NCERT Solutions Class 7 Mathematics Integers Ex 1.4

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(–206) ÷ _______=1

(i) [(–6)+5] ÷ [(–2)+1]

Answer :

(a)(–30) ÷ 10 = (–30) × ¹⁄₁₀ = ^{(-30 x 1)}⁄₁₀ = –3

(b) 50 ÷ (–5) = 50 × (^-1⁄₅) = ^(50x(-1))⁄₅=–10

(c) (–36) ÷ (–9) = (–36) × (^-1⁄₉) = ^{(-36) x (-1)}⁄₉ = ³⁶⁄₉ = 4

(d) (–49) ÷ 49 = (–49) × ¹⁄₄₉ = ^-49⁄₄₉ = –1

(e) 13 ÷ [(–2) + 1] = 13 ÷ (–1) = 13 × (^-1⁄₁) = –13

(f) 0 ÷ (–12) = 0 × (^-1⁄₁₂) = ⁰⁄₁₂ = 0

(g) (–31) ÷ [(–30) ÷ (–1)] = (–31) ÷ (–30–1) = (–31) ÷ (–31) = (–31) × (^-1⁄₃₁) = ³¹⁄₃₁ = 1

(h) [(–36) ÷ 12] ÷ 3 = [(–36) × ^-1⁄₁₂] × ¹⁄₃ = (^-36⁄₁₂) × ¹⁄₃ = (–3) × ¹⁄₃ = ^-3⁄₃ = –1

(i) [(–6) + 5] ÷ [(–2) + 1] = (–6 + 5) ÷ (–2 + 1) = (–1) ÷ (–1) = (–1) × ^(-1)⁄₁ = 1

Question: 2.Verify that a ÷ (b+c)≠(a ÷ b)+(a ÷ c) for each of the following values of a,b and c.

(a) a = 12, b = –4, c = 2

(b) a = (–10), b = 1, c = 1

Answer :

(a) Given: a ÷ (b + c)≠(a ÷ b) + (a ÷ c)

a = 12, b = –4, c = 2

Putting the given values in L.H.S. = 12 ÷ (–4 + 2)

= 12 ÷ (–2) = 12 ÷ (^-1⁄₂) = ^-12⁄₂ = –6

Putting the given values in R.H.S. = [12 ÷ (–4)] + (12 ÷ 2)

= (12 × –14) + 6 = –3 + 6 = 3

Since, L.H.S. ≠ R.H.S.

Hence verified.

(b) Given: a ÷ (b + c)≠(a ÷ b) + (a ÷ c)

a = –10, b = 1, c = 1

Putting the given values in L.H.S. = –10 ÷ (1 + 1)

= –10 ÷ (2) = –5

Putting the given values in R.H.S. = [–10 ÷ 1] + (–10 ÷ 1)

= –10 – 10 = –20

Since, L.H.S. ≠ R.H.S.

Hence verified.

Question: 3.Fill in the blanks:

(a) 369 ÷ _______ = 369

(b) (–75) ÷ _______ = (–1)

(c) (−206) ÷ _______ = 1

(d) (–87) ÷ _______ = 87

(e) _______ ÷ 1 = –87

(f) _______ ÷ 48 = –1

(g) 20 ÷ _______ = –2

(h) _______ ÷ (4) = –3

Answer :

(a) 369 ÷ 1 = 369

(b) (–75) ÷ 75 = (–1)

(c) (–206) ÷ (–206) = 1

(d) (–87) ÷ (–1) = 87

(e) (–87) ÷ 1 = –87

(f) (–48) ÷ 48 = –1

(g) 20 ÷ (–10) = –2

(h) (–12) ÷ (4) = –3

Question: 4.Write five pairs of integers (a,b) such that a ÷ b=–3. One such pair is (6,–2) because 6 ÷ (–2)=(–3).

Answer :

(i) (–6) ÷ 2 = –3

(ii) 9 ÷ (–3) = –3

(iii) 12 ÷ (–4) = –3

(iv) (–9) ÷ 3 = –3

(v) (–15) ÷ 5 = –3

Question: 5.The temperature at noon was 10^oC above zero. If it decreases at the rate of 2^oC per hour until mid-night, at what time would the temperature be 8^oC below zero? What would be the temperature at mid-night?

Answer :

Following number line is representing the temperature:

The temperature decreases 2^oC = 1 hour

The temperature decreases 1^oC = 12 hour

The temperature decreases 18^oC = 12 × 18 = 9 hours

Total time = 12 noon + 9 hours = 21 hours = 9 pm

Thus, at 9 pm the temperature would be 8^oC below 0^oC.

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 incorrect it?

Answer :

(i) Marks given for one correct answer = 3

Marks given for 12 correct answers = 3 x 12 = 36

Radhika scored 20 marks.

Therefore, Marks obtained for incorrect answers = 20 – 36 = –16

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = (–16) ÷ (–2)=8

Thus, Radhika has attempted 8 incorrect Questions.

(ii) Marks given for seven correct answers = 3 x 7 = 21

Mohini scores = –5

Marks obtained for incorrect answers = –5 – 21 = –26

Now, marks given for one incorrect answer = –2

Therefore, number of incorrect answers = (–26) ÷ (–2)=13

Thus, Mohini has attempted 13 incorrect Questions.