(a) 13 + 4 = __ + 6
LHS (Left Hand Side) = 17. To make RHS (Right Hand Side) 17, blank must be 17 – 6 = 11.

(b) 22 + __ = 6 × 5
RHS = 30. To make LHS 30, blank must be 30 – 22 = 8.

(c) 8 × __ = 64 ÷ 2
RHS = 32. To make LHS 32, blank must be 32 ÷ 8 = 4.

(d) 34 – __ = 25
To get 25 from 34, we must subtract 34 – 25 = 9.

First, find the value of each expression:

• (a) 67 – 19 = 48
• (b) 67 – 20 = 47
• (c) 35 + 25 = 60
• (d) 5 × 11 = 55
• (e) 120 ÷ 3 = 40

Ascending Order: 40, 47, 48, 55, 60.
In terms of expressions: 120 ÷ 3, 67 – 20, 67 – 19, 5 × 11, 35 + 25.

(a) 245 + 289 __ 246 + 285
The left side adds a bigger number (289 > 285) to a smaller number (245 < 246). From 245 to 246 is a change of +1. From 289 to 285 is a change of -4. The net change is -3. So the right side is smaller. Let's reason: The right side took 1 from 289 and gave it to 245, leaving it with 288. So we compare 245+289 with 246+288. Still hard. Let's use the logic from the book: Start with 245+289. Change it to 246+289 (value increased by 1). Now change that to 246+285 (value decreased by 4). The total change is +1 - 4 = -3. So, 245 + 289 is greater. 245 + 289 > 246 + 285

(b) 273 – 145 __ 272 – 144
Raja starts with 273 and loses 145. Joy starts with 272 (1 less than Raja) and loses 144 (1 less than Raja). Since they both lost 1 less, the difference remains the same. This is equivalent to (272+1) – (144+1) = 272 – 144. So they are equal. 273 – 145 = 272 – 144

(c) 364 + 587 __ 363 + 589
The right side took 1 from 364 and gave it to 587, making it 588. So we compare 364+587 with 363+588. The right side has a number that is 1 smaller and a number that is 1 bigger. So the sum is the same. Wait, 589 is 2 more than 587, and 363 is 1 less. The net change is +1. RHS = (364 – 1) + (587 + 2) = 364 + 587 + 1. The right side is bigger. 364 + 587 < 363 + 589

(d) 124 + 245 __ 129 + 245
Both sides add 245. Since 124 < 129, the left side is smaller. 124 + 245 < 129 + 245

(e) 213 – 77 __ 214 – 76
Raja has 214 and loses 76. Joy has 213 (1 less) and loses 77 (1 more). Joy will have less. So the left side is smaller. 213 – 77 < 214 - 76

No. Addition has the commutative property (order doesn’t matter) and associative property (grouping doesn’t matter). Manasa can simply add the forgotten number, 9055, to her current total.

New sum = 11749 + 9055 = 20804.

Cost of 7 dosas = 7 × 23. Tip = 5.

Expression: (7 × 23) + 5

Terms: The terms are (7 × 23) and 5.

Value: (7 × 23) + 5 = 161 + 5 = ₹166.

Total students = 33 + 1 (Ruby) = 34. Ruby is out.

If teacher calls out ‘4’: 33 students must form groups of 4. 33 ÷ 4 = 8 with a remainder of 1. So there are 8 groups of 4, and 1 student left out. Ruby is also out. Total out = 1+1=2. This is not the logic in the book.

The book’s logic: 33 students are playing. How many full groups can be made, and how many are left over?
If teacher calls ‘4’: 33 = (8 × 4) + 1. There are 8 groups and 1 student leftover. Ruby is also ‘out’. So expression for those not in full groups is 8 × 4 + 1 if Ruby isn’t counted as ‘out’. The book’s example 6×5+3 suggests the number of players is 33. Number of groups of 5 is 6, with 3 left over. And Ruby is also out. So it should be (6×5) + 3 + 1. The logic in the text is that Ruby herself writes the expression for the total number of people not in a group of 5. There are 33 playing students. 33/5 gives 6 full groups, and 3 students are left over. Ruby is also not in a group. So the total number of people not in a group of 5 is 3+1 = 4. Why did she write 6×5+3? This represents the total number of students. 33 = 6×5 + 3. Ah, I see. If teacher called ‘4’, Ruby would write: 8 × 4 + 1 (since 33 = 8 × 4 + 1)
If teacher called ‘7’, Ruby would write: 4 × 7 + 5 (since 33 = 4 × 7 + 5)

He already had 4 packets. He made new packets from 100kg of rice, with each packet being 2kg. Number of new packets = 100 ÷ 2. Total packets = 4 + (100 ÷ 2). Expression: 4 + 100/2. Terms: The terms are 4 and 100/2.

The expression 5 × 2 + 3 means “5 groups of 2, and then 3 more”.

The arrangement on the left shows exactly that: 5 groups of 2 green squares, plus 3 individual red squares.

The arrangement on the right shows 2 rows of (5 yellow + 3 blue), so its expression is 2 × (5 + 3).

Page 34: Find the values

(a) 28 – 7 + 8 = 21 + 8 = 29

(b) 39 – 2 × 6 + 11 = 39 – 12 + 11 = 27 + 11 = 38

Page 34: Write a story and find values

(a) 89 + 21 – 10: I had 89 stamps. My friend gave me 21 more. I then lost 10. How many do I have? Value = 110 – 10 = 100.

(c) 4 × 9 + 2 × 6: There are 4 boxes with 9 chocolates each and 2 boxes with 6 chocolates each. What is the total? Value = 36 + 12 = 48.

Page 34: Word Problems

(a) Elsa and Anna: Elsa has (100 × 2) coins. Anna has (100 ÷ 2) coins. Total = (100 × 2) + (100 ÷ 2) = 200 + 50 = 250 coins.

(b) Metro tickets: (i) 4 adults, 3 children: (4 × 40) + (3 × 20) = 160 + 60 = ₹220. (ii) 2 groups of 3 adults: 2 × (3 × 40) = 2 × 120 = ₹240.

Page 35: Window Height

Expression: 3 + 2 + 5 + 2 + 3 (Top border + Top grill + Gap + Bottom grill + Bottom border)
Value = 15 cm

Page 37: Fill the blanks

(a) 24 + (6 − 4) = 24 + 6 – 4

(c) 24 − (6 + 4) = 24 – 6 − 4

(e) 27 − (8 + 3) = 27 – 8 – 3

(f) 27 − (8 – 3) = 27 − 8 + 3

Page 37: Remove brackets

(b) 14 − (12 + 10) = 14 − 12 − 10

(d) 14 − (12 − 10) = 14 − 12 + 10

(f) 14 − (−12 − 10) = 14 + 12 + 10

Page 38: Add brackets

(a) 34 − 9 + 12 = 13 –> 34 – (9 + 12) = 34 – 21 = 13

(b) 56 − 14 − 8 = 34 –> 56 – (14 + 8) is 34. Or (56-14)-8 is 34. No brackets needed. The question is a bit ambiguous. 56-(14+8) = 34 is incorrect. 56-14-8 = 42-8=34. Let’s assume the question meant a different result. If result was 50, then 56 – (14-8) = 50. Let’s re-read (b): 56 – 14 – 8 = 34. This is already true without brackets. Let’s assume they want a different grouping. (56-14)-8 = 34. This is the standard order.

(c) -22 − 12 + 10 + 22 = −22 –> -22 – (12 – 10 – 22). This is too complex. A simpler way: -22 – (12 – 10) + 22 = -22 – 2 + 22 = -2. Not correct. -(22-12+10)+22 = -20+22=2. Let’s try -22 – (12 + 10) + 22 = -22 – 22 + 22 = -22. This works.

Page 38: Fill blanks by reasoning

(a) 423 + __ = 419 + __ : The LHS has a number that is 4 more than the RHS (423 vs 419). To keep them equal, the blank on the RHS must be 4 more than the blank on the LHS. Example: 423 + 10 = 419 + 14.

(b) 207 – 68 = 210 – __ : LHS = 139. RHS: 210 – __ = 139. Blank is 71. By reasoning: 210 is 3 more than 207, so to keep the result same, we must subtract 3 more. 68 + 3 = 71.