By using Ganita Prakash Class 7 Solutions and Part 1 Chapter 2 Arithmetic Expressions Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Chapter 2 Arithmetic Expressions Solutions
Ganita Prakash Class 7 Chapter 2 Solutions
Class 7 Maths Ganita Prakash Chapter 2 Solutions Arithmetic Expressions
2.1. Simple Expressions
Page: 24
Question 1.
Choose your favourite number and write as many expressions as you can having that value.
Answer:
My favourite number is ‘ 7 ‘.
Following are the few expressions using my favourite number ‘ 7 ‘:
7 + 20, 7 – 3, 7 + (7 – 3) + 21 ÷ 3, 7 × 4 + 3 × (2 – 7), (7 + 3) × 25, etc.
Page : 25
Question 2.
Figure it Out :
1. Fill in the blanks to make the expressions equal on both sides of the = sign:
(a) 13 + 4 = ________ + 6
Answer:
LHS = 13 + 4 = 17
Let the blank space be filled with ‘x’.
Then, RHS = x + 6
Since, LHS = RHS
Hence, 17 = x + 6
⇒ x = 17 – 6
⇒ x = 11
(b) ________ = 6 × 5
Answer:
Here, 22 + ? = 30
⇒ ? = 30 – 22
= 8
(c) ________ = 64 ÷ 2
Answer:
⇒ 8 × ? = 32
⇒ ? = 32 ÷ 8
= 4
(d) ________ = 25
Answer:
⇒ 34 – ? = 25
(Since ? is preceded by negative sign)
⇒ 34 – 25 = ?
9 = ?
∴ ? = 9
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
Answer:
(a) 67 – 19 = 48
(b) 67 – 20 = 47
(c) 35 + 25 = 60
(d) 5 × 11 = 55
(e) 120 ÷ 3 = 40
Ascending order of their values: –
(e) 40 < (b) 47 < (a) 48 < (d) 55 < (c) 60.
2.2. Reading and Evaluating Complex Expressions
Page: 26
Question 3.
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.
Answer:
(a) 245 + 289 [>] 246 + 285
LHS = 245 + 289 = 534
RHS = 246 + 285 = 531
LHS > RHS
If we compare numbers on both the sides one by one, we find numbers in LHS are 1 less in Ist and 4 greater in IInd than RHS, So, resultant is positive with LHS. So, LHS > RHS.
(b) 273 – 145 [ = ] 272 – 144
LHS = 273 – 145 = 128
RHS = 272 – 144 = 128
LHS = RHS
Here, 273 > 272 and 145 > 144, in both cases difference is 1.
(c) 364 + 587 [<] 363 + 589
LHS = 364 + 587 = 951
RHS = 363 + 589 = 952
So, LHS < RHS
Here, 364 is 1 more than 363 but 587 is 2 less than 589.
(d) 124 + 245 [<] 129 + 245
We can do it in some different way in which we will ignore the equal terms and compare only the unequal terms on both sides.
Here, 124 < 129
∴ LHS < RHS
(e) 213 – 77 [<] 214 – 76
LHS = 213 – 77 = 136
RHS = 214 – 76 = 138
So, LHS < RHS
Here, 213 is 1 less than 214 but 77 is one more than 76.
Page: 28
Question 4.
Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.
Answer:
(i) 76 – 14 = 76 + (- 14)
62 = 62
So, LHS = RHS
(ii) 104 – 24 = 104 + (- 24)
80 = 80
So, LHS = RHS
(iii) 87 – 16 = 87 + (- 16)
71 = 71
So, LHS = RHS
(iv) 102 – 62 = 102 + (- 62)
40 = 40
So, LHS = RHS
So, we can conclude that replacing subtraction by addition in this way does not change the value of the expression.
Question 5.
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?
Answer:
Let us subtract 6 from 14 by using the Token Model:
Page: 31
Question 6.
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.
Answer:
Examples: –
(i) (12 + 7) + 11 = 12 + (17 + 11)
→ 29 + 11 = 12 + 28
→ 40 = 40
(ii) (20 + 4) + 7 = 20 + (4 + 7)
→ 24 + 7 = 20 + 11
→ 31 = 31
(iii) (72 + 36) + (8 + 14) = 72 + (36 + 8) + 14
→ (108 + 22)
→ 130 = (72 + 44) + 14
→ 130 = 130
Question 7.
Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Answer:
Let us take an example as:
(4 + 3) + 2 = 4 + (3 + 2)
LHS = RHS
and they evaluate to ‘9’.
Question 8.
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?
Answer:
No, she need not to – start it all over again as she can do it by adding the fourth number ‘9055’ to the result she has obtained which is ‘11749’ to obtain the final sum as ‘20804’.
Note: This is due to ‘Commutative’ and “Associative” property of addition of numbers.
Page: 31 and 32
Question 9.
Manasa is going outside to play. Her mother says, “Wear your hat and shoes!” Which one should she wear first? She can wear her hat first and then her shoes. Or she can wear her shoes first and then her hat.
Manasa will look exactly the same in both cases. Imagine a different situation: Manasa’s mother says “Wear your socks and shoes!” Now the order matters. She should wear socks and then shoes. If she wears shoes and then socks, Manasa will feel very uncomfortable and look very different.
Answer:
In the first case, order does not matter as hat and shoes are worn at different parts of the body.
In the second case, order does matter to a great extent as socks and shoes are worn at one part that is our foot and there is a proper order of wearing them.
Page: 32
Question 9.
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.
Answer:
If the total number of friedns goes up to 7 and the tip remains same, the expression for this situation will be:
It’s terms are 7 × 23 and 5.
Page: 32
Question 10.
Think and discuss why she wrote this. The expression written as a sum of terms is –
Answer:
She has written the expression as there are 6 groups of 5 students and one group of 3 students.
Note : The expression evaluates to 33 meaning there are 33 students in all.
Page: 33
Question 11.
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.
Answer:
Question 12.
Identify the terms in the two expressions above.
Answer:
Two expression and given above are:
(i) 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1
(ii) 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1
Terms in the expression (i) are:
4 × 100, 1 × 20, 1 × 10, 2 × 1
and terms in the expression (ii) are:
8 × 50, 1 × 10, 4 × 5, 2 × 1
Question 13.
Can you think of some more ways of giving ₹432 to someone?
Answer:
Yes, 4 × 100 + 3 × 10 + 2 × 1,
Meaning: 4 notes of ₹100, 3 notes’of ₹10 and 2 notes of ₹1.
Another way can be:
2 × 200 + 1 × 20 + 1 × 10 + 2 × 1
Meaning: 2 notes of ₹200, 1 note of ₹20, 1 note of ₹10, 2 notes of ₹1.
Note: There can be many more ways of giving ₹432 to someone.
Page: 34
Question 14.
What is the expression for the arrangement in the right making use of the number of yellow and blue squares?
Answer:
The expressions for the given arrangement can be :
(5 + 3) + (5 + 3)
or, 2 × (5 + 3)
or, 2 × 5 + 2 × 3
Page: 34, 35
Question 15.
Figure it Out :
1. Find the values of the following expressions by writing the terms in each case.
(a) 28 – 7 + 8
Answer:
(b) 39 – 2 × 6 + 11
Answer:
(c) 40 – 10 + 10 + 10
Answer:
(d) 48 – 10 × 2 + 16 ÷ 2
Answer:
(e) 6 × 3 – 4 × 8 × 5
Answer:
2. Write a story/situation for each of the following expressions and find their values.
(a) 89 + 21 – 10
(b) 5 × 12 – 6
(c) 4 × 9 + 2 × 6
Answer:
(a) 89 + 21 – 10
Aarya was going to market. Her mother gave her ₹89 and her grand mother gave her ₹21. In the market she spant ₹10. Now, how much money Aarya has now?
To solve: 89 + 21 – 10, we write
So, Aarya has ₹100 left with her now.
(b) Shreyash purchased 5 dozens of pens and gave half dozen to his sister Sonu. How many pens are left with Shreyas now?
Hence, Shreyash has 54 pens with him now.
(c) Aadya gave 4 toffees each to her 9 friends and 2 toffees each to 6 of her friends. How many toffees Aadya distributed in all?
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.
Answer:
Expression describing the given situation is: Gold coins Princess Elsa and Princess Anna together have :
2 × 100 + \(\frac{1}{2}\) × 100
(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?
Answer:
Total cost of tickets for four adults and three children
(ii) for two groups having three adults each?
Answer:
Total cost of tickets for two groups having three adults each = 3 × 40
Its terms are: –
(c) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.
Answer:
Here, number of Borders = 1 top + 1 bottom = 2,
number of Gaps = 7,
number of Grills = 6,
width of 1 Border = 3 cm,
width of 1 Gap = 5 cm and
width of 1 Grill = 2 cm
An expression to represent the height of the window = 2 × 3 + 7 × 5 + 6 × 2
∴ total height of the window = 53 cm.
Page: 37
Answer:
Page: 347
Question 16.
Figure it Out :
1. Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.
Answer:
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)
Answer:
(a) 14 + (12 + 10) = 14 + 12 + 10
(b) 14 – (12 + 10) = 14 – 12 – 10
(c) 14 + (12 – 10) = 14 + 12 – 10
(d) 14 – (12 – 10) = 14 – 12 + 10
(e) – 14 + 12 – 10 = – 14 + 12 – 10
(f) 14 – (- 12 – 10) = 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)
Answer:
(a) My guess: They have the same values.
(6 + 10) – 2 and 6 + (10 – 2)
→ 16 – 2 and 6 + 8
→ 14 and 14
Yes, the two expressions are equal.
(b) My guess: They have different values.
16 – (8 – 3) and (16 – 8) – 3
→ 16 – 5 and 8 – 3
→ 11 and 5
No, the two expressions are not equal.
(c) My guess: They have same values.
27 – (18 + 4) and 27 + (- 18 – 4)
→ 27 – 22 and 27 – 22
→ 5 and 5
Yes, the two expressions are equal.
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
Answer:
319 – 537 and – 537 + 319, have the same value.
(b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 + 109, 87 – (46 + 109), (87 – 46) + 109
Answer:
87 – 46 + 109 and (87 – 46) + 109, have 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
Answer:
(a) 34 – (9 + 12) = 13
(b) 56 – (14 + 8) = 34
(c) – 22 – (12 + 10) + 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 – ______
Answer:
(a) 423 + 419 = 419 + 423
(Commutative Property)
(b) 207 – 68 = 210 – 71
(210 is 3 more than 207, so to get equal values on both sides we increase 68 by 3)
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.
Answer:
2 + 3 + 5 = 10
-2 + 3 + 5 = 6
-2 – 3 + 5 = 0
-2 – 3 – 5 = – 10
-2 + 3 – 5 = – 4
2 + 3 – 5 = 0
2 – 3 – 5 = – 6
2 – 3 + 5 = 4
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.
Answer:
(a) Yes, I think she will always get the correct answer because “subtracting 10 and then adding 1 to it gives as – 9 added to the number or 9 subtracted from the number”.
(b) Yes, there can be many more similar strategies as in part (a). For examples –
(i) to subtract 8 from any number we can, subtract ’10’ from the number and add ‘2’ to the result.
(ii) to subtract 5 from any number we can subtract ‘ 10 ‘ from the number and add ‘5’ to the result, the same trouble though!
(iii) to add 11 to any number we can add ‘ 10 ‘ to the number first and then add ‘ 1 ‘ to the result.
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)
(a) 73 – (14 + 1)
Answer:
From the given collection (b) and (c) are same as (a). All evaluate to 60.
(b) 73 – (14 – 1)
Answer:
From the given collection (a) and (d) are same as (b). All evaluate to 58.
Page: 38
Question 17.
What about the total amount they have to pay? Can it be described by the expression: 2 × 43 + 24 ?
Answer:
No, the total amount they have to pay cannot be described by the expression:
2 × 43 + 24
By writing the expression as 2 × (43 + 24), using bracket we can get the total amount they have to pay.
Page: 39
Question 18.
If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid?
Answer:
After joining in by Songmu, the expression for the total amount to be paid will be:
3 × (43 + 24)
Note: The above expression can also be written as: 3 × 43 + 3 × 24
or, (43 + 24) + (43 + 24) + (43 + 24)
or, (43 + 43 + 43) + (24 + 24 + 24), etc.
Page: 40
Question 19.
Is 5 × 4 + 3 ≠ 5 ×(4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5 ?
Answer:
In the expression: 5 × 4 + 3 ≠ 5 ×(4 + 3)
Clearly, 23 ≠ 35
Hence, 5 × 4 + 3 ≠ 5 ×(4 + 3)
In the expression:
5 ×(4 + 3) = 5 ×(3 + 4) = (3 + 4) × 5
LHS = 5 ×(4 + 3) = 5 × 7 = 35
RHS = (3 + 4) × 5 = 7 × 5 = 35
Middle term is 5 × (3 + 4) = 5 × 7 = 35
Since, LHS, RHS and the middle terms evaluate to the same value 35
Hence, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5.
Note: The multiple of a sum (difference) is the same as the sum (difference) of the multiples.
Page: 41
Question 20.
Use this method to find the following products:
(a) 95 × 8
(b) 104 × 15
(c) 49 × 50
Is this quicker than the multiplication procedure you use generally?
Answer:
(a) 95 × 8 = 100 × 8 – 5 × 8
= 800 – 40
= 760
(b) 104 × 15 = 100 × 15 + 4 × 15
= 1500 + 60
= 1560
(c) 49 × 50 = 50 × 50 – 1 × 50
= 2500 – 50
= 2450
Yes, this is a quicker method.
Question 21.
Which other products might be quicker to find like the ones above?
Answer:
This is the quickest method at this point.
Page: 41, 42
Question 22.
Figure it Out :
1. Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.
Answer:
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.
Answer:
(a) (8 – 3) × 29 > (3 – 8) × 29
My analysis: (8 – 3) > (3 – 8), the other number multiplied with these numbers is same which is 29.
(b) 15 + 9 × 18 < (15 + 9) × 18
My analysis: 15 is added to the product of 9 and 18 will be smaller than the number 15 × 18 added to the product of 9 and 18.
(c) 23 ×(17 – 9)>23 × 17 + 23 × 9
My analysis: 23 ×(17 – 9) is same as 23 × 8, which is smaller than any of the terms in the RHS and the terms on the RHS are being added.
(d) (34 – 28) × 42>34 × 42 – 28 × 42
My analysis: If we open LHS to get the same expression in the RHS.
3. Here is one way to make 14 : 2 × (1 + 6) = 14. Are there other ways of getting 14? Fill them out below:
Answer:
(a) 7 × (1 + 1) = 14
(b) 2 × (2 + 5) = 14
(c) 2 × (3 + 4) = 14
(d) 7 × (2 + 0) = 14
Note: There can be more ways.
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.
Answer:
(i)
(ii)
Note: there can be many other ways.
Page: 42, 43, 44
Question 23.
Figure it Out :
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.
Answer:
(a) Mangoes supplied by Rahim each day = 9 kg
Mangoes supplied by Shyam each day = 11 kg
Total mangoes supplied by both on each day = 9 + 11 = 20 kg
Number of days in a week the local district market operates = 7 (all days in a week)
∴ Total amount of mangoes supplied by them in a week to the Local district market = 7 × 20
(number of days it operates × amount of mangoes supplied each day)
= 140 kg
(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?
Answer:
(b) Money earned by Binu in a month = ₹20,000
Money she spends on :
Rent = ₹ 5,000
Food = ₹ 5,000
Others = ₹ 2,000
∴ , Total spends per month = ₹ 12,000
The amount Binu will save by the end of a year
= Number of months × money saved each month
= 12 × (Total earning – Total spends)
= 12 × (20000 – 12000)
= 12 × 8000
= 96000
So, Binu will save ₹ 96,000 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?
Answer:
(c) Height climbed by a snail each day = 3 cm
∴ , It slips down by 2 cm each day.
∴ , height it climbed each day = (3 – 2) = 1 cm
The height of the post where treat is kept = 10 cm
So, number of days required by the snail to get the treat = 7 + 1
(At a height of 7 cm , the snail will climb 3 cm in day time to rise to 10 cm)
Hence, the snail will get the treat in 8 days.
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
Answer:
Melvin reads a two – page story for 5 days in a week. (He does not read the story on Tuesdays and Saturdays)
∴ , number of stories completed by him in 8 weeks = 8 × 5 = 40
Among the given options, (b) (7 – 2) × 8, represents this scenario.
(g) 7 × 8 – 2 × 8, also represents the same scenario.
So, (b) (7 – 2) × 8 and (g) 7 × 8 – 2 × 8, both represent the given scenario.
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
Answer:
(a)
Another way can be:
(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Another way can be:
Note: There can be many more ways. Students are advised to do more ways.
4. Compare the following pairs of expressions using ‘<‘, ‘>’ or ‘ = ‘ by reasoning.
(a) 49 – 7 + 8, 49 – 7 + 8
(b) 83 × 42 – 18,83 × 40 – 18
(c) 145 – 17 × 8,145 – 17 × 6
(d) 23 × 48 – 35,23 × (48 – 35)
(e) (16 – 11) × 12, – 11 × 12 + 16 × 12
(f) (76 – 53) × 88, 88 × (53 – 76)
(g) 25 × (42 + 16), 25 × (43 + 15)
(h) 36 × (28 – 16), 35 × (27 – 15)
Answer:
(a) 49 – 7 + 8 [=] 49 – 7 + 8
Reason: Both expressions are same.
(b) 83 × 42 – 18 [>] 83 × 40 – 18
Reason: 42 > 40, same number is subtracted from the numbers obtained.
(c) 145 – 17 × 8 [<] 145 – 17 × 6
Reason: 8 > 6 and 17 × 8 and 17 × 6 are being subtracted from the same number. Bigger number subtracted from the same number gives the smaller number.
(d) 23 × 48 – 35 [>] 23 × (48 – 35)
Reason: In the LHS, 35 is being subtracted from 23 × 48 but in the RHS, 23 × 35 is being subtracted from 23 × 48.
(e) (16 – 11) × 12 [=] – 11 × 12 + 16 × 12
Reason: Expression in LHS = 5 × 12
Expression in RHS = (- 11 + 16) × 12
∴ Both will have the same value.
(f) (76 – 53) × 88 [>] 88 × (53 – 76)
Reason: 76 – 53 > 53 – 76 and they both are multiplied with the same number 88, so the expression in LHS will remain greater than the expression in RHS.
(g) 25 × (42 + 16) [>] 25 × (43 + 15)
Reason: 42 + 16 is equal to 43 + 15 and both are multiplied by the same number 25.
(h) 36 × (28 – 16) [>] 35 × (27 – 15)
Reason: (28 – 16) = (27 – 15), 36 > 35
So, LHS > RHS
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
(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
Answer:
(a) 83 – 37 – 12 (given)
(i) 84 – 38 – 12 (same, as 84 is one more than 83 and 38 is one more than 37)
(ii) 84 – (37 + 12) = 84 – 37 – 12 (not same, as 84 is 1 more than 83, other numbers and their signs are same.
(iii) 83 – 38 – 13 (not same, numbers subtracted are increased by 1, each, resulting in a decrease of 2 overall)
(iv) – 37 + 83 – 12 = 83 – 37 – 12, (Tinkering the terms)
(same as the given expression) Therefore, (i) and (iv) evaluate to same value.
(b) 93 + 37 × 44 + 76
(i) 37 + 93 × 44 + 76 (not same, as by tinkering the terms we cannot get the given expression)
(ii) 93 + 37 × 76 + 44 (not same, as by tinkering the terms we cannot get the given expression)
(iii) (93 + 37) ×(44 + 76) = 93 × (44 + 76) + 37 × (44 + 76) (not same)
(iv) 37 × 44 + 93 + 76 = 93 + 37 × 44 + 76 (Tinkering the terms gives us the same expression as that is given)
So, (iv) evaluates to the same value as the given expression evaluates to.
6. Choose a number and create ten different expressions having that value
Answer:
Let us take a number ‘100’.
Ten different expressions having value ‘100’ are:
(i) 10 × (15 – 5)
(ii) 4 × 5 + 8 × 10
(iii) 6 × (10 + 5) + 8 + 2
(iv) 3 × (20 + 10) + 7 + 3
(v) 4 × (12 + 8) + 20
(vi) 2 × (25 + 5) + (12 + 8) × 2
(vii) 20 + 30 + 50
(viii) 5 × 9 + 5 × 11
(ix) 9 × 11 + 1
(x) 10 × 8 + 4 × 4 + 2 × 2