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Integers Class 7 Extra Questions Maths Chapter 1, Extra Questions for Class 7 Maths Chapter 1 Integers, Integers Class 7 Extra Questions Very Short Answer Type, Question 1., Fill in the blanks using < or >., (a) -3 …… -4, (b) 6 ……. -20, (c) -8 …… -2, (d) 5 …… -7, Solution:, (a) -3 > -4, (b) 6 > -20, (c) -8 < -2, (d) 5 > -7, Question 2., Solve the following:, (i) (-8) × (-5) + (-6), (ii) [(-6) × (-3)] + (-4), (iii) (-10) × [(-13) + (-10)], (iv) (-5) × [(-6) + 5], Solution:, (i) (-8) × (-5) + (-6), = (-) × (-) × [8 × 5] + (-6), = 40 – 6, = 34, (ii) [(-6) × (-3)] + (-4), = (-) × (-) × [6 × 3] + (-4), = 18 – 4, = 14, (iii) (-10) × [(-13) + (-10)], = (-10) × (-23), = (-) × (-) × [10 × 23], = 230, (iv) (-5) × [(-6) + 5], = (-5) × (-1), = (-) × (-) × 5 × 1, = 5, Question 3., Starting from (-7) × 4, find (-7) × (-3), Solution:, (-7) × 4 = -28, (-7) × 3 = -21 = [-28 + 7], (-7) × 2 – -14 = [-21 + 7], (-7) × 1 = -7 = [-14 + 7], (-7) × 0 = 0 = [-7 + 7], (-7) × (-1) = 7 = [0 + 7], (-7) × (-2) = 14 = [7 + 7], (-7) × (-3) = 21 = [14 + 7], Question 4., Using number line, find:, (i) 3 × (-5), (ii) 8 × (-2), Solution:, (i) 3 × (-5), , From the number line, we have, (-5) + (-5) + (-5) = 3 × (-5) = -15, (ii) 8 × (-2), , From the number line, we have, (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) = 8 × (-2) = -16, Question 5., Write five pair of integers (m, n ) such that m ÷ n = -3. One of such pair is (-6, 2)., Solution:, (i) (-3, 1) = (-3) ÷ 1 = -3, (ii) (9, -3) = 9 ÷ (-3) = -3, (iii) (6, -2) = 6 ÷ (-2) = -3, (iv) (-24, 8) = (-24) ÷ 8 = -3, (v) (18, -6) = 18 ÷ (-6) = -3, Integers Class 7 Extra Questions Short Answer Type, Question 6., Solve the following:, (i) (-15) × 8 + (-15) × 4, (ii) [32 + 2 × 17 + (-6)] ÷ 15, Solution:, (i) (-15) × 8 + (-15) × 4, = (-15) × [8 + 4], = (-15) × 12, = -180, (ii) [32 + 2 × 17 + (-6)] ÷ 15, = [32 + 34 – 6] ÷ 15, = [66 – 6] ÷ 15, = 60 ÷ 15, = 4, Question 7., The sum of two integers is 116. If one of them is -79, find the other integers., Solution:, Sum of two integers = 116, One integer = -79, Other integer = Sum of integer – One of integer = 116 – (-79) = 116 + 79 = 195, Question 8., If a = -35, b = 10 cm and c = -5, verify that:, (i) a + (b + c) = (a + b) + c, (ii) a × (b + c) = a × b + a × c, Solution:, (i) Given that a = -35, b = 10, c = -5, LHS = a + (b + c) = (-35) + [10 + (-5)] = (-35) + 5 = -30, RHS = (a + b) + c = [(-35) + 10] + (-5) = (-25) + (-5) = -(25 + 5) = -30, LHS = RHS, Hence, verified., (ii) a × (b + c) = a × b + a × c, LHS = a × (b + c) = (-35) × [10 + (-5)] = (-35) × 5 = -175, RHS = a × b + a × c = (-35) × 10 + (-35) × (-5) = -350 + (-) × (-) × (35 × 5) = -350 + 175 = -175, LHS = RHS, Hence, verified., Question 9., Write down a pair of integers whose, (i) sum is -5, (ii) difference is -7, (iii) difference is -1, (iv) sum is 0, Solution:, (i) (-2) + (-3) = -5, Hence, the required pair of integers = (-2, -3), (ii) -10 – (-3) = -10 + 3 = -7, Hence, the required pair of integers = (-10, -3), (iii) (-3) – (-2) = -1, Hence, the required pair of integers = (-3, -2), (iv) (-4) + (4) = 0, Hence, the required pair of integers = (-4, 4), Question 10., You have ₹ 500 in your saving account at the beginning of the month. The record below shows all of your transactions during the month. How much money is in your account after these transactions?, , Solution:, Amount in the beginning of the month in the account = ₹ 500, Amount deposited in the account for Jal Board = ₹ 200, Amount paid to Jal Board = ₹ 120, Amount left in the account after the above transactions = ₹ (500 + 200 – 120) = ₹ (700 – 120) = ₹ 580, Amount deposited for LIC India = ₹ 150, Amount paid to LIC India = ₹ 240, Amount left after this transactions = ₹ (580 + 150 – 240) = ₹ (730 – 240) = ₹ 490, Question 11., The given table shows the freezing points in °F of different gases at sea level. Convert each of these into °C to the nearest integral value using the relations and complete the table, C = 59 [F – 32], , Solution:, Freezing point of Hydrogen = -435°F, C = 59 [-435 – 32], = 59 [-467], = 5 × (-51.9), = 259.5°C or 259°C, For Krypton, freezing point = -251°F, C = 59 [-251 – 32], = 59 [-283], = 5 × (-31.4)], = -157°C, For Oxygen, freezing point = -369°F, C = 59 [-369 – 32], = 59 [-401], = 5 × (44.56), = 222.80° C or 223°C, Hence, the required freezing points at sea level in °C for Hydrogen = -259°C, Krypton = -157°C, Oxygen = -223°C., Question 12., Taking today as zero on the number line, if the day before yesterday is 17 January, what is the date on 3 days after tomorrow? [NCERT Exemplar], Solution:, , The date before yesterday = 17 January, The date of yesterday = 17 + 1 = 18 January, Today’s date = 18 + 1 = 19 January, Tomorrow’s date = 19 + 1 = 20 January, Date on 3 days after tomorrow = (20 + 3) = 23rd January