Category: Class 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.1

Questions 1.
Express the following into exponential form:.
(i) 14 × 14 × 14
Answer:
14 × 14 × 14 = 143

(ii) 25 × 25 × 25 × 25 × 25
Answer:
25 × 25 × 25 × 25 × 25 = 25

(iii) ab × ab × ab × ab
Answer:
ab × ab × ab × ab = (ab)⁴

(iv) 7 × p × p × q
Answer:
7 × p × p × q = 7 × p² × q

Question 2.
Express the following into expanded form:
(i) 27⁶
Answer:
27⁶ = 27 × 27 × 27 × 27 × 27 × 27

(ii) 101⁵
Answer:
1015 = 101 × 101 × 101 × 101 × 101

(iii) (2b)⁴
Answer:
2b⁴ = 2b × 2b × 2b × 2b

(iv) 3a⁸
Answer:
3a⁸ = 3 × a× a × a × a × a × a × a × a

Question 3.
Express the following In exponential form through prime factorisation: .
(i) 81
Answer:

81 = 3 × 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
∴ 81 = 3⁴

(ii) 125
Answer:

125 = 5 × 25
= 5 × 5 × 5
∴ 125 = 5³

(iii) 324
Answer:

324 = 2 × 162
= 2 × 2 × 81
= 2 × 2 × 3 × 11
= 2 × 2 × 3 × 3 × 9
= 2 × 2 × 3 × 3 × 3 × 3
∴ 324 = 2² × 3⁴

(iv) 1080
Answer:

1080 = 2 × 540
= 2 × 2 × 270
= 2 × 2 × 2 × 135
= 2 × 2 × 2 × 3 × 45
= 2 × 2 × 2 × 3 × 3 × 15
= 2 × 2 × 2 × 3 × 3 × 3 × 5
∴ 1080 = 2³ × 3¹ × 5¹

Question 4.
Compute and identify the greater num-ber in the following pairs : 7
(i) 2⁵ or 5²
Answer:
2⁵ = 2 × 2 × 2 × 2 × 2 = 32
5² = 5 × 5 = 25
32 > 25
Therefore, 2⁵ > 5²

(ii) 7³ or 3⁷
Answer:
7³ = 7 × 7 × 7 = 343
3⁷ = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187
2187 > 343
Therefore, 3⁷ > 7³

(iii) 2³ or 3²
Answer:
2³ = 2 × 2 × 2 = 8
3² = 3 × 3 = 9
9 > 8
Therefore, 3² > 2³

Question 5.
Expand 3³ × 4² and 4³ × 3². Are they equal? Justify.
Answer:
3³ × 4² = 3 × 3 × 3 × 4 × 4
= 27 × 16 = 432

4³ × 3² = 4 × 4 × 4 × 3 × 3
= 64 × 9 = 576

432 × 576 (They are not equal)
Therefore, 576 > 432
So, 4³ × 3² > 3³ × 4²

Question 6.
Express the following numbers in exponential form with the given base.
(i) 1000, base 10
Answer:

∴ 1000 = 10 × 10 × 10 = 10³

(ii) 512 base 2
Answer:

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ 512 = 2⁹

(iii) 243 base 3
Answer:

243 = 3 × 3 × 3 × 3 × 3
∴ 243 = 3⁵

Question 7.
If a = 2, b = 3 find the value of
(i) a^a + b^b
Answer:
Given a = 2, b = 3
a^a + b^b = 2² + 3³
= (2 × 2) + (3 × 3 × 3)
= 4 + 27 = 31
Therefore, if a = 2, b = 3 then a^a + b^b = 31

(ii) a^b + b^a
Answer:
Given a = 2, b = 3
a^b + b^a = 2³ + 3²
= (2 × 2 × 2) + (3 × 3).
= 8 + 9 = 17
Therefore, if a = 2, b = 3
then a^b + b^a = 17

(iii) (a + b)^b
Answer:
Given a = 2, b = 3
(a + b)^b = (2 + 3)³ = 5³
= 5 × 5 × 5 = 125
Therefore, if a = 2, b == 3
then (a + b)^b =125

Question 8.
Write the following in Exponential form:
(i) The speed of light in vacuum is about 30,00,00,000 m/sec.
Answer:
Given the speed of light = 30,00,00,000 m/sec
= 3 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
∴ Speed of light = 3 × 10⁸ m/sec.

(ii) The population of India is about 121,00,00,000 as per 2011 census.
Answer:
Given population of India
= 121,00,00,000
= 121 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 11 × 11 × 10⁷
∴ Population of India = 11² × 10⁷ as per 2011 census.

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.4

Question 1.
Find the value of the expression
2x² – 4x + 5 when
(i) x = 1
(ii) x = – 2
(iii) x = 0.
Answer:
Given expression is 2x² – 4x + 5

(i) When x = 1, then
= 2(1)² – 4(1) + 5
= 2 × 1 – 4 + 5
= 2 – 4 + 5 = 3
When x = 1, then 2x² – 4x + 5 = 3

(ii) When x = – 2, then
= 2(- 2)² – 4(- 2) + 5
= 2(4) + 8 + 5
= 8 + 8 + 5 = 21
When x = – 2, then 2x² – 4x + 5 = 21

(iii) When x = 0, then
= 2(0)² – 4(0) + 5
= 2(0) – 0 + 5
= 0 – 0 + 5 = 5
When x = 0, then 2x² – 4x + 5 = 5

Question 2.
Find the value of Expressions when m = 2, n = – 1.
(i) 2m + 2n
Answer:
Given expression is 2m + 2n
If m = 2, n = – 1, then
2m + 2n = 2(2) + 2(- 1) = 4 – 2 = 2
∴ If m = 2, n = – 1, then 2m + 2n = 2

(ii) 3m – n
Answer:
Given expression is 3m – n
If m = 2, n = – 1, then
3m – n = 3(2) -(-1) = 6 + 1 = 7
∴ If m = 2, n = – 1, then 3m – n = 7

(iii) mn – 2.
Answer:
Given expression is mn – 2
If m = 2, n = – 1, then
mn – 2 = 2 × (-1)-2
= – 2 – 2 = – 4
∴ If m = 2, n = – 1, then mn – 2 = – 4

Question 3.
Simplify and find the value of the expression 5x² – 4 – 3x² + 6x + 8 + 5x – 13 when x = – 2.
Answer:
Given expression is
5x² – 4 – 3x² + 6x + 8 + 5x – 13
= (5x² – 3x²) + (6x + 5x) + (-4 + 8 – 13)
= (5 – 3)x² + (6 + 5)x + (- 9)
= 2x² + 11x – 9
If x = – 2, then 2x² + 11x – 9
= 2(- 2)² + 11 (- 2) – 9
= 2(4) – 22 – 9
= 8 – 22 – 9
∴ If x = – 2, then 2x² + 11x – 9 = – 23

Question 4.
Find the length of the line segment PQ when a = 3 cm.

Answer:
From the figure,
Given PR = 3a and RQ = 2a
PQ = PR + RQ
= 3a + 2a
= (3 + 2)a
PQ = 5a
If a = 3 cm, then PQ = 5(3)
∴ PQ = 15 cm

Question 5.
The area of a square field of side ‘s’ meters is s2 sq. m. Find the area of square field, when
(i) s = 5m
(ii) s =12m
(iii) s = 6.5m

Answer:
From the figure,
Area of square = s2 sq.m.
(i) If s = 5m, then
s² = (5)² = 5 × 5 = 25 sq.m

(ii) If s = 12 m, then
s² = (12)² = 12 × 12 = 144 sq.m

(iii) If s = 6.5m, then
s² = (6.5)² = 6.5 × 6.5 = 42.25 sq.m

Question 6.
The area of triangle is given by \(\frac{1}{2}\) ∙ b ∙ h and if b = 12 cm, h = 8 cm, then find the area of triangle.
Answer:
Given area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h
If b = 12 cm, h = 8 cm

= 6 × 8 = 48 sq.cm.

Question 7.
Simple interest is given by I = \(\frac{\text { PTR }}{100}\), If P = ₹ 900, T = 2 years and R = 5%, then find the simple interest.
Answer:
Given simple interest I = \(\frac{\text { PTR }}{100}\)
If P = ₹ 900, T= 2 and R = 5% then

= 9 × 2 × 5 = 90

Question 8.
Find the errors and correct them in the following:
The value of following when a = – 3.
(i) 3 – a = 3 – 3 = 0
Answer:
3 – a = 3 – (- 3) (when a = – 3)
= 3 + 3 = 6
(Error is – (- 3) = – 3)

(ii) a² + 3a = (- 3)² + 3(- 3) = 9 + 0 = 9
Answer:
a² + 3a = (- 3)² + 3(- 3) (when a = – 3)
= (- 3 × – 3) – 9
(Error is 3(- 3) = 0)
= 9 – 9 = 0

(iii) a² – a – 6 = (- 3)² – (- 3) – 6 = 9 – 3 – 6 = 0
Answer:
a² – a – 6
= (- 3)² – (- 3) – 6 (when a = – 3)
= (- 3 × – 3) + 3 – 6
(Error is – (- 3) = – 3)
= 9 + 3 – 6
= 12 – 6 = 6

(iv) a² + 4a + 4 = (- 3)² + 4(-3) + 4 = 9 + 1 + 4 = 14
Answer:
a² + 4a + 4
= (- 3)² + 4(- 3) + 4 (when a = – 3)
= (- 3 × – 3)- 12 + 4
(Error is 4 (- 3) = 1)
= 9 – 12 + 4 = 13 – 12 = 1

(v) a³ – a² – 3 = (- 3)³ – (-3)² – 3 = – 9 + 6 – 3 = – 6
Answer:
a³ – a² – 3
= (- 3)³ – (- 3)² – 3 (when a = – 3)
= (- 3 × – 3 × – 3) – (- 3 × – 3) – 3
= – 27 – (9) – 3
(Error is (- 3)³ = – 9 and (- 3)² = 6)
= – 27 – 9 – 3 = – 39

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.3

Question 1.
Write standard form and additive inverse of the following expressions.
(i) – 6a
Answer:
Additive inverse of – 6a = – (- 6a) = 6a

(ii) 2 + 7c²
Answer:
Standard form of 2 + 7c² = 7c² + 2
Additive inverse of 7c² + 2
= – (7c² + 2)
= – 7c² – 2

(iii) 6x² + 4x – 5
Answer:
Given expression is in standard form.
Additive inverse of 6x² + 4x – 5
= – (6x² + 4x – 5)
= – 6x² – 4x + 5

(iv) 3c + 7a – 9b
Answer:
Standard form of 3c + 7a – 9b = 7a – 9b + 3c
Additive inverse of 7a – 9b + 3c
= – (7a – 9b + 3c)
= – 7a + 9b – 3c

Question 2.
Write the following expressions in standard form:
(i) 6x + x² – 5
Answer:
Standard form of 6x + x² – 5
= x² + 6x – 5

(ii) 3 – 4a² – 5a
Answer:
Standard form of 3 – 4a² – 5a
= – 4a² – 5a + 3

(iii) – m + 6 + 3m²
Answer:
Standard form of – m + 6 + 3m²
= 3m² – m + 6

(iv) c³ + 1 + c + 2c²
Answer:
Standard form of
c³ + 1 + c + 2c² = c³ + 2c² + c + 1

(v) 9 – p²
Answer:
Standard form of 9 – p² = – p² + 9

Question 3.
Add the following algebraic expressions using both horizontal and vertical methods. Did you get the same answer with both the methods? Verify.
(i) 2x² – 6x +3; 4x² + 9x + 5
Answer:

(ii) a² + 6ab + 8; – 3a² – ab – 2
Answer:

(iii) – p² + 2p – 10; 4 – 5p – 2p²
Answer:

Question 4.
Subtract the second expression from the first expression:
(i) 2x + y , x – y
Answer:
Let A = 2x + y and B = x – y
A – B = A + (- B) Additive inverse of B is
-B = – (x – y) = – x + y
∴ A – B = A + (- B)
= 2x + y + (- x + y)
= 2x + y – x + y
= 2x – x + y + y
= (2 – 1)x + (1 + 1)y
∴ A – B = x + 2y

(ii) a + 2b + c, – a – b – 3c
Answer:
Let A = a + 2b + c and B = – a – b – 3c
A – B = A + (- B)
Additive inverse of B is
– B = – (- a – b – 3c)
= a + b + 3c
∴ A – B = A + (- B)
= a + 2b + c + (a + b + 3c)
= a + 2b + c + a + b + 3c
= (a + a) + (2b + lb) + (c + 3c)
∴ A – B = 2a + 3b + 4c

(iii) 2l² – 3lm + 5m², 3l² – 4lm + 6m²
Answer:
Let A = 2l² – 3lm + 5m² and
B = 3l² – 4lm + 6m²
A – B = A + (- B)
Additive inverse of B is
– B = – (3t² – 4lm + 6m²)
= – 3l² + 4lm – 6m²
∴ A – B = A + (- B)
= (2l² – 3lm + 5m²) + (- 3l² + 4lm – 6m²)
= 2l² – 3lm + 5m² – 3l² + 4lm – 6m²
= 2l² – 3l² – 3lm + 4lm + 5m² – 6m²
= (2 – 3)l² + (- 3 + 4)lm + (5 – 6)m²
= (- 1) l² + 1 lm + (- 1)m²
∴ A – B = – l² + lm – m²

(iv) 7 – x – 3x², 2x² – 5x – 3
Answer:
Let A = 7 – x – 3x² and B = 2x² – 5x – 3
Write the given expressions in standard form.
∴ A = – 3x² – x + 7 and B = 2x² – 5x – 3
A – B = A + (- B)
Additive inverse of B is
– B = – (2x² – 5x – 3)
= – 2x² + 5x + 3
∴ A – B = A + (- B)
= (- 3x² – x + 7) + (- 2x² + 5x + 3)
= – 3x² – x + 7 – 2x² + 5x + 3
= (- 3x² – 2x²) + (- x + 5x) + (7 + 3)
= (- 3 – 2)x² + (- 1 + 5)x + 10
∴ A – B = – 5x² + 4x + 10

(v) 6m³ + 4m² + 7m – 3, 2m³ + 4
Answer:
Let A = 6m³ + 4m² + 7m – 3 and
B = 2 m² + 4
A – B = A + (- B)
Additive inverse of B is
– B = – (2m³ + 4)
= – 2m³ – 4
∴ A – B = A + (- B)
= (6m³ + 4m² + 7m – 3) + (- 2m³ – 4)
= 6m³ + 4m² + 7m – 3 – 2m³ – 4
= (6m³ – 2m³) + 4m² + 7m + (- 3 – 4)
= (6 – 2)m³ + 4m² + 7m + (- 7)
∴ A – B = 4m³ + 4m² + 7m – 7

Question 5.
Find the perimeter of the beside rect¬angle whose length is 6x + y and breadth is 3x – 2y.

Answer:
Given length of rectangle l = 6x + y
breadth b = 3x – 2y
Perimeter of Rectangle = 2 (l + b)
= 2[(6x + y) + (3x – 2y)]
= 2[6x + y + 3x – 2y]
= 2[(6 + 3)x + (1 – 2)y]
= 2[9x + (- 1) y]
= 2[9x – 1y]
= 2 × (9x) – 2 × (1y)
∴ Perimeter of rectangle = (18x – 2y) units.

Question 6.
Find the perimeter of triangle whose sides are a + 3b, a – b and 2a – b.

Answer:
Let the sides of triangle are
x = a + 3b, y = a – b and z = 2a – b
Perimeter of triangle = x + y + z
= (a + 3b) + (a – b) + (2a – b)
= a + 3b + a – b + 2a – b
= (a + a + 2a) + (3b – b – b)
= (1 + 1 + 2)a + (3 – 1 – 1)b
Perimeter of triangle.
= (4a + b) units.

Question 7.
Subtract the sum of x² – 5xy + 2y² and y² – 2xy – 3x² from the sum of 6x² – 8xy – y² and 2xy – 2y² – x².
Answer:
Given expressions are
x² – 5xy + 2y² and y² – 2xy – 3x² and 6x² – 8xy – y² and 2xy – 2y² – x²
Write the given expressions in the standard form
x² – 5xy + 2y² and – 3x² – 2xy + y² and 6x² – 8xy – y² and – x² + 2xy – 2y²
Let the Sum
A = (x² – 5xy + 2y²) + (- 3x² – 2xy + y²)
= x² – 5xy + 2y² – 3x² – 2xy + y²
= x² – 3x² – 5xy – 2xy + 2y² + y²
= (1 – 3)x² + (- 5 – 2)xy + (2 + 1)y²
A = – 2x² – 7xy + 3y²

Let the Sum
B = (6x² – 8xy – y²) + (- x² + 2xy – 2y²)
= 6x² – 8xy – y² – x² + 2xy – 2y²
= (6 – 1 )x² + (- 8 + ²)xy + (- 1 – 2)y²
B = 5x² – 6xy – 3y²
B – A = B + (- A)
Additive inverse of A is
– A = – (A)
= – (- 2x² – 7xy + 3y²)
∴ – A = 2x² + 7xy – 3y²
B – A = B + (- A)
= (5x² – 6xy – 3y²) + (2x² + 7xy – 3y²)
= (5 + 2)x² + (- 6 + 7)xy + (- 3 – 3)y²
∴ B – A = 7x² + xy – 6y²

Question 8.
What should be added to 1 + ²p – 3P² to get p² – p – 1 ?
Answer:
Given expressions are
1 + 2p – 3p² and p² – p – 1
Write the given expressions in the standard form.
– 3p² + ²p + 1 and p² – p – 1
Let A should be added to B to get C.
i.e. A + B = C
∴ A = C – B
Let B = – 3p² + 2p + 1 and
C = p² – p – 1
A = C + (- B )
Additive inverse B is
– B = – (- 3p² + 2p + 1) .
– B = 3p² – 2p – 1 .
A = (p² – p – 1) + (3p² – 2p – 1)
= p² – p – 1 + 3p² – 2p – 1
= (1 + 3)p² + (- 1 – 2)p + (- 1 – 1)
∴ A = 4p² – 3p – ²
∴ 4p² – 3p – ² is added to 1 + 2p – 3p² to get p² – p – 1.

Question 9.
What should be taken away from 3a² – 4b² + 5ab + 20 to get – a² – b² + 6ab + 3 ?
Answer:
Given expressions are
3a² – 4b² + 5ab + 20 and – a² – b² + 6ab +3
Let A be taken away from B to get C. that is A = B – C = B + (- C)
Let B = 3a² – 4b² + 5ab + 20 and C = – a² – b² + 6ab + 3 Additive inverse of C is
(- C) = – (- a² – b² + 6ab + 3)
= a² + b² – 6ab – 3
A = B + (- C)
= (3a² – 4b² + 5ab + 20) + (a² + b² – 6ab – 3)
= 3a² – 4b² + 5ab + 20 + a² + b² – 6ab – 3
= 3a² + a² – 4b² + b² + 5ab – 6ab + 20 – 3
= (3 + 1)a² + (- 4 + 1)b² + (5 – 6) ab + (20 – 3)
A = 4a² – 3b² – 1 ab + 17
So, 4a² – 3b² – 1 ab + 17 is taken away from 3a² – 4b² + 5ab + 20 to get – a² – b² + 6ab + 3.

Question 10.
If A = 4x² + y² – 6xy;
B = 3y² + 12x² + 8xy;
C = 6x² + 8y² + 6xy then,
find(i) A + B + C (ii) (A – B) – C
Answer:
Given A = 4x² + y² – 6xy
B = 3y² + 12x² + 8xy
C = 6x² + 8y² + 6xy
Write the given expressions in standard form.
A = 4x² – 6xy + y²
B = 12x² + 8xy + 3y²
C = 6x² + 6xy + 8y²

(i) A + B + C = (4x² – 6xy + y²) + (12x² + 8xy + 3y²) + (6x² + 6xy + 8y²)
= 4x² – 6xy + y² + 12x² + 8xy + 3y² + 6x² + 6xy + 8y²
= (4x² + 12x² + 6x²) + (- 6xy + 8xy + 6xy) + (y² + 3y² + 8y²)
= (4 + 12 + 6) x² + (- 6 + 8 + 6) xy + (1 + 3 + 8)y²
∴ A + B + C = 22x² + 8xy + 12y²

(ii) (A – B) – C
A + (- B) + (- C)
Additive inverse of B is
– B = – (12x² + 8xy + 3y²)
∴ – B = – 12x² – 8xy – 3y²
Additive inverse of C is
– C = -(6x² + 6xy + 8y²)
∴ – C = – 6x² – 6xy – 8y²
A + (- B) + (- C)
= (4x² – 6xy + y²) + (- 12x² – 8xy – 3y²) + (- 6x² – 6xy – 8y²)
= 4x² – 6xy + y² – 12x² – 8xy – 3y² – 6x² – 6xy – 8y²
= (4x² – 12x² – 6x²) + (- 6xy – 8xy – 6xy) + (y² – 3y² – 8y²)
= (4 – 12 – 6)x² + (- 6 – 8 – 6)xy + (1 – 3 – 8)y²
∴ (A – B) – C = – 14x² – 20xy – 10y²

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.2

Question 1.
State True or False and give reasons for your answer.
(i) 7x² and 2x are unlike terms.
Answer:
True.
7x² and 2x have different algebraic factors and terms contain variables with different exponents.

(ii) pq² arid – 4pq² are like terms.
Answer:
True.
pq² and – 4pq² are contain variables with same exponents. So, they are like terms.

(iii) xy, – 12x²y and 5xy² are like terms.
Answer:
False.
xy, – 12x²y and 5xy² are contain variables with different exponents. So, they are unlike terms.

Question 2.
Write like terms in the following :
(i) a², b², 2a², c²
Answer:
Given a², b², 2a², c²
Like terms: a², 2 a².

(ii) 5x, yz, 3xy, \(\frac{1}{9}\)yz
Answer:
Given 5x, yz, 3xy, \(\frac{1}{9}\)yz
Like terms : yz, \(\frac{1}{9}\)yz.

(iii) 4m²n, n²p, – m²n, m²n²
Answer:
Given 4m²n, n²p, – m²n, m²n²
Like terms: 4m²n, – m²n.

(iv) acb², 2c²ab, 5b²ac, 3cab²
Answer:
Given acb², 2c²ab, 5b²ac, 3cab²
Like terms: acb², 5b²ac, 3cab².

Question 3.
Write number of terms and name of the expression for the following algebraic expressions.
(i) p²q²p
(ii) 2020
(iii) 3ab – \(\frac{a}{2}\) + \(\frac{b}{5}\)
Answer:

Algebraic expression	Number of terms	Name of the expression
(i) p2q + q2p	2	Binomial
(ii) 2020	1	Monomial
(iii) 3ab – \(\frac{a}{2}\) + \(\frac{b}{5}\)	3	Trinomial

Question 4.
Classify the following into monomials, binomilas and trinomials:
(i) 8a + 7b²
(ii) 15xyz
(iii) p + q – 2r
(iv) l²m²n²
(v) cab²
(vi) 3t – 5s + 2u
(vii) 1000
(viii) \(\frac{\mathbf{c d}}{2}\) + ab
(ix) 5ab – 9a
(x) 2p²q² + 4qr³
Answer:

Algebraic expression	Name of the expression
(i) 8a + 7b2	Binomial
(ii) 15xyz	Monomial
(iii) p + q – 2r	Trinomial
(iv) l2m2n2	Monomial
(v) cab2	Monomial
(vi) 3t – 5s + 2u	Trinomial
(vii) 1000	Monomial
(viii) \(\frac{\mathbf{c d}}{2}\) + ab	Binomial
(ix) 5ab – 9a	Binomial
(x) 2p2q2 + 4qr3	Binomial

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.1

Question 1.
Write numerical and algebraic coefficients of the following terms.
(i) 4xy
Answer:
Given term is 4xy
Numerical coefficient: 4
Algebraic coefficient: xy

(ii) – 7a² b³ c
Answer:
Given term is – 7a² b³ c
Numerical coefficient: – 7
Algebraic coefficient: a²b³c

(iii) \(\frac{p q}{2 r}\)
Answer:
Given terra is \(\frac{p q}{2 r}\)
Numerical coefficient: \(\frac{1}{2}\)
Algebraic coefficient: \(\frac{p q}{r}\)

(iv) – 6mn
Answer:
Given term is – 6mn
Numerical coefficient: – 6
Algebraic coefficient: mn

Question 2.
Write the number of terms in each of the following expressions and write ’ them:
(i) 5 – 3t²
(ii) 1 + t² + t³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p²q² + 7pq
Answer:

Question 3.
In – 5ab²c write the coefficients of
(i) b²c
(ii) – b²c
(iii) – 5abc
(iv) 5ac
(v) ab²
(vi) – 5ab
Answer:
Given term is – 5ab²c
(i) Coefficient of b²c is – 5a
(ii) Coefficient of – b²c is 5a
(iii) Coefficient of – 5abc is b
(iv) Coefficient of 5ac is – b²
(v) Coefficient of ab² is – 5c
(vi) Coefficient of – 5ab is bc

Question 4.
Write term containing x and coefficient of x for the following algebraic expressions.
(i) 2x + 5y
(ii) -x + y + 3
(iii) 6y² – 7xy
Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Review Exercise

Question 1.
Identify constants and variables in the following terms:
0, – x, 3t, – 5, 5ab, – m, 700, – n, 2pqr, – 1, ab, 10, – 6z
Answer:
Given, 0, – x, 3t, – 5, 5ab, – m, 700, – n, 2pqr, – 1, ab, 10, – 6z
Constants = 0, – 5, 700, – 1, 10
Variables = – x, 3t, 5ab, – m, – n, 2pqr, ab, – 6z.

Question 2.
Observe the pattern the side and express the pattern in the form of an algebraic expression.

Answer:
In row = 1, 2(1) + 1 = 3
In row = 2, 2(2) + 1 = 5
In row = 3, 2(3) + 1 = 7
In row = 4, 2(4) + 1 = 9
In row = n, 2(n) + 1 = 2n + 1

Question 3.
Write the following statements as expressions
(i) x reduced by 5
Answer:
Given x reduced by 5 = x – 5

(ii) 8 more than twice of k
Answer:
Given 8 more than twice of k = 2k + 8

(iii) Half of y
Answer:
Given Half of y = \(\frac{1}{2}\) of y = \(\frac{y}{2}\)

(iv) One fourth of product of b and c
Answer:
Given One fourth of product of b and c
= \(\frac{1}{4}\) of b × c
= \(\frac{1}{4}\) × bc = \(\frac{b c}{4}\)

(v) One less than three times of p
Answer:
Given One less than three times of p
= 3 times of p – 1
= 3 ∙ p – 1

Question 4.
Write the following expressions as statements :
(i) s + 3
Answer:
Given s + 3 = 3 added to s.

(ii) 3p + 10
Answer:
Given 3p + 10
= 10 more than thrice of p.

(iii) 5c – 8
Answer:
Given 5c – 8 = 8 less than 5 times of c.

(iv) 10z
Answer:
Given 10z = 10 times of z.

(v) \(\frac{b}{9}\)
Answer:
Given \(\frac{b}{9}\) = one nineth of b.

Question 5.
Write the following situations into algebraic expressions :
(i) Cost of one pen is double the cost of pencil.
Answer:
Let the cost of pencil = ₹x
then the cost of pen = double the cost of pencil = 2 × ₹ x
∴ The cost of pen = ₹ 2x

(ii) Age of John is 10 more than age of Yusuf.
Answer:
Let the age of Yusuf = a years
then the age of John = (a + 10) years

(iii) Height of Siri is 15 cm less than height of Giri.
Answer:
Let the height of Giri = x cm
then the height of Siri = (x – 15) cm

(iv) Length of a rectangle is 2 more than three times of it’s breadth.
Answer:
Let the breadth = b units
then the length of rectangle
= 3 times b + 2
= (3 ∙ b + 2) units.

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling InText Questions

[Page No. 108]

Where does the Arithmetic Mean lie ?
The marks obtained by Sarala, Bindu, Geeta and Rekha in Telugu, Hindi and English are given below.

Answer:
Now, let us calculate the average marks obtained by the students in each subject.

Answer:

(i) What do you observe from the above table ?
Answer:
Mean always lie between the maximum and minimum values/observations of the data.

(ii) Does the mean lies between maximum and minimum values in each case ?
Answer:
Yes, Its always true.
i.e., Mean always lies between the maximum and minimum values of the data.

Check Your Progress [Page No: 109]

Question 1.
Find the Arithmetic Mean of first three multiples of 5 ?
Answer:
Given, first three multiples of 5 are: 5, 10, 15.
Sum of observations = 5 + 10 + 15 = 30
Number of observations = 3
Arithmetic mean =\(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{30}{3}\) = 10
∴ Arithmetic mean of first three multiples of 5 is 10.

Let’s Explore [Page No. 109]

Collect the information about the weights of any ten students of your class in kilograms arid answer the following:

Answer:

Name of the student	Weight in kilograms
Aditya	38
Kishore	39
Balu	40
Sreeja	36
Sreekari	35
Kairavi	3.7
Swathi	36
Kirshna	41
Ram	39
Prasad	39

Question 1.
What are the greatest and smallest weights ?
Answer:
Krishna = 41 kg – Greatest weight
Sreekari = 35 kg – Smallest weight

Question 2.
Find Arithmetic mean of collected data.
Answer:
Given observations: 38, 39, 40, 36, 35, 37, 36, 41, 39, 39.
Sum of observations = 38 + 39 + 40 + 36 + 35 + 37 + 36 + 41 + 39 + 39 = 380
Number of observations = 10
Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{380}{10}\) = 38 kg.
Number of observations 10
∴ Arithmetic mean weight of 10 students of our class is 38 kg.

Question 3.
Verify whether Arithmetic Mean lies between greatest and smallest observations or not.
Answer:
Yes, Arithmetic mean 38 kg lies between greatest weight 41 kg and smallest weight 35 kg.

Check Your Progress [Page No. 110]

Question 1.
What is the range of first ten whole numbers?
Answer:
First ten wh4’e numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Maximum value = 9
Minimum value = 0
Range = Maximum Value – Minimum Value = 9 – 0 = 9
∴ Range = 9

Check Your Progress [Page No. 111]

Find the mode of 10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 and 2.
Answer:
Given data : 10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 and 2.
By arranging the numbers with same values together.
2, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 10, 12
As ’10’ occurs more frequently than other observations.
∴ Mode = 10

Let’s Explore [Page No. 111]

Take a dice, roll it 20 times and record the numbers you got on its top face. Find the ‘Mode’ of resulting numbers.
Answer:
A dice rolled 20 times and the numbers are
2, 4, 5, 3, 1, 6, 5, 4, 2, 1, 3, 5, 4, 2, 6, 2, 2, 5, 1, 3.
By arranging the numbers with same values together
1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6.
As 2 occurs more frequently than other observations.
Mode = 2.

Let’s Think [Page No: 111]

The following data shows that the number of hours spent by students for study. Find the mode:

Answer:
In the given data more number of students spend 1 hour for study.
So, mode of the data = 4.

Check Your Progress [Page No: 114]

What is the median of first 7 prime numbers ?
Answer:
First 7 prime numbers are 2, 3, 5, 7, 11,13, 17.
Arrange the given observations in ascending order

In seven observations the 4th observation 7 is the middle most value.
∴ Median = 7.

Let’s Explore [Page Mo. 114]

Question 1.
Collect 10th class pass percentage for last six years of your school (or) your near by school. Find median of the data.
Answer:
10th Class pass percentage for last six years of our school are: 100%, 98%, 93%, 95%, 96%, 97%.
Arrange the given observations in ascending order.

In six observations the 3rd and 4th observations are 96% and 97%.
Here, we have two middle most values 96% and 97%.
Median = Average of two middle most values
= \(\frac{96+97}{2}\) = 96.5%
∴ Median of the data = 96.5%

Project Work [Page No: 114]

Visit any vegetable market along with your parent. Collect the information about the costs of different vegetables. By using this data fill the following table.

Answer:

[Page No. 115]

Prasanna wanted to buy a mobile phone. He selected two mobile phones of different companies having same features. Now he would like to know which mobile phone has good performance. He collected the information of star rating from different magazines and news papers.

Question 1.
What information is given in the table ?
Answer:
The table is showing information about the mobile phones.

Question 2.
Is the above information being useful to Prasanna ?
Answer:
Yes.

Question 3.
Which mobile phone you will suggest to Prasanna ?
Answer:
Mobile A. (21 + \(\frac{1}{2}\) + \(\frac{1}{2}\) rating)

Check Your Progress [Page No. 116]

Observe the adjacent bar graph and answer the following.

Question 1.
Which fruit most of the people like ?
Answer:
Most of the people liked Apple.

Question 2.
How many people likes banana?
Answer:
10.

Observe the following graph. [Page No. 116]

Observe the above double bar graph and answer the following:

Question 1.
In which year the sales of both mobile phone companies are equal ?
Answer:
2018.

Question 2.
In 2017, which mobile phone company has more sales ?
Answer:
Mobile phone – A.

Examples

Question 1.
Mid Day Meal (MDM) taken by the students in six days are 132, 164, 145, 182, 163 and 114, Find Arithmetic Mean of students who took MDM per day.
Answer:
Given, number of students who took MDM 132, 164, 145, 182, 163, 114.
Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{132+164+145+182+163+114}{6}\)
= \(\frac{900}{6}\)
= 150

Question 2.
Ages of students (in years) are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4 7, 6, 7, 6, 5 8 and 6 find the mode of given data.
Answer:
Given, ages of students are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4, 7, 6, 7, 6, 5, 8, 6.
By arranging the numbers with same values together
4, 5, 5, 5, 5,5,6, 6,6,6, 6, 6,7,7, 7, 8, 8.
As 6’ occurs more frequently than other observations.
∴ Mode = 6

Question 3.
Find the mode of the data of Grades A, B, E, A, C, E, B, C, D, A, D, C, F, A and C.
Answer:
Given, A, B, E, A, C, E, B, C, D, A, D, C, F, A, C.
By arranging the letters of same type together
A, A, A, A, B, B, C, C, C, C, D, D, E, E, F
As ‘A’ and ‘C’ occurs most frequently in the data.
∴ Mode = A and C

Question 4.
Find the median of data 32, 43, 25, 67, 46, 71 and 182.
Answer:
Given 32, 43, 25, 67, 46, 71, 182
Arrange the given observations in ascending order.

In ‘7’ observations the 4th observation is the middle most value.
∴ Median = 46.

Question 5.
The monthly incomes of 8 members are ₹8000, ₹9000, f8200, ₹7900, ₹8500, ₹8600, ₹7700 and ₹60000. Find the median income.
Answer:
Given that, the monthly incomes of 8 members are:
₹8000, ₹9000, ₹8200, ₹7900, ₹8500, ₹8600, ₹7700, ₹60000
Arrange the given observations in ascending order.

Here, we have two middle most values 8200 and 8500.
In this case, Median = Average of two middle most values
8200 + 8500 16700
= \(\frac{8200+8500}{2}\) = \(\frac{16700}{2}\) = ₹8,350

Question 6.
The number of CFL bulbs and LED bulbs by a seller every month from March to August are given below. Draw by vertical double bar graph.

Month	CFL Bulbs	LED Bulbs
March	70	75
April	35	30
May	65	75
June	90	100
July	22	35
August	50	50

Answer:
Steps in drawing a Double bar graph:

1. Draw X-axis (horizontal line) and Y – axis (vertical line) on the graph paper and mark their intersection point as ‘O’.
2. Take ‘months’ on X – axis.
3. Take number of CFL and LED bulbs on Y ~ axis.
4. Take an appropriate scale on Y-axis so that number of both the bulbs can be shown easily. Here the maximum value to be plotted on Y – axis as 100, so let us take 1 cm =10 bulbs on Y – axis.
5. Find the length of each bar by dividing the value by 10 (as scale is 1 cm = 10 bulbs).

Eg: Length of bar represents 70 CFL bulbs = \(\frac{70}{10}\) = 7 cm
Length of bar represents 75 LED bulbs = \(\frac{75}{10}\) = 7.5 cm

Question 6.
Draw the bars of uniform width representing ‘CFL bulb’ and ‘LED bulb’ side by side of every month.

Observe the adjacent figure:

Question 1.
Largest portion of the circle is shaded by which colour ?
Answer:
Red.

Question 2.
Are the portions of circle shaded by blue and pink are in same size ?
Answer:
No.

Question 3.
Smallest portion of the circle is shaded by which colour ?
Answer:
Yellow.

Question 2.
Observe the below picture which shows various expenses of Manasa’s family:

(i) For which item highest amount spent?
Answer:
Food.

(ii) For which items same amount spent?
Answer:
Charity and Education.

(iii) For which item least amount spent?
Answer:
Others.

Question 7.
In a school, there are 100 students in class VII and every student is a member of any one of the club. The following table shows the number of students in various clubs, then Construct the pie chart to following data:

Club	Number of members
Mathematics	50
Science	30
Social Studies	40
English	40
Arts	20

Answer:
The angle of each sector will depend on the ratio between the number of students in club and total number of students.
Angle of sector = \(\frac{\text { Value of the item }}{\text { Sum of the values of all items }}\) × 360°

Steps of construction:

1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct ∠AOB = 100° to represent angle of the sector for Maths club.
4. Construct ∠BOC = 60° to represent angle of the sector for Science club.
5. Construct ∠COD = 80° to represent angle of the sector for Social Studies club.
6. Construct ∠DOE = 80° to represent angle of the sector for English club.
7. Now ∠EOA = 40° to represent angle of the sector for Arts club.

Practice Questions [Page No: 124]

In the following alphabetical series, a term (next term) is missing. Choose the missing term from the options.

Question 1.
B, F, J, N, R,’V, …..
(a) Z
(b) W
(c) X
(d) Y
Answer:
(a) Z

Explaination:

So, answer is Z.

Question 2.
A, C, E, G, I, K, …..
(a) P
(b) O
(c) N
(d) M
Answer:
(d) M

Explaination:

So, answer is M.

Question 3.
M, O, R, T, …….
(a) W
(b) U
(c) V
(d) Q
Answer:
(a) W

Explaination:

So, answer is W.

Question 4.
U, S, P, L, …….
(a) F
(b) G
(c) H
(d) I
Answer:
(b) G

Explaination:

So, answer is G.

Question 5.
ZA, YB, XC, WD,
(a) UE
(b) EV
(c) VE
(d) SH
Answer:
(c) VE

Explaination:

So, answer is VE.

Question 6.
AM, BO, CQ, DS, EU, …………
(a) WF
(b) FU
(c) GV
(d) FW
Answer:
(d) FW

Explaination:

In the series next word is FW.

Question 7.
ZY, XV, UR, QM, …..
(a) LG
(b) LI
(c) LH
(d) KJ
Answer:
(a) LG

Explaination:

So, in the series next word is LG.

Question 8.
AC, DF, GI, JL, …..
(a) NO
(b) MO
(c) MN
(d) NP
Answer:
(b) MO

Explaination:

So, in the series next word is MO.

Question 9.
DN, EM, FL, GK, HJ, …..
(a) IK
(b) GI
(c) IJ
(d) II
Answer:
(d) II

Explaination:

So, in the series next word is II.

Question 10.
CBA, STU, FED, VWX,
(a) IHG
(b) GHI
(c) IJK
(d) YZA
Answer:
(a) IHG

Explaination:

So, in the series next word is IHG.

Question 11.
AZC, DYF, GXI, JWL
(a) OVM
(b) UNV
(c) MVO
(d) MNO
Answer:
(c) MVO

Explaination:

So, in the series next word is MVO.

Question 12.
ABK, CDL, EFM, GHN,
(a) JIO
(b) IJO
(c) MNO
(d) ONM
Answer:
(b) IJO

Explaination:

So, in the series next word is IJO.

Question 13.
A2C, D5F, G8I, J11L, …..
(a) Ml40
(b) Ml20
(c) N15P
(d) N12P
Answer:
(a) Ml40

Explaination:

So, in the series next word is M140.

Question 14.
A, CD, HIJ, PQRS, …………..
(a) ZABCD
(b) ZYXW
(c) ABCDE
(d) RSTUV
Answer:
(c) ABCDE

Explaination:

So, in the series next word is ABCDE.

Question 15.
A, BC, DEF, GHIJ,
(a) KLMNP
(b) LMNOP
(c) KLMNO
(d) JKLMN
Answer:
(c) KLMNO

Explaination:
A BC DEF GHIJ KLMNO
So, in the series next word is KLMNO.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion InText Questions

[Page No. 1]

There are several situations in our daily life where we use ratio and proportion. Let us look at the following pictures and answers to the given questions:
Question 1.
Can we say the ratio of speeds of Cheetah to Man?

Answer:
Speed of Cheetah = 120 kmph
Speed of a man = 20 kmph
∴ Ratio of speed of Cheetah to a man
= 120 : 20
= 6 : 1

Question 2.
What will be the ratio of heights of HematoAmir?

Answer:
Height of Hema = 150 cm
Height of Amir 75 cm
∴ Ratio of heights 150 : 75
= 2 : 1

Question 3.

Answer:
Cost of 2 Hands of banana = ₹ 80
∴ Cost of 3 Hands of banana
= \(\frac{3}{2}\) × 80 = ₹ 120

Question 4.
Do you know which aspect ratio makes constructions more attractive and beautiful?
Answer:
Golden ratio is more beautiful. It is the ratio of a line segment cut in to two pieces of different lengths such that the ratio of whole segment to that of the longer segment is equal to that of longer segment to the shorter segment.

Check your Progress [Page No. 4]

Question 1.
Write the compound ratio of the following given ratios.
(i) 3 : 5 and 4 : 3
Answer:
Given ratios are 3 : 5 and 4 : 3.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes.
that is a × c : b : d
3 × 4 : 5 × 3
12 : 15
∴ Compound ratio = 4 : 5

(ii) 8 : 3 and 6 : 5
Answer:
Given ratios are 8 : 3 and 6 : 5.
If a : b and c : d are ratios, then their compound ratio is product of antecedents : product of extremes, that is
a × c : b : d
8 × 6 : 3 × 5
∴ Compound ratio = 16 : 5

(iii) 2: 1 and 8: 7
Answer:
Given Ratios are 2 : 1 and 8 : 7.
If a : b and c: d are ratios, then their compound ratio is product of antecedents : product of extremes that is a × c : b : d
2 × 8 : 1 × 7
16 : 7
∴ Compound ratio= 16 : 7

Question 2.
Fill in the boxes with correct answers.

Answer:

[Page No. 6]

Discuss and answer the following questions related to real-life situations:
Question 1.
If the cost of 3 ball pens is ₹ 15, then what is the cost of 6 such pens?
Answer:
3 : 15 : : 6 : cost
Cost x 3 = 15 × 6
∴ Cost = \(\frac{15 \times 6}{3}\) = ₹ 30

Question 2.
In a school, for implementing ‘Mid day Meal’ scheme for class VII of 40 students they need 6 kgs of rice for one day.
(i) How much rice is needed for 6 days?
Answer:
Rice required for 40 students for one day = 6 kgs
Rice required for 6 days = 6 × 6 = 36 kg

(ii) How much rice is needed for 10 days?
Answer:
Rice required for 10 days = 10 × 6 = 60 kg

(iii) How much rice is needed for 5 days?
Answer:
Rice required for 5 days = 5 × 6 = 30 kg

Let’s Explore [Page No. 6]

Question 1.
Given below the number of workers working for construction of a house and their wages in total.

Answer:

Let’s Think [Page No. 7]

Question 1.
If the cost of 4 note books is 80. What would be the cost of7 note books? The above problem can be solved by using unitary method. Can you solve? How? Think.
Answer:
Given the cost of 4 note books = ₹ 80
Cost of 1 note book = 80 ÷ 4 = ₹ 20
Cost of 7 note books = ₹ 20 × 7
= ₹ 140

Check Progress [Page No 7]

Question 1.
Fill in the blanks, if the given quantities are in direct proportion.

Answer:

Let’s’ Do Activity [Page No. 8]

Take a clock and fix Its minute hand at a particular number on clock (if it is 12 It will be easy). Then find and note the angles made by minute hand in every 15 minutes interval of time:


Check whether they are in direct proportion or not?
Answer:

15 : 30 and 90 : 180
Here 1 :2 = 1 :2
If ratios are equal, then they are in direct proportion.
So, time passed and angle are in direàt proportion.

Let’s Think [Page No. 8]

Question 1.
What is the angle made by minutes hand in a minute?
Answer:
We know angle made by minutes hand in 15 m = 90°
Let angle made by minutes hand in 1 m = x°
Then 15 : 1 = 90°: x°
If the ratios are equal,
The product of means = The product of extremes
⇒ 15 × x = 1 × 90°
⇒ \(\frac{15 x}{15}=\frac{90^{\circ}}{15}\)
⇒ x = 6°
∴ Angle made by minutes hand in 1 minute = 6°

Question 2.
What is the angle made by hours hand in one minute?
Answer:
We know angle made by hours hand in 1 hour (60 minutes) = 30°
Let angle made by hours hand 1 minute = x°
Then 60 : 1 = 30°: x°
If the ratios are equal,
The product of means = The product of the extremes
⇒ 60 × x = 1 × 30
⇒ \(\frac{60 x}{60}=\frac{30}{60}=\frac{1}{2}\)
⇒ x = \(\frac{1}{2}\) (or) 0.5°
∴ Angle made, by hours hand in 1 minute = \(\frac{1}{2}\)° (or) 0.5°

Check your Progress [Page No. 9]

Question 1.
Fill in the blanks if the given quantities are in inverse proportion.

Answer:

Check your Progress [Page No. 13]

Question 1.
Analyse how the three quantities given below are related and find ‘x’.

Answer:
From the above,

• Ration and the number of days are in inversely proportional.
• Ration and number of men are in directly proportional.
• Number of men is depends on both Ration and number of days.

So we have to take compound ratio of 108 : 70 and 25 : 15
i. e., 108 × 25 : 70 × 15
∴ 18 : x = 108 × 25 : 70 × 15
If the ratios are equal.
The Product of means = The Product of extremes
⇒ 108 × 25 × x = 18 × 70 × 15

∴ x = 7

Let’s Think [Page No. 18]

Question 1.
A man buys chocolates 10 for 10 rupees and sells them at 10 for ₹ 12. Does this result profit or loss ? What percent?
Answer:
Given cost price of 10 chocolates = ₹ 10
Selling price of 10 chocolates = ₹ 12
SP > CP, then man gets profit.
Profit = S.P – C.P
= 12 – 10 = ₹ 2
We know profit percent

Profit percentage = 20%

Question 2.
If a shopkeeper bought sofa sets and increases their prices by 50% and sell at 50% less, is it a loss or gain?
Answer:
Let the cost price of sofa set = ₹ 100

= ₹ 5o
Total printed price = ₹ 100 + ₹ 50
= ₹ 150
Loss percent = 50%

Loss = ₹ 75
Selling price Printed Price – Loss
= 150 – 75 = ₹ 75
C.P= ₹ 100; S.P = ₹ 75
and C.P > S.P
So, the shopkeeper will get loss of 25%.

Check Your Progress [Page No. 20]

Answer:

Check Your Progress [Page No. 24]

Answer:

Let’s Think [Page No. 24]

Question 1.
At what rate per annum will the principle doubles in 10 years?
Answer:
Method 1:
Let the principle = 100
Interest rate = x% ; Time = 10 years

Amount = Principle + Interest
= Double the principle
= 100 + 10x = 200
⇒ 100 + 10x – 100 = 200 – 100
⇒ 10x = 100
⇒ \(\frac{10 x}{10}=\frac{100}{10}\)
⇒ x = 10%
∴ Rate of interest = 10%

Method 2:
Let the principle =?
Rate of Interest = R%
T = 10 years
Given that A = 2P = P + P [∵ A = P + I]
∴ I = P
∴ \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}\) = P
⇒ \(\frac{\mathrm{P} \times 10 \times \mathrm{R}}{100}\) = P
R = 10%
∴ The principle become double in 10 years at 10% rate of interest.

Question 2.
At what rate per annum will the principle be 4 times in 15 years?
Answer:
Method 1:
Let the principle = ₹ 100
Interest rate = x %
Tirne = 15 years
Interest = \(\frac{\mathrm{PTR}}{100}\)

Amount = Principle + Interest
= 4 times the principle
= 100 + 15x = 400
⇒ 100 + 15x – 100 = 400 – 100
⇒ 15x = 300
⇒ \(\frac{15 x}{15}=\frac{300}{15}\) = 20%
∴ x = 20%

Method 2:
Let the principle = P
Time = 15 years
A = 4timesP = 4P
⇒ A = P + 3P
But A = P + I
So I = 3P

We know that I = \(\frac{\text { PTR }}{100}\)
⇒ 3P = \(\frac{P \times 15 \times R}{100}\)
⇒ R = \(\frac{100 \times 3}{15}\) = 20%
∴ The principle become 4 times in 15 years at 20% rate of interest.

Examples:

Question 1.
Find the compound ratio of 8 : 7 and 9 : 13.
Answer:
Given ratio = 8 : 7 :: 9 : 13
Compound ratio = 8 × 9 : 7 × 13 = 72 : 91
(∵ Compound ratio product of antecedent product consequents).

Question 2.
Two friends Prabhu and Suresh started a business with ₹ 1,00,000 each. After 3 months Suresh left the business. At the end of the year, there was a profit of ₹ 20,000. Calculate the profits shared by Prabhu and Suresh?
Answer:
Here Prabhu and Suresh started a business with ₹ 1,00,000.
Prabhu continued till the end of the year.

Suresh continued only for 3 months.
The ratio of the contributions
= 100000: 100000 = 1 : 1

Ratio of their period of business
= 12 : 3 = 4 : 1
So, their profits sho’uld be divided on the basis of compound ratio
Compound ratio = 1 × 4 : 1 × 1
Profit = ₹ 20,000
Total parts= 4 + 1 = 5
Suresh’s profit = 20000 × \(\frac{1}{5}\)
= ₹ 4000
Prabhus profit = 20000 – 4000
= ₹ 16000

Question 3.
Rani started a beauty parlour with an amount ₹ 75,000. After 4 months, Vani also joined with Rani with an amount ₹ 50,000. After one year, they got a profit of ₹ 52,000. Calculate the profits shared by Rani and Vani.
Answer:
Rani’s investment = ₹ 75,000
Rani’s period in business = 1 year = 12 months
Vani’s investment = ₹ 56,000
Vani’s period in business = 8 months
The ratio of investments of Rani to Vani = 75,000 : 50,000 = 3:2
The ratio of periods of business of Rani to Vani =12:8 = 3:2

The profit should be distributed on the, basis of compound ratio.
Compound ratio = 3 ×. 3 : 2 × 2
= 9 : 4
Profit = ₹ 52,000

Total parts = 9 + 4 = 13
Rani’s profit = 52,000 × \(\frac{9}{13}\)
= ₹ 36000
Vani’s profit = 52,000 – 36,000 = ₹ 16,000

Question 4.
If 3 : 4 and 9 : x are in direct proportion, then what is the value of x?
Answer:
If 3 : 4 and 9 : x are in direct proportion, then their ratio is constant.
∴ \(\frac{3}{4}=\frac{9}{x}\)
⇒ 3 × x = 4 × 9
⇒ x = \(\frac{4 \times 9}{3}\) = 12

Question 5.
If the cost of 4 note books is ₹ 80. What would be the cost of 7 note books?
Answer:
We know that as number of note books increases, the cost also increases such that the ratio of number of note books and the ratio of their costs will remain the same. That means here number of note books and the cost are in direct proportion.

Let the cost of 7 note books be ‘x’.
Then, 4 : 80 = 7 : x
If the ratios are equal, the product of means = The product of the extremes
4 × x = 80 × 7
⇒ x = \(\frac{80 \times 7}{4}\) = ₹ 140
Thus, the cost of 7 note book is equal to ₹ 140.

Question 6.
The scale of a map is given as 1 : 30000. If two cities are 20 cm apart on the map, then o-j, . find the actual distance between them.
Answer:
Let the actual distance be ‘x’ cm. Since the distance on the map is directly proportional to the actual distance.
1 : 30000 = 20 : x
If the ratios are equal, the product of means = The product of the extremes
∴ 1 × x = 30,000 × 20
⇒ x = 6,00,000 cm = 6 km. [1,00,000 cm = 1000 m = 1 km.]
Thus, two cities which are 20cm apart on the map are actually 6 km away from each other.

Question 7.
If 4, 7 and 2, x are inverse proportion, then what is the value of x?
Answer:
4, 7 and 2, x are in inverse proportion . Therefore, 4 × 7 = 2 × x
⇒ x = \(\frac{4 \times 7}{2}\) = 14

Question 8.
If 18 workers can build a wall in 12 days, how many days will eight workers take to build the same wall?

Answer:
If the number of workers decreases, the time taken to build the wall increases in the same proportion.
So, number of workers and the number of days to complete work are in inverse proportion.

Let the number of days to complete the work be ‘x’.

Number of workers	Number of days
18	12
8	x

By taking inverse proportion,
18 : 8 = x : 12
Then,
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days
∴ Eight workers can complete the will in 27 days.

Observe this:
In inverse Proportion, product is always constant.
∴ 18 × 12 = 8 × x
⇒ 8 × x = 18 × 12
⇒ x = \(\frac{18 \times 12}{8}\) = 27days

Question 9.
4 Pumps are required to fill a tank in 1 hr 30 min. How long will it take if only 3 pumps of the same type are used?

Answer:
Let the time be ‘x’.
1 hr 30min = 60 + 30 = 90 minutes ,
If the number of pumps are decrease, the time taken to fill the tank increases.
So, the number of pumps and time taken to fill the tank are in inverse proportion.

Number of pipes	Time taken to fill (min)
4	90
3	x

By taking inverse proportion,
4 : 3 = x : 90
Thus,
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min
∴ 3 pumps will fill the tank in 120 min or 2 hrs.

Observe this:
In inverse proportion, product is always constant
∴ 4 × 90 = 3 × x
⇒ 3 × x = 4 × 90
⇒ x = \(\frac{4 \times 90}{3}\) = 120 min

Question 10.
If 30 persons use 40 kg. of sugar in 10 days, find in how many days 80 persons will use 320 kg. of sugar? How can we solve?
Answer:
Here, the 3 quantities are persons, weight and number of days.

From the above comparison mustbe in between unknown form to know form.

• Here days and persons are in inverse we denote it with Tmark.
• Here days and sugar are in direct proportion we denote it with 4r mark.

Here number of days is depends on both persons and weight of sugar. So we have to take compound ratio of 80 : 30 and 40 : 320.
∴ 10 : x = 80 × 40 : 30 × 320 = 3200 : 9600
Since the ratios are equal, the product of extremes is equal to product of means.
10 : x = 3200 : 9600 ⇒ 10 × 9600 = x × 3200
⇒ x × 3200 = 10 × 9600 ⇒ x = \(\frac{10 \times 9600}{3200}\) = 30

Question 11.
8 painters can paint a wall of 160m long in 5 days. How many painters are required to paint 240kn. wall in 10 days?
Answer:
Here we have three quantities – number of painters, length of wall and number of days.

Number of painters, is directly proportional to length of the wall.
Number of painters is inversely proportional to the number of days.

Since the number of painters depends both on length of wall and number of days. We will take the compound ratio of 160 : 240 and 10 : 5 1
∴ 8 : x = 160 × 10 : 240 × 5
Since the ratios are equal, product of means is equal to product of extremes
8 : x = 160 × 10 : 240 × 5
⇒ x × 160 × 10 = 8 × 240 × 5
⇒ x = \(\frac{8 \times 240 \times 5}{160 \times 10}\) = 6
Hence the required number of painters = 6

Question 12.
195 men working 10 hours a day can finish a job in 20 days. How many men are employed to finish the job in 15 days if they work 13 hours a day?
Answer:
Here we have three quantities – number of workers, number of hours and number of days.

Here the number of workers inversely proportional to number of hours per day.
Here the number of workers inversely proportional to number of days.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of 13 : 10 and 15 : 20.
Since the ratios are equal, the product of means is equal to product of extremes.
195 : x = 13 × 15 : 20 × 10
⇒ x × 13 × 15 = 195 × 20 × 10
⇒ x = \(\frac{195 \times 20 \times 10}{13 \times 15}\) = 200
Hence the required number of workers = 200

Question 13.
Express the following percentages as fraction, decimal and In ratio.
(i) 45%
Answer:
45% = \(\frac{45}{100}=\frac{9}{20}\)(Fraction)
= 0.45 (decimal form)
= 9:20 (ratio)

(ii) 62%
Answer:
62% = \(\frac{62}{100}=\frac{31}{50}\) = (Fraction)
= 0.62 (decimal form)
= 31:50 (ratio)

Question 14.
Find 24% of 150 and àlso the remaining of that number.
Answer:
24% of 150 = \(\frac{24}{100}\) × 150 = 36
The remaining of that number
= 150 – 36 = 114

Question 15.
Raghu bought pens for 400 and he sold them for ₹ 480 what Is his profit or loss percent?
Answer:
Jyothi solved It this way:
Pen C.P of = ₹ 400, S.P = ₹ 480
S.P > C.P.
So, Raghu gets a profit P = 480 – 400 = ₹ 80
Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Anwar solved it this way :
Raghu’s Profit = S.P. – C.P = 480 – 400 = ₹ 80
The ratio of profit and cost or the fraction is \(\frac{80}{400}\)
∴ Profit Percentage = \(\frac{80}{400}\) × 100 = 20%

Suresh solved it using proportion :
When C.P is ₹ 400, the profit is ₹ 80 C.P. is ₹ 100, let the profit be ₹ x.
Here C.P and profit are in direct proportion.
x : 80 = 100 : 400 ⇒ \(\frac{x}{80}=\frac{100}{400}\)
⇒ x × 400 = 100 × 80
⇒ x = \(\frac{100 \times 80}{400}\) = 20%
Profit = 20 per 100
Profit Percentage = 20%

Question 16.
Ramana bought a cycle for ₹ 1200 and sold it to his friend Rehman for ₹ 900, then what is Ramanas profit or loss percentage?
Answer:
Here Ramana’s Cost price = ₹ 1200
Selling price = ₹ 900
S.P < C.P, then Ramana got loss
Loss = C.P – S.P = 1200 – 900 = ₹ 300
Loss percentage = \(\frac{300}{1200}\) × 100 = 25%

Question 17.
If John buys a ear for ₹ 1,50,000 and gains 10% on selling it. Then find the selling price.

Answer:
Cost Price = ₹ 1,50,000
Gain % = 10%
Profit = 10% of ₹ 1,50,000
= \(\frac{10}{100}\) × 150000 = ₹ 15000
Selling Price = C.P. + Profit
= ₹ 1,50,000 + ₹ 15,000
= ₹ 1,65,000
This can be solved using proportion:
Gain 10% means
If CP is ₹ 100,the gain is ₹ 10
Thus S.P = 100 + 10 = 110
Now, here C.P = ₹ 1,50,000

Let S.P = x
CP and SP are directly proportional.
\(\frac{110}{100}=\frac{x}{1,50,000}\)
\(\frac{x}{1,50,000}=\frac{110}{100}\)
x = \(\frac{1,50,000 \times 110}{100}\) = ₹ 1,65,000

Question 18.
Kiran sold a refrigerator for ₹ 16800 at a gain of 12%. Then what is the cost price of it?

Answer:
Roopa did the problem uslnj unitary method:
S.P = ₹ 16800 ; Gain% = 12%
If CP is ₹ 100, then Profit is ₹ 12 and thus S.P = ₹ 112
So when S.P is ₹ 112, then C.P is ₹ 100
When S.P is One Rùpee, C.P is \(\frac{110}{112}\) .
Here S.P is ₹ 16800
So, C.P = \(\frac{110}{112}\) × 16800 = ₹ 15000

Sneha solve did the prot1em using proportion as follows:
Gain % = 12% ;S.P = ₹ 16800
If C.P is ₹ 100 then profit is ₹ 12, thus SP = ₹ 12
here S.P = ₹ 16800
C.P be x
C.P and S.P are directly proportional.
∴ \(\frac{x}{16800}=\frac{100}{112}\)
⇒ x = \(\frac{100 \times 16800}{112}\) = ₹ 15000

Question 19.
The cost of an article goes down every year by 10% of its previous value. Find its original cost, if its cost after 2 years is ₹ 32400.
Answer:
Let the cost at the beginning of 1st year be ₹ 100. At the beginning of,2nd year i.e at the end of 1st year it will be decreased by 10% means cost will be ₹ 90.
At the end of 2nd year i.e at the beginning of 3rd year it will be reduced by 10%
i. e. ₹ 90 is reduced by 10%.
90 – 9 = ₹ 81
If the cost of object is ₹ 100 at the beginning, then after 2 years it’s cost will be ₹ 81.
Let the cost of the object ₹ ‘x’ at the beginning, after 2 years it’s cost is ₹ 32,400 .
Thus ratio of original costs = ratio of costs after 2 years
⇒ x : 100 = 32400 : 81
⇒ \(\frac{x}{100}=\frac{32400}{81}\)
⇒ x = \(\frac{32400 \times 100}{81}\) = ₹ 40000

Question 20.
Find discount if,
(i) marked price is ₹ 450, selling price = ₹ 415
Answer:
Discount = marked price – selling price ⇒ 450 – 415 = ₹ 35.

(ii) marked price = ₹ 810, selling price = ₹ 765.
Answer:
Discount = marked price – selling price ⇒ 810 – 765 = ₹ 45.

Question 21.
If discount is ₹ 40, marked price is ₹ 400, then find discount percentage.
Answer:
Discount = ₹ 40, marked price = ₹ 400,
then discount percentage = \(\frac{40}{400}\) × 100 = 10

Question 22.
A shopkeeper marks his goods 20% above the cost price and allows a discount of 10% on them. What percent does he gain?
Answer:
Let the cost price be ₹ 100.
Then the marked price = 100 + 20 = ₹ 120
Discount = 10%, so discount = x 120 = 12% /
SP = Marked price – Discount = ₹ 120 – ₹ 12 = ₹ 108
Gain = \(\frac{8}{100}\) × 100 = 8%
The shopkeeper gains 8% after discount.

Question 23.
Calculate simple interest and the total amount, if
(i) principal = ₹ 5000, time = 2 years, rate = 10%
Answer:
Simple interest
I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{5000 \times 2 \times 10}{100}\) = ₹ 1000
Total amount = principal + interest = 5000 + 1000 = ₹ 6000

(ii) principal = ₹ 25000, time = 3 years, rate = 12%
Answer:
Simple interest = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 12}{100}\) = ₹ 9000
Total amount = principal + interest = 25000 + 9000 = ₹ 34000

Question 24.
If Raheem borrowed a sum of ₹ 25000 at a rate 10% per annum, what is the simple interest and total amount he has to pay for 3 years?
Answer:
Rajesh did like this :
Principal = ₹ 25000 ; Time = 3 years ; Rate of interest = 10%
Simple interest = I = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{25000 \times 3 \times 10}{100}\) = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Sangeetha did like this :
For 1 year we have to pay 10%,
for 3 years we have to pay 3 × 10 = 30% as interest.
Interest = \(\frac{30}{100}\) × 25000 = ₹ 7500
Total amount = ₹ 25000 + ₹ 7500 = ₹ 32500

Question 25.
What sum will yield an interest of ₹ 6000 at 9% per annum in 3 years 4 months?
Answer:
S.I = ₹ 6000 ; R = 9%

Question 26.
At what rate per annum will ₹ 70000 yield an interest of ₹ 14000 in 2\(\frac{1}{2}\) years?
Answer:
Principal = ₹ 70000 ; Time = 2\(\frac{1}{2}\)years = \(\frac{5}{2}\) years ; Simple interest = ₹ 14000

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Simple Equations Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Simple Equations Unit Exercise

Question 1.
Runs made by two bats men in 3 matches are given below.
Kohli: 49, 98, 72
Rohit: 64, 45, 83, then find average of runs scored by Kohli and Rohit . Whose average is higher ?
Answer:
Given runs made by Kohli: 49, 98, 72.
Average runs of Kohli = \(\frac{\text { Sum of the observations }}{\text { Number of observations }}\) = \(\frac{49+98+72}{3}\) = \(\frac{219}{3}\) = 73
Runs made by Rohit: 64, 45, 83.
Average runs of = \(\frac{64+45+83}{3}\)
= \(\frac{192}{3}\)= 64
73 > 64.
Kohli average is higher than Rohit.

Question 2.
Find mode of 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 36 and 47. Verify whether it is Unimodal or Bimodal data.
Answer:
Given data : 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47.
Arrange the given observations in ascending order.

As 38 and 43 occurs most frequently than other observations in the data.
∴ Mode = 38 and 43.
So, given data is Bimodal data.

Question 3.
The temperature in different places are 0, – 5, 7, 10, 13, – 1 and 41 in degree Celsius. Find the Median. If another observation, ‘4°C’ is added to the given data, is there any change in value of Median ? Explain.
Answer:
Given data : 0, – 5′, 7, 10, 13, – 1, 41.
Arrange the given observations in ascending order.

In seven observations the fourth observation 7 is the middle most value.
∴ Median = 7 .
If 4°C is added to the given data,
0, – 5, 7, 10, 13, – 1, 41 and 4.
Arrange the given observations in ascending order.

In eight observations the fourth and fifth observations 4 and 7 are the middle most values in the data.
Median = Average of the two middle most values.
= \(\frac{4+7}{2}\) = \(\frac{11}{2}\) = 5.5
So, if we added a new observation, then the median also changed (decreased). Median is decreased from 7 to 5.5.

Question 4.
If the range of observation 7x, 5x, 3x, 2x, x (x > 0) is 12, then find value of ‘x’ and express all the observations in numerical form.
Answer:
Given data.: 7x, 5x, 3x, 2x, x (x > 0) and Range = 12.
Maximum value = 7x; Minimum value = x
Range = Maximum value – Minimum value
= 7x – x – 12
⇒ 6x = 12 ⇒ \(\frac{6 x}{6}\) = \(\frac{12}{6}\)
∴ x = 12

Question 5.
Birth and death rates of different states in 2015 are given below. Approximately draw a double bar graph for the given data.

Answer:

Question 6.
The following data relates to the cost of construction of a house.

Draw a Pie diagram to represent the above data.
Answer:
The angle of each sector will depend on the ratio between the number of students each item and total expenditure (%).
Angle of sector = \(\frac{\text { Value of the item }}{\text { Sum of the value of all items }}\) × 360°

Steps of construction:

1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct ∠AOB = 108° to represent angle of the sector of Cement.
4. Construct ∠BOC = 36° to represent angle of the sector of Steel.
5. Construct ∠COD = 36° to represent angle of the sector of Bricks.
6. Construct ∠DOE = 54° to represent angle of the sector of Timber.
7. Construct ∠EOF = 72° to represent angle of the sector of Labour.
8. Now ∠FOA = 54° to represent angle of the sector of Miscellaneous.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Unit Exercise

Question 1.
If the cost of 7 toys is ₹ 1575, then what would be the cost of 6 such toys?
Answer:
We know that as number of toys decreases, the cost also decreases such that the ratio of number of toys and the ratio of their costs will remain the same. That means here number of toys and the cost are in direct proportion.
Let the cost of 6 toys be x.
Then 7 : 1575 = 6 : x
If the ratios are equal, the product of extremes = Product of means
⇒ 7 × x = 1575 x 6

⇒ x = ₹ 1350
∴ Cost of 6 toys is ₹ 1350.

Question 2.
A boy went to a hotel and wants to buy 5 plates of Idly worth ₹ 24 each. After, going to Hotel he observed that rates of Idly were increased to ₹ 30. Now if the boy wants to buy idlies, how many plates of idlies he can buy with the same amount?
Answer:
If the cost of idly increases number of plates decreases.
So, cost of idly and number of plates are in inverse proportion.
Let number of plates at ₹ 30 for the same amount is x.

Cost of each Idly Plate	Number of Plates
24	5
30	x

By taking inverse proportion, 24 : 30 = x : 5
Then, Product of means = Product of extremes

Number of Plates at ₹ 30 for the same amount is 4.

Question 3.
Raju covers a distance of 28 kilometers in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Answer:
We know that as number of kilometers (distance) increases the time taken is also increases with the same speed. That means distance and the time taken are in direct proportion.

Let the time taken to cover the 56 km distance is x hours.
Then 28 : 2 = 56 : x

If the ratios are equal, the product of extremes = the product of means

The time taken to cover the 56 km distance is 4 hours.

Question 4.
24 men working at 8 hours per day can do a piece of work in 15 days. In how many days 20 men working 9 hours per day do the same work?
Answer:
Here, we have three quantities number of workers, number of hours and number of days.

Number of Men	Hours	Number of Days
24	8	15
20	9	x

Here the number of worker’s inversely proportional to number of hours perday.
Here the number of workers inversely proportional to number of days.

Let the number of days be ‘x’.
Since number of workers depends on both number of hours and number of days, we will take compound ratio of
24 : 20 and 8 : 9 is 24 × 8 : 20 × 9

Since the ratios are equal,
The product of means = The product of extemes

∴ Number of days required to complete 20 men at 9 hours per day is 16.

Question 5.
Out of 15000 voters in a constituency, 60% of the voters voted. Find the number of people not voted in the constituency.
Answer:
Given number of voters in the constituency = 15000 Percentage of voters voted = 60 %
Number of people voted = 60% of 15000

Number of people voted in the constituency = 9000
Number of people not voted = 15000 – 9000
∴ Number of people not voted = 6000

Question 6.
A shopkeeper bought a suitcase for ₹ 950 and sold it for ₹ 1200. Find its profit or loss percentage.
Answer:
Given cost price of suitcase = ₹ 950
Selling price of suitcase = ₹ 1200
S.P > C.P
So, shopkeeper got profit.
Profit = S.P – C.P = 1200 – 950 = ₹ 250

∴ Profit percent = 26\(\frac{6}{19}\)%.

Question 7.
On selling a mobile for ₹ 4500, a shop-keeper losses 10%. For what amount should be sell it to gain of 15%?
Answer:
Method 1 :
Given selling of mobile = ₹ 4500
Loss = 10%

S.P = C.P – Loss

∴ x = 5000

Method 2:
Let the C.P of mobile = 100%
S.P at a loss of 10% = (100 – 10)%
= ₹ 4500
90% of C.P = ₹ 450O
S.P at a gain of 15% = (100 + 15)%.
= 115%
So, if 90% of CP = ₹ 4500
115% of CP = ?
s.p= \(\frac{115}{90}\) × 4500= ₹ 5750
∴ Cost price of mobile = ₹ 5000
If he gain 15%.

Gain = 15% of 5000

S.P = C.P + Gain = 5000 + 750
∴ Selling price of Mobile = ₹ 5750

Question 8.
A carpenter allows 15% discount on his goods. Find the marked. price of a chair which is sold by him for ₹ 680.
Answer:
Given selling price = ₹ 680
Discount percent = 15%
Let marked price = x

Note:
85% of M.P = ₹ 680
100% of M.P = \(\frac{100}{85}\) × 680
= ₹ 800.

Question 9.
What Is the simple Interest accrued on asum of ₹ 75000 at the rate of 11% for 3 years? Find the total amount.
Answer:
Given Principle = ₹ 7500
Rate of Interest = 11%
Time = 3 years

Total amount = Principle + Interest = ₹ 75000 + ₹ 24750
∴ Total amount = ₹ 99,150

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.4

Question 1.
In the adjacent Double bar graph, number of students of a class are shown according to academic year.
Answer the following questions on the basis of this Double bar graph,

(i) In which academic year, number of girls wrere more than number of boys in the school ?
Answer:
In the year 2016-17 number of girls were more than number of boys in the school.

(ii) In which academic year, number of both girls and boys in the school were equal ?
Answer:
In the year 2015-16 number of bot.fi’ girls and boys in the school were equal.

(iii) What was the total number of students in the school in the academic year 2013-14 ?
Answer:
In the academic year 2013-14
Number of boys = 70
Number of girls = 50 .

Total = 120
Total number of students in the school in the academic year 2013-14 is 120.

Question 2.
The marks in Mathematics and Science of five students of class 7 are given in the table. Exhibit these by the vertical double bar graph by given information.

Answer:

Question 3.
The adjacent Pie chart gives the expenditure on various items during a month for a family. In the figure angles made by each sector at the center are given then answer the following : i) If the expenditure on rent is ?3000, then how much amount they saved ?

Answer:
Angle of sector on Rent = 90°
Angle of sector on Savings = 90°
Angle of sectors are equal. So, their amounts are also equal. Expenditure on rent is ₹3000.

(ii) On which item the expenditure is minimum ?
Answer:
Smallest angle of sector on Education is the item of expenditure is minimum.
So, expenditure on education is minimum.

(iii) On which item the expenditure is maximum ?
Answer:
Largest angle of sector on Food is the item of expenditure is maximum.
So, expenditure on food is maximum.

Question 4.
The following data shows the number of students opting different subjects in college.

Construct a Pie diagram to represent the above data.
Answer:
The angle of each sector will depend on the ratio between the number of students each subject and total number of students.
Angle of sector = \(\frac{\text { Value of the item }}{\text { Sum of the values of all items }}\) × 360°

Steps of construction:

1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct ∠AOB = 90° to represent angle of the sector of Botany.
4. Construct ∠BOC = 120° to represent angle of the sector of Mathematics.
5. Construct ∠COD = 40° to represent angle of the sector of Physics.
6. Construct ∠DOE = 60° to represent angle of the sector of Chemistry.
7. Construct ∠EOF = 20° to represent angle of the sector of Economics.
8. Now ∠FOA = 30° to represent angle of the sector of Commerce.

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.3

Question 1.
Find the Median of the following:
(i) 7, 3, 15, 0, 1, 71, 19, 4, 17.
Answer:
Given data : 7, 3, 15. 0, 1, 71, 19, 4, 17.
Arrange the given observations in ascending order.

In nine observations the fifth observation 7 is the middle most value.
∴ Median = 7

(ii) 12, 23, 11, 18, 15, 20, 86, 27.
Answer:
Given data : 12, 23, 11, 18, 15, 20, 86, 27.
Arrange the given observations in ascending order:

In eight observations the 4th and 5th observations are 18 and 20.
Here, we have two middle most values 18 and 20.
Median = Average of the two middle most values.
= \(\frac{18+20}{2}\) = \(\frac{38}{2}\) = 19
∴ Median of the data = 19.

Question 2.
The number of pages in text books of different subjects are 421, 175, 128, 117, 150, 145, 147 and 113 find median of given data.
Answer:
Given data : 421, 175, 128, 117, 150, 145, 147, 113
Arrange the given observations in ascending order

In eight observations the 4th and 5th observations are 145 and 147.
Here, we have two middle most values 145 and 147.
Median = Average of the two middle most values
= \(\frac{145+147}{2}\) = \(\frac{292}{2}\) = 146
∴ Median of the data = 146.

Question 3.
The weekly sales of motor bikes in a showroom for the past 14 weeks are 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4 and 7. Find the Median of the data.
Answer:
Given data: 10, 6, 8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4, 7
Arrange the given observations in ascending order.

In fourteen observations the 7th and 8th observations are 7 and 7.
Here we have two middle most values 7 and 7.
Median = Average of the two middle most values
= \(\frac{7+7}{2}\) = \(\frac{14}{2}\) = 7
∴ Median of the data = 7

Question 4.
Find the Median of 0.3, 0.25, 0.32, 0.147, 0.19, 0.2 and 7.1.
Answer:
Given data: 0.3, 0.25, 0.32, 0.147, 0.19, 0.2, 7.1.
Arrange the given observations in ascending order.

In seven observations the 4th observation 0.25 is the middle most value.
∴ Median. = 0.25

Question 5.
If the Median of observations 2x, 3x, 4x, 5x, 6x, (x > 0) is 28, then find the value of ‘x’.
Answer:
Given data : 2x, 3x, 4x, 5x, 6x and median is 28.
Data is arranged in the ascending order

In five observations the third observation 4x is the middle most value. .
∴ Median – 4x = 28
⇒ \(\frac{4 x}{4}\) = \(\frac{28}{4}\) = 7
∴ x = 7