GSEB Class 7 Maths Solutions Chapter 1 Integers Exercise 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
(i) [(-6) + 5] ÷ [(- 2) + 1]
Answer:
(a) \( (-30) \div 10 = \frac {-30 }{ 10 } = -3 \)
(b) \( 50 \div (-5) = \frac {50}{ (- 5) } = -10 \)
(c) \( (-36) \div (-9) = \frac {-36 }{ (-9) } = 4 \)
(d) \( (-49) \div 49 = \frac {(-49) }{ 49 } = -1 \)
(e) \( 13 \div [(-2) + 1] = 13 \div [-1] = \frac { 13 }{ (-1) } = -13 \)
(f) \( 0 \div (-12) = 0 \)
(g) \( (-31) \div [(-30) + (-1)] = (-31) \div (-31) = \frac { -31 }{ (-31) } = 1 \)
(h) \( [(-36) \div 12] \div 3 = [\frac { (-36) }{ 12 }] \div 3 = [-3] \div 3 = -1 \)
(i) \( [(-6) + 5] \div [(-2) + 1] = [-1] \div [-1] = \frac { (-1) }{ (-1) } = 1 \)
In simple words: For each part, we are performing division with integers. Remember that dividing a negative number by a positive one gives a negative result, a positive by a negative gives a negative, and two negatives divided by each other give a positive result. Dividing zero by any non-zero number always results in zero.

Exam Tip: Always pay attention to the signs of the integers when performing division. Simplify any expressions inside brackets first, following the order of operations.

Question 2. Verify that \( a \div (b + c) \neq (a \div b) + (a \div c) \) 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 that \( a = 12 \), \( b = -4 \) and \( c = 2 \)
L.H.S. \( = a \div (b + c) = 12 \div (-4 + 2) \)
\( = 12 \div (-2) = -6 \)
R.H.S. \( = (a \div b) + (a \div c) \)
\( = [12 \div (-4)] + (12 \div 2) \)
\( = [-3] + 6 = 3 \)
Since \( -6 \neq 3 \)
Thus, L.H.S. \( \neq \) R.H.S.
Hence, \( a \div (b + c) \neq (a \div b) + (a \div c) \)

(b) Given that \( a = -10 \), \( b = 1 \), \( c = 1 \)
L.H.S. \( = a \div (b + c) \)
\( = (-10) \div (1 + 1) = (-10) \div 2 = -5 \)
R.H.S. \( = (a \div b) + (a \div c) \)
\( = [(-10) \div 1] + [(-10) \div 1] \)
\( = (-10) + (-10) = -20 \)
Since \( -5 \neq -20 \)
Thus, L.H.S. \( \neq \) R.H.S.
Hence, \( a \div (b + c) \neq (a \div b) + (a \div c) \)
In simple words: This problem asks us to check if the distributive property works for division over addition. We calculate both sides of the inequality: \( a \div (b + c) \) and \( (a \div b) + (a \div c) \). The results prove that division is not distributive over addition for these values.

Exam Tip: When verifying inequalities, always calculate both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) separately and then compare the final values. Be careful with integer operations.

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) \( 369 \div 1 = 369 \)
(b) \( (-75) \div 75 = -1 \)
(c) \( (-206) \div (-206) = 1 \)
(d) \( -87 \div (-1) = 87 \)
(e) \( -87 \div 1 = -87 \)
(f) \( -48 \div 48 = -1 \)
(g) \( 20 \div (-10) = -2 \)
(h) \( -12 \div 4 = -3 \)
In simple words: To fill in these blanks, you need to use the rules of integer division. Any number divided by 1 is itself. Any number divided by itself is 1. A number divided by its negative is -1. Also, a positive number divided by a negative number gives a negative result, and a negative number divided by a positive number gives a negative result.

Exam Tip: Remember the basic properties of division: dividing any number by 1 yields the number itself, and dividing any number by itself yields 1. Pay close attention to the signs to determine the correct missing integer.

Question 4. Write five pairs of integers (a, b) such that \( a \div b = -3 \). One such pair is (6, -2) because \( 6 \div (-2) = -3 \).
Answer:
I. Since, \( 3 \div (-1) = -3 \). Comparing it with \( a \div b = -3 \), we have \( a = 3 \) and \( b = (-1) \). The required pair of integers is \( (3, -1) \).
II. Since, \( -3 \div 1 = -3 \). Comparing it with \( a \div b = -3 \), we have \( a = (-3) \) and \( b = 1 \). The required pair of integers is \( (-3, 1) \).
III. Since, \( 9 \div (-3) = -3 \). Comparing it with \( a \div b = -3 \), we have \( a = 9 \) and \( b = -3 \). The required pair of integers is \( (9, -3) \).
IV. Since, \( -12 \div 4 = -3 \). Comparing it with \( a \div b = -3 \), we have \( a = -12 \) and \( b = 4 \). The required pair of integers is \( [-12, 4] \).
V. Since, \( 12 \div (-4) = -3 \). Comparing it with \( a \div b = -3 \), we have \( a = 12 \) and \( b = -4 \). The required pair of integers is \( [12, (-4)] \).
In simple words: We need to find pairs of whole numbers (a, b) where dividing 'a' by 'b' always gives us -3. We can achieve this by making one number positive and the other negative. For example, a positive number divided by a negative number results in a negative quotient.

Exam Tip: To find pairs (a, b) where \( a \div b = -3 \), choose an integer for 'b' and then multiply it by -3 to get 'a'. Always ensure that 'b' is not zero, as division by zero is undefined.

Question 5. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour, until mid-night, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
Answer:
Temperature at 12 noon \( = +10^\circ C \)
Rate of change in temperature \( = -2^\circ C \) per hour
Number of hours from 12 noon to mid-night \( = 12 \)
Change in temperature in 12 hours \( = 12^\circ \times (-2) = -24^\circ C \)
Temperature at mid-night (i.e., 12 hours after 12 noon) \( = +10^\circ C + (-24^\circ C) = -14^\circ C \)
Thus, temperature at mid-night \( = -14^\circ C \)
Now, temperature difference between \( +10^\circ C \) and \( -8^\circ C = +10^\circ C - (-8^\circ C) = 18^\circ C \)
Time taken for a temperature change of \( 18^\circ C = \frac {18}{ 2 } = 9 \) hours.
Therefore, temperature change of \( 18^\circ C \) will take place in 9 hours from 12 noon.
Time after 9 hours from 12 noon \( = 9 \text{ p.m.} \)
Thus, the temperature \( 8^\circ C \) below \( 0^\circ \) (i.e., \( -8^\circ C \)) would be at 9 p.m.
In simple words: We start with 10 degrees Celsius at noon. The temperature goes down by 2 degrees every hour. We found that the temperature will drop to -8 degrees Celsius after 9 hours, which means it will be 9 p.m. Also, by midnight, the temperature will be -14 degrees Celsius.

Exam Tip: When dealing with temperature problems, define positive values for "above zero" and negative values for "below zero." Clearly calculate the total temperature change over time and the total difference between two temperatures.

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?
Answer:
Marks for every correct answer \( = +3 \)
Marks for every incorrect answer \( = -2 \)

(i) Total marks scored by Radhika \( = 20 \)
Marks scored for correct answers \( = 12 \times 3 = 36 \)
Marks given for incorrect answers \( = 20 - 36 = -16 \)
Number of incorrect answers \( = (-16) \div (-2) = 8 \)
Thus, Radhika attempted 8 questions incorrectly.

(ii) Marks obtained by Mohini \( = -5 \)
Marks obtained for 7 correct answers \( = 7 \times 3 = 21 \)
Marks obtained for incorrect answers \( = -5 - 21 = -26 \)
Number of incorrect answers \( = -26 \div (-2) = 13 \)
Thus, Mohini attempted 13 questions incorrectly.
In simple words: This problem involves calculating marks in a test where right answers get positive marks and wrong answers get negative marks. We use the total marks and marks from correct answers to figure out the marks lost due to incorrect answers, and then calculate how many questions were answered incorrectly.

Exam Tip: Clearly list the marks for correct and incorrect answers. When solving such problems, always subtract the marks obtained from correct answers from the total score to find marks for incorrect answers, then divide by the mark for one incorrect answer.

Question 7. An elevator descends into a mine 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?
Answer:
Present position of the elevator is at 10 m above the ground level.
Distance to be moved by the elevator below the ground level \( = 350 \text{ m} \)
Total distance to be moved by the elevator \( = 350 \text{ m} + 10 \text{ m} = 360 \text{ m} \)
The rate of descent \( = 6 \text{ m/min} \)
Time taken \( = \frac{\text{Total Distance}}{\text{Rate of descent}} \)
\( = \frac{360 \text{ m}}{6 \text{ m/min}} = 60 \text{ min} \)
Required time \( = 60 \) minutes (or one hour).
In simple words: The elevator starts 10 meters above the ground and needs to go down to 350 meters below the ground. So, it has to cover a total distance of 360 meters. Since it moves at 6 meters per minute, it will take 60 minutes, or one hour, to reach the bottom.

Exam Tip: When calculating total distance for vertical movement, remember to add the initial height above ground to the depth below ground. Always use the formula: Time = Distance / Speed.