Students can use Class 4 Maths Solutions and Maths Mela NCERT Class 4 Maths Chapter 10 Equal Groups Question Answer Solutions to explore alternative problem-solving methods.
Class 4 Maths Chapter 10 Elephants, Tigers, and Leopards Question Answer Solutions
Elephants, Tigers, and Leopards Class 4 Maths Solutions
Class 4 Maths Chapter 10 Solutions
NCERT Textbook Pages 149-150 NIM Game (2 Player Game)
You have played a version of this game in the chapter ‘Vacation with my Nani Maa’ in Grade 3. We will add either 1 or 2 each time to reach the target number 10.
Can you win the game if
(а) the other player has reached the total of 6 and it is your turn?
(b) the other player has reached the total of 7 and it is your turn?
(c) the other player has reached the total of 8 and it is your turn?
Solution:
(a) Yes, we can win the game. Add 1 brincging the total to 7. On other player’s tun n, the opponent can add either 1 or 2. T-f he/she adds 1, the total becomes 8 and or.i our turn, we can add 2 to reach 10 and win. If he/ she adds 2, the total becomes- 9 and on our turn, we can add 1 to reach 10 and win.
(b) No, we cannot win. We can add either 1 or 2. If we add 1, the total becomes 8 and the other player can a dd 2 to reach 10 and win. If we add 2, the; total becomes 9 and the other player can add 1 to reach 10 and win.
(c) Yes, we can win. Add 2, bringing the total to 10 and we will win.
Play the game to reach other target numbers (like 10, 11 or 12) by adding 1 or 2 each time.
Can you find a number in each case when you are sure that you can win?
Solution:
Suppose, if other player has reached the total 9 and its our turn, we can add 1 to reach 10 and win. Similarly, we can find some numbers in each case when we are sure that we can win if the target number is 11 or 12. For example, if the target number is 11 and the other player has reached the total 7 and its our turn. Then we can add 1 to reach 8. The other player can either add 1 or 2. If they add 1 the total becomes 9 and in our turn we can add 2 to win.
If the player add 2, the total becomes 10 and in our turn we can add 1 to reach 11 and win. Similarly, if the target number is 12 and the other player has reached the total 8 and its our turn, then we can be sure to win.
NCERT Textbook Pages 150-151 Addition Chart
Look at the table given below and discuss how the table is made.
Question 1.
Identify some patterns in the table.
Solution:
Following are some patterns that can be observed in the table.
- Any number plus 0 remains the same.
- The sum increases by 2 when moving diagonally.
- Each row and column increase by 1 as we move right or down, respectively.
- The table mirrors itself across the main diagonal.
Question 2.
Observe the cells where the number 9 appears in the table. How many times do you see number 9?
What about other numbers?
Solution:
There are ten 9’s in the table.
Following are the patterns of appearance of the numbers.
0-one time, 1-two times, 2-three times, …, 12-thirteen times, 13-twelve times, 14- eleven times, 15-ten times, …, 24-one time.
Each number appears one more time than its value till the number 12 and then the appearances of numbers start decreasing symmetrically.
Question 3.
Are there any rows or columns that contain only even numbers or only odd numbers? Explain your observation.
Solution:
Every row and column in the table has both odd and even numbers. This happens because adding two even numbers or two odd numbers gives an even number as resulf while adding one odd and one even number gives an odd number as result.
Question 4.
Look at the window frame highlighted in red colour in the table.
(a) Find the sum of the two numbers in each row.
(b) Find the sum of the two numbers in each column. What do you notice?
(c) Now, find the sum of the numbers in each of the two diagonals marked by arrows. What do you notice?
(d) Now, put the red window frame in other places and find the sums as above. What do you notice?
Solution:
(a) Sum of 10 and 11 = 10 +11 = 21 Sum of 11 and 12 = 11 + 12 = 23
(b) Sum of 10 and 11 = 21 (in column)
Sum of 11 and 12 = 23
Again we get numbers 21 and 23 as result.
(c) Sum of 10 and 12 = 22 Sum of 11 and 11 = 22
The sum of the numbers in two diagonals is same.
(d) If we put the window frame in any two consecutive numbers in row and column, the sum of rows and columns will change but the difference between the sum of any two rows, columns and diagonals will remain the same.
Question 5.
Identify some patterns and relationships among the numbers in the blue window frame.
Solution:
In the blue window frame, numbers in each row and in each column are same and the difference between the sum of numbers of each row and each column is 3.
NCERT Textbook Page 151 – Reverse and Add
(a) Take a 2-digit number say, 27. Reverse its digits (72). Add them (99). Repeat for different 2-digit numbers.
Solution:
Some other such examples are as follows:
10 + 01 = 11
13 + 31 = 44
35 + 53 = 88
37 + 73 = 110, etc.
(Answer may vary)
(b) What sums can we get when we add a 2-digit number with its reverse?
Solution:
Let’s add some 2-digit numbers and their reverse to identify any pattern.
10 + 01 = 11 = 1 × 11
11 + 11 = 22 = 2 × 11
12 + 21 = 33 = 3 × 11
13 + 31 = 44 = 4 × 11
.
.
.
.
.
18 + 81 = 99 = 9 × 11
99 + 99 = 198 = 18 × 11
We can observe that, when we add a 2-digit number with its reverse, we get a number that can be obtained in the times-11 table.
(c) List down all numbers which when added to their reverse gives
(i) 55
(ii) 88
Solution:
(i) 14 + 41 = 55, 32 + 23 = 55, 50 + 05 = 55
Thus, all the numbers that, when added to their reverse give 55 are 14, 23, 32, 41 and 50.
(ii) 17 + 71 = 88, 26 + 62 = 88, 35 + 53 = 88 44 + 44 = 88, 80 + 08 = 88
Thus, all the numbers that, when added to their reverse give 88 are 17, 26, 35, 44, 53, 62, 71, and 80.
(d) Can we get a 3-digit sum? What is the smallest 3-digit sum that we can get?
Solution:
Yes, we can get a 3-digit sum by adding a 2-digit number and its reverse.
The smallest such 3-digit number is 110. As, 19 + 91 = 110.
Fill the blanks with appropriate numbers.
Solution:
(a) Since 2590 – 2540 = 50, each number is obtained by adding 50 to the previous number.
(b) Since 1354 – 354 = 1000, each number is obtained by adding 1000 to its previous number.
(c) Since 7670 – 7645 = 25, each number is obtained by adding 25 to the previous number.
NCERT Textbook Pages 152-154 – How Many Animals?
India is rich in biodiversity. It is home to some of the endangered wildlife, like elephants, tigers and leopards.
\(\frac{3}{4}\) of the world’s tiger population and \(\frac{3}{5}\) of the Asiatic elephant population is in India.
Question 1.
The population of elephants in Karnataka is 6049 and in Kerala is 3054. How many total elephants are there in these two states? Estimate the answer.
Solution:
6000 + 3000 = 9000.
So, approximately 9000 elephants are there in these two states.
Question 2.
The highest number of leopards are found in three states. Gujarat has 1355, Karnataka has 1131 and Madhya Pradesh has 1817.
How many total leopards are there in these states? Estimate the answer.
Solution:
1000 + 1000 + 2000 = 4000.
So, there are approximately 4000 leopards in these states.
Question 3.
Maharashtra has 444 tigers. Madhya Pradesh has 341 more tigers than Maharashtra. Uttarakhand has 116 tigers more than Maharashtra.
(a) How many tigers does Madhya Pradesh have?
Solution:
So, Madhya Pradesh has 785 tigers.
(b) How many tigers does Uttarakhand have?
Solution:
Therefore, there are 560 tigers in Uttarakhand.
(c) How many tigers does Madhya Pradesh and Uttrakhand have?
Solution:
So, there are 1345 tigers in Uttarakhand and Madhya Pradesh.
(d) How many tigers are there in total across the three states?
Solution:
So, there are 1789 tigers across the three states.
NCERT Textbook Pages 155-156 – More or Less?
Question 1.
Assam has 5719 elephants. It has 3965 more elephants than Meghalaya. How many elephants are there in Meghalaya? Estimate the answer.
Solution:
6000 – 4000 = 2000
So, there are approximately 2000 elephants in Meghalaya.
Question 2.
The population of leopards as per the 2022 census was 8820 in the Central India and the Eastern Ghats. It had increased by 749 in comparison to the number of leopards in 2018 in the same region. How many leopards were there in 2018? Estimate the answer.
__________ leopards were there in 2018.
Solution:
There were approximately 8100 leopards in 2018.
So, 8071 leopards were there in 2018.
Write the number of animals on this map based on the data from the problems in the previous pages.
Solution:
NCERT Textbook Pages 157-160 – Let Us Do
Question 1.
The board in the ticket office in the Kaziranga National Park shows the following:
(a) How many more visitors came in December than in November?
(b) The number of visitors in November is 1587 more than October. How many visitors were there in October?
Solution:
(a) Number of visitors in December = 8591
Number of visitors in November = 6415
The difference between the number of visitors in these two months = 8591 – 6415 = 2176
Therefore, 2176 more visitors came in December than in November.
(b) Number of visitors in November = 6415
Since the number of visitors in November is 1587 more than October.
Therefore, the number of visitors in October = 6415 – 1587 = 4828.
Thus, there were 4828 visitors in October.
Question 2.
In a juice making factory, women make different types of juices as given below:
(a) The number of bottles of guava juice is 759 more than the number of bottles of pineapple juice. Find the number of bottles of guava juice.
(b) The number of bottles of orange juice is 1257 more than the number of bottles of guava juice and 1417 less than the number of bottles of passion fruit juice. How many bottles of orange juice are made in a month?
(c) Is the total number of bottles of guava juice and orange juice more or less than the number of bottles of passion fruit juice? How much more or less?
Solution:
(a) Number of bottles of pineapple juice = 1348.
Number of bottles of guava juice is 759 more than number of bottles of pineapple juice.
Number of bottles of guava juice = 1348 + 759 = 2107
Therefore, the number of bottles of guava juice is the number of bottles of guava juice.
(b) The number of bottles of guava juice = 2107.
Number of bottles of orange juice is 1257 more than the number of bottles of guava juice.
Number of bottles of orange juice = 2107 + 1257 = 3364
Therefore, 3364 bottles of orange juice is packed in a month.
(c) Total number of bottles of guava juice and orange juice = 2107 + 3364 = 5471
Number of bottles of passion fruit juice = 4781
∵ 5471 -4781 = 690
Thus, the number of bottles of guava juice and orange juice is 690 bottles more than passion fruit juice.
Question 3.
In a small town, the following vehicles were registered in the year 2022. Find the number of vehicles as per the conditions given below.
(a) The number of buses are 253 more than the number of jeeps. How many buses are there in the town?
(b) The number of tractors are 5247 less than the number of buses. How many tractors are in the town?
(c) The number of taxis is 1579 more than the number of tractors? How many taxis are there?
(d) Arrange the number of each type of vehicle from lowest to highest.
Solution:
(a) Number of jeeps = 6304
Number of buses = 253 more than 6304 = 253 + 6304 = 6557
Therefore, there are 6557 buses in the town.
(b) Number of buses = 6557
Number of tractors = 5247 less than 6557 = 6557 – 5247 = 1310
Therefore, there are 1310 tractors.
(c) Number of tractors = 1310
Number of taxis = 1579 more than 1310 = 1579 + 1310 = 2889
Therefore, there are 2889 taxis.
(d) We have, Jeeps: 6304; Buses: 6557; Tractors: 1310; Taxis: 2889.
The order of these numbers from the lowest to the highest is: 1310, 2889, 6304, 6557.
Question 4.
Solve
a) 1459 + 476
b) 3863 + 4188
c) 5017 + 899
d) 4285 + 2132
e) 3158 + 1052
f) 7293 – 2819
g) 3105 – 1223
h) 8006 – 5567
i) 5000 – 4124
j) 9018 – 487
Solution:
(a) 1459 + 476 = 1935
(b) 3863 + 4188 = 8051
(c) 5017 + 899 = 5916
(d) 4285 + 2132 = 6417
(e) 3158 + 1052 = 4210
(f) 7293-2819 = 4474
(g) 3105 – 1223 = 1882
(h) 8006-5567 = 2439
(i) 5000-4124 = 876
(j) 9018-487 = 8531
Question 5.
The children in a school in Chittoor are planning to organise a Baal Mela in their school.
Raju, Rani and Roja decided to raise some money to make arrangements for the mela. The money is availabe in notes of 500, 100, 50, 10 and coins of 5, 2 and 1. They decide to put the money in the School Panchayat Bank.
Help each of the children fill the deposit slip given below.
Different combinations of notes can give the same amount. Can you guess a possible combination of notes they might have? Fill in the amounts appropriately.
Solution:
Another combination for the same amount is: four 500-rupee notes, four 10-rupee notes and one 5-rupee coin.
Another possible combination for the same amount is: five 500-rupee notes, nine 100-rupee notes, three 50-rupee notes and five 5-rupee coins and three 1-rupee coins.
Another possible combination is:,one 500-rupee notes, two 100-rupee notes, ten 50-rupee notes and four 10-rupee notes.. (Answer may vary for the possible > combinations)
NCERT Textbook Page 161 – Let Us Solve
Question 1.
Solve Sol.
Solution:
Solution:
Solution:
Solution:
Question 2.
Arrange the following in columns and solve in your notebook.
a) 3683 – 971
Solution:
b) 8432-46
Solution:
c) 4011 -3666
Solution:
d) 5203 – 2745
Solution:
e) 1465 + 632
Solution:
f) 3567 + 77
Solution:
g) 8263 + 3737
Solution:
h) 5429 + 3287
Solution:
NCERT Textbook Page 162 – Let Us Solve
Question 1.
Find easy ways to solve the following problems. Write the answers in the given space. Share your thinking with the grade.
a) 8787 – 99 =__________
Solution:
8787 – 99 (Subtract 100, then add 1)
8787 – 100 = 8687
8687 + 1 = 8688
Therefore, 8787 – 99 = 8688
b) 4596 + 104=__________
Solution:
4596 + 104 (Add 100, then add 4)
4596 + 100 = 4696
4696 + 4 = 4700
Therefore, 4596 + 104 = 4700
c) 3459 + 21 =__________
Solution:
3459 + 21 (Add 1, then add 20)
3459 + 1 = 3460
3460 + 20 = 3480
Therefore, 3459 + 21 = 3480
d) 5010 + 95 =__________
Solution:
5010 + 95 (Add 100, then subtract 5)
5010 + 100 = 5110 5110-5 = 5105
Therefore, 5010 + 95 = 5105
e) 4990 + 310 =__________
Solution:
4990 + 310 (Add 10, then add 300)
4990 + 10 = 5000 5000 + 300 = 5300
Therefore, 4990 + 310 = 5300
f) 7844 – 15 =__________
Solution:
7844 – 15 (Subtract 20, then add 5)
7844 – 20 = 7824 7824 + 5 = 7829
Therefore, 7844 — 15 = 7829
g) 260 + 240 =__________
Solution:
260 + 240 (Add 200, then add 40)
260 + 200 = 460 460 + 40 = 500
Therefore, 260 + 240 = 500
h) 1575 – 125 =__________
Solution:
1575 – 125 (Subtract 100, then subtract 25)
1575 – 100 = 1475
1475 – 25 = 1450
Therefore, 1575 – 125 = 1450
i) 3999 + 290 =__________
Solution:
3999 + 290 (Add 1, then add 289)
3999 + 1 = 4000
4000 + 289 = 4289
Therefore, 3999 + 290 = 4289
Question 2.
Use the signs <, =, > as appropriate to compare the following without actually calculating. Try to reason them out and share in grade.
Solution:
Bigger the number added to the same number, bigger the sum.
Same number subtracted from the bigger number is bigger.
Bigger the number subtracted from the same number, smaller the difference.
Question 3.
Use the given information to find the values. Share your reasoning with the grade.
Solution:
NCERT Textbook Page 163
Question 1.
Add
a) 2783 + 378
Solution:
b) 8948 + 97
Solution:
c) 7006 + 367
Solution:
d) 8009 + 485
Solution:
e) 6062 + 3809
Solution:
f) 3792 + 2688
Solution:
g) 4999 + 3888
Solution:
h) 5005 + 4895
Solution:
i) 5768 + 4053
Solution:
j) 3480 + 479
Solution:
Question 2.
Subtract.
a) 4456 – 2768
Solution:
b) 5300 – 467
Solution:
c) 8067 – 4546
Solution:
d) 5302 – 1034
Solution:
e) 8004 – 3107
Solution:
f) 3400 – 897
Solution:
g) 9382 – 4857
Solution:
h) 7561 – 2933
Solution:
i) 6478 – 5986
Solution:
j) 3444 – 2555
Solution:
Question 3.
Fill the squares with the numbers 1-9. The difference between any two neighbouring squares (connected by a line) must be odd.
Can you find other ways to fill the squares? Solution:Yes, we can fill the squares in many ways. For example, –
Solution:
Can you do the same thing such that the difference between any two neighbouring squares is even?
Solution:
No, it is not possible to fill the squares such that the difference between any two neighbouring squares is even. This happens because the difference of either two odd numbers is even or two even numbers is even, means we cannot fill any even number that is connected to an odd number and vice versa, that is not possible in the given connected squares.
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