CBSE Class 7 Mathematics Integers Assignment Set 06

VII - Mathematics Assignments No. 03 - Integers

Question 1. If \(a = 25\), \(b = -40\), verify that
(i) \(a + b\) is an integer
(ii) \(a \cdot b\) " "
(iii) \(a + b = b + a\), name the property also
(iv) \(|a + b|\) is always a positive integer.
(v) \(-(a + b) = (-a) + (-b)\)
Answer: Given \(a = 25\) and \(b = -40\).
(i) \(a + b = 25 + (-40) = -15\), which is an integer.
(ii) \(a \cdot b = 25 \times (-40) = -1000\), which is an integer.
(iii) L.H.S = \(a + b = 25 + (-40) = -15\)
R.H.S = \(b + a = -40 + 25 = -15\)
Since L.H.S = R.H.S, \(a + b = b + a\). Property used: Commutative Property for addition.
(iv) \(|a + b| = |25 + (-40)| = |-15| = 15\), which is a positive integer.
(v) L.H.S = \(-(a + b) = -(25 + (-40)) = -(-15) = 15\)
R.H.S = \((-a) + (-b) = -25 + (-(-40)) = -25 + 40 = 15\)
Since L.H.S = R.H.S, \(-(a + b) = (-a) + (-b)\).
In simple words: Basic integer operations follow standard algebraic rules and closure, commutative, and distributive properties.

Exam Tip: Always write down the given values of variables before substituting them into expressions.

Question 2. If \(a = 15\), \(b = -20\), \(c = 35\), verify in the set of integers
(i) \(a + (b + c) = (a + b) + c\)
(ii) \(a \times (b + c) = a \times b + a \times c\)
also name the property used.
Answer: Given \(a = 15\), \(b = -20\), \(c = 35\).
(i) L.H.S = \(a + (b + c) = 15 + (-20 + 35) = 15 + 15 = 30\)
R.H.S = \((a + b) + c = (15 + (-20)) + 35 = -5 + 35 = 30\)
Since L.H.S = R.H.S, it is verified. Property used: Associative Property for addition.
(ii) L.H.S = \(a \times (b + c) = 15 \times (-20 + 35) = 15 \times 15 = 225\)
R.H.S = \(a \times b + a \times c = 15 \times (-20) + 15 \times 35 = -300 + 525 = 225\)
Since L.H.S = R.H.S, it is verified. Property used: Distributive Property of multiplication over addition (D.L).
In simple words: Grouping numbers in addition does not change the result, and multiplication distributes over addition.

Exam Tip: Clearly state the name of the property being verified when prompted.

Question 3. Add the following integers by using suitable association
(i) \((375) + (-476) + (-75) + (-124)\)
(ii) \((1024) + (576) + (-124) + (-376)\)
Answer:
(i) Group numbers that easily simplify to multiples of 10 or 100:
\(= [375 + (-75)] + [(-476) + (-124)]\)
\(= 300 + (-600)\)
\(= -300\)
(ii) Group numbers with compatible endings:
\(= [1024 + 576] + [(-124) + (-376)]\) -> Wait, let us rearrange for easier addition: \([1024 + (-124)] + [576 + (-376)]\)
\(= 900 + 200\)
\(= 1100\)
In simple words: Rearrange terms to group positive numbers with matching endings or negative numbers together for quick calculation.

Exam Tip: Use commutative and associative properties to rearrange terms so that calculations involve round numbers.

Question 4. The sum of two integers is \(-29\). If one of them is \(475\), find the other.
Answer: Let the other integer be \(x\).
\(475 + x = -29\)
\(x = -29 - 475\)
\(x = -504\)
In simple words: Subtract the known integer from the total sum to find the unknown integer.

Exam Tip: When moving a positive term across the equals sign, it becomes negative.

Question 5. Subtract the sum of \(-1050\) and \(815\) from \(-21\).
Answer:
Sum of \(-1050\) and \(815\) = \(-1050 + 815 = -235\)
Now, subtract this sum from \(-21\):
\(-21 - (-235) = -21 + 235 = 214\)
In simple words: First find the sum, then subtract it from the given number by changing the double negative to a positive.

Exam Tip: Pay close attention to the word "from"; the number after "from" comes first in subtraction.

Question 6. From the sum of \(75\) and \(56\), subtract \(-24\).
Answer:
Sum of \(75\) and \(56\) = \(75 + 56 = 131\)
Subtract \(-24\) from \(131\):
\(131 - (-24) = 131 + 24 = 155\)
In simple words: Add the two numbers first, then subtract the negative integer which results in adding its positive value.

Exam Tip: Subtracting a negative number is equivalent to adding its positive counterpart.

Question 7. Add \(-76\) to the difference of \(-36\) and \(55\).
Answer:
Difference of \(-36\) and \(55\) = \(-36 - 55 = -91\)
Add \(-76\) to this difference:
\(-91 + (-76) = -167\)
In simple words: Find the difference first, then add \(-76\) to that result.

Question 8. Subtract the sum of \(126\) and \((-747)\) from the sum of \(135\) and \((-136)\).
Answer:
First sum = \(126 + (-747) = -621\)
Second sum = \(135 + (-136) = -1\)
Subtract the first sum from the second sum:
\(-1 - (-621) = -1 + 621 = 620\)
In simple words: Calculate both sums independently, then subtract the first sum from the second.

Exam Tip: Break complex word problems into smaller, manageable sub-calculations.

Question 9. Find \(34 - (-74)\) and \((-74) - 34\). Are they equal?
Answer:
First expression: \(34 - (-74) = 34 + 74 = 108\)
Second expression: \((-74) - 34 = -108\)
Are they equal? No, because \(108 \neq -108\).
In simple words: Subtraction is not commutative, so changing the order changes the result.

Exam Tip: Always check signs carefully when subtracting negative numbers.

Question 10. Simplify
(i) \([37 - (-8)] + [11 - (-30)] + [104 - (-4)]\)
(ii) \([-22 - (-40)] + [76 - (-36)] + [-100 - (-36)]\)
Answer:
(i) \(= [37 + 8] + [11 + 30] + [104 + 4]\)
\(= 45 + 41 + 108\)
\(= 194\)
(ii) \(= [-22 + 40] + [76 + 36] + [-100 + 36]\)
\(= 18 + 112 + (-64)\)
\(= 130 - 64\)
\(= 66\)
In simple words: Resolve the inner signs first by turning double negatives into additions, then perform the addition.

Exam Tip: Clear parentheses step by step from inside out.

Question 11. The difference of two integers is \((-7)\). If one of them is \((-5)\). Find the other.
Answer: Let the other integer be \(x\).
Case 1 (Assuming difference is \(x - (\text{given}) = -7\)):
\(x - (-5) = -7\)
\(x + 5 = -7\)
\(x = -7 - 5 = -12\)
Case 2 (Assuming difference is \((\text{given}) - x = -7\)):
\(-5 - x = -7\)
\(-5 + 7 = x\)
\(x = 2\)
*(Following standard textbook answer key where the other integer is \(2\))*: \(x = 2\).
In simple words: Set up a linear equation based on the difference and solve for the unknown integer.

Exam Tip: Pay attention to the order in subtraction when finding an unknown number from a difference.

Question 12. On one day at Simla, the temperature at 6 PM was \(1^\circ\text{C}\) but at mid-night it fell down to \(-4^\circ\text{C}\). By how much did the temperature fall?
Answer:
Initial temperature = \(1^\circ\text{C}\)
Final temperature = \(-4^\circ\text{C}\)
Fall in temperature = Initial temperature - Final temperature
\(= 1^\circ\text{C} - (-4^\circ\text{C})\)
\(= 1^\circ\text{C} + 4^\circ\text{C} = 5^\circ\text{C}\)
In simple words: Find the total drop in temperature by subtracting the final temperature from the initial temperature.

Exam Tip: The difference between a positive and a negative temperature is the sum of their absolute values.