Laxmi

Students can Download Maths Chapter 3 Geometry Ex 3.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3

Question I.
Construct the following trapeziums with the given measures and also find their area.

Question 1.
AIMS with \(\overline { AI } \) ∥ \(\overline { SM } \), AI = 6 cm, IM = 5 cm, AM = 9 cm and MS 6.5 cm.
Solution:

Given
AI = 6cm
IM = 5cm
AM = 9cm, and \(\overline { AI } \) ∥ \(\overline { SM } \)
MS = 6.5 cm

Construction:
Steps:

1. Draw a line segment AI = 6cm.
2. With A and I as centres, draw arcs of radii 9 cm and 5 cm respectively and let them cut at M
3. Join AM and IM.
4. Draw MX parallel to AI
5. With M as centre, draw an arc of radius 6.5 cm cutting MX at S.
6. Join AS AIMS is the required trapezium.

Calculation of Area:
Area of the trapezium AIMS = \(\frac{1}{2}\) x h x (a + b) sq.units
= \(\frac{1}{2}\) x 4.6 x (6 + 6.5) = \(\frac{1}{2}\) x 4.6 x 12.5
= 28.75 Sq.cm

Question 2.
BIKE with \(\overline { BI } \) ∥ \(\overline { EK } \), BI = 4 cm, IK = 3.5 cm, BK = 6 cm and BE = 3.5 cm
Solution:

Given:
In the trapezium BIKE,
BI = 4 cm
IK = 3.5 cm
BK = 6 cm
BE = 3.5 cm and \(\overline { BI } \) ∥ \(\overline { EK } \)

Construction:
Steps:

1. Draw a line segment BI = 4 cm.
2. With B and I as centres, draw arcs of radii 6 cm and 3.5 cm respectively and let them cut at K.
3. Join BK and IK
4. Draw KX parallel to BI
5. With B as centre, draw an arc of radius 3.5 cm.cutting KX at E
6. Join BE. BIKE is the required trapezium.

Calculation of area:
Area of the trapezium BIKE = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 3.5 x (4 + 4.2)
= \(\frac{1}{2}\) x 3.5 x 8.2 = 14.35 sq.cm

Question 3.
CUTE with \(\overline { CD } \) ∥ \(\overline { ET } \), CU = 7 cm, ∠UCE = 80°, CE = 6 cm and TE = 5 cm.
Solution:

Given:
In the trapezium CUTE,
CU = 7 cm, ∠UCE = 80°,
CE = 6 cm, TE = 5 cm and \(\overline { CD } \) ∥ \(\overline { ET } \)

Construction:
Steps:

1. Draw a line segment CU = 7 cm.
2. Construct an angle ∠UCE = 80° at C
3. With C as centre, draw an arc of radius 6 cm cutting CY at E
4. Draw EX parallel to CU
5. With E as centre, draw an arc of radius 5 cm cutting EX at T
6. 6. Join UT. CUTE is the required trapezium.

Calculation of area:
Area of the trapezium CUTE = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 5.9 x (7 + 5) sq. units
= 35.4 sq.cm –

Question 4.
DUTY with \(\overline { DU } \) ∥ \(\overline { YT } \), DU = 8 cm, ∠DUT = 60°, UT = 6 cm and TY = 5 cm.
Solution:

Given:
In the trapezium DUTY
DU = 8 cm, ∠DUT = 60°,
UT = 6 cm, TY = 5 cm and \(\overline { DU } \) ∥ \(\overline { YT } \)

Construction:
Steps:

1. Draw a line segment DU = 8 cm.
2. Construct an angle ∠DUT = 60° at U
3. With U as centre, draw an arc of radius 6 cm cutting UA at T.
4. Draw TX parallel to DU
5. With T as centre, draw an arc of radius 5 cm cutting TX at Y
6. Join DE. DUTY is the required trapezium.

Calculation of area:
Area of the trapezium DUT Y = \(\frac{1}{2}\) x h x (a + b) sq. units= \(\frac{1}{2}\) x 5.2 x (8 + 5) sq. units = \(\frac{1}{2}\) x 5.2 x 13
= 33.8 sq.cm ,

Question 5.
ARMY with \(\overline { AR } \) ∥ \(\overline { YM } \), AR = 7 cm, RM = 6.5 cm ∠RAY = 100° and ∠ARM = 60° 5
Solution:

Given:
In the trapezium ARMY
AR = 7 cm, RM = 6.5 cm,
∠RAY = 100° and ARM = 60°, \(\overline { AR } \) ∥ \(\overline { YM } \)

Construction:
Steps:

1. Draw a line segment AR = 7 cm.
2. Construct an angle ∠RAX = 100° at A
3. Construct an angle ∠ARN = 60° at R
4. With R as centre, draw an arc of radius 6.5 cm cutting RN at M
5. Draw MY parallel to AR
6. ARMY is the required trapezium.

Calculation of area:
Area of the trapezium ARMY = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 5.6 x (7 + 4.8) sq. units
= \(\frac{1}{2}\) x 5.6 x 11.8 = 33.04 sq.cm

Question 6.
BELT with \(\overline { BE } \) ∥ \(\overline { TL } \), BT = 7 cm ∠EBT = 85° and ∠BEL = 110°
Solution:

Given:
In the trapezium BELT
BE = 10 cm, BT = 7cm,
∠EBT = 85°, ∠BEL = 110° and \(\overline { BE } \) ∥ \(\overline { TL } \)

Construction:
Steps:

1. Draw a line segment BE = 10 cm.
2. Construct two angles ∠TBE = 85° and ∠BEL =110° respectively at the points B and E.
3. With B as centre, draw an arc of radius 7 cm cutting BX at T.
4. Draw TL ∥ BE
5. BELT is the required trapezium

Question 7.
CITY with \(\overline { CI } \) ∥ \(\overline { YT } \) Cl = 7 cm, IT = 5.5 cm, TY = 4 cm and YC = 6 cm.
Solution:

Given:
In the trapezium CITY,
Cl = 7 cm
IT = 5.5 cm
TY = 4 cm
YC = 6 cm, and \(\overline { CI } \) ∥ \(\overline { YT } \)

Construction:
Steps:

1. Draw a line segment Cl = 7 cm.
2. Mark a point D on Cl such that CD = 4cm
3. With D and I as centres, draw arcs of radii 6 cm and 5.5 cm respectively. Let them cut at T. Join DT and IT.
4. With C as centre, draw an arc of radius 6 cm.
5. Draw TY parallel to CL Let the line cut the previous arc at Y.
6. Join CY. CITY is the required trapezium.

Calculation of area:
Area of the trapezium CITY = \(\frac{1}{2}\) x h x (a + b) sq. units
= \(\frac{1}{2}\) x 5.5 x (7 + 4) sq. units = \(\frac{1}{2}\) x 5.5 x 11
= 30.25 sq.cm

Question 8.
DICE with \(\overline { DI } \) ∥ \(\overline { EC } \), DI = 6 cm, IC = ED = 5 cm and CE = 3 cm. Solution:

Given:
In the trapezium DICE,
DI = 6 cm
IC = ED = 5 cm
CE = 3 cm and \(\overline { DI } \) ∥ \(\overline { EC } \)

Construction:
Steps:

1. Draw a line segment DI = 6 cm.
2. Mark a point M on DI such that DM = 3cm
3. With D and I as centres, draw arcs of radii 5 cm each Let them cut at C. Join MC and IC.
4. Draw CX parallel to DI
5. With D as centre, draw an arc of radius 5 cm. Let it cut CX at E
6. Join DE. DICE is the required trapezium.

Calculation of area:
Area of the trapezium DICE = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 3.8 x (6 + 3) sq. units
= \(\frac{1}{2}\) x 3.8 x 9 = 17.1 sq. cm

Students can Download Maths Chapter 1 Number System 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 3 Chapter 1 Number System Additional Questions

Additional Questions and Answers

Exercise 1.1

Question 1.
Match the following:

Solution:
1 – (v)
2 – (iv)
3 – (i)
4 – (iii)
5 – (ii)

Question 2.
Round 89.357 to the nearest whole number.
Solution:
Underlining the digit to be rounded 89.357. Since the digit next to the underlined digit 3 which is less than 5, the underlined digit remains the same.
∴ The nearest whole number 89.357 rounds to 89.

Question 3.
Round 110.929 to the nearest tenths place.
Solution:
Underlining the digit to be rounded 110.929. Since the digit next to the underlined digit is 2 which is less than 5.
∴ The underlined digit 9 remains the same. Hence the rounded number is 110.9

Question 4.
Round 87.777 upto 2 places of decimal.
Solution:
Rounding 87.777 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.777 gives 87.777. Since the digit after the hundredth place value is 7 which is more than 5, we add 1 to the underlined digit. So the rounded value of 87.777 upto 2 places of decimal is 87.78

Exercise 1.2

Question 1.
If Sheela bought 2.083 kg of grapes and 3.752 kg of orange. What is the total weight of fruits
Solution:

Weight of grapes = 2.083 Kg
Weight of orange = 2.752 Kg
Total weight = (2.083 + 2.752) Kg = 4.835 Kg

Question 2.
Kathir bought 8.72 kg of sugar, 7.302 kg of grains. His carry bag can contain only 15 kg of weight. What is the remaining weight of items bought?
Solution:

Question 3.
Use place value grid to add 7.357 and 13.92.
Solution:
Let as use place value grid.

Exercise 1.3

Question 1.
Cost of 1m cloth is ₹ 6.75. Find the cost of 14.75m correct to two places of decimal.
Solution:

Cost of 1 m cloth = ₹ 6.75
Cost of 14.75 m cloth = 14.75 × 6.75
= ₹ 99.5625
= ₹ 99.56

Question 2.
Length of a side of a square is 18.35 cm. Find its Area.
Solution:

Side of a square = 18.35 cm
Area of a square = (Side × Side) sq.units
= 18.35 × 18.35 cm²
= 336.7225 cm²

Exercise 1.4

Question 1.
A wire of length 363.987m is cut into 30 pieces. What is the length of each piece?
Solution:
Length of the wire = 363.987m
i.e Total length of 30 pieces = \(\frac { 363987 }{ 1000 } \) m

∴ Length of 1 piece
= 12132.9 × \(\frac { 1 }{ 1000 } \)
Length of 1 piece of wire = 12.1329 m

Question 2.
A cake of 50kg needs 23.4 kg sugar. Find the weight of cake made by 1 kg of sugar.
Solution:
Weight of cake made using 23.4 kg sugar = 50 kg
Weight of cake made using 1 kg sugar = \(\frac { 50 }{ 23.4 } \)
= \(\frac { 50 }{ 23.4 } \) x \(\frac { 10 }{ 10 } \) = \(\frac { 500 }{ 234 } \) = 2.1367 Kg
= 2.14 Kg
Weight of cake made using 1 kg sugar = 2.14 Kg

Question 3.
A pack of 20 pencils cost ₹ 94.4. What is the cost of each pencil?
Solution:
Cost of 20 pencils = ₹ 94.4

∴ Cost of 1 pencil = ₹ 4.72

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.4 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 3 Chapter 2 Percentage and Simple Interest Ex 2.4

Question 1.
Find the simple interest on ₹ 35,000 at 9% per annum for 2 years?
Solution:

Principal P = ₹ 35,000
Rate of interest r = 9 % Per annum
Time (n) = 2 years
Simple Interest I = \(\frac { Pnr }{ 100 } \) = \(\frac{35000 \times 2 \times 9}{100}\) = ₹ 6300
Simple intrest I = ₹ 6300

Question 2.
Aravind borrowed a sum of ₹ 8,000 from Akash at 7% per annum. Find the interest and amount to be paid at the end of two years.
Solution:
Here Principal P = ₹ 8,000
Rate of interest r = 7% Per annum
Time (n) = 2 Years
Simple Interest (I) = \(\frac { Pnr }{ 100 } \) = \(\frac{8000 \times 2 \times 7}{100}\)
I = ₹ 1120
Amount = P + I
I = ₹ 8000 + 1120 = 9120
Interest to be paid = ₹ 1,120
Amount to be paid = ₹ 9,120

Question 3.
Sheela has paid simple interest on a certain sum for 4 years at 9.5% per annum is ₹ 21,280. Find the sum.
Solution:
Let the Principal be ₹ P
Rate of interest r = 9.5% per annum
Time (n) = 4 years
Simple Interest I = \(\frac { Pnr }{ 100 } \)
Given I = ₹ 21,280

∴ Sum of money Sheela bought = ₹ 56,000

Question 4.
Basha borrowed ₹ 8,500 from a bank at a particular rate of simple interest. After 3 years, he paid ₹ 11,050 to settle his debt. At what rate of interest he borrowed the money?
Solution:
Let the rate of interest be r% per annum
Here Principal P = ₹ 8,500
Time n = 3 years
Total amount paid = ₹ 11,050
A = P + 1 = ₹ 11,050
i.e. 8,500 + 1 = ₹ 11,050
I = ₹ 11,050 – ₹ 8,500 = ₹ 2,550

Question 5.
In What time will ₹ 16,500 amount to ₹ 22,935 at 13% per annum?
Solution:
Rate of interest r = 13% per annum
Here Amount A = ₹ 22,935
Principal P = ₹ 16,500
A = P + I
22935 = 16,500 + I
∴ Interest I = 22935 – 16,500 = ₹ 6,435
Simple Interest I = \(\frac { pnr }{ 100 } \)

6435 = \(\frac{16500 \times n \times 13}{100}\)
n =\(\frac{6435 \times 100}{16500 \times 13}\)
n = 3 years
Required time n = 3 years

Question 6.
In what time will ₹ 17800 amount to ₹ 19936 at 6% per annum?
Solution:
Let the require time be n years
Here Principal P = ₹ 17,800
Rate of interest r = 6% per annum
Amount A = ₹ 19,936
A = P + I
19936 = 17800 + 1
19936 – 17800 = I
2136 = I
Simple Interest (I) = \(\frac { pnr }{ 100 } \)
2136 = \(\frac{17800 \times n \times 6}{100}\)
n = \(\frac{2136 \times 100}{17800 \times 6}\)

n = 2 Years
Required time = 2 years

Question 7.
A sum of ₹ 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was ₹ 55,560. Find the rate of interest per year.
Solution:
Given Principal P = ₹ 48,000
Time n = 2 years 3 months
= 2 + \(\frac { 3 }{ 12 } \) years = 2 + \(\frac { 1 }{ 4 } \) years
= \(\frac { 8 }{ 4 } \) + \(\frac { 1 }{ 4 } \) years = \(\frac { 9 }{ 4 } \) years
Amount A = ₹ 55,660
A = p + 1
55660 = 48000 + I
I = 55660 – 48000 = ₹ 7660
∴ Interest for \(\frac { 9 }{ 4 } \) years = ₹ 7660
Simple intrest = \(\frac { pnr }{ 100 } \)
7660 = 48000 × \(\frac { 9 }{ 4 } \) × \(\frac { r }{ 100 } \)
r = \(\frac{7660 \times 4 \times 100}{9 \times 48000}\) = 7.09 % = 7 %
Rate of interest = 7 % Per annum

Question 8.
A principal becomes ₹ 17,000 at the rate of 12% in 3 years. Find the principal.
Solution:
Given the Principal becomes ₹ 17,000
Let the principle initially be P
Rate of Interest r Time = 12 % Per annum
Time n = 3 years
According to the problem given I = 17000 – P = \(\frac{P \times 3 \times 12}{100}\)
17000 = \(\frac { 36 }{ 100 } \) p + p
17000 = p(\(\frac { 36 }{ 100 } \) + 1)
17000 = p(\(\frac { 136 }{ 100 } \))
p = \(\frac{17000 \times 100}{136}\) = 12,500
∴ Principal P = ₹ 12,500

Objective Type Questions

Question 9.
The interest for a principle of? 4,500 which gives an amount of? 5,000 at end of certain period is
(i) ₹ 500
(ii) ₹ 200
(iii) 20%
(iv) 15%
Hint: Interest = Amount – Principle = ₹ 5000 – ₹ 4500 = ₹ 500
Answer:
(i) ₹ 500

Question 10.
Which among the following is the simple interest for the principle of ₹ 1,000 for one year at the rate of 10% interest per annum?
(i) ₹ 200
(ii) ₹ 10
(iii) ₹ 100
(iv) ₹ 1,000
Hint: Intrest = \(\frac { pnr }{ 100 } \) = \(\frac{1000 \times 1 \times 10}{100}\) = ₹ 100
Answer:
(iii) ₹ 100

Question 11.
Which among the following rate of interest yields an interest of ₹ 200 for the principle of ₹ 2,000 for one year.
(i) 10%
(ii) 20%
(iii) 5%
(iv) 15%
Hint: r = \(\frac{I \times 100}{P \times n}\) = \(\frac{200 \times 100}{2000 \times 1}\) = 10 %
Answer:
(i) 10%

Students can Download English Lesson 4 The Midnight Visitor Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English 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 12th English Solutions Supplementary Chapter 4 The Midnight Visitor

Warm up

Study the title of the story ‘The Midnight Visitor’. Discuss in groups what the story is all about.
Student activity:

Question (a)
Certain professionals can be identified by their appearance.
Answer:
What comes to your mind first when you think of a ‘pilot’ or a ‘traffic policeman?
Discuss in pairs and share your thoughts with the class:
A pilot controls and steers an airplane. He operates the directional flight controls. He wears milk white uniform and golden stripes on his shoulders. He wears a shiny black cap. A traffic policeman wears white stripes on his shoulders in a khaki uniform. In some states, a traffic policeman wears a white and white uniform also. He regulates traffic, fines people who violate traffic rules. He prevents accidents by monitoring over speeding vehicles and by discouraging drunken driving.

Question (b)
Let us try to picturise people in a few interesting professions (based on common perception there can be exceptions).
Answer:
Form groups of four and draw a picture of one or two of the following:

• scientist
• soldier
• journalist

In your attempt to sketch you may include the following:

• typical dress
• hair style
• accessories

Samacheer Kalvi 12th English The Midnight Visitor Textual Questions

1. Answer in a sentence or two the following questions.

Question (a)
Mention two features of Ausable that were uncharacteristic of a detective.
Answer:
Ausable’s fat body and American accent were the two uncharacteristic features of a detective.

Question (b)
What was Ausable waiting for?
Answera;
Ausable was waiting for an important report.

Question (c)
Who was the Midnight Visitor? What was the purpose of his visit?
Answer:
Max, a gunman was the midnight visitor. His purpose was to snatch the report on new missiles which was about to be delivered to Ausable

Question (d)
How had Max actually entered the room?
Answer:
Max had used a duplicate key to enter Ausable’s room.

Question (e)
Did Max’s presence alarm Ausable?
Answer:
No, he was a little startled. But he was not alarmed. He just expressed his surprise seeing Max who should have been in Berlin.

Question (f)
How did Ausable describe the balcony and the manner in which one could get into his room, through it?
Answer:
It was not Ausable’s balcony. It belonged to the next apartment. It extends under his window now. One can get on to it two doors down as someone did last month. The hotel management, in spite of his protest, has not blocked the balcony till now.

Question (g)
Where did Max try to hide himself?
Answer:
Max tried to hide himself in the balcony described by Mr. Ausable.

Question (h)
Who was Henry? Why had he visited Ausable’s room?
Answer:
Henry was the waiter who entered Ausable’s room with his drinks.

Question (i)
What happened to Max finally?
Answer:
Max foolishly, believing in the presence of a balcony, jumped from the 6th floor to his sure death.

Additional Question

Question (a)
How does Ausable say he got in?
Answer:
Ausable wants to confuse Max. So he says that this is the second time in a month that somebody has got into his room through that balcony next to the window.

Question (b)
Was Max deserving to get a chance to accomplish a risky task?
Answer:
I do not believe that Max deserved a chance to accomplish a risky task. He was not intelligent enough to understand that he was being tricked by the detective. Also, his appearance did not startle the detective.

Question (c)
What did so many people risk their lives for?
Answer:
Ausable was waiting for a report. It contained certain important information about new missiles. This report was so important that several men and women had risked their lives to get it.

Question (d)
What did Mr. Ausable tell Max when he heard the knock?
Answer:
Ausable told Max that it would be the police. He said that he had requested the police for extra protection as he was receiving an important information that night.

Question (e)
What did Ausable tell Fowler when he said that Max would soon come back from the balcony?
Answer:
Ausable told Fowler that Max would never return as he knew that there was no balcony outside. He said that Max has fallen down to the ground and met his end.

2. Answer the following questions in about three to four sentences each.

Question (a)
Who was Fowler? Why did he meet Ausable?
Answer:
Fowler was a journalist. He was assigned to write a column about a private detectives. He was disappointed on seeing a very fat man with American accent living in a small room in the 6th ’ floor. The hotel was not even well-lit.

Question (b)
Why was Fowler initially disillusioned with Ausable?
Answer:
Fowler had a romantic notion of a private detective. Ausable did not look like Mr. Bond. He was very fat and had American accent. He lived in a small room. So, he was initially disappointed.

Question (c)
Fowler was thrilled when he entered Ausable’s room. Why?
Answer:
As soon as Ausable closed the door behind and switched on the light, Fowler had his first authentic thrill of the day. Half way across the room a man with a small automatic pistol in hand stood.

Question (d)
How, according to Ausable, had Max entered the room?
Answer:
Ausable did not disclose the real method Max must have adopted to enter his room. He invented a story on the spot that it was the second time in a month someone had entered his room through the neighbouring building’s balcony which extended just below his window. He made Max believe that there was another possible route of escape through the window.

Question (e)
How did the three men react to the knocking at the door?
Answer:
Fowler jumped at the sudden knocking at the door. Ausable smiled and said it must be the police who he had asked to check on him to ensure a little extra protection. Max bit his lip nervously. He rushed to the window so that he could hide in the balcony and come back after sometime.

Question (f)
Was Ausable really waiting for the police? Give reasons.
Answer:
No, Ausable was not waiting for the police. He had ordered his drink with waiter Henry. He was expecting him only. As the knock was heard, he decided to use the opportunity to scare away Max by casually informing that it must be the police. He added the police may fire if they see Max there.

Additional Questions

Question (a)
Do you think that Ausable was a good secret agent? How?
Answer:
Yes, I think that Ausable was a good secret agent. When he saw Max in his room with an automatic pistol, he didn’t get frightened at all. He diverted his attention. He made story of balcony next to the window. By his description, he made Max believe his story. All these traits show that he was a good secret agent.

Question (b)
Pride before a fall befits Max, who arrived at Ausable’s room to steal the important documents. Did he acquire it? What lesson can one learn from this?
Answer:
Max was an overconfident and proud spy. He thought to himself that wielding a gun would give a further edge over the apparently slow Ausable. However, his pride and confidence, made Max utterly unsuccessful. Ausable conveniently outwitted him without lifting a finger. Max, jumped to his own death. Thus, one learns that one should not be too proud, and must be vigilant before attempting to do anything.

Question (c)
How did Max enter the room? Why did he tell this to Ausable?
Answer:
Max entered the room through the door. He had a passkey. He told this to Ausable because Ausable told him that it was the second time in a month that somebody had got into his room through the balcony. He told that he had no idea about balcony.

3. Answer in a paragraph of about 150 words the following questions.

Question (a)
How did Ausable outwit Max?
Answer:
Ausable was a shrewd private detective. He did not become panicky on seeing Max with an automatic gun in his room. Even if he were, he did not show it off. Instead, he expressed surprise that he expected him to be in Berlin. He cooked up a nice story that it was the second time someone had broken into his room through the balcony of the neighbouring apartment which reaches down under his window. He expressed his displeasure that he would raise hell with the hotel management for not blocking that balcony. This gave a strong suggestion to Max, that he could have used the balcony instead of the pass key to enter Ausable’s room.

Being a criminal or spy, a person always looks for various routes of escape in times of danger. Very rarely he starts direct encounter risking his life. When Henry, the waiter who arrived with the pre-ordered drinks, knocked, Mr. Ausable simply smiled. When Max asked who it was, he told a blatant lie that it was the police who had come for his extra protection and wouldn’t hesitate to fire as the door wasn’t locked but just closed. Max, assuming that there is a balcony extending below Ausable’s window, jumped. He never knew that it was a suicidal jump from the 6th floor. Thus, Ausable outwitted the spy, Max.

Question (b)
Describe the significance of the balcony.
Answer:
On seeing Max, the spy, with a loaded automatic gun in his room, Ausable expressed surprise. Max disclosed his plan very clearly. He had come to receive the report on the new missile which was likely to be handed over to Mr. Ausable in a short while. Ausable, without being shocked sat heavily on an armchair. He grimly stated that he would raise hell with the hotel management because this was the second time that someone had sneaked into his room through the nuisance of the unblocked balcony. Max asks with disbelief, “balcony?” Max remarked that he did not enter through the balcony but with a pass key. Ausable explained that it was the balcony which extended from the living room of the next apartment just below his room. One could walk through two doors and enter his room.

He was not happy with the management as they had failed to block it. When there was a knock at the door, both Max and Fowler got perturbed. But Ausable smiled and said casually it must be the police whom he had informed to check on him for extra protection. Max was confused for a moment. Ausable said as the door was just closed and not locked, the police could enter even by force at any moment and fire at him. This gave Max no time to think. He jumped through the window believing he will end up in the non-existent balcony, but fell like a stone from the 6th floor with a scream. The vividly portrayed balcony led to the suicidal jump of Max, the spy who had a gun but was not smart enough to tackle Ausable’s ploy.

Question (c)
Ausable planned to get rid of Max the very moment he noticed him. Explain with supporting evidence from the story.
Answer:
Ausable spun a story on seeing Max. Max was armed, and he was unarmed. Any wrong move would cost his life. So, he decided to be cool throughout. He feigned anger against the hotel management. His vivid word picture of a balcony extending just below his window from the next apartment makes both Fowler and Max believe that Ausable is angry and irritated. This makes Max happy for learning another possible way of escape in case any threat comes through the door. He tries to add little pieces of evidence like “the management promised to block it” to make Max believe that someone had already used the same balcony to break into

Ausable’s room earlier. Ausable knew that Henry would arrive soon with the drinks he had ordered. So, he timed his narrative in such a way that the arrival of Henry takes place soon after his vivid word portrait of the balcony to Max and Fowler. His ingenious idea of relating the knock of the waiter to a non-existent police officer is a stroke of genius. Because the fear of encountering police and a possible gun fight only goads Max to jump to his death from the 6th floor of the hotel. He thought he was jumping down on to the balcony but he was outwitted. These facts indicate that Ausable had planned to get rid of Max soon after he saw him in his room.

Question (d)
Sketch the character of Ausable.
Answer:
Ausable does not look very handsome, worthy of being called a secret agent or a detective. He is the central character of the story “The Midnight Visitor”. He is not physically very strong. There is nothing elegant or mysterious about him. He is an American who is unable to cover up his American accent when he speaks French and German though he has lived in France for over 20 years. He is practical and shrewd. He is a well balanced individual. Even at gun point he keeps his cool and instantly cooks up a story to trap the villain, spy Max. He outwits Max without moving from his armchair. He uses his presence of mind. He tells two lies which not only save him and Fowler but also give the momentum for Max to kill himself in an attempt to hide in the non-existent balcony.

Question (e)
Do you think physical appearance matters most for a secret agent? Answer giving reasons in the context of the story ‘The Midnight Visitor.’
Answer:
Physical appearance is important for heroes like James Bond who acts in movies or plays. They need sophisticated cars, a royal life style to flaunt about. But in reality, a detective or a secret agent is not much different from an ordinary citizen at least in appearance. He is an ordinary person who thinks and acts with extraordinary intelligence. When it comes to the question of survival, sharpness of wit and handsome looks would help a person, the disappointment of Fowler, the journalist is very obvious. The young and romantic writer envisioned mysterious figures in the night, the crack of pistols, etc.

The writer must have cherished the idea of beauties with dark eyes passing on secret notes. But he has witnessed nothing but a dull music in a French hotel with a sloppy man who made a prosaic appointment only in a prosaic telephone call. Ausable raises the expectations of both the readers and Fowler when he mentions the important paper he waited which many men and women had risked their lives to possess. The drama that ensues in the room testifies the fact that there is no correlation whatsoever with sharpness of wit and the physical appearance of a person. Only after Fowler witnesses how Ausable had outwitted Max to choose his own death without moving from his armchair, he realizes the truth.

Question (f)
The unexpected presence of a criminal wielding a gun triggers different reactions in the two men who entered the room. In this light, discuss the appropriacy of the title.
Answer:
A visitor usually comes during the wakeful hours. Nocturnal visitors are usually thieves. They don’t inform one in advance because the purpose of their visit is never noble enough to inform in advance. People involved in shady businesses only choose midnight to break into someone’s house. In this story a spy is after an important report about missiles, a secretly guarded report, which is expected to be delivered to the private detective Ausable at 12.30 am. The spy gets the secret information from his trusted connections. So, he arrives earlier than Ausable and sneaks into his room using a pass key. The spy is brandishing his pistol to coerce the private detective to pass on the report as and when it arrives. Ausable outsmarts him by cooking up a story about a non-existent balcony beneath his window which extends from the neighbouring apartment. He connects brilliantly the knock of Henry the waiter to that of police who might fire at Max as he is armed. So, the title of the story, “The Midnight Visitor” is very pertinent.

Additional Questions

Question (а)
What impression do you form about Ausable as a secret agent after reading the story ‘The Midnight Visitor’?
Answer:
Ausable is a secret agent. But his appearance is not appropriate to his profession. He does not look smart and intelligent. He is very fat. But he is a very active person. He proves it throughout the story. He never takes decision in a hurry. He works with a cool mind. He is good at talking. He understands Fowler’s internal views about himself. He does not lose his temper when he finds Max in his room with pistol. He sits into an armchair and cooks up a quick story about balcony. It is his style of conversation that he easily makes Max believe about balcony. Thus, we find that Ausable is fearless, clever and fit for the job of a secret agent.

Question (b)
Why was Fowler disappointed after meeting Ausable? Did he change his idea at last?
Answer:
Fowler was a young writer. He wrote for a magazine. He had read in the books that secret agents are mysterious and smart. So he wanted to see all these things in Ausable, a secret agent. But Ausable was a fat man. It appeared that he was not fit for a secret agent’s job. So Fowler was disappointed after meeting Ausable.

But Ausable showed his presence of mind, when he saw armed Max in his room. He misled Max and told him that there was a balcony below the window. Max came there to take an important paper relating to missiles. After this, once again Ausable showed his intelligence, when Henry knocked at the door. He told Max that it might be police to protect him. They might shoot him at sight because he is armed. Without examining the truth of Ausable’s statement, Max jumped from the window to hide himself in the balcony. But it was the end of Max.
Now Fowler was very happy to see Ausable’s intelligence.

4. Look at the following expressions used in the story. Match them with their meanings.

Answer:

5. Based on your understanding of the story, complete the Graphic Organiser (GO) suitably.

Answer:

6. Given below are pictures of fictitious detective characters in English & Tamil short stories. Match them with the authors who created them.

1. Agatha Christie – Hercule Poirot
2. Sujatha – Ganesh, Vasanth
3. Sir Arthur Conan Doyle – Sherlock Holmes & Dr. Watson
4. Devan – Sambu
5. Tamizhvanan – Sankar Lai

The Midnight Visitor About The Author

Robert A. Arthur, Jr. was a mystery and speculative fiction writer known for “The Mysterious Traveller” radio series and his “Three Investigators” series of novels. He was born on November 10, 1909. Arthur was a graduate from the University of Michigan. Between 1930 and 1940, his stories were published in Amazing Stories, Argosy All-Story Weekly, Black Mask, etc. He wrote a number of mystery books for children. Arthur, along with his writing partner David Kogan, was twice honoured by the Mystery Writers of America with an Edgar award for best radio drama. Robert Arthur, Jr. died in Philadelphia in 1969.

The Midnight Visitor Summary in English

The detective Ausable:
Ausable was a detective. But he was very fat and he did not look like a detective. He had a room on the sixth floor in a French Hotel and it was the top floor. It was a cheap accommodation unworthy of a detective’s station in life.

Fowler meets Ausable:
Fowler was a writer. He wanted to write a book on detectives. So he came to meet Ausable. But after meeting Ausable, he was quite disappointed as he did not possess the qualities like a detective like James Bond. Ausable could speak French and German. But he had an American accent. Contrary to his expectations, Ausable told Fowler that there were no beautiful girls around him. Talking to each other, they reached the room in the hotel and opened the door.

Max with a pistol:

After entering the room, Ausable told Fowler that he would see an important paper that could change the course of History. Several men and women were after it. As soon as Ausable switched on the light, they saw a man with an automatic pistol. Seeing him, Ausable said that he was shocked to see him there. He thought that he was in Berlin. At this time Fowler was much frightened.

Spinning a Romance:
To confuse Max, Ausable made a false story of balcony next to the window. He sat in an armchair and started saying that it was the second time in a month that somebody had got into his room through the balcony. It is an extension of the neighbour’s balcony reaching just below his window. The hotel management had failed to close it despite his complaint. Ausable showed anger and disappointment. Max believed him and he told that he had come to take the report about missiles.

The sound of knocking at the door:
Just then, they heard a knock at the door. Ausable immediately made a story and told that the police might have come to provide him security due to this important paper. He told Max that the police would enter the room, if he did not open the door. They might fire if they found Max armed.

Max drops from the balcony:

Max believed Ausable and he went towards the window. He caught the door frame with his free hand and put his gun over Ausable and Fowler. Then he moved his other leg up and over the window sill. The doorknob turned. Max freed himself and dropped in the balcony. He cried loudly.

A waiter enters:
After this a waiter entered the room with a bottle and two glasses. It was ordered by Ausable. Fowler was very surprised at this. He asked Ausable about Max. Ausable replied that he (Max) would never return. Thus Ausable had proved himself a true detective.

The Midnight Visitor Summary in Tamil

துப்பறியும் நிபுணர் அவுசபில்:
அவுசபில் ஒரு துப்பறியும் நிபுணர். அவர் பிரஞ் ஹோட்டலின் கடைசி மாடியில் அதாவது ஆறாவது மாடியின் அறையில் தங்கி இருந்தார். அவர் பருமனானவராய் துப்பறியும் நிபுணருக்கேற்ற தோற்றம் அற்றவராகத் தெரிந்தார். அது துப்பறியும் நபர் தன் வாழ்நாளில் தங்கக் கூடிய அறையாக இல்லாமல் மிகவும் எளிமையானதாகக் காணப்பட்டது.

பவுலர். அவுசபிலை சந்திக்கிறார்:
பவுலர் ஒரு எழுத்தாளர். அவர் துப்பறிவதை குறித்து நூல் எழுத விரும்புகிறார். ஆதலால், அவுசபிலைக் காண வருகிறார். ஆனால் அவுசபிலை பார்த்த பின்னர் அவர் ஜேம்ஸ் பாண்ட்டை | போல் எந்த விதத்திலும் தோன்றவில்லையே என | ஏமாற்றமடைந்தார். அவுசபில் பிரஞ்சு மற்றும் ஜெர்மன் மொழிகளை பேசினார். ஆனால் அமெரிக்கர்கள் பேசும் வண்ணம் பேசினார். பவுலர் எதிர்பார்த்ததிற்கு மாறாக, அவுசபில் தன்னை சுற்றி அழகான பெண்கன் எவரும் இல்லை என்றார். இருவரும் பேசிக்கொண்டே தங்கும் விடுதியை அடைந்து அறையைத் திறந்தனர்.

கையில் துப்பாக்கியுடன் மாக்ஸ்:

அறையை அடைந்த உடன் அவுசபில், | பவுலரிடம் சரித்திரத்தையே மாற்றக் கூடிய ஒரு | காகிதத்தை அவர் காணப் போவதாக கூறினார். எத்தனையோ, ஆண்களும், பெண்களும் அதன் பின்னே அலைந்தனர் என்றார். அறையின் உள்ளே நுழைந்து மின்விளக்குப் பொத்தானைத் தட்டியவுடன் ஒருவன் தானியங்கி கைத் துப்பாக்கியோடு நிற்பதை அவர்கள் கண்டனர். தான் அதிர்ந்து போய்விட்டதாக அவனிடம் அவுசபில் தெரிவித்தார். அவன் பெர்லினில்

கற்பனைக் கதை ஒன்றை புனைதல்:
மாக்ஸ்சை குழப்புவதற்காக ஜன்னலருகே பால்கனி இருப்பதாக ஒரு பொய் கதையை அவுசபில் கூறினார். சாய்வு நாற்காலியில் அமர்ந்த வண்ணம், இவ்வாறாக இரண்டு முறை அந்த பால்கனி வழியாக தன் அறை உள்ளே எவரோ நுழைந்துள்ளனர் என விவரித்தார்.
மேலும், விடுதி உரிமையாளரிடம் ஏற்கனவே இருமுறை புகார் செய்தும் பயனில்லை எனக் கோபமாகச் சொன்னார். அதை உண்மை என மாக்ஸ் நம்பினான். தான் ஆயுதங்களைப் பற்றிய குறிப்புத் | தாள்களை எடுத்துப் போக வந்ததாக உரைத்தான்.

கதவை தட்டும் சத்தம்:
அப்போது யாரோ கதவை தட்டும் சத்தம் கேட்டது. உடனே ஒரு புனைக்கதையாக, அவுசபில் தன்னிடம் ஆயுத குறிப்பு தாள்கள் உள்ளதால் போலீஸ் பாதுகாப்பு தர வேண்டி வந்திருக்கக் கூடும் என வினவினார். தான் கதவை திறக்காவிட்டால் போலீஸ் உள்ளே நுழைந்து விடுவார்கள் என மாக்ஸிடம் தெரிவித்தார். கையில் துப்பாக்கியோடு நின்றிருக்கும் மாக்சை சுட்டு விடக் கூடும் என கூறினார்.

பால்கனியிலிருந்து மாக்ஸ் குதிக்கிறான்:

அவுசபில் கூறியதை நம்பிக் கொண்டு ஜன்னல் அருகே மாக்ஸ் சென்றான். துப்பாக்கி குறி அவுசபில் மற்றும் பவுலரை நோக்கியிருக்க ஜன்னல் சட்டத்தை எட்டிப் பிடித்தான் மற்ற காலை ஊன்றி ஜன்னல் விளிம்பை எட்டிப் பிடித்தான். கதவின் கைப்பிடி அகன்றது. மாக்ஸ் தன் இடது கைப் பிடியை தளர்த்தி பால்கனி மேல் குதித்தான். குதிக்கும் போது ஒரே ஒரு முறை சத்தமாக அலறினான்.

விடுதி ஊழியன் உள்ளே நுழைகிறான்:
கதவு திறந்ததும் மதுபானம் மற்றும் இரண்டு டம்ளர்களுடன் ஹென்றி உள்ளே நுழைகிறான். அதை அவுசபில் வரவழைத்திருந்தார். இதைக் கண்டு பவுலர் மிகவும் வியந்து போனார். பால்கனியில் நிற்பவன் மீண்டும் வந்து விட்டால் என வினவினார். அவன் வரப்போவதில்லை என்றார். ஏனென்றால், அங்கு பால்கனியே இல்லை என்றார். அவுசபில் தான் ஒரு உண்மையான துப்பறியும் நிபுணர் என நிரூபித்து விட்டார்

The Midnight Visitor Glossary

Textual:

Additional:

Students can Download Maths Chapter 1 Number System 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 3 Chapter 1 Number System Intext Questions

Exercise 1.1
Try These (Text book Page No. 1)

Question 1.
Represent the fraction \(\frac { 1 }{ 4 } \) in decimal form
Solution:
\(\frac { 1 }{ 4 } \) = \(\frac{1 \times 25}{4 \times 25}\) = \(\frac { 25 }{ 100 } \) = 0.25

Question 2.
What is the place value of 5 in 63.257.
Solution:
Place value of 5 in 63.257 is 5 hundredths (Hundreth place)

Question 3.
Identify the digit in the tenth place of 75.036.
Solution:
0

Question 4.
Express the decimal number 3.75 as a fraction.
Solution:
3.75 = \(\frac { 375 }{ 100 } \) = \(\frac { 15 }{ 4 } \)

Question 5.
Write the decimal number for the fraction 5 \(\frac { 1 }{ 5 } \)
Solution:
5 \(\frac { 1 }{ 5 } \) = \(\frac { 26 }{ 5 } \) = \(\frac{26 \times 2}{5 \times 2}\) = \(\frac { 52 }{ 10 } \) = 5.2

Question 6.
Identify the biggest number : 0.567 and 0.576.
Solution:
Comparing the digits of 0.567 and 0.576 from left to right, we have the tenths place same comparing the hundredths place 7 > 6.
⇒ 0.576 > 0.567

Question 7.
Compare 3.30 and 3.03 and identify the smaller number.
Solution:
The whole number is equal in both the numbers.
Now comparing the tenths place we have 3 > 0
⇒ 3.03 < 3.30 Smaller number is 3.03

Question 8.
Put the appropriate sign (<, >, =). 2.57 [ ] 2.570
Solution:
2.57 [=] 2.570

Question 9.
Arrange the following decimal numbers in ascending order.
5.14, 5.41, 1.54, 1.45, 4.15, 4.51.
Solution:
Comparing the numbers from left to right. Ascending order : 1.45, 1.54, 4.15, 4.51, 5.14, 5.41

Exercise 1.2
Try These (Text book Page No. 6)

Question 1.
Find the following using grid models:
(i) 0.83 + 0.04
(ii) 0.35 – 0.09
Solution:
(i) 0.83 + 0.04
0.83 = \(\frac { 83 }{ 100 } \) and 0.04 = \(\frac { 4 }{ 100 } \)

Shading the regions
0.83 and 0.04
The sum is the total shaded region.
S = 0.83 + 0.04 = 0.87

(ii) 0.35 – 0.09
0.35 = \(\frac { 35 }{ 100 } \) and 0.09 = \(\frac { 9 }{ 100 } \)

Shading the regions 0.35 by shading 35 boxes out of 100. Striking off 9 boxes out of 35 shaded boxes to subtract 0.09 from 0.35.
The left over shaded boxes represent the required value.
∴ 0.35 – 0.09 = 0.26

Try These (Text book Page No. 7)

Question 1.
Using the area models solve the following
(i) 1.2 + 3.5
(ii) 3.5 – 2.3
Solution:
(i) 1.2 + 3.5

Here 1.2 is represented in blue colour and 3.5 is represented in Green colour. Sum of 1.2 and 3.5 is 4.7.

(ii) 3.5 – 2.3

Representing 3.5 using 3 squares and 5 rectangular strips. Crossing out 2 squares from 3 squares and 3 rectangular strips from 5 to get the difference. So 3.5 – 2.3 = 1.2.

Try These (Text book Page No. 9)

Question 1.
Complete the magic square in such a way that rows, columns and diagonals give the same sum 1.5.

Solution:

Exercise 1.3
Think (Text book Page No. 13)

Question 1.
How are the products 2.1 × 3.2 and 21 × 32 alike? How are they different.
Solution:
2.1 × 3.2 = 6.72 and 21 × 32 = 672.
In both the cases the digits ambers are the same. But the place value differs.

Try These (Text book Page No. 13)

Question 1.
Shade the grid to multiply 0.3 × 0.6.

Solution:

3 rows of Yellow represent 0.3, 6 columns of Red colour represent 0.6 Double shaded 18 squares of orange colour represent.
∴ 0.3 × 0.6 = 0.18

Question 2.
Use the area model to multiply

Solution:

Here each row contains 1 whole and 2 tenths. Each column contains 2 wholes and 5 tenths. The entire area model represents 2 wholes 9 tenths and 10 hundredths ( = 1 tenths). So 1.2 × 2.5 = 3.

Try These (Text book Page No. 14)

Question 1.
Complete the following table

Solution:

Try These (Text book Page No. 15)

Question 1.
Find:

1. 9.13 × 10
2. 9.13 × 100
3. 9.13 × 1000

Solution:

1. 9.13 × 10 = 91.3
2. 9.13 × 100 = 913
3. 9.13 × 1000 = 9130

Try These (Text book Page No. 16)

Question 1.
Complete the following table

Solution:

Exercise 1.4
Try These (Text book Page No. 19)

Question.

Solution:

Try These (Text book Page No. 19)

Question 1.
Divide the following
(i) 17.237 ÷ 10
(ii) 17.237 ÷ 100
(iii) 17.237 ÷ 1000
Solution:
(i) 17.237 ÷ 10
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 17237 }{ 1000 } \)
= 1.7237

(ii) 17.237 ÷ 100
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 17237 }{ 100000 } \)
= 0.17237

(iii) 17.237 ÷ 1000
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 17237 }{ 1000000 } \)
= 0.017237

Try These (Text book Page No. 21)

Question 1.
Find the value of the following:
(i) 46.2 ÷ 3 = ?
(ii) 71.6 ÷ 4 = ?
(iii) 23.24 ÷ 2 = ?
(iv) 127.35 ÷ 9 = ?
(v) 47.201 ÷ 7 = ?
Solution:
(i) 46.2 ÷ 3
= \(\frac { 462 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 1 }{ 10 } \) × \(\frac { 462 }{ 3 } \)
= \(\frac { 1 }{ 10 } \) × 15.4
= \(\frac { 154 }{ 10 } \)
= 15.4

(ii) 71.6 ÷ 4
= \(\frac { 716 }{ 10 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 1 }{ 10 } \) × \(\frac { 716 }{ 4 } \)
= \(\frac { 1 }{ 10 } \) × 179
= 17.9

(iii) 23.24 ÷ 2
= \(\frac { 2324 }{ 100 } \) × \(\frac { 1 }{ 2 } \)
= \(\frac { 2324 }{ 2 } \) × \(\frac { 1 }{ 100 } \)
= 1162 × \(\frac { 1 }{ 100 } \)
= \(\frac { 1162 }{ 100 } \)
= 11.62

(iv) 127.35 ÷ 9
= \(\frac { 12735 }{ 100 } \) × \(\frac { 1 }{ 9 } \)
= \(\frac { 12735 }{ 9 } \) × \(\frac { 1 }{ 100 } \)
= 1415 × \(\frac { 1 }{ 100 } \)
= \(\frac { 1415 }{ 100 } \)
= 14.15

(v) 47.201 ÷ 7
= \(\frac { 47201 }{ 1000 } \) × \(\frac { 1 }{ 7 } \)
= \(\frac { 47201 }{ 7 } \) × \(\frac { 1 }{ 1000 } \)
= 6743 × \(\frac { 1 }{ 1000 } \)
= \(\frac { 6743 }{ 1000 } \)
= 6.743

Try These (Text book Page No. 22)

Question 1.
Divide the following
(i) \(\frac { 9.25 }{ 0.25 } \)
(ii) \(\frac { 8.6 }{ 4.3 } \)
(iii) \(\frac { 44.1 }{ 0.21 } \)
(iv) \(\frac { 9.6 }{ 1.2 } \)
Solution:
(i) \(\frac { 9.25 }{ 0.25 } \)

(ii) \(\frac { 8.6 }{ 4.3 } \)

(iii) \(\frac { 44.1 }{ 0.21 } \)

(iv) \(\frac { 9.6 }{ 1.2 } \)

Think (Text book Page No. 22)

Question 1.
The price of a tablet strip containing 30 tablets is 22.63 Then how will you find the price of each tablet?
Solution:
Price of 30 tablets = ₹ 22.63 = ₹ \(\frac { 2263 }{ 100 } \)
∴ Price of 1 tablet

= \(\frac { 2263 }{ 100 } \) × \(\frac { 1 }{ 30 } \)
= \(\frac { 2263 }{ 30 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 2263 }{ 3 } \) × \(\frac { 1 }{ 1000 } \)
= 754.33 × \(\frac { 1 }{ 1000 } \)
= \(\frac { 754.33 }{ 1000 } \)
= 0.75433
Price of each tablet is ₹ 0.7543

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.3 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 3 Chapter 2 Percentage and Simple Interest Ex 2.3

Question 1.
14 out of the 70 magazines at the bookstore are comedy magazines. What percentage of the magazines at the bookstore are comedy magazines?
Solution:
Total number of magazines in the bookstore = 100 m
Number of comedy magazines = 14
Percentage of comedy magzines = \(\frac { 14 }{ 70 } \) × 100% = 20%
20% of the magazines are comedy magazines.

Question 2.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water will fill the tank, so that it is 50% full?
Solution:
Capacity of the tank = 50 litres
Amount of water filled = 30% of 50 litres = \(\frac { 30 }{ 100 } \) × 50 = 15 litres
Amount of water to be filled = 50 – 15 = 35 litres

Question 3.
Karun bought a pair of shoes at a sale of 25%. If the amount he paid was ₹ 1000, then find the marked price.
Solution:
Let the marked price of the raincoat be ₹ P
Amount he paid at a discount of 25% = ₹ 1000
(Marked Price) – (25% of P) = 1000
P – (\(\frac { 25 }{ 100 } \) × P) = 1000
P – \(\frac { 1 }{ 4 } \) × P = 1000
P (1 – \(\frac { 1 }{ 4 } \)) = 1000
\(\frac { 3 }{ 4 } \) P = 1000
P = 1000 × \(\frac { 4 }{ 3 } \)
= \(\frac { 4000 }{ 3 } \)
P = 1333.33
∴ Marked price of the shoes = ₹ 1333

Question 4.
An agent of an insurance company gets a commission of 5% on the basic premium he collects. What will be the commission earned by him if he collects ₹ 4800?
Solution:
Premium collected = ₹ 4800
Commission earned = 5% of basic premium
Commission earned for ₹ 4800 = 5% of 4800
= \(\frac { 5 }{ 100 } \) × 4800
= ₹ 240
Commission earned = ₹ 240

Question 5.
A biology class examined some flowers in a local Grass land. Out of the 40 flowers they saw, 30 were perennials. What percentage of the flowers were perennials?
Solution:
Number of flowers examined = 40
Number of perennials = 30
Percentage = \(\frac { 30 }{ 40 } \) × 100%
= 75%
75% of the flowers were perennials.

Question 6.
Ismail ordered a collection of beads. He received 50 beads in all. Out of that 15 beads were brown. Find the percentage of brown beads?
Solution:
Number of beads received = 50
Number of brown beads = 5
Percentage of brown beads = \(\frac { 15 }{ 50 } \) × 100 %
= 10 %
10% of the beads was brown

Question 7.
Ramu scored 20 out of 25 marks in English, 30 out of 40 marks in Science and 68 out of 80 marks in mathematics. In which subject his percentage of marks is best?
Solution:
Ramu’s score in English = 20 out of 25
Percentage scored in English = \(\frac { 20 }{ 25 } \) × 100 % = 80 %
Ramu’s Score in Science = 30 out of 40
Percentage scored in Science = \(\frac { 30 }{ 40 } \) × 100 % = 75%
Ramu’s score in Mathematics = 68 out of 80
Percentage scored in Maths = \(\frac { 68 }{ 80 } \) × 100 % = 85 %
85% > 80% > 75%.
∴ In Mathematics his percentage of marks is the best.

Question 8.
Peter requires 50% to pass. If he gets 280 marks and falls short by 20 marks, what would have been the maximum marks of the exam?
Solution:
Peters score = 280 marks
Marks needed for a pass = 20
∴ Total marks required to get a pass = 280 + 20 = 300
i.e. 50% of total marks = 300
\(\frac { 50 }{ 100 } \) × Total marks = 300
\(\frac { 1 }{ 2 } \) × Total Marks = 300
Total Marks = 300 × 2 = 600
Total marks of the exam = 600

Question 9.
Kayal scored 225 marks out of 500 in revision test 1 and 265 out of 500 marks in revision test 2. Find the percentage of increase in her score.
Solution:
Marks scored in revision I = 225
Marks scored in revision II = 265
Change in marks = 265 – 225 = 40

Percentage of increase in marks = 8%

Question 10.
Roja earned ₹ 18,000 per month. She utilized her salary in the ratio 2 : 1 : 3 for education, savings and other expenses respectively. Express her usage of income in percentage.
Solution:
Amount of Salary = ₹ 18,000
(i) Total number of parts of salary = 2 + 1 + 3 = 6
Salary is divided into 3 portions as \(\frac { 2 }{ 6 } \),\(\frac { 1 }{ 6 } \) and \(\frac { 3 }{ 6 } \)
Portion of salary used for education = \(\frac { 2 }{ 6 } \)
Salary used for education = \(\frac { 2 }{ 6 } \) × 18,000 = ₹ 6,000
Percentage for Education = \(\frac { 6000 }{ 18000 } \) × 100 = 33.33%

(ii) Usage of salary for savings = \(\frac { 1 }{ 6 } \) × 18,000 = ₹ 3,000
Percentage for savings = \(\frac { 3000 }{ 18000 } \) × 100 = 16.67 %

(iii) Usage of salary for other expenses = \(\frac { 3 }{ 6 } \) × 18,000 = ₹ 9,000
Percentage for other expenses = \(\frac { 9000 }{ 18000 } \) × 100 = 50 %

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.2 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 3 Chapter 2 Percentage and Simple Interest Ex 2.2

Question 1.
Write each of the following percentage as decimal.
(i) 21 %
(ii) 93.1 %
(iii) 151 %
(iv) 65 %
(v) 0.64 %
Solution:
(i) 21 %
= \(\frac { 21 }{ 100 } \) = 0.21

(ii) 93.1 %
= \(\frac { 93.1 }{ 100 } \) = 0.931

(iii) 151 %
= \(\frac { 151 }{ 100 } \) = 1.51

(iv) 65 %
= \(\frac { 65 }{ 100 } \) = 0.65

(v) 0.64 %
= \(\frac { 0.64 }{ 100 } \) = 0.0064

Question 2.
Convert each of the following decimal as percentage
(i) 0.282
(ii) 1.51
(iii) 1.09
(iv) 0.71
(v) 0.858
Solution:
(i) 0.282
= 0.282 × 100% = \(\frac { 282 }{ 1000 } \) × 100 %
= 28.2 %

(ii) 1.51
= \(\frac { 151 }{ 100 } \) × 100 %
= 151 %

(iii) 1.09
= \(\frac { 109 }{ 100 } \) × 100 %
= 109 %

(iv) 0.71
= \(\frac { 71 }{ 100 } \) × 100 %
= 71 %

(v) 0.858
= \(\frac { 858 }{ 1000 } \) × 100 %
= 85.8 %

Question 3.
In an examination a student scored 75% of marks. Represent the given the percentage in decimal form?
Solution:
Student’s Score = 75% = \(\frac { 75 }{ 100 } \) = 0.75

Question 4.
In a village 70.5% people are literate. Express it as a decimal.
Solution:
Percentage of literate people = 70.5%
= \(\frac { 70.5 }{ 100 } \)
= 0.705

Question 5.
Scoring rate of a batsman is 86%. Write his strike rate as decimal.
Solution:
Scoring rate of the batsman = 86%
= \(\frac { 86 }{ 100 } \)
= 0.86

Question 6.
The height of a flag pole in school is 6.75m. Write it as percentage.
Solution:
Height of flag pole = 6.75m
= \(\frac { 675 }{ 100 } \)
= 6.75%

Question 7.
The weights of two chemical substances are 20.34 g and 18.78 g. Write the difference in percentage?
Solution:
Weight of substance 1 = 20.34g
Percentage of substance 1 = \(\frac { 2034 }{ 100 } \) = 2034 %
Weight of substance 2 = 18.78g
Percentage of substance 2 = \(\frac { 1878 }{ 100 } \) = 1878 %
Their difference = 2034 – 1878 = 156%

Question 8.
Find the percentage of shaded region in the following figure.

Solution:
Total region = 4 parts
Shaded region = 1 part
Fraction of shaded region = \(\frac { 1 }{ 4 } \)
Percentage of shaded region = \(\frac { 1 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 4 } \) × 100 %
= 25 %

Objective Type Questions

Question 1.
Decimal value of 142.5% is
(i) 1.425
(ii) 0.1425
(iii) 142.5
(iv) 14.25
Hint:
142.5 % = \(\frac { 1425 }{ 10 } \) %
= \(\frac { 1425 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= 1.425
Answer:
(i) 1.425

Question 2.
The percentage of 0.005 is
(i) 0.005 %
(ii) 5 %
(iii) 0.5 %
(iv) 0.05 %
Hint:
0.005 = \(\frac { 5 }{ 1000 } \)
= \(\frac { 5 }{ 1000 } \) × \(\frac { 100 }{ 100 } \)
= 0.5 %
Answer:
(iii) 0.5 %

Question 3.
The percentage of 4.7 is
(i) 0.47 %
(ii) 4.7 %
(iii) 47 %
(iv) 470 %
Hint:
4.7 = \(\frac { 47 }{ 10 } \)
= \(\frac { 47 }{ 10 } \) × \(\frac { 100 }{ 100 } \)
= 470 %
Answer:
(iv) 470 %

Students can Download Maths Chapter 1 Number System Ex 1.5 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 3 Chapter 1 Number System Ex 1.5

Miscellaneous Practice problems

Question 1.
Malini bought three ribbon of lengths 13.92 m, 11.5 m and 10.64 m. Find the total length of the ribbons?
Solution:
Length of ribbon 1 = 13.92 m
Length of ribbon 2 = 11.50 m
Length of ribbon 3 = 10.64 m

Total Length of the ribbons = 13.92 m + 11.5 m + 10.64 m = 36.06 m
Totla length of the ribbons = 36.06 m

Question 2.
Chitra has bought 10 kg 35 g of ghee for preparing sweets. She used 8 kg 59 g of ghee. How much ghee will be left?
Solution:
Total weight of ghee bought = 10 kg 35 g
Weight of ghee used = 8 kg 59 g
Weight of ghee left = 10.35 kg – 8.59 kg = 1.76 kg

∴ Weight of ghee left= 1 kg 76 g = 1.76 kg

Question 3.
If the capacity of a milk can is 2.53 l, then how much milk is required to fill 8 such cans?
Solution:
Capacity of 1 milk can= 2.53 l
∴ Capacity of 8 milk cans= 2.53 l × 8 = 20.24 l

To fill 8 cans 20.24 l of milk is required.

Question 4.
A basket of orange weighs 22.5 kg. If each family requires 2.5 kg of orange, families can share?
Solution:
Total weight of orange = 22.5 kg
Weight of orange required for 1 family = 2.5 kg
∴ Number of families sharing orange = 22.5 kg ÷ 2.5 kg
= \(\frac { 22.5 }{ 2.5 } \) = \(\frac { 22.5 }{ 2.5 } \) × \(\frac { 10 }{ 10 } \) = \(\frac { 225 }{ 25 } \) = 9
∴ 9 families can share the oranges.

Question 5.
A baker uses 3.924 kg of sugar to bake 10 cakes of equal size. How much sugar is used in each cake?
Solution:
For 10 cakes sugar required = 3.924 kg
For 1 cake sugar required = 3.924 ÷ 10 = \(\frac { 3.924 }{ 10 } \) = 0.3924 kg
For 1 cake sugar required = 0.3924 kg.

Question 6.
Evaluate:
(i) 26.13 × 4.6
(ii) 3.628 + 31.73 – 2.1
Solution:
(i) 26.13 × 4.6
26.13 × 4.6 = 120.198

(ii) 3.628 + 31.73 – 2.1 = 33.258

Question 7.
Murugan bought some bags of vegetables. Each bag weighs 20.55 kg. If the total weight of all the bags is 308.25 kg, how many bags did he buy?
Solution:
Total weight of all bags = 308.25 kg
Weight of 1 bag = 20.55 kg

Question 8.
A man walks around a circular park of distance 23.761 m. How much distance will he cover in 100 rounds?
Solution:
In 1 round distance covered = 23.761 m
∴ In 100 rounds distance = 23.761 × 100
= 2376.1 m
∴ In 100 round he covers 2376.1 m.

Question 9.
How much 0.0543 is greater than 0.002?
Solution:

∴ Required answer is 0.0523

Question 10.
A printer can print 15 pages per minute. How many pages can it print in 4.6 minutes?
Solution.
In 1 minute the pages printed = 15
In 4.6 minutes the pages printed = 15 × 4.6
= 69

The printer prints 69 pages.

Challenge Problems

Question 1.
The distance travelled by Prabhu from home to Yoga centre is 102 m and from Yoga centre to school is 165 m. What is the total distance travelled by him in kilometres (in decimal form)?
Solution:

∴ 267 metres = \(\frac { 267 }{ 1000 } \) km = 0.267 km
∴ Total distance travelled = 0.267 km

Question 2.
Anbu and Mala travelled from A to C in two different routes. Anbu travelled from place A to place B and from there to place C. A is 8.3 km from B and B is 15.6 km from C. Mala travelled from place A to place D and from there to place C. D is 7.5 km from A and C is 16.9 km from D. Who travelled more and by how much distance?
Solution:
Distance travelled by Anbu:

From place A to place B = 8.3 km
Distance from place B to place C = 15.6 km
∴ Total distance travelled by Anbu = 8.3 + 15.6
= 23.9 km

Distance travlled by Mala:

Distance travelled place A to D = 7.5 km
Distance from place D to place C = 16.9 km
Total distance travelled by mala = (7.5 + 16.9) km = 24.4 km
24.4 > 23.9
∴ Mala travelled more distance. She travelled (24.4 – 23.9) km more i.e she travelled 0.5 km more

Question 3.
Ramesh paid ₹ 97.75 per hour for a taxi and he used 35 hours in a week. How much he has to pay totally as taxi fare for a week?
Solution:
Payment for the taxi for an hour = ₹ 97.75
Total hours the taxi was used = 35 hrs
∴ Total payment for the taxi for the week

= 97.75 × 35
= 3421.25
Total payment for a week = ₹ 3421.25

Question 4.
An Aeroplane travelled 2781.20 kms in 6 hours. Find the average speed of the aeroplane in Km/hr.
Solution:
In 6 hours the distance travelled = 2781.20 km
In 1 hour the distance travelled = \(\frac { 2781.20 }{ 6 } \) km

Average speed of the aroplane = 463.53 km/hr.

Question 5.
Kumar’s car gives 12.6 km mileage per litre. If his fuel tank holds 25.8 litres then how far can he travel?
Solution.
Distance travelled with 1 litre fuel = 12.6 km
∴ with 25.8 litres distance travelled = 12.6 × 25.8
= 325.08 km

The car can travel 325.08 km

Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.1 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 3 Chapter 2 Percentage and Simple Interest Ex 2.1

Question 1.
In each of the following grid, find the numbers of coloured squares and express it as a fraction, decimal and percentage.

Solution:
Number of coloured square = 58
Total number of squares = 100
∴ Fraction : \(\frac { 58 }{ 100 } \)
Decimal : 0.58
Percentage : 58%

(ii) Number of coloured square = 53
Total number of squares = 100
∴ Fraction : \(\frac { 53 }{ 100 } \)
Decimal : 0.53
Percentage : 53%

(iii) Number of coloured square = 25
Total number of squares = 50
∴ Fraction : \(\frac { 25 }{ 50 } \)

Decimal : \(\frac { 25 }{ 50 } \) × \(\frac { 2 }{ 2 } \)
= \(\frac { 50 }{ 100 } \)
= 0.50

Percentage : \(\frac { 25 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 25 }{ 50 } \) × 100% = 50%

(iv) Number of coloured square = 17
Total number of squares = 25
∴ Fraction : \(\frac { 17 }{ 25 } \)
Decimal : \(\frac { 17 }{ 25 } \) × \(\frac { 4 }{ 4 } \)
= \(\frac { 68 }{ 100 } \) = 0.68
Percentage : \(\frac { 17 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 17 }{ 25 } \) × 100%
= 68%

(v) Number of coloured square = 15
Total number of squares = 30
∴ Fraction : \(\frac { 15 }{ 30 } \)
Decimal : \(\frac { 15 }{ 30 } \)
= \(\frac { 1 }{ 2 } \) × \(\frac { 50 }{ 50 } \)
= \(\frac { 50 }{ 100 } \) = 0.50
Percentage : \(\frac { 15 }{ 30 } \)
= \(\frac { 15 }{ 30 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 15 }{ 30 } \) × 100%
= 50 %

Question 2.
A picture of chess board is given.
(i) Find the percentage of the white coloured squares
(ii) Find the percentage of gray coloured squares
(iii) Find the percentage of the squares that have the pieces and
(iv) The squares that do not have the pieces.

Solution:
(i) Total number of squares in the chess board = 64
Number of white coloured squares = 32
Percentage = \(\frac { 32 }{ 64 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 32 }{ 64 } \) × 100 %
= 50%

(ii) Grey coloured squares = 64
Percentage = \(\frac { 32 }{ 64 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 32 }{ 64 } \) } [/latex] × 100 %
= 50 %

(iii) Number of squares having pieces = 20
Total number of squares = 64

(iv) Number of squares do not have pieces = 44

Question 3.
A picture of dart board is given. Find the percentage of white coloured portion and black coloured portion.

Solution:
Total sector = 20
White coloured sector = 10
Black coloured sector = 10
Percentage of white : \(\frac { 10 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
Decimal : \(\frac { 10 }{ 20 } \) × 100 %
= 50 %
Percentage of black colour : \(\frac { 10 }{ 20 } \) × \(\frac { 100 }{ 100 } \)
Decimal : \(\frac { 10 }{ 20 } \) × 100 %
= 50 %

Question 4.
Write each of the following fraction as percentage.
(i) \(\frac { 36 }{ 50 } \)
(ii) \(\frac { 81 }{ 30 } \)
(iii) \(\frac { 42 }{ 56 } \)
(iv) 2 \(\frac { 1 }{ 4 } \)
(v) 1 \(\frac { 3 }{ 5 } \)
Solution:
(i) \(\frac { 36 }{ 50 } \)
= \(\frac { 36 }{ 50 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 36 }{ 50 } \) × 100 %
= 72 %

(ii) \(\frac { 81 }{ 30 } \)
= \(\frac { 81 }{ 30 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 81 }{ 30 } \) × 100 %
= 270 %

(iii) \(\frac { 42 }{ 56 } \)
= \(\frac { 42 }{ 56 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 42 }{ 56 } \) × 100 %
= \(\frac { 21 }{ 28 } \) × 100 %
= 75 %

(iv) 2 \(\frac { 1 }{ 4 } \)
= \(\frac { 9 }{ 4 } \)
= \(\frac { 9 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 9 }{ 4 } \) × 100 %
= 225 %

(v) 1 \(\frac { 3 }{ 5 } \)
= \(\frac { 8 }{ 5 } \)
= \(\frac { 8 }{ 5 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 8 }{ 5 } \) × 100 %
= 160 %

Question 5.
Anbu scored 436 marks out of 500 in his exams. What was the percentage he scored?
Answer:
Total marks = 500
Anbu’s Score = 436
Percentage = \(\frac { 436 }{ 500 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 436 }{ 500 } \) × 100 %
= 87.2 %
Anbu’s Score = 87.2 %

Question 6.
Write each of the following percentage as fraction,
(i) 21%
(ii) 93.1 %
(iii) 151 %
(iv) 65 %
(v) 0.64 %
Solution:
(i) 21%
= \(\frac { 21 }{ 100 } \)

(ii) 93.1 %
= \(\frac { 93.1 }{ 100 } \)
= \(\frac{93.1 \times 10}{100 \times 10}\)
= \(\frac { 931 }{ 1000 } \)

(iii) 151 %
= \(\frac { 151 }{ 100 } \)

(iv) 65 %
= \(\frac { 65 }{ 100 } \)
= \(\frac { 13 }{ 20 } \)

(v) 0.64 %
= \(\frac { 0.64 }{ 100 } \)
= \(\frac{0.64 \times 100}{100 \times 100}\)
= \(\frac { 64 }{ 10000 } \)
= \(\frac { 4 }{ 625 } \)

Question 7.
Iniyan bought 5 dozen eggs. Out of that 5 dozen eggs, 10 eggs are rotten. Express the number of good eggs as percentage.
Solution:
1 dozen eggs = 12
5 dozen = 5 × 12
Total eggs = 60 eggs
Rotten eggs = 10 Good
eggs = 60 – 10 = 50
Fraction of good eggs = \(\frac { 50 }{ 60 } \)
Percentage of good eggs = \(\frac { 50 }{ 60 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 50 }{ 60 } \) × 100 %
= \(\frac { 5 }{ 6 } \) × 100 %
= 83.33 %
Percentage of good eggs = 83.33 %

Question 8.
In an election, Candidate X secured 48% of votes. What fraction will represent his votes?
Solution:
Percentage of votes x secured = 48% = \(\frac { 48 }{ 100 } \)
Fraction of votes x secured = \(\frac { 12 }{ 25 } \)

Question 9.
Ranjith total income was ₹ 7,500. He saved 25% of his total income. Find the amount saved by him.
Solution:
Total income of Ranjith = ₹ 7500
His savings = 25 % of 7500
= \(\frac { 25 }{ 100 } \) of 7500
= \(\frac { 25 }{ 100 } \) × 7500
= ₹ 1,875
∴ Amount saved by Ranjith = ₹ 1,875

Objective Type Questions

Question 1.
Thendral saved one fourth of her salary. Her savings percentage is
(i) \(\frac { 3 }{ 4 } \)
(ii) \(\frac { 1 }{ 4 } \) %
(iii) 25 %
(iv) 1 %
Hint: \(\frac { 1 }{ 4 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 1 }{ 4 } \) × 100 %
= 25 %
Answer:
(iii) 25 %

Question 2.
Kavin scored 15 out of 25 in a test. The percentage of his marks is
(i) 60%
(ii) 15%
(iii) 25%
(iv) 15/25
Hint: \(\frac { 15 }{ 25 } \) × \(\frac { 100 }{ 100 } \)
= \(\frac { 15 }{ 25 } \) × 100 %
= 60 %
Answer:
(i) 60%

Question 3.
0.07% is
(i) \(\frac { 7 }{ 10 } \)
(ii) \(\frac { 7 }{ 100 } \)
(iii) \(\frac { 7 }{ 1000 } \)
(iv) \(\frac { 7 }{ 10,000 } \)
Hint: 0.07 %
= 0.07%
= \(\frac { 0.07 }{ 100 } \)
= \(\frac{7}{\frac{100}{100}}\)
= \(\frac{7}{100 \times 100}\)
= \(\frac { 7 }{ 10,000 } \)
Answer:
(iv) \(\frac { 7 }{ 10,000 } \)

Students can Download Maths Chapter 1 Number System Ex 1.4 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 3 Chapter 1 Number System Ex 1.4

Question 1.
Simplify the following
(i) 0.6 ÷ 3
(ii) 0.90 ÷ 5
(iii) 4.08 ÷ 4
(iv) 21.56 ÷ 7
(v) 0.564 ÷ 6
(vi) 41.36 ÷ 4
(vii) 298.2 ÷ 3
Solution:
(i) 0.6 ÷ 3
= \(\frac { 6 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= 2 × \(\frac { 1 }{ 10 } \)
= \(\frac { 2 }{ 10 } \)
= 0.2

(ii) 0.90 ÷ 5
= \(\frac { 90 }{ 100 } \) × \(\frac { 1 }{ 5 } \)
= \(\frac { 90 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= 18 × \(\frac { 1 }{ 100 } \) = \(\frac { 18 }{ 100 } \)
= 0.18

(iii) 4.08 ÷ 4
= \(\frac { 408 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 408 }{ 4 } \) x \(\frac { 1 }{ 100 } \)
= 102 × \(\frac { 1 }{ 100 } \)
= \(\frac { 102 }{ 100 } \)
= 1.02

(iv) 21.56 ÷ 7
= \(\frac { 2156 }{ 100 } \) × \(\frac { 1 }{ 7 } \)
= \(\frac { 2156 }{ 7 } \) × \(\frac { 1 }{ 100 } \)
= 308 × \(\frac { 1 }{ 100 } \)
= \(\frac { 308 }{ 100 } \)
= 3.08

(v) 0.564 ÷ 6
= \(\frac { 564 }{ 1000 } \) × \(\frac { 1 }{ 6 } \)
= \(\frac { 564 }{ 6 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 94 }{ 1000 } \)
= 0.094

(vi) 41.36 ÷ 4
= \(\frac { 4136 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 4136 }{ 4 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 1034 }{ 100 } \)
= 10.34

(vii) 298.2 ÷ 3
= \(\frac { 2982 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 2982 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 994 }{ 10 } \)
= 99.4

Question 2.
Simplify the following.
(i) 5.7 ÷ 10
(ii) 93.7 ÷ 10
(iii) 0.9 ÷ 10
(iv) 301.301 ÷ 10
(v) 0.83 ÷ 10
(vi) 0.062 ÷ 10
Solution:
(i) 5.7 ÷ 10
= \(\frac { 57 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 57 }{ 100 } \)
= 0.57

(ii) 93.7 ÷ 10
= \(\frac { 937 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 937 }{ 100 } \)
= 9.37

(iii) 0.9 ÷ 10
= \(\frac { 9 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 9 }{ 100 } \)
= 0.09

(iv) 301.301 ÷ 10
= \(\frac { 301301 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 301301 }{ 10000 } \)
= 30.1301

(v) 0.83 ÷ 10
= \(\frac { 83 }{ 100 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 83 }{ 1000 } \)
= 0.083

(vi) 0.062 ÷ 10
= \(\frac { 62 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 62 }{ 1000 } \)
= 0.0062

Question 3.
Simplify the following.
(i) 0.7 ÷ 100
(ii) 3.8 ÷ 100
(iii) 49.3 ÷ 100
(iv) 463.85 ÷ 100
(v) 0.3 ÷ 100
(vi) 27.4 ÷ 100
Solution:
(i) 0.7 ÷ 100
= \(\frac { 7 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 7 }{ 1000 } \)
= 0.007

(ii) 3.8 ÷ 100
= \(\frac { 38 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 38 }{ 1000 } \)
= 0.038

(iii) 49.3 ÷ 100
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493

(iv) 463.85 ÷ 100
= \(\frac { 46385 }{ 100 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 46385 }{ 1000 } \)
= 4.6385

(v) 0.3 ÷ 100
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 3 }{ 1000 } \)
= 4.6385

(vi) 27.4 ÷ 100
= \(\frac { 274 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 274 }{ 1000 } \)
= 0.274

Question 4.
Simplify the following.
(i) 18.9 ÷ 1000
(ii) 0.87 ÷ 1000
(iii) 49.3 ÷ 1000
(iv) 0.3 ÷ 1000
(v) 382.4 ÷ 1000
(vi) 93.8 ÷ 1000
Solution:
(i) 18.9 ÷ 1000
= \(\frac { 189 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 189 }{ 10000 } \)
= 0.0189

(ii) 0.87 ÷ 1000
= \(\frac { 87 }{ 100 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 87 }{ 100000 } \)
= 0.00087

(iii) 49.3 ÷ 1000
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493

(iv) 0.3 ÷ 1000
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3 }{ 10000 } \)
= 0.0003

(v) 382.4 ÷ 1000
= \(\frac { 3824 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3824 }{ 10000 } \)
= 0.3824

(vi) 93.8 ÷ 1000
= \(\frac { 938 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 938 }{ 10000 } \)
= 0.0938

Question 5.
Simplify the following.
(i) 19.2 ÷ 2.4
(ii) 4.95 ÷ 0.5
(iii) 19.11 ÷ 1.3
(iv) 0.399 ÷ 2.1
(v) 5.4 ÷ 0.6
(vi) 2.197 ÷ 1.3
Solution:
(i) 19.2 ÷ 2.4

(ii) 4.95 ÷ 0.5

(iii) 19.11 ÷ 1.3

(iv) 0.399 ÷ 2.1

(v) 5.4 ÷ 0.6

(vi) 2.197 ÷ 1.3

Question 6.
Divide 9.55 kg of sweet among 5 children. How much will each child get?
Solution:
Weight of the sweet = 9.55 Kg
Weight of sweet for 5 children = \(\frac { 955 }{ 100 } \) Kg
Weight of sweet for 1 child = \(\frac{\left(\frac{955}{100}\right)}{5}\) = \(\frac { 955 }{ 100 } \) × \(\frac { 1 }{ 5 } \) = \(\frac { 955 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 191 }{ 100 } \) = 1.91
Each child will get 1.91 kg sweet.

Question 7.
A vehicle covers a distance of 76.8 km for 1.2 litre of petrol. How much distance will it cover for one litre of petrol?
Solution:
For 1.2 litre of petrol the distance covered = 76.8 Km = \(\frac { 768 }{ 10 } \) Km

For 1 litre of petrol distance covered = 64 Km

Question 8.
Cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075. Find the area of the land.
Solution:
Cost of levelling the entire land = ₹ 10,075
Cost of levelling 1 sq. ft = ₹ 15.50

∴ Area of the land = 650 sq.ft.

Question 9.
The cost of 28 books are ₹ 1506.4. Find the cost of one book.
Solution:
Cost of 28 books = ₹ 1506.4

Cost of 1 book = ₹ 53.80

Question 10.
The product of two numbers is 40.376. One number is 14.42. Find the other number.
Product of two numbers = 40.376
One number = 14.42

Objective Type Questions

Question 1.
5.6 ÷ 0.5 = ?
(i) 11.4
(ii) 10.4
(iii) 0.14
(iv) 11.2
Answer:
(iv) 11.2
Hint:
\(\frac { 5.6 }{ 0.5 } \) = \(\frac { 56 }{ 5 } \)
= 11.2

Question 2.
2.01 ÷ 0.03 = ?
(i) 6.7
(ii) 67.0
(iii) 0.67
(iv) 0.067
Answer:
(ii) 67.0
Hint: \(\frac { 2.01 }{ 0.03 } \) = \(\frac { 201 }{ 3 } \)
= 67

Question 3.
0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Answer:
(ii) 0.1
Hint: