Solutions For All Chapters Maths Class 7
Ex 1.3 Class 7 Maths Question 1.
 Evaluate each of the following:
(a) (-30) ÷ 10
(b) 50 ÷ (-5)
(c) (-36) ÷ (-9)
(d) (-49) ÷ (49)
(e) 13 ÷ [(-2) + 1]
(f) 0 ÷ (-12)
(g) (-31) ÷ [(-30) + (-1)]
(h) [(-36) ÷ 12] ÷ 3
(i) [(-6) + 5] ÷ [(-2) + 1]
Solution:
(a) (-30) ÷ 10 = -30/10 = -3
 (b) 50 ÷ (-5) = 50/-5 = -10
 (c) (-36) ÷ (-9) = -36/-9 = 4
 (d) (-49) ÷ (49) = -49/49 = -1
 (e) 13 ÷ [(-2) + 1] = 13 ÷ -1 = 13/-1 = -13
 (f) 0 ÷ (-12) = 0/-12 = 0
 (g) (-31) ÷ [(-30) + (-1)] = (-31) ÷ (-31) = -31/-31 = 1
 (h) [(-36) + 12] ÷ 3 = [-36/12] ÷ 3 = -3 ÷ 3
 = -3/3 =-1
 (i) [(-6) + 5] + [(-2) + 1] = (-1) – (-1) = -1/-1 = 1
Ex 1.3 Class 7 Maths Question 2.
 Verify that: a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(а) a = 12, b = – 4, c = 2
(b) a = (-10), b = 1, c = 1
Solution:
(a) a = 12, 6 = – 4, c = 2
 a ÷ (b + c) = 12 ÷ [(-4) + 2]
 = 12 ÷ (-2) = 12/-2 = -6
 (a ÷ b) + (a ÷ c) = [12 ÷ (-4)] + [12 ÷ 2]
 =12/-4 + 12/2=-3+6=3
 Since, (-6) + 3
 Hence, a ÷ (b + c) + (a ÷ b) + (a ÷ c)
(b) a = (-10), b = 1, c = 1
 a ÷ (b + c) = (-10) ÷ (1 + 1)
 =(-10) ÷ 2 = -10/2 = -5
 (a ÷ b) + (a ÷ c)
 =[(-10) ÷ 1] + [(-10) ÷ 1]
 =(-10)/1+(-10)/1
 = (-10) + (-10) = -20
 Since (-5) ≠ (-20)
 Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Ex 1.3 Class 7 Maths Question 3.
 Fill in the blanks:
(a) 369 ÷ ___ = 369
(b) (-75) ÷ ___ = -1
(c) (-206) ÷ ___ =1
(d) -87 ÷ ___ = -87
(e) ___ ÷ 1 = -87
(g) 20 ÷ ___ = -2
 (h) _____ ÷ (4) = –3
 
Solution:
(a) 369 ÷ ___ = 369 = 369 ÷ 1 = 369
 (b) (-75) ÷ ___ = -1 = (-75) ÷ 75 = -1
 (c) (-206) ÷ ___ = 1 = (-206) ÷ (-206) = 1
 (d) 87 ÷ ___ = 87 = -87 ÷ (-1) = 87
 (e) ___ ÷ 1 = -87 = -87 ÷ 1 = -87
 (f) ___ ÷ 48 = -1 = (-48) ÷ 48 = -1
 (g) 20 + ___ = -2 = 20 ÷ (-10) = -2
 (h) ___ + (4) = -3 = (-12) ÷ (4) = -3
Ex 1.3 Class 7 Maths Question 4.
 Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 + (-2) = -3.
Solution:
(a) (24, -8) because 24 ÷ (-8) = -3
 (b) (-12, 4) because (-12) ÷ 4 = -3
 (c) (15, -5) because 15 ÷ (-5) = -3
 (d) (18, -6) because 18 ÷ (-6) = -3
 (e) (60, -20) because 60 ÷ (-20) = -3
Ex 1.3 Class 7 Maths Question 5.
 The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?
Solution:
Temperature at 12 noon was 10°C above zero i.e. +10°C
 Rate of decrease in temperature per hour = 2°C
 Number of hours from 12 noon to midnight = 12
 ∴ Change in temperature in 12 hours
 = 12 × (-2°C) = -24°C
 ∴ Temperature at midnight
 = +10°C + (-24°C) = -14°C
 Hence, the required temperature at midnight =-14°C
 Difference in temperature between + 10°C and -8°C
 = +10°C – (-8°C) = +10°C + 8°C = 18°C
 Number of hours required = 18°C/2°C= 9 hours
 ∴ Time after 9 hours from 12 noon = 9 pm.
Ex 1.3 Class 7 Maths Question 6.
 In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question:
 (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
 (ii) Mohini scores -5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
Given that:
 +3 marks are given for each correct answer. (-2) marks are given for each incorrect answer. Zero marks for not attempted questions.
 (i) Marks obtained by Radhika for 12 correct answers = (+3) × 12 = 36
 Marks obtained by Radhika for correct answers = 12 × 3 = 36
 Total marks obtained by Radhika = 20
 ∴ Marks obtained by Radhika for incorrect answers = 20 – 36 = -16
 Number of incorrect answers
 =(-16)/(-2)=(-16)/(-2)=8
 Hence, the required number of incorrect answers = 8
(ii) Marks scored by Mohini = -5
 Number of correct answers = 7
 ∴ Marks obtained by Mohini for 7 correct answers = 7 × (+3) – 21
 Marks obtained for incorrect answers
 = -5 – 21 = (-26)
 ∴ Number of incorrect answers
 = (-26) ÷ (-2) = 13
 Hence, the required number of incorrect answers – 13.
Ex 1.3 Class 7 Maths Question 7.
 An elevator descends into a nine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m.
Solution:
The present position of the elevator is at 10 in above the ground level.
 Distance moved by the elevator below the ground level = 350 m
 ∴ Total distance moved by the elevator = 350 m + 10 m = 360 m
 Rate of descent = 6 m/min.
 Total time taken by the elevator
 Total time taken by the elevator
 = 360 m/6 m\min
 = 60 minutes = 1 hour
 Hence, the required time = 1 hour.

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