Practice Set 40 Class 7 Answers Chapter 10 Bank and Simple Interest Maharashtra Board

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

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

Std 7 Maths Practice Set 40 Solutions Answers

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)

Maharashtra Board Class 7 Maths Solutions Chapter 10 Banks and Simple Interest Practice Set 40 1

  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

Class 7 Maths Solution Maharashtra Board

Practice Set 38 Class 7 Answers Chapter 9 Direct Proportion and Inverse Proportion Maharashtra Board

Direct Proportion and Inverse Proportion Class 7 Maths Chapter 9 Practice Set 38 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 38 Answers Solutions Chapter 9 Direct Proportion and Inverse Proportion.

Std 7 Maths Practice Set 38 Solutions Answers

Question 1.
Five workers take 12 days to weed a field. How many days would 6 workers take? How many would 15 take?
Solution:
Let 6 workers take x days and 15 workers take y days to weed the field.
The number of workers and the time required to weed the field are in inverse proportion.
∴ 6 × x = 5 × 12
∴ \(x=\frac{5 \times 12}{6}\)
∴ x = 10 days
Also, 15 × y = 5 × 12
∴ \(y=\frac{5 \times 12}{15}\)
= 4 days
∴ 6 workers will take 10 days and 15 workers will take 4 days to weed the field.

Question 2.
Mohanrao took 10 days to finish a book, reading 40 pages every day. How many pages must he read in a day to finish it in 8 days?
Solution:
Let Mohanrao read x pages every day to finish the book in 8 days.
The number of pages read per day and the days required to finish the book are in inverse proportion.
∴ 8 × x = 10 × 40
∴ \(x=\frac{10 \times 40}{8}\)
= 50
∴ Mohanrao will have to read 50 pages every day to finish the book in 8 days.

Question 3.
Mary cycles at 6 km per hour. How long will she take to reach her Aunt’s house which is 12 km away? If she cycles at a speed of 4 km/hr, how long would she take?
Solution:
Speed of the cycle = 6 km / hr
Distance travelled to reach her Aunt’s house = 12 km
∴ \(\text { Time required }=\frac{\text { Distance travelled }}{\text { Speed }}\)
= \(\frac { 12 }{ 6 }\)
= 2 hours
Let the time required when the speed of the cycle is 4 km/hr be x hours.
The speed of the cycle and the time required to travel the same distance are in inverse proportion.
∴ 4 × x = 6 × 2
∴ \(x=\frac{6 \times 2}{4}\) = 3 hours
∴ Mary will require 2 hours if she is cycling at 6 km/hr and 3 hours if she is cycling at 4 km/hr to reach her Aunt’s house.

Question 4.
The stock of grain in a government warehouse lasts 30 days for 4000 people. How many days will it last for 6000 people?
Solution:
Let the stock of grain last for x days for 6000 people.
The number of people and the days for which stock will last are in inverse proportion.
∴ 6000 × x = 4000 × 30
∴ \(x=\frac{4000 \times 30}{6000}=20\)
∴ The stock of grain will last for 20 days for 6000 people.

Maharashtra Board Class 7 Maths Chapter 9 Direct Proportion and Inverse Proportion Practice Set 38 Intext Questions and Activities

Question 1.
Students of a certain school went for a picnic to a farm by bus. Here are some of their experiences. Say whether the quantities in each are in direct or in inverse proportion.
(Textbook pg. no. 65 and 66)
i. Each student paid Rs 60 for the expenses.
As there were 45 students,___rupees were collected.
Had there been 50 students,___rupees would have been collected.
The number of students and money collected are in___proportion.

ii. The sweets shop near the school gave 90 laddoos for the picnic.
If 45 students go for the picnic, each will get___laddoos.
If 30 students go for the picnic, each will get___laddoos.
The number of students and that of laddoos each one gets are in___proportion.

iii. The farm is 120 km away from the school.
The bus went to the farm at a speed of 40 km per hour and took___hours.
On the return trip, the speed was 60 km per hour. Therefore, it took___hours.
The speed of the bus and the time it takes are in___proportion.

iv. The farmer picked 180 bors from his trees.
He gave them equally to 45 students. Each student got___bors.
Had there been 60 students, each would have got___bors.
The number of students and the number of bors each one gets are in___proportion.
Solution:
i. Rs 2700, Rs 3000, direct
ii. 2,3, inverse
iii. 3,2, inverse
iv. 4,3, inverse

Class 7 Maths Solution Maharashtra Board

Practice Set 39 Class 7 Answers Chapter 9 Direct Proportion and Inverse Proportion Maharashtra Board

Direct Proportion and Inverse Proportion Class 7 Maths Chapter 9 Practice Set 39 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 39 Answers Solutions Chapter 9 Direct Proportion and Inverse Proportion.

Std 7 Maths Practice Set 39 Solutions Answers

Question 1.
Suresh and Ramesh together invested Rs 144000 in the ratio 4 : 5 and bought a plot of land. After some years they sold it at a profit of 20%. What is the profit each of them got?
Solution:
Total investment = Rs 144000
Profit earned = 20%
∴ Total profit = 20% of 144000 = \(\frac{20}{100} \times 144000\) = Rs 28800
Proportion of investment of Suresh and Ramesh = 4:5
Let the profit of Suresh be Rs 4x and that of Ramesh be Rs 5x.
4x + 5x = 28800
∴ 9x = 28800
∴ \(x=\frac { 28800 }{ 9 }\)
= 3200
∴ Suresh’s profit = 4x = 4 × 3200 = Rs 12800
Ramesh’s profit = 5x = 5 × 3200 = Rs 16000
∴ The profit earned by Suresh and Ramesh are Rs 12800 and Rs 16000 respectively.

Question 2.
Virat and Samrat together invested Rs 50000 and Rs 120000 to start a business. They suffered a loss of 20%. How much loss did each of them incur?
Solution:
Total investment = Rs 50000 + Rs 120000 = Rs 170000
Loss incurred = 20%
∴ Total loss = 20% of 170000 = \(\frac{20}{100} \times 170000\) = Rs 34000
Proportion of investment = 50000 : 120000
= 5 : 12 …. (Dividingby 10000)
Let the loss incurred by Virat be Rs 5x and that by Samrat be Rs 12x.
5x + 12x = 34000
∴ 17x = 34000
∴ \(x=\frac { 34000 }{ 17 }=2000\)
∴ Virat’s loss = 5x = 5 × 2000 = Rs 10000
Samrat’s loss = 12x = 12 × 2000 = Rs 24000
∴ The loss incurred by Virat and Samrat are Rs 10000 and Rs 24000 respectively.

Question 3.
Shweta, Piyush and Nachiket together invested Rs 80000 and started a business of selling sheets and towels from Solapur. Shweta’s share of the capital was Rs 30000 and Piyush’s Rs 12000. At the end of the year they had made a profit of 24%. What was Nachiket’s investment and what was his share of the profit?
Solution:
Total investment = Rs 80000
Nachiket’s investment = Total investment – (Shweta’s investment + Piyush’s investment)
= 80000 – (30000+ 12000)
= 80000 – 42000 = Rs 38000
Profit earned = 24%
∴ Total profit = 24% of 80000 = \(\frac { 24 }{ 100 }\) x 80000 = Rs 19200
Proportion of investment = 30000 : 12000 : 38000
= 15 : 6 : 19 …. (Dividing by 2000)
Let the profit of Shweta, Piyush and Nachiket be Rs 15x, Rs 6x and Rs 19x respectively.
15x + 6x + 19x = 19200
∴ 40x = 19200
∴ \(x=\frac { 19200 }{ 40 }=480\)
∴ Nachiket’s profit = 19x = 19 × 480 = Rs 9120
∴ Nachiket’s investment is Rs 38000 and his profit is Rs 9120.

Question 4.
A and B shared a profit of Rs 24500 in the proportion 3 : 7. Each of them gave 2% of his share of the profit to the Soldiers’ Welfare Fund. What was the actual amount given to the Fund by each of them?
Solution:
Proportion of share = 3:7
Let the profits of A and B be Rs 3x and Rs 7x respectively.
3x + 7x = 24500
∴ 10x = 24500
∴ \(x=\frac { 24500 }{ 10 }=2450\)
Profit earned by A = 3x = 3 × 2450 = Rs 7350
Amount given by A = 2% of his profit
= \(\frac { 2 }{ 100 }\) × 7350 = Rs 147
Profit earned by B = 7x = 7 × 2450 = Rs 17150
Amount given by B = 2% of his profit
= \(\frac { 2 }{ 100 }\) × 17150 = Rs 343
∴ The amount given by A and B to the Soldiers’ Welfare Fund are Rs 147 and Rs 343 respectively.

Question 5.
Jaya, Seema, Nikhil and Neelesh put in altogether Rs 360000 to form a partnership, with their investments being in the proportion 3 : 4 : 7 : 6. What was Jaya’s actual share in the capital? They made a profit of 12%. How much profit did Nikhil make?
Solution:
Total investment = Rs 360000
Profit earned = 12%
∴ Total profit = 12% of 360000
= \(\frac{12}{100} \times 360000\) = Rs 43200
Proportion of investment = 3 : 4 : 7 : 6
Let the investment of Jaya, Seema, Nikhil and Neelesh be Rs 3x, Rs 4x, Rs 7x and Rs 6x respectively.
3x + 4x + 7x + 6x = 360000
∴ 20x = 360000
∴ \(x=\frac { 360000 }{ 20 }\)
= 18000
∴ Jaya’s investment = 3x = 3 x 18000 = Rs 54000
Also, profit made by them is Rs 43200
∴ 3x + 4x + 7x + 6x = 43200
∴ 20x = 43200
∴ \(x=\frac { 43200 }{ 20 }\)
= 2160
∴ Nikhil’s profit = 7x = 7 x 2160 = Rs 15120
∴ Jaya’s share in the capital was Rs 54000 and the profit made by Nikhil was Rs 15120.

Maharashtra Board Class 7 Maths Chapter 9 Direct Proportion and Inverse Proportion Practice Set 39 Intext Questions and Activities

Question 1.
Saritaben, Ayesha and Meenakshi started a business by investing Rs 2400, Rs 5200 and Rs 3400. They made a profit of 50%. If they reinvested all their profit by adding it to the capital, find out the proportions of their shares in the capital during the following year. (Textbook pg. no. 67)
Solution:
Total investment = Rs 2400 + Rs 5200 + Rs 3400 = Rs 11000
Total profit = 50% of 11000 = \(\frac{50}{100} \times 11000\) = Rs 5500
Proportion of shares = 2400 : 5200 : 3400
= 12 : 26 : 17 …. (Dividingby 200)
Let the profit of Saritaben, Ayesha and Meenakshi be Rs 12x, Rs 26x and Rs 17x respectively.
12x + 26x + 17x = 5500
∴ 55x = 5500
∴ x = 100
∴ Saritaben’s profit = 12x = 12 × 100 = Rs 1200
Ayesha’s profit = 26x = 26 × 100 = Rs 2600
Meenakshi’s profit = 17x = 17 × 100 = Rs 1700
∴ Saritaben’s new investment = 2400 + 1200 = Rs 3600
Ayesha’s new investment = 5200 + 2600 = Rs 7800
Meenakshi’s new investment = 3400 + 1700 = Rs 5100
∴ New proportion of shares = 3600 : 7800 : 5100
= 12 : 26 : 17 …. (Dividing by 300)
∴ The proportion of the shares in the capital during the following year is 12 : 26 :17

Question 2.
Are the amount of petrol filled in a motorcycle and the distance traveled by it, in direct proportion? (Textbook pg. no. 63)
Solution:
Yes.
If amount of petrol filled in the motorcycle is less, it will travel less distance and if the amount of petrol filled is more, it will travel more distance.
Hence, the amount of petrol filled in the motorcycle and the distance traveled by it are in direct proportion.

Question 3.
Can you give examples from science or everyday life of quantities that vary in direct proportion? (Textbook pg. no. 63)
Solution:

  1. Number of chairs and the number of spectators.
  2. Quantity (litres) of water and number of vessels required to store the water.

Class 7 Maths Solution Maharashtra Board

Practice Set 37 Class 7 Answers Chapter 9 Direct Proportion and Inverse Proportion Maharashtra Board

Direct Proportion and Inverse Proportion Class 7 Maths Chapter 9 Practice Set 37 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 37 Answers Solutions Chapter 9 Direct Proportion and Inverse Proportion.

Std 7 Maths Practice Set 37 Solutions Answers

Question 1.
If 7 kg onions cost Rs 140, how much must we pay for 12 kg onions?
Solution:
Let the cost of 12 kg onions be Rs x.
The quantity of onions and their cost are in direct proportion.
∴ \(\frac{7}{140}=\frac{12}{x}\)
∴ 7x = 12 × 140 ….(Multiplying both sides by 140x)
∴ x = \(\frac { 12\times 140 }{ 7 }\)
= 240
We must pay Rs 240 for 12 kg onions.

Question 2.
If Rs 600 buy 15 bunches of feed, how many will Rs 1280 buy?
Solution:
Let the bunches of feed bought for Rs 1280 be x.
The quantity of feed bought and their cost are in direct proportion.
∴ \(\frac{600}{15}=\frac{1280}{x}\)
∴ 600x = 1280 × 15 …. (Multiplying both sides by 15x)
∴ \(x=\frac{1280 \times 15}{600}=32\)
∴ 32 bunches of feed can be bought for Rs 1280.

Question 3.
For 9 cows, 13 kg 500 g of food supplement are required every day. In the same proportion, how much will be needed for 12 cows?
Solution:
Let the food supplement required for 12 cows be x kg.
The quantity of food supplement required and the number of cows are in direct proportion.
∴ \(\frac{13 \mathrm{kg} 500 \mathrm{gram}}{9}=\frac{x \mathrm{kg}}{12}\)
∴ \(\frac{13.5}{9}=\frac{x}{12}\) ….(13 kg 500 gram = 13.5 kg)
∴ 13.5 × 12 = 9x ….(Multiplying both sides by 9 x 12)
∴ \(\frac{13.5 \times 12}{9}=x\)
∴ x = 18
∴ The food supplement required for 12 cows is 18 kg.

Question 4.
The cost of 12 quintals of soyabean is Rs 36,000. How much will 8 quintals cost?
Solution:
Let the cost of 8 quintals of soyabean be Rs x.
The quantity of soyabeans and their cost are in direct proportion.
∴ \(\frac{12}{36000}=\frac{8}{x}\)
∴ 12x = 8 × 36000 ….(Multiplying both sides by 36000x)
∴ \(x=\frac{8 \times 36000}{12}=24000\)
∴ The cost of 8 quintals of soyabean is Rs 24000.

Question 5.
Two mobiles cost Rs 16,000. How much money will be required to buy 13 such mobiles ?
Solution:
Let the cost of 13 mobiles be Rs x.
The quantity of mobiles and their cost are in direct proportion.
∴ \(\frac{2}{16000}=\frac{13}{x}\)
∴ 2x = 13 × 16000 ….(Multiplying both sides by 16000x)
∴ \(x=\frac{13 \times 16000}{2}=104000\)
∴ Rs 104000 will be required to buy 13 mobiles.

Class 7 Maths Solution Maharashtra Board

Practice Set 35 Class 7 Answers Chapter 8 Algebraic Expressions and Operations on them Maharashtra Board

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 35 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 35 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Std 7 Maths Practice Set 35 Solutions Answers

Question 1.
Multiply:
i. 16xy × 18xy
ii. 23xy² × 4yz²
iii. (12a + 17b) × 4c
iv. (4x + 5y) × (9x + 7y)
Solution:
i. 16xy × 18xy
= 16 × 18 × xy × xy
= 288x²y²

ii. 23xy² × 4yz²
= 23 × 4 × xy² × yz²
= 92xy³z²

iii. (12a + 17b) × 4c = 12a × 4c + 17b × 4c
= 48ac + 68bc

iv. (4x + 5y) × (9x + 7y)
= 4x × (9x + 7y) + 5y × (9x + 7y)
= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²

Question 2.
A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area. Solution:
Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
∴ Area of the rectangle = length × breadth
= (8x + 5) × (5x + 3)
= 8x × (5x + 3) + 5 × (5x + 3)
= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15
∴ The area of the rectangle is (40x² + 49x + 15) sq. cm.

Class 7 Maths Solution Maharashtra Board

Practice Set 36 Class 7 Answers Chapter 8 Algebraic Expressions and Operations on them Maharashtra Board

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 36 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 36 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Std 7 Maths Practice Set 36 Solutions Answers

Question 1.
Simplify (3x – 11y) – (17x + 13y) and choose the right answer.
(A) 7x – 12y
(B) -14x – 54y
(C) -3(5x + 4y)
(D) -2(7x + 12y)
Solution:
(D) -2(7x + 12y)

Hints:
(3x – 11y) – (17x + 13y) = 3x – 11y – 17x – 13y
= – 14x – 24y
= – 2 × 7x – 2 × 12y
= – 2(7x + 12y)

Question 2.
The product of (23x²y³z) and (-15x³yz²) is __
(A) -34x5y4z3
(B) 34x2y3z5
(C) 145x3y2z
(D) 170x3y2z3
Solution:
(A) -34x5y4z3

Question 3.
Solve the following equations:
i. \(4 x+\frac{1}{2}=\frac{9}{2}\)
ii. 10 = 2y + 5
iii. 5m – 4 = 1
iv. 6x – 1 = 3x + 8
v. 2(x – 4) = 4x + 2
vi. 5(x + 1) = 74
Solution:
i. \(4 x+\frac{1}{2}=\frac{9}{2}\)
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 1

ii. 10 = 2y + 5
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 2

iii. 5m – 4 = 1
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 3

iv. 6x – 1 = 3x + 8
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 4

v. 2(x – 4) = 4x + 2
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 5

vi. 5(x + 1) = 74
Maharashtra Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 6

Question 4.
Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?
Solution:
Let the age of Rakesh be x years.
∴ Sania’s age = (x + 5) years.
According to the given condition,
x + (x + 5) = 27
∴ 2x + 5 = 27
∴ 2x = 27 – 5
∴ 2x = 22
∴ \(x=\frac { 22 }{ 2 }=11\)
Sania’s age = x + 5 = 11 + 5 = 16 years
∴ The ages of Rakesh and Sania are 11 years and 16 years respectively.

Question 5.
When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted?
Solution:
Let the number of jambhul trees planted be x.
∴ Number of ashoka trees = x – 60
According to the given condition, x + x – 60 = 200
∴ 2x = 200 + 60
∴ 2x = 260
∴ \(x=\frac { 260 }{ 2 }=130\)
∴ 130 jambhul trees were planted.

Question 6.
Shubhangi has twice as many 20-rupee notes as she has 50-rupee notes. Altogether, she has 2700 rupees. How many 50-rupee notes does she have?
Solution:
Let the number of 50-rupee notes with shubhangi be x.
∴ Number of 20-rupee notes = 2x
∴ Total amount with Shubhangi = Number of 50-rupee notes × 50 + Number of 20-rupee notes × 20
= x × 50 + 2x × 20
= 50x + 40x
= 90x
According to the given condition,
90x = 2700
∴ \(x=\frac { 2700 }{ 90 }=30\)
∴ Shubhangi has 30 notes of 50 rupees.

Question 7.
virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make?
Solution:
Let the runs made by Rohit be x.
∴ Runs made by Virat = 2x
According to the given condition,
x + 2x = 200 – 2
∴ 3x = 198
∴ \(x=\frac { 198 }{ 3 }=66\)
∴ Runs made by Virat = 2x = 2 × 66 = 132
∴ The runs made by Virat and Rohit are 132 and 66 respectively.

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 Intext Questions and Activities

Question 1.
Solve the following equations. (Textbook pg. no. 59)
i. x + 7 = 4
ii. 4p = 12
iii. m – 5 = 4
iv. \(\frac { t }{ 3 }=6\)
Solution:
i. x + 7 = 4
∴ x + 7 – 7 = 4 – 7 ….(Subtracting 7 from both sides)
∴ x + 0 = -3
∴ x = -3

ii. 4p = 12
∴ \(\frac{4 p}{4}=\frac{12}{4}\) ….(Dividing both sides by 4)
∴ p = 3

iii. m – 5 = 4
∴ m – 5 + 5 = 4 + 5
…. (Adding 5 to both sides)
∴ m + 0 = 9
∴ m = 9

iv. \(\frac { t }{ 3 }=6\)
∴ \(\frac { t }{ 3 }\) × 3 = 6 × 3 …. (Multiplying both sides by 3)
∴ t = 18

Class 7 Maths Solution Maharashtra Board

Practice Set 34 Class 7 Answers Chapter 8 Algebraic Expressions and Operations on them Maharashtra Board

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 34 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 34 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Std 7 Maths Practice Set 34 Solutions Answers

Question 1.
Subtract the second expression from the first.
i. (4xy – 9z); (3xy – 16z)
ii. (5x + 4y + 7z); (x + 2y + 3z)
iii. (14x² + 8xy + 3y²); (26x² – 8xy – 17y²)
iv. (6x² + 7xy + 16y²); (16x² – 17xy)
v. (4x + 16z); (19y – 14z + 16x)
Solution:
i. (4xy – 9z) – (3xy – 16z)
= 4xy – 9z – 3xy + 16z
= (4xy – 3xy) + (16z – 9z)
= xy + 7z

ii. (5x + 4y + 7z) – (x + 2y + 3z)
= 5x + 4y + 7z – x – 2y – 3z
= (5x – x) + (4y – 2y) + (7z – 3z)
= 4x + 2y + 4z

iii. (14x² + 8xy + 3y²) – (26x² – 8xy – 17y²)
= 14x² + 8xy + 3y² – 26x² + 8xy + 17y²
= (14x² – 26x²) + (8xy + 8xy) + (3y² + 17y²)
= -12x² + 16xy + 20y²

iv. (6x² + 7xy + 16y²) – (16x² – 17xy)
= 6x² + 7xy + 16y² – 16x² + 17xy
= (6x² – 16x²) + (7xy + 17xy) + 16y²
= -10x² + 24xy + 16y²

v. (4x + 16z) – (19y— 14z + 16x)
= 4x + 16z – 19y + 14z – 16x
= (4x – 16x) – 19y + (16z + 14z)
= -12x – 19y + 30z

Class 7 Maths Solution Maharashtra Board

Practice Set 32 Class 7 Answers Chapter 8 Algebraic Expressions and Operations on them Maharashtra Board

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 32 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 32  Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Std 7 Maths Practice Set 32 Solutions Answers

Question 1.
Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
i. 7x
ii. 5y – 7z
iii. 3x³ – 5x² – 11
iv. 1 – 8a – 7a² – 7a³
v. 5m – 3
vi. a
vii. 4
viii. 3y² – 7y + 5
Solution:
i. Monomial
ii. Binomial
iii. Trinomial
iv. Polynomial
v. Binomial
vi. Monomial
vii. Monomial
viii. Trinomial

Class 7 Maths Solution Maharashtra Board

Practice Set 33 Class 7 Answers Chapter 8 Algebraic Expressions and Operations on them Maharashtra Board

Algebraic Expressions and Operations on them Class 7 Maths Chapter 8 Practice Set 33 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 33 Answers Solutions Chapter 8 Algebraic Expressions and Operations on them.

Std 7 Maths Practice Set 33 Solutions Answers

Question 1.
Add:
i. 9p + 16q; 13p + 2q
ii. 2a + 6b + 8c; 16a + 13c + 18b
iii. 13x² – 12y²; 6x² – 8y²
iv. 17a²b² + 16c; 28c – 28a²b²
v. 3y² – 10y + 16; 2y – 7
vi. – 3y² + 10y – 16; 7y² + 8
Solution:
i. (9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q

ii. (2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

iii. (13x² – 12y²) + (6x² – 8y²)
= (13x² + 6x²) + [(-12y²) + (-8y²)]
= 19x² + (-20y²)
= 19x² – 20y²

iv. (17a²b² + 16c) + (28c – 28a²b²)
= [17a²b² + (-28a²b²)] + (16c + 28c)
= -11a²b² + 44c

v. (3y² – 10y + 16) + (2y – 7)
= 3y² + (-10y + 2y) + (16 – 7)
= 3y² – 8y + 9

vi. (-3y² + 10y – 16) + (7y² + 8)
= (-3y² + 7y²) + (10y) + (-16 + 8)
= 4y² + 10y – 8

Maharashtra Board Class 7 Maths Chapter 8 Algebraic Expressions and Operations on them Practice Set 33 Intext Questions and Activities

Question 1.
Answer the following questions. (Textbook pg. no. 57)

  1. 3x + 4y = How many?
  2. 3 guavas + 4 mangoes = 7 guavas.
  3. 7m – 2n = 5m.

Solution:

  1. 3x and 4y are unlike terms. Hence, they cannot be added, further to get a single term.
  2. No. Guava and mango are different fruits. Hence, 3 guavas + 4 mangoes & 7 guavas.
  3. No. 7m and 2n are unlike terms. Hence, 7m – 2n ≠ 5m.

Class 7 Maths Solution Maharashtra Board

Think Before You Speak Poem Questions and Answers Class 7 English Chapter 3.6 Maharashtra Board

Balbharti Maharashtra State Board Class 7 English Solutions Chapter 3.6 Think Before You Speak Notes, Textbook Exercise Important Questions and Answers.

Std 7 English Lesson 3.6 Think Before You Speak Question Answer Maharashtra Board

Class 7 English Chapter 3.6 Think Before You Speak Textbook Questions and Answers

1. The same word can be used as a noun in some sentences and as a verb in others.

Question 1.
For example,
1. Many are the empty remarks …………….. Noun.
2. A wise man once remarked, ………….. Verb.
Make two sentences of your own with each of the words given below, using the same word as a noun in one and as a verb in another.
1. change
2. show throw
3. return
4. benefit
Answer:
1. Change:

  • Picnics are a welcome change from the daily routine. (Noun)
  • Change the way you think. (Verb)

2. Show:

  • The students put up a grand musical show. (Noun)
  • You need not show me your tickets. (Verb)

3 Throw:

  • That was an easy throw. (Noun)
  • You must throw garbage in the trash can. (Verb)

4 Return:

  • She promised to take me for a movie on her return from the market. (Noun)
  • Please return my book, as I need to complete it. (Verb)

5 Benefit:

  • I advice you for your own benefit. (Noun)
  • Children benefit from parents’ advice. (Verb)

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

2. Complete the following.

Question 1.
Our ears are like funnels because …………..
Answer:
they are open all the time and there is no door with which you can close them.

Question 2.
The two rows of teeth are like a fence because ……………..
Answer:
if we wish to speak our words have to pierce through them.

Question 3.
The two lips are like fence because ……………….
Answer:
before a word is spoken, it has to pass through it.

Question 4.
Harsh words are like scattered bits of paper carried away by the wind because ………………
Answer:
once you have spoken them aloud, it is very difficult to take them back.

3. Write a brief summary of the story of the young man and his spiritual teacher, making the young man the narrator.

Question 1.
You may begin as given below.
“I went quickly to my spiritual teacher for advice because I had ………….”
Answer:
I went quickly to my spiritual teacher for advice because I had hurt and insulted my dear friend with unkind and harsh words. When I asked him for the solution, he gave me a fresh sheet of blank paper and a pen and instructed me to write down on that paper all the harsh things I had told my friend. I obeyed. He then asked me to tear it into as many bits as I could and throw the bits out of the window.

The tiny bits scattered far and wide in no time. But to my surprise, he then asked me to collect as many bits as possible. I ran to collect them, but in vain. I couldn’t get hold of even in a single bit of paper. I returned all exhausted. He then revealed to me that spoken words are like those bits of paper, easy to scatter but difficult to take back. What was done, was done, and could not be altered. I learnt that I should think before I speak.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

4. Language study.

Adverbial:
An adverbial is something that is used as an adverb. An adverbial is often one word, an adverb, as in the following example:
1.You have run fast.
But, it can also be a phrase or a clause.
2. We played on the playground.
3. I will go home when the bell rings.
In sentence 2, the phrase ‘on the playground’ is used as an adverb. In
sentence 3, the clause ‘when the bell rings’ Is used as an adverb.

Class 7 English Chapter 3.6 Think Before You Speak Additional Important Questions and Answers

Answer the following questions.

Question 1.
Was Disraeli trying to give a scientific reason?
Answer:
No, Disraeli wasn’t trying to give a scientific reason, but a logical one. He just wanted to draw our attention to the fact, that human physiology suggests that we should speak less and listen more.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 2.
Was he only trying to give a message in a light-hearted but effective way?
Answer:
Yes, he was only trying to give a message in a light-hearted but effective way.

Question 3.
Have you ever passed an empty remark or win statement that might hurt someone? What can you do to avoid it again?
Answer:
Yes, I have passed a vain statement about a classmate’s dressing sense only to realize later that she came from a poor financial background. I was lucky she didn’t hear it or else she would have been hurt. From that day I decided not to pass such remarks because often we do not know the complete background.

Answer the following in one or two sentences.

Question 1.
What did the wise teacher ask the young man to do?
Answer:
The wise teacher asked the young man to write down on a paper all the harsh things he had told his friend.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 2.
What did the wise teacher ask the young man to do with the sheet of paper?
Answer:
The wise teacher asked the young man to tear the sheet of paper into a hundred tiny pieces and throw the bits out of the window.

Question 3.
What is the similarity between the bits of paper scattered in the wind and spoken words?
Answer:
Just as bits of paper scattered in the wind are difficult to gather, similarly spoken words are impossible to take back.

Reading Skills, Vocabulary and Grammar.

Simple Factual Questions.

Question 1.

  1. Benjamin Disraeli was a great
  2. To speak eve one single words, it must
  3. We must think at least

Answer:

  1. Britsh Statesman
  2. pass through two walls – two fences,
  3. twice before we utter a word.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Complex Factual Questions.

Question 1.
How do we become slaves of words spoken by us?
Answer:
Once we have spoken a word we become slaves as we cannot take it back, no matter how hard we try. You cannot retrieve it.

Question 2.
What are unspoken words?
Answer:
Unspoken words are things one wants to say, but remain unsaid as thoughts in the mind.

Question 3.
How do spoken words make you a slave?
Answer:
Spoken words are meant to be honoured which means we have to stand by it. We say something and do not follow it, then we lose our credibility. Hence we need to think before we speak and become slave to our own words.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 4.
Two rows of teeth’. Thinks of similar expressions.
Answer:

  1. Two sets of five fingers.
  2. Two sets of five toes.
  3. A pairs of eyes
  4. A pair of ears are similar expression.

Vocabulary.

Question 1.
Use the words ‘left’ in separate sentences and show the difference in the meaning.
Answer:
My’sister uses her left hand efficiently. After the function, a lot of food was left behind. She lift without telling anybody.

Question 2.
Man was meant to listen more and talk less similarly make a sentence using words of opposite meaning.
Answer:
We breathe in oxygen and breath out carbon dioxide.

Grammar.

Question 1.
The wise teacher gave him a fresh sheet of blank paper and pen. (Rewrite ending with ‘the wise teacher’)
Answer:
He was given a fresh sheet of blank paper and pen by the wise teacher.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 2.
You cannot change them or control them. (Make affirmative)
Answer:
You can hardly change them or control them.

Question 3.
Spoken – unspoken
Write two similar antonyms using a prefix.
Answer:
1. able – unable
2. happy – unhappy

Form adjectives.

Question 1.

  1. thought
  2. remark
  3. nature
  4. man
  5. time

Answer:

  1. thoughtful/thoughtless
  2. remarkable
  3. natural
  4. manual/ manly
  5. timely

Personal Response.

Question 1.
Do you remember someone else speaking to you angrily, without thinking? What did you do on that occasion? Did you also speak angrily?
Answer:
Yes, I do remember a few occasions. When people have spoken to me angrily without a valid reason. Though I was angry, I did not speak angrily as I did not want an angry exchange in public.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Simple Factual Questions.

Question 1.
1. Socrates had influenced the lives of many youths for the better.
2. According to Socrates, one must never open one’s mouth to speak.
Answer:
1. True
2. False.

Complex Factual Questions.

Question 1.
What are three questions one needs to ask before speaking?
Answer:
Before speaking one needs to ask three questions such as – ‘Is it true?’, ‘Is it pleasant?’, ‘Is it useful?’.

Question 2.
Why should one ask the question ‘Is it true?’ before speaking?
Answer:
One must ask the question ‘Is it true?’ before speaking because if we are not sure about the truth of what we are saying, it is better we do not speak. When we speak words carelessly, we become transmitters of the untruth.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 3.
What is the importance of asking the question ‘ ‘Is it useful?’
Answer:
It is important to ask ‘Is it useful?’ before speaking because only if our words benefit the listener and comfort someone, they should be spoken.

Vocabulary.

Question 1.
Pick out a word from the extract that means ‘useless’.
Answer:
vain

Question 2.
Give antonyms.
1. ancient
2. affirmative
Answer:
1. recent / contemporary
2. negative

Grammar.

Question 1.
Socrates was one of the wise men of the ancient world. (Rewrite using ‘as … as’)
Answer:
Very few men of the ancient world were as wise as Socrates.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 2.
“O wise one, how may we know when it is right to speak?” they asked him. (Write in indirect speech)
Answer:
Addressing him as a wise one, they asked him how they might know when it was right time to speak.

Personal Response.

Question 1.
Do you agree that thinking before speaking saves one from many troubles? Explain.
Answer:
Yes, I definitely agree that thinking before speaking saves one from many troubles as we get time to analyse our thoughts and decide whether we should speak or not. Many a times when we are angry, if we think before speaking we will not have to regret what we have said. This saves many relationships.

Do as directed.

Question 1.
Write down on this paper all the harsh things you said to him. (Add a question tag)
Answer:
Write down on this paper all the harsh things you said to him, will you?

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 2.
“Throw the bits out of this window” the teacher told him. (Write in indirect speech)
Answer:
The teacher instructed him to throw the bits out of that window.

Question 3.
It will be difficult indeed. (Make negative)
Answer:
It will not be easy indeed.

Question 4.
It will be difficult indeed, but do give it a try. (Rewrite using although)
Answer:
Although it will be difficult, give it a try.

Question 5.
Speak only when absolutely necessary. (Frame a Wh-question)
Answer:
When should one speak?

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 6.
The young man went out. (Rewrite in question form)
Answer:
Didn’t the young man go out?

Question 7.
He returned half an hour later. (Rewrite using modal auxiliary of compulsion)
Answer:
He must return half an hour later.

Question 8.
Now tear up this sheet of paper into as many small bits as you can. (Rewrite beginning with ‘Let’)
Answer:
Let this sheet of paper be torn into as many small bits as you can.

Question 9.
Fill in the blanks with the appropriate phrases / idioms given below. Change the form where necessary to fill in the blanks. (far and wide, to make amends, taken aback, bring comfort)

  1. His reassuring words ………….. to my troubled mind.
  2. People come to see flamingos from …………… .
  3. Mishti ……………. by his brother’s arrogance.
  4. She ………….. for her rudeness by apologizing.

Answer:

  1. brought comfort
  2. far and wide
  3. was taken aback
  4. made amends.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Question 10.
Underline the adverbial in the following sentence.

  1. The birds flew over the trees.
  2. She speaks in a self tone.
  3. The workers left without permission
  4. The king promised to give then food to eat.
  5. Seema walked behind them briskly.
  6. The young man did as he was told.

Answer:

  1. over the trees
  2. food to eat
  3. without permission
  4. behind them briskly
  5. behind them briskly
  6. as he was told.

Think Before You Speak Summary in English

The great British statesment Benjamin Disraeli argued, that as man is endowed with two ears and one mouth it indicates that we should listen more and talk less. We are masters of unspoken words and slaves of the spoken ones.

The lesson narrates a story of a young man who had hurt his friend with his harsh words. Worried that he might lose his friend, he goes to a wise man who teaches him a lesson that words spoken are like scattered bits of paper thrown out in the wind. Easy to blow but difficult to collect. The advice of Socrates in the end nails the message quite forcefully. Socrates once told his disciples that when you wish to speak, ask three questions. If the answer to all of them is ‘Yes’, then go ahead and speak. The questions are – ‘Is it true?’, ‘Is it pleasant?’ and ‘Is it useful?’.

The lesson is a practical tip on how one can earn respect, happy relationships and peace of mind by speaking less and only when needed.

Introduction:

The lesson ‘Think Before You Speak!’ carries the message “speak only when necessary”.

Maharashtra Board Class 7 English Solutions Chapter 3.6 Think Before You Speak

Glossary:

  1. statesman (n) – a person experienced in the art of governance
  2. endowed (v) – to be provided with something
  3. funnels (n) – a tube or pipe that is wide at the top and narrow at the bottom
  4. pierce (v) – go into something
  5. stammer (v) – speak with sudden involuntary pause and repeat some letters
  6. exhausted (adj) – very tired
  7. counselled (v) – advised, guided
  8. affirmative (adj) – answer which is ‘yes’
  9. veracity (n) – truth
  10. transmitters (n) – persons who spread something, in this case, untruth
  11. vain (adj) – useless

7th Std English Balbharati Textbook Solutions