Prasanna

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.3

Miscellaneous Practice Problems

Question 1.
If the cost of 7 kg of onions is ₹ 84 find the following :
(i) Weight of the onions bought for ₹ 180
(ii) The cost of 3 kg of onions
Solution:
(i) For ₹ 84 weight of onion bought
for ₹ 1 weight of onion bought
∴ For ₹ 180 weight of onion bought w
∴ For ₹ 180 weight of onion bought

(ii) Cost of 7 kg of onions = 15 kg

Question 2.
If C = kd
(i) what is the relation between C and d ?
(ii) Find k when C = 30 and d = 6
(iii) Find C, when d = 10
Solution:

As C increases d also increases
∴ It is direct proportion

Question 3.
Every 3 months Tamilselvan deposits ₹ 5000 as savings in his bank account. In how many years he can save ₹ 1,50,000.
Solution:
Let the number of years required be x.

No. of years and deposit are direct proportion as they both increases simultaneously.

He can save ₹ 1,50,000 in \(7 \frac{1}{2}\) years.

Question 4.
A printer, prints a book of 300 pages at the rate of 30 pages per minute. Then, how long will it take to print the same book if the speed of the printer is 25 pages per minute?
Solution:
Let the required time taken to print be x
As the speed increases time taken to print decreases
∴ They are in inverse proportion
Time taken to print 30 pages = 1 min

Question 5.
If the cost of 6 cans of juice in ₹ 210, then what will be the cost of 4 cans of juice?
Solution:
Let the cost required be x

As number of cans increases cost also increases.
∴ They are in direct proportion

x = 140
Cost of 4 cans of juice = 140

Question 6.
x varies inversely as twice of y. Given that when y = 6, the value of x is 4. Find the value of x when y = 8.
Solution:
Given x varies inversely as twice of y.

Question 7.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required to cover a distance of 1650 km?
Solution:
Let the required distance be x

As the distance increases fuel quantity also increases.
∴ They are direct proportion.

∴ The diesel required = 300 liters

Challenge Problems

Question 8.
If the cost of a dozen soaps is ₹ 396, what will be the cost of 35 such soaps?
Solution:
1 dozen = 12
Cost of 12 soaps = ₹ 396

Question 9.
In a school there is 7 periods a day each of 45 minutes duration. How long each period is if the school has 9 periods a day assuming the number of hours to be the same?
Solution:
Number of periods increases as duration decreases, since the number of hours is same.
Let the duration of each period be x.

Question 10.
Cost of 105 note books is ₹ 2415. How many notebooks can be bought for ₹ 1863?
Solution:
For 2415 number of notebooks bought = 105

Question 11.
10 farmers can plough a field in 21 days. Find the number of days reduced if 14 farmers ploughed the same field?
Solution:
Let the required number of days if 14 farmers ploughed = x

As number of farmers increases, number of days decreases.
∴ They are in inverse proportion

Initially the farmers worked for 21 days. Now they worked for 15 days.
∴ The number of days reduced = 21 – 15 = 6 days

Question 12.
A flood relief camp has food stock by which 80 people can be benefited for 60 days. After 10 days 20 more people have joined the camp. Calculate the number of days of food shortage due to the addition of 20 more people?
Solution:

As number of people increases food last for less number of days.

Remaining food is to be used for 50 days.
But it only last for 40 days.
No. of days shortage = 50 – 40 = 10 days.
∴ 10 days of food shortage due to the addition of 20 more people.

Question 13.
Six men can complete a work in 12 days. Two days later, 6 more men joined them. How many days will they take to complete the remaining work?
Solution:

As the number of men increases number of days increases.
∴ They are inversely proportional

x = 5 days
∴ Remaining work will be complete in 5 days

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.2

Question 1.
Fill in the blanks
(i) 16 taps can fill a petrol tank in 18 minutes. The time taken for 9 taps to fill the same tank will be ___ minutes.
(ii) If 40 workers can do a project work in 8 days, then ____ workers can do it in 4 days.
Solutions:
(i) 32
(ii) 80

Question 2.
6 pumps are required to fill a water sump in 1 hr 30 minutes. What will be the time taken to fill the sump if one pump is switched off?
Solution:
Let x be the required time taken

Time taken in minutes 1 hr. 48m

Question 3.
A farmer has enough food for 144 ducks for 28 days. If he sells 32 ducks how long will the food last?
Solution:
Let the required number of days be x.

As the number of ducks decreases the food will last for more days.
∴ They are in inverse proportion. x₁y₁ = x₂y₂

The food lasts for 36 days

Question 4.
It takes 60 days for 10 machines to dig a hole. Assuming that all machines work at the same speed, how long will it take 30 machines to dig the same hole?
Solution:
Let the number of days required be x.

As the number of machines increases it takes less days to complete the work
∴ They are in inverse proportion, x₁y₁ = x₂y₂

It takes 20 days to dig the hole

Question 5.
Forty students stay in a hostel. They had food stock for 30 days. If the students are doubled then for how many days the stock will last?
Solution:
Let the required number of days be x.

As the number of students increases the food last for less number of days
∴ They are in inverse proportion.

The food stock lasts for 15 days

Question 6.
Meena had enough money to send 8 parcels each weighing 500 grams through a courier service. What would be the weight of each parcel, if she has to send 40 parcel for the same money?
Solution:
Let the required weight of the parcel be x grams.

As the number of parcels increases weight of a parcel decreases.
∴ They are in inverse proportion.

Weight of each parcel = 100 grams

Question 7.
It takes 120 minutes to weed a garden with 6 gardeners. If the same work is to be done in 30minutes, how many more gardeners are needed?
Solution:
Let the, number of gardeners needed be x.

As the number of gardeners increases the time decreases. They are in inverse proportion,
x₁y₁ = x₂y₂

∴ To complete the work in 30 min gardeners needed = 24
Already existing gardeners = 6
∴ More gardeners needed = 24 – 6 = 18
18 more gardeners are needed

Question 8.
Neelaveni goes by bicycle to her school every day. Her average speed is 12km/hr and she reaches school in 20 minutes. What is the increase in speed, If she reaches the school in 15 minutes?
Solution:
Let the speed to reach school in 15 min be x

∴ They are in inverse proportion x₁y₁ = x₂ y₂

If she reaches in 15 min the speed = 16 km/hr
Already running with 12 km / hr
∴ Increased speed = 16 – 12 = 4km / hr
Increase in speed = 4 km / hr

Question 9.
A toy company requires 36 machines to produce car toys in 54 days. How many machines would be required to produce the same number of car toys in 81 days?
Solution:
Let the required number of machines be x

As the number of machines increases number of days required decreases.

∴ 24 machines would be required

Objective Type Questions

Question 10.
12 cows can graze a field for 10 days. 20 cows can graze the same field for ____ days
(i) 15
(ii) 18
(iii) 6
(iv) 8
Solution:
(iii) 6
Hint:

Question 11.
4 typists are employed to complete a work in 12 days. If two more typists are added, they will finish the same work in days
(i) 7
(ii) 8
(iii) 9
(iv) 10
Solution:
(ii) 8

Students can Download Maths Chapter 4 Direct and Inverse Proportion Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Ex 4.1

Fill in the blanks.

(i) If the cost of 8 apples is 56 then the cost of 12 apples is ____.
(ii) If the weight of one fruit box is \(3 \frac{1}{2}\) kg, then the weight of 6 such boxes is ____.
(iii) A car travels 60 km with 3 liters of petrol. If the car has to cover the distance of
200 km, it requires ___ liters of the petrol.
(iv) If 7 m cloth costs ₹ 294, then the cost of 5m of cloth is ____.
(v) If a machine in a cool drinks factory fills 600 bottles in 5 hrs, then it will fill _____ bottles in 3 hours.
Solutions:
(i) 84
(ii) 21 kg
(iii) 10
(iv) ₹ 210
(v) 360

Question 2.
Say True or False
(i) Distance travelled by a bus and time taken are in direct proportion.
(ii) Expenditure of a family to number of members of the family are in direct proportion.
(iii) Number of students in a hostel and consumption of food are not in direct proportion.
(iv) If Mallika walks 1km in 20 minutes, then she can convert 3km in 1 hour.
(v) If 12 men can dig a pond in 8 days, then 18 men can dig it in 6 days.
Solutions:
(i) True
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
A dozen bananas costs ₹ 20. What is the price of 48 bananas ?
Solution:
Let the required price be ₹ x. As the number of bananas increases price also increases
∴ Number of bananas and cost are in direct proportion.

Question 4.
A group of 21 students paid ₹ 840 as the entry fee for a magic show. How many students entered the magic show if the total amount paid was ₹ 1680?
Solution:
Let the required number of students be x.

As the number of students increases the entry fees also increases.
∴ They are in direct proportion .

∴ The number of students entered magic show = 42

Question 5.
A birthday party is arranged in third floor of a hotel. 120 people take 8 trips in a lift to go to the party hall. If 12 trips were made how many people would have attended the party?
Solution:
Let the number of people attended the party be x.

As the number of trips increases, number of people also increases.
∴ They are in direct proportion.


180 people attend the party in 12 trips

Question 6.
The shadow of a pole with height of 8m is 6m. If the shadow of another pole measured at the same time is 30m, find the height of the pole?
Solution:
Let the required height of the pole be ‘x’ m.

Height of the pole and its shadow are in direct proportion

∴ Height of the pole x = 40m.

Question 7.
A postman can sort out 738 letters in 6 hours. How many letters can be sorted in 9 hours?
Solution:
Let the required number of letters be x.

They are in direct proportion.

In 9 hours 1107 letters can be sorted.

Question 8.
If half a meter of cloth costs ₹ 15. Find the cost of \(8 \frac{1}{3}\) meters of the same cloth.
Solution:
Let the cost of cloth required be x.

Cost and length are in direct proportion.

Question 9.
The weight of 72 books is 9 kg. What is the weight of 40 such books (using unitary method)
Solution:
Weight of 72 books = 9 kg = 9000 g
∴ Weight of 1 book = \(\frac{9000}{72}\) = 125 g
∴ Weight of 40 books = 125 × 40 g = 5000 g = 5 kg.
Weight of 40 books = 5 kg

Question 10.
Thamarai pages ₹ 7500 as rent for 3 months. With the same rate how much does she have to pay for 1 year (using unitary method).
Solution:
Rent paid by Thamarai for 3 months = ₹ 7500
∴ Rent paid for 1 month = \(\frac{7500}{3}\) = 2500
Rent paid for 1 year or 12 moths = 2500 × 12 = ₹ 30,000
For 1 year rent to be paid = ₹ 30,000

Question 11.
If 30 men can reap a field in 15 days, then in how many days can 20 men reap the same field? (using unitary method).
Solution:

∴ 20 men can reap the field in 10 days.

Question 12.
Valli purchase 10 pens for ₹ 180 and Kamala boys 8 pens for ₹ 96. Can you say who bought the pen cheaper (using unitary method).
Solution:

∴ Kamala bought the pen cheaper.

Question 13.
A motorbike requires 2 liters of petrol to cover 100 kilometres. How many liters of petrol will be required to cover 250 kilometers? (using unitary method).
Solution:
To cover 100 km quantity of petrol required = 2 litres

5 litres of petrol required to cover 250 km

Objective Type Questions

Question 14.
If the cost of 3 books is ₹ 90, then find the cost of 12 books.
(i) ₹ 300
(ii) ₹ 320
(iii) ₹ 360
(iv) ₹ 400

Solution:
(iii) ₹ 360

Question 15.
If Mani buys 5 kg of potatoes for ₹ 75 then he can buy ₹ 105.
(i) 6
(ii) 7
(iii) 8
(iv) 5

Solution:
(ii) 7

Question 16.
35 cycles were produced in 5 days by a company then ___ cycles will be produced in 21 days.
(i) 150
(ii) 70
(iii) 100
(iv) 147

Solution:
(iv) 147

Question 17.
An aircraft can accommodate 280 people in 2 trips. It can take ______ trips to take 1400 people.
(i) 8
(ii) 10
(iii) 9
(iv) 12
Solution:
(ii) 10

Question 18.
Suppose 3 kg of sugar is used to prepare sweets for 50 members, then ___ kg of sugar is required for 150 members.
(i) 9
(ii) 10
(iii) 15
(iv) 6

Solution:
(i) 9

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Exercise 2.1

Parallelogram

(Try These Text book Page No. 33)

Question 1.
Find the missing values for the following:

Solution:
(i) Given length l = 12 m; Breadth b = 8 cm
∴ Area of rectangle = l × b sq. units = 12 × 8 m² = 96 m²
Perimeter of the rectangle = 2 × (1 + b) units = 2 × (12 + 8)m = 2 × 20 = 40m

(ii) Given Length l = 15 cm ; Area of the rectangle = 90 sq. cm
l × b = 90; 15 × 6 = 90; b = \(\frac{90}{15}\) = 6 cm
Perimeter of the rectangle = 2 × (l+ b) units = 2 × (15 + 6) cm = 2 × 21 cm = 42 cm

(iii) Given Breadth of rectangle = 50 mm ; Perimeter of the rectangle = 300 mm
2 × (l + b) = 300
2 × (l + 50) = 300
l + 50 = \(\frac{300}{2}\) = 150
l = 150 – 50
l = 100
Area = l × b sq. untis = 100 × 50 mm² = 5000 mm²

(iv) Length of the rectangle = 12 cm ; Perimeter = 44 cm
2(l + b) = 44
2(12 + b) = 44
12 + b = \(\frac{44}{2}\)
12 + b = 22 ; b = 22 – 12; b = 10 cm
Area = l × b sq. units
= 12 × 10 cm² = 120 cm²

Question 2.

Solution:
(i) Given side a = 60 cm
Area of the square = a × a sq.units = 60 × 60 cm² = 3600 cm²
Perimeter of the square = 4 × a units = 4 × 60 cm = 240 cm

(ii) Given area of a square = 64 sq. m
a × a = 64
a × a = 8 × 8
a = 8m
Perimeter = 4 × a
= 4 × 8
= 32 m

(iii) Given perimeter of the square = 100 mm
4 × a = 100
a = \(\frac{100}{4}\) mm
a = 25 mm
Area = a × a sq. units
= 25 × 25 mm²
= 625 mm²

Question 3.

Solution:
(i) Given base of the right angled triangle = 13 m ; height = 5 m

(iii) Given height h = 6 mm ; Area = 84 sq. mm
\(\frac{1}{2}\) × b × h = 84 ; \(\frac{1}{2}\) × b × 6 = 84
b = \(\frac{{84} \times 2}{6}\); b = 28 mm

(Try This Textbook Page No. 35)

Question 1.
Explain the area of the parallelogram as sum of the areas of the two triangles.
Solution:
ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.

Area of the parallelogram ABCD = base × height
= AB × DE

Question 2.
A rectangle is a parallelogram but a parallelogram is not a rectangle. Why?
Solution:
(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to 90°. But for parallelogram opposite angles are equal.
∴ All rectangles are parallelograms. But all parallelograms are not rectan¬gles as their angles need not be equal to 90°.

(Try These Textbook Page No. 36)

Question 1.
Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).

(a) ______ sq. units
(b) ______ sq. units
(c) ______ sq. units
(d) ______ sq. units
Solution:
Converting the given parallelograms into rectangles we get.

(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Question 2.
Draw the heights for the given parallelograms and mark the measure of their bases and find the area. Analyze your result.

Solution:
(a) Area of the parallelogram = b × h sq. units
= 4 × 2 sq. units = 8 sq. units

By counting the small squares also we get number of full
squares + number of square more than half = 6 + 2 = 8 sq. units.

(b) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

(c) Area of the parallelogram = base × height = 4 × 2 = 8 sq. units

Also area = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) Number of half squares = 4 + 4 = 8 sq. units

(d) Area of the parallelogram = (base × height) sq. units = 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + \(\frac{1}{2}\) [Number of half squares] + Number of squares more than half = 4 + 0 + 4 = 8sq. units

(e) Area of parallelogram = (base × height) sq. units
= 4 × 2 sq. units = 8 sq. units

Also area of the parallelogram = Number of full squares + Number of squares more than half + \(\frac{1}{2}\) [Number of half squares] = 2 + 6 = 8sq. units

Question 3.
Find the area o the following parallelograms by measuring their base and height, using formula.

(a) _____ sq. units
(b) _____ sq. units
(c) _____ sq. units
(d) _____ sq. units
(e) _____ sq. units
Solution:
(a) Area of the rectangle = (base × height) sq. units
base = 5 units
height = 5 units
∴ Area = (5 × 5 ) = sq. units = 25 sq. units

(b) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 1 units
∴ Area = (4 × 1 ) = sq. units = 4 sq. units

(c) Area of the rectangle = (base × height) sq. units
base = 2 units
height = 3 units
∴ Area = (2 × 3 ) = sq. units = 6 sq. units

(d) Area of the rectangle = (base × height) sq. units
base = 4 units
height = 4 units
∴ Area = (4 × 4 ) = sq. units = 16 sq. units

(e) Area of the parallelogram = (base × height) sq. units
base = 7 units
height = 5 units
= 7 × 5 = 35 sq. units

Question 4.
Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.
Solution:

Area of parallelogram (a), (b) or (c) = 20 sq. units

Exercise 2.2

Rhombus

(Try These Textbook Page No. 41)

Question 1.
Observe the figure and answer the following questions.
(i) Name two pairs of opposite sides.
(ii) Name two pairs of adjacent sides.
(iii) Name the two diagonals.

Solution:
(i) (a) PQ and RS (b) QR and PS
(ii) (a) PQ and QR (b) PS and RS
(iii) (a) PR and Question are diagonals.

Question 2.
Find the area of the rhombus given in (i) and (ii).

Solution:
(i) Area of the rhombus = \(\frac{1}{2}\) (d₁ + d₂) sq. units = \(\frac{1}{2}\)(11 + 13) sq. units
= \(\frac{1}{2}\) × (24) cm² = 12 cm²

(ii) Base = 10 cm ; Height = 7 cm
Area of the rhombus = b × h sq. units = 10 × 7 cm² = 70 cm²

Question 3.
Can you find the perimeter of the rhombus?
Solution:
If we know the length of one side we can find the perimeter using 4 × side units.

Question 4.
Can diagonals of a rhombus be of the same length?
Solution:
When the diagonals of a rhombus become equal it become a square.

Question 5.
A square is a rhombus but a rhombus is not a square. Why?
Solution:
In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to 90°.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.

Question 6.
Can you draw a rhombus in such a way that the side is equal to the diagonal.
Solution:
Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.

Exercise 2.3

(Try These Textbook Page No. 46)

Question 1.
Can you find the perimeter of the trapezium? Discuss.
Solution:
If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.

Question 2.
In which case a trapezium can be divided into two equal triangles?
Solution:
If two parallel sides are equal in length. Then it can be divided into two equal triangles.

Question 3.
Mention any three life situations where the isosceles trapeziums are used?
Solution:
(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
In the following figure, PQRS is a parallelogram find x and y.

Solution:
We know that in a parallelogram opposite sides are equal.
∴ 3x = 18
x = \(\frac{18}{3}\)
x = 6 and
3y – 1 = 26
3y = 26 + 1 = 27
y = \(\frac{27}{9}\)
y = 9

Question 2.
Two adjacent sides of a parallelogram are 5 cm and 7 cm respectively. Find its perimeter.
Solution:

Perimeter = AB + BC + CD + AD [∵ AB = DC & AD = BC]
= 7 cm + 5 cm + 7 cm + 5 cm = 24 cm

Question 3.
The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the length of the sides of the parallelogram.
Solution:

Given perimeter = 150 cm
Let one side of the parallelogram be ‘b’ cm
Then the other side = b + 25 cm
b + (b + 25) + b + (b + 25) = 150
b + b + 25 + b + b + 25 = 150
4h + 50 = 150 =4b = 100
b = \(\frac{100}{4}\) = 25
∴ One side b = 25 cm
Other side b + 25 = 50 cm

Exercise 2.2

Question 1.
ABCD is a rhombus. Find x, y, and z.

Solution:
We know that all sides of rhombus are equal and its diagonals bisect each other.
∴ x = 5, y = 12 and z = 13.

Question 2.
Find the altitude of the rhombus whose area is 315 cm² and its perimeter is 180 cm.
Solution:
Given perimeter of the rhombus = 180 cm
∴ One side of the rhombus = \(\frac{180}{4}\) = 45 cm
Given area of the rhombus = 315 cm²
b × h = 315
45 × h = 315 = \(\frac{315}{45}\)
h = 7 cm
Altitude of the rhombus = 7 cm

Question 3.
The floor of a building consists of 2000 titles which are rhombus shaped and each of its diagonals are 40 cm and 25 cm. Find the total cost of polishing the floor, if the cost per m² = ₹ 5.
Solution:
Area of each title = \(\frac{1}{2}\) × d₁ × d₂ sq. units
= \(\frac{1}{2}\) × 40 × 25 cm² = 500 cm²
∴ Area of 2000 titles = 500 × 20,000 = 10,000 cm² = 100 m²
Cost of polishing 1 m² = ₹ 5
∴ Cost of polishing 100 m² = 5 × 100 = ₹ 500

Students can Download Maths Chapter 4 Direct and Inverse Proportion Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Intext Questions

Try These (Text Book Page No. 72)

Question 1.
Find the ratio of the number of circles to number of squares.

Solution:
Number of circles = 4 ;
Number of squares = 6
Number of circles : Number of squares = 4 : 6 = 2 : 3

Question 2.
Find the ratio (i) 555 g to 5 kg
(ii) 21 km to 175 m.
Solution:
(i) 1 kg = 1000 g
∴ 5 kg = 5000 g
∴ 555 g : 5 kg = 555 g : 5000 g = 111 : 1000

(ii) 21 km to 175m
1 km = 1000 m
21 km = 21 × 1000 = 21,000m
∴ 21 km : 175 m = 21000 : 175 = 120 : 1

Question 3.
Find the value of x in the following proportions :
(i) 110 8: 88
(ii) x : 26 :: 5: 65
Solution:
(i) Given 110 : x :: 8 : 88
Product of the means = Product of the extremes
x × 8 = 110 × 88

(ii) x : 26 :: 5 : 65
Product of the means = Product of the extremes

Try this (Text Book Page No. 74)

Question 1.
The number of chocolates to be distributed to the number of children. Is this statement in direct proportion?
Solution:
Let the number of students be x and the number of chocolates be y. As the valve of x increases also correspondingly increases.
ie \(\frac{x}{y}=k\)
∴ x and y are in direct proportion.

Try This (Text book Page No. 74)

Question 2.
Observe the following 5 squares of different sides given in the graph sheet.

The measures of the sides are recorded in the table given below. Find the corresponding perimeter and the ratios of each of these with the sides given and complete the table.

From the information so obtained state whether the side of a square is in direct proportion to the perimeter of the square.
Solution:
Perimeter of the square y = 4x, units where x is the side of a square

Completing the table

From the above table we find that as x increases y also increased in such a way that \(\frac{x}{y}=\frac{1}{4}\), constant.
∴ Side of a square is in direct proportion to the perimeter of the square.

Try this (Text book Page No. 75)

Question 3.
When a fixed amount is deposited for a fixed rate of interest, the simple interest changes proportionally with the number of years it is being deposited. Can you find any other examples of such kind.
Solution:
Some other examples of such kind are
(i) Cost of book and number of books
(ii) Distance and time to travel
(iii) Men workers and wages.

Exercise 4.2

Try this (Text book Page No. 78)

Question 1.
Think of an example in real life where two variable are inversely proportional.
Solution:
Example of inverse proportion are
(i) Men working and the amount of work.
(ii) Speed and time to travel

Try These (Text book Page No. 78)

Question 1.
Complete the table given below and find the type of proportion.

Solution:

Question 2.
Read the following examples and group them in two categories.

Solution:

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
The base of the parallelogram is 16 cm and the height is 7 cm less than its base. Find the area of the parallelogram.
Solution:
In a parallelogram
Given base b = 16 cm; height h = base – 7 cm = 16 – 7 = 9 cm
Area of the parallelogram = (base × height) sq. units
= 16 × 9 cm² = 144 cm²
Area of the parallelogram = 144 cm²

Question 2.
An agricultural field is in the form of a parallelogram, whose area is 68.75 sq. hm. The distance between the parallel sides is 6.25 cm. Find the length of the base.
Solution:
Height of the parallelogram = 6.25 hm
Area of the parallelogram = 68.75 sq. hm
b × h = 68.75
b × 6.25 = 68.75
b = \(\frac{68.75}{6.25}=\frac{6875}{625}\) = 11 km
Length of the base = 11 km.

Question 3.
A square and a parallelogram have the same area. If the side of the square is 48m and the height of the parallelogram is 18 m. Find the length of the base of the parallelogram.
Solution:
Given side of the square is 48 m
Area of the square = (side × side) sq. unit = 48 × 48 m²
Height of the parallelogram = 18 m
Area of the parallelogram = ‘bh’ sq. units = b × 18 m²
Also area of the parallelogram = Area of the square
b × 18 = 48 × 48
b = \(\frac{{48} \times 48}{18 }\) = 8 × 16 = 128 m
Base of the parallelogram = 128 m

Question 4.
The height of the parallelogram is one fourth of its base. If the area of the parallelogram is 676 sq. cm, find the height and the base.
Solution:
Let the base of the parallelogram be ‘b’ cm
Given height = \(\frac{1}{4}\) × base ; Area of the parallelogram = 676 sq. cm
b × h = 676
b × \(\frac{1}{4}\)b = 676
b × b = 676 × 4
b × b = 13 × 13 × 4 × 4
b = 13 × 4 cm = 52 cm
Height = \(\frac{1}{4}\) × 52 cm = 13 cm
Height = 13 cm, Base 52 cm

Question 5.
The area of the rhombus is 576 sq. cm and the length of one of its diagonal is half of the length of the other diagonal then find the length of the diagonal.
Solution:
Let one diagonal of the rhombus = d₂ cm
The other diagonal d₂ = \(\frac{1}{2}\) × d₁ cm
Area of the rhombus = 576 sq. cm
\(\frac{1}{2}\) × (d₁ × d₂) = 576
\(\frac{1}{2}\) × (d₁ × \(\frac{1}{2}\) d₁) = 576
d₁ × d₁ = 576 × 2 × 2 = 6 × 6 × 4 × 4 × 2 × 2
d₁ × d₁ = 6 × 4 × 2 × 6 × 4 × 2
d₁ = 6 × 4 × 2
d₁ = 48 cm
d₂ = \(\frac{1}{2}\) × 48 = 24 cm
∴ Length of the diagonals d₁ = 48 cm and d₂ = 24 cm.

Question 6.
A ground is in the form of isoceles trapezium with parallel sides measuring 42 m and 36 m long. The distance between the parallel sides is 30 m. Find the cost of levelling it at the rate of ₹ 135 per sq. m.
Solution:
Parallel sides of the trapezium a = 42 m; b = 36 m
Also height h = 30 m
Area of the trapezium = \(\frac{1}{2}\) × h × (a + b) sq. unit
= \(\frac{1}{2}\) × 30 × (42 + 36) m²
= \(\frac{1}{2}\) × 30 × 78 m²
Area = 1,170 m²
Cost of levelling 1 m² = ₹ 135
∴ Cost of levelling 1170 m² = ₹ 1170 × 135 = ₹ 1,57,950
Cost of levelling the ground = ₹ 1,57,950

Challenge Problems

Question 7.
In a parallelogram PQRS (See the diagram) PM and PN are the heights corresponding to the sides QR and RS respectively. If the area of the parallelogram is 900 sq. cm and the length of PM and PN are 20 cm and 36 cm respectively, find the length of the sides QR and SR.

Solution:
Considering QR as base of the parallelogram height h₁ = 20 cm
Area of the parallelogram = 900 cm²
b₁ × h₁ = 900 ; b₁ × 20 = 900
b₁ = \(\frac{900}{20}\) = 45 cm
Again considering SR as base height = 36 cm ; Area = 900 cm²
b₂ × h₂ = 900 ; b₂ × 36 = 900
b₂ = \(\frac{900}{36}\)
b₂ = 25 cm
SR = 25 cm; QR = 45 cm ; SR = 25 cm

Question 8.
If the base and height of a parallelogram are in the ratio 7:3 and the height is 45 cm, then fixed the area of the parallelogram.
Solution:
Given base; height = 7 : 3
Let base = 7x cm
height = 3x cm
also given height = 45 cm
3x = 45 cm 45 .
x = \(\frac{45}{3}\) = 15
Now base = 7x cm = 7 × 15 cm = 105 cm
Area of the parallelogram = b × h sq. unit
= 105 × 45 = 4725 cm²
= 4725 cm²

Question 9.
Find the area of the parallelogram ABCD if AC is 24 cm and BE = DF = 8 cm.

Solution:
Area of the parallelogram ABCD Area of the triangle =Area of the triangle ABC + Area of the triangle ADC

Area of the parallelogram ABCD = 96 + 96 = 192 cm²

Question 10.
The area of the parallelogram ABCD is 1470 sq. cm. If AB = 49 cm and AD = 35 cm then, find the height, DF and BE.

Solution:
Area of the parallelogram = 1470 sq. cm
Considering AB = base = 49 cm
height = DF
Area = base × height
49 × DF = 1470
DF = \(\frac{1470}{49}\)
DF = 30 cm
Now considering AD as base
Base = AD = 35 cm ; height = BE
Base × Height = 1470
35 × BE = 1470 : BE = \(\frac{1470}{35}\)
BE = 42 cm ; DF = 30 cm ; BE = 42 cm

Question 11.
One of the diagonals of a rhombus is thrice as the other. If the sum of the length of the diagonals is 24 cm, then find the area of the rhombus.
Solution:
Let one of the diagonals of rhombus be ‘d₁’ cm and the other be d₂ cm.
Give d₁ = 3 × d₂
Also d₁ + d₂ = 24 cm
⇒ 3d₂ + d₂ = 24
4d₂ = 24
d₂ = \(\frac{24}{4}\)
d₂ = 6 cm
d₁ = 3 × d₂ = 3 × 6
d₁ = 18 cm
∴ Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq. units
= \(\frac{1}{2}\) × 18 × 6 cm² = 54 cm²
Area of the rhombus = 54 cm²

Question 12.
A man has to build a rhombus shaped swimming pool. One of the diagonal is 13 m and the other is twice the first one. Then find the area of the swimming pool and also find the cost of cementing the floor at the rate of ₹ 15 per sq. cm.
Solution:
Let the first diagonal d₁ = 13 m
d₂ = 2 × 13 m = 26 m
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂ sq. units
= \(\frac{1}{2}\) × 13 × 26 m² = 169m²
Cost of cementing 1 m² = ₹ 15
Cost of cementing 169 m² = ₹ 169 × 15 = ₹ 2,535
Cost of cementing = ₹ 2,535

Question 13.
Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. cm.
Solution:
Let the height be ‘A’ and base be ‘h’ units
Given b = 4 × h
Area of the parallelogram = 576 sq. cm
b × h = 576
4h × h = 576
h × h = \(\frac{576}{4}\) = 144
h × h = 12 × 12
h = 12 cm
Height = 12 cm; base = 4 × 12 = 48 cm

Question 14.
The table top is in the shape of a trapezium with measurements given in the figure. Find the cost of the glass used to cover the table at the rate of ₹ 6 per 10 sq. cm.

Solution:
Length of the parallel sides a = 200 cm
b = 150 cm
Height h = 50 cm
Area of the trapezium = \(\frac{1}{2}\) × h (a + b) sq. units
= \(\frac{1}{2}\) × 50 (200+ 150) cm²
= \(\frac{1}{2}\) × 50 × 350 cm² = 8750 cm²
Cost for 10 sq. cm glass = ₹ 68
∴ Cost of 8750 cm² glass = \(\frac{8750}{10}\) × 6 = ₹ 5250
Cost of glass used = ₹ 5,250

Question 15.
Arivu has a land ABCD with the measurements given in the figure. If a portion ABED is used for cultivation (where E is the midpoint of DC). D Find the cultivated area.

Solution:
From the figure given ABED is a trapazium with height h = 18 m
One of the parallel side a = 24 m
Since E is the midpoint of D.
Other parallel side b = \(\frac{24}{2}\) = 12 m
Area of the cultivated ADEB = \(\frac{1}{2}\) × h(a + b) m² = \(\frac{1}{2}\) × 18 (24 + 12)
= 9 × 36 m² = 324 m²
Area of cultivation = 324 m²

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 6 Information Processing Intext Questions

Exercise 6.1

Try this (Text book Page No. 111)

Question 1.
Use the given five tetrominoes only once and create the shape given below.

Solution:
Using the given five tetrominoes in the proper places we can make the given shape as follows.

Try these (Text Book Page No. 113)

Question 1.
Complete the rectangle given below using five tetrominoes only once

Solution:
The given rectangle is halfly filled with the five tetrominoes.
Using the five tetrominoes only once we can fill the rectangles as follows:

Question 2.

Solution:

Exercise 6.2

Try these (Text Book Page No. 119)

Question 1.
Observe the pictures and answer the following.
(i) Find all the possible routes from house to school via fire station.
(ii) Find all the possible routes between central park and school with distance. Mention the shortest route?
(iii) Calculate the shortest distance between bank and school.
Solution:
(i) (a) House ➝ Fire station ➝ Library ➝ Central Park ➝ Hotel ➝ Fruit shop ➝ School.
(b) House ➝ Fire station ➝ Library ➝ Fruit shop ➝ School.
(c) House ➝ Fire station ➝ Library ➝ School.

(ii) Possible routes between Central park and school and their distances are
(a) School ➝ Fruit shop ➝ Hotel ➝ Central park.
Distance ➝ (150 + 300 + 100)m = 550 m

(b) School ➝ Fruit shop ➝ Library ➝ Central park Distance ➝(150 + 100 + 200)m
= 450 m

(c) School ➝ Library ➝ Central park
Distance = (20 + 200 m)
= 220 m

(d) School ➝ Library ➝ Fire station ➝ House ➝ central park Distance = (20 + 50 + 300 + 150) m
= 520 m

(e) School ➝ Emit shop ➝ Hotel ➝ Bank ➝ House ➝ Central park
Distance ➝ (150 + 300 + 150 + 200 + 150) m
= 950 m

(f) School ➝ fruit shop ➝ Hotel ➝ Bank ➝ House ➝ Fire station ➝ Library ➝ central park
Distance = (150 + 300 + 150 + 200 + 300 + 50 + 200)m = 1350 m
∴ Route (c) is the shortest path (i.e.,) School ➝ Library ➝ Central park

(iii) Shortest distance between school and Bank is calculated as follows Bank ➝ Hotel ➝ Central park ➝ Library ➝ school.
Distance = (150 + 100 + 200 + 20)m
= 470 m

Question 2.
A School has planned for a trip to Ooty. Using the route map, the school decides to visit the places such as Boat House Adam Fountain and Botanical Garden.

(i) How much distance you have to travel to Botanical Garden from Ooty Boat House?
(ii) Find the shortest route to Botanical from the Ooty main Bus stand.
(iii) Mention the direction of Botanical Garden from Adam Foundation.
(iv) In what direction, Ooty Boat House is situated from Ooty Main Bus Stand. Complete the following route map from Ooty Main Bus Stand to Botanical Garden.

Solution:
(i) (a) 700 m + 8.1 km = 700 m + (8 km + 100 m) = 8 km + 800 m = 8.8 km
(b) 700 m+ 1.7 km + 1.5 km = 700 m + (1 km, 700 m) + 1 km 500 m
= 2 km + 1900 m
= 2 km + 1 km + 900 m
= 3 km + 900 m
= 3.9 km
From the Ooty Boat House, the Botanical Garden is at a distance of 3.9 km(shortest)

(ii) The route from Botanical Garden from Ooty main Bus stand are
(a) Ooty main Bus stand ➝ Boat house ➝ Government Botanical garden Distance 1.5 km + 8.1 km + 700 m
= 1 km 500 m + 8 km 100 m + 700 m
= 9 km 1300 m
= 9 km + 1 km +300 m
= 10 km 300 m
= 10.3 km.

(b) Another route.
Ooty Main Bus stand ➝ Adam Fountain ➝ Botanical garden.
Distance = 1.7 + 700 m
= 1 Km 700 m + 700m
= 1 Km 1400 m
= 1 Km+ 1 Km 400 m
= 2 Km 400 m
= 2.4Km
∴ Shortest Route is Ooty main Bus stand ➝ Adams Foundation ➝ Botanical garden

iii. Botanical garden is north of Adam Foundation.

iv. Ooty Boat House is situated to the west of Ooty main bus bus stand.

v.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks
(i) The addition of – 7b and 2b is _______
(ii) The subtraction of 5m from -3m is ______
(iii) The additive inverse of -37xyz is _____
Solution:
(i) -5b
(ii) -8m
(iii) 37xyz

Question 2.
Say True or False
(i) The expressions 8x + 3y and 7x + 2y cannot be added
(ii) If x is a natural number, then x + 1 is its predecessor.
Hint: x – 1 is its predecessor.
(iii) Sum of a – b + c and -a + b – c is zero
Solution:
(i) False
(ii) False
(iii) True

Question 3.
Add: (i) 8x, 3x
(ii) 7mn, 5mn
(iii) -9y, 11y, 2y
Solution:
(i) 8x + 3x = (8 + 3) x = 11x
(ii) 7mn + 5mn = (7 + 5)mn = 12mn
(iii) -9y + 11y + 2y =(-9 + 11 + 2 )y = (2 + 2)y = 4y

Question 4.
Subtract:
(i) 4k from 12k
(ii) 15q from 25q
(iii) 7xyz from 17xyz
Solution:
(i) 4k from 12k
12k – 4k = (12 – 4) k = 8k
(ii) 15q from 25q
25q – 15q = (25 – 15)q = 10q
(iii) 7xyz from 17xyz
17xyz – 7xyz = (17 – 7)xyz = 10xyz

Question 5.
Find the sum of the following expressions
(i) 7p + 6q, 5p – q, q + 16p
Solution:
(7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p
= (7p + 5p + 16p) + (6q – q + q)
= (7 + 5 + 16) p + (6 – 1 + 1) q
= (12 + 16) p + 6q = 28p + 6q

(ii) a + 5b + 7c, 2a + 106 + 9c
Solution:
(a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c
= a + 2a + 5b + 10b + 7c + 9c
= (1 + 2)a + (5 + 10)b + (7 + 9)c
= 3a + 15b + 16c

(iii) mn + t, 2mn – 2t, – 3t + 3mn
Solution:
(mn + t) + (2mn – 2t) + (-3t + 3mn)
= mn + t + 2mn – 2t + (-3t) + 3mn
= (mn + 2mn + 3mn) + (t – 2t – 3t)
= (1 + 2 + 3) mn + (1 – 2 – 3) t
= 6mn + (1 – 5)t
= 6mn + (- 4) t
= 6mn – 4t

(iv) u + v, u – v, 2u + 5v, 2u – 5v
Solution:
(u + v) + (u – v) + (2u + 5v) + (2u – 5v)
= u + v + u – v + 2u + 5v + 2u – 5v
= u + u + 2u + 2u + v – v + 5v – 5v
= (1 + 1 + 2 + 2) u +(1 – 1 + 5 – 5)v = 6u + 0v
= 6u

(v) 5xyz – 3xy, 3zxy – 5yx
Solution:
5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy
= (5 + 3) xyz + [(-3) + (-5)] xy = 8xyz + (-8) xy
= 8xyz – 8xy

Question 6.
Subtract
(i) 13x + 12y – 5 from 27x + 5y – 43
Solution:
27x + 5y – 43 – (13x + 12y – 5) = 27z + 5y – 43 + (-13x – 12y + 5)
= 27x + 5y – 43 – 13x – 12y + 5
= (27 – 13) x + (5 – 12)y + (- 43) + 5
= 14x + (- 7) y + (- 38) = 14x – 7y – 38

(ii) 3p + 5 from p – 2q + 7
Solution:
p – 2q + 7 – (3p + 5) = p – 2q + 7 + (- 3p – 5)
= p – 2q + 7 – 3p – 5 = p – 3p – 2q + 7 – 5
= (1 – 3)p – 2q + 2 = -2p – 2q + 2

(iii) m + n from 3m – 7n
Solution:
3m – 7n – (m + n) = 3m – 7n + (-m – n)
= 3m – 7n – m – n = (3m – m) + (-7n – n)
= (3 – 1 )m + (-7 – 1) n = 2m + (-8) n
= 2m – 8n

(iv) 2y + z from 6z – 5y
Solution:
6z – 5y – (2y + z) = 6z – 5y + (-2y – z)
= 6z – 5y – 2y – z = 6z – z – 5y – 2y
= (6 – 1) z + (-5 -2) y = 5z + (-7) y
= 5z – 7y = -7y + 5z

Question 7.
Simplify
(i) (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)
Solution:
(x + y – z) + (3x – 5y + 7z) – (14x – 7y – 6z)
= (x + y – z) + (3x – 5y + 7z) + (-14x – 7y + 6z)
= (x + 3x – 14x) + (y – 5y – 7y) + (-z + 7z + 6z)
= (1 + 3 – 14) x + (1 – 5 – 7)y + (-1 + 7 + 6) z
= – 10x – 11y + 12z

(ii) p + p + 2 + p + 3 + p – 4 – p – 5 + p + 10
Solution:
p + p + 2 + 3 – p – 4 – p – 5 + p + 10 = (p + p + p – p – p + p) + (2 + 3 – 4 – 5 + 10)
= (1 + 1 + 1 – 1 – 1 + 1) p + 6 = 2p + 6

(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
Solution:
n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
= n + m + 1 + n + 2 + m + 3 + n + 4 + m + 5
= n + n + n + m + m + m + 1 + 2 + 3 + 4 + 5
= (1 + 1 + 1)n + (1 + 1 + 1)m + 15
= 3n + 3m + 15 = 3m + 3n + 15

Objective Type Questions

Question 8.
The addition of 3mn, -5mn, 8mn and – 4mn is
(i) mn
(ii) – mn
(iii) 2mn
(iv) 3mn
Solution:
(iii) 2mn
Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn

Question 9.
When we subtract ‘a’ from ‘-a’, we get ______
(i) a
(ii) 2a
(iii) -2a
(iv) -a
Solution:
(iii) -2a
Hint: – a – a = – 2a

Question 10.
In an expression, we can add or subtract only _____
(i) like terms
(ii) unlike terms
(iii) all terms
(iv) None of the above
Solution:
(i) like terms

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Additional Questions

Exercise 4.1

Question 1.
The amount of extension in an elastic spring varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?
Solution:
To produce 2.9 cm extension weight needed = 150 gm

Question 2.
Reeta types 540 words during half on hour. How many words would she type in 12 minutes?
Solution:
In \(\frac{1}{2}\) an hour number of words typed = 540
i.e., In 30 min No. of words typed = 540

= 18
In 12 minutes number of words typed = 18 × 12
= 216
216 words can be typed in 12 min

Question 3.
A call taxi charges ₹ 130 for 100 km. How much would one travel for ₹ 390?
Solution:

Exercise 4.2

Question 1.
In the following table find out x and y vary directly or inversely?

Solution:
From the table itself we observe that as x increases y decreases.
∴ x and y are inversely proportional
∴ xy = 8 × 32 = 16 × 16 = 32 × 8 = 256 × 1 = 256

Question 2.
If x and y vary inversely as each other and x = 10 when y = 6. Find y when x = 15.
Solution:
Since x and y vary inversely as each other
xy = constant
10 × 6 = 15 xy
60 = 15y

Question 3.
If x and y vary inversely and if y = 35 find x when constant of variation is 7.
Solution:
Given x andy are inversely proportional
xy = constant
when y = 35 and constant = 7 ; x × 35 = 7

Exercise 4.3

Question 1.
Sumathi sweeps 600 m long road in 2\(\frac{1}{2}\) hrs. Ramani sweeps \(\frac{2}{3}\) rd of same road in 1\(\frac{1}{2}\) hrs. Who sweeps more speedily?
Solution:

Question 2.
Suma weaves 25 baskets in 35 days. In how many days will she weave 110 baskets?
Solution: