GSEB Class 7 Maths Solutions Chapter 1 પૂર્ણાંક સંખ્યાઓ Exercise 1.4

Question 1. નીચે આપેલ દરેકના જવાબ લખોઃ
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) \( 0 \div (-12) \)
\( = \frac { 0 }{ (-12) } \)
\( = 0 \)
(g) \( (-31) \div [(-30) + (-1)] \)
\( = (-31) \div (-31) \)
\( = \frac { (-31) }{ (-31) } \)
\( = 1 \)
(h) \( [(-36) \div 12] \div 3 \)
\( = [\frac { (-36) }{ 12 }] \div 3 \)
\( = \frac { (-3) }{ 3 } \)
\( = (-1) \)
(i) \( [(-6) + 5] \div [(-2) + 1] \)
\( = [(-1) \div (-1)] \)
\( = \frac { (-1) }{ (-1) } \)
\( = 1 \)
In simple words: To find the answer, just divide the numbers as shown in each part. Remember that a negative number divided by a positive number gives a negative result, and a negative number divided by a negative number gives a positive result.

Exam Tip: Pay close attention to the signs of the numbers (positive or negative) when performing division, as this will determine the sign of your answer. Also, follow the order of operations (PEMDAS/BODMAS) when brackets are involved.

Question 2. નીચેના દરેક a, b અને c ની કિંમતો માટે, \( a \div (b + c) \neq (a \div b) + (a \div c) \) ને ચકાસો:
(a) \( a = 12, b = (-4), c = 2 \)
(b) \( a = (-10), b = 1, c = 1 \)
Answer:
(a) Given values are \( a = 12, b = (-4) \), and \( c = 2 \). We will verify \( a \div (b + c) \neq (a \div b) + (a \div c) \).
LHS (Left Hand Side) \( = a \div (b + c) \)
\( = 12 \div (-4 + 2) \)
\( = 12 \div (-2) \)
\( = (-6) \)
RHS (Right Hand Side) \( = (a \div b) + (a \div c) \)
\( = [12 \div (-4)] + [12 \div 2] \)
\( = (-3) + 6 \)
\( = 3 \)
Since \( (-6) \neq 3 \), we can see that LHS is not equal to RHS.
Therefore, \( a \div (b + c) \neq (a \div b) + (a \div c) \) is verified.

(b) Given values are \( a = (-10), b = 1 \), and \( c = 1 \). We will verify \( a \div (b + c) \neq (a \div b) + (a \div c) \).
LHS \( = a \div (b + c) \)
\( = (-10) \div (1 + 1) \)
\( = (-10) \div 2 \)
\( = (-5) \)
RHS \( = (a \div b) + (a \div c) \)
\( = [(-10) \div 1] + [(-10) \div 1] \)
\( = (-10) + (-10) \)
\( = (-20) \)
Since \( (-5) \neq (-20) \), we can see that LHS is not equal to RHS.
Therefore, \( a \div (b + c) \neq (a \div b) + (a \div c) \) is verified.
In simple words: We are checking if division is distributive over addition for integers. This means we check if dividing by a sum is the same as dividing by each part and then adding. The calculations show that it is not the same.

Exam Tip: Always calculate both sides of the inequality (LHS and RHS) separately before comparing them. Ensure all calculations within brackets are performed first, following the order of operations.

Question 3. ખાલી જગ્યા પૂરોઃ
(a) \( 369 \div ...... = 369 \)
(b) \( (-75) \div ...... = (-1) \)
(c) \( (-206) \div ...... = 1 \)
(d) \( (-87) \div ...... = 87 \)
(e) \( ...... \div 1 = (-87) \)
(f) \( ...... \div 18 = (-1) \)
(g) \( 20 \div ...... = (-2) \)
(h) \( ...... \div 4 = (-3) \)
Answer:
(a) To get 369 when dividing 369, the blank must be 1. So, \( 369 \div 1 = 369 \).
(b) To get (-1) when dividing (-75), the blank must be 75. So, \( (-75) \div 75 = (-1) \).
(c) To get 1 when dividing (-206), the blank must be (-206). So, \( (-206) \div (-206) = 1 \).
(d) To get 87 when dividing (-87), the blank must be (-1). So, \( (-87) \div (-1) = 87 \).
(e) To get (-87) when dividing by 1, the blank must be (-87). So, \( (-87) \div 1 = (-87) \).
(f) To get (-1) when dividing by 18, the blank must be (-18). So, \( (-18) \div 18 = (-1) \). The source shows -48/48, which is also -1, but to make it simpler and match the question style, (-18) is more direct. If the source had (-48) in the question, then (-48) would be correct.
(g) To get (-2) when dividing 20, the blank must be (-10). So, \( 20 \div (-10) = (-2) \).
(h) To get (-3) when dividing by 4, the blank must be (-12). So, \( (-12) \div 4 = (-3) \).
In simple words: We fill in the missing numbers using the rules of division for integers. If a number divided by itself (or its negative) gives 1 or -1, or if a number divided by 1 gives itself, these principles help us solve.

Exam Tip: Remember the division rules for signs: positive ÷ positive = positive; negative ÷ negative = positive; positive ÷ negative = negative; negative ÷ positive = negative. This will help you determine the correct sign for the missing number.

Question 4. પૂર્ણાંક સંખ્યા (a, b)ની પાંચ જોડ લખો જેથી, \( a \div b = (-3) \) થાય. આવી એક જોડ \( (6, -2) \) છે કારણ કે \( 6 \div (-2) = (-3) \).
Answer: We need to find five pairs of integers (a, b) such that when 'a' is divided by 'b', the result is (-3). This implies that 'a' must be a multiple of 'b', and their signs should be opposite to yield a negative result. Also, the absolute value of 'a' must be three times the absolute value of 'b'.
Here are five such pairs:
(i) If we compare \( a \div b = (-3) \) with \( 3 \div (-1) = (-3) \), then \( a = 3 \) and \( b = (-1) \). So, a pair is \( (3, -1) \).
(ii) We know that \( 12 \div 4 = 3 \). Thus, \( 12 \div (-4) = (-3) \). So, \( a = 12 \) and \( b = (-4) \). Another pair is \( (12, -4) \).
(iii) We know that \( 15 \div 5 = 3 \). Thus, \( 15 \div (-5) = (-3) \). So, \( a = 15 \) and \( b = (-5) \). Another pair is \( (15, -5) \).
(iv) Following the same pattern, a pair could be \( (18, -6) \), since \( 18 \div (-6) = (-3) \).
(v) Following the same pattern, a pair could be \( (21, -7) \), since \( 21 \div (-7) = (-3) \).
In simple words: We are looking for numbers that, when divided, give -3. This means one number must be positive, and the other negative. Also, the first number should be three times bigger than the second one (ignoring the negative sign). We just list five such pairs.

Exam Tip: To find such pairs, pick any integer for 'b' (except 0), and then multiply it by -3 to get 'a'. This method ensures \( a \div b = -3 \). For example, if \( b = 7 \), then \( a = (-3) \times 7 = -21 \), so \( (-21, 7) \) is a valid pair.

Question 5. બપોરે 12 વાગ્યાનું તાપમાન શૂન્યથી ઉપર \( 10 \text{°C} \) છે. જો એ \( 2 \text{°C} \) પ્રતિ કલાકના દરે મધ્યરાત્રિ સુધી ઓછું થતું જાય, તો કયા સમયે તાપમાન શૂન્યથી નીચે \( 8 \text{°C} \) હોય? મધ્યરાત્રિનું તાપમાન શું હોય?
Answer: The temperature at 12 noon is \( +10 \text{°C} \).
The temperature decrease rate is \( 2 \text{°C} \) per hour.
We need to find when the temperature will be \( -8 \text{°C} \).
The total change in temperature needed is from \( +10 \text{°C} \) to \( -8 \text{°C} \).
Temperature difference \( = 10 \text{°C} - (-8 \text{°C}) = 10 \text{°C} + 8 \text{°C} = 18 \text{°C} \).
Time taken to drop \( 18 \text{°C} = \frac { 18 \text{°C} }{ 2 \text{°C/hour} } = 9 \text{ hours} \).
So, 9 hours after 12 noon (12:00 PM), the time will be 9:00 PM. At 9:00 PM, the temperature will be \( -8 \text{°C} \).

Now, we need to find the temperature at midnight (12:00 AM).
The total hours from 12 noon to midnight is 12 hours.
Total decrease in temperature over 12 hours \( = 12 \text{ hours} \times (-2 \text{°C/hour}) = -24 \text{°C} \).
Temperature at midnight \( = \text{Initial temperature} + \text{Total decrease} \)
\( = 10 \text{°C} + (-24 \text{°C}) \)
\( = (10 - 24) \text{°C} \)
\( = -14 \text{°C} \).
So, the temperature at midnight will be \( -14 \text{°C} \).
In simple words: The temperature starts at 10 degrees positive and drops 2 degrees every hour. It will reach 8 degrees below zero after 9 hours, which is 9 PM. By midnight, which is 12 hours later, the temperature will be 14 degrees below zero.

Exam Tip: Clearly define the starting temperature and the rate of change. When calculating the time to reach a target temperature, determine the total temperature drop and divide by the hourly rate. For midnight temperature, calculate the total drop over the full duration.

Question 6. વર્ગપરીક્ષામાં \( (+ 3) \) ગુણ દરેક સાચા જવાબ માટે અને \( (-2) \) ગુણ દરેક ખોટા જવાબ માટે આપવામાં આવે છે અને કોઈ પણ સવાલના જવાબ માટે જો પ્રયત્ન ન કરવામાં આવે, તો તેનો એક પણ ગુણ આપવામાં આવતો નથી.
(i) રાધિકાએ કુલ 20 ગુણ મેળવ્યા. જો તેણે 12 સાચા જવાબો આપ્યા હોય, તો તેણે કેટલા ખોટા જવાબો લખ્યા?
(ii) મોહિનીએ આ પરીક્ષામાં \( (-5) \) ગુણ મેળવ્યા. જોકે તેના 7 સાચા જવાબો હતા, તો તેણે કેટલા ખોટા જવાબો લખ્યા?
Answer:
Marks for each correct answer \( = +3 \)
Marks for each incorrect answer \( = -2 \)
Marks for not attempting a question \( = 0 \)

(i) For Radhika:
Total marks obtained by Radhika \( = 20 \)
Number of correct answers given by Radhika \( = 12 \)
Marks obtained for correct answers \( = 12 \times 3 = 36 \)
Marks obtained for incorrect answers \( = \text{Total marks} - \text{Marks for correct answers} \)
\( = 20 - 36 \)
\( = (-16) \)
Number of incorrect answers \( = \frac { \text{Marks for incorrect answers} }{ \text{Marks for each incorrect answer} } \)
\( = \frac { (-16) }{ (-2) } \)
\( = 8 \)
So, Radhika answered 8 questions incorrectly.

(ii) For Mohini:
Total marks obtained by Mohini \( = (-5) \)
Number of correct answers given by Mohini \( = 7 \)
Marks obtained for correct answers \( = 7 \times 3 = 21 \)
Marks obtained for incorrect answers \( = \text{Total marks} - \text{Marks for correct answers} \)
\( = (-5) - 21 \)
\( = (-26) \)
Number of incorrect answers \( = \frac { \text{Marks for incorrect answers} }{ \text{Marks for each incorrect answer} } \)
\( = \frac { (-26) }{ (-2) } \)
\( = 13 \)
So, Mohini answered 13 questions incorrectly.
In simple words: For both students, we first calculated the marks they got for their correct answers. Then, we subtracted this from their total score to find out how many marks they lost due to wrong answers. Finally, dividing the lost marks by -2 gave us the number of incorrect answers.

Exam Tip: Be careful with positive and negative signs when calculating marks. Always calculate marks for correct answers first, then find the difference to determine marks from incorrect answers, and finally divide by the mark per incorrect answer to get the count.

Question 7. એક લિફ્ટ (એલિવેટર) 6 મીટર પ્રતિ મિનિટના દરે ખાણમાં ઊતરે છે. જો તે જમીનથી 10 મીટર ઉપરથી નીચે ઊતરતી હોય, તો \( (-350) \) મીટર સુધી પહોંચતાં તેને કેટલો સમય લાગશે?
Answer: The lift's descent rate into a mine is 6 meters per minute.
The lift starts 10 meters above the ground level. We can represent this as \( +10 \) m.
The target depth the lift needs to reach is \( (-350) \) meters below ground level.
The total distance the lift needs to cover is from \( +10 \) m to \( -350 \) m.
Total distance \( = \text{Starting height} - \text{Final depth} \)
\( = 10 - (-350) \)
\( = 10 + 350 \)
\( = 360 \text{ meters} \).
Time taken to cover this distance \( = \frac { \text{Total distance} }{ \text{Speed} } \)
\( = \frac { 360 \text{ meters} }{ 6 \text{ meters/minute} } \)
\( = 60 \text{ minutes} \).
Since 60 minutes is equal to 1 hour, the lift will take 1 hour to reach \( (-350) \) meters.
In simple words: The lift starts 10 meters above the ground and needs to go down to 350 meters below the ground. This means it has to travel a total of 360 meters. Since it goes down 6 meters every minute, it will take 60 minutes (or 1 hour) to reach the bottom.

Exam Tip: When dealing with heights above and below ground, treat above ground as positive and below ground as negative. The total distance traveled is the absolute difference between the initial and final positions. Ensure units are consistent (meters and minutes in this case).