate of $$10\frac { 1 }{ 2 }$$ p.c.p.a. After two years how much compound interest will she have to pay?
Solution:
Here, P = Rs 8000, N = 2 years and  I = Amount (A) – Principal (P)
= 9768.20 – 8000
= Rs 1768.20
∴ After two years Shalaka will have to pay Rs 1768.20 as compound interest.

## Maharashtra State Board Class 8 Maths Solutions Chapter 2 Parallel Lines and Transversals Practice Set 2.2

Question 1.
Choose the correct alternative.
i. In the given figure, if line m || line n and line p is a transversal, then find x. (A) 135°
(B) 90°
(C) 45°
(D) 40°
Solution:
(C) 45°

Hint: line m || line n and line p is a transversal.
∴ m∠BFG + m∠FGD = 180°
…[Interior angles]
∴ 3x + x = 180°
∴ 4x = 180°
∴ x = $$\frac { 180 }{ 4 }$$
∴ x = 45°

ii. In the given figure, if line a || line b and line l is a transversal, then find x. (A) 90°
(B) 60°
(C) 45°
(D) 30°
Solution:
(D) 30°

Hint: line a || line b and line l is a transversal.
∴ m∠UVS = m∠PUV
…[Alternate angles]
= 4x
m∠UVS + m∠WVS = 180°
… [Angles in a linear pair]
∴ 4x + 2x = 180°
∴ 6x = 180°
∴ x = $$\frac { 180 }{ 6 }$$
∴ x = 30°

Question 2.
In the given figure, line p || line q. Line t and line s are transversals. Find measure of ∠x and ∠y using the measures of angles given in the figure. Solution: i. Consider ∠z as shown in figure.
line p || line q and line t is a transversal.
∴ m∠z = 40° …(i) [Corresponding angles]
m∠x + m∠z = 180° …[Angles in a linear pair]
∴ m∠x + 40o = 180° …[From(i)]
∴ m∠x= 180° – 40°
∴ m∠x = 140°

ii. Consider ∠w as shown in the figure.
m∠w + 70° = 180° …[Angles in a linear pair]
∴ m∠w = 180° – 70°
∴ m∠w = 110° …(ii)
line p || line q and line s is a transversal.
∴ m∠y = m∠w …[Alternate angles]
∴ m∠y =110° …[From (ii)]
∴ m∠x = 140°, m∠y = 110°

Question 3.
In the given figure, line p || line q, line l || line m. Find measures of ∠a, ∠b and ∠c, using the measures of given angles. Justify your answers. Solution:
i. line p || line q and line l is a transversal.
∴ m∠a + 80° = 180° …[Interior angles]
∴ m∠a= 180° – 80°
∴ m∠a= 100°

ii. line l || line m and line p is a transversal.
∴ m∠c = 80° …(i) [Exterior alternate angles]

iii. line p || line q and line m is a transversal.
∴ m∠b = m∠c … [Corresponding angles]
m∠b = 80° …[From (i)]
∴ m∠a = 100°, m∠b = 80°, m∠c = 80°

Question 4.
In the given figure, line a || line b, line l is a transversal. Find the measures of ∠x, ∠y, ∠z using the given information. Solution:
line a || line b and line l is a transversal.
∴ m∠x = 105° …(i) [Corresponding angles]

ii. m∠y = m∠x … [Vertically opposite angles]
∴ m∠y = 105° …[From (i)]

iii. m∠z + 105° = 180° …[Angles in a linear pair]
∴ m∠z = 180°- 105°
∴ m∠z = 75°
∴ m∠x = 105°, m∠y = 105°, m∠z = 75°

Question 5.
In the given figure, line p || line l || line q. Find ∠x with the help of the measures given in the figure. Solution: line p || line l and line IJ is a transversal.
m∠IJN = m∠JIH … [Alternate angles]
∴ m∠IJN = 40° …(i)
line l || line q and line MJ is a transversal.
m∠MJN = m∠JMK … [Alternate angles]
∴ m∠MJN = 30° …(ii)
Now, m∠x = m∠IJN + m∠MJN
= 40° + 30° …[From (i) and (ii)]
∴ m∠x = 70°

Maharashtra Board Class 8 Maths Chapter 2 Parallel Lines and Transversals Practice Set 2.2 Intext Questions and Activities

Question 1.
When two parallel lines are intersected by a transversal eight angles are formed. If the measure of one of these eight angles is given, can we find measures of remaining seven angles? (Textbook pg, no. 9)
Solution:
Yes, we can find the measures of the remaining angles. In the given figure, line m || line n and line l is a transversal.
m∠a = 60°(say) …(i)
i. m∠a + m∠b = 180° …[Angles in a linear pair]
∴ 60° + m∠b =180° … [From (i)]
∴ m∠b = 180° – 60°
∴ m∠b = 120° …(ii)

ii. m∠c = m∠b …[Vertically opposite angles]
∴ m∠c = 120° .. .(iii) [From (ii)]

iii. m∠d = m∠a …[Vertically opposite angles]
∴ m∠d = 60° …(iv) [From (i)]

iv. m∠e = m∠d …[Alternate angles]
∴ m∠e = 60° … [From (iv)]

v. m∠f = m∠c …[Alternate angles]
∴ m∠f = 120° …[From (iii)]

vi. m∠g = m∠d …[Corresponding angles]
∴ m∠g = 60° … [From (iv)]

vii. m∠h = m∠c … [Corresponding angles]
∴ m∠h = 120° …[From (iii)]

Question 2.
As shown in the figure (A), draw two parallel lines and their transversal on a paper. Draw a copy of the figure on another blank sheet using a trace paper, as shown in the figure (B). Colour part Land part II with different colours. Cut out the two parts with a pair of scissors. Place, part I and part II on each angle in the figure A and answer the following questions. (Textbook pg. no. 9)

1. Which angles coincide with part I?
2. Which angles coincide with part II? Solution:

1. ∠d, ∠f and ∠h coincide with part I.
2. ∠c, ∠e and ∠g coincide with part II.

## Maharashtra State Board Class 7 Maths Solutions Chapter 10 Banks and Simple Interest Practice Set 41

Question 1.
If the interest on Rs 1700 is Rs 340 for 2 years, the rate of interest must be__.
(A) 12%
(B) 15%
(C) 4%
(D) 10%
Solution:
(D) 10%

Hint:
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$340=\frac{1700 \times R \times 2}{100}$$
∴ R = 10% Question 2.
If the interest on Rs 3000 is Rs 600 at a certain rate for a certain number of years, what would the interest be on Rs 1500 under the same conditions?
(A) Rs 300
(B) Rs 1000
(C) Rs 700
(D) Rs 500
Solution:
(A) Rs 300

Hint:
The interest on Rs 3000 at certain rate of interest is Rs 600.
Let us suppose the interest on Rs 1500 at the same rate is x.
∴ $$\frac{600}{3000}=\frac{x}{1500}$$
∴ x = Rs 300 Question 3.
Javed deposited Rs 12000 at 9 p.c.p.a in a bank for some years, and withdrew his interest every year. At the end of the period, he had received altogether Rs 17,400. For how many years had he deposited his money?
Solution:
Here, P = Rs 12000, R = 9 p.c.p.a and amount = Rs 17400
Amount = Principal + Interest
∴17400 = 12000 + Interest
∴Interest = 17400 – 12000 = Rs 5400
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
$$5400=\frac{12000 \times 9 \times \mathrm{T}}{100}$$
∴ $$\frac{5400 \times 100}{12000 \times 9}=\mathrm{T}$$
∴ T = 5 years
∴ Javed had deposited the amount for 5 years.

Question 4.
Lataben borrowed some money from a bank at a rate of 10 p.c.p.a interest for $$2\frac { 1 }{ 2 }$$ years to start a cottage industry. If she paid Rs 10250 as total interest, how much money had she borrowed?
Solution:
Here, R = 10 p.c.p.a, T = 2.5 years, I = Rs 10250 ∴ P = Rs 41000
∴ Lataben had borrowed an amount of Rs 41000 from the bank.

Question 5.
Fill in the blanks in the table.

 Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years ii. 6% 4 years Rs 1200 iii. Rs 8000 5% Rs 800 iv. 5% Rs 6000 Rs 18000 v. $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400

Solution:
i. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
= $$\frac{4200 \times 7 \times 3}{100}$$
= Rs 882
Amount = Principal + interest
= 4200 + 882
= Rs 5082

ii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$1200=\frac{\mathrm{P} \times 6 \times 4}{100}$$
∴ $$\frac{1200 \times 100}{6 \times 4}=\mathrm{P}$$
∴ P = Rs 5000
Amount = Principal + interest
= 5000 + 1200
= Rs 6200

iii. $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$800=\frac{8000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{800 \times 100}{8000 \times 5}=\mathrm{T}$$
∴ T = 2 years
Amount = Principal + interest
= 8000 + 800
= Rs 8800

iv. Amount = Principal + interest
∴ 18000 = Principal + 6000
∴ Principal = Rs 12000
$$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$6000=\frac{12000 \times 5 \times \mathrm{T}}{100}$$
∴ $$\frac{6000 \times 100}{12000 \times 5}=\mathrm{T}$$
∴ T = 10 years

v. R = $$2\frac { 1 }{ 2 }$$ % = 2.5 %
∴ $$\text { Total interest }=\frac{P \times R \times T}{100}$$
∴ $$2400=\frac{\mathrm{P} \times 2.5 \times 5}{100}$$
∴ $$2400=\frac{P \times 25 \times 5}{100 \times 10}$$
∴ $$\frac{2400 \times 10 \times 100}{25 \times 5}=P$$
∴ P = Rs 19200
Amount = Principal + interest
= 19200 + 2400
= Rs 21600

 Principal Rate of interest (p.c.p.a.) Time Interest Amount i. Rs 4200 7% 3 years Rs 882 Rs 5082 ii. Rs 5000 6% 4 years Rs 1200 Rs 6200 iii. Rs 8000 5% 2 years Rs 800 Rs 8800 iv. Rs 12000 5% 10 years Rs 6000 Rs 18000 v. Rs 19200 $$2\frac { 1 }{ 2 }$$ % 2 5 years Rs 2400 Rs 21600

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 41 Intext Questions and Activities

Question 1.
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

Question 2.
Visit different banks and find out the rates of the interest they give for different types of accounts. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

Question 3.
With the help of your teachers, start a Savings Bank in your school and open an account in it to save up some money. (Textbook pg. no. 74)
Solution:
(Students should attempt the above activities with the help of their parent / teacher.)

## Maharashtra State Board Class 9 Maths Solutions Chapter 7 Co-ordinate Geometry Practice Set 7.1

Question 1.
State in which quadrant or on which axis do the following points lie.
i. A(-3, 2)
ii. B(-5, -2)
iii. K(3.5, 1.5)
iv. D(2, 10)
V. E(37, 35)
vi. F(15, -18)
vii. G(3, -7)
viii. H(0, -5)
ix. M(12, 0)
x. N(0, 9)
xi. P(0, 2.5)
xii. Q(-7, -3)
Solution: Question 2.
In which quadrant are the following points?
i. whose both co-ordinates are positive.
ii. whose both co-ordinates are negative.
iii. whose x co-ordinate is positive and the y co-ordinate is negative.
iv. whose x co-ordinate is negative and y co-ordinate is positive.
Solution:

Question 3.
Draw the co-ordinate system on a plane and plot the following points.
L(-2, 4), M(5, 6), N(-3, -4), P(2, -3), Q(6, -5), S(7, 0), T(0, -5)
Solution: Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.1 Intext Questions and Activities

Question 1.
Plot the points R(-3,-4), S(3,-l) on the same co-ordinate system. (Textbook pg. no. 93)
Steps for plotting the points:
i. Draw X-axis and Y-axis on the plane. Show the origin.
ii. Draw a line parallel to Y-axis at a distance of 3 units in the -ve direction of X-axis.
iii. Draw another line parallel to X-axis at a distance of 4 units in the -ve direction of Y-axis.
iv. Intersection of these lines is the point R (-3, -4).
v. The point S can be plotted in the same manner. ## Maharashtra State Board Class 8 Maths Solutions Chapter 15 Area Practice Set 15.5

Question 1.
Find the areas of given plots. (All measures are in meters.) Solution:
i. Here, ∆QAP, ∆RCS are right angled triangles and ☐QACR is a trapezium.
In ∆QAP, l(AP) = 30 m, l(QA) = 50 m
A(∆QAP)
= $$\frac { 1 }{ 2 }$$ x product of sides forming the right angle
= $$\frac { 1 }{ 2 }$$ x l(AP) x l(QA)
= $$\frac { 1 }{ 2 }$$ x 30 x 50
= 750 sq. m
In ☐QACR, l(QA) = 50 m, l(RC) = 25 m,
l(AC) = l(AB) + l(BC)
= 30 + 30 = 60 m
A(☐QACR)
= $$\frac { 1 }{ 2 }$$ x sum of lengths of parallel sides x height
= $$\frac { 1 }{ 2 }$$ x [l(QA) + l(RC)] x l(AC)
= $$\frac { 1 }{ 2 }$$ x (50 + 25) x 60
= $$\frac { 1 }{ 2 }$$ x 75 x 60
= 2250 sq.m
In ∆RCS, l(CS) = 60 m, l(RC) = 25 m A(∆RCS)
= $$\frac { 1 }{ 2 }$$ x product of sides forming the right angle
= $$\frac { 1 }{ 2 }$$ x l(CS) x l(RC)
= $$\frac { 1 }{ 2 }$$ x 60 x 25
= 750 sq. m
In ∆PTS, l(TB) = 30 m,
l(PS) = l(PA) + l(AB) + l(BC) + l(CS)
= 30 + 30 + 30 + 60
= 150m
A(∆PTS) = $$\frac { 1 }{ 2 }$$ x base x height
= $$\frac { 1 }{ 2 }$$ x l(PS) x l(TB)
= $$\frac { 1 }{ 2 }$$ x 150 x 30
= 2250 sq. m
∴ Area of plot QPTSR = A(∆QAP) + A(☐QACR) + A(∆RCS) + A(∆PTS)
= 750 + 2250 + 750 + 2250
= 6000 sq. m
∴ The area of the given plot is 6000 sq.m.

ii. In ∆ABE, m∠BAE = 90°, l(AB) = 24 m, l(BE) = 30 m
∴ [l(BE)]² = [l(AB)]² + [l(AE)]²
…[Pythagoras theorem]
∴ (30)² = (24)² + [l(AE)]²
∴ 900 = 576 + [l(AE)]²
∴ [l(AE)]² = 900 – 576
∴ [l(AE)]² = 324
∴ l(AE) = √324 = 18 m
…[Taking square root of both sides]
A(∆ABE)
= $$\frac { 1 }{ 2 }$$ x product of sides forming the right angle
= $$\frac { 1 }{ 2 }$$ x l(AE) x l(AB)
= $$\frac { 1 }{ 2 }$$ x 18 x 24
= 216 sq. m
In ∆BCE, a = 30m, b = 28m, c = 26m ∴ Area of plot ABCDE
= A(∆ABE) + A(∆BCE) + A(∆EDC)
= 216 + 336 + 224
= 776 sq. m
∴ The area of the given plot is 776 sq.m.
[Note: In the given figure, we have taken l(DF) = 16 m]

## Maharashtra State Board Class 8 Maths Solutions Chapter 11 Statistics Practice Set 11.1

Question 1.
The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.

 No. of saplings (Scores) xi No. of students (frequency) fi fi × xi 1 4 4 2 6 __ 3 12 __ 4 8 __ N = __ ∑ fi × xi = __ Solution:

 No. of saplings (Scores) xi No. of students (frequency) fi fi × xi 1 4 4 2 6 12 3 12 36 4 8 32 N = __ ∑ fi × xi = 84 Question 2.
The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.

 Electricity used (Units) xi No. of families (frequency) fi fi × xi 30 4 45 6 60 12 75 8 90 3 N = __ ∑ fi × xi =

i. How many families use 45 units electricity?
ii. State the score, the frequency of which is 5.
iii. Find N, and ∑ fi × xi .
iv. Find the mean of electricity used by each family in the month of May.
Solution:

 Electricity used (Units) xi No. of families (frequency) fi fi × xi 30 7 210 45 2 90 60 8 480 75 5 375 90 3 270 N = 25 ∑ fi × xi = 1425

i. 2 families used 45 units of electricity.
ii. The score for which the frequency is 5 is 75
iii. N = 25 and ∑ fi × xi = 1425
iv. The mean of electricity used by each.

Question 3.
The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2,3, 5, 6, 4, 2. Prepare a frequency table and And the mean of members of 40 families.
Solution:  ∴ The mean of the members of 40 families is 3.9.

Question 4.
The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is:
2, 3 ,4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:  ∴ The mean of the given data is 2.75.

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.1 Intext Questions and Activities

Question 1.
The number of pages of a book Ninad read for five consecutive days were 60, 50, 54, 46, 50. Find the average number of pages he read everyday. (Textbook pg. no. 67)
Solution:
$$\frac{60++++50}{}=\frac{260}{}=$$
∴ Average number of pages read daily is 52

## Maharashtra State Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.2

Compound Interest class 8 practice set 14.2 Question 1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Solution:
Here, P = Initial number of workers = 320
R = Increase in the number of workers per year = 25%
N = 2 years
A = Number of workers after 2 years ∴ The number of workers after 2 years would be 500.

Question 2.
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Solution:
Here, P = Present number of sheeps = 200
R = Increase in number of sheeps per year = 8%
N = 3 years
A = Number of sheeps after 3 years = $$\frac{0.32}{25} \times 27 \times 27 \times 27$$
= 0.0128 × 27 × 27 × 27
= 251.9424
= 252
∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).

8th Class Math Practice Set 14.2 Question 3.
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Solution:
Here, P = Present number of trees in the forest = 40,000
R = Increase in the number of trees per year = 5%
N = 3 years
A = Number of trees after 3 years = 5 × 21 × 21 × 21
= 5 × 9261
= 46,305
∴ The expected number of trees in the forest after 3 years is 46,305.

Std 8 Maths Practice Set 14.2 Question 4.
The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Solution:
Here, P = Cost price of machine = Rs 2,50,000
R = Rate of depreciation per year = 10%
N = 2 years
A = Depreciated price of the machine after 2 years = 2,500 × 81
= Rs 2,02,500
Depreciation in price = Cost price (P) – Depreciated price (A)
= 2,50,000 – 2,02,500
= Rs 47,500
∴ The depreciation in price of the machine after 2 years would be Rs 47,500.

Question 5.
Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.
Solution:
Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years
i. $$\mathbf{A}=\mathbf{P}\left[1+\frac{\mathbf{R}}{100}\right]^{N}$$ ii. Interest = Amount (A) – Principal (P)
= 4036.80 – 3000
= Rs 1036.80
∴ The compound interest after 2 years would be Rs 1036.80.

Question 6.
A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Solution:
Here, P = Rs 15,000, R = 12 p.c.p.a, and
N = 3 years ∴ The amount required to settle the loan after 3 years is Rs 21,073.92.

Practice Set 14.2 Class 8 Question 7.
A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Solution:
Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years ∴ P = 4 x 50 x 50
∴ P = Rs 10,000
∴ The principal is Rs 10,000.

Question 8.
The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.
Solution:
Here, P = Population of a suburb = 16,000
N = 2 years
A = Increase in the population after 2 years = 17,640
R = Rate of increase in population  ∴5 = R
i.e., R = 5%
∴The rate of increase in the population is 5 p.c.p.a.

Compound Interest Practice Set 14.2 Question 9.
In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.
Solution:
Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847 ∴Rs 700 will amount to Rs 847 in 2 years.

Practice Set 14.2 Question 10.
Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.
Solution:
Here, P = Rs 20,000, R = 8 p.c.p.a.,
N = 2 years
i. Simple interest (I) Simple interest (I) = Rs 3200

ii. Compound Interest (I):  = 32 × 27 × 27
= Rs 23,328
Compound interest (I)
= Amount (A) – Principal (P)
= 23,328 – 20,000
= Rs 3328 ,..(ii)

iii. Difference
= Compound interest – Simple interest
= 3328 – 3200 … [Form (i) and (ii)]
= Rs 128
∴ The difference between compound interest and simple interest is Rs 128.
[Note: The question is modified as per the answer given in the textbook.]

Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities

8th Standard Maths Practice Set 14.2 Question 1.
Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90)
Solution:
(Students should attempt this activity at their own.)

## Maharashtra State Board Class 8 Maths Solutions Chapter 11 Statistics Practice Set 11.2

practice set 11.2 8th class Question 1.
Observe the following graph and answer the questions. i. State the type of the graph.
ii. How much is the savings of Vaishali in the month of April?
iii. How much is the total of savings of Saroj in the months March and April?
iv. How much more is the total savings of Savita than the total savings of Megha?
v. Whose savings in the month of April is the least?
Solution:
i. The given graph is a subdivided bar graph.
ii. Vaishali’s savings in the month of April is Rs 600.
iii. Total savings of Saroj in the months of March and April is Rs 800.
iv. Savita’s total saving = Rs 1000, Megha’s total saving = Rs 500
∴ difference in their savings = 1000 – 500 = Rs 500.
Savita’s saving is Rs 500 more than Megha.
v. Megha’s savings in the month of April is the least.

practice set 11.2 Question 2.
The number of boys and girls, in std 5 to std 8 in a Z.P. School is given in the table. Draw a subdivided bar graph to show the data. (Scale : On Y axis, 1cm = 10 students)

 Standard 5th 6th 7th 8th Boys 34 26 21 25 Girls 17 14 14 20

Solution:

 Standard 5th 6th 7th 8th Boys 34 26 21 25 Girls 17 14 14 20 Total 51 40 35 45 Statistics class 8 practice set 11.1 Question 3.
In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

 Year\Town karjat Wadgaon Shivapur Khandala 2016 150 250 200 100 2017 200 300 250 150

Solution:

 Year\Town karjat Wadgaon Shivapur Khandala 2016 150 250 200 100 2017 200 300 250 150 Total 350 550 450 250 Statistics class 8 Question 4.
In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)

 Means of commutation\Town Paithan Yeola Shahapur Cycle 3250 1500 1250 Bus and auto 750 500 500 On foot 1000 1000 500

Solution:

 Means of commutation\Town Paithan Yeola Shahapur Cycle 3250 1500 1250 Bus and auto 750 500 500 On foot 1000 1000 500 Total 5000 3000 2250 ## Maharashtra State Board Class 8 Maths Solutions Chapter 10 Division of Polynomials Practice Set 10.2

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y2 + 10y + 24) ÷ (y + 4)
ii. (p2 + 7p – 5) ÷ (p + 3)
iii. (3x + 2x2 + 4x3) ÷ (x – 4)
iv. (2m3 + m2 + m + 9) ÷ (2m – 1)
v. (3x – 3x2 – 12 + x4 + x3) ÷ (2 + x2)
vi. (a4 – a3 + a2 – a + 1) ÷ (a3 – 2)
vii. (4x4 – 5x3 – 7x + 1) ÷ (4x – 1)
Solution:
i. (y2 + 10y + 24) ÷ (y + 4) ∴ Quotient = y + 6
Remainder = 0

ii. (p2 + 7p – 5) ÷ (p + 3) ∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x2 + 4x3) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x ∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m3 + m2 + m + 9) ÷ (2m – 1) ∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x2 – 12 + x4 + x3) ÷ (2 + x2)
Write the dividend in descending order of their indices.
(x4 + x3 – 3x2 + 3x – 12) ÷ (x2 + 2) ∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a4 – a3 + a2 – a + 1) ÷ (a3 – 2) ∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x4 – 5x3 – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x4 – 5x3 – 7x + 1) = (4x4 – 5x3 + 0x2 – 7x + 1) ∴ Quotient = $$x^{3}-x^{2}-\frac{x}{4}-\frac{29}{16}$$
Remainder = $$\frac { -13 }{ 16 }$$