Category: Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions InText Questions

Question 1.
In the expressions given below identify all the terms. (Page No. 194)
i) 5x² + 3y + 7
ii) 5x²y + 3
iii) 3x²y
iv) 5x – 7
v) 5x + 8 – 2(-y)
vi) 7x² – 2x
Solution:
i) 5x² + 3y + 7 is a trinomial
ii) 5x²y + 3 is a binomial
iii) 3x²y is a monomial
iv)  5x – 7 is a binomial
v) 5x + 8 – 2 (-y) is a trinomial
vi) 7x² – 2x is a binomial

Question 2.
Write the following expressions in statements. (Page No. 195)
12x, 12, 25x, -25, 25y, 1, x, 12y, y, 25xy, 5x²y, 7xy², 2xy, 3xy², 4x²y.
Solution:
Like terms Groups → {12x, 25x, x}
→ {-25, 12, 1}
→ {25y, 12y, 1}
→ {25xy, 2xy}
→ {5x²y,4x²y}
→ {7xy², 3xy²}

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Question 1.
i) What is the numerical coefficient of ‘x’ ? (Page No. 195)
Solution:
The numerical coefficient of ‘x’ is 1.

ii) What is the numerical coefficient of -‘y’ ?
Solution:
The numerical coefficient of -y is -1.

iii) What is the literal coefficient of ‘-3z’ ?
Solution:
z.

iv) Is a numerical coefficient a constant ?
Solution:
Yes.

v) Is a literal coefficient always a variable ?
Solution:
Yes.

Question 2.
Write 3 algebraic expressions with 3 terms each. (Page No. 196)
Solution:
i) ax² + bx + c
ii) px + qy + rz
iii) x² + y² + z²

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Question 1.
State true or false and give reasons for your answer. (Page No. 195)
i) 7x² and 2x are unlike terms.
ii) pq² and – 4pq² are like terms.
iii) xy, – 12x²y and 5xy² are like terms.
Solution:
i) 7x² and 2x are unlike terms is true. Since the power of the variable x is not same in both the terms.
ii) pq² and – 4pq² are like terms is true. Since both the terms are having same variables and same exponents.
iii) xy, -12x²y and 5xy² are like terms is false. Since all the terms are not contains same exponents.

Question 2.
How many terms are there in each of the following expressions ?
i) x + y
ii) 11x – 3y – 5,
iii) 6 x² + 5x – 4
iv) x²z + 3
v) 5x²y
vi) x + 3 + y
vii) x – \(\frac{11}{3}\)
viii) \(\frac{3 x}{7 y}\)
ix) 2z – y
x) 3x + 5 (Page No. 196)
Solution:
One term – (v) 5x²y, viii) \(\frac{3 x}{7 y}\)
Two terms –  (i) x + y , (iv) x²z + 3, (vii) x – \(\frac{11}{3}\), (ix) 2z – y (x) 3x + 5
Three terms – (ii) 11x – 3y – 5, (iii) 6x² + 5x – 4, (vi) x + 3 + y

Question 3.
Give two examples for each type of algebraic expression. ( Page No. 197)
Solution:
Monomial : i) 5x²y, ii) \(\frac{3}{2}\) xyz
Binomial :i) ax + by, ii) 2z – 5
Trinomial : i) ax + by + cz, ii) p² + q² + r²
Polynomial: i) 5x⁴ – 2x² + x – 1, ii) 6 + 5x – 4x² + 3y³ – 2z⁴

Question 4.
Identify the expressions given below as monomial, binomial, trinomial, and multinomial. (Page No. 197)
i) 5x² + y + 6
ii) 3xy
iii) 5x²y + 6x
iv) a + 4x – xy + xyz
Solution:
i) 5x² + y + 6     →  is a trinomial.
ii) 3xy → is a monomial.
iii) 5x²y + 6x →  is a binomial.
iv) a + 4x – xy + xyz →  is a multinomial.

Question 5.
Find the sum of the like terms. (Page No. 200)
i) 5x, 7x
ii) 7x²y, -6x²y
iii) 2m, 11m
iv) 18ab, 5ab, 12ab
v) 3x², -7x², 8x²
vi) 4m², 3m², -6m², m²
Solution:
i) 5x + 7x = 12x

ii) 7x²y + (-6x²y) = (7 – 6) x²y = x²y

iii) 2m + 11m = (2 + 11)m = 13m
iv) 18ab + 5ab + 12ab = (18 + 5 + 12) ab
= 35ab

v) 3x + (-7x) + 8x  = (3 – 7 + 8) x²
= (11 – 7) x²
= 4x²

vi) 4m” + 3m² +(-6m²) + m² = (4 + 3 – 6 + 1) m²
= (8 – 6) m²
= 2m²

vii) 18pq + (-15 pq) + 3pq = (18 -15 + 3) pq
= (21 – 15) pq
= 6pq

Question 6.
Subtract the first term from the second term. (Page No. 200)
i) 2xy, 7xy
ii) 5a², 10a²
iii) 12y, 3y
iv) 6x²y, 4x²y
v) 6xy, -12xy
Solution:
i) 2xy, 7xy
7xy – 2xy = (7 – 2) xy = 5xy
ii) 5a², 10a²
10a -5a = (10-5) a² = 5a²

iii)  12y, 3y
3y – 12y = (3 – 12)y = -9y

iv) 6x²y, 4x²y
4x²y – 6x²y = (4 – 6) x²y = -2x²y

v) 6xy,-12xy
(-12xy) – 6xy = (-12 – 6) xy = -18xy

Question 7.
Simplify the following. (Page No. 201)
i) 3m + 12m -5m
ii) 25yz – 8yz – 6yz
iii) 10m² – 9m + 7m – 3m²
iv) 9x² – 6 + 4x + 11 – 6x² – 2x + 3x² – 2
v) 3a² – 4a²b + 7a² – b² – ab
vi) 5x² + 10 + 6x + 4 + 5x + 3x² + 8
Solution:
i) 3m + 12m -5m =(3+12-5)m
= (15 – 5) m
= 10m

ii) 25yz – 8yz – 6yz = (25 – 8 – 6) yz
= (25 – 14) yz
= 11 yz

iii) 10m² – 9m + 7m – 3m² – 5m – 8 = (10m² – 3m²) + (-9m + 7m – 5m) – 8
= (10 – 3)m² + (-9 + 7 – 5) m – 8
= 7m² + (-7m) – 8
= 7m² – 7m – 8

iv) 9x² – 6 + 4x + 11- 6x² – 2x + 3x² – 2
= (9x²-6x² + 3x²) + (4x-2x) + (-6 + 11 -2)
= (9 – 6 + 3) x² + (4 – 2) x + (11 – 8)
= (12 – 6) x² + 2x + 3
= 6x² + 2x + 3

v) 3a² + 4a²b – 7a² – b² – ab = (3a² + 7a²) – 4a²b – b² – ab
= 10a² – b² – 4a²b – ab

vi) 5x² + 10 + 6x + 4 + 5x + 3X² + 8 = (5x² + 3x²) + (6x + 5x) + (10 + 4 + 8)
= 8x² + 1 lx + 22

Question 8.
Write the following expressions in standard form. (Page No. 202)
Solution:
Expression  – Standard form
i) 3x + 18 + 4x² → 4x² + 3x + 18
ii) 8 – 3x + 4x → -3x² + 4x + 8
iii) -2m + 6 – 3m^{2 →} -3m² – 2m + 6
iv) y³ + 1 + y + 3 → y³ + 3y² + y + 1
Question 9.
Identify the expressions that are in standard form. (Page No. 202)
i) 9x² + 6x + 8
ii) 9x² + 15 + 7x
iii) 9x² + 7
iv) 9x³ + 15x + 3
v) 15x² + x³ + 3x
vi) x²y + xy + 3
vii) x³ + x²y² + 6xy
Solution:
(i), (iii), (iv), (vi) are in standard form.

Question 10.
Write 5 different expressions in standard form. (Page No. 202)
Solution:
i) ax² + bx + c
ii) ax + b
iii) 4x³ + 5x² – 6x + 2
iv) 5x⁴ – 3x³ – 2x – 2
v) px³ + qx² + r

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Question 1.
Find the value of the expression ‘-9x’ if x = -3. (Page No. 203)
Solution:
The value of -9x when x = -3
-9x = -9 (-3) = + 27

Question 2.
Write an expression’ whose value is equal to -9, when x = -3. (Page No. 203)
Solution:
When x = -3, then the value of an expression 3x is -9.
∴ 3x = 3 (-3)= -9.

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Question 1.
Answer the following expressions (Page No. 206)
i) x – 2y, 3x + 4y
ii) 4m² – 7n² + 5 mil, 3m² + 5n² – 2mn
iii) 3a – 4b, 5c – 7a + 2b
Solution:
i) x – 2y, 3x + 4y = (x – 2y) + (3x + 4y)
= (x + 3x) + (-2y + 4y) = 4x + 2y

ii) 4m² – 7n² + 5mn, 3m² + 5n² – 2mn = (4m² – 7n² + 5mn) + (3m² + 5n² – 2mn)
= (4m² + 3m²) + (-7n² + 5n²) + (5mn – 2mn)
= 7 m² + (-2n²) + 3mn = 7 m² – 2n² + 3mn

iii) 3a – 4b, 5c – 7a + 2b = (3a – 4b) + (5c – 7a + 2b) = (3a – 7a) + (-4b + 2b) + 5c = -4a – 2b + 5c

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

AP State Syllabus 7th Class Maths Solutions 9th Lesson Construction of Triangles InText Questions

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Question 1.
Construct a triangle with the same measurements given in above example taking PQ as base. Are the triangles congruent ? (Page No. 183)
Solution:
PQ = 4 cm, QR = 5 cm. RP = 7cm.


Step -1 : Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2 : Draw a line segment PQ of length 4 cm.
Step – 3 : With centre P, drawn an arc of radius 7 cm.
Step – 4 : Since R is at a distance of
5 cm from Q. draw another arc from Q with radius 5 cm such that it intersects first arc at R.
Step – 5 : Join P, R and Q, R. The required APQR is constructed.
The two triangles are congruent by Side – Side – Side (SSS) criterion for congruence.

Question 2.
Construct a ΔPET, PE = 4.5 cm, ET = 5.4 cm and TP = 6.5 cm in your notebook. Now construct ΔABC, AB = 5.4 cm, BC = 4.5 cm and CA = 6.5 cm on a piece of paper. Cut it out and place it on the figure you have constructed in your notebook. Are the triangles congruent ? Write your answer using mathematical notation. (Page No. 183)
Solution:
ΔPET, PE = 4.5 cm, ET = 5.4 cm, TP = 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 4.5 cm.
Step – 3 : With centre P, draw an arc of radius 6.5 cm.
Step – 4 : Draw another arc from E with radius – 5. 4 cm such that it intersets first arc at T.
Step – 5 : Join P, T and E, T. The required APET is constructed.

ΔABC, AB = 5.4 cm, BC = 4.5 cm and CA = 6.5
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 5.4 cm.
Step – 3 : With centre A, draw an arc of radius 6.5 cm.
Step – 4 : Draw another arc from B
with radius 4.5 cm such that it intersects first arc at C.
Step – 5 : Join A, C and B,C. The required AABC is constructed.
If we place the ΔABC on ΔPET, the triangles are congruent. This is because,
AB = TE
AC = PT
BC = PE
∴ ΔABC ≅ ΔTEP (SSS criteria)

Question 3.
Sushanth prepared a problem. Construct ΔXYZ in which XY = 2 cm, YZ = 8 cm and XZ = 4 cm. He also draw the , rough sketch as shown in figure (i).

Reading the problem, Srija told Sushanth that it would not be possible to draw a triangle with the given measurements. However, Sushanth started to draw the diagram as shown in figure (ii).Check whether Sushanth can draw the triangle. If not, why ?
Discuss with your firends. What property of triangles supports Srija’s idea ? (Page No. 184)

Solutuion:
XY = 2 cm, YZ = 8 cm, XZ = 4 cm
Sushanth can not draw the triangle. This is because, in a triangle, the sum of any two sides of a triangle is greater than its third side. But in this case, sum of the two sides is not greater than the third side.
That is XY + XZ ≯ YZ
2 + 4 ≯ YZ


So, it is not possible to construct a triangle with the given measurements.

Question 4.
Construct a triangle with angles 105° and 95° and a side of length of your choice. Could you construct the triangle ? Discuss and justify.  (Page No. 187)
Solution:
Let PQ = 5 cm, ∠P = 105°, ∠Q = 95°

We cannot construct a triangle with these measurements. This is because, the sum of three angles in a triangle is 180°. But the sum of the given two angles itself is greater than 180°.
That is 105° + 95° = 200° > 180°
So, it is not possible to construct a triangle with angles 105° and 95°.

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

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles InText Questions

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Question 1.
Here are some shapes. See whether all the shapes given in row are congruent to each other or not. You can trace the figures and check. (Page No. 164 )

Solution:
All are congruent.

Question 2.
Which of the following pairs of figures are congruent ? (Page No. 164 )


Solution:
Both is (i) and (iii) are congruent.

Question 3.
ΔEFG ≅ ΔLMN (Page No. 166)

Write the corresponding vertices, angles and sides of the two triangles.
Solution:
Corresponding Sides EF = LM; FG = MN; EG = LN
Angles ∠E = ∠L, ∠F = ∠M, ∠G = ∠N
Vertices E = L, F = M, G = N

Question 4.
If ΔABC ≅ ΔDEF, write the parts of AABC that correspond to (Page No. 166)
i) DE
ii) ∠E
iii) DF
iv) EF
v) ∠F
Solution:
i) \(\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AB}}\)
ii) ∠E = ∠B
iii) \(\overrightarrow{\mathrm{DF}}=\overline{\mathrm{AC}}\)
iv) \(\overline{\mathrm{EF}}=\overline{\mathrm{BC}}\)
ii) ∠F = ∠C

Question 5.
Name the congruent triangles in each of the following pairs. Write the statement using ≅. (Page No. 166)

Solution:
i) ΔRJK ≅ ΔSUT
ii) ΔBIE ≅ ΔCIE

Question 6.
Name the congruent angles and sides for each pair of congruent triangles. (Page No. 166)
i) ΔCDG ≅ ΔRSW
Solution:
i) ΔTUV ≅ ΔXYZ
Sides: TU = XY
UV = YZ
TV = XZ
Angles :∠T = ∠X,
∠U = ∠Y,
∠V =∠Z

ii) ΔCDG ≅ ΔRSW
Sides : CD = RS
DG = SW
CG = RW
Angles : ∠C = ∠R,
∠D = ∠S,
∠G = ∠W

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Question 1.
Is the following pair of triangles congurent ? Give reason to support your answer. (Page No. 174)

Solution:
The triangles are not congruent since the corresponding parts are not equal.

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Question 1.
In the figures given below, measures of some parts of triangles are given. By applying RHS congruence rule state, which pairs of triangle are congruent. In case of congruent tri¬angles, write the result in symbolic form. (Page No. 177)

Solution:
i) Triangles are not congruent.
ii) ΔABC ≅ ΔBAD or ΔACB ≅ ΔBDA
iii) ΔABC ≅ ΔADC
iv) ΔPQS ≅ ΔPRS

Question 2.
It is to be established by R.H.S congruence rule that ΔABC = ΔRPQ. What additional information is needed, if it is given that ∠B = ∠P = 90° and AB = RP ? (Page No. 177)
Solution:
∠B = ∠P (Right angle)
AB = RP (Side)
So we need AC = RQ

Question 3.
In the adjacent figure, BD and CE are altitudes of ΔABC such that BD = CE.
i) State the three pairs of equal parts in ΔCBD andΔBCE. (Page No. 177)
Solution:
∠CDB ≅ ∠BEC (Right angle)
BD = CE (Side)
BC = BC common/hypotenuse

ii) Is ΔCBD ≅ ΔBCE ? Why or why not ?
Solution:
Yes, ΔCBD ≅ ΔBCE by R.H.S congruence.

iii) Is ∠DBC = ∠EBC ? Why or why not ?
Solution:
No, ∠DBC ≠ ∠EBC

Question 4.
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes. Page No. 177
i) State the three pairs of equal parts in ΔADB and ΔADC.
Solution:
∠ADB = ∠ADC Right angle
AB = AC hypotenuse; AD = AD common side

ii) Is ΔADB = ΔADC ? Why or why not ?
Solution:
Yes, by R.H.S congruence

iii) Is ∠B = ∠C ? Why or why not ?
Solution:
Yes by c.p.c.t

iv) Is BD ≅ CD ? Why or why not ?
Solution:
Yes, by c.p.c.t

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

AP State Syllabus 7th Class Maths Solutions 7th Lesson Data Handling InText Questions

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Question 1.
In a unit test Amar secured 20, 18, 23, 21, 24 and 22 marks in Telugu, Hindi, English, Mathematics, Science and Social Science respectively. Peter got 23, 21, 20, 19, 24 and 17 marks in the above subjects respectively. Interpret the data in an organized manner. (Page No. 143)
Solution:

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Question 1.
The ages (in years) of players are in a team of 16,16,16,14,17,18. Then find the following:
i) Age of the youngest and the oldest player, (Page No. 145)
ii) Mean age of the players.
Solution:
i) The age of youngest player = 14 years
The age of oldest player = 18 years
(ii) Mean of the players = \(\frac{\text { Sum of the ages of all }}{\text { Number of players }}\)
= \(\frac{16+16+16+14+17+18}{6}=\frac{97}{6}\) = 16.16

Question 2.
What is the average number of glasses of water that you drink per day ? In a week. How did you find the average.
Solution:
Student Activity.

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Question 1.
A data of 10 observations has a minimum value 15 and maximum value 25. What is the mean of the data? (Page No. 147)
(i) 12 (ii) 15 (iii) 21 (iv) 27
Solution:
(iii) 21

Question 2.
Observations of a data are 23, 45, 33, 21, 48, 30, 34, 36 and 35. Without actual calculation choose the mean of the data. (Page No. 147)
(i) 20
(ii) 35
(iii) 48
(iv) 50
Solution:
(ii) 35

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Question 1.
One situation where mean would be an appropriate representative value. (Page No. 152)
Solution:
i) The quantity of rice required per day for conducting mid-day- meal programme,
ii) The daily income of a vegetable vendor.

Question 2.
One situation where mode would be an appropriate representative value. (Page No. 152)
Solution:
i) The size of the shoe required for to place the order by a shopkeeper,
ii) The brand of the notebook for a stationary .seller to place the order.

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Question 1.
Find the modes of the following data. (Page No.150)
i) 5, 6, 3, 5,4, 9, 5, 6, 4, 9, 5
ii) 25, 14, 18, 15,17, 16, 19, 13, 12, 24
iii) 10,15, 20,15, 20, 10, 15, 20,10
i) 5, 6, 3, 5,4, 9, 5, 6, 4, 9, 5 – Mode 5
ii) 25, 14, 18, 15,17, 16, 19, 13, 12, 24 – No mode
iii) 10,15, 20,15, 20, 10, 15, 20,10 – No data

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

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications InText Questions

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Question
Think of some real life situations in which you have to compare quantities in the form of a ratio. (Page No. 111)
Solution:

1. Boys and girls ratio in a class.
2. Number of matches won and lost by India at World Cup.
3. Number of students come by bycycle and on foot.

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Question 1.
40 benches are required to seat 160 Sfciidents. How many benches will be required to seat 240 students at the same rate ?  (Page No. 116)
Solution:
Number of benches required to seat 160 students= 40
On one bench, the number of students that can be seated = \(\frac{160}{40}\) = 4 . 40
∴ For 240 students the number of benches required = \(\frac{240}{4}\)= 60

Question 2.
When a Robin bird flies, it flaps wings 23 times in 10 sec. How many times will it flap its wings in 2 minutes ? (Page No. 116)
Solution:
In 10 sec. Robin bird flaps for 23 times.
In 1 sec Robin bird flaps \(\frac{23}{10}\) times
In 2 minutes = 2 x 60 = 120 sec Robin bird flaps for = \(\frac{23}{10}\) x 120 = 276 times

Question 3.
The average human heart beats at 72 times per minute. How maiiy times does it beat in 15 seconds ? How many in 1 hour ? How many in a day ? (Page No. 116)
Solution:
Number of beats per 1 minute (60 seconds) = 72 times
Number of beats per 1 second = \(\frac{72}{60}=\frac{6}{5}\) times
Number of beats per 15 seconds = \(\frac{6}{5}\) x 15 =18 times
∴ Number of beats per 1 hour = \(\frac{6}{5}\) x 60 x 60 = 4320 times
∴ Number of beats per 1 day = 4320 x 24 = 103680 times.

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Question 1.
Population of our country as per 2011 census is about 12 x 10⁸ (120,00,00,000). If the popu-lation of our country increases by 3% every year what will be the population by 2012 ? (Page No. 126)
Solution:
Population in 2011 = 12 x 10⁸
It increases by 3%
Population in 2012 = (100 + 3) % of 2011
= 103% of 12 x 10⁸
= \(\frac{103 \times 12 \times 10^{8}}{100}\)
= \(\frac{1236 \times 10^{8}}{100}\)
= 12.36 x 10⁸ .

Question 2.
i) Can you eat 75% of a dosa ? (Page No. 126)
Solution:
Yes.

ii) Can the price of an item go up by 90% ?
Solution:
Yes.

iii) Can the price of an item go up by 100% ?
Solution:
Yes.

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Question 1.
Find the interestonasum of825O for3yearsat the rate of 8%perannum. (Page No. 138)
P = ₹ 8250 R = 8% T = 3 years
Interest =P x R% x T
I = 8250 x \(\frac{8}{100}\) x 3 .
Interest = \(\frac{198000}{100}\) = 1980

Question 2.
₹ 3000 is lent out at 9% rate of interest. Find the Interest which will be reèieved at the end of 2½ years. (Page No. 138)
Sol. P = ₹3000 R = 9% T = 2½ years = \(\frac{5}{2}\),
Interest = P x R% x T
3000 x \(\frac{9}{100} \times \frac{5}{2}\) = ₹675

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Question
Given below are various grids of 100 squares. Each has a different number of squares coloured.
In each case, write the coloured and white part in the form of a 1) Percentage 2) FractIon and 3) Decimal (Page No. 121)

Solution:
I) Coloured squares = 21
as a percentage 21%
as a fraction = \(\frac{21}{100}\)
asadecimal = \(\frac{21}{100}\) =0.21

ii) Coloured squares = 50
as a percentage = 50%
asafraction = \(\frac{50}{100}=\frac{1}{2}\)
as a decimal = \(\frac{50}{100}\) = 0.50

iii) Coloured squares = 69
as a percentage = 69%
as a fraction = \(\frac{69}{100}\)
as a decimal = \(\frac{69}{100}\) = 0.69

iv) Coloured squares = 8
as a percentage = 8%
as a fraction = \(\frac{8}{100}=\frac{2}{25}\)
as a decimal = \(\frac{8}{100}\) = 0.08

v) Coloured squares = 70
as a percentage = 70%
as a fraction = \(\frac{70}{100}\)
as a decimal = \(\frac{70}{100}\) = 0.70

vi) Coloured squares = 99
as a percentage = 99%
as a fraction = \(\frac{99}{100}\)
as a decimal =\(\frac{99}{100}\) = 0.99

Question 2.
Look at the grid paper given below. It Is shaded In various designs. Find the percentage of each design (Page No. 122)


Solution:
(i) 19%
(ii) 40%
(iii) 34%
(iv) 7%

Question 3.
The strength particular of a school are given below. Express the strength of each class as a fraction, percentage of total strength of the school. (Page No.122)

Solution:

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Question 1.
The C.P of 12 mangoes is equal to the selling price (S.P) of 15 mangoes. Find the loss percent. (Page No. 128)
Solution:
C.P. of 12 = S.P of 15
∴ Loss = 3
Loss percentage = \(\frac{3}{15}\) x 100% = 20%

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

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties InText Questions

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Question 1.
The lengths of two sides of a triangle are 6 cm and 9 cpi. Write all the possible lengths of the third side. (Page No. 92)
Solution:
The possible length of the third side must greater than (9— 6) and less than (9 + 6).
∴ The third side may be 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm, 10 cm, 11 cm, 12 cm, 13 cm, 14 cm.

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Question 1.
Uma felt that a triangle can be formed with three collinear points. Do you agree ? Why? Draw diagrams to justify your answer. (Page No. 88)
Solution:
Uma ¡s not correct.
A triangle cant be formed with three collinear points.

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Question 1.
Classify the following triangles according to their (i) sides and (ii) angles. (Page No. 89)
Solution:

Two sides AC and BC are equal.
∴ ∆ABC is an isosceles triangle

As ∠E = 90°;
∆NET is a right angled triangle.

In ∆MNL,
MN = NL and ∠N = 90°
∆ MNL is right angled isosceless triangle.

Question 2.
Write the six elements (i.e., the 3 sides and 3 angles) of ∆ABC.
Solution:
The three sides \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CA}}\) and the three angles ∠A, ∠B and ∠C are the six elements of ∆ABC.

Question 3.
Write the side opposite to vertex Q in ∆PQR.
Solution:
The side opposite to the vertex Q is \(\text { PR }\) .

Question 4.
Write the angle opposite to side \(\text { LM }\) in ∆LMN.
Solution:
Angle opposite to side \(\text { LM }\) is ∠N.

Question 5.
Write the vertex opposite to side \(\text { RT }\) in ∆RST.
Solution:
Vertex opposite to \(\text { RT }\) is S.

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Question 1.
Rashmi claims that no triangle can have more than one right angle. Do you agree with her. Why ?
Solution:
Yes, Rashmi is correct. A triangle can’t have more than one right angle. As the sum of two right angles is 180° there will be no scope for 3rd angle.

Question 2.
Kamal claims that no triangle can have more than two acute angles. Do you agree with him. Why?
Solution:
No. Kamal is not right. A triangle can have all the three acute angles.
(60°, 60°, 60°), (30°, 70°, 80°), (40°, 60°, 80°) etc ………….

Question 3 i)
Draw altitudes from P to \(\text { QR }\) for the following triangles. Also, draw altitudes from the other vertices.

ii) Will an altitude always lie in the interior of a triangle ?
iii) Can you think of a triangle in which the two altitudes of a triangle are two of its sides ?
Solution:
i)

ii) An altitude need not necessarily be in the Interior of a triangle.
iii) Yes. In a right triangle the two sides containing the right angle are the two altitudes of it.

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Question 1.
Draw ∆ABC and fofm an exterior ∠ACD. Now take a protractor and measure ∠ACD, ∠A and ∠B. Find the sum ∠A + ∠B and compare it with the measure ∠ACD. Do you observe that ZACD is equal (or nearly equal) to ∠A + ∠B ? (Page No. 102)
Solution:
For instance in the adjacent figure
∠A = 68° ; ∠B = 57°
∠A + ∠B = 68° + 57° = 125 ∠ACD = 125°
∴ ∠ACD = ∠A + ∠B

Question 2.
Copy each of the following triangles. In each case verify that an exterior angle of a triangle^ is equal to the sum of the two interior opposite angles. (Page No. 103)

Solution:
From the figures
exterior ∠B = ∠A + ∠C
∠130° = 65° + 65°
∠PQS = ∠P + ∠R also ∠YZO = ∠X
90° = 40° + 50° 120° = 90° + 30°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 4 Lines and Angles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles InText Questions

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Question 1.
How many transversals can be drawn for two distinct lines? (Page No.77)
Solution:
Infinite number of transversals can be drawn for two distinct lines.

DoThis:

Question 1.
Identify the transversal in figure (1) and (ii). Identify the exterior and interior angles and fill the table given below. (Page No. 78)

Solution:

Question 2.
Consider the following lines. Which line ¡s a transversal. Number and list all the angles formed. Which are the exterior angles and which are the interior angles (Page No.79)
Solution:

For fig. (i) P is transversal for l, m and n.
∠3, ∠4, ∠5, ∠6 are interior angles for l, m.
∠7, ∠8, ∠9, ∠10 are Interior angles for m, n.
∠1, ∠2, ∠7, ∠8 are exterior angles for l, m.
∠1, ∠2, ∠11, ∠12 are exterior angles for l, n.
for fig (ii),
line d is transversal for pair of lines b and c.
∠3, ∠4, ∠5, ∠6 are interior angles.
∠1, ∠2, ∠7, ∠8 are exterior angles.

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Question 1.
Name the pairs of angles in each figure by their property. (Page no. 80)

Question 2.
Fill the table with the measures of the corresponding angles. (Page no. 83)

1. Find out in which figure the pairs of corresponding angles are equal
Answer:
In fig. (I) pairs of corresponding angles are equal. q

2. What can you say about the lines ‘1’ and ‘m’?
Answer:
l is parallel to m, fig (ii)

3. What can you say about the lines ‘p’ and ‘q’?
Answer:
In fig. (ii) pairs of corresponding angles are not equal and hence p is not parallel to q.

4. Which pair of lines are parallel?
Answer:
L and M are parallel.

Fill the table with the measures of the interior alternate angles. (Page no. 84)

Table – 2

1. Find out in which the pair of interior ¿ ternate angles are equal?
Answer:
In fig (i) the pairs of alternate interior angles are equal.

2. What can you say about lines ‘L’ and ‘M’?
Answer:
L||m.

3. What can you say about the lines ‘p’ and ‘q’?
Answer:
In fig (ii) the pair of alternate interior angles are not equal. Therefore p ∦ q

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Question 1.
Write any five pairs of supplementary angles of your choice. (Page no 72)
Solution:
(80°, 100°), (60°, 120°), (108°, 72°) (140°, 40°). 30°, 150°)

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Question 1.
Draw five pairs of complementary angles of your choice. (Page no. 71)
Solution:
The following are the pairs of complementary angles

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Question 1.
Draw an angle ∠AOB = 40°. With the same vertex 0’ draw ∠BOC = 50°, taking \(\overrightarrow{\mathbf{O B}}\) Initial ray as shown In the figure. Since the sum of these angles Is 90°, they together form a right angle. Take another pair 60° and 50° and join in the same way. Do they form complementary angles? Why? Why not? (Page No. 71)
Solution:
∠AOB = 40°, ∠BOC = 50°

∠POQ = 60°
∠QOR = 50°

These two do not form pair of complementary angles since their sum is 60° + 50° = 110° ≠ 90°

Question 2.
Draw an angle ∠AOB = 100° with the same vertex O, draw ∠BOC = 80° such that \(\overline{\mathrm{OB}}\) is common to two angles. (Page No. 72)
Solution:
∠AOB = 100°
∠BOC = 80°
∠AOB +∠BOC = 180°

Question 3.
Are 130° and 70° supplementary angles? Why? Why not?
∠POQ = 130°, ∠QOR = 70°
∠POQ + ∠QOR = 130° + 70° = 200° ≠ 180°
Hence they are not supplementary angles.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers InText Questions

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Question 1.
Write five examples, each of proper, improper, mixed fractions. ? (Page No. 27)
Solution:
Proper fractions
\(\frac{1}{5}, \frac{2}{3}, \frac{4}{7}, \frac{3}{8}, \frac{4}{9}\)

Improper fractions
\(\frac{7}{2}, \frac{3}{2}, \frac{9}{4}, \frac{11}{5}, \frac{8}{3}\)

Mixed fractions
\(1 \frac{2}{3}, 2 \frac{3}{5}, 4 \frac{1}{7}, 8 \frac{6}{7}\)

Question 2.
Write five equivalent fractions for. i) \(\frac{3}{5}\) ii) \(\frac{4}{7}\) (Page No. 28)
Solution:

Do This

Question 1. (Page No. 31)
i) 4 x \(\frac { 2 }{ 7 }\)
ii) 4 x \(\frac { 3 }{ 5 }\)
iii) 7 x \(\frac { 1 }{ 3 }\)
Solution:
i) 4 x \(\frac{2}{7}=\frac{4 \times 2}{7}=\frac{8}{7}\)
ii) 4 x \(\frac{3}{5}=\frac{4 \times 3}{5}=\frac{12}{5}\)
iii) 7 x \(\frac{1}{3}=\frac{7 \times 1}{3}=\frac{7}{3}\)

Question 2.
Find i) 5 x \(\frac { 3 }{ 2 }\) =
ii) 4 x \(\frac { 7 }{ 5 }\) =
iii) 7 x \(\frac { 8 }{ 3 }\) = (Page No. 31)
Solution:
i) 5 x \(\frac{3}{2}=\frac{5 \times 3}{2}=\frac{15}{2}=7 \frac{1}{2}\)
ii) 4 x \(\frac{7}{5}=\frac{4 \times 7}{5}=\frac{28}{5}=5 \frac{3}{5}\)
iii) 7 x \(\frac{8}{3}=\frac{7 \times 8}{3}=\frac{56}{3}=18 \frac{2}{3}\)

Question 3.
Find the following: (Page No. 32)
i) 3 x 2 \(\frac{2}{7}\)
ii) 5 x 2\(\frac{1}{3}\)
iii) 8 x 4\(\frac{1}{7}\)
iv) 4 x 1\(\frac{2}{9}\)
v) 5 x 1\(\frac{1}{3}\)
Solution:

Do These

Question 1.
Fill in these boxes. (Page No. 35)
i) \(\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}\) = ……………
ii) \(\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=\) = …………………
Solution:
i) \(\frac{1}{5} \times \frac{1}{7}=\frac{1 \times 1}{5 \times 7}=\frac{1}{35}\)
ii) \(\frac{1}{2} \times \frac{1}{6}=\frac{1 \times 1}{2 \times 6}=\frac{1}{12}\)

Do This

Question 1.
Find (Page No. 39)
i) 2 ÷ \(\frac{1}{4}\)
ii) 7 ÷ \(\frac{1}{2}\)
iii) 3 ÷ \(\frac{1}{5}\) (Page No. 39)
Solution:
i) 2 ÷ \(\frac{1}{4}=2 \times \frac{4}{1}=\frac{8}{1}\) = 8
ii) 7 ÷ \(\frac{1}{2}=7 \times \frac{2}{1}\) = 14
iii) 3 ÷ \(\frac{1}{5}=3 \times \frac{5}{1}\) = 15

Question 2.
Find (Page No. 41)
i) 9 ÷ \(\frac{2}{5}\)
ii) 3 ÷ \(\frac{4}{7}\)
iii) 2 ÷ \(\frac{8}{9}\)
Solution:
i) 9 ÷ \(\frac{2}{5}\) = \(9 \times \frac{5}{2}=\frac{45}{2}\)
ii) 3 ÷ \(\frac{4}{7}\) = \(3 \times \frac{7}{4}=\frac{21}{4}\)
iii) 2 ÷ \(\frac{8}{9}\) = \(2 \times \frac{9}{8}=\frac{9}{4}\)

Question 3.
Find (Page No. 41)
i) 7 ÷ 5\(\frac{1}{3}\)
ii) 5 ÷ 2\(\frac{4}{7}\)
Solution:
i) 7 ÷ 5\(\frac{1}{3}\) = \(7 \div \frac{16}{3}=7 \times \frac{3}{16}=\frac{21}{16}\)
ii) 5 ÷ 2\(\frac{4}{7}\) = \(5 \div \frac{18}{7}=5 \times \frac{7}{18}=\frac{35}{18}\)

Question 4.
Find (Page No. 42)
i) \(\frac{3}{5} \div \frac{1}{2}\)
ii) \(\frac{1}{2} \div \frac{3}{5}\)
iii) \(2 \frac{1}{2} \div \frac{3}{5}\)
iv) \(5 \frac{1}{6} \div \frac{9}{2}\)
Solution:

Do This

Question
Find (Page No. 45)
i) 0.25 + 5.30
ii) 29.75 – 25.97
Solution:
i) 0.25 + 5.30

ii) 29.75 – 25.97

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Question 1.
Find (Page No. 48)
i) 1.7 x 3
ii)2.0 x 1.5
iii) 2.3 x 4.35
Solution:
i)1.7 x 3 = 5.1
ii) 2.0 x 1.5 = 3.00
iii) 2.3 x 4.35 = 10.005

Question 2.
Arrange the products obtained in (I) In descending order.
Solution:
Arranging above answers in descending order 10.005 > 5.1 > 3.00

Question 3.
Find (Page No. 50)
i) 35.7 ÷ 3
ii) 25.5 ÷ 3
Solution:

Do These

Question 1.
Find the greatest and the smallest numbers among the following groups. (Page No. 52)
i) 2, -2, -3, 4, 0, -5
ii) -3, -7, -8,0,-5,-2
Solution:
i) 2, -2, -3, 4, 0, -5 : greatest number = 4; smallest number = -5
ii) -3, -7, -8, 0, -5, -2: greatest number = 0; smallest number = -8

Question 2.
Write the following numbers In ascending order. (Page No. 52)
i) -5,-75,3,-2,4, \(\frac{3}{2}\)
ii) \(\frac{2}{3}, \frac{3}{2}\), 0, -1, -2, 5
Solution:
i) -5, -75, 3, -2, 4, \(\frac{3}{2}\)
Ascending order = -75 <-5 <-2 < \(\frac{3}{2}\) <3 < 4 or -75, -5, -2, \(\frac{3}{2}\), 3, 4

ii) \(\frac{2}{3}\),\(\frac{3}{2}\), 0, -1, -2, 5
Ascending order = -2, -1, 0, \(\frac{2}{3}\), \(\frac{3}{2}\), 5

Question 3.
Write 5 equlvalent rational numbers to (i) \(\frac{5}{2}\) (Page No. 56)
(ii) \(\frac{-7}{8}\)
(iii) \(\frac{-3}{7}\)
Solution:

Do These

Question 1.
Which is bigger \(\frac{5}{8}\) or \(\frac{3}{5}\) ? (PageNo.28)
Solution:
\(\frac{5}{8} \times \frac{5}{5}=\frac{25}{40}\) and \(\frac{3}{5} \times \frac{8}{8}=\frac{24}{40}\)
As \(\frac{24}{40}<\frac{25}{40}\) \(\frac{5}{8}\) is bigger than \(\frac{3}{5}\)

Question 2.
Determine if the following pairs are equal by writing each in their simplest form. (Page No. 28)
i) \(\frac{3}{8}\) and \(\frac{375}{1000}\)
Solution:

ii) \(\frac{18}{54}\) and \(\frac{23}{69}\)
Solution:

iii) \(\frac{6}{10}\) and \(\frac{600}{1000}\)
Solution:

iv) \(\frac{17}{27}\) and \(\frac{25}{45}\)
Solution”

Do These

Question 1.
Identify the equivalent rational number is each question
i) \(\frac{-1}{2}, \frac{-3}{4}, \frac{-2}{4}, \frac{-4}{8}\)
Solution:
\(\frac{-1}{2}=\frac{-2}{4}=\frac{-4}{8}\)

ii) \(\frac{1}{4}, \frac{3}{4}, \frac{5}{3}, \frac{10}{6}, \frac{2}{4}, \frac{20}{12}\)
Solution:
\(\frac{5}{3}=\frac{20}{12}=\frac{10}{6}\)

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Question 1.
You have seen that the product of two natural numbers is one or more than one is bigger than each of the two natural numbers. For example 3 x 4 = 12; 12 > 4 and 12 > 3. What happens to the value of the product when we multiply two proper fractions? (Page No.37)
Fill the following table and conclude your observations.

Solution:

Question 2.
Will the reciprocal of a proper fraction be a proper fraction? (Page No. 40)
Solution:
No. Reciprocal of a proper fraction is always an improper fraction.

Question 3.
Will the reciprocal of an Improper fraction be an Improper fraction?
Solution:
No. The reciprocal of an improper fraction is always a proper fraction.

Question 4.
Look at the following table and fill up the blank spaces. (Page No. 44)

Solution:

Question 5.
WrIte the following numbers in their expanded form. (Page No.44)
i) 30.807
ii) 968.038
iii) 8370. 705
Solution:
i) 30.8O7 = 10 x 3 + 1 x 0 + \(\frac{1}{10}\) x 8 +\(\frac{1}{100}\) x 0 + \(\frac{1}{1000}\) x 7 = 30 + \(\frac{8}{10}+\frac{7}{1000}\)

ii) 968.038 = 100 x 9 + 10 x 6 + 1 x 8 + \(\frac{1}{10}\) x 0 + \(\frac{1}{100}\) x 3 + \(\frac{1}{1000}\) x 8 = 900 + 60 + 8 + \(\frac{3}{100}+\frac{8}{1000}\)

iii) 8370.705 = 1000 x 8 + 100 x 3 + 10 x 7 + \(\frac{1}{10}\) x 7 + \(\frac{1}{1000}\) x 5 = 8000 + 300 + 70 + \(\frac{7}{10}+\frac{5}{1000}\)

Question 6.
Take any5 Integers and make all possible rational numbrs with them. (Page No. 54)
Solution:
Consider 2, 3, 4, 5 and 7
Ratlonalnumberare \(\frac{2}{3}, \frac{2}{4}, \frac{2}{5}, \frac{2}{7}, \frac{3}{4}, \frac{3}{5}, \frac{3}{7}, \frac{3}{2}, \frac{4}{2}, \frac{4}{3}, \frac{4}{5}, \frac{4}{7}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{7}, \frac{7}{2}, \frac{7}{3}, \frac{7}{4}, \frac{7}{5}\)

Question 7.
Consider any 5 rational numbers. Find out which itegers constitute them? (Page No.54)
Solution:
Take \(\frac{3}{4}, \frac{5}{8}, \frac{6}{11}, \frac{2}{7}\) and \(\frac{1}{5}\).
The integers are 1, 2, 3, 4, 5, 6, 7, 8 and 11.

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Question 1.
Find (i) 50 paise = ₹………….. (ii) 22 g = ………….. kg (iii) 80 cm = ………………m (Page No. 44)
Solution:
(i) 50 paise =₹\(\frac{50}{100}\) = ₹ 0.5
(ii) 22 g = \(\frac{22}{1000}\) kg = 0.022 kg
(iii) 80 cm = \(\frac{80}{100}\) m = 0.8 m

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Question 1.
Represent \(\frac{3}{4}\) and \(\frac{1}{4}\) in different ways using different figures. Justify your representation. Share, and check it with your friends. (Page No.27)
Solution:

Question 2.
Represents 2 1/4 pictorially. How many units are needed for this. (Page No. 27)
Solution:

We need 3 units to represent 2½

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Question 1.
Represent pictorially 2 x \(\frac{1}{5}=\frac{2}{5}\) (Page No. 32)
Solution:

Do These

Question 1.
Find \(\frac{1}{2} \times \frac{1}{5}\) and \(\frac{1}{5} \times \frac{1}{2}\) using diagram check whether \(\frac{1}{2} \times \frac{1}{5}=\frac{1}{5} \times \frac{1}{2}\) (Page No. 35)
Solution:

Do These 

Question 1.
Write 5 more fractions between (i) 0 and 1 (ii) 1 and 2. (Page No. 52)
Solution:
i) Fractions between 0 and 1 are \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\)
ii) Fractions between land 2 are \(\frac{8}{7}, \frac{9}{7}, \frac{10}{7}, \frac{11}{7}, \frac{12}{7}, \frac{13}{7}\)

Question 2.
Where does 4\(\frac{3}{5}\) lie on the number line? (Page No. 52)
Solution:
4\(\frac{3}{5}\) lies between 4 and 5 on the number line.

Question 3.
On the number line given below represent the following numbei. (Page No.53)
i) \(\frac{-7}{2}\)
ii) \(\frac{3}{2}\)
iii) \(\frac{7}{4}\)
iv) \(\frac{-7}{4}\)
v) \(\frac{-1}{2}\)
vi) \(\frac{1}{4}\)

Solution:

Question 4.
Consider the following numbers on a number line. (Page No. 53)
27, \(\frac{-7}{8}, \frac{11}{943}, \frac{54}{17}\), -68, -3, \(\frac{-9}{6}, \frac{7}{2}\)
i) Which of these are to the left of a) 0
Solution:
Left to zero are negative numbers
∴ \(\frac{-7}{8}\), -68, -3, \(\frac{-9}{6}\)

b) -2
Left to – 2 are less than – 2.
∴ -3, -68

c) 4
Left to 4 are less than 4.

d) 2
Left to 2 are less than 2.
∴ \(\frac{-7}{8}, \frac{11}{943}-68,-3, \frac{-9}{6}\)

ii) Which of these would be to the right of
a) 0
Right to zero are positive number.
∴ \(27, \frac{11}{943}, \frac{54}{17}, \frac{7}{2}\)

b) -5
Right to -5 are greater than -5.
\(27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17}-3, \frac{-9}{6}, \frac{7}{2}\)

c) 3\(\frac{1}{2}\)
Right to 3\(\frac{1}{2}\) are more than 3\(\frac{1}{2}\).
∴ 27

d) \(\frac{-5}{2}\)
Right to \(\frac{-5}{2}\) are more than \(\frac{-5}{2}\)
∴ \(-27, \frac{-7}{8}, \frac{11}{943}, \frac{54}{17} \frac{-9}{6}, \frac{7}{2}\)

Try These

Question 1.
Write three more equivalent fractions of \(\frac{3}{4}\) and mark them on the number line. What do you observe? (Page No. 55)
Solution:
Equivalent fractions of \(\frac{3}{4}\) lie on the same mark.

Question 2.
Do all equivalent fractions of \(\frac{6}{7}\) represent the same point on the number line.
(Page No. 55)
Solution:
Yes.

Question 3.
Are \(\frac{-1}{2}\) and \(\frac{-3}{6}\) represent same point on the number line? (Page No. 55)
Solution:
Yes.

Question 4.
Are \(\frac{-2}{3}\) and \(\frac{-4}{6}\) equivalent? . (Page No. 55)
Solution:
Yes.

Question 5.
Mark the following rational numbers on the number line. (In Ex 7,3)
(i) \(\frac{1}{2}\)
(ii) \(\frac{3}{4}\)
(iii) \(\frac{3}{2}\)
(iv) \(\frac{10}{3}\)
Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers InText Questions

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Question 1. (Page No. 3)
9 + (-3) =
9 + (- 2) =
9 + C- 1) =
9 + 0 =
9 + 1 =

9 + 2 =
9 + 3 =
9 + 4 =
9 + 5 =
9 + 6 = 15
9 + 7 = 16
Solution:
9 + (- 3) = 6
9 + (- 2) = 7
9 + (- 1) = 8
9 + 0 =9
9 + 1 =10

9 + 2 = 11
9 + 3= 12
9 + 4 = 13
9 + 5 = 14
9 + 6 = 15
9 + 7 = 16

i) When you added a positive integer, in which direction did you move on the number line ? (Page No. 4)
Solution:
When you add a positive integer you move right side on the number line.

ii) When you added a negative integer in which direction did you move on the number line ? (Page No. 4)
Solution:
When you add a negative integer you move left side on the number line.

Question 2.
Sangeetha said that each time you add two integers, the value of the sum is greater than the numbers. Is Sangeetha right ? Give reasons for your answer. (Page No. 4)
Solution:
Sangeetha is always not correct. When we add – 8 and 3 the sum is (- 8) + 3 = – 5 which is less than 3. Therefore when we add two positive integers then only the sum is greater than the numbers.

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Complete the pattern. (Page No. 5)

Question 1.
8-6 = 2
8-5 = 3
8-4 =
8-3 =
8-2 =
8-1 =

8 – 0 =
8 – (- 1) =
8 – (- 2) =
8 – (- 3) =
8 – (- 4) =
Solution:
8-6 = 2
8-5 = 3
8-4 = 4
8-3 = 5
8-2 = 6
8-1 = 7

8-0 =8
8 – (- 1) =9
8 – (-2) =10
8 – (- 3) = 11
8 – (-4) =12

i) When you subtract a positive integer in which direction do you move on the number line ? (Page No. 5)
Solution:
When you subtract a positive integer, you move left side on the number line.

ii) When you subtract a negative integer, in which direction do you move on the number line ? (Page No. 5)
Solution:
When you subtract a negative integer, you move right side on the number line.

Question 2.
Richa felt that each time you subtract an integer from another integer, the value of the difference is lesser than the two numbers. Is Richa right ? Give reasons for your answer. (Page No. 5)
Solution:
Richa may not be correct all times. Richa is correct when we subtract a positive integer, but when we subtract a negative integer the difference is greater than the two numbers.

E.g: 8 – (- 3) = 11 11 > 8 and 11 > – 3
5-(-6) = 11 11 > 5 and 11 >-6
7-(-8) = 15 15 > 7 and 15 > – 8

Do this

Question 1.
Solve the following. (Page No. 7)
i) 2 x – 6 =
Solution:
i) 2 x -6 = -12

ii) 5 x – 4 =
Solution:
ii) 5 x – 4 = – 20

iii) 9 x – 4 =
Solution:
iii) 9 x – 4 = – 36

Activity

Question
Fill the grid by multiplying each number in the first column with each number in the first row. (Page No. 9)

i) Is the product of two positive integers always a positive integer ?
ii) Is the product of two negative integers always a positive integer ?
iii) Is the product of a negative and positive integer always a negative integer ?
Solution:

i) Yes
ii) Yes
iii) Yes

Try This

(Page No. 10)

a) (- 1) x (- 1) =
b) (- 1) x (- 1) x (- 1) =
c) (- 1) x (- 1) x (- 1) x (- 1) =
d) (- 1) x (- 1) x (- 1) x (- 1) x (- 1) =
Solution
a) (- 1) x (- 1) = 1
b) (- 1) x (- 1) x (- 1) = – 1
c) (- 1) x (- 1) x (- 1) x (- 1) = 1
d) (- 1) x (- 1) x (- 1) x (- 1) x (- 1) = – 1

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Solve the following. (Page No. 12)
i) (- 100) ÷ 5 =
Solution:
-20

ii) (-81)-9 =
Solution:
-9

iii)(- 75) – 5 =
Solution:
– 15

iv) (- 32) – 2 =
Solution:
– 16

v) 125-(-25) =
Solution:
– 5

vi) 80-(-5) =
Solution:
– 16

vii) 64 ÷ (-16)
Solution:
-4

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Question 2.
Compute the following. (Page No. 13)

i) -36 ÷ (-4) =
Solution:
9

ii) (-201) ÷ (-3) =
Solution:
67

iii) (-325) ÷ (- 13)=
Solution:
25

Try This

Question 1.
Can we say that (- 48) + 8 = 48 + (- 8) ? (Page No. 12)
Solution:
Yes.

Check whether
i) 90 ÷ (- 45) and (- 90) ÷ (45)
Solution:
90 ÷ – 45 = – 2 and – 90 45 = – 2
∴ 90 ÷ (-45) = -90 ÷ 45 .

ii) (- 136) ÷ 4 and 136 ÷ (- 4)
Solution:
(- 136) ÷ 4 = – 34 and 136 ÷ (- 4) = – 34
∴ (- 136) ÷ 4 = (136) ÷ (-4)

Question 2.
i) (2 + 5) + 4 = 2 + (5 + 4)
ii) (2 + 0) + 4 = 2 + (0 + 4) (Page No. 15)
Does the associative property hold for whole numbers ? Take two more examples and write your answer.
Solution:
Yes. The associative property holds for whole numbers.
Ex : i) 5 + (6 + 8) = (5 + 6) + 8
ii) 3 + (5 + 4) = (3 + 5) + 4

Question 3.
Add the following. (Page No. 15)
i) 2 + 0
ii) 0 + 3 =
iii) 5 + 0 =
Solution:
i) 2
ii) 3
iii) 5

Question 4.
Similarly, add zero to as many whole numbers as possible. Is zero the additive identity for whole numbers ? (Page No. 15)
Solution:
4+0=4
8+0=8
0+7=7
1+0=1
Therefore zero is the additive identity for whole numbers.

Question 5.
Multiply the following. (Page No. 15)
i) 2 x 3 = …………..
ii) 5 x 4 = …………..
iii) 3 x 6 = …………..
Solution:
i) 6
ii) 20
iii) 18

Question 6.
Similarly, multiply any two whole numbers of your choice. (Page No. 17)
Is the product of two whole numbers always a whole number ?
Solutlion:
3 x 5 = 15
8 x 2 = 16
4 x 3 = 12
Yes, the product of two whole number is always a whole number.

Question 7.
(5 x 2) x 3 = 5 x (2 x 3)  (Page No. 18)
Is the associative property true for whole number. Take many more examples and verify.
i) 4 x (3 x 2) = (4 x 3) x 2
4 x 6 = 12 x 2
24 = 24

ii) (8 x 5) x 2 = 8 x (5 x 2)
40 x 2 = 8 x 10
80 = 80

iii) (3 x 6) x 5 = 3 x (6 x 5)
18 x 5 = 3 x 30
90 = 90

Do This

Question 1.
Compute the following :  (Page No. 13)
i) -36 ÷ (-4) = \(\frac{-36}{-4}\) = 9
ii)(-201) ÷ (-3) = \(\frac{-201}{-3}=\frac{201}{3}\) = 67
iii) (-325) ÷ (-13) = \(\frac{-325}{-13}\) = 25

Question 2.
Is [(- 5) x 2] x 3 = (- 5) x [(2 x 3)] ? (Page No. 18)
Solution:
LHS : [- 10] x 3 = – 30
RHS : (- 5) x (6) =-30
∴ L.H.S = R.H.S

Question 3.
Is [(- 2) x 6] x 4 = (- 2) x [(6 x 4)] ? (Page No. 18)
Solution:
LHS: [-12] x 4 = -48
RHS : (-2) x (24) =-48.
∴ L.H.S = R.H.S

Try This

Question 1.
Take at least 5 different pairs of Integers and see if subtraction is commutative.

i) Consider 5 and -3 (Page No. 20)
5 – ( – 3) = 5 + 3 = 8
(-3) – 5 = -8
Therefore 5 – (-3) ≠ (-3) – 5

ii) Consider 7 and -5
7-(-5) = 7 + 5 = 12
(-5) – 7 = -12
Therefore 7 – (-5) ≠ (- 5) – 7

iii) Consider 2 and – 6
2 – (-6) = 2 + 6 = 8
(-6)- 2 = -8
Therefore 2 – (-6) ≠ (- 6) – 2

iv) Consider -4 and 5
(- 4) – (5) = -9
5- (-4) = 5 + 4 = 9
∴ (-4) – 5 ≠ 5 – (-4)

v) -3 and 8
(-3)- 8 = – 11
8 – (-3) = 8 + 3 = 11
∴(-3)-8 ≠ 8-(-3)
Hence subtraction ¡s not commutative over integers.

Question 2.
Take at least five pairs of integers and check whether they are closed under division. (Page No. 21)
Statement — Inference
i) 12 +- (- 2) = – 6 — Result is an integer

Try This

Question 1.
Show 8 – 6, 8 – 1, 8 – 0, 8 – ( -2), 8 – ( – 4) on the number line. (Page No. 5)
Solution:

∴ 8 – 6 = 2; 8 – 1 = 7; 8 – 0 = 8; 8 – (-2) = 10; 8 – (-4) = 12

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 4

Question 1.
Some english alphabets have fascinating symmetrical structures. Which capital letters have only one line of symmetry (like E))’? Which capital letters have rotational symmetry of order 2 (like I)’?
Fill the following table, thinking along such lines.

Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 3

Question 1.
Which of the following figures have rotational symmetry of order more than 1?

Solution:
The above figure (i), (ii), (iii), (iv), (v) have rational symmetry of order more than 1.

Question 2.
Give the order of rotational symmetry for each figure.

Solution:
(i) Order 2
(ii) Order 4
(iii) Order 3
(iv) Order 4
(v) Order 4
(vi) Order 5
(vii) Order 6
(viii) Order 3

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 15 Symmetry Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 15th Lesson Symmetry Exercise 2

Question 1.
In the figures given below fmd the axes of symmetry such that on folding along the axis the two dots fall on each other.


Solution:

Question 2.
Given the line of symmetry, find the other dot.

Solution:

Question 3.
In the following incomplete figures, the mirror line (i.e. the line of symmetry) is given as a dotted line. Complete each figure, pcrlorming reflection on the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete’?

Solution:

Question 4.
State whether the following statements are true or false.
(i) Every closed figure has an axis of symmetry. ( )
(ii) A figure with at least one axis of symmetry is called a symmetric figure. ( )
(in) A regular polygon of 10 sides will have 12 axes of symmetry. ( )
Solution:
(i) False
(ii) True
(iii) False

Question 5.
Draw a square and construct all its axes of symmetry. Measure the angles between each pair of successive axes of symmetry. What do you notice? Does the same rule apply for other regular polygons’?

Solution:
Angle between successive axes of
Symmetry = 45° = \(\frac{360^{\circ}}{2 \mathrm{n}}\)
This is true for all regular polygons.