Ap board solutions class 7 maths

(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 :...

Andhra Pradesh SCERT AP State Board Syllabus 7th Class Maths Study Material Guide Pdf free download, 7th Class Maths Textbook Solutions in English Medium and Telugu Medium are part of Students can also go through AP State Board Syllabus 7th Class Maths Textbook Solutions Study Material Guide AP 7th Class Maths Guide in English Medium New Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers • • • • • • • AP State 7th Class Maths Textbook Solutions Chapter 2 Fractions, Decimals and Rational Numbers • • • • • Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.4 • Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.5 • • AP 7th Class Maths Study Material Pdf Chapter 3 Simple Equations • • • • • • 7th Class Maths Guide AP State Syllabus Pdf Chapter 4 Lines and Angles • • • • • • • 7th Class Maths AP State Syllabus Chapter 5 Triangles • • • • • • Chapter 5 Triangles Ex 5.5 • Chapter 5 Triangles Ex 5.6 • • AP Board 7th Class Maths Solutions Chapter 6 Data Handling • • • • • • 7th Class Maths AP State Syllabus Chapter 7 Ratio and Proportion • • • • • • • • • • AP Board 7th Class Maths Solutions Chapter 8 Exponents and Powers • • • • • 7th Class Maths AP State Syllabus Chapter 9 Algebraic Expressions • • • • • • • 7th Class Maths Guide AP State Syllabus Pdf Chapter 10 Construction of Triangles • • • • • • AP 7th Class Maths Study Material Pdf Chapter 11 Area of Plane Figures • • • • • • • AP State 7th Class Maths Textbook Solutions Chapter 12 Symmetry • • • • • •...

SCERT AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Unit Exercise Question 1. Answer the following. (i) The exponential form 14 9 should read as Answer: 14 is raised to the power of 9. (ii) When base is 12 and exponent is 17, it’s exponential form is _________ Answer: 12 17. (iii) The value of (14 × 21) 0 is Answer: We know a 0 = 1 So, (14 × 21) 0 = 1 Question 2. Express the following numbers as a product of powers of prime factors : (i) 648 Answer: Given 648 = 2 × 324 = 2 × 2 × 162 = 2 × 2 × 2 × 81 = 2 × 2 × 2 × 3 × 27 = 2 × 2 × 2 × 3 × 3 × 9 = 2 × 2 × 2 × 3 × 3 × 3 × 3 ∴ 648 = 2 3 × 3 4 (ii) 1600 Answer: Given 1600 = 2 × 800 = 2 × 2 × 400 = 2 × 2 × 2 × 200 = 2 × 2 × 2 × 2 × 100 = 2 × 2 × 2 × 2 × 2 × 2 × 25 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 ∴ 1600 = 2 6 × 5 2 (iii) 3600 Answer: Given 3600 = 2 × 1800 = 2 × 2 × 900 = 2 × 2 × 2 × 450 = 2 × 2 × 2 × 2 × 225 = 2 × 2 × 2 × 2 × 3 × 75 = 2 × 2 × 2 × 2 × 3 × 3 × 25 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 ∴ 3600 = 2 4 × 3 2 × 5 2 Question 3. Simplify the following using laws of exponents. (i) a 4 × a 10 Answer: a 4 × a 10 We know a m × a n = a m+n = a 4+10 ∴ a 4 × a 10 = a 14 (ii) 18 18 ÷ 18 14 Answer: 18 18 ÷ 18 14 We Know a m ÷ a n = a m-n = 18 18-14 ∴ 18 18 ÷ 18 14 = 18 4 (iii) (x m) 0 Answer: (x m) 0 We Know (a m) n = a m.n = x m×n = x 0(∵ a 0 = 1) ∴ (x m) 0 = 1 (iv) (6 2 × 6 4) ÷ 6 3 Answer: (6 2 X 6 4) ÷ 6 3 We Know a m × a n = a m+n = (6 2+4) ÷ 6 3 = 6 6 ÷ 6 3 We Know a m ÷ a n = a m-n = 6 6-3 ∴ (6 2 × 6 4...

SCERT 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 × \(\frac\) = ₹ 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 :...

AP State Syllabus 7th Class Maths Textbook Solutions Study Material Guide Pdf Free Download AP 7th Class Maths Guide in English Medium New Syllabus AP Board 7th Class Maths Solutions Chapter 1 Integers • • • • • • • AP State 7th Class Maths Textbook Solutions Chapter 2 Fractions, Decimals and Rational Numbers • • • • • Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.4 • Chapter 2 Fractions, Decimals and Rational Numbers Ex 2.5 • • AP 7th Class Maths Study Material Pdf Chapter 3 Simple Equations • • • • • • 7th Class Maths Guide AP State Syllabus Pdf Chapter 4 Lines and Angles • • • • • • • 7th Class Maths AP State Syllabus Chapter 5 Triangles • • • • • • Chapter 5 Triangles Ex 5.5 • Chapter 5 Triangles Ex 5.6 • • AP Board 7th Class Maths Solutions Chapter 6 Data Handling • • • • • • 7th Class Maths AP State Syllabus Chapter 7 Ratio and Proportion • • • • • • • • • • AP Board 7th Class Maths Solutions Chapter 8 Exponents and Powers • • • • • 7th Class Maths AP State Syllabus Chapter 9 Algebraic Expressions • • • • • • • 7th Class Maths Guide AP State Syllabus Pdf Chapter 10 Construction of Triangles • • • • • • AP 7th Class Maths Study Material Pdf Chapter 11 Area of Plane Figures • • • • • • • AP State 7th Class Maths Textbook Solutions Chapter 12 Symmetry • • • • • • • AP 7th Class Maths Guide in Telugu Medium New Syllabus AP Board 7th Class Maths Solutions Chapter 1 పూర్ణ సంఖ్యలు • • • • • • • AP State 7th Class Maths Textbook Solutions Chapter 2 భిన్నాలు మరియు ద...

AP State Syllabus AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions InText Questions Question 1. In the expressions given below identify all the terms. (Page No. 194) i) 5x 2 + 3y + 7 ii) 5x 2y + 3 iii) 3x 2y iv) 5x – 7 v) 5x + 8 – 2(-y) vi) 7x 2– 2x Solution: i) 5x 2 + 3y + 7 is a trinomial ii) 5x 2y + 3 is a binomial iii) 3x 2y is a monomial iv) 5x – 7 is a binomial v) 5x + 8 – 2 (-y) is a trinomial vi) 7x 2– 2x is a binomial Question 2. Write the following expressions in statements. (Page No. 195) 12x, 12, 25x, -25, 25y, 1, x, 12y, y, 25xy, 5x 2y, 7xy 2, 2xy, 3xy 2, 4x 2y. Solution: Like terms Groups → \) xyz Binomial :i) ax + by, ii) 2z – 5 Trinomial : i) ax + by + cz, ii) p 2 + q 2 + r 2 Polynomial: i) 5x 4– 2x 2 + x – 1, ii) 6 + 5x – 4x 2 + 3y 3– 2z 4 Question 4. Identify the expressions given below as monomial, binomial, trinomial, and multinomial. (Page No. 197) i) 5x 2 + y + 6 ii) 3xy iii) 5x 2y + 6x iv) a + 4x – xy + xyz Solution: i) 5x 2 + y + 6 → is a trinomial. ii) 3xy → is a monomial. iii) 5x 2y + 6x → is a binomial. iv) a + 4x – xy + xyz → is a multinomial. Question 5. Find the sum of the like terms. (Page No. 200) i) 5x, 7x ii) 7x 2y, -6x 2y iii) 2m, 11m iv) 18ab, 5ab, 12ab v) 3x 2, -7x 2, 8x 2 vi) 4m 2, 3m 2, -6m 2, m 2 Solution: i) 5x + 7x = 12x ii) 7x 2y + (-6x 2y) = (7 – 6) x 2y = x 2y iii) 2m + 11m = (2 + 11)m = 13m iv) 18ab + 5ab + 12ab = (18 + 5 + 12) ab = 35ab v) 3x + (-7x) + 8x = (3 – 7 + 8) x 2 = (11 – 7) x 2 = 4x 2 vi)...