Category: Class 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.2

Question 1.
Find mode of the following data.
(i) 2, 3, 7, 5, 3, 2, 6, 7, 1,2.
Answer:
Given data : 2, 3, 7, 5, 3, 2, 6, 7, 1,2.
By arranging the numbers with same values together
1, 2, 2, 2, 3, 3, 5, 6, 7, 7.
As 2 occurs more frequently than other observations in the data.
∴ Mode = 2

(ii) K, A, B, C, B, C, D, K, B, D, B, K, A, K.
Answer:
Given data : K, A, B, C, B, C, D, K, B, D, B, K, A, K
By arranging the letters in the alphabetical order of same type together.
A, A, B, B, B, B, C, C, D, D, K, K, K, K.
As B and K occurs most frequently than other observations in the data.
∴ Mode = B and K.

(iii) First ten natural numbers.
Answer:
First 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
In the given observations there is no repeated number.
So, the given data has no mode.

(iv) 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8.
Answer:
Given data : 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8.
In the given observations, data is repeated an equal number of times. .
So, the given data has no mode.

Question 2.
20 students were participated in ‘SWATCH BHARAT ABHIYAN’ campaign. The number of days each student participated were 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4 and 5. Find mode of the data.
Answer:
Given data 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4, 5.
By arranging the numbers with same value together
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4; 4, 5, 5, 5.
As 2 occurs more frequently than other observations in the data.
∴ Mode = 2.

Question 3.
The number of goals scored by a 3, 2, 4, 6, 1, 3, 2, 4, 1 and 6. Find the mode of data.
Answer:
Given data: 3, 2, 4, 6, 1, 3, 2, 4, 1, 6.
By arranging the numbers with same values together.
1, 1, 2, 2, 3, 3, 4, 4, 6, 6.
In the given observation, data is repeated an equal number of times.
So, the given data has no mode.

Question 4.
Find the mode of letters in the adjacent figure. Verify whether it is Unimodal or Bimodal Data.

Answer:
In the figure data is: S, A, H, S, A, M, S, T, M, T, H, % A, T, S, M, H, M, A, S, T, M, A, T, S, T, H, M.
By arranging the letters, of same type together.
A, A, A, A, A, H, H, H, H, M, M, M, M/M, M, S, S, S, S, S, S, T, T, T, T, T, T, T.
As T occurs most frequently in the data.
∴ Mode = T
Data having only one mode is known as unimodal data.
So, given data is unimodal data.

SCERT AP 7th Class Maths Solutions Pdf Chapter 6 Data Handling Ex 6.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Data Handling Ex 6.1

Question 1.
Find Arithmetic Mean of the following.
(i) 4, 5, 11, 8
Answer:
Given data: 4,5, 11, 8.
Sum of observations = 4 + 5 + 11 + 8 = 28
Number of observations = 4
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{28}{4}\) = 7.

(ii) 10, 15, 21, 12, 17
Answer:
Given data: 10, 15, 21, 12, 17
Sum of observations 10 + 15 + 21 + 12 + 17 = 75
Number of observations = 5
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{75}{5}\) = 15

(iii) \(\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \frac{3}{2}, \frac{5}{4}\)
Answer:
Given data: \(\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \frac{3}{2}, \frac{5}{4}\)
Sum of observations = \(\frac{1}{4}+\frac{1}{2}+\frac{3}{4}+\frac{3}{2}+\frac{5}{4}\) = \(\frac{1+2+3+6+5}{4}\) = \(\frac{17}{4}\)
Number of observations = 5
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{\frac{17}{4}}{5}=\frac{17}{4} \div \frac{5}{1}=\frac{17}{4} \times \frac{1}{5}=\frac{17}{20}\)
∴ Arithmetic Mean = \(\frac{17}{20}\)

Question 2.
Amounts donated by eight students to ‘NIVAR’ cyclone effected people are ₹300, ₹450, ₹700, ₹650, ₹400, ₹750, ₹900 and ₹850. Find the Arithmetic Mean of amounts donated.
Answer:
Given data, ₹300, ₹450, ₹700, ₹650, ₹400, ₹750, ₹900 and ₹850.
Sum of observations = 300 + 450 + 700 +650 +400 + 750 + 900 + 850 = 5000
Number of observations = 8
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{5000}{8}\) = ₹625
∴ Arithmetic Mean of amount donated = ₹625

Question 3.
The number of passengers who travelled in APSRTC bus from Eluru to Rangapuram in 5 trips in a day are 35, 42, 28, 41 and 44. What is the average of number of passengers travelled per trip ?

Answer:
Given data are 35, 42, 28, 41, 44.
Sum of observations = 35 + 42 + 28 + 41 + 44 = 190
Number of observations = 5
∴ Arithmetic Mean (or) Average = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{190}{5}\) = 38
∴Average of number of passengers travelled per trip = 38.

Question 4.
Find Arithmetic mean of factors of 24.
Answer:
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Sum of (factors) observations = 1 + 2 + 3 + 4 + 6 + 8+ 12 + 24 = 60
Number of observations = 8
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = \(\frac{60}{8}\) = 7.5
∴ Arithmetic Mean of factors of 24 is 7.5

Question 5.
Find the Arithmetic Mean of x, x + 1 and x + 2.
Answer:
Given data x, x + 1, x + 2.
Sum of observations = x + x + 1 + x + 2 = 3x + 3
Number of observations = 3
∴ Arithmetic Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{3 x+3}{3}\) = \(\frac{3(x+1)}{3}\)
∴ Arithmetic mean = x + 1

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.4

Question 1.
Find the height of the statue from the adjacent diagram.

Answer:
From the given diagram
Height of the statue (BC) = x m
Height of the Pedastal (AB) = 1.9 m
Total Height (AC) = 3.6 m
BC + AB = 3.6 m
⇒ x + 1.9 m = 3.6 m
⇒ x + 1.9 – 1.9 = 3.6 – 1.9 (Subtract 1.9 on both sides)
⇒ x = 1.7 m
∴ Height of the statue = 1.7 m

Question 2.
The sum of twice a number and 4 is 80, find the number.
Answer:
Let the number be x.
Twice the number = 2x
Sum of twice a number and 4 = 80
2x + 4 = 80
⇒ 2x + 4 – 4 = 80 – 4
(Subtract 4 on both sides)
⇒ 2x = 76
⇒ \(\frac{2 x}{2}\) = \(\frac{76}{2}\) (Divide by 2 on both sides)
⇒ x = 38
∴ Number = 38

Question 3.
The difference between a number and one-fourth of itself is 24, find the number.
Answer:
Let the number be a.
One-fourth of a number = \(\frac{1}{4}\) of a = \(\frac{a}{4}\)
Difference of number and one-fourth of itself = 24

⇒ a = 32
∴ The number a = 32

Question 4.

Find the value of x from the adjacent diagram.
Answer:

From the above diagram
AB = 12 cm; CD = x cm
EF = 5 cm; PQ = 24 cm
We know,
AB + CD + EF = PQ
⇒ 12 + x + 5 = 24
⇒ x + 17 = 24
⇒ x + 17 – 17 = 24 – 17
(Subtract 17 on both sides)
⇒ x = 7 cm

Question 5.
To convert temperature from Fahrenheit to Centigrade, we use the formula (F – 32) = \(\frac{9}{5}\) × C. If C = – 40° C, then find F.
Answer:
Given formula (F – 32) = \(\frac{9}{5}\) × C and C = – 40° C

⇒ F – 32 = 9(- 8)
⇒ F – 32 = – 72
⇒ F – 32 + 32 = – 72 + 32 (Add 32 on both sides)
⇒ F = – 40° F
Therefore we conclude that
– 40° C = – 40° F.

Question 6.
Rahim has ₹x, from which he spends ₹6. If twice of the money left with him is ₹86, then find x.
Answer:
Rahim has money = ₹x
Money after spends ₹6 = x – 6
Twice the money left with Rahim
= 2(x – 6) = ₹86
⇒ – 2(x – 6) = 86
⇒ \(\frac{2(x-6)}{2}\) = \(\frac{86}{2}\) (Divide by 2 on both sides)
⇒ x – 6 = 43
⇒ x – 6 + 6 = 43 + 6 (Add 6 on both sides)
⇒ x = 49
Therefore money at Rahim = ₹49.

Question 7.
The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.
Answer:
Let the smaller number is x.
Larger number = x + 7
Six times the smaller = 6.x
Six times smaller plus the larger = 77
⇒ 6x + x + 7 = 77
⇒ 7x + 7 = 77

⇒ 7x = 70
⇒ \(\frac{7 \mathrm{x}}{7}\) = \(\frac{70}{7}\) (Divide by 7 on both sides) ⇒x = 10
∴ Smaller number x = 10
Larger number = x + 7 = 10 + 7 = 17
∴ The numbers are 10 and 17.

Question 8.
The sum of three consecutive even numbers is 54. Find the numbers.
Answer:
Let the three consecutive even numbers are 2x, 2(x +1), 2(x + 2).
Sum of three consecutive even numbers = 54
⇒ 2x + 2(x + 1) + 2(x + 2) = 54
⇒ 2x + 2x + 2 + 2x + 4 = 54
⇒ 6x + 6 = 54

⇒ 6x = 48
⇒ \(\frac{6 x}{6}\) = \(\frac{48}{6}\) (Divide by 6 on both sides)
⇒ x = 8
Numbers are, 2x, 2(x + 1), 2(x + 2)
2(8), 2(8 + 1), 2(8 + 2)
∴ The numbers are 16, 18, 20.

Question 9.
In a class of 48 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.
Answer:
Let the number of boys = x
The number of girls = \(\frac{1}{3}\) of boys
= \(\frac{1}{3}\) of x = \(\frac{x}{3}\)
Boys + Girls = Total students
⇒ \(\frac{x}{1}+\frac{x}{3}\) = 48
⇒ \(\frac{3 x+x}{3}\) = 48
⇒ \(\frac{4 x}{3}\) = 48
⇒ \(\frac{4 x}{3}\) × 3 = 48 × 3
(Multiply by 3 on both sides)
⇒ 4x = 144
⇒ \(\frac{4 x}{4}\) = \(\frac{144}{4}\) (Divide by 4 on both sides)
⇒ x = 36
∴ Number of boys = 36
Number of girls = \(\frac{x}{3}\) = \(\frac{36}{3}\) = 12
Boys = 36 and girls = 12

Question 10.
The present ages of Mary and Joseph are in the ratio 5 : 3. After 3 years sum of their ages will be 38. Find their present ages.
Answer:
Ratio of present ages of Mary and Joseph = 5:3 = 5x : 3x
After 3 years their ages
= 5x + 3 : 3x + 3
Sum of their ages = 38
⇒ 5x + 3 + 3x + 3 = 38
⇒ 8x + 6 = 38
⇒ 8x + 6 – 6 = 38 – 6 (Subtract 6 on both sides)
⇒ 8x = 32
⇒ \(\frac{8 x}{8}\) = \(\frac{32}{8}\)
⇒ x = 4
⇒ Ratio of present ages = 5x : 3x
= 5(4) : 3(4) = 20 : 12
Present ages of Mary and Joseph are 20 years and 12 years.

Question 11.
A sum of ₹500 is in the form of notes of denominations of ₹5 and ₹10. If the total number of notes is 90, then find the number of notes of each type.

Answer:
Let, number of f5 notes = x
Given sum of number of ₹5 notes and ₹10 notes is 90.
So, number of ₹10 notes = 90 – x
Sum of ₹5 notes = ₹5 × x = 5x
Sum of ₹ 10 notes = ₹ 10 × (90 – x)
= 900 – 10x
⇒ 5x + 900 – 10x = 500
⇒ – 5x + 900 = 500
⇒ – 5x + 900 – 900 = 500 – 900 (Subtract 900 on both sides)
⇒ – 5x = – 400
⇒ \(\frac{-5 x}{-5}\) = \(\frac{-400}{-5}\)
(Divide by – 5 on both sides)
⇒ x = 80
∴ Number of ₹5 notes x = 80
Number of ₹10 notes
= 90 – x = 90 – 80 = 10.

Question 12.
John and Ismail donated some money to the Relief Fund. The amount paid by Ismail is ₹85 more than twice that of John. If the total money paid by them is ₹4000, then find the amount of money donated by John.
Answer:
Let the amount donated by John = ₹x
The amount donated by Ismail is ₹85 more than twice of John.
So, the amount donated by Ismail = ₹(2x + 85)
Total amount = John amount + Ismail amount = ₹4000
⇒ x + 2x + 85 = 4000
⇒ 3x + 85 = 4000
⇒ 3x + 85 – 85 = 4000 – 85 (Subtract 85 on both sides)
⇒ 3x = 3915
⇒ \(\frac{3 x}{3}\) = \(\frac{3915}{3}\) (Divide by 3 on both sides)
⇒ x = 1305
∴ Money donated by John = x = ₹ 1305
∴ Money donated by Ismail = 2x + 85
= 2(1305) + 85
= ₹2695

Question 13.
Length of a rectangle is 4 m less than 3 times its breadth. If the perimeter of rectangle is 32 m. Then find its length and breadth.
Answer:
Let breadth of a rectangle (b) = a m
Length of rectangle (l) = 4 m less than 3 times of breadth
= (3 ∙ a – 4) m
Given perimeter = 2(l + b) = 32 m
⇒ 2[3a – 4 + a] = 32
⇒ 2[4a – 4] = 32
⇒ (8a – 8) = 32 (Distributive property)
⇒ 8a – 8 + 8 = 32 + 8 (Add 8 on both sides)
⇒ 8a = 40
⇒ \(\frac{8 \mathrm{a}}{8}\) = \(\frac{40}{8}\) (Divide by 8 on both sides) o o
⇒ a = 5
∴ breadth (b) = 5 m
length (l) = 3a – 4
= 3(5) – 4 = 15 – 4 = 11 m
∴ length = 11 m and breadth = 5 m.

Question 14.
A bag contains some number of white balls, twice the number of white balls are blue balls, thrice the number of blue balls are the red balls. If the total number of balls in the bag are 27. Then calculate the number of balls of each colour present in the bag.
Answer:
Let the number of white balls = x
Number of blue balls = twice the number of white balls
= 2(x) = 2x
Number of red balls
= thrice the number of blue balls = 3(2x) = 6x
Total balls = White + Blue + Red = 27
⇒ x + 2x + 6x = 27
⇒ 9x = 27
⇒ \(\frac{9 x}{9}\) = \(\frac{27}{9}\) (Divide by 9 on both sides) 9 9 .
⇒ x = 3
Therefore white balls x = 3
Blue balls = 2x = 2(3) = 6
Red balls = 6x = 6(3) = 18

Question 15.

Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.3

Question 1.
Solve the following equations and check the result.
(i) 5x – 17 = 18
Answer:
Given 5x – 17 = 18
⇒ 5x – 17 + 17 = 18 + 17 (Add 17 on both sides)
⇒ 5x = 35
⇒ \(\frac{5 x}{5}\) = \(\frac{35}{5}\) (Divide by 5 on both sides)
⇒ x = 7

Check:
Substitute x = 7 in 5x – 17 = 18
LHS = 5x – 17
= 5(7) – 17
= 35 – 17 = 18 = RHS
Hence verified.

(ii) 29 – 7y = 1
Answer:
Given 29 – 7y = 1
⇒ 29 – 7y – 29 = 1 – 29 (Subtract 29 on both sides)
⇒ – 7y = -28 .
⇒ \(\frac{-7 y}{-7}\) = \(\frac{-28}{-7}\) (Divide by – 7 on both sides)
⇒ y = 4

Check:
Substitute y = 4 in
29 – 7y = 1
LHS = 29 – 7y
= 29 – 7(4)
= 29 – 28 = 1 = RHS
Hence verified.

(iii) a – 2.3 = 1.5
Answer:
Given a – 2.3 = 1.5
⇒ a – 2.3 + 2.3 = 1.5 + 2.3 (Add 2.3 ort both sides)
⇒ a = 3.8

Check:
Substitute a = 3.8 in a – 2.3 =1.5
LHS = a – 2.3
= 3.8 – 2.3
= 1.5 = RHS
Hence verified.

(iv) b + 3\(\frac{1}{2}\) = \(\frac{7}{4}\)
Answer:

(v) \(\frac{7 p}{10}\) + 9 = 15
Answer:

(vi) 6(q – 5) = 42.
Answer:
Given 6(q – 5) = 42
⇒ \(\frac{6(q-5)}{6}\) = \(\frac{42}{6}\) (Divide by 6 on both sides)
⇒ q – 5 = 7
⇒ q – 5 + 5 = 7 + 5 (Add 5 on both sides)
⇒ q = 12

Check:
Substitute q = 12 in
6(q – 5) = 42
LHS = 6(q – 5)
= 6(12 – 5)
= 6(7) = 42 = RHS
Hence verified.

(vii) – 3(m + 5) + 1 = 13
Answer:
Given – 3(m + 5)4-1 = 13
⇒ – 3(m + 5)+ 1 – 1 = 13 – 1 (Subtract 1 on both sides)
⇒ – 3(m + 5) = 12
⇒ \(\frac{-3(m+5)}{-3}\) = \(\frac{12}{-3}\) (Divide by – 3 on both sides)
⇒ m + 5 = – 4
⇒ m + 5 – 5 = – 4 – 5 (Subtract 5 on both sides)
⇒ m = – 9

Check:
Substitute m = – 9 in – 3(m + 5) + 1 = 13
LHS = – 3(m + 5) + 1
= – 3(- 9 + 5) + 1 = (- 3 × – 4) + 1
= + 12 + 1 = 13 = RHS
Hence verified.

(viii) \(\frac{n}{2}+\frac{n}{3}+\frac{n}{5}\) = 31
Answer:

Question 2.
Solve the following equations and check the result.

(i) 3(p – 7) – 4 = 5
Answer:
Given 3(p – 7) – 4 = 5
⇒ 3(p – 7) – 4 + 4 = 5 + 4 (Add 4 on both sides)
⇒ 3(p – 7) = 9
⇒ \(\frac{3(p-7)}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)
⇒ p – 7 = 3
⇒ p – 7 + 7 = 3 + 7 (Add 7 on both sides)
⇒ p = 10

Check:
Substitute p = 10 in
3(p – 7) – 4 = 5
LHS = 3(p – 7) – 4.
= 3(10 – 7) – 4.
= 3 × 3 – 4
= 9 – 4 = 5 = RHS
Hence verified.

(ii) 5(q – 3) – 3(q – 2) = 0
Answer:
Given 5(q – 3) – 3(q – 2) = 0
⇒ 5q – 15 – 3q + 6 = 0 (Distributive property)
⇒ 2q – 9 = 0 .
⇒ 2q – 9 + 9 = 0 + 9 (Add 9 on both sides)
⇒ 2q = 9
⇒ \(\frac{2 \mathrm{q}}{2}\) = \(\frac{9}{2}\) (Divide by 2 on both sides)
⇒ q = \(\frac{9}{2}\)

Check : Substitute q = \(\frac{9}{2}\) in
5(q – 3) – 3(q – 2) = 0
LHS = 5(q – 3) – 3(q – 2)
= 5\(\left(\frac{9}{2}-3\right)\) – 3\(\left(\frac{9}{2}-2\right)\)
= 5\(\left(\frac{9-6}{2}\right)\) – 3\(\left(\frac{9-4}{2}\right)\)
= 5 × \(\frac{3}{2}\) – 3 × \(\frac{5}{2}\)
= \(\frac{15}{2}-\frac{15}{2}\) = 0 = RHS
Hence verified.

(iii) 4x – 0.3x – 1.2 = 0.6
Answer:
Given 0.4x – 0.3x – 1.2 = 0.6
⇒ 0.1 x – 1.2 = 0.6
⇒ 0.1 x – 1.2 + 1.2 = 0.6 + 1.2 (Add 1.2 on both sides)
⇒ 0.1 x = 1.8
⇒ \(\frac{1 \mathrm{x}}{10}\) = \(\frac{18}{10}\)

(Multiply by 10 on both sides)
⇒ x = 18

Check:
Substitute x = 18 in
0.4x – 0.3x – 1.2 = 0.6
LHS = 0.4x – 0.3x – 1.2
= 0.4(18) – 0.3(18) – 1.2
= 7.2 – 5.4 – 1.2
= 7.2 – 6.6 = 0.6 = RHS
Hence verified.

(iv) 4(3y + 4) = 7.6
Answer:
Given 4(3y + 4) = 7.6.
⇒ \(\frac{4(3 y+4)}{4}\) = \(\frac{7.6}{4}\) (Divide by 4 on both sides)
⇒ 3y + 4 = 1.9
⇒3y + 4 – 4= 1.9 – 4 (Subtract 4 on both sides)
⇒ 3y = – 2.1
⇒ \(\frac{3 y}{3}\) = – \(\frac{2.1}{3}\) (Divide by 3 on both sides)
⇒ y = – 0.7

Check:
Substitute y = – 0.7 in
4(3y + 4) = 7.6
LHS = 4(3y + 4)
= 4[3 (- 0.7) + 4]
= 4[- 2.1 + 4]
= 4 × 1.9 = 7.6 = RHS
Hence verified.

(v) 20 – (2r – 5) = 25
Answer:
Given 20 – (2r – 5) = 25

(Subtract 20 on both sides)
⇒ – (2r – 5) = 5
⇒ – (2r -5) × – 1 = 5 × – 1
(Multiply by – 1 on both sides)
⇒ 2r – 5 = 5
⇒ 2r – 5 + 5= – 5 + 5 (Add 5 on both sides)
⇒ 2r = 0
⇒ \(\frac{2 \mathrm{r}}{2}\) = \(\frac{0}{2}\) (Divide by 2 on both sides)
⇒ r = 0

Check:
Substitute r = 0 in
20 – (2r – 5) = 25
LHS = 20 – (2r – 5)
= 20 – [2(0) – 5]
= 20 – [0 – 5]
= 20 – (- 5)
= 20 + 5 = 25 = RHS
Hence verified.

(vi) 3(5 – t) – 2(t – 2) = – 1
Answer:
Given 3(5 – t) – 2(t – 2) = – 1
⇒ 15 – 3t – 2t + 4 .= – 1 (Distributive property)
⇒ 19 – 5t = – 1

(Subtract 19 on both sides)
⇒ – 5t = – 20
⇒ \(\frac{-5 t}{-5}\) = \(\frac{-20}{-5}\) (Divide by – 5 on both sides)
⇒ t = 4

Check:
Substitute t = 4 in
3(5 -1) – 2(t – 2) = – 1
LHS = 3(5 – t) – 2(t – 2)
= 3(5 – 4) – 2(4 – 2)
= 3(1) – 2(2)
= 3 – 4 = – 1 = RHS
Hence verified.

(vii) 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
Answer:
Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16
⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property)
⇒ 14k – 12 = 16
⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides)
⇒ 14k = 28
⇒ \(\frac{14 \mathrm{k}}{14}\) = \(\frac{28}{14}\) (Divide by 14 on both sides)
⇒ k = 2

Check:
Substitute k = 2 in
3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k)
= 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2]
= 3[4 + 1] – 2(- 3) – 5(5 – 4)
= 15 + 6 – 5 – 16 = RHS
Hence verified.

(viii) \(\frac{3 \mathrm{~m}}{4}\) – 5m – \(\frac{3}{4}\)
Answer:
⇒ – 17m – 3 = 48
⇒ – 17m – 3 + 3 =48 + 3 (Add 3 on both sides)
⇒ – 17m = 51
⇒ \(\frac{-17 \mathrm{~m}}{-17}\) = \(\frac{51}{-17}\) (Divide by – 17 on both sides)
⇒ m = – 3

(ix) \(\frac{4 n}{5}+\frac{n}{4}-\frac{n}{2}=\frac{11}{10}\)
Answer:

(x) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)
Answer:

Question 3.
Write any three equivalent equations having the solution x = 2.
Answer:

Question 4.
Write any three equivalent equations having the solution a = – 5.
Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.7 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.7

Question 1.
Calculate the simple interest accrued on a sum oR 12600, at the rate of 9% per annum for 2 years.
Answer:
Given principle = ₹ 12600
Time = 2 years
Rate of Interest = 9%

∴ Simple Interest = ₹ 2268

Question 2.
Calculate the simple interest accrued for 3 years and the total amount on a sum of ₹ 85000, at the rate of 11% per annum.
Answer:
Given Principle = ₹ 85000
Time = 3 years
Rate of Interest = 11%

Simple Interest = ₹ 28,050
Total amount = Principle + Interest
= 85000 + 28050
∴ Total amount = ₹ 1,13,050

Question 3.
In what time will ₹ 45000 amounts to ₹ 63000, If the simple Interest Is calculated 10% per annum?
Answer:
Method 1:
Given principle = ₹ 45000
Rate of Interest = 10%
Amount = ₹ 63000
Let the time = x years

Interest = ₹ 4500 x
Amount = Principle + Interest
= 45000 + 4500x = 63000
= 4500x = 63000 – 45000 = 18000
⇒ x = \(\frac{18000}{4500}\) = 4
∴ Time = 4 years

Method 2:
Given A = ₹ 63000;
P = ₹ 45000; R = 10%; T=?
So I = A – P = 63000 – 45000 = ₹ 18000

But I = \(\frac{\text { PTR }}{100}\)
⇒ 18000 = \(\frac{45000 \times T \times 10}{100}\)
∴ T = \(\frac{18000 \times 100}{45000 \times 10}\) = 4 years

Question 4.
On a certain amount at the rate of 12% per annum for time 3 years the toti1 interest becomes ₹ 18000. What is the principal amount?
Answer:
Given Rate of Interest = 12%
Time.= 3 years
Interest = ₹ 18000
Let Principle = ₹ x

∴ Principle = ₹ 50,000

Question 5.
In what time the simple interest accrued on a sum of?35000 at the rate of 13% per annum becomes?27300?
Answer:
Given Principle = ₹ 35000 ;
Rate of Interest = 13% ;
Time = x years
Interest = ₹ 27300 .

Time = 6 years

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.6 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.6

Question 1.
A shop selling sewing machines, offers 3% discount on purchases. If the marked price is ₹ 6500. then what is the selling price?
Answer:
Method 1:
Given the marked price = ₹ 6500
bis count = 3%
Disount calculated on marked price.
Discount = 3% of ₹ 6500

∴ Discount = ₹ 195

Selling price
= Marked price – Discount = 6500 – 195
∴ Selling price of sewing machine
= ₹ 6305

Method 2 :
M.P = 100% = ₹ 6500
S.P at a discount of 3%
= (100 – 3)% =?
So, if 100% = 6500
97% = ?
S.P = \(\frac{97}{100}\) × 6500 = ₹ 6305

Question 2.
The marked price of a ceiling fan is ₹ 720 during off season, it is sold for ₹ 684. Determine the discount percentage.

Answer:
Given marked price = ₹ 720
Selling price = ₹ 684
Discount = M.P – S.P = 720 – 684
= ₹ 36
Discount percent = \(\frac{\text { Discount }}{\mathrm{MP}}\) × 100

= 5%
∴ Discount percent = 5%

Question 3.
A publisher gives 32% discount on the printed price of books to book sellers. If the printed price is ₹ 275, then what amount does the seller has to pay to publisher?
Sol. Given printed (marked) price = ₹ 275
Discount percent = 32%

Selling price = M.P – Discount = 275 – 88
∴ Selling price of books = ₹ 187

Question 4.
Rohit buys an item at 25% discount on the marked price. If he bought it for ₹ 660, what is the marked price?
Answer:
Method 1 :
Given selling price = ₹ 660
Discount percentage = 25%
Let the marked price = ₹ x

Marked price – Selling price = Discount
x – 660 = \(\frac{x}{4}\)

∴ Marked price = ₹ 880

Method 2 : M.P = 100% (say)
S.P at a discount of 25% means
S.P = (100 – 25) = 75%
∴ If 75% = ₹ 660 then 100% =?
⇒ SP = \(\frac{100}{75}\) × 660 = ₹ 880

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.5 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.5

Question 1.
Rekha bought a wrist watch for ₹ 2250 and sold it for ₹ 1890. Then find her loss or gain percentage?

Answer:
Given cost price of wrist watch = ₹ 2250
Selling price of wrist watch = ₹ 1890
C.P > S.P, then Rekha got loss.
Loss = C.P – S.P = 2250 – 1890 = ₹ 360

We know loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100

∴ Loss percentage 16%.

Question 2.
A shopkeeper buys a toy for ₹ 250 and sells it for ₹ 300. Find his gain or loss percentage?

Answer:
Given cost price of a toy = ₹ 250
Selling price of a toy = ₹ 300
C.P < S.P, then Shopkeeper got profit.
Profit = S.P – C.P
= 300 – 250 = ₹ 50

We know profit percent

∴ Profit percentage = 20%

Question 3.
The cost price of a chair ₹ 480. 1f he sold at a profit of 10%. what would be selling price of it’?
Answer:
Method: 1
Given cost price of a chair = ₹ 480
Profit percent = 10%
Profit = 10% of ₹ 480

Selling price = C.P + Profit
= ₹ 480 + ₹ 48
∴ Selling price = ₹ 528

Method: 2
C.P = 100% = ₹ 480
Selling it 10% gain.
= S.P = 100% + 10% =?
z 100% = ₹ 480
110% =?
S.P = \(\frac{110}{100}\) × 480= ₹ 528
∴ Selling price = ₹ 528

Question 4.
If Sharma purchased a car for ₹ 350000. After two years he sold at a loss of 12%. Find its selling price’?
Answer:
Given cost price of a car = ₹ 350000
Loss = 12 %
Loss 12% of ₹ 350000

Loss = ₹ 42000
Selling price = CP – Loss
= 350000 – 42000
∴ Selling price of a car = ₹ 3,08,000

Question 5.
A shopkeeper buys wooden tables each at ₹ 2800 and expends ₹ 400 on each table for painting. If he sells it at a cost of ₹ 4000. Find his profit percentage?

Answer:
Given cost price of wooden table = ₹ 2800
Amount spent on painting = ₹ 400
Effective cost price of wooden table
= ₹ 2800 + ₹ 400 = ₹ 3200

Selling price wooden table = 4000
C.P < SP, then shopkeeper got profit.
Profit = S.P – C.P
= ₹ 4000 – ₹ 3200 = ₹ 800

We know profit percentage

∴ Profit percentage = 25%

Question 6.
In a garment shop a saree is sold at a cost of ₹ 1800 after a profit of ₹ 600. Find the profit percentage and cost price?
Answer:
Given selling price of a saree = ₹ 1800
Profit = ₹ 600
Cost price of a saree = S.P – Profit
= 1800 – 600
= ₹ 1200.
We know that profit percentage

∴ Profit percentage = 50%

Question 7.
After incurring a loss of ₹ 258, Jean pant was sold at ₹ 1750. Then find the cost price and loss percentage?
Answer:
Given selling price of Jean pant = ₹ 1750
Loss = ₹ 258
Cost price of Jean pant = S.P + Loss
= 1750 + 258
= ₹ 2008
Loss percent = \(\frac{\text { Loss }}{\text { C.P }}\) × 100
⇒ \(\frac{258}{2008}\) × 100
∴ Loss percentage = 12.85%

Question 8.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percentage?
Answer:
Method 1:
Let Cost price of 10 articles be ₹ x.
Cost price of 10 articles = 10 x
Selling price of 9 articles = 10 x
Cost price of 9 articles 9x
Profit = S.P — C.P = 10x – 9x
= ₹ x
Profit percentage = \(\frac{\text { Profit }}{\text { C.P }}\) × 100

∴ Profit percentage = \(\frac{100}{9}\)% (or)
11.11% (or) 11\(\frac{1}{9}\)%

Method 2:
By selling 9 articles, there is a gain of 1 article.
∴ Profit Percentage = \(\frac{1}{9}\) × 100
= 11\(\frac{1}{9}\)%

Question 9.
By selling a book for ₹ 258, a bookseller gains 20%. For how much should he sell it to gain 30%?
Answer:
Method 1:
Given selling priceof a book = ₹ 258
Gain = 20%
Cost price of book = ₹ x

∴ Cost price of a book = ₹ 215

If profit percent = 30%

Profit = ₹ 64.50
∴ S.P = C.P + Profit
= 215 + 64.50 = 279.50
To get 30% profit he has to sold at ₹ 279.50.

Method 2:
Let the C.P of the book = 100%
Selling at a gain of 20% means (100 + 20)% = ₹ 258
120% = ₹ 258
Selling at a gain of 30% means (100+30)%=?
If 120% = ₹ 258
then 130% =?
∴ SP = \(\frac{130}{120}\) × 258 = ₹ 279.5

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.4

Question 1.
If 5 men take 8 days to type 10 books, then how many days it takes for 8 men to type 2 books?
Answer:

Number of Men	Number of books typed	Number of Days
5	10	8
8	2	x

Here the 3 quantities are men, books typed and number of days.
Let the number days taken to type 2 books by 8 men be x.

From the above,

• Number of days and men are in inverse proportion.
• Number of days and typed books are in direct proportion.

Here number of days depends on both men and number of books typed.

So, we have to take compound ratio of 8 : 5 and 10:2.
8 : x = 8 × 10 : 5 × 2
Since the ratios are equal,
The product of means = The product of extremes
⇒ 8 : x = 80 : 10
⇒ x × 80 = 8 × 10
⇒ \(\frac{80 x}{80}=\frac{80}{80}\)
⇒ x= 1 day
Or
x = \(\frac{5}{8} \times \frac{2}{10}\) × 8 = 1

∴ Number of days taken to type 2 books by 8 men is 1.

Question 2.
If 5 men take 9 days to plough 18 acres of land, then find how long it takes 25 men for to plough 30 acres?

Answer:
Here we have three quantities number of men, jj number of days and acres of land ploughed.

Number of Men	Number of Land	Number of Days
5	18	9
25	30	X

Let the number of days taken to plough the 30 acres of land by 25 men is x.

From the above,

• Number of days and number of men are in increase proportion.
• Number of days and acres of land ploughed in direct proportion.

Here number of days depends on both men and number of acres of land ploughed.

So, we have to take compound ratio of 25 : 5 and 18 : 30 is
25 × 18 : 5 × 30
∴ 9 : x = 25 × 18 : 5 × 30
Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 25 × 18 = 9 × 5 × 30

Or
x = \(\frac{5}{25} \times \frac{30}{18}\) × 9 = 3 days
∴ Number of days taken to plough 30 acres of land by 25 men is 3 days.

Question 3.
Rice cost ₹ 480 is needed for 8 members for 20 days. What is the cost of rice required for 12 members for 15 days?
Answer:
Here we have three quantities cost of rice, needed members and number of days.

Number of Members	Number of Days	Cost of Rice
8	20	₹ 480
12	15	X

Let the Cost of rice needed for 12 members for 15 days is x.

From the above,
• Number of members and number of days are in inverse proportion.
• Number of days and cost of rice are in direct proportion.

Here the cost of rice depends on both number of member needed and number of days.
So, we have to take compound ratio of 8 : 12 = 20 : 15 is
8 × 20: 12 × 15
480 : x = 8 × 20 : 12 × 15 Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 8 × 20 = 480 × 12 × 15

∴ The cost of rice needed for 12 members for 15 days is ₹ 540.

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

Number of Men	Working hours	Number of Days
24	8	15
20	9	x

Let the number of days needed for 20 men working at 9 hours is x.
From the above,
• Number of men and members are in inversely proportional:
• Number of men and their working hours per day are in direct proportion.
Here the number of days depends on both number of men and working hours per day.
So, we have to take compound ratio of 24 : 20 = 8 : 9 is
24 × 8 : 20 × 9
∴ x : 15 = 24 × 8 : 20 × 9

Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 20 × 9 = 15 × 24 × 8

∴ The number of days needed fur 20 men working at 9 hours is 16 days.

Question 5.
12 painters can paint a wall of 180 meters long in 3 days. How many painters are required to paint 200 meters long wall in 5 days?
Answer:
Here we have three quantities number of painters, length of wall and number of days required.

Length of wall	Number of working days	Number of Painters
180	3	12
200	5	X

Let the number of painters needed to paint 200 m long wall in 5 days is x.

From the above,

• Number of working days and number of painters are in inversely proportional.
• Number of working days and length of wall are in direct proportion.

Here the number of painters depends on both number of working days and length of wall.
So, we have to take compound ratio of 180 :,200 = 5 : 3 is 180 × 5 : 200 × 3
∴ 12 : x = 180 × 5 : 200 × 3
Since the ratios are equal,
The product of means = The product of extremes
⇒ x × 180 × 5 = 12 × 200 × 3

⇒ x = 8
The number of painters needed to paint 200 m long wall in 5 days is 8.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.3

Question 1.
Find out whether the given quantities vary directly or inversely.
(i) Time taken to cover a distance, speed.
Answer:
Inversely proportional.

(ii) Area of land and its cost.
Answer:
Direct proportion.

(iii) Number of men for work, time taken to complete the work.
Answer:
Inverselv proportional.

(iv) Number of people, quantity of food grains each one gets (total quantity remains same).
Answer:
Inversely proportional.

(v) The length of a journey by bus and price of the ticket.
Answer:
Direct proportion.

Question 2.
If 24 men can construct a wall in. 10 days. In how many days will 15 men do it?
Answer:
If the number of men decreases working days will increase in the same proportion.
So, number of men and the working days are in inverse proportion.

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

Number of mens	Number of days
24	10
15	x

By taking inverse proportion,
24 : 15 = x : 10

Then, product of means = Product of the extremes
⇒ 15 × x = 24 × 10

⇒ x = 16
∴ 15 men will complete the wall in 16 days.

Question 3.
In a hostel there are food provisions for 50 girls for 40 days. If 30 more girls join the hostel, how long will the provisions last?
Answer:
If the number of girls increases, then number of days will decrease in the same proportion.
So, number of girls and the days for food provisions lost are in inverse proportion.
Let the number of food provisions days be ’x’.

Number of girls	Number of days
50	40
50 + 30 = 80	x

By taking inverse proportion,
50 : 80 = x : 40
Then, product of means = Product of extremes
⇒ 80 × x = 50 × 40

⇒ x = 25
∴ Number of days required to com-plete the food provisions for 80 girls is 25 days.

Question 4.
Suman travels a distance for 5 hours with a speed of 48 kilometres per hour. If he wants to travel the same in 4 hours at what speed he should travel?
Answer:
If the time decreases, then the speed will increase in the same proportion. Let the speed to cover the distance in 4 hours is ’x’.

Time	Speed (kmph)
5	48
4	x

By taking inverse proportion,
5 : 4 = x : 48
Then, Product of means = Product of extremes
⇒ 4 × x = 48 × 5

⇒ x = 60 kmph.
∴ Distance covered in 4 hours with the speed of 60 kmph.

Question 5.
A person has money to buy 8 bicycles of worth ₹ 4500 each. If the cost of the bicycle is decreased by ₹ 500, then how many bicycles cap he buy with the amount he has?

Answer:
If the cost of the bicyble decreases, then the number of cycles will increase in the same proportion.
Let the number of bicycles he can buy each with ₹ 4000 (4500 – 500) be ‘x’.

Cost of the bicycle	Number of bicycles
4500	8
(4500-500) = 4000	x

By taking inverse proportion,
4500 : 4000 = x : 8
Then, Product of means = Product of extremes
⇒ 4000 × x = 4500 × 8
⇒ \(\frac{4000 x}{4000}=\frac{4500 \times 8}{4000}\) = 9
⇒ x = 9
∴ Number of bicycles bought with the same amount is 9.

Question 6.
2 pumps are required to fill a tank in 1 hour. How many pumps of the same type are used to fill the tank in 24 minutes?
Answer:
If the filling time decreases, then the number of pumps will increase in the same proportion.
Let the number of pumps to fill the tank be x.

Time taken to fill the tank	Number of Pumps
1 hour (60 minutes)	2
24 minutes	x

By taking inverse proportion,
60 : 24 = x : 2

Then, Product of means = Product of extremes
⇒ 24 × x = 60 × 2

⇒ x = 5
∴ Number of pumps required to fill the tank in 24 minutes is 5.

Question 7.
18 men can reap a field in 10 days. For reaping the same field in 6 days, how many more men are required?

Answer:
If the reaping days decreases, then the number of men increases in the same proportion.
Let the number of men required to reap the field in 6 days is x.

Number of mens	Number of reaping days
18	10
x	6

By taking inverse proportion, 18 : x = 6 : 10
Then, Product of means = Product of extremes

∴ Number of men required to reap the field in 6 days = 30
Number of more men = 30 – 18 = 12
Therefore 12 more men required to reap the field in 6 days.

Question 8.
1200 soldiers in a checkpost had enough food for 28 days. After 4 days some soldiers were transferred to another checkpost and thus remaining food is sufficient for 32 more days. How many soldiers left the checkpost?

Answer:
If the number of soldiers decreases, then the number of days food lost will increases in the same proportion.
Let the number of soldiers after transfer is x.

After 4 days,

Number of Soldiers	Enough Food days
1200	28 – 4 = 24
x	32

By taking inverse proportion, 1200 : x = 32 : 24
Then, Product of means = Product of extremes

∴ Number of Soldiers remained in the checkpost = 900
Number of Soldiers transferred = 1200 – 900 = 300
So, 300 soldiers left the checkpost.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.2

Question 1.
Find out whether the given quantities are in direct proportion or not.
(i) Cost of pens, number of pens.
Answer:
Direct proportion. (As the number of pens increases, their total cost also increases)

(ii) Number of people, food required to them.
Answer:
Direct proportion. (As the number of people increases, food required for them also increases)

(iii) Speed of a car, time taken to reach destination.
Answer:
Not in direct proportion. (As the speed increases, time taken by the car reduces)

(iv) Time taken, distance covered.
Answer:
Direct proportion. (As the time increases, distance covered will be increased)

(v) Cost of the vegetables, number of vegetable bags.
Answer:
There is no relation between cost of vegetables and number of vegetable bags.

Question 2.
Five people went to a Park and paid ₹ 580 for tickets. If three people went to the Park, how much money did they have to pay?

Answer:
We know that if the number of people decreases the amount of the tickets also decreases such that the ratio of number of people and the ratio of their tickets amount will remain the same. That means here number of people and the amount of their tickets are in proportion.

Amount paid for tickets by 5 people = ₹ 580
Let the amount paid for tickets by 3 people be x.
Then, 5 : 3 = 580 : x

If the ratios are equal, the product of means = the product of the extremes.
⇒ 5 × x = 3 × 580
⇒ \(\frac{5 x}{5}=\frac{3 \times 580}{5}\)
⇒ x = 3 × 116
⇒ x = 348
∴ Money paid by three persons for tickets = ₹ 348.

Question 3.
A map drawn with a scale of 1 cm re-presents 26 km. If the original distance between two stations 1404 km., then what would be the distance between the stations in the map?
Answer:
We know that if the distance in the map increases the original distance also increases such that the ratio of distance in the map end ratio of their original distance will remain the same. That means here the distance in map and original distance are in proportion.

Distance 26 km scale represented in the map = 1 cm.
Let the distance 1404 km in the map be equal to x cm.
Then, 26 : 1404 = 1 : x
If the ratio are equal, the product of means = the product of the extremes
⇒ 26 × x = 1404 × 1
⇒ \(\frac{26 x}{26}=\frac{1404}{26}\)
⇒ x = 54 cm
∴ Distance 1404 km between the stations in the map = 54 cm.

Question 4.
The weight of 72 pipes is 180 kg. Then what is the weight of 90 such pipes?

Answer:
We know that if the number of pipes increases their weight also increases, such that the ratio of number of pipes and the ratio of their weights will remain the same. That means here number of pipes and their weights are in proportion.
Given weight of 72 pipes 180 kg
Let the weight of 90 pipes = x kg ,
Then 72 : 90 = 180 : x
If the ratios are equal, the product of means = The product of extremes
⇒ 72 × x = 90 × 180

⇒ x = 225 kg
∴ weight of 90 pipes = 225 kg

Question 5.
A motorbike requires 3 liters of petrol to cover 135 km. on average. How many litres of petrol will be required to cover 495km?
Answer:
We know that if the distance increases quantity of petrol also increases, such that the ratio of distances and the ratio of the required quantity of petrol will remain the same. That means here distance and required quantity of petrol are in proportion.
Given quantity of petrol required to cover 135 km = 3l
Let the quantity of petrol required to cover 495 km = xl
Then 135 : 495 = 3 : x

If the ratios are equal, the product of means = The product of the extremes
⇒ 135 × x = 495 × 3

⇒ x = 11 l
∴ The quantity of petrol required to cover 495 km = 11 l.

Question 6.
The shadow of a pole with the height of 10 m. is 6 m. Then find the height of another pole whose shadow is 9 m. at the same time.
Answer:
We know that if the length of shadow increases the height of the pole also increases, such that the ratio of lengths of shadows and the ratio of heights of poles will remain the same.
That means here lengths of shadows and heights of poles are in proportion.

Given height of the pole when its shadow’s length is 6 m = 10 m
Let the height of the pole when its shadow’s length is 9 m = x m
Then 6 : 9 = 10 : x If the ratios are equal,
The product of means = The product of the extremes
⇒ 6 × x = 9 × 10

⇒ x = 3 × 5 = 15m
∴ The height of the pole when its shadows length is 9 m = 15 m.

SCERT AP 7th Class Maths Solutions Pdf Chapter 7 Ratio and Proportion Ex 7.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 7th Lesson Ratio and Proportion Ex 7.1

Question 1.
Pavan and Roshan started a business with ₹ 1,50,000 and ₹ 2,00,000 respectively. After nine months Roshan left from the business. At the end of the year they got a profit of ₹ 45,000. Then find the profits shared by Pavan and Roshafi.
Answer:
Given Pavan’s investment =₹ 150000
Pavan’s period in business = 1 year = 12 months
Roshan’s investment = ₹ 200000
Rohan’s period in business = 9 months

The ratio of investments of Pavan to Roshan = 150000 : 200000 = 3 : 4
The ratio of periods of business of Pavan to Roshan = 12 : 9 = 4 : 3
The profit should be distributed on the basis of compound ratio of investments and period = 3 : 4 and 4 : 3 is
3 × 4 : 4 × 3 = 12 : 12

Therefore, compound Ratio =1:1 (i.e.,) they should share profit equally. Profit = ₹ 45,000
Total parts = 1 + 1 = 2
Pavan’s profit = 45000 × \(\frac{1}{2}\) = ₹ 22500
Roshan’s profit = 45000 – 22500 = ₹ 22,500

Question 2.
Salman started a hotel with an amount of ₹ 75,000. After 5 months Deepak joined with an amount of ₹ 80,000. At the end of the year they earned a profit of ₹ 73,000. How will they share their profit?
Answer:
Given Salman’s investment = ₹ 75000
Salman’s period in business = 1 year = 12 months
Deepak’s investment = ₹ 80000 Deepak’s period in business = 7 months
The ratio of investments of Salman to Deepak = 75000 : 80000 = 15 : 16
The ratio of periods of business of Salman to Deepak = 12 : 7

The profit should be distributed on the basis of compound ratio of 15 : 16 and 12 : 7 is
15 × 12 : 16 × 7 = 180 : 112
= 45 : 28
Therefore, compound ratio = 45 : 28
Profit = ₹ 73,000
Total parts = 45 + 28 = 73

∴ Salman’s profit = 73000 × \(\frac{45}{73}\)
= ₹ 45000

Deepak’s profit = 73000 × \(\frac{28}{73}\)
= ₹ 28,000

Question 3.
Ramayya took a grass field for rent for grazing his 24 cows. After 5 months, Somayya also joined with him for his 40 cows. At the end year they paid a rent of ₹ 35,500. What would be the rent paid by each of them?

Answer:
Given number of cows grazed by Ramayya = 24
Grazed months 1 year =12 months.
Number of cows grazed by Somayya = 40
Grazed months = 7 months.

The ratio number of grazed cows
= 24 : 40 = 3 : 5

The ratio of months grazed = 12 : 7
The rent should be distributed on the basis of compound ratio of 3 : 5 and 12 : 7 is
3 × 12 : 5 × 7 = 36 : 35

Therefore, compound ratio = 36 : 35
Annual Rent = ₹ 35,500
Total parts = 36 + 35 = 71

Rent paid by Rarnavya = 35,500 × \(\frac{36}{71}\)
= ₹ 18,000
Rent paid by Somaa = 35,500 × \(\frac{35}{71}\)
= ₹ 17,500

Question 4.
Ravi started a business with ₹ 2,10,000. After a few months, Prakash joined in the business with an amount of ₹ 3,60,000. At the end of the year if they got a profit of ₹ 1,20,000 each, then find after how many months did Prakash join in the business?
Answer:
Given Ravi’s investment = ₹ 2,10,000
Ravi’s period in business = 1 year = 12 months
Prakash’s investment = ₹ 3,60,000

Let, Prakash’s period in business = x months
The Ratio of investments of Ravi and Prakash = 210000 : 360000 = 7 : 12
The ratio of periods of business of Ravi to Prakash = 12 : x

The profit should distributed on the basis of compound ratio of 7 : 12 and 12 : x is
7 × 12 : 12 × x = 84 : 12x
= 7 : x
Therefore, compound ratio = 7 : x
They got equal profits.

So, Total profit = 120000 + 120000 = ₹ 2,40,000
Total parts = 7 + x
So, x = 7
So Prakash joined after (12 – 7) = 5 months
Prakash’s period in business = x
= 7 months
So, Prakash joined in business after (12 – 7) = 5 months.

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

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

Question 1.
Find the ratio of the following :
(i) 5,8
Answer:
5 : 8

(ii) ₹ 10, ₹ 15
Answer:
10 : 15
2 : 3
∴ ₹ 10 : ₹15 = 2 : 3

(iii) 25 kg., 20. kg.
Answer:
25 kg : 20 kg
25 : 20
5 : 4
∴ 25 kg : 20 kg = 5 : 4

(iv) 5l, 500ml
Answer:
5l : 500 ml
5000 ml: 500 ml (∵ 1l = 1000 ml, 5l = 5000 ml)
5000 : 500
10 : 1
∴ 5l : 500 ml = 10 : 1

(v) 2 km. 500 m, 1 km. 750 m
Answer:
2 km. 500 m : 1 km. 750 m
2000 m + 500 m : 1000 m + 750 m (∵ 1 km = 1000m; 2 km = 2000 m)
2500 : 1750
10 : 7 (∵ HCF = 25)
∴ 2 km. 500 m : 1 km. 750 m = 10 : 7

(vi) 3 hrs, 1 hr. 30 min.
Answer:
3 hrs : 1 hr. 30 min
3 × 60 : (1 × 60) + 30 (∵ 1 hr = 60 min; 3h = 3 × 60 = 180 min)
180 min : 60 + 30 min

2 : 1
∴ 3 hrs : 1 hr. 30 min = 2:1

(vii) 40 days, 1 year
Answer:
40 days : 1 year
40 days : 365 days (∵ 1 year = 365 days)
40 : 365 (∵ HCF = 5)
8 : 73

∴ 40 days : 1 year = 8 : 73
(Or)
40 days : 366 days (∵ 1 year = 366 days for leap year)
20 : 183 (∵ HCF = 2)

Question 2.
Express the following ratios in the simplest form:
(i) 120 : 130
Answer:
Given 120 : 130

12 : 13
∴ 120 : 130 = 12 : 13

(ii) 135:90
Answer:
Given 135 : 90
27 : 18
3 : 2
∴ 135 : 90 = 3 : 2

(iii) 48 : 144
Answer:
Given 48: 144
12 : 36
1 : 3
∴ 48 : 144 = 1:3

(iv) 81 : 54
Answer:
Given 81 : 54
9 : 6
3 : 2
∴ 81 : 54 = 3 : 2

(v) 432 : 378
Answer:
Given 432 : 378
216 : 189
24 : 21
8 : 7
∴ 432 : 378 = 8 : 7

Question 3.
Check whether the two ratios given below are in proportion.

(i) 10 : 20, 25 : 50
Answer:
Given 10 : 20, 25 : 50
a :b = 10 : 20
So, a : b = 1 : 2

c : d = 25 : 50
also, c : d = 1 : 2

If a : b = c: d, then a, b, c, d are in proportion.
a : b = c : d = 1 : 2
Therefore, 10 : 20, 25 : 50 are in proportion.

(ii) 18 : 12, 15 : 10
Answer:
Given 18 : 12, 15 : 10
a : b = 18 : 12
= 6 : 4

c : d = 15 : 10
also c : d = 3 : 2
So, a : b = 3 : 2

If a : b = c : d, then a, b, c, d are in proportion.
a : b = c : d = 3 : 2
Therefore, 18 : 12, 15 : 10 are in proportion.

(iii) 25 : 20, 16 : 14
Answer:
Given 25 : 20, 16 : 14
a : b = 25 : 20
So, a : b = 5 : 4
5 : 4 ≠ 8 : 7
∴ a : b ≠ c : d

c : d = 16 : 14
c : d = 8 : 7

If a : b = c : d, then a, b, c, dare in propottion.
Here a : b ≠ c : d
So, a, b, c, d are not in proportion.

(iv) 54 : 27, 18 : 9
Answer:
Given 54 : 27, 18 : 9
a : b = 54 : 27.
= 6 : 3
So, a : b = 2 : 1

c : d = 18 : 9
also, c : d = 2 : 1

If a : b = c : d, then a, b, c, d are in proportion.
a : b = c : d = 2 : 1

Therefore, 54: 27, 18 : 9 are in proportion.

Question 4.
Find the missing number in each of the following problems:
(i) 15 : 19 = 45 : _______
Answer:
Given 15 : 19 = 45 : x
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extremes is equal to the product of means that is a . d = b . c
⇒ 15 × x = 19 × 45

⇒ x = 19 × 3
∴ x = 57

(ii) 9 : 13 = ____ : 65
Answer:
Given 9 : 13 = x : 65
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extremes is equal to the product of means that is a . d = b . c
⇒ 9 × 65 = 13 × x

⇒ 45 = x
∴ x = 45

(iii) 8 : ______ = 72 : 63
Answer:
Given 8 : x = 72 : 63
Let the missing term be ‘x’.
If a : b = c : d, then a, b, c and d are in proportion, then the product of extre¬mes is equal to the product of means that is a . d = b . c
⇒ 8 × 63 = x × 72

⇒ 7 = x
∴ x = 7

Question 5.
Fill in the boxes with equivalent ratio of 3 : 4.

Answer:
Given a : b = 3 : 4 (or) \(\frac{3}{4}\)
\(\frac{a}{b}=\frac{3}{4}\)

Equivalent ratios

Therefore, a : b = 3 : 4 = 6 : 8
= 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24

Question 6.
Pick out any four numbers from below and arrange them so that they are in proportion 2, 3, 10, 12> 15, 18.
Ex : 2 : 10 = 3 : 15
(i) : __________
Answer:
2 : 3; 12 : 18
a : b = 2 : 3, c : d = 12: 18
If a, b, C, d are in proportion, then the product of extremes is equal to the pro¬duct of means.

a . d = b . c
2 × 18 = 3 × 12
36 = 36
So, 2 : 3 = 12 : 18

(ii) : __________
Answer:
10 : 12 = 15 : 18
a : b = 10 : 12
c : d = 15 : 18
If a, b, c, d are in proportion, then the product of extremes is equal to the product of means.

a . d = b . c
10 × 18 = 15 × 12
180 = 180
So, 10 : 12 = 15 : 18

Question 7.
Divide ₹ 1500 into two parts such that they are in the ratio of 7 : 3.
Answer:
Given amount to be divide = ₹ 1500 .
Ratio to be divide = 7:3
Sum of the terms in the ratio = 7 + 3 = 10

[Or 1500 – 1050 = 450]

Question 8.
If there are 20 chocolates in a packet, Rajani and Ragini share them and Rajani takes 12 chocolates, then what is the ratio of chocolates taken by Rajani to Ragini?
Answer:
Total number of Chocolates = 20
Number of chocolates Rajani taken = 12
Number of chocolates Ragini taken = 20 – 12 = 8
Ratio of Rajani to Ragini = 12 : 8
Ratio of chocolates taken by Rajani to Ragini = 3 : 2

Question 9.
A pipe is cut into two parts in such a ratio that the first part to second part is 7:8. If the length of the 2nd part is 48cm, then what is the length of the first part? What is the total length of the pipe before cutting?
Answer:
Given ratio of two parts of pipe = 7:8
Length of the second part = 48 cm
Length of the first part = x cm
Ratio of the two parts = x : 48

Product of extremes = Product of means
⇒ 8 × x = 48 × 7

⇒ x = 42
∴ Length of first part of the pipe = x = 42 cm
Length of the pipe = 42 + 48 = 90 cm