NCERT Solutions for Class 3 Maths Chapter 07 Raksha Bandhan

Chapter 7: Raksha Bandhan

Question. What do you find interesting here?
Answer: Gopal's family is tidying up and decorating their home for their favorite festival. Everyone is very happy because Atya is coming to visit them.
In simple words: The family is cleaning and decorating their house to welcome their relatives for the festival.

Exam Tip: Notice the decorations like the toran and sweet plates in the picture when describing how the family is preparing.

Question. Find and count the number of each of these objects and write.
Leaves [ ] Glasses [ ] Pomegranate [ ] Flowers [ ]
Answer: After counting the objects in the picture, we find:
* Leaves: 24
* Glasses: 30
* Pomegranates: 24
* Flowers: 20
In simple words: We can count the items on the festival table to find there are 24 leaves, 30 glasses, 24 pomegranates, and 20 flowers.

Exam Tip: Group the glasses by tens as you count to make finding the total of 30 quick and easy.

Question. For each Rakhi we need one flower, ........ threads and ........ beads.
Answer: To make one Rakhi, we need one flower, 2 threads, and 4 beads.
In simple words: Each handmade Rakhi is made using one flower, two long threads, and four colorful beads.

Exam Tip: Look closely at the single Rakhi drawing to identify the number of parts needed to make one unit.

Question. We need to make 5 Rakhis. How many threads, flowers and beads do they need?
Answer: To make 5 Rakhis, we will need 5 flowers, 10 threads, and 20 beads in total.
In simple words: We can multiply the parts of one Rakhi by five to find that we need 5 flowers, 10 threads, and 20 beads.

Exam Tip: This is a multiplication problem. Multiply each item needed for one Rakhi (1 flower, 2 threads, 4 beads) by the total number of Rakhis (5).

Question. Dhara takes: 1 + 1 + 1 + 1 + 1 = ......
This can also be said as: 5 times 1
or 5 x 1 = 5
For 5 Rakhis, we need ........ flowers.
Answer: We can find the number of flowers by repeating addition or using multiplication:
* We can add them up: 1 + 1 + 1 + 1 + 1 = 5.
* This is the same as: 5 times 1.
* Using multiplication: 5 × 1 = 5.
* So, we need 5 flowers to make 5 Rakhis.
In simple words: Adding one flower five times is the same as multiplying 5 by 1, which gives five flowers.

Exam Tip: Repeated addition of the same number is the same as multiplication.

Question. Dhara takes: 2 + 2 + 2 + 2 + 2 = ......
or 5 times 2
or 5 x 2
For 5 Rakhis, we need ........ threads.
Answer: We can count the threads:
* We can add them up: 2 + 2 + 2 + 2 + 2 = 10.
* This is the same as: 5 times 2.
* Using multiplication: 5 × 2 = 10.
* So, we need 10 threads to make 5 Rakhis.
In simple words: Adding two threads five times is the same as 5 times 2, which gives ten threads in total.

Exam Tip: Practice skip-counting by 2s (2, 4, 6, 8, 10) to quickly solve 5 times 2.

Question. 4 + 4 + 4 + 4 + 4 = ......
or 5 times 4
or 5 x 4
For 5 Rakhis, we need ........ beads.
Answer: We can count the beads:
* We can add them up: 4 + 4 + 4 + 4 + 4 = 20.
* This is the same as: 5 times 4.
* Using multiplication: 5 × 4 = 20.
* So, we need 20 beads to make 5 Rakhis.
In simple words: Adding four beads five times is the same as 5 times 4, which gives twenty beads in total.

Exam Tip: Use the times-4 table to quickly find the total number of beads (5 × 4 = 20).

Question. How would we count the laddoos in this box?
3 + 3 + 3 = .....
or, three times three equals 9
or, 3 x 3 = ......
There are ...... laddoos in this box.
Answer: To count the laddoos in the box:
* By adding: 3 + 3 + 3 = 9.
* This means: three times three equals 9.
* Using multiplication: 3 × 3 = 9.
* So, there are 9 laddoos in this box.
In simple words: The box has three rows of three laddoos. This is written as 3 times 3, which equals nine laddoos.

Exam Tip: Count the rows and the columns of items to easily write the multiplication sentence.

Question. Please give me 2 boxes of laddoos.
9 + 9 = 18
Two times nine equals 18.
2 x 9 = 18
There are 18 laddoos.
3 + 3 + 3 + 3 + 3 + 3 = ........
Or, 6 times 3 equals 18.
Or, 6 x 3 = 18
There are 18 laddoos.
Answer: Let us add groups of three:
* 3 + 3 + 3 + 3 + 3 + 3 = 18.
* This is the same as: 6 times 3.
* Using multiplication: 6 × 3 = 18.
* So, there are 18 laddoos.
In simple words: Adding three six times is the same as six times three, which gives eighteen laddoos.

Exam Tip: Notice that 2 times 9 and 6 times 3 both give the same total of 18 laddoos.

Question. After Atya and children come, we will be 9 people in the house. When we distribute 18 laddoos equally to all, how many will each of us get?
Answer: Let us share the laddoos: 18 ÷ 9 = 2. Each person gets exactly 2 laddoos.
In simple words: Sharing eighteen sweets equally among nine people means everyone gets two sweets.

Exam Tip: Equal sharing is called division. Divide the total sweets (18) by the number of people (9) to get the share.

Question. Imagine yourself to be Dhara. Distribute 18 laddoos equally among nine of your friends. Let's see how Dhara has done it.
Answer: Let us share the laddoos step-by-step:
* First, we give 1 laddoo to each of the 9 friends. We have 9 laddoos left in the box.
* Next, we give one more laddoo to each friend. Now, we have 0 laddoos left in the box.
* This means: when 18 laddoos are shared equally among 9 people, each person gets 2 laddoos.
* In numbers: 18 ÷ 9 = 2.
In simple words: We share eighteen laddoos by giving them out in rounds. Each friend gets two sweets, and none are left in the box.

Exam Tip: Subtracting 9 in each round (18 - 9 = 9, and then 9 - 9 = 0) is called repeated subtraction, which is division.

Question. Try it Out! Look at the figure carefully. Estimate the number of kaju katlis. Total number of kaju katlis = ........ Distribute all kaju katlis equally among 5 people. You can do it by drawing kaju katlis on the plates. How many will each get? Compare your work with Dhara's work.
Answer: Let us work out the kaju katli sharing:
* The total number of kaju katlis in the tray is 20.
* If we share 20 sweet pieces equally among 5 people, each person gets 4 pieces (20 ÷ 5 = 4).
* At the start, we have 20 pieces. After giving 1 piece to each of the 5 people, we have 15 pieces left.
In simple words: Sharing twenty kaju katlis equally among five friends means everyone gets four pieces.

Exam Tip: Try skip-counting by 5s up to 20 (5, 10, 15, 20) to find that 4 groups of 5 make 20.

Question 1. Let us Do. Distribute all the kaju katlis equally among 4 people. How many kaju katlis will each get? Let us do this in the picture given below. Strike out the kaju katlis from the tray and draw them in the plates. The first step has been done for you. 16 / 4 = ........
Answer: Let us share the 16 pieces step-by-step:
* Start with 16. Give 1 to each of the 4 people. There are 12 pieces left.
* Give a second piece to each person. There are 8 pieces left.
* Give a third piece to each person. There are 4 pieces left.
* Give the last piece to each person. There are 0 pieces left.
* So, each person gets 4 kaju katlis.
* In numbers: 16 ÷ 4 = 4.
In simple words: We distribute sixteen sweets among four plates by giving them out in rounds. Each plate gets four sweets in total.

Exam Tip: Keep subtracting 4 in each round (16, 12, 8, 4, 0) until the tray is empty to show your working clearly.

Question 2. Distribute all the 15 pedas in plates equally among 5 people. How many pedas will each get? 15 equally shared by 5 is ........ each. 15 / 5 = ........
Answer: Let us share the 15 pedas among 5 plates:
* First round: 15 - 5 = 10 pedas left.
* Second round: 10 - 5 = 5 pedas left.
* Third round: 5 - 5 = 0 pedas left.
* So, 15 equally shared by 5 is 3 pedas each.
* In numbers: 15 ÷ 5 = 3.
In simple words: Sharing fifteen yellow pedas equally among five plates means each plate gets exactly three sweets.

Exam Tip: Use the times-5 table to verify your answer, as 3 × 5 is equal to 15.

Question 1. Let us Think. Each cycle needs 2 wheels. How many cycles can be fitted with 12 wheels? 12 equally divided by 2 is ........ 12 / 2 = ........
Answer: Let us find how many cycles we can build:
* 12 equally divided by 2 is 6.
* In numbers: 12 ÷ 2 = 6.
In simple words: Since each bicycle needs two wheels, twelve wheels are enough to build six bicycles.

Exam Tip: Group the 12 wheels in pairs (2s) to easily find that there are 6 pairs of wheels.

Question 2. Look at the picture carefully. Count the number of jalebis. There are ........ jalebis. How did you count? Discuss with your friends. Counting in groups, we see there are six groups of four jalebis each, or, 4 + 4 + 4 + 4 + 4 + 4 = ........ or ........ x 4 = ........ jalebis.
Answer: Let us count the sweet jalebis:
* There are 24 jalebis in total.
* By adding: 4 + 4 + 4 + 4 + 4 + 4 = 24.
* Using multiplication: 6 groups of 4 is 24, which is written as 6 × 4 = 24.
In simple words: There are six plates with four jalebis on each. We can multiply 6 by 4 to get twenty-four jalebis in total.

Exam Tip: Counting in groups of 4 (4, 8, 12, 16, 20, 24) is much faster than counting each jalebi one-by-one.

Question. Are there enough jalebis for everyone in Dhara's family to have four each?
Answer: No, there are not enough jalebis for everyone to have 4. Since there are 9 members, they would need more than 24 pieces to give 4 to each person.
In simple words: No, they only have 24 jalebis, which is not enough to give four to each of the nine family members.

Exam Tip: Multiply the family members (9) by the share (4) to find the total jalebis needed (36), and compare it with 24.

Question. How many jalebis should Dhara buy so that everyone can get four each?
Answer: Let us calculate what is needed:
* With 9 family members, getting 4 jalebis each means: 9 × 4 = 36 jalebis.
* Since they already have 24 jalebis, Dhara needs to buy: 36 - 24 = 12 more jalebis.
In simple words: They need thirty-six jalebis in total. Since they have twenty-four, Dhara needs to buy twelve more.

Exam Tip: Subtract the jalebis they already have (24) from the total needed (36) to find how many more to buy.

Question. Let us write the times-2 table. (1 times 2 to 10 times 2)
Answer: Here is the completed times-2 table:

In simple words: This table displays the multiplication of numbers by two, where we add 2 at each next step.

Exam Tip: Practice writing tables in both words and equations to solve multiplication problems quickly.

Question. Dhara is SKIP JUMPING BY 3. Fill in the blank spaces. (1 Jump to 10 Jumps)
Answer: Here is the completed skip-jumping table for 3:

In simple words: Jumping by three means skip-counting by 3, which is the same as the times-3 multiplication table.

Exam Tip: Use the addition of 3 (for example, 27 + 3 = 30) to easily find the next landing spot on your track.

Question 1. Let us Do. Guess and write the next number she will jump onto. (3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = ........ )
Answer: Adding 11 threes together gives us exactly 33 (which is written as 11 × 3 = 33).
In simple words: Taking eleven jumps of three steps each brings us to thirty-three on the number line.

Exam Tip: 11 times 3 is equal to 33. You can find this by adding 3 to the 10th jump (30 + 3 = 33).

Question 2. Is there a pattern in these numbers: 3, 6, 9, ...?
Answer: Yes, there is a clear pattern. Every next number increases by adding exactly 3.
In simple words: Yes, the numbers go up by three at each step. This is the skip-counting pattern of 3.

Exam Tip: When describing a number pattern, always state the constant value that is being added or subtracted.

Question 3. How many steps forward is Dhara jumping each time?
Answer: Dhara is jumping forward by 3 steps each time.
In simple words: Dhara takes three steps forward with every jump she makes.

Exam Tip: Check the difference between any two adjacent landing points (such as 6 - 3 = 3) to find the size of the jump.

Question 4. Continue Skip Jumping by 6 by drawing the jumps on the number track. (0 to 25)
Answer: When jumping by 6, the counter lands on: 6, 12, 18, and 24.
In simple words: Draw loops on the number track connecting 0 to 6, 6 to 12, 12 to 18, and 18 to 24.

Exam Tip: Be sure to stop at 24. Since the track only goes to 25, you cannot make another full jump of 6.

Question 5. Can this skip jumping be used to form times-6 table? Write times-6 table in your notebook.
Answer: Yes, we can write the times-6 table using these jumps:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
In simple words: Yes, skip-counting by six is the exact same as writing the multiplication table of 6.

Exam Tip: Practice writing your tables in a neat column to help you remember the sequence easily.

Question 6. Is there repeated addition happening? Make time-4 table using repeated addition in the picture given below.
Answer: Yes, repeated addition is happening. We can write the times-4 table by adding four each time:
* 1 × 4 = 4
* 2 × 4 = 4 + 4 = 8
* 3 × 4 = 4 + 4 + 4 = 12
* 4 × 4 = 4 + 4 + 4 + 4 = 16
* 5 × 4 = 4 + 4 + 4 + 4 + 4 = 20
* 6 × 4 = 4 + 4 + 4 + 4 + 4 + 4 = 24
* 7 × 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28
* 8 × 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32
* 9 × 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36
* 10 × 4 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 40
In simple words: Yes, adding four repeatedly is the same as multiplying by 4. For example, three fours add up to twelve.

Exam Tip: Repeated addition is the basic concept of multiplication. Learn both to solve word problems easily.

Question 7. Gopal is doing skip jumping of ...... steps. After 27 he will jump on ........., .........
Answer: Based on Gopal's jumps:
* Gopal is skip-jumping by 9 steps each time.
* After landing on 27, his next two jumps will be on 36 and then 45.
In simple words: Gopal is jumping by nines. After twenty-seven, he will land on thirty-six and then forty-five.

Exam Tip: Add 9 to the last number (27 + 9 = 36, and 36 + 9 = 45) to find his next landing spots.

Question 8. What times table can you construct from Gopal's jumps? Make it in your notebook.
Answer: We can build the times-9 table from Gopal's jumps:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, and 99.
In simple words: Gopal's jumps represent the multiplication table of nine, where we add 9 at each step.

Exam Tip: A simple trick for the times-9 table is that the digits of each result (like 1+8, 2+7, 3+6) always add up to 9.

Question 9. Dhara also skip jumps. Gopal notes down the jumps but he misses the first few numbers. [ ][ ][ ] 32, 40, 48, 56. By what numbers was Dhara skip jumping? Construct the times in your notebook.
Answer: We can find the missing numbers by counting backward by 8:
* The first three missing numbers are 8, 16, and 24.
* Dhara was skip-jumping by 8 steps each time.
In simple words: Dhara is jumping by eights. The first three numbers on her path are eight, sixteen, and twenty-four.

Exam Tip: To find the skip-number, subtract two known numbers (like 40 - 32 = 8). Then subtract 8 from 32 to go backward.

Question 10. Let us Play. Atya places a flower on 12. Skip jump with equal steps to reach the flower. No direct jumping to the flower is allowed. The one who reaches the flower in the smallest number of jumps wins. What skip jumping number will you choose? Are there numbers that can be reached only through skip jumping by 1?
Answer:
* To reach 12 in the fewest jumps, we should choose to skip-jump by 6 steps (which takes only 2 jumps: 6 and then 12).
* Yes, prime numbers like 3, 7, and 13 can only be reached from 0 if we skip-jump by 1 step at a time.
In simple words: Jumping by sixes is the fastest way to reach twelve. Some prime numbers cannot be shared into equal groups, so we can only reach them by counting by ones.

Exam Tip: To find the fastest jump size, look for the largest factor of the target number (like 6 for 12) that is smaller than the number itself.

Question 11. Fun Way of Writing Tables. Let's try making a 5 times table with sticks. Complete the times-5 table using sticks.
Answer: Here is the completed times-5 table using the stick intersection method:

In simple words: This table displays the times-five table. We can find the answers by counting where the sticks cross each other.

Exam Tip: The intersection method is a fun physical way to learn multiplication tables by counting the overlapping points of crossing sticks.

Question 12. Give 3 examples of numbers that when taken 5 times gives an answer ending with
(i) 0 ........ ........ ........
(ii) 5 ........ ........ ........
Answer: Based on the properties of the times-5 table:
* (i) Ending with 0: If we multiply 5 by even numbers like 2 (2 × 5 = 10), 6 (6 × 5 = 30), or 10 (10 × 5 = 50), the answer ends in 0.
* (ii) Ending with 5: If we multiply 5 by odd numbers like 3 (3 × 5 = 15), 5 (5 × 5 = 25), or 7 (7 × 5 = 35), the answer ends in 5.
In simple words: Multiplying 5 by an even number gives a total ending in 0, and multiplying by an odd number gives a total ending in 5.

Exam Tip: Remember this rule: odd multiples of 5 always end in 5, and even multiples of 5 always end in 0.

Question 13. Without finding the answer, can you tell the last digits of 18 x 5, 23 x 5, 32 x 5, 50 x 5?
Answer: Yes, we can easily find the last digits by looking at the ones place of each number:
* For 18 × 5: The ones digit is 8 (even), so the last digit will be 0.
* For 23 × 5: The ones digit is 3 (odd), so the last digit will be 5.
* For 32 × 5: The ones digit is 2 (even), so the last digit will be 0.
* For 50 × 5: The ones digit is 0 (even), so the last digit will be 0.
In simple words: Since 18, 32, and 50 are even, their answers will end in 0. Since 23 is odd, its answer will end in 5.

Exam Tip: You only need to check if the number is even or odd to find if its product with 5 ends in 0 or 5.

Question 14. Let us Do. Question 1(a) : There are 5 jars with 4 cookies in each jar. How many cookies are there?
Answer: Let us calculate the cookies: 4 × 5 = 20 cookies. (On the line, we take five jumps of 4 steps each to land on 20).
In simple words: Five jars with four biscuits each contain twenty biscuits in total.

Exam Tip: Write down the multiplication equation (5 × 4 = 20) to show your work clearly in the exam.

Question 15. Question 1(b) : An idli vessel contains 6 idli plates. In each plate we can make 4 idlis. How many idlis can be cooked in one go?
Answer: Let us multiply: 6 plates × 4 idlis = 24 idlis in total. (On the line, we take six jumps of 4 steps each to land on 24).
In simple words: We can steam twenty-four idlis at once using six plates that hold four idlis each.

Exam Tip: This is a simple multiplication of 6 groups of 4. Use the times-4 table to quickly find the answer.

Question 16. Question 1(c) : 30 cookies are to be distributed among 5 children equally. How many cookies will each child get?
Answer: Let us divide the cookies: 30 ÷ 5 = 6 cookies. Each of the five children gets 6 cookies.
In simple words: Sharing thirty biscuits equally among five children means everyone gets six biscuits.

Exam Tip: This is a division problem. Use your division sign (30 ÷ 5 = 6) to write the math sentence.

Question 17. Question 1(d) : Roro starts from 0 and takes 6 jumps to reach 18. All his jumps are of the same size. What is the size of Roro's jump? The size of Roro's jump = 18 / 6 = ........
Answer: To find the jump size: 18 ÷ 6 = 3 steps. Each jump of Roro is 3 steps long.
In simple words: Since Roro reaches eighteen in six equal jumps, each of his jumps is exactly three steps long.

Exam Tip: Divide the total distance (18) by the number of jumps (6) to find the size of a single jump.

Question 1(e) : Toto does not take jumps of the same size and still reaches 18 in 6 jumps. How did Toto jump?
Answer: Toto can take jumps of different sizes that still add up to 18. For example, he could jump 2, 2, 3, 3, 4, and 4 steps (2 + 2 + 3 + 3 + 4 + 4 = 18).
In simple words: Toto can reach eighteen in six jumps by taking different sized hops, like jumping two steps, then three, then four.

Exam Tip: There are many different ways to solve this. Just make sure the six numbers you choose add up to exactly 18.

Question 1(f) : Suma saves Rs 8 every day. After how many days will she have Rs 56?
Answer: Let us divide the total savings: Rs. 56 ÷ Rs. 8 = 7 days. It will take Suma exactly 7 days to save Rs. 56.
In simple words: Since Suma saves eight rupees daily, it will take her seven days to collect fifty-six rupees.

Exam Tip: Since 7 × 8 = 56, dividing 56 by 8 gives the exact number of days needed.

Question 1(g) : Mary has 63 sea shells. She gives 7 sea shells to each of her 5 friends. How many does she have left?
Answer: Let us solve this step-by-step:
* Shells given away: Mary gave 7 shells to each of her 5 friends, which is: 7 × 5 = 35 shells.
* Shells left: She started with 63, so she has: 63 - 35 = 28 shells left.
In simple words: Mary gave thirty-five shells to her friends, which means she has twenty-eight shells left for herself.

Exam Tip: This is a two-step problem. First multiply to find the total shells given away, then subtract that from the initial total.

Question 2. Solve the following problems. Try constructing a word problem.
a. 4 x 9 = ........
b. 32 / 8 = ........
c. 6 x 7 = 42
d. 45 / 5 = ........
Answer: Here are the completed calculations:
* a. 4 × 9 = 36: Four groups of nine equals thirty-six.
* b. 32 ÷ 8 = 4: Thirty-two divided into eight equal groups is four.
* d. 45 ÷ 5 = 9: Forty-five divided into five equal groups is nine.
In simple words: We can solve these multiplication and division sums using our tables. For example, 4 times 9 is 36.

Exam Tip: Use your times tables to solve division quickly. Since 9 × 5 = 45, dividing 45 by 5 gives 9.

Question. Help Bhim! Bhim will need ........ spokes. Think and share with your friends how you found the answer. Let us see how Bhim did it. 10 wheels will need: 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 10 x 5 = ........ spokes. Another 10 wheels will need ........ x ........ = ........ spokes. So, the total number of spokes needed is ........ + ........ = ........ spokes.
Answer: Let us calculate the spokes needed for the wheels:
* For the first 10 wheels, Bhim needs: 10 × 5 = 50 spokes.
* For another 10 wheels, he needs: 10 × 5 = 50 spokes.
* So, the total number of spokes needed for all 20 wheels is: 50 + 50 = 100 spokes.
In simple words: Ten wheels need fifty spokes. To make twenty wheels, Bhim needs one hundred spokes in total.

Exam Tip: Double the spokes of 10 wheels (50 × 2) to quickly find the total spokes needed for 20 wheels.

Question. Try these. 30 x 5 = ........
(Hint: You can find this by counting the spokes in 30 wheels.)
First 10 wheels will have ........ spokes
Next 10 wheels will have ........ spokes
Next 10 wheels will have ........ spokes
Total = ........ spokes
30 x 5 = ........ spokes
Complete the following:
40 x 5 = ........ | 70 x 5 = ........ | 100 x 5 = ........
50 x 5 = ........ | 80 x 5 = ........
60 x 5 = ........ | 90 x 5 = ........
Answer: Let us solve these large multiples of 5:
* First 10 wheels: 50 spokes
* Next 10 wheels: 50 spokes
* Next 10 wheels: 50 spokes
* Total: 50 + 50 + 50 = 150 spokes
* So, 30 × 5 = 150 spokes.
Completed Calculations:
* 40 × 5 = 200
* 50 × 5 = 250
* 60 × 5 = 300
* 70 × 5 = 350
* 80 × 5 = 400
* 90 × 5 = 450
* 100 × 5 = 500
In simple words: We can multiply large tens by 5 easily. For example, 40 times 5 is 200, and 100 times 5 is 500.

Exam Tip: To multiply a number ending in 0 by 5, multiply the non-zero digit first (like 4 × 5 = 20) and then add the 0 at the end (200).

Question. Dhara collected 45 spokes. How many wheels can she make?
Answer: Based on the spoke sharing:
* Dhara does not have enough spokes to make 10 wheels because 10 wheels require 50 spokes, and she only has 45.
* With 60 spokes, we can make exactly 12 wheels (60 ÷ 5 = 12).
In simple words: Dhara cannot make ten wheels with 45 spokes. With sixty spokes, we can build twelve wheels.

Exam Tip: Since each wheel needs 5 spokes, divide the total spokes (60) by 5 to find the number of wheels.

Question 1. Let us Do. A spider has 8 legs. 5 spiders will have ........ legs. 10 spiders will have ........ legs. 15 spiders will have ........ legs.
Answer: Let us multiply to find the total legs:
* 5 spiders will have: 5 × 8 = 40 legs.
* 10 spiders will have: 10 × 8 = 80 legs.
* 15 spiders will have: 15 × 8 = 120 legs.
In simple words: Since one spider has eight legs, ten spiders have eighty legs, and fifteen spiders have one hundred and twenty legs.

Exam Tip: Use the times-8 multiplication table to quickly solve these questions.

Question 2. How many legs will 23 spiders have?
Answer: Let us multiply: 23 spiders × 8 legs = 184 legs in total.
In simple words: Twenty-three spiders have a total of one hundred and eighty-four legs.

Exam Tip: To multiply 23 by 8, split it into 20 and 3: (20 × 8 = 160) + (3 × 8 = 24) = 184.

Question 3. A group of spiders have 32 legs. How many spiders are there in the group?
Answer: Let us divide the total legs: 32 ÷ 8 = 4 spiders in the group.
In simple words: Since each spider has eight legs, a group with thirty-two legs must have four spiders.

Exam Tip: This is a division problem because you are grouping the total legs into sets of 8.

Question 4. Here is a 3-wheeled auto rickshaw. How many wheels are there in:
a. 18 auto rickshaws?
b. 34 auto rickshaws?
Answer: Let us multiply to find the total wheels:
* a. 18 auto rickshaws will have: 18 × 3 = 54 wheels.
* b. 34 auto rickshaws will have: 34 × 3 = 102 wheels.
In simple words: Since each auto has three wheels, eighteen autos have fifty-four wheels, and thirty-four autos have one hundred and two wheels.

Exam Tip: Multiply the number of vehicles by 3 because each auto rickshaw has exactly three wheels.

Question 5. Auto rickshaws in a garage have a total of 36 wheels. How many auto rickshaws are there in the garage?
Answer: Let us divide the total wheels: 36 ÷ 3 = 12 auto rickshaws in the garage.
In simple words: Since each auto has three wheels, thirty-six wheels belong to twelve auto rickshaws.

Exam Tip: Division is the opposite of multiplication. Since 12 × 3 = 36, dividing 36 by 3 gives 12.

Question 6. There is a line of 55 ants (one ant has 6 legs). What is the total number of legs in the line?
Answer: Let us multiply: 55 ants × 6 legs = 330 legs in total.
In simple words: A line of fifty-five ants has a total of three hundred and thirty crawling legs.

Exam Tip: To multiply 55 by 6, you can calculate: (50 × 6 = 300) + (5 × 6 = 30) = 330.

Question 7. Micky, the mouse, can see 48 legs of cows in the shed. How many cows are there in the shed?
Answer: Since each cow has 4 legs, we divide: 48 ÷ 4 = 12 cows in the shed.
In simple words: Since cows have four legs each, forty-eight legs belong to twelve cows inside the shed.

Exam Tip: Division helps you group. Grouping 48 legs into sets of 4 gives exactly 12 cows.

Question 8. Karry, the crow, can see 24 horns of cows in the shed. What is the total number of legs in the shed?
Answer: Let us work this out in two steps:
* Number of cows: Since each cow has 2 horns, we divide: 24 ÷ 2 = 12 cows.
* Total legs: Since each cow has 4 legs, we multiply: 12 cows × 4 legs = 48 legs.
In simple words: Twenty-four horns belong to twelve cows. These twelve cows have forty-eight legs in total.

Exam Tip: First use the horns to count the cows, then use the number of cows to calculate the total legs.

Question 1. Let us Think. A frog is at 0. It takes jumps of only 7. What would be the largest number that the frog will reach before crossing 50?
Answer: The frog will make jumps of 7 (7, 14, 21, 28, 35, 42, 49). The largest number it lands on before crossing 50 is 49.
In simple words: When jumping by sevens, the biggest number the frog hits before fifty is forty-nine.

Exam Tip: Write down the multiples of 7 (7, 14, 21, 28, 35, 42, 49, 56) and select the highest one that is less than 50.

Question 2. A frog wants to jump backwards from 50. It continues to take jumps of 7. What is the number after which it is not possible for the frog to make a jump of 7?
Answer: The frog will jump backward (50 -> 43 -> 36 -> 29 -> 22 -> 15 -> 8 -> 1). After landing on 1, it is not possible to make another backward jump of 7.
In simple words: After jumping back by sevens, the frog lands on 1. It cannot jump back seven more steps from 1.

Exam Tip: Keep subtracting 7 from 50 until you get a number that is smaller than 7.

Question 3. What numbers should the frog start from to reach 0, taking jumps of 7 each time? What do you observe?
Answer: The frog must start from any multiple of 7, such as 7, 14, 21, 28, 35, 42, or 49. We observe that jumping backward by equal steps is the same as subtraction.
In simple words: The frog must start from any number in the times-7 table to land exactly on zero when jumping backward.

Exam Tip: To reach exactly 0, the starting number must be completely divisible by the jump size (7).

Question 1. Puri Beach. One wall-hanging costs Rs. 42. How much do two wall hangings cost? The cost of the two wall hangings: ............
Answer: Let us calculate the total cost: Rs. 42 + Rs. 42 = Rs. 84 (or 2 × Rs. 42 = Rs. 84).
In simple words: Two wall decorations cost eighty-four rupees in total.

Exam Tip: Use the rupee symbol (Rs. or ₹) in your final answer to show that you are writing a cost.

Question 2. One Rabdi cup costs Rs. 75. Preeti buys 5 cups of Rabdi. She has her mother's purse which has only Rs. 100 notes. How many Rs. 100 notes should she give the shopkeeper? How much will the shopkeeper then return to Preeti? What is the total cost of 5 cups of Rabdi?
Answer: Let us work this out in three simple steps:
* Total cost of 5 Rabdi cups: 5 × Rs. 75 = Rs. 375.
* Notes to give: Since the total is Rs. 375, Preeti must give 4 notes of Rs. 100 (which is Rs. 400) to the shopkeeper.
* Change returned: The shopkeeper will return: Rs. 400 - Rs. 375 = Rs. 25.
In simple words: Five cups of rabdi cost Rs. 375. Preeti gives four hundred-rupee notes and gets twenty-five rupees back as change.

Exam Tip: Break the cost down: 5 × 70 = 350, and 5 × 5 = 25. Adding them together gives Rs. 375.

Question. How many shells are left now? (Nandini: 112, Chirag: 28)
Answer: After making two necklaces, there are exactly 56 shells left.
In simple words: After making two necklaces using twenty-eight shells each, fifty-six shells remain in the bucket.

Exam Tip: Subtract 28 twice (112 - 28 = 84, and 84 - 28 = 56) to find the remaining shells.

Question. Then he took shells for the third necklace. So he was left with ............ shells.
Answer: After making the third necklace, he is left with: 56 - 28 = 28 shells.
In simple words: After making three necklaces, twenty-eight shells are left in the bucket.

Exam Tip: Subtracting 28 from 56 leaves exactly 28 because 28 × 2 is equal to 56.

Question. Are the shells enough for making necklaces for all his friends? (There are 3 friends)
Answer: Yes, the shells are enough because he has exactly 28 shells left, which is just enough to make one final necklace for his fourth friend.
In simple words: Yes, twenty-eight shells are left, which is exactly what is needed to make a final necklace.

Exam Tip: Since 112 divided by 28 is exactly 4, he can make exactly 4 complete necklaces without any leftovers.

Question 1. Try these. Kannu makes a necklace of 17 sea-shells. How many such necklaces can be made using 100 sea-shells?
Answer: We can make 5 necklaces of 17 shells each. There will be 15 unused shells left over (5 × 17 = 85, and 100 - 85 = 15).
In simple words: One hundred shells are enough to make five necklaces, and fifteen single shells are left over.

Exam Tip: Use repeated subtraction of 17 from 100 to find how many groups of 17 can fit into 100.

Question 2. While searching for sea-shells, Dhruv also finds 127 shiny pebbles. He distributes them equally to his 3 friends. How many will each get?
Answer: Each of his three friends will get exactly 42 pebbles, and there will be 1 leftover pebble (42 × 3 = 126, and 127 - 126 = 1).
In simple words: Sharing 127 pebbles among three friends means everyone gets forty-two, and one pebble is left in the bag.

Exam Tip: Perform a long division of 127 by 3 to find the quotient (42) and the remainder (1).

Question 3. Preeti has a Rs 500 note and wants to exchange it for lower denomination notes. How many notes will she get if she wants -
(i) All 50 rupees notes?
(ii) All 20 rupees notes?
(iii) All 10 rupees notes?
Answer: Based on division calculations:
* (i) If she wants Rs. 50 notes: She will get: Rs. 500 ÷ Rs. 50 = 10 notes.
* (ii) If she wants Rs. 20 notes: She will get: Rs. 500 ÷ Rs. 20 = 25 notes.
* (iii) If she wants Rs. 10 notes: She will get: Rs. 500 ÷ Rs. 10 = 50 notes.
In simple words: A five hundred rupee note can be exchanged for ten Rs. 50 notes, twenty-five Rs. 20 notes, or fifty Rs. 10 notes.

Exam Tip: Divide 500 by the note value (50, 20, or 10) to find the total number of notes received.

Question. Let us Explore. There are ten number cards from 1-10. There are five sealed envelopes. Each has two cards. On the top of each envelope the multiplication of the numbers contained in it is written. Identify the number cards inside each of the envelopes. (7, 18, 32, 20, 45)
Answer: Let us find the correct card pairs for each envelope:
* For the envelope with 7: The cards inside are 1 and 7 (1 × 7 = 7).
* For the envelope with 18: The cards inside are 3 and 6 (3 × 6 = 18).
* For the envelope with 32: The cards inside are 4 and 8 (4 × 8 = 32).
* For the envelope with 20: The cards inside are 2 and 10 (2 × 10 = 20).
* For the envelope with 45: The cards inside are 5 and 9 (Given).
In simple words: We can find the two cards inside each envelope by looking for pairs of numbers from 1 to 10 that multiply to the number written on top.

Exam Tip: Write down all the factors of the envelope number first, then choose the pair of single digits between 1 and 10.