## Maharashtra Board Practice Set 40 Class 7 Maths Solutions Chapter 10 Bank and Simple Interest

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 40 Answers Solutions Chapter 10 Bank and Simple Interest.

## Bank and Simple Interest Class 7 Practice Set 40 Answers Solutions Chapter 10

Question 1.
If Rihanna deposits Rs 1500 in the school fund at 9 p.c.p.a for 2 years, what is the total amount she will get?
Solution:
Here, P = Rs 1500, R = 9 p.c.p.a , T = 2 years
∴ Total interest = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
= $$\frac{1500 \times 9 \times 2}{100}$$
= 1500 x 9 x 2
= Rs 270
∴ Total amount = Principal + Interest
= 1500 + 270
= Rs 1770
∴ Rihanna will get a total amount of Rs 1770.

Question 2.
Jethalal took a housing loan of Rs 2,50,000 from a bank at 10 p.c.p.a. for 5 years. What is the yearly interest he must pay and the total amount he returns to the bank?
Solution:
Here, P = Rs 250000, R = 10 p.c.p.a., T = 5 years
∴ Total interest = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
= $$\frac{250000 \times 10 \times 5}{100}$$
= 2500 x 10 x 5
= Rs 1,25,000
∴ Yearly interest = Total interest ÷ Time = 1,25,000 ÷ 5 = Rs 25000
Total amount to be returned = Principal + Total interest
= 250000 + 125000 = Rs 375000
∴ The yearly interest is Rs 25,000 and Jethalal will have to return Rs 3,75,000 to the bank.

Question 3.
Shrikant deposited Rs 85,000 for $$2\frac { 1 }{ 2 }$$ years at 7 p.c.p.a. in a savings bank account. What is the total
interest he received at the end of the period?
Solution:
Here, P = Rs 85000, R = 7 p.c.p.a., T = $$2\frac { 1 }{ 2 }$$ years = 2.5 years
∴ Total interest = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
= $$\frac{85000 \times 7 \times 2.5}{100}$$
= $$\frac{85000 \times 7 \times 25}{100 \times 10}$$
= 85 x 7 x 25
= Rs 14875
∴ The total interest received by Shrikant at the end of the period is Rs 14875.

Question 4.
At a certain rate of interest, the interest after 4 years on Rs 5000 principal is Rs 1200. What would be the interest on Rs 15000 at the same rate of interest for the same period?
Solution:
The interest on Rs 5000 after 4 years is Rs 1200.
Let us suppose the interest on Rs 15000 at the same rate after 4 years is Rs x.
Taking the ratio of interest and principal, we get
∴ $$\frac{x}{15000}=\frac{1200}{5000}$$
∴ $$x=\frac{1200 \times 15000}{5000}$$
= Rs 3600
∴ The interest received on Rs 15000 is Rs 3600.

Question 5.
If Pankaj deposits Rs 1,50,000 in a bank at 10 p.c.p.a. for two years, what is the total amount he will get from the bank?
Solution:
Here, P = 150000, R = 10 p.c.p.a., T = 2 years
∴ Total interest = $$\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$$
= $$\frac{150000 \times 10 \times 2}{100}$$
= Rs 30000
∴ Total amount = Principal + Total Interest
= 150000 + 30000
= Rs 180000
∴ Pankaj will receive Rs 180000 from the bank.

Maharashtra Board Class 7 Maths Chapter 10 Banks and Simple Interest Practice Set 40 Intext Questions and Activities

Question 1.
Observe the entries made in the page of a passbook shown below and answer the following questions. (Textbook pg. no. 70)

1. On 2.2.16 the amount deposited was Rs__and the balance Rs__.
2. On 12.2.16, Rs__were withdrawn by cheque no. 243965. The balance was Rs__
3. On 26.2.2016 the bank paid an interest of Rs__

Solution:

1. 1500, 7000
2. 3000, 9000
3. 135

Practice Set 40 Class 7 Question 2.
Suvidya borrowed a sum of Rs 30000 at 8 p.c.p.a. interest for a year from her bank to buy a computer. At the end of the period, she had to pay back an amount of Rs 2400 over and above what she had borrowed.
Based on this information fill in the blanks below. (Textbook pg. no. 70)

1. Principal = Rs__
2. Rate of interest =__%
3. Interest = Rs__
4. Time =__year.
5. The total amount returned to the bank = 30,000 + 2,400 = Rs__

Solution:

1. 30000
2. 8
3. 2400
4. 1
5. Rs 32400

## Maharashtra Board Practice Set 12 Class 7 Maths Solutions Chapter 3 HCF and LCM

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 12 Answers Solutions Chapter 3 HCF and LCM.

## HCF and LCM Class 7 Practice Set 12 Answers Solutions Chapter 3

Question 1.
i. 25, 40
ii. 56, 32
iii. 40, 60, 75
iv. 16, 27
v. 18, 32,48
vi. 105, 154
vii. 42, 45, 48
viii. 57, 75, 102
ix. 56, 57
x. 777, 315, 588
Solution:
i. 25, 40

∴ 25 = 5 × 5
40 = 2 × 2 × 2 × 5
∴ HCF of 25 and 40 = 5

ii. 56, 32

∴ 56 = 2 × 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
∴ HCF of 56 and 32 = 2 × 2 × 2
∴ HCF of 56 and 32 = 8

iii. 40, 60, 75

∴ 40 = 2 × 2 × 2 × 5
60 = 2 × 2 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 40, 60 and 75 = 5

iv. 16, 27

∴ 16 = 2 × 2 × 2 × 2 × 1
27 = 3 × 3 × 3 × 1
∴ HCF of 16 and 27 = 1

v. 18, 32,48

∴ 18 = 2 × 3 × 3
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18, 32 and 48 = 2

vi. 105, 154

∴ 105 = 3 × 5 × 7
154 = 2 × 2 × 11
∴ HCF of 105 and 154 = 7

vii. 42, 45, 48

∴ 42 = 2 × 3 × 7
45 = 3 × 3 × 5
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 42,45 and 48 =3

viii. 57, 75, 102

∴ 57 = 3 × 19
75 = 3 × 5 × 5
102 = 2 × 3 × 17
∴ HCF of 57, 75 and 102 = 3

ix. 56, 57

∴ 56 = 2 × 2 × 2 × 7 × 1
57 = 3 × 19 × 1
∴ HCF of 56 and 57 = 1

x. 777, 315, 588

∴ 777 = 3 × 7 × 37
315 = 3 × 3 × 5 × 7
588 = 2 × 2 × 3 × 7 × 7
∴ HCF of 777, 315 and 588 = 3 × 7
HCF of 777, 315 and 588 = 21

Question 2.
Find the HCF by the division method and reduce to the simplest form:
i. $$\frac { 275 }{ 525 }$$
ii. $$\frac { 76 }{ 133 }$$
iii. $$\frac { 161 }{ 69 }$$
Solution:
i. $$\frac { 275 }{ 525 }$$

ii. $$\frac { 76 }{ 133 }$$

iii. $$\frac { 161 }{ 69 }$$

Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 12 Intext Questions and Activities

Question 1.
In each of the following examples, write all the factors of the numbers and find the greatest common divisor. (Textbook pg. no. 17)
i. 28, 42
ii. 51, 27
iii. 25, 15, 35
Solution:
i. Factors of 28 = 1,2,4, 7, 14, 28
Factors of 42 = 1,2, 3, 6, 7, 14, 21, 42
∴ HCF of 28 and 42 = 14

ii. Factors of 51 = 1, 3, 17, 51
Factors of 27 = 1, 3, 9, 27
∴ HCF of 51 and 27 = 3

iii. Factors of 25 = 1, 5, 25
Factors of 15 = 1, 3, 5, 15
Factors of 35 = 1, 5, 7, 35
∴ HCF of 25, 15 and 35 = 5

## Maharashtra Board Practice Set 1 Class 7 Maths Solutions Chapter 1 Geometrical Constructions

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

## Geometrical Constructions Class 7 Practice Set 1 Answers Solutions Chapter 1

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.

Line AB is the perpendicular bisector of seg PQ.

ii.

Line UV is the perpendicular bisector of seg ST.

iii.

Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°

ii. 55°

iii. 90°

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:

The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:

The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
2. Is l(PM) = l(SM)?

Solution:

1. Here, m∠CMS = 90°
2. Also, l(PM) = l(SM) = 2cm
∴ line CD is the perpendicular bisector of seg PS.

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

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 31 Answers Solutions Chapter 7 Joint Bar Graph.

## Joint Bar Graph Class 7 Practice Set 31 Answers Solutions Chapter 7

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 Board Practice Set 30 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 30 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 30 Answers Solutions Chapter 6

Question 1.
Find the square root:
i. 625
ii. 1225
iii. 289
iv. 4096
v. 1089
Solution:
i. 625

∴ 625 = 5 x 5 x 5 x 5
∴ √625 = 5 x 5 = 25

ii. 1225

∴ 1225 = 5 x 5 x 7 x 7
∴ √1225 = 5 x 7 = 35

iii. 289

∴ 289 = 17 x 17
∴ √289 = 17

iv. 4096

∴ 4096 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
∴ √4096 = 2 x 2 x 2 x 2 x 2 x 2
= 64

v. 1089

∴ 1089 = 3 x 3 x 11 x 11
∴ √1089 = 3 x 11
= 33

Maharashtra Board Class 7 Maths Chapter 6 Indices Practice Set 30 Intext Questions and Activities

Question 1.
Try to write the following numbers in the standard form. (Textbook pg. no. 48)
i. The diameter of Sun is 1400000000 m.
ii. The velocity of light is 300000000 m/sec.
Solution:
i. 1400000000 m = 1.4 x 109 m
ii. 300000000 m/s = 3.0 x 108 m/sec.

Question 2.
The box alongside shows the number called Googol. Try to write it as a power of 10. (Textbook pg. no. 48)

Solution:
1 x 10100

## Maharashtra Board Practice Set 29 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 29 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 29 Answers Solutions Chapter 6

Question 1.
Simplify:
i. $$\left[\left(\frac{15}{12}\right)^{3}\right]^{4}$$
ii. (34)-2
iii. $$\left[\left(\frac{1}{7}\right)^{-3}\right]^{4}$$
iv. $$\left[\left(\frac{2}{5}\right)^{-2}\right]^{-3}$$
v. (65)4
vi. $$\left[\left(\frac{6}{7}\right)^{5}\right]^{2}$$
vii. $$\left[\left(\frac{2}{3}\right)^{-4}\right]^{5}$$
viii. $$\left[\left(\frac{5}{8}\right)^{3}\right]^{-2}$$
ix. $$\left[\left(\frac{3}{4}\right)^{6}\right]^{7}$$
x. $$\left[\left(\frac{2}{5}\right)^{-3}\right]^{2}$$
Solution:
i. $$\left[\left(\frac{15}{12}\right)^{3}\right]^{4}$$
$$=\left(\frac{15}{12}\right)^{3 \times 4}=\left(\frac{15}{12}\right)^{12}$$

ii. (34)-2
= 34×(-2)
= 3-8

iii. $$\left[\left(\frac{1}{7}\right)^{-3}\right]^{4}$$
$$=\left(\frac{1}{7}\right)^{(-3) \times 4}=\left(\frac{1}{7}\right)^{-12}$$

iv. $$\left[\left(\frac{2}{5}\right)^{-2}\right]^{-3}$$
$$=\left(\frac{2}{5}\right)^{(-2) \times(-3)}=\left(\frac{2}{5}\right)^{6}$$

v. (65)4
= 65×4
= 620

vi. $$\left[\left(\frac{6}{7}\right)^{5}\right]^{2}$$
$$=\left(\frac{6}{7}\right)^{5 \times 2}=\left(\frac{6}{7}\right)^{10}$$

vii. $$\left[\left(\frac{2}{3}\right)^{-4}\right]^{5}$$
$$=\left(\frac{2}{3}\right)^{(-4) \times 5}=\left(\frac{2}{3}\right)^{-20}$$

viii. $$\left[\left(\frac{5}{8}\right)^{3}\right]^{-2}$$
$$=\left(\frac{5}{8}\right)^{3 \times(-2)}=\left(\frac{5}{8}\right)^{-6}$$

ix. $$\left[\left(\frac{3}{4}\right)^{6}\right]^{7}$$
$$=\left(\frac{3}{4}\right)^{6 \times 1}=\left(\frac{3}{4}\right)^{6}$$

x. $$\left[\left(\frac{2}{5}\right)^{-3}\right]^{2}$$
$$=\left(\frac{2}{5}\right)^{(-3) \times 2}=\left(\frac{2}{5}\right)^{-6}$$

Question 2.
Write the following numbers using positive indices:
i. $$\left(\frac{2}{7}\right)^{-2}$$
ii. $$\left(\frac{11}{3}\right)^{-5}$$
iii. $$\left(\frac{1}{6}\right)^{-3}$$
iv. $$(y)^{-4}$$
Solution:
i. $$\left(\frac{7}{2}\right)^{2}$$
ii. $$\left(\frac{3}{11}\right)^{5}$$
iii. $$6^{3}$$
iv. $$\frac{1}{y^{4}}$$

## Maharashtra Board Practice Set 28 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 28 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 28 Answers Solutions Chapter 6

Question 1.
Simplify:
i. a6 ÷ a4
ii. m5 ÷ m8
iii. p3 ÷ p13
iv. x10 ÷ x10
Solution:
i. a6 ÷ a4
= a6-4
= a2

ii. m5 ÷ m8
= m5-8
= m-3

iii. p3 ÷ p13
= p3-13
= p-10

iv. x10 ÷ x10
= x10-10
= x0
= 1

Question 2.
Find the value of:
i. (-7)12 ÷ (-7)12
ii. 75 ÷ 73
iii. $$\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}$$
iv. 47 ÷ 45
Solution:
i. (-7)12 ÷ (-7)12
= (-7)12-12
= (-7)0
= 1

ii. 75 ÷ 73
= 75-3
= 72
= 49

iii. $$\left(\frac{4}{5}\right)^{3} \div\left(\frac{4}{5}\right)^{2}$$
$$=\left(\frac{4}{5}\right)^{3-2}=\frac{4}{5}$$

iv. 4 ÷ 4
= 47-5
= 42
= 16

## Maharashtra Board Practice Set 27 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 27 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 27 Answers Solutions Chapter 6

Question 1.
Simplify:
i. 74 × 72
ii. (-11)5 × (-11)2
iii. $$\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}$$
iv. $$\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}$$
v. a16 × a7
vi. $$\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}$$
Solution:
i. 74 × 72
= 74+2
= 76

ii. (-11)5 × (-11)2
= (-11)5+2
= (-11)7

iii. $$\left(\frac{6}{7}\right)^{3} \times\left(\frac{6}{7}\right)^{5}$$
$$=\left(\frac{6}{7}\right)^{3+5}=\left(\frac{6}{7}\right)^{8}$$

iv. $$\left(-\frac{3}{2}\right)^{5} \times\left(-\frac{3}{2}\right)^{3}$$
$$=\left(-\frac{3}{2}\right)^{5+3}=\left(-\frac{3}{2}\right)^{8}$$

v. a16 × a7
= a16+7
= a23

vi. $$\left(\frac{\mathrm{P}}{5}\right)^{3} \times\left(\frac{\mathrm{P}}{5}\right)^{7}$$
$$=\left(\frac{\mathrm{P}}{5}\right)^{3+7}=\left(\frac{\mathrm{P}}{5}\right)^{10}$$

## Maharashtra Board Practice Set 26 Class 7 Maths Solutions Chapter 6 Indices

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 26 Answers Solutions Chapter 6 Indices.

## Indices Class 7 Practice Set 26 Answers Solutions Chapter 6

Question 1.
Complete the table below:

 Sr. No. Indices (Numbers in index form) Base Index Multiplication form Value i. 34 3 4 3 x 3 x 3 x 3 81 ii. 163 iii. (-8) 2 iv. $$\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}$$ $$\frac { 81 }{ 2401 }$$ v. (-13)4

Solution:

 Sr. No. Indices (Numbers in index form) Base Index Multiplication form Value i. 34 3 4 3 x 3 x 3 x 3 81 ii. 163 16 3 16 x 16 x 16 4096 iii. (-8)² (-8) 2 -8 x -8 64 iv. $$\left(\frac{3}{7}\right)^{4}$$ $$\frac { 7 }{ 7 }$$ 4 $$\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}$$ $$\frac { 81 }{ 2401 }$$ v. (-13)4 -13 4 (-13) x (-13) x (-13) x (-13) 28561

Question 2.
Find the value of.
i. 210
ii. 53
iii. (-7)4
iv. (-6)3
v. 93
vi. 81
vii. $$\left(\frac{4}{5}\right)^{3}$$
viii. $$\left(-\frac{1}{2}\right)^{4}$$
Solution:
i. 210
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024

ii. 53
= 5 × 5 × 5
= 125

iii. (-7)4
= (-7) × (-7) × (-7) × (-7)
= 2401

iv. (-6)3
= (-6) × (-6) × (-6)
= -216

v. 93
= 9 × 9 × 9
= 729

vi. 81
= 8

vii. $$\left(\frac{4}{5}\right)^{3}$$
$$=\frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}=\frac{64}{125}$$

viii. $$\left(-\frac{1}{2}\right)^{4}$$
$$=\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)=\frac{1}{16}$$

## Maharashtra Board Practice Set 25 Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 25 Answers Solutions Chapter 5 Operations on Rational Numbers.

## Operations on Rational Numbers Class 7 Practice Set 25 Answers Solutions Chapter 5

Question 1.
Simplify the following expressions.
i. 50 x 5 ÷ 2 + 24
ii. (13 x 4) ÷ 2 – 26
iii. 140 ÷ [(-11) x (-3) – (-42) ÷ 14 – 1)]
iv. {(220 – 140) + [10 x 9 + (-2 x 5) ]} – 100
v. $$\frac{3}{5}+\frac{3}{8} \div \frac{6}{4}$$
Solution:
i. 50 x 5 ÷ 2 + 24 = 250 ÷ 2 + 24
= 125 + 24
= 149

ii. (13 x 4) = 2 – 26
= 52 ÷ 2 – 26
= 26 – 26
= 0

iii. 140 ÷ [(-11) x (-3) – (-42) ÷ 14 – 1)]
= 140 ÷ [33 + 42 ÷ 14 – 1]
= 140 ÷ [33 + 3 – 1]
= 140 ÷ 35
= 4

iv. {(220 – 140) + [10 x 9 + (-2 x 5) ]} – 100
= {80 + [90 – 10]} – 100
= {80 + 80} – 100
= 160 – 100
= 60

v. $$\frac{3}{5}+\frac{3}{8} \div \frac{6}{4}$$
$$=\frac{3}{5}+\frac{3}{8} \times \frac{4}{6}$$
$$=\frac{3}{5}+\frac{1}{4}$$
$$=\frac{12}{20}+\frac{5}{20}=\frac{12+5}{20}=\frac{17}{20}$$

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 25 Intext Questions and Activities

Question 1.
Use the signs and numbers in the boxes and form an expression such that its value will be 112. (Textbook pg. no. 42)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[+ x ÷ -]
Solution:
{3 + (6 x 7) + (9 ÷ 3)} + {- 8 + 8 x 9}
Note: The above problem has many solutions. Students may write solution other than the one given.