Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 2

Question 1.
Find the missing in the following proportion in the table given below.

Solution:

Question 2.
Write true or false.
(i) 15 : 30 : : 30 : 40
(ii) 22 : 11 : : 12 : 6
(iii) 90 : 30 : : 36 : 12
(iv) 32 : 64 : : 6 : 12
(v) 25 : 1 : : 40 : 160
Solution:
i) Product of extremes = 15 x 40 600
Prod uct of means 30 x 30 = 900
600 ≠ 900

ii) 22 11:; 12:6 -True
Product oi extremes 22 x 6 = 132
Product of means = 11 x 12 = 132

iii) 90 : 30 :: 36 : 12 -True
Product oi extremes = 90 x 12 = 1080
(5 x 2 = 10 M)
Product of means = 30 x 36 = 1080
1080 = 1080

iv) 32 : 64 :: 6 : 12 -True
Product of extremes = 32 x 12 = 384
Product of means = 64 x 6 = 384

v) 25 : 1 :: 40 : 1.60 – True
Product of extremes = 25 x 1.60 = 40.00
Product of means = 1 x 40 = 40

Question 3.
Madhu buys 5 kg of potatoes at the market. If the cost of 2 kg is ₹ 36, how much will Madhu pay?
Solution:
Cost of 2 kg potatoes = ₹.36
∴ Cost of 1 kg potatoes = \(\frac{36}{2}\) = ₹.18
Cost of 5kg potatoes @ ₹.18 = 5 × ₹.18 = ₹.90
∴ Madhu pays ₹.90
(OR)
Let Madhu pays ₹.x
2 : 36 :: 5 : x
x = \(\frac{36 \times 5}{2}\) = ₹ 90

Question 4.
Physics tells us that weights of an object on the moon is proportional to its weight on Earth.
Suppose a 90 kg man weighs 15 kg on the moon what will a 60 kg woman weigh on the moon?
Solution:
Man’s weight on earth = 90 kg
Man’s weight on moon = 15 kg
Woman’s weight on earth = 60 kg
Let the womans weight on moon = x kg
then 90 : 15 :: 60 : x
Product of means Product of extremes
90x = 15 × 60
x = \(\frac{15 \times 60}{90}\) = 10
∴ Womans weight on the moon = 10 kg

Question 5.
A disaster relief team consists of engineers and doctors in the ratio of 2 : 5.
(i) If there are 18 engineers, find the number of doc tors.
(ii) If there are 65 doctors, find the number of engineers.
Solution:
i) Ratio of engineers and doctors= 2 : 5
The number of engineers = 18
Let the number of doctors = x
Now 2 : 5 :: 18 : x .
By the rule of proportion
2x = 5 × 18
= \(\frac{5 \times 18}{2}\) = 45
∴ The number of doctors 45

ii) Numbers of doctors = 65
Let the number of engineers = x
then 2 : 5 :: x : 65
By the rule of proportion
5x = 2 × 65
x = \(\frac{2 \times 65}{2}\) = 26
∴ The number of engineers = 26

Question 6.
The ratio of two angles is 3: 1. Find the
(i) larger angle if the smaller is 180°
(ii) smaller angle if the larger is 63°.
Solution:
i) Given that the ratio of angles = 3 : 1
Smaller angle is given as = 180°
Let the larger angle be x°
then 3: 1 :: x : 180
∴ By rule of proportion
1 . x = 180° × 3
x = 540°
∴ The required larger angle = 540°

ii) The larger angle is given as = 63°
Let the smaller angle be y°
then 3 : 1 :: 63°: y
∴ By rule of proportion
3 × y = 63 × 1
y = \(\frac{63}{3}\) = 21°
∴ The required smaller angle = 21°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 9 Construction of Triangles Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Construction of Triangles Exercise 1

Question 1.
Construct ∆ ABC in which AB = 5.5 cm, BC = 6.5 cm and CA = 7.5 cm.
Solution:
AB = 5.5 cm, BC = 6.5 cm, CA = 7.5 cm.

Step – 1: Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2: Draw a line segment AB of A length 5.5 cm.
Step – 3: With A as centre, draw an arc of radius 7.5 cm.
Step – 4: With B as centre, draw another arc of radius 6.5 cm such that it intersects first arc at C.
Step – 5: Join A, C and B, C. The required ABC is constructed.

Question 2.
Construct ∆ NIB in which NI = 5.6 cm, IB = 6 cm and BN = 6 cm. What type of triangle is this?
Solution:
NI = 5.6 cm, IB = 6 cm, BN = 6 cm

Step – 1: Draw a rough sketch of the trIangle and label It with the given measurements.
Step – 2: Draw a line segment NI of length 5.6 cm. ,
Step – 3: With N as centre, draw an arc of radius 6 cm.
Step – 4: With I as centre, draw another arc of radius 6 cm such that it intersects first arc at B.
Step – 5: JoIn N, B and I, B. The required triangle ∆ NIB is constructed.
This triangle ¡s an Isosceles triangle.

Question 3.
Construct an equilateral ∆ APE with side 6.5 cm.
Solution:

AP = FE = EA : 6.5 cm.
Step – 1: Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2: Draw a line segment PE of length 6.5 cm.
Step – 3: With P as centre, draw an arc of radius 6.5 cm.
Step – 4: WIth E as centre, draw another arc of radius 6.5 cm such that it intersects first arc at A.
Step – 5: Join P, A and E, A. The required ∆AFE is constructed.

Question 4.
Construct a ∆ XYZ in which XY = 6 cm, YZ = 8 cm and ZX = 10 cm. Using protractor find the angle at X. What type of triangle is this?
Solution:
XY = 6cm, YZ = 8cm, ZX = 10cm.


Step – 1: Draw a rough sketch of the triananle and label it with the given measurements.
Step – 2: Draw a line segment YZ of length of 8 cm.
Step – 3: With Y as centre, draw an arc of radius 6 cm.
Step – 4: With Z as centre, draw another arc of radius 10 cm such that it intersects first arc at X.
Step – 5: Join Y, X and Z, X. The required XYZ is constructed ∠X = 52°
This triangle is an acute angled triangle.

Question 5.
Construct ∆ ABC in which AB =4 cm, BC = 7 cm and CA= 3 cm. Which type of triangle is this?
Solution:
AB = 4 cm, BC = 7 cm, CA = 3 cm.

Step – 1: Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2: Draw a line segment AB of length 4 cm.
Step – 3: With A as centre, draw an arc of radius 3 cm.
Step – 4: With B as centre, draw another arc of radius 7 cm such that it intersects first arc at C.
Step – 5: Join C, A and C, B. The required ∆ABC is constructed.
This triangle Is an obtuse angled triangle.

Question 6.
Construct ∆ PEN with PE = 4 cm, EN =5 cm and NP =3 cm. If you draw circles instead of arcs how many points of intersection do you get? How many triangles with given measurements are possible? Is this true in case of every triangle?
Solution:

PE = 4cm,EN = 5cm, NP = 3crn.
Step – 1: Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2: Draw a line segment EN of length 5 cm.
Step – 3: With E as centre, draw on arc (circle) of radius 4 cm.
Step – 4: With N as centre, draw another arc (circle) of radius 3 cm such that it intersects first arc at P.
Step – 5: Join E, P and N, P. The required ∆PEN is constructed,

if we draw circles instead of arcs, we get two points of intersection. Then, we can draw two triangles with the given measurements.
Yes, this is true in case of every triangle.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3

Question 1.
Write the information given in the picture in the form of an equation. Also, find ‘x’ in the following figure.

Solution:
From the figure x + 11 = 15
∴ x = 15 – 11 (transposing +11)
∴ x = 4cm

Question 2.
Write the information given in the picture in the form of an equation. Also, find ‘y in the following figure.

Solution:
From the figure y + 8 = 13
y = 13 – 8 (transposing 8)
∴ y = 5cm

Question 3.
If we add 7 to twice a number, we get 49. Find the number.
Solution:
Let the number be x
Then twice the number = 2x
Onaddingi = 2x + 7
By problem, 2x + 7 = 49
2x = 49 – 7 (transposing + 7)
2x = 42
x = \(\frac { 42 }{ 2 }\) (transposing x 2)
x = 21

Question 4.
If we subtract 22 from three times a number, we get 68. Find the number.
Solution:
Let the number be x
Then three times the number = 3x
On subtracting 22 ⇒ 3x – 22
By problem, 3x – 22 = 68
3x = 68 + 22 (transposing -22)
3x = 90
x = \(\frac { 90 }{ 3 }\)(transposingx3)
x = 30
∴ The required number 30

Question 5.
Find a number which when multiplied by 7 and then reduced by 3 is equal to 53.
Solution:
Let the number be x
Multiplied by 7 ⇒ 7x
Then reduced by 3 ⇒ 7x – 3
By problem, 7x – 3 = 53
7x = 53 + 3 (transposing – 3)
7x = 56
x = \(\frac { 56 }{ 7 }\) (transposing x 7)
x = 8 .
∴ The required number = 8

Question 6.
Sum of two numbers is 95. 1f one exceeds the other by 3, find the numbers.
Solution:
Let one number be = x.
Then the other number x – 3
Sumofthenumbers x + x – 3 = 2x – 3
By problem, 2x – 3 = 95
2x = 95 + 3 (transposing – 3)
2x = 98
x = \(\frac { 98 }{ 2 }\) (transposing x 2)
x = 49
∴if one number x 49 then the other number x – 3 = 49 – 3 = 46

Question 7.
Sum of three consecutive integers is 24. Find the integers.
Solution:
Let the three integers be = x, x + 1, x 2
Sumoftheintegers = x + x + 1 + x + 2 = 3x + 3
By problem, 3x + 3 = 24’
3x = 24 – 3 (Transposing + 3)
3x = 21
x = \(\frac { 21 }{ 3 }\) (transposing x 3)
x = 7
∴ The integers x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9

Question 8.
Find the length and breadth of the rectangle given below if its perimeter is 72m.

Solution:
Length of the rectangle = 5x + 4
Breadth of the rectangLe = x – 4
Perimeter of the rectangle = 2 x (length + breadth)
= 2x[(5x + 4)+(x – 4)]
= 2[5x + 4 + x – 4]
= 2(6x)
= 12x
Byproblem, 12x = 72
x = \(\frac { 72 }{ 12 }\) (transposing x 12)
x = 6
∴ Length of the rectangle = 5x + 4 = 5 x 6.4 = 34cm
Breadth of the rectangÌe = x – 4 = 6 – 4 = 2cm

Question 9.
Length of a rectangle exceeds its breadth by 4 m. 1f the perimeter of the rectangle is 84 m, find its length and breadth.
Solution:
Let the breadth be x
Then its length = x + 4
Perimeter of the rectangle = 2 x (length + breadth)
=2[x + 4 + x]
= 2 (2x + 4)
= 4x + 8
By problem, 4x + 8 = 84
4x = 84 – 8 (transposing + 8)
4x = 76
x = \(\frac{76}{4}\) (transposing x 4)
x = 19
∴ Length of the rectangle =x+4= 19+4 = 23m
Breadth of the rectangle = 19m

Question 10.
After 15 years, Hema’s age will become four times that of her present age. Find her present age.
Solution:
Let the preser age of Hema be x years
After 15 years Hema Age = 4x
By problem, x + 15 = 4x
x + 15 – 4x = 4x – 4x (Subtracting 4x from both sides)
– 3x + 15 = 0
– 3x = – 15
x = \(\frac{-15}{-3}\) transposing x ( – 3)]
x = 5
∴ Her present age is 5 years.
∴ Her present age is 5 years.

Question 11.
A sum of ₹.3000 is to be given in the form of 63 prizes. Ifthe prize money is either ₹. 100 or.25. Find the number of prizes of each type.
Solution:
Let the number of ₹ 100 prizes be x
Then the number of ₹ 25 prizes be = 63 – x
Value of the prizes = 100x + (63 – x) x 25
= 100x+ 1575 – 25x
= 75x + 1575
By problem, 75x + 1575 = 3000
75x = 3000 – 1575
75x = 1425
x = \(\frac { 1425 }{ 75 }\)
x = 19
∴ ₹ 100 prizes = 19
₹25 prizes= 63 – x =63 – 19 = 44

Question 12.
A number is divided into two parts such that one part is 10 more than the other. Ifthe two parts are in the ratio 5:3, find the number and the two parts.
Solution:
Let one part be x
Then the other part = x + 10
Ratio of these two parts = x + 10 : x
Byproblem, x + 10 : x=5:3
∴ \(\frac{x+10}{x}=\frac{5}{3}\)
3(x+ 10)=5x
3x + 30 = 5x
30 = 5x – 3x
2x = 30
x = \(\frac{30}{2}\)
x = 15
If one part is 15 then the other part must be x + 10 = 15 + 10 = 25
∴ The number is 15 + 25 40

Question 13.
Suhana said, “multiplying my number by 5 and adding 8 to it gives the same answer as subtracting my number from 20”. Find Suhana’s numbers.
Solution:
Let Suhana’s number be x
Muhtplying by S and adding 8 to that number 5x + 8
Subtracting that number from 20 = 20 – x
By problem above two are equal.
i.e. 5x + 8 = 20 – x
5x + x = 20 – 8
6x = 12
x = \(\frac{12}{6}\) = 2
x = 2
∴ Suhanas number is 2

Question 14.
The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest mark is 87. What is the lowest mark?
Solution:
Let the Lowest mark of the class = x
Twice the least mark = 2x
On adding 7 = 2x + 7
By problem; 2x + 7 = 87
2x = 87 – 7
2x = 80
x = \(\frac{80}{2}\) = 40
x = 40
∴ The lowest mark = 40

Question 15.
In adjacent figure find the magnitude of each of the three angles formed?
(Hint: Sum of all angles at a point on a line is 180°)
Solution:
(Hint : Sum of all angles at a point on a line is 180°)

We know sum of angles at a point = 180°
∴ x° + 2x° + 3x° = 180
6x° = 180°
x = \(\frac{180}{6}\) = 30
∴ The angles are x = 30°
2x = 2 x 30 = 60°
3x° = 3 x 30° = 90°

Question 16.
Solve the following riddle:
I am a number
Tell my identity
Take me two times over
And add a thirty six.
To reach a century
Solution:
Let the number be x
Two times the number = 2x
On adding 36 = 2x 36
By problem, 2x + 36 = 100 – 4
2x 36 = 96
2x = 96 – 36
2x = \(\frac{60}{2}\)
x = 30

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 1

Question 1.
What is the ratio of ₹ 100 and ₹ 10? Express your answer in the simplest form.
Solution:
Ratio of ₹ 100 and ₹ 10 = 100 : 10 = \(\frac{100}{10}=\frac{10}{1}\) = 10 : 1
Simplest form = 10 : 1.

Question 2.
Sudha has ₹ 5. Money with Radha is 3 times the money with Sudha. How much money does Radha have?
(i) What is the ratio of Radha’s money and Sudha’s money?
(ii) What is the ratio of Sudha’s money and Radha’s money?
Solution:
Money with Sud ha = 5
∴ MoneywltbRadha = 3 times Sudha
= 3 × ₹5 = ₹ 15

i) Radha : Sudha 15 : 5
\(\frac{15}{5}=\frac{3}{1}\)
= 3 : 1

ii) Sudha:Radha = ₹ 5 : ₹ 15
= 5 : 15
= \(\frac{5}{15}\)
= 1 : 3

Question 3.
Divide 96 chocolates between Raju and Ravi in the ratio 5 : 7
Solution:
Given ratio = 5 : 7
Sum of the terms of the ratio = 5 + 7 = 12
TotaL chocolates = 96
∴ Rajus share \(\frac{5}{12}\) x 96 = 40
Ravis share \(\frac{7}{12}\) x 96 = 56

Question 4.
The length of a line segment AB is 38 cm. A point X on it divides it in the ratio 9: 10. Find the lengths of the line segments AX and XB.

Solution:

Given ratio AX : XF = 9 : 10
Sum of the terms of the ratio = 9 + 10 = 19
Total length of the line segment = 38 cm
∴ \(\overline{\mathrm{AX}}=\frac{9}{19}\) × 38 = 18cm
\(\overline{\mathrm{XB}}=\frac{10}{19}\) × 38 = 20cm

Question 5.
A sum of ₹ 160,000 is divided in the ratio of 3 : 5. What is the smaller share’?
Solution:
Given that the sum is divided in the ratio = 3: 5
Sum divided = ₹ 1,60,000
Sum of the terms of the ratio 3 + 5 = 8
∴ Smaller share = \(\frac{3}{8}\) × 1,60.000 = ₹ 60,000

Question 6.
To make green paint, a painter mixes yellow paint and blue paint in the ratio of 3 :2. If he
used twelve liters of yellow paint, how much blue paint did he use ?
Solution:
Ratio of yellow paint and blue paint = 3: 2
Quantity of yellow paint = 3 parts = 12 litres
Quantity of blue paint = 2 parts = \(\frac{2}{3}\) × 12 = 8 litres

Question 7.
A rectangle measures 40 cm at its length and 20 cm at its width. Find the ratio of the length
to the width.
Solution:
Length of the rectangle = 40cm
Width of the rectangle = 20cm
Ratio of length and width = 40 cm : 20 cm
= \(\frac{40}{20}=\frac{2}{1}\) = 2 : 1

Question 8.
The speed of a Garden-Snail is 50 meters per hour and that of the Cheetah is 120 kilometers per hour. Find the ratio of the speeds.
Solution:
Speed of the Garden Snell = 50 m/hour
= \(\frac{50}{1000}\) kmph
= \(\frac{1}{20}\) kmph
Speed of the Cheetah = 120 kmph
Ratio of their speeds = \(\frac{1}{20}\) : 120
= \(\frac{20}{20}\) : 20 x 120
= 1 : 2400

Question 9.
Find (i) The ratio of boys and girls in your class.
(ii) The ratio of number of doors and number of windows of your classroom.
(ii) The ratio of number of text books and number of note books with you

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 4

Question 1.
Which congruence criterion do you use in the following?
(i) Given :AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF

(ii) Given: ZX = RP
RQ = ZY
∠PRQ ≅ ∠XZY
So, ΔPQR ≅ ΔXYZ

(iii) Given: ∠MLN ≅∠FGH
∠NML ≅ ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH

(iv) Given: EB = DB . D
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB

Solution:
(i) S.S.S congruence
(ii) S.A.S congruence
(iii) A.S.A congruence
(iv) By R.H.S congruence

Question 2.
You want to show that ΔART ≅ ΔPEN,
(i) If you have to use SSS criterion, then you need to show
(a) AR= (b) RT = (c) AT=


Solution:
(i) (a) AR = PE
(b) RT = EN
(c) AT = PN

(ii) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(a) RT = ? and (ii) PN = ?
Solution:
(a) RT = EN and
(ii) PN = AT

(iii) If it is given that AT = PN and you arelo use ASA criterion, you need to have
(a)? (b)?
Solution:
a) ∠A = ∠P
b) ∠T = ∠N

Question 3.
You have to show that ΔAMP ≅ ΔAMQ. In the following proof. supply the missing reasons.

Solution:

Question 4.
In ΔABC’, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40°and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is hejustified? Why or why not?
Solution:
The student Is not correct.
AAA is not a congruency criteria.
Triangles with same corresponding angles can have different sizes.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT? ≅ ?

Solution:
ΔRAT ≅ ΔWON

Question 6.
Complete the congruence statement.

Solution:
ΔABC ≅ ?
ΔABC ≅ ΔABT

ΔQRS ≅ ?
ΔQRS ≅ ΔTPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
Solution:

i) ΔABC ≅ ΔPQR
Since AB = PQ
∠B = ∠Q
BC = QR (Perimeter is same)

ii) Area of ΔXYZ = \(\frac { 1 }{ 2 }\) x 10 x 4 : 20 sq.units.
Areas of ΔLMN= \(\frac { 1 }{ 2 }\) x 8 x 5=2Osq. units
But ΔXYZ and ΔLMN are not congruent.
(Perimeter is different)

Question 8.
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:
We need to know that BC = QR. Here we use A.S.A criterion.

Question 9.
Explain, why
ΔABC ≅ ΔFED.

Solution:
∠B = ∠E
BC = ED
∠C = ∠D by angle sum property.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 2

Question 1.
Solve the following equations without transposing and check your result.
(i) x + 5 = 9
(ii) y – 12 = -5
(iii) 3x + 4 = 19
(iv) 9z = 81
(v) 3x + 8 = 5x + 2
(vi) 5y + 10 = 4y – 10
Solution:
(i) x + 5 = 9
Solution:
i) x + 5 = 9
x + 5 – 5 9 – 5 (subtract 5 from both sides)
x = 4
Check
LHS = x + 5
(substituting x = 4)
= 4 + 5 = 9
RHS = 9
∴ L.H.S = R.H.S

ii) y – 12 = – 5
y – 12 = – 5
y – 12 + 12= – 5 + 12 (add l2onbothsides)
y = 7
Check
LHS = y – 12
= 7 – 12= – 5
RHS = -5
∴ L.H.S = R.H.S

iii) 3x+4= 19
3x + 4 = 19
3x + 4 – 4 = 19 – 4
(subtract 4 from both sides)
3x = 15
\(\frac{3 x}{3}=\frac{15}{3}\) (Divide both sides by3)
x = 5
Check
LHS = 3x + 4
= 3 x 5 + 4
= 15 + 4 = 19
RHS = 19
∴ L.,H.S = R.H.S

iv) 9z = 81
\(\frac{9 z}{9}=\frac{81}{9}\) (Divide both sides by 9)
z = 9
Check
LHS = 9z = 9 x 9 = 81
RHS = 81
∴ LHS = RHS

v) 3x + 8 = 5x + 2
3x + 8 = 5x + 2
3x + 8 – 8 = 5x + 2 – 8
(Adding -8 on both sides)
3x = 5x – 6
3x – 5x = 5x – 6 – 5x
(Subtract 5x from both sides)
-2x = -6
\(\frac{-2 x}{-2}=\frac{-6}{-2}\)(Divide both sides by -2)
x = 3
Check
LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17
RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17
∴ LHS = RHS

(vi) 5y + 10 = 4y – 10
5y + 10 = 4y – 10
5y + 10 – 1o = 4y – 10 – 10
(Subtract 10 from both sides)
5y =4y – 20
5y – 4y = 4y – 20 – 4y
(Substract ty from both sides)
y = – 20
Check
LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90
RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90
∴ LHS = RHS

Question 2.
Solve the following equations by transposing the terms and check your result.
(i) 2 + y = 7
Solution:
y= 7 – 2 (transposlng+2)
y = 5
Check:
LHS = 2 + y = 2 + 5 = 7
RHS = 7
∴ L.H.S = R.H.S

(ii) 2a – 3 = 5
2a – 3 = 5
2a = 5 + 3 (transposing – 3)
2a = 8
(transposing x 2)
a = 4
Check
LHS = 2a – 3 = 2 x 4 – 3 = 8 – 3 = 5
RHS =5
∴ L.H.S = R.H.S

(iii) 10 – q = 6
10 – q = 6
– q = 6 – 10(transposing + 10)
– q – 4
q = \(\frac{-4}{-1}\) = 4 (transposing x ( – 1)
Check
LHS= 10 – q= 10 – 4= 6
RHS = 6
∴ L.H.S = R.H.S

(iv) 2t – 5 = 3
2t – 5 = 3
2t – 5 = 3 (transposing – 5)
2t = 3 + 5
2t = 8
t = \(\frac{8}{2}\) (transposing x (2))
Check
LHS=2t – 5= 2 x 4 – 5 = 8 – 5 = 3
RHS = 3
∴ L.H.S = R.H.S

(v) 14 = 27 – x
14 = 27 – x
0 = 27 – x – 14 (transposing + 14)
0 = 13 – x (transposIng – x)
x = 13
Check
LHS = 14
RHS = 27 – x = 27 – 13 = 14
∴ L.H.S = R.FIS

(vi) 5(x + 4) = 35
5(x + 4) = 35
x + 4 = \(\frac{35}{5}\) (lransposingx5)
x + 4 = 7
x = 7 – 4 (transposing + 4)
x = 3
Check
LHS = 5(3 + 4) = 5 x 7 = 35
RHS = 35
∴ L.H.S = R.H.S

(vii) -3x = 15
– 3x= 15
x = \(\frac{15}{-3}\) (transposingx( – 3))
x= – 5
Check
LHS = – 3x = -3x( – 5)= 15
RHS= 15
∴ L.H.S = R.H.S

(viii) 5x – 3 = 3x – 5
5x – 3 = 3x – 5
5x = 3x – 5 + 3(transposing – 3)
5x = 3x – 2
5x – 3x = – 2(trarisposing + 3x)
2x = – 2
x = \(\frac{-2}{2}\) (transposingx2)
x= – 1
C heck
LHS = 5x – 3 = 5x( – 1) – 3 = – 5 – 3 = – 8
RHS = 3x – 5 = 3x( – 1) – 5 = – 3 – 5 = – 8
∴ L.H.S = R.H.S

(ix) 3y + 4 = 5y – 4
3y + 4 = 5y – 4
3y = 5y – 4 – 4 (transposing + 4)
3y – 5y = – 8 (transposing + 5y)
– 2y = – 8
y = \(\frac{-8}{-2}\) =4 (transposingx( – 2)
y = 4
Check
LHS = 3y + 4 = 3 x (4) + 4 = 12 + 4 = 16
RHS = 5y – 4 = 5 x (4) – 4 = 20 – 4= 16
∴ L.H.S = R.H.S

(x) 3(x-3)=5(2x + 1)
3(x – 3)=5(2x+ 1)
3(x – 3)= 5(2x ÷ 1)
3x – 9= 10x+5
3x = 10x + 5 ÷ 9 (transposing – 9)
3x = 10x + 14
3x – 10x = 14(transposing+ lOx)
– 7x =14
x = \(\frac{14}{-7}\)(transposing x ( – 7))
x = – 2
Check
LHS = 3(x – 3) = 3[( – 2) – 3] = 3x( – 5) = – 15
RHS = 5(2x + 1) = 5 x [2( – 2) + 1] = 5 x [ – 4 + 1]
= 5 x ( – 3)= – 15
∴ LH.S = R.H.S

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 3

Question 1.
In following pairs of triangles, fmd the pairs which are congruent? Also, write the criterion of congruence.

Solution:

AB = PR
∠B = ∠P
∠C = ∠Q
Now by angle sum property
∠A = ∠R
∴ By A.S.A, ΔABC ≅ ΔRPQ

ΔADB ≅ ΔCBD by A.S.A.
∠ABD = ∠CDB
DB = DB

AB = CD
∠A = ∠D
∠B = ∠C (by angle sum properly
By A.S.A
ΔABO ≅ ΔDCO

We can say whether the triangles are congruent or not, at least one pair of sides should be given equal.

Question 2.
In the adjacent figure.
(i) Are ΔABC and ΔDCB congruent?
(ii) Are ΔAOB congruent to ΔDOC’?
Also identify the relation between corresponding elements and give reason for your answer.

Solution:
i) yes.
∠ACB = ∠DBC
BC = BC
∠ABC = ∠DCB (by angle sum property)
∴ ΔABC ≅ ΔDCB (A.S.A)

ii) yes
∠A = ∠D
∠ABO = ∠DCO (angle sum property)
AB = DC [From (i)]
∴ ΔAOB ≅ ΔDOC (A.S.A)
Otherwise ΔAOB and ΔDOC are similar by A.A.A. In congruent triangles corresponding parts
are equal.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 2

Question 1.
What additional information do you need to conclude that the two triangles given here under are congruent using SAS rule?

Solution:
We need
HG = TS
HJ = TR

Question 2.
The map given below shows five different villages. Village M lies exactly halfway between the two pairs of villages A and B as well as and P and Q. What is the distance between village A and village P. (Hint: check if ΔPAM ≅ ΔQBM)

Solution:
Given that
AM = MB
PM = MQ
also ∠PMA = ∠QMB
∴ By S.A.S
ΔPMA ≅ ΔQMB
Now PA = BQ
(∵ Corresponding parts of congruent triangles are equal)
∴ PA = 4 km
i.e., Distance between the villages
A and P is 4 km.

Question 3.
Look at the pairs of triangles given below. Are they congruent ? If congruent write the corresponding parts.

Solution:

AB=ST
AC = SR (given)
∠A = ∠S
∴ ΔABC ≅ ΔSTR (S.A.S)
Also ∠A = ∠S, ∠B = ∠T, ∠C = ∠R

From the figure,
FO = RO
OQ = OS
∠POQ = ∠ROS
∴ ΔPOQ ≅ ΔROS (SAS)
Also ∠P = ∠R, ∠Q = ∠S, and PQ = RS

From the figure, IAPBoardSoIJ
WD = OR
∠W= ∠R
WO = RD
∴ ΔDWO ≅ ΔORD
Also ∠EO = ∠OE; ∠WDO = ∠ROD; ∠WOD = ∠RDO

Here the two triangles ΔABC and ΔCDA are not congruent.

Question 4.
Which corresponding sides do we need to know to prove that the triangles are congruent using the SAS criterion?

Solution:
i) We need to know that AB = QR.
ii) We need to know that AD = AB.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 1

Question 1.
Decide whether the SSS congruence is true with the following figures. Give reasons

Solution:
i) From the figures,
BE = CA
EN = AR
BN = CR
∴ S.S.S congruency is true.

ii) From the figure,
AL = SD = 4 cm
LD = LD = common side
AD ≠ LS (3 cm ≠ 2.5 cm)
∴ S.S.S congruency is not ture.

Question 2.
For the following congruent triangles, and the pairs of corresponding angles.

Solution:
i) ∠P = ∠R
∠T = ∠S
∠TQP = ∠SQR

ii) ∠P=∠S
∠Q = ∠R
∠POQ = ∠SOR

Question 3.
In the adjacent figure, choose the correct answer!
(i) ΔPQR ≅ ΔPQS
(ii) ΔPQR ≅ ΔQPS
(iii) ΔPQR ≅ ΔSQP
(iv) ΔPQR ≅ ΔSPQ

Solution:
(ii) ΔPQR ≅ ΔQPS is correct.

Question 4.
In the figure given below, AB = DC and AC = DB. Is ΔABC ≅ ΔDCB.

Solution:
Given that AB = DC
AC = DB
Also from the figure ∠B = ∠C
∴ ΔABC ≅ ΔDCB

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 3

Question 1.
Find the value of the unknown ‘x’ in the following triangles.

Solution:
i) In ΔABC

∠A + ∠B + ∠C = 180° (angle sum property
x° + 50° + 60° = 180°
x° + 110° = 180°
x° = 180° – 110°

ii) In ΔPQR,

∠P + ∠Q + ∠R = 180° (angle – sum property)
90° + 30° + x° = 180°
120° + x° – 180°
x° = 180° – 120°
∴ x° = 60°

iii) In ΔXYZ,

∠X + ∠Y + ∠Z = 180° (angle – sum property)
30°+110° + x° =180°
140° + x° = 180°
x° = 180° – 140°
∴ x° = 40°

Question 2.
Find the values of the unknowns ‘x ‘and ‘y ‘in the following diagrams.
Solution:
i) In ΔPQR,

x° + 50° = 120° (exterior angle property)
x°= 120°- 50°
x°= 70°
Also
∠P + ∠Q +∠R = 180° (angle – sum property)
70° + 50° + y° = 180°
120° + y° = 180°
y° = 180° – 120°
y° = 60°

(OR)

y° + 120° = 1800 (linear pair of angles)
y° = 180°- 120°
∴ y° = 60°

ii) In the figure ΔRST,
x° = 80° (vertically opposite angles)
also ∠R + ∠S + ∠T = 180° (angle – sum property)
80° + 50°+ y°= 180°
130° + y° = 180°
y° = 180° – 130°
∴ y° = 50°

iii) m ΔMAN,
x° = ∠M + ∠A (exterior angle property)
x° = 50° + 60°
x° = 110°
Also x° + y° = 180°
110°+ y°= 180°
y° = 180° – 110°
y° = 70°

(OR) in ΔMAN,
∠M + ∠A + ∠N = 180° (angle – sum property )
50° + 60° + y° = 180°
110° + y° = 180°
y° = 180°- 110°
∴ y° = 70°

iv) In the figure ΔABC,
x° = 60° (vertically opposite angles)
∠A + ∠B + ∠ACB = 180° (angle – sum property)
y° + 30° + 60° = 180°
y° + 90° = 180°
y° = 180° – 90°
∴ y° = 90°

v) In the figure ΔEFG,
y° = 90° (vertically opposite angles)
Also in ΔEFG;
∠F + ∠E + ∠G = 180° (angle – sum property)
∴ x° + x° + y° = 180
2x° + 90° = 180°
2x° = 180°- 90°
2x° = 90°
x° = \(\frac{90^{\circ}}{2}\)
∴ x° = 45°

vi) In the figure ΔLET,
∠L = ∠T = ∠E = x° (vertically opposite angles)
Also in ΔLET
∠L + ∠E + ∠T = 180° (angle – sum property)
x° + x° + x° = 180°
3x° = 180°
x° = \(\frac{180^{\circ}}{3}\)
x° = 60°

Question 3.
Find the measure of the third angle of triangles whose two angles are given below:
(i) 38° , 102°
(ii) 116°, 30°
(iii) 40°, 80°
Solution:
(i) 38° , 102°
Let the third angle be x° then
38° + 102° + x° = 180° (angle – sum property)
140° + x° = 180°
x° = 180° – 140° = 40°

(ii) 116°, 30°
Let the third angle be x°
then 116° + 300 + x° = 180° (angle – sum property)
146° + x = 180°
x= 180°- 146° = 34°

(iii) 40°, 80°
Let the third angle be x° then
40C + 80° + x° = 180° (angle-sum property)
120°+x° = 180°
x° = 180 – 120° – 60°.

Question 4.
In a right-angled triangle, one acute angle is 30°. Find the other acute angle.
Solution:
Given triangle is right angled triangle.
Let the third angle be x°
then 90° +30° + x° = 180° (angle – sum property)
120° + x° – 180°
x° = 180° – 120° = 60°

Question 5.
State true or false for each of the following statements.
(i) A triangle can have two right angles.
(ii) A triangle can have two acute angles.
(iii) A triangle can have two obtuse angles.
(iv) Each angle of a triangle can be less than 60°.
Solution:
(i) A triangle can have two right angles. – False
(ii) A triangle can have two acute angles. – True
(iii) A triangle can have two obtuse angles. – False
(iv) Each angle of a triangle can be less than 60°. – False

Question 6.
The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.
Solution:
Given that ratio of the angle = 1: 2: 3
Sum of the terms of the ratio – 1 + 2 + 3 = 6
Sum of the angles of a triangle = 1800
∴ 1st angle = \(\frac { 1 }{ 6 }\) x 180° = 30° (angle-sum property)
∴ 2nd angle = \(\frac { 2 }{ 6 }\) x 180° = 60° (angle – sum property)
∴ 3rd angle = \(\frac { 3 }{ 6 }\) x 180° = 90° (angIe – sum property)

Question 7.
In the figure, \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\) , ∠A = 300 and ∠B = 50°. Find the values of x, y and z.
Solution:

y° = 50° (correspondIng angles)
x° = z° (corresponding angles)
Also in MBC;
∠A + ∠B +∠C = 180° (angIe – sum property)
30° + 50° + z° = 180°
80° + z° = 180°
z° = 180° – 80°
∴ z° = 100°
∴ x° = 100°; y° = 50°; z° = 100°

Question 8.
In the figure, ∠ABD = 3 ∠DAB and ∠BDC = 96°. Find ∠ABD.
Solution:

In the figure
∠ABD + ∠DAB = 96° (exterior angle property)
3∠DAB + ∠DAB = 96° (given)
4∠DAB = 96°
∠DAB = \(\frac{96^{\circ}}{4}\) = 24°
∠ABD = 3 x 24° = 72°

Question 9.
In ΔPQR ∠P= 2 ∠Q and 2 ∠R =3 ∠Q , calculate the angles of ΔPQR.
Solution:
In ΔPQR
∠P + ∠Q + ∠R = 180° also
∠P:∠Q:∠R = 2∠Q:∠Q: \(\frac { 3 }{ 2 }\)∠Q = 4 : 2 : 3
Sum of the term of the ratio = 4 + 2 + 3 = 9
∠P = \(\frac { 4 }{ 9 }\) x 180° = 80°
∠Q = \(\frac { 2 }{ 9 }\) x 180° = 40°
∠R = \(\frac { 3 }{ 9 }\) x 180° = 60°

Question 10.
If the angles of a triangle are in the ratio 1:4: 5, find the angles.
Solution:
Given that ratio of the angles = 1 : 4 : 5
Sum of the terms of the ratio = 1 + 4 + 5 = 10
Sum of the angles 180°
∴ 1st angIe = \(\frac { 1 }{ 10 }\) x 180° = 18°
2nd angle \(\frac { 4 }{ 10 }\) x 180° = 72°
3rd angle = \(\frac { 5 }{ 10 }\) x 180°= 90°

Question 11.
The acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle.
Solution:
Given that ratio of acute angles = 2 : 3
Sum of the terms of the ratio = 2 + 3 = 5
Sum of the acute angles = 90°
∴ 1st acute angle x 900 = 36°
2nd acute angle = x 900 = 540
∴ Angles of the triangle = 36°, 54° and 90°

Question 12.
In the figure, ∆PQR is right angled at Q, \(\overline{\mathrm{ML}} \| \overrightarrow{\mathrm{RQ}}\) and ∠LMR = 130°. Find ∠LPM, ∠PML and ∠PRQ.

Solution:
∠PRQ + ∠LMR = 180° (int, angles on the same side of the transversal)
∠PRQ + 130° = 18 0°
∴ ∠PRQ = 180° – 130°= 50°
In ∆PRQ,
∠P + ∠R +∠Q = 180° (angle – sum property)
∠LPM + 50° + 90° = 180°
∠LPM + 140° = 180°
∴ ∠LPM = 180° – 140° = 40°
Also ∠PML + ∠LMR = 180° (Linear pair of angles)
∴ ∠PML + 130° = 180°
∠PML = 180° – 130° = 50°
(or) ∠PML = ∠PRQ = 50° (corresponding angles)

Question 13.
In Figure ABCDE, find ∠1 +∠2 + ∠3 + ∠4 + ∠5.

Solution:
In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° ………….(1)
In ∆ACD, ∠CAD + ∠ADC + ∠ACD = 180° ……………. (2)
In ∆ADE, ∠DAE + ∠ADE + ∠AED = 180° …………(3)
Adding (1), (2) and (3) we get
(∠BAC + ∠ACB + ∠ABC) + ∠CAD+ (∠ADC + ∠ACD + ∠DAE + ∠ADE + ∠AED = 180°+ 180° 180°
(∠BAC + ∠CAD + ∠DAE) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC + ∠ADE) + ∠AED = 180° + 180° + 180°
Hene
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 540°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 7 Data Handling Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling Exercise 4

Question 1.
Draw a bar graph for the following data.
Population of India in successive census years-

Solution:

Question 2.
Draw a pie chart for the following data.

Solution:

Question 3.
Draw a double bar graph for the following data.
Birth and Death rates of different states in 1999.

Solution:

Question 4.
Draw a pie chart for the following data.
Time spent by a child during a day-

Solution:

Question 5.
The adjoining pic chart gives the expenditure on various items during a month for a family.
(The numbers written around the pie chart tell us the angles made by each sector at the centre.)

Answer the following –
(i) On which item is the expenditure minimum?
(ii) On which item is the expenditure maximum?
(iii) If the monthly income of the family is ₹ 9000, what is the expenditure on rent?
(iv) If the expenditure on food is ₹ 3000, what is the expenditure on education of children?
Solution:
i) Education
ii) Food
iii) Total income is represented by 360° = Rs. 9,000
food represented by 120 = \(\frac{120}{360}\) × 9000 = Rs. 3000
iv) The expenditure on food is represented by 120 = Rs. 3000
Then expenditure on education is represented by \(\frac{60}{120}\) × 3000 = Rs. 1500

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 7 Data Handling Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling Exercise 3

Question 1.
Say true or false and why?
(i) The difference between the largest and smallest observations in a data set is called the
mean.
(ii) In a bar graph, the bar which has greater length indicates mode.
(iii) Value of every observation in the data set is taken into account when median is calculated.
(iv) The median of a set of numbers is always one of the numbers
Solution:
i) False
ii) True
iii) False.
We take only the mid value when the observations are arranged either in ascending / descending order.
iv) False
It may or may not be in the data set.

Question 2.
The monthly income (in rupees) of 7 households in a village are 1200, 1500, 1400, 1000, 1000, 1600, 10000. (i) Find the median income of the house holds. (ii) If one more household with monthly income of 1500 is added, what will the median income be?
Solution:
Given that
The household incomes in Rs. are 1200, 1500. 1400, 1000, 1000, 1600, 10,000
Arranging these observations in ascending order 1000, 1000, 1200, 1400, 1500, 1600, 10,000
Medían is the mid value = Rs. 1400
If one more household with income Rs. 1500 ¡s added them the data becomes
1000, 1000, 1200. 1400, 1500, 1500, 1600, 10,000
Now the median = average of 1400 and 1500
= \(\frac{1400+1500}{2}=\frac{2900}{2}\) = Rs.1450

Question 3.
Observations of a data are 16, 72,0, 55, 65, 55, 10, and 41. Chaitanya calculated the mode and median without taking the zero into consideration. Did Chaitanya do the right thing?
Solution:
Mode is correct but median is wrong.

Question 4.
How many distinct sets of three positive integers have a mean of 6, a median of 7, and no mode?
Solution:
It is not possible

Question 5.
Four integers arc added to a group of integers 3,4,5,5 and 8 and the mean, median, and mode of the data increases by 1 each. What is the greatest integer in the new group of integers?
Solution:
Given set of integers = 3, 4, 5, 5 and 8
The mean of the given data = \(\frac{\text { Sum of the integers }}{\text { Number of integers }}=\frac{3+4+5+5+8}{5}=\frac{25}{5}\) = 5

The median of the given data 3. 4. 5, 5. 8 (already in ascending order) = 5 (the mid
Mode of the given data 3. 4. 5. 58 iS 5.

After adding 4 integers the mean, mode and median increased by 1
New
Mean = 5 + 1 = 6
Median = 5 + 1 = 6
Mode = 5 + 1 = 6

Now sum of the (5 + 4 = 9) numbers = Number of integers × average
Sum = 9 × 6 = 54
But sum of the given set of 5 numbers = 25
∴ Sum of the newly added 4 numbers = 54 – 25 = 29
Mode = 6 means 6 should appear for 3 times.
But sum of four numbers = 29
29 = 6 + 6 + 6 + fourth number
29 = 18 + fourth number
∴ fourth number = 29 – 18 = 11