Question Answer For All Chapters – Ganita Prakash Class 7th
2.1 Simple Expressions (Page 24)
Choose your favourite number and write as many expressions as you can having that value.
Ans: We can write different arithmetic expressions for this number as follows:
12 + 8 = 20
4 × 5 = 20
40 ÷ 2 = 20
25 – 5 = 20
2 × 10 = 20
100 ÷ 5 = 20
Thus, many different expressions can represent the value 20.
Figure it Out (Page 25)
1. Fill in the blanks to make the expressions equal on both sides of the = sign:
(a) 13 + 4 = ____ + 6 (b) 22 + ____ = 6 × 5
(c) 8 × ____ = 64 ÷ 2 (d) 34 – ____ = 25
Ans:
(a) 13 + 4 = ____ + 6
13 + 4 = 17
So, ____ + 6 = 17
____ = 17 – 6 = 11
13 + 4 = 11 + 6
(b) 22 + ____ = 6 × 5
6 × 5 = 30
So, 22 + ____ = 30
____ = 30 – 22 = 8
22 + 8 = 6 × 5
(c) 8 × ____ = 64 ÷ 2
64 ÷ 2 = 32
So, 8 × ____ = 32
____ = 32 ÷ 8 = 4
8 × 4 = 64 ÷ 2
(d) 34 – ____ = 25
So, ____ = 34 – 25 = 9
34 – 9 = 25
2. Arrange the following expressions in ascending (increasing) order of their values.
(a) 67 – 19 (b) 67 – 20 (c) 35 + 25 (d) 5 × 11 (e) 120 ÷ 3
Ans: (a) 67 – 19 = 48
(b) 67 – 20 = 47
(c) 35 + 25 = 60
(d) 5 × 11 = 55
(e) 120 ÷ 3 = 40
Now, arrange them in ascending (increasing) order:
40, 47, 48, 55, 60
Page 26
Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.
Ans:
(a) Compare 245 + 289 and 246 + 285.
→ If we increase 245 by 1, it becomes 246, but we increase 285 by 1 it becomes 286.
(245 + 1) + 289 ______ 246 + (285 +1)
246 + 289 ______ 246 + 286
246 + 289 > 246 + 286
So left side is 3 more than right side.
∴ LHS > RHS
(b) Compare 273 – 145 and 272 – 144.
We add 1 to both sides
273 + 1 – 145 ______ 272 + 1 – 144
274 – 145 ______ 273 – 144
Both sides are same.
∴ LHS = RHS
(c) Compare 364 + 587 and 363 + 589.
→ If we increase 587 by 1, it becomes 588, but increase 363 by 1, it becomes 364.
364 + (587 + 1) ______ (363 + 1 )+ 589
364 + 588 ______ 364 + 589
364 + 588 ______ 364 + 589
So left side is 2 less than right side.
∴ LHS < RHS
(d) Compare 124 + 245 and 129 + 245.
→ If we increase 124 by 5, it becomes 129. The second term is same (245).
(124 + 5) + 245 ______ 129 + 245
So,
124 + 245 < 129 + 245
Because left side was 5 less.
∴ LHS < RHS
Now, (e) Try To Do yourself.
Terms in Expressions (Pages 28-29)
(Q.) Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.
Ans: Expression: 15 – 7
Replace subtraction: 15 + (–7)
Now, 15 – 7 = 8
15 + (–7) = 8
Same value.
(Q.) Complete the Table:-
Ans:
Does changing the order in which the terms are added give different values?
For example, in the expression 7+(−3)+8:
If we add in the given order: 7+(−3)+8=4+8=12
If we change the order: (−3)+8+7=5+7=12
Swapping and Grouping (Pages 29-31)
(Q.) Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.
Ans: Even if the numbers are negative, changing their order does not affect the sum. For example:
(−3)+(−2)=−5 and if we swap them: (−2)+(−3)=−5
The result is the same in both cases, showing that the sum remains unchanged.
(Q.) Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Ans: 6+(−4)+2 = 2+2 = 4
Swapped order: (−4)+2+6 = −2+6 = 4
Another order: 2+6+(−4) = 8+(−4) = 4
The sum remains 4 in all cases.
(Q.) Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.
Ans: Yes, this rule holds true even when negative numbers are included. Let’s test it with some examples:
(−7)+4+3
Original order: (−7)+4+3 = −3+3 = 0
Changed order: 4+3+(−7) = 7+(−7) = 0
Sum is the same.
(Q.) Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Ans: 5+(−3)
Represented as 5 blue tokens and 3 red tokens.
Pair each red with a blue → 3 pairs cancel.
Leftover = 2 blue tokens = +2.
Now, if we swap: (−3)+5
Start with 3 red tokens and add 5 blue tokens.
After cancellation, again 2 blue tokens remain.
Order does not change the result.
(Q.) 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.
Ans: Yes
(Q.) Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Ans: Do it yourself.
(Q.) 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?
Ans: she doesn’t have to start again. She can just add the missing number to her result.
fourth number is 9055
11749 + 9055 = 20804
More Expressions and Their Terms (Pages 32-33)
(Q.) If the total number of friends goes up to 7 and the tip remains the same, how much will they have to pay? Write an expression for this situation and identify its terms.
Ans: If the total number of friends goes up to 7, then the number of dosas ordered = 7.
the cost of each dosa = ₹ 23
Cost of 7 dosas = 7×23
Adding the tip of ₹5, the total cost = 7×23+5
Now, 7×23 = 161
161+5 = 166
So, they will have to pay ₹166.
(Q.) Think and discuss why she wrote this.
The expression written as a sum of terms is—
Ans: The part 6×5 means she is calculating the cost of 6 items, each costing ₹5.
The +3 means she is adding an extra amount of 3 (maybe a tip or some fixed charge).
So, she has broken down the total into two parts:
6×5 → the cost of 6 items.
3 → the additional fixed amount.
The expression written as a sum of terms is (6×5)+3, where the terms are 6×5 and 3.
(Q.) For each of the cases below, write the expression and identify its terms:
If the teacher had called out ‘4’, Ruby would write ____________
If the teacher had called out ‘7’, Ruby would write ____________
Write expressions like the above for your class size.
Ans: If teacher says 4 → Ruby writes 8 × 4 + 1
Terms: 8 × 4 and 1
If teacher says 7 → Ruby writes 4 × 7 + 5
Terms: 4 × 7 and 5
(Q.) Identify the terms in the two expressions above.
Ans: First expression: 432 = 4×100+1×20+1×10+2×1
Terms: 4×100,1×20,1×10,2×1
Second expression: 432 = 8×50+1×10+4×5+2×1
Terms: 8×50,1×10,4×5,2×1
(Q.) Can you think of some more ways of giving ₹432 to someone?
Ans: Way 1
432 = 2×200+3×10+2×1
2 notes of ₹200, 3 notes of ₹10, 2 coins of ₹1.
Way 2
432 = 4×100+3×10+2×1
4 notes of ₹100, 3 notes of ₹10, 2 coins of ₹1.
Figure it Out (Pages 34-35)
1. Find the values of the following expressions by writing the terms in each case.
(a) 28 – 7 + 8
(b) 39 – 2 × 6 + 11
(c) 40 – 10 + 10 + 10
(d) 48 – 10 × 2 + 16 + 2
(e) 6 × 3 – 4 × 8 × 5
Ans: (a) 28 – 7 + 8
Terms: 28, – 7, +8
28−7+8 = 21+8 = 29
Value = 29
(b) 39 – 2 × 6 + 11
Terms: 39, –2×6, +11
39−(12)+11 = 39−12+11 = 27+11 = 38
Value = 38
(c) 40 – 10 + 10 + 10
Terms: 40, – 10, + 10, + 10
40−10+10+10 = 30+10+10 = 50
Value = 50
(d) 48 – 10 × 2 + 16 + 2
Terms: 48, – 10 × 2, + 16, + 2
48−(20)+16+2 = 28+16+2 = 46
Value = 46
(e) 6 × 3 – 4 × 8 × 5
Terms: 6 × 3, – 4 × 8 × 5
18−(160) = −142
Value = -142
2. Write a story/situation for each of the following expressions and find their values.
(а) 89 + 21 – 10
(b) 5 × 12 – 6
(c) 4 × 9 + 2 × 6
Ans: (a) 89 + 21 – 10
Story:
A shopkeeper had ₹89 in his cash box. A customer paid him another ₹21. Later, he spent ₹10 to buy tea and snacks.
Expression: 89 + 21 – 10
110−10 = 100
Value = ₹100
(b) 5 × 12 – 6
Story:
There are 5 baskets, each containing 12 mangoes. Out of the total mangoes, 6 got spoiled.
Expression: 5 × 12 – 6
60−6 = 54
Value = 54 mangoes
(c) 4 × 9 + 2 × 6
Story:
A teacher arranged 4 rows of 9 chairs each for boys, and 2 rows of 6 chairs each for girls.
Expression: 4 × 9 + 2 × 6
= 36+12 = 48
Value = 48 chairs
3. For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(a) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(b) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(i) for four adults and three children?
(ii) for two groups having three adults each?
(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.
Ans: (a) Elsa had 100 coins, and doubled them → 2×100
Anna had 100 coins, and spent half → \(\frac{1}{2}\)×100
Expression: 2×100+\(\frac{1}{2}\)×100
Terms: 2×100, \(\frac{1}{2}\)×100
200+50 = 250
Together they have 250 gold coins.
(b) Metro Train Tickets
(i) Four adults and three children
Adult ticket = ₹40
Child ticket = ₹20
Expression: 4×40+3×20
Terms: 4×40, 3×20
160+60 = 220
Total = ₹220
(ii) Two groups with three adults each
One group of 3 adults = 3×40 = 120
Two groups = 2×120
Expression: 2×(3×40)
Terms: 2, (3×40)
2×120 = 240
Total = ₹240
(c) There are 7 gaps, each of 5 cm → 7×5
There are 6 grills, each of 2 cm → 6×2
There are 2 borders, each of 3 cm → 2×3
Expression: 7×5+6×2+2×3
Terms: 7×5, 6×2, 2×3
35+12+6=53
So, the total height of the window = 53 cm.
Figure it Out (Pages 36-37)
1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
Ans: (a) 24 + (6 – 4) = 24 + 6 – 4
(b) 38 + (9 – 4) = 38 + 9 – 4
(c) 24 – (6 + 4) = 24 – 6 – 4
(d) 24 – 6 – 4 = 24 – 6 – 4
(e) 27 – (8 + 3) = 27 – 8 – 3
(f) 27 – (8 – 3) = 27 – 8 + 3
2. Remove the brackets and write the expression having the same value.
(a) 14 + (12 + 10) (b) 14 – (12 + 10)
(c) 14 + (12 – 10) (d) 14 – (12 – 10)
(e) –14 + 12 – 10 (f) 14 – (–12 – 10)
Ans: (a) 14 + 12 + 10
(b) 14 – 12 – 10
(c) 14 + 12 – 10
(d) 14 – 12 + 10
(e) -14 + 12 – 10
(f) 14 + 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)
(b) 16 – (8 – 3) and (16 – 8) – 3
(c) 27 – (18 + 4) and 27 + (-18 – 4)
Ans: (a) (6+10)–2 and 6+(10–2)
LHS: (6+10)–2 = 16–2 = 14
RHS: 6+(10–2) = 6+8 = 14
Both are equal (14).
(b) 16 – (8 – 3) and (16 – 8) – 3
LHS:16–(8–3) = 16–5 = 11
RHS: (16–8)–3 = 8–3 = 5
They are not equal (11 ≠ 5).
(c) 27–(18+4) and 27+(–18–4)
LHS: 27–(18+4) = 27–22 = 5
RHS: 27+(–18–4) = 27–22 = 5
Both are equal (5).
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
(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109
Ans:
Hence, the three expressions 87+46–109 are equal in value.
Similarly, the expressions 87–46+109 and (87–46)+109 also represent the same value.
5. Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(a) 34 – 9 + 12 = 13
(b) 56 – 14 – 8 = 34
(c) -22 – 12 + 10 + 22 = – 22
Ans: (a) 34 – 9 + 12 = 13
= 34 – (9 + 12) = 34 – 21
= 13
(b) 56 – 14 – 8 = 34
= 56 – (14 + 8)
= 56 – 22
= 34
(c) –22 – (12 + 10) + 22
= –22 – 22 + 22
= –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 + ________
(b) 207 – 68 = 210 – ________
Ans: (a) 423 + 419 = 419 + 423
(b) 207 – 68 = 210 – 71
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. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.
Ans: 2–(3–5) = 2–(–2) = 4
2–(5–3) = 2–2 = 0
3–(2–5) = 3–(–3) = 6
3–(5–2) = 3–3 = 0
5–(2–3) = 5–(–1) = 6
8. Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1.
(a) Do you think she always gets the correct answer? Why?
(b) Can you think of other similar strategies? Give some examples.
Ans: (a) Yes, because subtracting 9 is the same as subtracting 10 and then adding 1.
For example: 25−9 = (25−10)+1 = 15+1 = 16
(b) In the same way, adding 99 can be thought of as adding 100 and then subtracting 1.
For example, to add 99 to 176: 176+99 = (176+100)−1 = 276−1 = 275
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.
(a) 73 – (14 + 1)
(b) 73 – (14 – 1)
(c) 73 + (-14 + 1)
(d) 73 + (-14 – 1)
Ans: (a)73–(14+1) = 73–15 = 58 → not equal
(b) 73–(14–1) = 73–13 = 60 → equal
(c) 73+(–14+1) = 73–13 = 60 → equal
(d) 73+(–14–1) = 73–15 = 58 → not equal
Removing Brackets—II (Pages 39-40)
If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid?
Ans: One person’s share = 43+24
Two people’s share = 2×(43+24)
If Sangmu (third friend) also orders the same items, then three people’s share = 3×(43+24)
3×(43+24) = 3×67 = 201
(Q.) 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?
Ans: 5 × (4 + 3) means: multiply 5 by the sum of 4 and 3.
5×(4+3) = 5×7 = 35
5 × 4 + 3 means: multiply 5 × 4 first, then add 3.
5×4+3 = 20+3 = 23
So clearly, 23 ≠ 35.
5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5
Addition is commutative, which means 4 + 3 = 3 + 4.
Multiplication is also commutative, so 5 × 7 = 7 × 5
5×(4+3) = 5×(7)=35
5×(3+4) = 5×7=35
(3+4)×5 = 7×5=35
Tinker the Terms II (Page 41)
(Q.) Use this method to find the following products:
(a) 95 × 8
(b) 104 × 15
(c) 49 × 50
Ans: (a) 95 × 8
Break 95 as 100 – 5
100 × 8 – 5 × 8
= 800 – 40
= 760
(b) 104 × 15
Break 104 = 100 + 4
= (100 + 4) × 15
= 100 × 15 + 4 × 15
= 1500 + 60
= 1560
(c) 49 × 50
Break 50 = 50 – 1
= (50 – 1) × 50
= 50 × 50 – 50 × 1
= 2500 – 50
= 2450
Figure it Out (Pages 41-42)
1. Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.
Ans:
(а) 3 × (6 + 7) = 3 × 6 + 3 × 7
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4
(c) 3 × (5 + 8) = 3 × 5 + 3 × 8
(d) (9 + 2) × 4 = 9 × 4 + 2 × 4
(e) 3 × (5 + 4) = 3 × 5 + 3 × 4
(f) (13 + 6) × 4 = 13 × 4 + 24
(g) 3 × (5 + 2) = 3 × 5 + 3 × 2
(h) (2 + 3) × 4 = 2 × 4 + 3 × 4
(i) 5 × (9 – 2) = 5 × 9 – 5 × 2
(j) (5 – 2) × 7 = 5 × 7 – 2 × 7
(k) 5 × (8 – 3) = 5 × 8 – 5 × 3
(l) (8 – 3) × 7 = 8 × 7 – 3 × 7
(m) 5 × (12 – 3) = 60 – 5 × 3
(n) (15 – 6) × 7 = 105 – 6 × 7
(o) 5 × (9 – 4) = 5 × 9 – 5 × 4
(p) (17 – 9) × 7 = 17 × 7 – 9 × 7
2. In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.
Ans:
(a) (8 – 3) × 29 ____ (3 – 8) × 29
LHS = (8–3)×29 = 5×29 = 145
RHS = (3–8)×29 = (–5)×29 = –145
Since 145>–145, therefore LHS > RHS
(b) 15 + 9 × 18 ____ (15 + 9) × 18
LHS = 15+9×18 = 15+162 = 177
RHS = (15+9)×18 = 24×18 = 432
Since 177<432,therefore LHS < RHS
(c) 23 × (17 – 9) ____ 23 × 17 + 23 × 9
LHS = 23×(17–19) = 23×(–2) = –46
RHS = 23×17+23×9 = 391+207 = 598
Since –46<598,therefore LHS < RHS
(d) (34 – 28) × 42 ____ 34 × 42 – 28 × 42
LHS = (34–28)×42 = 6×42 = 252
RHS = 34×42–28×42 = 1428–1176 = 252
Since both are equal, therefore LHS = RHS
3. Here is one way to make 14: 2 × ( 1 + 6 ) = 14. Are there other ways of getting 14? Fill them out below:
(a) _____× (_____+_____) = 14
(b) _____× (_____+_____) = 14
(c) _____× (_____+_____) = 14
(d) _____× (_____+_____) = 14
Ans: 2 × (6 + 1) = 14
(a) 2 × (2 + 5) = 14
2 × (3 + 4) = 14
1 × (12 + 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.
Ans:
First Way :
(4 × 5) + (8 × 4) = 20 + 32 = 52
Second way :
First row: 4 + 8 + 4 = 16
Second row: 8 + 4 + 8 = 20
Third row: 4 + 8 + 4 = 16
Total sum: 16 + 20 + 16 = 52
Ans:
First Way :
5 × 8 + 6 × 8 = 40 + 48 = 88
Second way :
8 × (5 + 6) = 8 × 11 = 88
Third way :
Each row: 5 + 6 + 6 + 5 = 22 (or equivalently for other rows: 6 + 5 + 5 + 6 = 22)
Total sum: 22 × 4 = 88
Figure it Out (Pages 42-44)
1. Read the situations given below. Write appropriate expressions for each of them and find their values.
(a) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(b) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(c) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?
Ans: (a) Mango supply
Rahim: 9kg per day × 7days = 9×7 = 63kg
Shyam: 11kg per day × 7days = 11×7 = 77kg
Total supply = 63+77 = 140kg
(b) Binu’s savings
Monthly income = ₹20,000
Monthly expenses = ₹5,000 + ₹5,000 + ₹2,000 = ₹12,000
Monthly savings = 20,000–12,000 = 8,000
Yearly savings = 8,000×12 = 96,000
(c) Snail climbing
Day climb = 3 cm, night slip = 2 cm → Net climb per day = 3–2 = 1cm
But on the 8th day the snail will climb 3 cm before slipping at night and directly reach the top without slipping.
Height of post = 10 cm.
After 7 days, net climb = 7 cm.
On the 8th day morning, snail climbs 3 cm more and reaches 10 cm.
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 × 2 × 8
(b) (7 – 2) × 8
(c) 8 × 7
(d) 7 × 2 × 8
(e) 7 × 5 – 2
(f) (7 + 2) × 8
(g) 7 × 8 – 2 × 8
(h) (7 – 5) × 8
Ans: Number of reading days in 1 week
= Total days in a week – Days he doesn’t read
= 7–2 = 5
Number of stories read in 8 weeks
= 5×8 = 40
Now, 5×8 can also be written as:
(7–2) × 8 or 7×8 – 2×8
Hence, the correct options are (b) and (g).
3. Find different ways of evaluating the following expressions:
(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Ans: (a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
Way 1: Pair terms
(1–2) + (3–4) + (5–6) + (7–8) + (9–10)
= –1+–1+–1+–1+–1 = –5
Way 2: Group positives and negatives separately
(1 + 3 + 5 + 7 + 9) + (-2 – 4 – 6 – 8 – 10)
25 + (-30) = -5
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Way 1: Pair consecutive terms
= (1–1)+(1–1)+(1–1)+(1–1)+(1–1)
= 0+0+0+0+0 = 0
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
= (1 + 1 + 1 + 1 + 1) + (-1 – 1 – 1 – 1 – 1)
= 5 + (-5)
= 0
4. Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning.
Ans: (a) (49 – 7) + 8
= 42 + 8 = 50
49 – (7 – 8)
= 49 – (-1)
= 49 + 1 = 50
50 = 50
(b)
83 × 42 – 18
= 3486 – 18 = 3468
83 × 40 – 18
= 3320 – 18 = 3302
3468 > 3302
(c) LHS = 145 – 17 × 8
= 145 – 136
= 9
RHS = 145 – 17 × 6
= 145 – 102
= 43
LHS < RHS
(d)
23 × 48 − 35
1104 − 35 = 1069
23 × (48 – 35)
= 23 × 13 = 299
1029 > 299
(e)
(16 – 11) × 12
= 5 × 12 = 60
-11 × 12 + 16 × 12
= -132 + 192 = 60
60 = 60
(f) (76 – 53) × 88
= 23 × 88 = 2024
RHS = 88 × (53 – 76)
= 88 × (-23) = -2024
2024 > −2024
(g) 25 × (42 + 16)
= 25 × 58 = 1450
25 × (43 + 15)
= 25 × 58
= 1450
1450 = 1450
(h) 36 × (28 – 16)
= 36 × 12 = 432
35 × (27 – 15)
= 35 × 12 = 420
432 > 420
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. There can be more than one expression which is equal to the given expression.
(a) 83 – 37 – 12
- (i) 84 – 38 – 12
- (ii) 84 – (37 + 12)
- (iii) 83 – 38 – 13
- (iv) – 37 + 83 –12
Ans: (a) Expression: 83 – 37 – 12
This means: (83–37)–12.
(i) 84 – 38 – 12
→ (84–38)–12 = 46–12 Same as original.
Equal
(ii) 84 – (37 + 12)
→ 84–49.Different.
Not equal
(iii) 83 – 38 – 13
→ (83–38)–13 = 45–13 Different.
Not equal
(iv) –37 + 83 – 12
→ Rearrangement of 83–37–12.
Equal
∴ (i) and (iv) give the same value.
(b) 93 + 37 × 44 + 76
- (i) 37 + 93 × 44 + 76
- (ii) 93 + 37 × 76 + 44
- (iii) (93 + 37) × (44 + 76)
- (iv) 37 × 44 + 93 + 76
Ans: (b) Expression: 93 + 37 × 44 + 76
By BODMAS: 93 + (37×44) + 76.
(i) 37 + 93 × 44 + 76
→ Multiplication with 93, not 37.
Not equal
(ii) 93 + 37 × 76 + 44
→ Multiplication with 76, not 44.
Not equal
(iii) (93 + 37) × (44 + 76)
→ Product of sums, not the same.
Not equal
(iv) 37 × 44 + 93 + 76
→ Just rearranged, exactly the same.
Equal
Hence, (iv) gives the same value.
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