NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 2 Arithmetic Expressions

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Section 2.1: Simple Expressions

1. Example 1: Mallika spends ₹25 every day for lunch at school. Write the expression for the total amount she spends on lunch in a week from Monday to Friday.
See SolutionMallika buys lunch on 5 days (Monday to Friday). The cost each day is ₹25.
The expression for the total amount is: 5 x 25.

2. Choose your favourite number and write as many expressions as you can having that value.
See SolutionLet’s choose the number 20. Here are some expressions that have the value 20:
– 15 + 5
– 30 – 10
– 4 x 5
– 100 / 5
– 2 x (7 + 3)
– (6 x 5) – 10
– 18 + (10 / 5)

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Figure it Out
1. Fill in the blanks to make the expressions equal on both sides of the = sign:
(a) 13 + 4 = __ + 6
See SolutionLHS (Left Hand Side) = 13 + 4 = 17.
RHS (Right Hand Side) = __ + 6.
We need __ + 6 = 17.
The blank is 17 – 6 = 11.

(b) 22 + __ = 6 x 5
See SolutionRHS = 6 x 5 = 30.
LHS = 22 + __.
We need 22 + __ = 30.
The blank is 30 – 22 = 8.

(c) 8 x __ = 64 / 2
See SolutionRHS = 64 / 2 = 32.
LHS = 8 x __.
We need 8 x __ = 32.
The blank is 32 / 8 = 4.

(d) 34 – __ = 25
See SolutionWe need 34 – __ = 25.
The blank is 34 – 25 = 9.

2. Arrange the following expressions in ascending (increasing) order of their values.
(a) 67 – 19
(b) 67 – 20
(c) 35 + 25
(d) 5 x 11
(e) 120 / 3
See SolutionFirst, find the value of each expression:
(a) 67 – 19 = 48
(b) 67 – 20 = 47
(c) 35 + 25 = 60
(d) 5 x 11 = 55
(e) 120 / 3 = 40
Now arrange these values (48, 47, 60, 55, 40) in ascending order: 40, 47, 48, 55, 60.
The corresponding expressions in ascending order are:
120 / 3, 67 – 20, 67 – 19, 5 x 11, 35 + 25.

Example 2: Which is greater? 1023 + 125 or 1022 + 128?
See SolutionThe chapter explains this using a comparison:
– Start with 1023 vs 1022 (Raja has 1 more).
– Then compare additions: +125 vs +128 (Joy gets 3 more).
– Joy started 1 behind but gained 3 more, so Joy ends up with 2 more.
Therefore, 1022 + 128 is greater than 1023 + 125.
(Calculation: 1023 + 125 = 1148; 1022 + 128 = 1150. 1150 > 1148).

Example 3: Which is greater? 113 – 25 or 112 – 24?
See SolutionThe chapter explains this using a comparison:
– Start with 113 vs 112 (Raja has 1 more).
– Compare subtractions: -25 vs -24 (Raja loses 1 more than Joy).
– Raja started 1 ahead but also lost 1 more.
Therefore, they end up with the same amount. 113 – 25 is equal to 112 – 24.
(Calculation: 113 – 25 = 88; 112 – 24 = 88. They are equal).

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1. Use ‘>’, ‘<‘ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

(a) 245 + 289 ___ 246 + 285
See Solution>
Reasoning: Compare LHS (Left Hand Side) to RHS (Right Hand Side). 245 is 1 less than 246. 289 is 4 more than 285. So, the LHS starts 1 less but adds 4 more compared to the RHS. The net difference is +3 for the LHS. Therefore, LHS > RHS.

(b) 273 – 145 ___ 272 – 144
See Solution=
Reasoning: 273 is 1 more than 272. 145 is 1 more than 144 (so -145 means subtracting 1 more than -144). The LHS starts 1 higher but also subtracts 1 more compared to the RHS. The net difference is 0. Therefore, LHS = RHS.

(c) 364 + 587 ___ 363 + 589
See Solution<
Reasoning: 364 is 1 more than 363. 587 is 2 less than 589. The LHS starts 1 higher but adds 2 less compared to the RHS. The net difference is -1 for the LHS. Therefore, LHS < RHS.

(d) 124 + 245 ___ 129 + 245
See Solution<
Reasoning: The second number (+245) is the same on both sides. Comparing the first numbers, 124 is less than 129. Therefore, LHS < RHS.

(e) 213 – 77 ___ 214 – 76
See Solution<
Reasoning: 213 is 1 less than 214. 77 is 1 more than 76 (so -77 means subtracting 1 more than -76). The LHS starts 1 lower and also subtracts 1 more compared to the RHS. The net difference is -2 for the LHS. Therefore, LHS < RHS.

Section 2.2: Reading and Evaluating Complex Expressions

(This section discusses ambiguity in language and mathematical expressions without specific questions needing answers on this page, using the “Shalini sat next to a friend with toys” example).

Example 4: Mallesh brought 30 marbles to the playground. Arun brought 5 bags of marbles with 4 marbles in each bag. How many marbles did Mallesh and Arun bring to the playground? Mallesh summarized this by writing the mathematical expression – 30 + 5 x 4.
(30+5)x4 = 140. Mallesh evaluated it as 30+(5×4) = 50. Mallesh was right based on the context (30 marbles + 5 bags of 4 marbles). But why did Purna get it wrong?
See SolutionPurna got it wrong because the expression 30 + 5 x 4, written without context or brackets, is ambiguous. It wasn’t clear whether the addition (30+5) should be done first or the multiplication (5×4) should be done first. Without rules or brackets, different calculation orders lead to different results.

2. Explanation of Brackets: How are brackets used to clarify the order of operations in the expression 30 + 5 x 4 for the marbles example?
See SolutionTo ensure the multiplication (Arun’s marbles: 5 bags of 4) is done before adding Mallesh’s 30 marbles, brackets are used:
30 + (5 x 4)
This means you must first calculate the value inside the brackets (5 x 4 = 20) and then perform the addition (30 + 20 = 50).

Example 5: Irfan bought a pack of biscuits for ₹15 and a packet of toor dal for ₹56. He gave the shopkeeper ₹100. Write an expression that can help us calculate the change Irfan will get back from the shopkeeper.
See SolutionFirst, find the total cost: 15 + 56.
Then, subtract the total cost from the amount paid (₹100). To ensure the addition is done first, use brackets.
Expression for change: 100 – (15 + 56)
Value = 100 – (71) = 29. Irfan will get back ₹29.

Why is the expression 100 – 15 + 56 incorrect for calculating Irfan’s change?
See SolutionIf calculated from left to right without considering the context or implied grouping:
100 – 15 = 85
85 + 56 = 141
This result (₹141 change from ₹100 paid) doesn’t make sense in the real world. The expression needs brackets, 100 – (15 + 56), to correctly represent subtracting the *total* cost.

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Terms in Expressions

1. Explanation: Terms are parts separated by ‘+’. Subtraction is converted to adding the inverse (e.g., 83 – 14 becomes (83) + (-14)).

Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.
See SolutionYes, replacing subtraction with adding the inverse gives the same value.
Example 1: 10 – 3 = 7. Also, 10 + (-3) = 7.
Example 2: 5 – 8 = -3. Also, 5 + (-8) = -3.
Example 3: -2 – 4 = -6. Also, -2 + (-4) = -6.

Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
See SolutionIn the Token Model:
– Subtracting a positive number (e.g., subtracting 3) means removing 3 positive tokens.
– Adding a negative number (e.g., adding -3) means adding 3 negative tokens. These 3 negative tokens cancel out 3 existing positive tokens (by forming zero pairs), which is equivalent to removing 3 positive tokens.
Since both actions result in removing the same number of positive tokens (or adding the same number of negative tokens if starting below zero), the final value is the same.

Try This: Complete the table (identifying terms).
– Expression: 5 + 6 x 3
Expression as sum of terms: (5) + (6 x 3)
Terms: 5, (6×3)

– Expression: 4 + 15 – 9
Expression as sum of terms: (4) + (15) + (-9)
Terms: 4, 15, -9

– Expression: 23 – 2 x 4 + 16
Expression as sum of terms: (23) + (-2 x 4) + (16)
Terms: 23, (-2×4), 16

– Expression: 28 + 19 – 8
Expression as sum of terms: (28) + (19) + (-8)
Terms: 28, 19, -8

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Does changing the order in which the terms are added give different values?
See SolutionNo, changing the order in which terms are added does not change the final value. This is due to the commutative and associative properties of addition which apply to integers (positive and negative numbers).

Swapping and Grouping

Example 6: Madhu is flying a drone from a terrace. The drone goes 6 m up and then 4 m down. Write an expression to show how high the final position of the drone is from the terrace. Will the sum change if we swap the terms?
See SolutionExpression: 6 – 4.
As sum of terms: (6) + (-4).
Value = 2.
Swapping the terms: (-4) + (6).
Value = 2.
The sum does not change if we swap the terms.

Will this (swapping terms does not change the sum) also hold when there are terms having negative numbers as well? Take some more expressions and check.
See SolutionYes, swapping terms works even with negative numbers because addition of integers is commutative.
Example 1: (-5) + 8 = 3. Swapped: 8 + (-5) = 3. Same value.
Example 2: (-2) + (-7) = -9. Swapped: (-7) + (-2) = -9. Same value.

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Can you explain why this (swapping terms does not change the sum) is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
See SolutionUsing the Token Model, addition represents combining collections of positive and negative tokens. Whether you combine collection A with collection B, or collection B with collection A, the final combined collection of tokens remains exactly the same. This illustrates the commutative property (swapping order) holds for integers.
Statement: Thus, in an expression having two terms, swapping them does not change the value. (Term 1) + (Term 2) = (Term 2) + (Term 1)
Discussion/Example: Consider the expression (-7) + 10 + (-11).
Adding the first two terms then the third:
((-7) + 10) + (-11)
= (3) + (-11)
= -8.
Adding the last two terms then the first:
(-7) + (10 + (-11))
= (-7) + (-1)
= -8.
The results are the same.

Will this (grouping terms differently does not change the sum) also hold when there are terms having negative numbers as well? Take some more expressions and check.
See SolutionYes, the way terms are grouped for addition does not change the sum, even with negative numbers.
This is the associative property of addition for integers.
Example:
(5 + (-3)) + (-4)
= (2) + (-4)
= -2.
Also,
5 + ((-3) + (-4))
= 5 + (-7)
= -2.
Same value.

Can you explain why this (grouping does not change the sum) is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
See SolutionIn the Token Model, adding three (or more) collections of tokens means ultimately combining all of them into one large pile. It doesn’t matter if you combine pile A and pile B first, and then add pile C, or if you combine pile B and pile C first, and then add pile A. The final combined pile will contain the same total number of positive and negative tokens regardless of the intermediate grouping.

Let us consider the expression (-7) + 10 + (-11) again. What happens when we change the order and add -7 and -11 first, and then add this sum to 10? Will we get the same sum as before?
See SolutionYes, we will get the same sum.
Calculation: Add -7 and -11 first:
(-7) + (-11)
= -18.
Then add this sum to 10:
(-18) + 10
= -8.
This is the same sum (-8) obtained by the other grouping methods.

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Does adding the terms of an expression in any order give the same value? Take some more expressions and check. Consider expressions with more than 3 terms also.
See SolutionYes, adding the terms of an expression in any order gives the same value.
Example:
10 + (-5) + 2 + (-3)
= 5 + 2 + (-3)
= 7 + (-3)
= 4.
Reordered:
10 + 2 + (-5) + (-3)
= 12 + (-5) + (-3)
= 7 + (-3)
= 4.
Same value.

Can you explain why this (addition in any order gives same value) is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
See SolutionUsing the Token Model, adding multiple terms means combining several collections of positive and negative tokens. The final result depends only on the total count of positive and negative tokens across all initial collections. It doesn’t matter in which sequence you combine these collections; the final combined pile will always have the same composition and thus represent the same value.

Example Evaluation: 30 + 5 x 4
See SolutionIdentify terms: (30) and (5 x 4).
Evaluate terms: (30) and (20).
Add evaluated terms: 30 + 20 = 50.

Example Evaluation: 5 x (3 + 2) + 7 x 8 + 3
See SolutionIdentify terms: (5 x (3 + 2)), (7 x 8) and (3).
Evaluate terms:
Term 1: 5 x (3 + 2) = 5 x 5 = 25.
Term 2: 7 x 8 = 56.
Term 3: 3.
Add evaluated terms:
25 + 56 + 3
= 81 + 3
= 84.

Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again?
See SolutionNo, she does not have to start all over again. Because addition is commutative and associative (terms can be added in any order or grouping), she can simply add the forgotten number (9055) to her previous sum (11749).
Calculation: 11749 + 9055 = 20804.

Swapping the Order of Things in Everyday Life:

Manasa is going outside to play. Her mother says, “Wear your hat and shoes!” Which one should she wear first?
See SolutionIt doesn’t matter which she wears first. Wearing the hat then shoes, or shoes then hat, results in the same final state. This situation is commutative. (Unlike wearing socks and shoes, where the order matters).

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More Expressions and Their Terms

Example 7: Amu, Charan, Madhu, and John (4 people) went to a hotel and ordered four dosas. Each dosa cost ₹23 and they wish to thank the waiter by tipping ₹5. Write an expression describing the total cost.
See SolutionCost of 4 dosas = 4 x 23 = 92.
Total cost including tip = Cost of dosas + Tip.
Expression: (4 x 23) + 5.
Value = 92 + 5 = 97. The total cost is ₹97.

If the total number of friends goes up to 7 (ordering 7 dosas at ₹23 each) and the tip remains the same (₹5), how much will they have to pay? Write an expression for this situation and identify its terms.
See SolutionCost of 7 dosas = 7 x 23.
Total cost including tip = Cost of 7 dosas + Tip.
Expression: (7 x 23) + 5.
Terms: The terms are (7 x 23) and 5.
Value: 7 x 23 = 161. So, 161 + 5 = 166.
They will have to pay ₹166.

Example 8: Children playing “Fire in the mountain”. 33 students playing, Ruby sitting out. Teacher calls out ‘5’. Students form groups of 5. Ruby wrote 6 x 5 + 3. Think and discuss why she wrote this.
See SolutionWhen 33 students try to form groups of 5:
– 33 divided by 5 is 6 with a remainder of 3 (33 = 6 x 5 + 3).
– This means they can form 6 complete groups of 5 students each.
– There are 3 students left over who couldn’t form a full group (they are “out”).
Ruby’s expression 6 x 5 + 3 represents the total number of students playing: (6 groups multiplied by 5 students per group) plus the 3 students left over.
6 x 5 + 3
= 30 + 3
= 33.

How is the expression 6 x 5 + 3 written as a sum of terms?
See SolutionThe terms are separated by ‘+’. Multiplication within a term is evaluated first.
Expression as sum of terms: (6 x 5) + (3).

If the teacher had called out ‘4’:
See Solution33 students forming groups of 4: 33 / 4 = 8 with a remainder of 1.
Expression: 8 x 4 + 1.
Terms: (8 x 4) and 1.
Value = 32 + 1 = 33.

If the teacher had called out ‘7’:
See Solution33 students forming groups of 7: 33 / 7 = 4 with a remainder of 5.
Expression: 4 x 7 + 5.
Terms: (4 x 7) and 5.
Value = 28 + 5 = 33.

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Example 9: Raghu bought 100 kg of rice from the wholesale market and packed them into 2 kg packets. He already had four 2 kg packets. Write an expression for the number of 2 kg packets of rice he has now and identify the terms.
See SolutionNumber of new packets from 100 kg = 100 / 2 = 50 packets.
Number of packets he already had = 4 packets.
Total number of packets = Old packets + New packets.
Expression: 4 + (100 / 2). (Or 4 + 100/2).
Terms: 4 and (100/2).
Value = 4 + 50 = 54 packets.

Example 10: Kannan has to pay ₹432 using coins of ₹1 and ₹5, and notes of ₹10, ₹20, ₹50 and ₹100. Two ways are shown:
Way 1: 4 x 100 + 1 x 20 + 1 x 10 + 2 x 1
Way 2: 8 x 50 + 1 x 10 + 4 x 5 + 2 x 1

Question: Identify the terms in the two expressions above.
See SolutionTerms in Way 1: (4 x 100), (1 x 20), (1 x 10), (2 x 1).
Terms in Way 2: (8 x 50), (1 x 10), (4 x 5), (2 x 1).

Can you think of some more ways of giving ₹432 to someone?
See SolutionYes, there are many other ways. Here are a couple of examples:
Example Way 3: 3 notes of ₹100, 6 notes of ₹20, 1 note of ₹10, 2 coins of ₹1.
(Expression: 3×100 + 6×20 + 1×10 + 2×1 = 300 + 120 + 10 + 2 = 432).
Example Way 4: 20 notes of ₹20, 1 note of ₹20, 1 note of ₹10, 2 coins of ₹1.
(Expression: 20×20 + 1×20 + 1×10 + 2×1 = 400 + 20 + 10 + 2 = 432).

Example 11: Here are two pictures (arrangements of squares). Which of these two arrangements matches with the expression 5 x 2 + 3?
See SolutionThe expression 5 x 2 + 3 written as sum of terms is (5 x 2) + (3).
This means “3 more than (5 multiplied by 2)”.
This matches the arrangement shown on the left in the document (which visually depicts two groups of 5 squares plus 3 additional squares).

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Example 11: What is the expression for the arrangement in the right making use of the number of yellow and blue squares?
See SolutionThe arrangement on the right shows 2 rows, each containing 5 yellow squares and 3 blue squares (total 8 squares per row).
Possible expressions given in the text:
– 2 x (5 + 3) (Meaning: 2 times the sum of 5 and 3)
– 5 + 3 + 5 + 3 (Meaning: Sum of squares in row 1 + sum of squares in row 2)
– 5 x 2 + 3 x 2 (Meaning: Total yellow squares + total blue squares)
All these expressions evaluate to 16.

Figure it Out

1. Write the following expressions as a sum of terms and identify the terms in each case.
(a) 28 – 7 + 8
See SolutionSum of terms: (28) + (-7) + (8). Terms: 28, -7, 8.
(b) 5 x 12 – 6
See SolutionSum of terms: (5 x 12) + (-6). Terms: (5 x 12), -6.
(c) 40 – 10 + 10 + 10
See SolutionSum of terms: (40) + (-10) + (10) + (10). Terms: 40, -10, 10, 10.
(d) 39 – 2 x 6 + 11
See SolutionSum of terms: (39) + (-2 x 6) + (11). Terms: 39, (-2 x 6), 11.
(e) 6 x 3 – 4 x 8 x 5
See SolutionSum of terms: (6 x 3) + (-4 x 8 x 5). Terms: (6×3), (-4 x 8 x 5).
(f) 48 – 10 x 2 + 16 / 2
See SolutionSum of terms: (48) + (-10 x 2) + (16 / 2). Terms: 48, (-10 x 2), (16/2).

2. Find the values of the expressions identified in (1) above.
See Solution(a) 28 – 7 + 8 = 21 + 8 = 29.
(b) 5 x 12 – 6 = 60 – 6 = 54.
(c) 40 – 10 + 10 + 10 = 30 + 10 + 10 = 50.
(d) 39 – 2 x 6 + 11 = 39 – 12 + 11 = 27 + 11 = 38.
(e) 6 x 3 – 4 x 8 x 5 = 18 – (32 x 5) = 18 – 160 = -142.
(f) 48 – 10 x 2 + 16 / 2 = 48 – 20 + 8 = 28 + 8 = 36.

3. For each situation, write the expression, identify terms, find value.
(a) Queen Alia gave 100 coins each to Elsa and Anna. Elsa doubled hers (2×100). Anna has half left (100/2). How many together?
See SolutionExpression: (2 x 100) + (100 / 2).
Terms: (2 x 100), (100 / 2).
Value: 200 + 50 = 250 coins.

(b) Metro: Adult=₹40, Child=₹20. Total cost for:
(i) four adults and three children?
See SolutionExpression: (4 x 40) + (3 x 20).
Terms: (4 x 40), (3 x 20).
Value: 160 + 60 = ₹220.

(ii) for two groups having three adults each?
See SolutionExpression: 2 x (3 x 40). (or (2 x 3) x 40 = 6 x 40)

(c): Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture. Identify its terms and find the value.
See SolutionThe picture shows these vertical sections for the window height:
– Top Border: 3 cm
– Grill: 12 cm
– Gap: 5 cm
– Grill: 12 cm
– Gap: 5 cm
– Grill: 12 cm
– Bottom Border: 3 cm (Assuming top and bottom borders are the same)
Expression for total height = Sum of all sections:
3 + 12 + 5 + 12 + 5 + 12 + 3
Alternatively, grouping similar parts: (2 x 3) + (3 x 12) + (2 x 5)
Terms (based on grouped expression): (2 x 3), (3 x 12), (2 x 5).
Value Calculation: 6 + 36 + 10 = 52 cm.
The total height of the window is 52 cm.

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Removing Brackets-I

Explanation/Example: Evaluating 200 – (40 + 3).
– Method 1 (Brackets first): 200 – (43) = 157.
– Method 2 (Removing brackets): When removing brackets preceded by a ‘-‘, change the sign of each term inside.
200 – (40 + 3) becomes 200 – 40 – 3.
Calculation: 200 – 40 = 160. Then 160 – 3 = 157.
– Observation: 200 – (40 + 3) = 200 – 40 – 3.

Example 12 (Irfan’s change): Calculating 100 – (15 + 56).
– Method 1 (Brackets first): 100 – (71) = 29.
– Method 2 (Step-by-step subtraction): 100 – 15 = 85 (after buying biscuits). Then 85 – 56 = 29 (after buying dal). This sequence is 100 – 15 – 56.
– Observation: 100 – (15 + 56) = 100 – 15 – 56.
Rule Summary: Notice how upon removing the brackets preceded by a negative sign, the signs of the terms inside the brackets change (e.g., +40 became -40, +3 became -3; +15 became -15, +56 became -56).

Example 13: Consider the expression 500 – (250 – 100). Is it possible to write this expression without the brackets?
See SolutionValue with brackets: First evaluate inside: 250 – 100 = 150. Then 500 – 150 = 350.
Removing brackets after ‘-‘: Changes signs inside -> 500 – 250 + 100.
Value without brackets (modified): 500 – 250 = 250. Then 250 + 100 = 350.
Conclusion: Yes, it can be written as 500 – 250 + 100.
Check: It is NOT equal to 500 – 250 – 100 (which equals 150).
Rule Recap: Removing brackets preceded by ‘-‘ changes the sign of each term inside (e.g., +250 becomes -250, -100 becomes +100).

Example 14: Hira has 28 coins in one bag and 35 in another. She gifts 10 coins from the second bag. Write an expression for the number of coins left with Hira.
See SolutionCoins left = Coins in bag 1 + (Coins in bag 2 – Gifted coins)
Expression: 28 + (35 – 10).
Removing brackets (not preceded by ‘-‘): The signs inside do not change.
Equivalent expression: 28 + 35 – 10.
Value: 28 + 25 = 53. Or 28 + 35 – 10 = 63 – 10 = 53 coins.

Tinker the Terms I

(Discussion on how expression value changes if terms change)
Complete the comparison table (fill blanks doing minimal computation):
Column 1:
– Given: 53 + (-16) = 37
– Calculate: 54 + (-16) = ?
See Solution38 (Since 54 is 1 more than 53, the sum is 1 more than 37).
– Calculate: 52 + (-16) = ?
See Solution36 (Since 52 is 1 less than 53, the sum is 1 less than 37).

Column 2:
– Given: 53 + (-16) = 37
– Calculate: 53 + (-15) = ?
See Solution38 (Since -15 is 1 more than -16, the sum is 1 more than 37).
– Calculate: 53 + (-17) = ?
See Solution36 (Since -17 is 1 less than -16, the sum is 1 less than 37).

Column 3:
– Given: -87 – 16 = -87 + (-16) = -103
– Calculate: -88 + (-15) = ?
See Solution-103 (Compare to -87 + (-16). -88 is 1 less. -15 is 1 more. Net change = -1 + 1 = 0. Value is same).
– Calculate: -86 + (-18) = ?
See Solution-104 (Compare to -87 + (-16). -86 is 1 more. -18 is 2 less. Net change = +1 – 2 = -1. Value is 1 less than -103).

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Figure it Out

1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
(a) 24 + (6 – 4) = 24 + 6 ___ 4
See Solution(Removing brackets after ‘+’ doesn’t change internal signs: 24 + 6 – 4)

(b) 38 + ( ___ ) = 38 + 9 – 4
See Solution19 (RHS = 38+9-4 = 47-4 = 43. Need 38 + ___ = 43. Blank = 43 – 38 = 5)

(c) 24 – (6 + 4) = 24 ___ 6 – 4
See Solution-, – (Removing brackets after ‘-‘ changes signs of terms inside: 24 – 6 – 4)

(d) 24 – 6 – 4 = 24 – 6 (__) ______
See Solution-, + (LHS = 24-10 = 14. Need RHS = 14. Try 24 – (6+4) = 24-10 = 14. So, 24 – (6 + 4))

(e) 27 – (8 + 3) = 27 ___ 8 ___ 3
See Solution-, – (Removing brackets after ‘-‘ changes signs: 27 – 8 – 3)

(f) 27 – ( ___ ) = 27 – 8 + 3
See Solution8 – 3 (or 5) (RHS = 27-8+3 = 19+3 = 22. Need 27 – ___ = 22. Blank = 5. Can write as 8-3)

2. Remove the brackets and write the expression having the same value.
(a) 14 + (12 + 10)
See Solution14 + 12 + 10
(b) 14 – (12 + 10)
See Solution14 – 12 – 10
(c) 14 + (12 – 10)
See Solution14 + 12 – 10
(d) 14 – (12 – 10)
See Solution14 – 12 + 10
(e) -14 + 12 – 10
See Solution-14 + 12 – 10 (No brackets to remove)
(f) 14 – (-12 – 10)
See Solution14 + 12 + 10

3. Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(a) (6 + 10) – 2 and 6 + (10 – 2)
See SolutionValues: (6 + 10) – 2 = 16 – 2 = 14.
6 + (10 – 2) = 6 + 8 = 14.
Guess: Equal. They are equal. This shows associativity seems to hold when structured this way.

(b) 16 – (8 – 3) and (16 – 8) – 3
See SolutionValues: 16-(8-3) = 16-5 = 11. (16-8)-3 = 8-3 = 5.
Guess: Not equal. They are not equal. This shows subtraction is not associative.

(c) 27 – (18 + 4) and 27 + (-18 – 4)
See SolutionValues: 27-(18 + 4) = 27 – 22 = 5. 27+(-18 – 4) = 27 + (-22) = 5.
Guess: Equal. They are equal. This shows subtracting a sum is the same as adding the sum of the negatives.

4. In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.
(a) 319 + 537, 319 – 537, 537 + 319, -537 + 319, 537 – 319
See SolutionGroup 1 (Equal): 319 + 537, 537 + 319 (Commutative property)
Group 2 (Equal): 319 – 537, -537 + 319 (Commutative property with negative term)
Group 3 (Unique): 537 – 319

(b) 87 + 46 – 109, -(46 + 109), 87 – (46 – 109), (87 – 46) + 109, 87 + (-46 – 109)
See SolutionExpression 1: 87 + 46 – 109
Expression 2: -(46 + 109) = -46 – 109
Expression 3: 87 – (46 – 109) = 87 – 46 + 109
Expression 4: (87 – 46) + 109 = 87 – 46 + 109
Expression 5: 87 + (-46 – 109) = 87 – 46 – 109
Groups with same value:
Group 1: { 87 – (46 – 109), (87 – 46) + 109 }
Other expressions are unique in this set.

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5. Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(a) 34 – 9 + 12 = 13
See Solution34 – (9 + 12) = 34 – 21 = 13.
Expression with brackets: 34 – (9 + 12)

(b) 56 – 14 – 8 = 34
See SolutionCalculation proceeds left-to-right by default: (56 – 14) – 8 = 42 – 8 = 34.
Expression with brackets: (56 – 14) – 8 (or no brackets needed as left-to-right works).

(c) -22 – 12 + 10 + 22 = -22
See SolutionNeed the last three terms to sum to zero when subtracted. 12 + 10 – 22 = 22 – 22 = 0.
Expression with brackets: -22 – (12 + 10 – 22)

6. Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal.
(a) 423 + ___ = 419 + ___
See Solution419 is 4 less than 423. To keep the sums equal, the number added to 419 must be 4 more than the number added to 423.
Example: 423 + 5 = 419 + 9 (where 9 is 4 more than 5).

(b) 207 – 68 = 210 – ___
See Solution71 (The first number on the RHS, 210, is 3 more than the first number on the LHS, 207. To maintain equality, we must subtract 3 more on the RHS than on the LHS. So, subtract 68 + 3 = 71).
Check: 207 – 68 = 139.
210 – 71 = 139.

7. Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘-‘ and brackets, as necessary, generate expressions to give as many different values as possible.
See SolutionExamples (other combinations possible):
– 2 + 3 + 5 = 10
– 5 + 3 – 2 = 6
– 3 – (2 – 5) = 3 – (-3) = 6
– (5 – 2) + 3 = 3 + 3 = 6
– 2 – 3 + 5 = -1 + 5 = 4
– 5 + 2 – 3 = 7 – 3 = 4
– 2 – (3 – 5) = 2 – (-2) = 4
– 5 – (2 + 3) = 5 – 5 = 0
– 3 + 2 – 5 = 5 – 5 = 0
– 3 – (5 – 2) = 3 – 3 = 0
– 5 – 3 – 2 = 2 – 2 = 0
– 3 – 2 – 5 = 1 – 5 = -4
– 2 – (3 + 5) = 2 – 8 = -6
– 2 – 3 – 5 = -1 – 5 = -6
Possible values found: 10, 6, 4, 0, -4, -6.

8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = (36 – 10) + 1 = 26 + 1 = 27.
(a) Do you think she always gets the correct answer? Why?
See SolutionYes, she always gets the correct answer. This is because -9 is mathematically equivalent to -10 + 1. Substituting (-10 + 1) for (-9) in an expression like N – 9 gives N + (-10 + 1) which equals (N – 10) + 1 by the associative property.
(b) Can you think of other similar strategies? Give some examples.
See SolutionYes.
– To subtract 8: Subtract 10, Add 2 (since -8 = -10 + 2). Ex: 55 – 8 = (55 – 10) + 2 = 45 + 2 = 47.
– To subtract 19: Subtract 20, Add 1 (since -19 = -20 + 1). Ex: 72 – 19 = (72 – 20) + 1 = 52 + 1 = 53.
– To add 9: Add 10, Subtract 1 (since +9 = +10 – 1). Ex: 46 + 9 = (46 + 10) – 1 = 56 – 1 = 55.
– To add 98: Add 100, Subtract 2 (since +98 = +100 – 2). Ex: 130 + 98 = (130 + 100) – 2 = 230 – 2 = 228.

9. Consider the two expressions: a) 73 – 14 + 1, b) 73 – 14 – 1. For each of these expressions, identify the expressions from the following collection that are equal to it.
Options: (i) 73 – (14 + 1) (ii) 73 – (14 – 1) (iii) 73 + (-14 + 1) (iv) 73 + (-14 – 1)
See SolutionEvaluate targets: (a) 73 – 14 + 1 = 59 + 1 = 60. (b) 73 – 14 – 1 = 59 – 1 = 58.
Evaluate options:
(i) 73 – (14 + 1) = 73 – 15 = 58. Matches (b).
(ii) 73 – (14 – 1) = 73 – 13 = 60. Matches (a).
(iii) 73 + (-14 + 1) = 73 + (-13) = 60. Matches (a).
(iv) 73 + (-14 – 1) = 73 + (-15) = 58. Matches (b).
Equal to (a) [Value 60]: (ii), (iii).
Equal to (b) [Value 58]: (i), (iv).

Removing Brackets-II

Example 15. Lhamo and Norbu went to a hotel. Each of them ordered a vegetable cutlet (₹43) and a rasgulla (₹24). Write an expression for the amount they will have to pay.
See SolutionCost per person = 43 + 24.
Total cost for two people = 2 x (Cost per person).
Expression: 2 x (43 + 24).

1. (Continuing Example 15 discussion) Why is the expression 2 x 43 + 24 incorrect for the total cost of two people each ordering a ₹43 cutlet and a ₹24 rasgulla?
See SolutionThe expression 2 x 43 + 24, when interpreted by evaluating terms first, becomes (2 x 43) + 24 = 86 + 24 = 110. This represents the cost of two cutlets plus only *one* rasgulla, which is wrong. The correct interpretation requires doubling the *total* cost per person (43 + 24).

2. How are brackets used correctly for Example 15 and what property does it illustrate?
See SolutionThe correct expression uses brackets to group the cost per person first: 2 x (43 + 24).
Value: 2 x (67) = 134.
The text also shows this is equivalent to calculating the cost of two cutlets plus two rasgullas: (2 x 43) + (2 x 24) = 86 + 48 = 134.
This demonstrates the distributive property: 2 x (43 + 24) = (2 x 43) + (2 x 24).

Example 15: If another friend, Sangmu, joins them (making 3 friends) and orders the same items (₹43 cutlet, ₹24 rasgulla), what will be the expression for the total amount to be paid?
See SolutionCost per person = 43 + 24.
Total cost for 3 people = 3 x (Cost per person).
Expression: 3 x (43 + 24).
Alternatively, using the distributive property: (3 x 43) + (3 x 24).
Value: 3 x (67) = 201. Or (3 x 43) + (3 x 24) = 129 + 72 = 201. The total cost is ₹201.

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Example 16: In the Republic Day parade, there are boy scouts (4 rows, 5 scouts/row) and girl guides (3 rows, 5 guides/row) marching together. How many scouts and guides are marching in this parade? (Two methods shown).
See SolutionMethod 1 (Calculate separately, then add):
– Number of scouts = 4 rows x 5 scouts/row = 20.
– Number of guides = 3 rows x 5 guides/row = 15.
– Total = Scouts + Guides = 20 + 15 = 35.
– Expression: (4 x 5) + (3 x 5).
Method 2 (Combine rows first, then multiply):
– Total rows = Scout rows + Guide rows = 4 + 3 = 7 rows.
– Children per row = 5.
– Total = Total rows x Children per row = 7 x 5 = 35.
– Expression: (4 + 3) x 5.
Conclusion from example: (4 x 5) + (3 x 5) = (4 + 3) x 5. This again illustrates the distributive property.
Calculation for Method 2: (4 + 3) x 5 = 7 x 5 = 35. (Matches Method 1).

Question: Why is 5 x 4 + 3 ≠ 5 x (4 + 3)? Can you explain why?
See SolutionLet’s evaluate both sides:
– LHS (Left Hand Side): 5 x 4 + 3 = (5 x 4) + 3 = 20 + 3 = 23. (Multiplication is done before addition).
– RHS (Right Hand Side): 5 x (4 + 3) = 5 x (7) = 35. (Operation inside brackets is done first).
Since 23 is not equal to 35, the expressions are not equal. The presence of brackets changes the order of operations, leading to a different result. The distributive property applies when multiplying a number by a sum *inside* brackets, like on the RHS, not when adding after a multiplication like on the LHS.

Question: Is 5 x (4 + 3) = 5 x (3 + 4) = (3 + 4) x 5?
See SolutionYes, all three expressions are equal.
– 5 x (4 + 3) = 5 x 7 = 35.
– 5 x (3 + 4) = 5 x 7 = 35. (Because 4+3 = 3+4, commutative property of addition).
– (3 + 4) x 5 = 7 x 5 = 35. (Because 5 x 7 = 7 x 5, commutative property of multiplication).

Generalization of Distributive Property (Examples):
– Addition: 10 x 98 + 3 x 98 = (10 + 3) x 98. (Sum of multiples = multiple of sum).
– Showing reverse: (10 + 3) x 98 = 10 x 98 + 3 x 98. (Multiple of sum = sum of multiples).
– Showing with number first: 98 x 10 + 98 x 3 = 98 x (10 + 3).
– Subtraction: 14 x 10 – 6 x 10 = (14 – 6) x 10. (Difference of multiples = multiple of difference).
– Showing reverse: (14 – 6) x 10 = 14 x 10 – 6 x 10. (Multiple of difference = difference of multiples).

Summary Statement: The multiple of a sum (difference) is the same as the sum (difference) of the multiples. (This describes the Distributive Property).

Page 40

Tinker the Terms II

Example 17: Given 53 x 18 = 954. Find out 63 x 18.
See SolutionUsing the distributive property:
63 x 18 = (53 + 10) x 18
= (53 x 18) + (10 x 18)
= 954 + 180 (Used the given value 53 x 18 = 954)
= 1134.

Example 18: Find an effective way of evaluating 97 x 25. Find this value.
See SolutionMethod using distributive property:
97 x 25 = (100 – 3) x 25
= (100 x 25) – (3 x 25)
= 2500 – 75
= 2425.

Question: Use this method (writing numbers as sum/difference from round numbers) to find the following products:
(a) 95 x 8
See Solution95 x 8 = (100 – 5) x 8 = (100 x 8) – (5 x 8) = 800 – 40 = 760.
(b) 104 x 15
See Solution104 x 15 = (100 + 4) x 15 = (100 x 15) + (4 x 15) = 1500 + 60 = 1560.
(c) 49 x 50
See Solution49 x 50 = (50 – 1) x 50 = (50 x 50) – (1 x 50) = 2500 – 50 = 2450.

Question: Is this quicker than the multiplication procedure you use generally?
See SolutionFor many people, yes, this method can be quicker, especially when multiplying mentally, as it breaks the problem down into simpler multiplications involving multiples of 10 or 100.

Math Talk Question: Which other products might be quicker to find like the ones above?
See SolutionThis method is useful when one of the numbers is close to a multiple of 10, 100, 1000, etc.
Examples:
– 38 x 7 = (40 – 2) x 7
– 102 x 9 = (100 + 2) x 9
– 19 x 15 = (20 – 1) x 15
– 998 x 4 = (1000 – 2) x 4

Figure it Out

1. Fill in the blanks with numbers and boxes by signs, so that the expressions on both sides are equal (using the distributive property).
(a) 3 x (6 + 7) = 3 x 6 + 3 x 7 (Already filled)
(b) (8 + 3) x 4 = 8 x 4 + 3 x 4 (Already filled)
(c) 3 x (5 + 8) = 3 x 5 [+] 3 x [8]
(d) (9 + 2) x 4 = 9 x 4 [+] [2] x 4
(e) 3 x ([5] + 4) = 3 x [5] + [3 x 4] (Assuming first blank requires a number)
(f) ([13] + 6) x 4 = 13 x 4 + [6 x 4]
(g) 3 x ([5] + [2]) = 3 x 5 + 3 x 2
(h) ([2] + [3]) x [4] = 2 x 4 + 3 x 4
(i) 5 x (9 – 2) = 5 x 9 – 5 x [2]
(j) (5 – 2) x 7 = 5 x 7 – 2 x [7]
(k) 5 x (8 – 3) = 5 x 8 [-] 5 x [3]
(l) (8 – 3) x 7 = 8 x 7 [-] [3] x 7
(m) 5 x (12 [-] [7]) = [5×12] [-] 5 x [7] (Following distributive pattern for subtraction)
(n) (15 – [6]) x 7 = [15×7] [-] 6 x 7 (Following distributive pattern for subtraction)
(o) 5 x ([9] – [4]) = 5 x 9 – 5 x 4 (Identifying numbers from the expanded form)
(p) ([17] – [9]) x [7] = 17 x 7 – 9 x 7 (Identifying numbers from the expanded form)

Page 42

2. In the boxes below, fill ‘<‘,’>’ or ‘=’ after analysing the expressions… Use reasoning… not by evaluating.
(a) (8 – 3) x 29 ___ (3 – 8) x 29
See Solution>
Reasoning:
LHS = (Positive 5) x 29, which is positive.
RHS = (Negative 5) x 29, which is negative.
Positive > Negative.

(b) 15 + 9 x 18 ___ (15 + 9) x 18
See Solution<
Reasoning:
LHS = 15 + (9 x 18).
RHS = (15 + 9) x 18
= 15 x 18 + 9 x 18.
Comparing LHS and RHS:
15 + (9×18) vs (15×18) + (9×18).
Since 15 is much smaller than 15×18, the LHS is smaller than the RHS.

(c) 23 x (17 – 9) ___ 23 x 17 + 23 x 9
See Solution<
Reasoning:
LHS = 23 x (8).
RHS = 23 x 17 + 23 x 9
= 23 x (17 + 9)
= 23 x (26).
Since 8 is much smaller than 26, the LHS is smaller than the RHS.

(d) (34 – 28) x 42 ___ 34 x 42 – 28 x 42
See Solution=
Reasoning:
The RHS is the expanded form of the LHS using the distributive property.
Therefore, they are equal.
LHS = 6 x 42.
RHS = (34-28) x 42
= 6 x 42.

3. Here is one way to make 14: 2 x (1 + 6) = 14. Are there other ways…? Fill them out below: (Using format _ x (_ + _) = 14 etc.)
See Solution(Assuming any numbers can be used)
(a) _ x (_ + _) = 14.
Example:
7 x (1 + 1) = 14 or 2 x (3 + 4) = 14
(b) _ x (_ + _) = 14.
Example:
1 x (10 + 4) = 14 or 7 x (0 + 2) = 14
(c) _ x (_ + _ + _) = 14.
Example:
2 x (1 + 2 + 4) = 14 or 1 x (5 + 6 + 3) = 14
(d) _ x (_ + _ + _) = 14.
Example:
7 x (1 + 1 + 0) = 14 or 2 x (2 + 3 + 2) = 14

4. Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.
(a) Picture 1 (5 squares of ‘4’, 4 circles of ‘8’):
See SolutionWay 1 (Group by number):
(Number of 4s x 4) + (Number of 8s x 8)
= (5 x 4) + (4 x 8)
= 20 + 32
= 52.
Way 2 (Sum rows): (Row 1) + (Row 2) + (Row 3)
= (4+8+4) + (8+4+8) + (4+8+4)
= 16 + 20 + 16
= 52.

(b) Picture 2 (8 circles of ‘5’, 8 circles of ‘6’):
See SolutionWay 1 (Group by number):
(Number of 5s x 5) + (Number of 6s x 6)
= (8 x 5) + (8 x 6)
= 40 + 48
= 88.
Way 2 (Use distributive property): From Way 1, (8 x 5) + (8 x 6)
= 8 x (5 + 6)
= 8 x 11
= 88.
Way 3 (Sum rows): Each row sums to 5 + 6 + 6 + 5 = 22  or   6 + 5 +5 + 6 = 22.
There are 4 rows.
Total = 4 x 22 = 88.
Expression: 4 x (5 + 6 + 6 + 5).

Figure it Out

1. Read the situations given below. Write appropriate expressions for each of them and find their values.
(a) The district market… operates 7 days/week. Rahim supplies 9 kg mangoes/day, Shyam supplies 11 kg/day. Find amount supplied by them in a week.
See SolutionMethod 1 Expression: (Total daily supply) x Days = (9 + 11) x 7. Value = 20 x 7 = 140 kg.
Method 2 Expression: (Rahim’s weekly) + (Shyam’s weekly) = (9 x 7) + (11 x 7). Value = 63 + 77 = 140 kg.
(Either expression is appropriate).

Read situations, write expressions, find values.
(b) Binu earns ₹20,000/month. Spends ₹5,000 rent, ₹5,000 food, ₹2,000 other. What is the amount Binu will save by the end of a year?
See SolutionMonthly Expenses = 5000 + 5000 + 2000 = 12000.
Monthly Savings = Monthly Earnings – Monthly Expenses = 20000 – 12000 = 8000.
Yearly Savings = Monthly Savings x 12.
Expression: (20000 – (5000 + 5000 + 2000)) x 12 OR 12 x (20000 – 5000 – 5000 – 2000).
Value = 12 x 8000 = ₹96,000.

(c) Snail climbs 3 cm up day, slips 2 cm night. Post is 10 cm high. In how many days will the snail get the treat?
See SolutionNet climb per full day-night cycle = 3 cm – 2 cm = 1 cm.
– End of Day 1: 3 cm. End of Night 1: 1 cm.
– End of Day 2: 1+3=4 cm. End of Night 2: 2 cm.
– End of Day 3: 2+3=5 cm. End of Night 3: 3 cm.
– End of Day 4: 3+3=6 cm. End of Night 4: 4 cm.
– End of Day 5: 4+3=7 cm. End of Night 5: 5 cm.
– End of Day 6: 5+3=8 cm. End of Night 6: 6 cm.
– End of Day 7: 6+3=9 cm. End of Night 7: 7 cm.
– During Day 8: Snail starts at 7 cm, climbs 3 cm, reaching 10 cm (the top).
It will get the treat on Day 8.

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2. Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?
(a) 5 x 2 x 8 (b) (7 – 2) x 8 (c) 8 x 7 (d) 7 x 2 x 8 (e) 7 x 5 – 2 (f) (7 + 2) x 8 (g) 7 x 8 – 2 x 8 (h) (7 – 5) x 8
See SolutionCalculation:
– Reading days per week = 7 days – 2 non-reading days = 5 days.
– Pages per week = 5 days/week x 2 pages/day = 10 pages/week.
– Stories per week (since story is 2 pages) = 10 pages/week / 2 pages/story = 5 stories/week.
– Stories in 8 weeks = 5 stories/week * 8 weeks = 40 stories.

Check expressions:
(a) 5 x 2 x 8 = 80 (Maybe total pages read?)
(b) (7 – 2) x 8 = 5 x 8 = 40 (Matches total stories)
(c) 8 x 7 = 56 (Total days in 8 weeks)
(d) 7 x 2 x 8 = 112 (Maybe total pages if read every day?)
(e) 7 x 5 – 2 = 35 – 2 = 33 (?)
(f) (7 + 2) x 8 = 9 x 8 = 72 (?)
(g) 7 x 8 – 2 x 8 = (Total days) – (Non-reading days) = 56 – 16 = 40 (Matches total reading days.
If multiplied by stories/day = 1, this would work, but it calculates reading days).
Or = (7-2)x8 = 5×8 = 40.
Yes, this uses distributive property: (Days/wk – NonReadDays/wk)*Wks.
This correctly calculates total stories read if interpreted as (Total days read over 8 weeks) * (1 story/day since story=2pg, read=2pg).
Okay, (g) also equals 40.
(h) (7-5)x8 = 2×8=16 (?)
Expressions describing the scenario (total stories = 40): (b) and (g).

3. Find different ways of evaluating the following expressions:
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
See SolutionWay 1 (Left-to-right):
= – 1 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= 2 – 4 + 5 – 6 + 7 – 8 + 9 – 10
= -2 + 5 – 6 + 7 – 8 + 9 – 10
= 3 – 6 + 7 – 8 + 9 – 10
= – 3 + 7 – 8 + 9 – 10
= 4 – 8 + 9 – 10
= – 4 + 9 – 10
= 5 – 10
= – 5.
Way 2 (Grouping pairs):
= (1-2) + (3-4) + (5-6) + (7-8) + (9-10)
= (-1) + (-1) + (-1) + (-1) + (-1)
= -5.
Way 3 (Positives and Negatives):
= (1+3+5+7+9) – (2+4+6+8+10)
= 25 – 30
= -5.

(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
See SolutionWay 1 (Left-to-right):
= 0 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= 0 + 1 – 1 + 1 – 1 + 1 – 1 = … = 0.
Way 2 (Grouping pairs):
= (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 + 0 + 0 = 0.

4. Compare the following pairs of expressions using ‘<‘, ‘>’ or ‘=’ or by reasoning.
(a) 49 – 7 + 8 ___ 49 – (7 + 8)
See Solution> (LHS = 42+8=50. RHS = 49-15=34. Subtracting 7 then adding 8 is different from subtracting the sum).
(b) 83 x 42 – 18 ___ 83 x 40 – 18
See Solution> (Comparing 83×42 vs 83×40. Since 42>40, 83×42 is greater. Subtracting 18 from both doesn’t change the inequality).
(c) 145 – 17 x 8 ___ 145 – 17 x 6
See Solution< (Comparing -17×8 vs -17×6. Since 17×8 > 17×6, subtracting a larger value results in a smaller overall value).
(d) 23 x 48 – 35 ___ 23 x (48 – 35)
See Solution> (LHS = (23 x 48) – 35. RHS = 23 x 13. Since 23 x 48 is much larger than 23 x 13, subtracting 35 won’t make LHS smaller than RHS).
(e) (16 – 11) x 12 ___ -11 x 12 + 16 x 12
See Solution= (RHS can be rewritten using distributive property as (16 – 11) x 12, which is identical to LHS).
(f) (76 – 53) x 88 ___ 88 x (53 – 76)
See Solution> (LHS = 23 x 88 (positive). RHS = 88 x (-23) (negative). Positive > Negative).
(g) 25 x (42 + 16) ___ 25 x (43 + 15)
See Solution= (LHS = 25 x 58. RHS = 25 x 58. The sums inside the brackets are equal).
(h) 36 x (28 – 16) ___ 35 x (27 – 15)
See Solution> (LHS = 36 x 12. RHS = 35 x 12. Since 36 > 35, the LHS is greater).

Page No. 44

5: Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets.

(a) Given: 83 – 37 – 12
Options: (i) 84 – 38 – 12, (ii) 84 – (37 + 12), (iii) 83 – 38 – 13, (iv) -37 + 83 – 12
See SolutionThe terms of the given expression are (83), (-37), and (-12).
Let’s examine the terms of the options:
(i) Terms: (84), (-38), (-12). Not the same.
(ii) 84 – (37 + 12) = 84 – 37 – 12. Terms: (84), (-37), (-12). Not the same.
(iii) Terms: (83), (-38), (-13). Not the same.
(iv) Terms: (-37), (83), (-12).
These are the same terms as the given expression, just rearranged.
Expression equal to the given one: (iv) -37 + 83 – 12.

(b) Given: 93 + 37 x 44 + 76
Options: (i) 37 + 93 x 44 + 76, (ii) 93 + 37 x 76 + 44, (iii) (93 + 37) x (44 + 76), (iv) 37 x 44 + 93 + 76
See SolutionThe terms of the given expression are (93), (37 x 44), and (76).
Let’s examine the terms of the options:
(i) Terms: (37), (93 x 44), (76). Not the same.
(ii) Terms: (93), (37 x 76), (44). Not the same.
(iii) This expression involves multiplying sums, different structure. Not the same.
(iv) Terms: (37 x 44), (93), (76).
These are the same terms as the given expression, just rearranged.
Expression equal to the given one: (iv) 37 x 44 + 93 + 76.

6. Choose a number and create ten different expressions having that value.
See SolutionLet’s choose the number 15. Here are ten different expressions with the value 15:
1. 10 + 5
2. 20 – 5
3. 3 x 5
4. 30 / 2
5. 7 + 8
6. 1 + (2 x 7)
7. (5 x 5) – 10
8. 45 / (1 + 2)
9. 12 + 6 / 2
10. 2 x 8 – 1