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

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

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

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

(e) $13÷[(-2)+1]$

        $=13÷-1$

        $={\displaystyle \frac{13}{-2}}$

        $=-13$

(f) $0÷(-12)={\displaystyle \frac{0}{-12}}=0$

(g) $(-31)÷[(-30)+(-1)]$

        $=(-31)÷(-31)$

        $={\displaystyle \frac{-31}{-31}}$

        $=1$

(h) $[(-36)+12]÷3$

        $=[{\displaystyle \frac{-36}{12}}]÷3$

        $=-3÷3$

        $={\displaystyle \frac{-3}{3}}$

        $=-1$

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

        $=(-1)÷(-1)$

        $={\displaystyle \frac{-}{-1}}$

        $=1$

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) $a=12,6=–4,c=2$

$a÷(b+c)=12÷[(-4)+2]$

            $=12÷(-2)$

            $={\displaystyle \frac{12}{-2}}$

            $=-6$

$(a÷b)+(a÷c)=[12÷(-4)]+[12÷2]$

                        $={\displaystyle \frac{12}{-4}}+{\displaystyle \frac{12}{2}}$

                        $=-3+6=3$

Since, $(-6)+3$

Hence, $a÷(b+c)+(a÷b)+(a÷c)$

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

$a÷(b+c)=(-10)÷(1+1)$

                    $=(-10)÷2$

                    $={\displaystyle \frac{-10}{2}}$

                    $=-5$

$(a÷b)+(a÷c)=[(-10)÷1]+[(-10)÷1]$

$={\displaystyle \frac{(-10)}{1}+{\displaystyle \frac{-10}{1}}}$

$=(-10)+(-10)=-20$

Since $(-5)\ne (-20)$

Hence, $a÷(b+c)\ne (a÷b)+(a÷c)$

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 = ${\displaystyle \frac{18°\text{C}}{2°\text{C}}}$= 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

=$(-16)÷(-2)={\displaystyle \frac{(-16)}{(-2)}}=8$

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 = ${\displaystyle \frac{360}{6}}$

                                                    = 60 minutes

                                                     = 1 hour

Hence, the required time = 1 hour