class 7 maths chapter 1 exercise 1.4 solutions

simple solutions for class 7 maths chapter 1 exercise 1.4 solutions are given here in the post. This class 7 maths chapter 1 exercise 1.4 solutions contains topics related to the raw data collection and its organisation. so we definitely want students of Class 7 to solve class 7 maths chapter 1 exercise 1.4 solutions to empower their basics and test the students capability of understanding the concepts. It also helps the students of CBSE Class 7 Maths students

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) \((-30) ÷ 10 = \displaystyle{-30 \over 10} = -3 \)

(b) \(50 ÷ (-5) = \displaystyle{50 \over -5} = -10\)

(c) \((-36) ÷ (-9) = \displaystyle{-36 \over -9} = 4\)

(d) \((-49) ÷ (49) =\displaystyle{-49 \over 49} = -1\)

(e) \(13 ÷ [(-2) + 1] \)

        \(= 13 ÷ -1 \)

        \(= \dfrac{13 }{ -2} \)

        \(= -13\)

(f) \(0 ÷ (-12) =  \dfrac{0 }{ -12}= 0\)

(g) \((-31) ÷ [(-30) + (-1)] \)

        \(= (-31) ÷ (-31) \)

        \(= \dfrac{-31 }{ -31}\)

        \(= 1\)

(h) \([(-36) + 12] ÷ 3 \)

        \(= [\dfrac{- 36}{ 12}] ÷ 3\)

        \( = -3 ÷ 3\)

        \(= \dfrac{- 3}{ 3}\)

        \(=-1\)

(i) \([(-6) + 5] + [(-2) + 1] \)

        \(= (-1) ÷ (-1) \)

        \(= \dfrac{- }{ -1} \)

        \(= 1\)

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) \(a = 12, 6 = – 4, c = 2\)

\(a ÷ (b + c) = 12 ÷ [(-4) + 2] \)

            \(= 12 ÷ (-2) \)

            \(=\dfrac{12}{-2}\)

            \( = -6\)

\((a ÷ b) + (a ÷ c) = [12 ÷ (-4)] + [12 ÷ 2] \)

                        \(= \dfrac{12}{ -4} +\dfrac{12}{ 2}\)

                        \( = -3+6 = 3\)

Since, \((-6) + 3\)

Hence, \(a ÷ (b + c) + (a ÷ b) + (a ÷ c)\)

(b) \(a = (-10), b = 1, c = 1\)

\(a ÷ (b + c) = (-10) ÷ (1 + 1) \)

                    \(= (-10) ÷ 2 \)

                    \(= \dfrac{-10}{ 2} \)

                    \(= -5\)

\((a ÷ b) + (a ÷ c)=[(-10) ÷ 1] + [(-10) ÷ 1]\)

\(=\displaystyle{(-10)\over 1}+\dfrac{-10}{ 1}\)

\(= (-10) + (-10) = -20\)

Since \((-5) \neq (-20)\)

Hence, \(a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)\)

Question 3.

Fill in the blanks:

(a) 369 ÷ ___ = 369

(b)  (-75) ÷ ___ = -1

(c) (-206) ÷ ___ =1

(d) -87 ÷ ___ = -87

(e) ___ ÷ 1 = -87

(g) 20 ÷ ___ = -2

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

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

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

= \(+10°\text{C} + (-24°\text{C}) = -14°\text{C}\)

Hence, the required temperature at midnight =\(-14°\text{C}\)

Difference in temperature between + 10°C and -8°C

= +10° C– (-8°C ) = +10°C + 8°C = 18°C

Number of hours required = \(\displaystyle{18°\text{C} \over  2°\text{C}}\)= 9 hours

∴ Time after \(9\) hours from \(12\) noon = \(9\) pm.

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

=\((-16) \div(-2)=\dfrac{(-16)}{(-2)}=8\)

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.

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 elevator = 350 m + 10 m 

                                                            = 360 m

Rate of descent = 6 m/min.

Total time taken by the elevator = \( \dfrac{360}{ 6}\)

                                                    = 60 minutes

                                                     = 1 hour

Hence, the required time = 1 hour