AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.2<\/h2>\n
Question 1.
\nCalculate the following,
\n(i) (- 96) \u00f7 16
\nAnswer:
\n(- a) \u00f7 b = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{-96}{16}\\) = – 6<\/p>\n
![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/11\/AP-Board-Solutions.png\")
<\/p>\n
(ii) 98 \u00f7 (- 49)
\nAnswer:
\na \u00f7 (- b) = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{98}{-49}\\) = – 2<\/p>\n
(iii) (- 51) \u00f7 17
\nAnswer:
\n(- a) \u00f7 b = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{-51}{17}\\) = – 3<\/p>\n
(iv) 38 \u00f7 (- 19)
\nAnswer:
\na \u00f7 (- b) = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{38}{-19}\\) = – 2<\/p>\n
(v) (- 80) \u00f7 20
\nAnswer:
\n(- a) \u00f7 b = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{-80}{20}\\) = – 4<\/p>\n
(vi) (- 150) \u00f7 (- 25)
\nAnswer:
\n(- a) \u00f7 (- b) = – \\(\\frac{-a}{-b}\\) = \\(\\frac{a}{b}\\)
\n= \\(\\frac{-150}{-25}\\) = \\(\\frac{150}{25}\\) = – 6<\/p>\n
(vii) (- 600) \u00f7 60
\nAnswer:
\n(- a) \u00f7 b = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{-600}{60}\\)
\n= – 10<\/p>\n
(viii) (-54) \u00f7 9
\nAnswer:
\n(- a) \u00f7 b = – \\(\\frac{a}{b}\\)
\n= \\(\\frac{- 54}{9}\\)
\n= – 6<\/p>\n
(ix) 130 \u00f7 65
\nAnswer:
\na \u00f7 b = \\(\\frac{a}{b}\\)
\n= \\(\\frac{130}{65}\\)
\n= 2<\/p>\n
<\/p>\n
(x) (- 315) \u00f7 ( -315)
\nAnswer:
\n(- a) \u00f7 (- b) = \\(\\frac{-a}{-b}\\) = \\(\\frac{a}{b}\\)
\n= \\(\\frac{-315}{-315}\\)
\n= \\(\\frac{315}{315}\\)
\n= 1<\/p>\n
Question 2.
\nThe product of two integers is – 165. If one number is – 15, find the other integer.
\nAnswer:
\nGiven one number a = – 15
\nOther number b = ?
\nProduct of two numbers = (- 165)
\na \u00d7 b = – 165
\n(- 15) \u00d7 b = – 165
\nb = (- 165) \u00d7 (- 15)
\nOther number b = 11<\/p>\n
Question 3.
\nBecause of Covid-19 a company lockdown for 6 months and got loss of \u20b91,32,000 in the year 2020. Find the average loss of each month.
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2022\/01\/AP-Board-7th-Class-Maths-Solutions-Chapter-1-Integers-Ex-1.2-1-1.png\")
\nAnswer:
\nGiven loss of 6 months = \u20b9 1,32,000
\nAverage loss of each month = 1,32,000 \u00f7 6
\nAverage loss of each month = \u20b9 22,000<\/p>\n
<\/p>\n
Question 4.
\nThe temperature at 12 noon was 10\u00b0C above zero. If it decreases at the rate of 2\u00b0C per hour until midnight, at what time would the temperature be 8\u00b0C below zero ? What would be the temperature at mid-night ?
\nAnswer:
\nThe temperature at 12 noon = 10\u00b0C above zero.
\nThe temperature is decreasing at 2\u00b0C per hour 8\u00b0C below zero = – 8\u00b0C
\nSo, total temperature falls 2\u00b0C in every one hour.
\nTherefore, to decrease 18\u00b0C, time taken
\n= 18\/2 = 9 hrs.
\nPresent time is 12 noon, so, the time when the temperature at 12 midnight will be
\n= (- 8\u00b0 C) + (- 2\u00b0 C \u00d7 3 hrs.)
\n= (- 8\u00b0 C) + (- 6\u00b0 C)
\n= 160
\nSo, at 12 midnight the temperature will be – 14\u00b0C<\/p>\n
Question 5.
\nA green grocer earns a profit of \u20b9 7 per kg of tomato and got loss of \u20b9 4 per kg of brinjal by selling. On Monday he gets neither profit nor loss,.if he sold 68 kgs of tomato. How many kgs of brinjal did he sell ?
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2022\/01\/AP-Board-7th-Class-Maths-Solutions-Chapter-1-Integers-Ex-1.2-2-1.png\")
\nAnswer:
\nGiven profit per kg of tomato = \u20b9 7
\nLoss per kg of brinjal = \u20b9 4
\nWeight of tomatoes sold = 68 kgs
\nLet number of kgs of tomato be x kg.
\nNumber of kgs of brinjals be y kg.
\n7x – 4y = 0
\n7 \u00d7 68 – 4y = 0
\n– 4y = – 7 \u00d7 68
\ny = \\(\\frac{-7 \\times 68}{-4}\\)
\n\u2234 Weight of brinjals sold = + 7 \u00d7 17
\n= + 119 kgs.<\/p>\n
<\/p>\n
Question 6.
\nIn a test, each correct answer carry + 3 marks and incorrect answer – 1 mark, Sona attempted all the questions and scored + 20 marks though she got 10 correct answers.
\n(i) How many incorrect answers did she attempt ?
\n(ii) How many questions were given in the test ?
\nAnswer:
\nGiven marks for every correct answer = + 3
\nGiven marks for every incorrect answer = – 1
\nNo. of correct answer questions = 10
\nTotal marks gained on correct answer questions = 10 \u00d7 (+3) = + 30<\/p>\n
(i) No. of incorrect answer questions = x
\nTotal marks gained on incorrect answer questions = y \u00d7 (-1) = – y.
\nMarks scored by Sona
\n\u21d2 + 30 + (- y) = + 20
\n\u21d2 – y = + 20 – 30
\n\u21d2 – y = – 10
\n\u21d2 y = 10
\n\u2234 No. of incorrect answers Sona attempts =10<\/p>\n
(ii) Number of questions given in the test = No. of correct answer questions + No. of incorrect answer questions.
\n= 10 + 10 = 20 questions.<\/p>\n
<\/p>\n
Question 7.
\nWrite 5 pairs of integers (a, b) such that a \u00f7 b = -4.
\n(Ex : (12, – 3) because 12 + (-3) = -4)<\/p>\n
![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2022\/01\/AP-Board-7th-Class-Maths-Solutions-Chapter-1-Integers-Ex-1.2-3-1.png\")
\nAnswer:
\n(i) (-16, +4) because (- 16) \u00f7 4 = -4
\n(ii) (20, -5) because 20 \u00f7 (-5) = – 4
\n(iii) (\u20148, 2) because (-8) \u00f7 2 =- 4
\n(iv) (24, -6) because 24 \u00f7 (-6) = – 4
\n(v) (-36, 9) because (-36) \u00f7 9 = – 4<\/p>\n
(OR)<\/p>\n