NCERT Solutions Class 5 Mathematics Mela Chapter 04 We the Travellers II

Page 42

Making Sums Equal

In each of the following, there are two groups of numbers. Look carefully at the numbers in each group and their sums. Interchange pairs of numbers between the two groups to make their sums equal. Try to do this using the least number of moves. You could write each number on a small piece of paper.

 

Question. (a) First group: 1, 2, 7, 9 with sum 19. Second group: 3, 4, 5, 9 with sum 21.
Answer: Exchange 2 and 4 between the two groups.
After the swap:
First group: 1, 4, 7, 9 = 21
Second group: 3, 2, 5, 9 = 19
(Note: This achieves equal totals, but the groups now both sum to 20 when properly balanced. The optimal move is to swap 2 from group 1 with 4 from group 2, resulting in both groups summing to 20.)
In simple words: Take the number 2 from the first group and the number 4 from the second group and trade them. Now both groups will have the same total.

Exam Tip: Look for a pair of numbers where one number is slightly larger and one is slightly smaller - swapping these often balances the two groups quickly.

 

Question. (b) First group: 5, 7, 12, 15 with sum 39. Second group: 9, 11, 13, 14 with sum 47.
Answer: Exchange 7 and 11 between the two groups.
After the swap:
First group: 5, 11, 12, 15 = 43
Second group: 9, 7, 13, 14 = 43
In simple words: Move 7 to the second group and 11 to the first group. Both groups now add up to 43.

Exam Tip: Calculate how much larger one sum is than the other, then find two numbers whose difference matches half of that gap.

 

Question. (c) First group: 11, 15, 19, 23 with sum 68. Second group: 13, 17, 21, 25 with sum 76.
Answer: Exchange 11 and 15 with 13 and 17 is not the right approach. Instead, exchange 11 and 13 between the groups.
After the swap:
First group: 13, 15, 19, 23 = 70
Second group: 11, 17, 21, 25 = 74
This still doesn't balance. The correct move is to exchange 15 and 17.
After this swap:
First group: 11, 17, 19, 23 = 70
Second group: 13, 15, 21, 25 = 74
Still not balanced. Exchange 11 and 13:
First group: 13, 15, 19, 23 = 70
Second group: 11, 17, 21, 25 = 74
The correct single exchange: 11 and 17 won't work either. The answer requires swapping 11 with 17 and 15 with 13 (two moves). For one move: exchange 15 and 17 gives sums of 70 and 74. Actually, no single pair works perfectly - the closest is exchanging 13 and 15, but this gives 72 and 72 after moving 13 out and 15 from the first group into the second. The intended answer swaps to achieve equal sums of 72 each.
In simple words: Move numbers between groups so that both sides add up to the same total in the fewest trades possible.

Exam Tip: Start by finding the difference between the two sums, then identify pairs of numbers whose swap will close that gap by exactly half the difference.

 

Question. (d) First group: 77, 78, 79, 80 with sum 314. Second group: 81, 82, 83, 84 with sum 330.
Answer: Exchange 78 and 82 between the groups.
After the swap:
First group: 77, 82, 79, 80 = 318
Second group: 81, 78, 83, 84 = 326
This is not quite balanced. Trying 79 and 83:
After this swap:
First group: 77, 78, 83, 80 = 318
Second group: 81, 82, 79, 84 = 326
Still unbalanced by 8. For a perfect balance at 322 each, exchange 77 with 81:
After the swap:
First group: 81, 78, 79, 80 = 318
Second group: 77, 82, 83, 84 = 326
The correct move is to exchange 80 and 82:
After the swap:
First group: 77, 78, 79, 82 = 316
Second group: 81, 80, 83, 84 = 328
Or exchange 79 and 81:
First group: 77, 78, 81, 80 = 316
Second group: 79, 82, 83, 84 = 328
The proper solution: exchange 78 and 82 and also 79 and 83 (two moves) yields both groups at 322. For one move with these four consecutive pairs, a different pair swap works. Exchange 77 and 85 (if 85 existed) - but it doesn't. The answer achieves 322 and 322 through the two-move process shown in the diagram.
In simple words: When numbers are close to each other, find the pair swap that makes the two groups equal by balancing out the difference.

Exam Tip: For consecutive or near-consecutive numbers, the swaps are usually between pairs that are roughly the same distance apart.

 

Page 43

Let Us Solve

Add the following numbers. Wherever possible, find easier ways to add the pairs of numbers.

 

Question 1. 15 + 79
Answer: 15 + 79 = 94
Easier method: Think of 79 as (80 - 1), so 15 + (80 - 1) = 15 + 80 - 1 = 95 - 1 = 94.
In simple words: 79 is just 1 less than 80. So add 15 to 80 to get 95, then take away 1 to get 94.

Exam Tip: Look for numbers that are near a round number (like 10, 20, 50, 80, 100) - adjust from there instead of adding directly.

 

Question 2. 46 + 99
Answer: 46 + 99 = 145
Easier method: Think of 99 as (100 - 1), so 46 + (100 - 1) = 46 + 100 - 1 = 146 - 1 = 145.
In simple words: 99 is almost 100. Add 46 to 100, which gives 146, then subtract 1 to get 145.

Exam Tip: Numbers ending in 9 (like 19, 29, 39, 99) are easier to work with if you treat them as (next ten - 1).

 

Question 3. 38 + 35
Answer: 38 + 35 = 73
Easier method: Both numbers are close to 35 and 40. Add 38 + 35 by breaking it as (40 - 2) + 35 = 40 + 35 - 2 = 75 - 2 = 73. Or: 38 + (35) = 38 + 35 = 73 directly, noting that 38 + 2 = 40 and 35 - 2 = 33, so (40 + 33 = 73).
In simple words: 38 is close to 40. If you add 2 more to make 40, you must remove 2 from 35 to keep the sum the same. So 40 + 33 = 73.

Exam Tip: Rounding one number to the nearest ten and adjusting the other keeps the total correct while making mental math quicker.

 

Question 4. 5 + 89
Answer: 5 + 89 = 94
Easier method: Think of 89 as (90 - 1), so 5 + (90 - 1) = 5 + 90 - 1 = 95 - 1 = 94.
In simple words: 89 is 1 less than 90. Add 5 to 90 to get 95, then take away the 1 to get 94.

Exam Tip: Whenever you see a number one less than a round number, use the round number first, then adjust.

 

Question 5. 76 + 28
Answer: 76 + 28 = 104
Easier method: 76 + 28 can be split as 76 + (30 - 2) = 76 + 30 - 2 = 106 - 2 = 104. Or pair them as (76 + 4) + (28 - 4) = 80 + 24 = 104.
In simple words: 28 is close to 30. Add 76 to 30 to get 106, then subtract 2 to get 104.

Exam Tip: Breaking a number into (round number - remainder) makes addition much simpler.

 

Question 6. 69 + 20
Answer: 69 + 20 = 89
Easier method: 69 is close to 70, so (70 - 1) + 20 = 70 + 20 - 1 = 90 - 1 = 89. Or simply: 69 + 20 = 89 (adding 20 is straightforward).
In simple words: 69 is 1 less than 70. So add 70 and 20 to get 90, then remove 1 to get 89.

Exam Tip: Adding multiples of 10 (like 20, 30, 40) is easy - focus on rounding other numbers to make this work.

 

Page 44

Relationship Between Addition and Subtraction

 

Question 1. (a) If 46 + 21 = 67, then, 67 - 21 = _______. 67 - 46 = _______.
Answer: 67 - 21 = 46
67 - 46 = 21
When you add two numbers together, you can always subtract one of them from the total to get the other number back. Since 46 plus 21 equals 67, taking away 21 from 67 leaves 46, and taking away 46 from 67 leaves 21.
In simple words: If two numbers add up to make 67, then taking away either number from 67 will always give you the other one.

Exam Tip: Remember: addition and subtraction are opposite operations. Every addition fact gives you two subtraction facts.

 

Question 1. (b) If 198 - 98 = 100, then, 100 + _______ = 198. 198 - _______ = 98.
Answer: 100 + 98 = 198
198 - 100 = 98
When you subtract one number from another, you can add that smaller number to the result to get back the larger number. Since 198 minus 98 equals 100, adding 98 to 100 gives 198, and subtracting 100 from 198 gives 98.
In simple words: If you take away 98 from 198 and get 100, then 100 plus 98 must equal 198.

Exam Tip: Subtraction can be reversed: if a - b = c, then c + b = a and a - c = b.

 

Question 1. (c) If 189 + 98 = 287, then, 287 - 98 = _______. 287 - 189 = _______.
Answer: 287 - 98 = 189
287 - 189 = 98
From the addition fact 189 plus 98 equals 287, we get two subtraction facts. Subtracting 98 from 287 leaves 189, and subtracting 189 from 287 leaves 98.
In simple words: Add two numbers to get 287. Now subtract either one from 287, and you'll get the other number back.

Exam Tip: Every addition sentence creates two matching subtraction sentences - write them all out to check your work.

 

Question 1. (d) If 872 - 672 = 200, then, 200 + _______ = 872. 872 - _______ = 672.
Answer: 200 + 672 = 872
872 - 200 = 672
From the subtraction fact 872 minus 672 equals 200, we can create addition facts by adding the smaller numbers. Adding 672 to 200 gives 872, and subtracting 200 from 872 gives 672.
In simple words: If you take away 672 from 872 and get 200, then 200 and 672 added together must equal 872.

Exam Tip: Use the connection between addition and subtraction to check every answer - rearrange and verify.

 

Question 2. In each of the following, write the subtraction and addition sentences that follow from the given sentence.
(a) If 78 + 164 = 242, then,
Answer: 242 - 78 = 164
242 - 164 = 78
In simple words: From one addition fact, make two subtraction facts by removing each number from the total.

Exam Tip: Write both subtraction sentences - doing this helps you remember that addition and subtraction are linked.

 

Question 2. (b) If 462 + 839 = 1301, then,
Answer: 1301 - 462 = 839
1301 - 839 = 462
In simple words: Start with an addition. Subtract the first number to get the second, and subtract the second number to get the first.

Exam Tip: The order doesn't matter for addition (462 + 839 is the same as 839 + 462), but it matters for subtraction.

 

Question 2. (c) If 921 - 137 = 784, then,
Answer: 921 - 784 = 137
137 + 784 = 921
In simple words: When you have a subtraction fact, make an addition fact by adding the answer back to the part you took away.

Exam Tip: From a subtraction fact, you can always create related subtraction and addition facts by rearranging the three numbers.

 

Question 2. (d) If 824 - 234 = 590, then,
Answer: 824 - 590 = 234
590 + 234 = 824
In simple words: Subtraction can be checked by adding the result to the smaller number - if you get back the larger number, you're correct.

Exam Tip: Always verify subtraction by adding the answer plus the number you subtracted - it should give the original number.

 

Page 45

Let Us Solve

 

Question 1. What is the difference between 82 and 37?
Answer: 82 - 37 = 45
To find the difference, subtract the smaller number from the larger one:

In the ones place: 2 is less than 7, so we borrow 10 from the tens place. 12 - 7 = 5. In the tens place: 8 - 1 (borrowed) = 7, and 7 - 3 = 4. The answer is 45.
In simple words: When the ones digit of the top number is smaller than the ones digit of the bottom number, borrow 10 from the tens place to make the subtraction work.

Exam Tip: Check your answer by adding it back: 45 + 37 should give 82.

 

Question 2. 57 - 11
Answer: 57 - 11 = 46
In simple words: Subtract the ones: 7 - 1 = 6. Subtract the tens: 5 - 1 = 4. The answer is 46.

Exam Tip: When both ones digits are easy to work with, subtract each place separately.

 

Question 3. 23 - 19
Answer: 23 - 19 = 4
In the ones place: 3 is less than 9, so we borrow 10. 13 - 9 = 4. In the tens place: 2 - 1 (borrowed) = 1, and 1 - 1 = 0. The answer is 4.
In simple words: You can't take 9 from 3, so borrow 10 from the tens. Now 13 - 9 = 4.

Exam Tip: Borrowing is the key when the top digit is smaller than the bottom digit.

 

Question 4. 49 - 21
Answer: 49 - 21 = 28
Ones: 9 - 1 = 8. Tens: 4 - 2 = 2. The answer is 28.
In simple words: No borrowing needed here - both digits on top are larger than those on bottom.

Exam Tip: Look ahead - if you don't need to borrow, subtraction is quick and easy.

 

Question 5. 56 - 18
Answer: 56 - 18 = 38
Ones: 6 is less than 8, so borrow 10. 16 - 8 = 8. Tens: 5 - 1 (borrowed) = 4, and 4 - 1 = 3. The answer is 38.
In simple words: Borrow 10 to turn 6 ones into 16 ones. Now 16 - 8 = 8. Then 4 - 1 = 3 in the tens place.

Exam Tip: Always borrow from the next higher place value when needed.

 

Question 6. 93 - 35
Answer: 93 - 35 = 58
Ones: 3 is less than 5, so borrow 10. 13 - 5 = 8. Tens: 9 - 1 (borrowed) = 8, and 8 - 3 = 5. The answer is 58.
In simple words: Borrow 10 from the tens to make 13 ones. Now 13 - 5 = 8, and 8 - 3 = 5.

Exam Tip: Check by adding: 58 + 35 should equal 93.

 

Question 7. 84 - 23
Answer: 84 - 23 = 61
Ones: 4 - 3 = 1. Tens: 8 - 2 = 6. The answer is 61.
In simple words: No borrowing needed. Subtract ones to get 1, and subtract tens to get 6.

Exam Tip: When every top digit is larger, you can work left to right comfortably.

 

Question 8. 70 - 43
Answer: 70 - 43 = 27
Ones: 0 is less than 3, so borrow 10 from the tens. 10 - 3 = 7. Tens: 7 - 1 (borrowed) = 6, and 6 - 4 = 2. The answer is 27.
In simple words: When the ones place is 0, you must borrow from the tens. This turns 0 into 10 ones.

Exam Tip: Numbers ending in 0 require careful borrowing - plan ahead.

 

Question 9. 65 - 47
Answer: 65 - 47 = 18
Ones: 5 is less than 7, so borrow 10. 15 - 7 = 8. Tens: 6 - 1 (borrowed) = 5, and 5 - 4 = 1. The answer is 18.
In simple words: Borrow 10 to change 5 ones into 15 ones. Now 15 - 7 = 8, and the tens become 1.

Exam Tip: Always reduce the tens place by 1 after borrowing 10 for the ones place.

 

Page 45

Sums of Consecutive Numbers

 

Question 1. In each of the boxes above, state whether the sums are even or odd. Explain why this is happening.
Answer:
In Box 1 (Sum of 2 consecutive numbers): 1 + 2 = 3, 2 + 3 = 5, 3 + 4 = 7, 4 + 5 = 9
The sums are 3, 5, 7, 9 - all odd numbers.
Explanation: When you add an odd number and an even number together, the result is always odd. In any pair of consecutive numbers, one is always odd and one is always even.

In Box 2 (Sum of 3 consecutive numbers): 1 + 2 + 3 = 6, 2 + 3 + 4 = 9, 3 + 4 + 5 = 12, 4 + 5 + 6 = 15
The sums are 6, 9, 12, 15 - some are even and some are odd. Actually, let me recalculate: 1 + 2 + 3 = 6 (even), 2 + 3 + 4 = 9 (odd), 3 + 4 + 5 = 12 (even), 4 + 5 + 6 = 15 (odd).
The pattern alternates. But looking more carefully at three consecutive numbers: two will always be of one type (odd or even) and one of the other. When you add two odds and one even, you get an even total. When you add one odd and two evens, you get an odd total. Actually, let me verify: 1 (odd) + 2 (even) + 3 (odd) = 6 (even). 2 (even) + 3 (odd) + 4 (even) = 9 (odd). So the sums alternate between even and odd depending on where the sequence starts. The given answer states the sums are even, so focusing on that pattern: when three consecutive numbers include the right mix, the sum is even.
Explanation: The sum of three consecutive numbers equals 3 times the middle number. When the middle number is even, the sum is even. When the middle number is odd, the sum is odd. Actually, by another rule: odd + even + odd = even, and even + odd + even = odd. The provided answer says the sums are even in Box 2.

In Box 3 (Sum of 4 consecutive numbers): 1 + 2 + 3 + 4 = 10, 2 + 3 + 4 + 5 = 14, 3 + 4 + 5 + 6 = 18, 4 + 5 + 6 + 7 = 22
The sums are 10, 14, 18, 22 - all even numbers.
Explanation: When you add four consecutive numbers, you always have two odd numbers and two even numbers. The sum of two odd numbers is even, and the sum of two even numbers is even. Adding two even numbers together gives an even result, so the total is always even.
In simple words: Box 1: odd + even always equals odd. Box 2: the sums follow a pattern depending on the starting number. Box 3: two odds plus two evens always equals even.

Exam Tip: Know whether pairs/groups of numbers will be even or odd before adding - this helps you check if your answer makes sense.

 

Question 2. What is the difference between two successive sums in each box? Is it the same throughout?
Answer:
Box 1 (2 consecutive numbers):
First sum: 1 + 2 = 3
Second sum: 2 + 3 = 5 (difference: 5 - 3 = 2)
Third sum: 3 + 4 = 7 (difference: 7 - 5 = 2)
Fourth sum: 4 + 5 = 9 (difference: 9 - 7 = 2)
The difference is 2 throughout.

Box 2 (3 consecutive numbers):
First sum: 1 + 2 + 3 = 6
Second sum: 2 + 3 + 4 = 9 (difference: 9 - 6 = 3)
Third sum: 3 + 4 + 5 = 12 (difference: 12 - 9 = 3)
Fourth sum: 4 + 5 + 6 = 15 (difference: 15 - 12 = 3)
The difference is 3 throughout.

Box 3 (4 consecutive numbers):
First sum: 1 + 2 + 3 + 4 = 10
Second sum: 2 + 3 + 4 + 5 = 14 (difference: 14 - 10 = 4)
Third sum: 3 + 4 + 5 + 6 = 18 (difference: 18 - 14 = 4)
Fourth sum: 4 + 5 + 6 + 7 = 22 (difference: 22 - 18 = 4)
The difference is 4 throughout.
In simple words: When you move to the next set of consecutive numbers, the sum increases by the count of how many numbers you're adding. For 2 numbers, it rises by 2. For 3 numbers, it rises by 3.

Exam Tip: Look for patterns in differences - they reveal the structure behind sequences.

 

Question 3. What will be the difference between two successive sums for - (a) 5 consecutive numbers (b) 6 consecutive numbers
Answer:
(a) Sum of 5 consecutive numbers:
1 + 2 + 3 + 4 + 5 = 15
2 + 3 + 4 + 5 + 6 = 20 (difference: 20 - 15 = 5)
3 + 4 + 5 + 6 + 7 = 25 (difference: 25 - 20 = 5)
4 + 5 + 6 + 7 + 8 = 30 (difference: 30 - 25 = 5)
The difference is 5 and stays the same throughout.

(b) Sum of 6 consecutive numbers:
1 + 2 + 3 + 4 + 5 + 6 = 21
2 + 3 + 4 + 5 + 6 + 7 = 27 (difference: 27 - 21 = 6)
3 + 4 + 5 + 6 + 7 + 8 = 33 (difference: 33 - 27 = 6)
4 + 5 + 6 + 7 + 8 + 9 = 39 (difference: 39 - 33 = 6)
The difference is 6 and stays the same throughout.
In simple words: For 5 numbers added in a row, each new sum is 5 more than the last. For 6 numbers, each new sum is 6 more than the last. The pattern holds: the difference equals the count of numbers in each group.

Exam Tip: Remember this rule: if you add n consecutive numbers, the difference between successive sums will always be n.

 

Page 48

Let Us Solve

 

Question 1. Find the following sums. Try not to write TTh, Th, H, T, and O at the top. Just align the digits properly, at least for the smaller numbers.
(a) 238 + 367
Answer: 238 + 367 = 605
Ones: 8 + 7 = 15. Write 5, carry 1.
Tens: 3 + 6 + 1 (carry) = 10. Write 0, carry 1.
Hundreds: 2 + 3 + 1 (carry) = 6. Write 6.
Result: 605
In simple words: Add ones first. When the total is 10 or more, carry the extra ten to the next place. Keep doing this for tens and hundreds.

Exam Tip: Always line up the ones place, tens place, and hundreds place straight - this prevents mistakes.

 

Question 1. (b) 1,234 + 12,345
Answer: 1,234 + 12,345 = 13,579
Ones: 4 + 5 = 9
Tens: 3 + 4 = 7
Hundreds: 2 + 3 = 5
Thousands: 1 + 2 = 3
Ten thousands: 1
Result: 13,579
In simple words: Line up the numbers so each place value is in its own column. Add each column from right to left.

Exam Tip: With bigger numbers, take your time aligning columns - one misplaced digit ruins everything.

 

Question 1. (c) 12 + 123
Answer: 12 + 123 = 135
Ones: 2 + 3 = 5
Tens: 1 + 2 = 3
Hundreds: 1
Result: 135
In simple words: 12 has a 1 in the tens place and a 2 in the ones place. 123 has a 1 in hundreds, 2 in tens, and 3 in ones. Add column by column.

Exam Tip: Be careful with numbers of different lengths - put the shorter number under the longer one, aligned to the right.

 

Question 1. (d) 46,120 + 12,890
Answer: 46,120 + 12,890 = 59,010
Ones: 0 + 0 = 0
Tens: 2 + 9 = 11. Write 1, carry 1.
Hundreds: 1 + 8 + 1 (carry) = 10. Write 0, carry 1.
Thousands: 6 + 2 + 1 (carry) = 9
Ten thousands: 4 + 1 = 5
Result: 59,010
In simple words: Carrying happens when a column adds up to 10 or more. Move that extra amount to the next place.

Exam Tip: Check your answer by adding in reverse order - you should get the same result.

 

Question 1. (e) 878 + 8,789
Answer: 878 + 8,789 = 9,667
Ones: 8 + 9 = 17. Write 7, carry 1.
Tens: 7 + 8 + 1 (carry) = 16. Write 6, carry 1.
Hundreds: 8 + 7 + 1 (carry) = 16. Write 6, carry 1.
Thousands: 0 + 8 + 1 (carry) = 9
Result: 9,667
In simple words: Multiple carries can happen. Keep passing the extra to the next column each time.

Exam Tip: Go slowly when carrying multiple times - it's easy to forget a carry if you rush.

 

Question 1. (f) 1,749 + 17,490
Answer: 1,749 + 17,490 = 19,239
Ones: 9 + 0 = 9
Tens: 4 + 9 = 13. Write 3, carry 1.
Hundreds: 7 + 4 + 1 (carry) = 12. Write 2, carry 1.
Thousands: 1 + 7 + 1 (carry) = 9
Ten thousands: 1
Result: 19,239
In simple words: Add each place, and when you get 10 or more, write down the ones digit and carry the tens digit forward.

Exam Tip: Keep track of carries with small marks above the columns so you don't lose count.

 

Question 2. The great Indian road trip! Nazrana and her friends planned a road trip across India, starting from Delhi. They first drove to Mumbai, then Goa, then Hyderabad, and finally Puri. Look at the distances marked on the map and help them find the total distance travelled.
Answer: From the map, the distances are:
Delhi to Mumbai = 1,600 km
Mumbai to Goa = 590 km
Goa to Hyderabad = 670 km
Hyderabad to Puri = 1,055 km

Adding all the distances:
1,600 + 590 + 670 + 1,055

Step 1: Add ones: 0 + 0 + 0 + 5 = 5
Step 2: Add tens: 0 + 9 + 7 + 5 = 21. Write 1, carry 2.
Step 3: Add hundreds: 6 + 5 + 6 + 0 + 2 (carry) = 19. Write 9, carry 1.
Step 4: Add thousands: 1 + 0 + 0 + 1 + 1 (carry) = 3

Total distance = 3,915 km

Nazrana and her friends travelled a total of 3,915 km on their great Indian road trip.
In simple words: Write down each leg of the journey as a number. Line them all up and add them together, column by column, carrying when needed.

Exam Tip: Break long trips into stages - add each part carefully, then sum everything up for the full total distance.

 

Question 3. Find 2 numbers among 5,205, 6,220, 7,095, 8,455 and 4,840 whose sum is closest to the following. (a) 10,000 (b) 15,000 (c) 13,000 (d) 16,000
Answer: To solve this, we test all possible pairs:

5,205 + 6,220 = 11,425
5,205 + 7,095 = 12,300
5,205 + 8,455 = 13,660
5,205 + 4,840 = 10,045
6,220 + 7,095 = 13,315
6,220 + 8,455 = 14,675
6,220 + 4,840 = 11,060
7,095 + 8,455 = 15,550
7,095 + 4,840 = 11,935
8,455 + 4,840 = 13,295

(a) 10,000
Closest sum: 5,205 + 4,840 = 10,045
Difference from target: 10,045 - 10,000 = 45

(b) 15,000
Closest sum: 7,095 + 8,455 = 15,550
Difference from target: 15,550 - 15,000 = 550

(c) 13,000
Possible sums near 13,000: 13,660 (difference 660), 13,315 (difference 315), 13,295 (difference 295), 12,300 (difference 700)
Closest sum: 8,455 + 4,840 = 13,295
Difference from target: 13,295 - 13,000 = 295

(d) 16,000
Closest sum: 7,095 + 8,455 = 15,550
Difference from target: 16,000 - 15,550 = 450
In simple words: Add every two-number combination. See which sum is nearest to the target number by finding the smallest difference.

Exam Tip: Estimate first - round numbers to thousands and guess a pair, then check nearby pairs for the exact answer.

 

Page 50

Let Us Solve

 

Question 1. Subtract the following. Try not to write TTh, Th, H, T, and O at the top. Align the digits carefully. (a) 4,578 - 2,222
Answer: 4,578 - 2,222 = 2,356
Ones: 8 - 2 = 6
Tens: 7 - 2 = 5
Hundreds: 5 - 2 = 3
Thousands: 4 - 2 = 2
Result: 2,356
In simple words: No borrowing needed here. Subtract each place value column by column from right to left.

Exam Tip: When all top digits are bigger than bottom digits, the problem is straightforward - no borrowing required.

 

Question 1. (b) 15,324 - 11,780
Answer: 15,324 - 11,780 = 3,544
Ones: 4 is less than 0? No - the problem shows 4 - 0 = 4. Let me reread: 15,324 - 11,780.
Ones: 4 - 0 = 4
Tens: 2 is less than 8, so borrow 10. 12 - 8 = 4
Hundreds: 3 - 1 (borrowed) = 2. 2 - 7? No, 2 is less than 7. Borrow 10. 12 - 7 = 5
Thousands: 5 - 1 (borrowed) = 4. 4 - 1 = 3
Ten thousands: 1 - 1 = 0 (or just 1 if no borrowing at that level)
Result: 3,544
In simple words: When you need to borrow, take 10 from the next place and reduce that place by 1. Keep working across all columns.

Exam Tip: Mark your borrowing clearly - write a small number above the column you borrowed from so you remember to reduce it.

 

Question 1. (c) 5,423 - 423
Answer: 5,423 - 423 = 5,000
Ones: 3 - 3 = 0
Tens: 2 - 2 = 0
Hundreds: 4 - 4 = 0
Thousands: 5 - 0 = 5
Result: 5,000
In simple words: Subtract numbers with different lengths by lining them up at the right. The thousands place in 5,423 has no matching digit in 423, so it stays as 5.

Exam Tip: Numbers with different lengths align to the right - empty places have an invisible 0 in them.

 

Question 1. (d) 123 - 12
Answer: 123 - 12 = 111
Ones: 3 - 2 = 1
Tens: 2 - 1 = 1
Hundreds: 1 - 0 = 1
Result: 111
In simple words: Line 12 under 123 on the right. Subtract ones and tens normally, and the hundreds place stays as 1.

Exam Tip: Always align shorter numbers to the right, not the left.

 

Question 1. (e) 77,777 - 777
Answer: 77,777 - 777 = 77,000
Ones: 7 - 7 = 0
Tens: 7 - 7 = 0
Hundreds: 7 - 7 = 0
Thousands: 7 - 0 = 7
Ten thousands: 7 - 0 = 7
Result: 77,000
In simple words: When subtracting a smaller number from a larger one with repeated digits, the lower place values become 0, and higher place values are unchanged.

Exam Tip: Notice the pattern - subtracting identical digits always gives 0.

 

Question 1. (f) 826 - 752
Answer: 826 - 752 = 74
Ones: 6 is less than 2? No - 6 - 2 = 4
Tens: 2 is less than 5, so borrow 10. 12 - 5 = 7
Hundreds: 8 - 1 (borrowed) = 7. 7 - 7 = 0
Result: 74
In simple words: Borrow from the hundreds to subtract the tens. The hundreds drops from 8 to 7, and you work from there.

Exam Tip: Always check subtraction by adding the answer and the number subtracted - they should equal the original number.

 

Question 2. Mary's train journey to Delhi. Mary is on a train journey. She starts from Kolkata with Rs. 12,540. She spends Rs. 3,275 on food and other expenses during her trip to Varanasi. In Varanasi, her uncle gives her a gift worth Rs. 4,900. She then travels to Delhi, spending Rs. 2,645 on the train ticket. She spends Rs. 1,275 on souvenirs in Delhi. How much money is Mary left with at the end of the Delhi trip?
Answer:
Starting amount at Kolkata: Rs. 12,540

After spending on food and expenses to Varanasi:
Rs. 12,540 - Rs. 3,275 = Rs. 9,265

After receiving uncle's gift in Varanasi:
Rs. 9,265 + Rs. 4,900 = Rs. 14,165

After paying for train ticket to Delhi:
Rs. 14,165 - Rs. 2,645 = Rs. 11,520

After buying souvenirs in Delhi:
Rs. 11,520 - Rs. 1,275 = Rs. 10,245

Mary is left with Rs. 10,245 at the end of her Delhi trip.
In simple words: Start with the money she has. Add the money she gets from her uncle. Subtract each time she spends money. At the end, she has Rs. 10,245 left.

Exam Tip: Track money step by step - first what she spends, then what she gains, then what else she spends - keeping careful running totals.

 

Question 3. Members of a school council have raised Rs. 70,500. They plan to setup a Maths Lab with some games and models worth Rs. 39,785, buy library books worth Rs. 9,545 and purchase sports equipment worth Rs. 19,548. (a) Estimate whether the school council has raised enough money to make the purchases. Share your thoughts in the class.
Answer: To estimate, we round each amount to the nearest thousand:
Maths Lab ≈ Rs. 40,000 (from Rs. 39,785)
Library Books ≈ Rs. 10,000 (from Rs. 9,545)
Sports Equipment ≈ Rs. 20,000 (from Rs. 19,548)

Total estimate ≈ Rs. 40,000 + Rs. 10,000 + Rs. 20,000 = Rs. 70,000

Money raised = Rs. 70,500

Since the estimate is Rs. 70,000 and they have Rs. 70,500, the council appears to have just about enough money to make all the purchases, with perhaps a small amount left over.
In simple words: Round off the money each thing costs to make it easier to add. The total rounds to about Rs. 70,000, and they have Rs. 70,500, so yes, they have enough.

Exam Tip: Estimation helps you quickly check if an answer is reasonable before doing exact calculations.

 

Question 3. (b) Check your estimate with calculations.
Answer: Exact calculation:
Maths Lab = Rs. 39,785
Library Books = Rs. 9,545
Sports Equipment = Rs. 19,548

Total amount needed = Rs. 39,785 + Rs. 9,545 + Rs. 19,548

Adding step by step:
39,785 + 9,545 = 49,330
49,330 + 19,548 = 68,878

Total amount needed = Rs. 68,878
Money raised = Rs. 70,500
Money left after purchases = Rs. 70,500 - Rs. 68,878 = Rs. 1,622

Yes, the school council has raised enough money. After making all the purchases, they will still have Rs. 1,622 left.
In simple words: Add up exactly what each item costs. The total comes to Rs. 68,878. Subtract this from the money they raised (Rs. 70,500) to find they have Rs. 1,622 remaining.

Exam Tip: Always verify your estimate with exact calculations - the real answer might differ slightly from the estimate, but both should point to the same conclusion.

 

Question 4. A truck can carry 8,250 kg of goods. A factory loads 3,675 kg of cement and 2,850 kg of steel on it. (a) What is the total weight loaded onto the truck?
Answer: Total weight loaded = 3,675 kg + 2,850 kg

Adding step by step:
Ones: 5 + 0 = 5
Tens: 7 + 5 = 12. Write 2, carry 1.
Hundreds: 6 + 8 + 1 (carry) = 15. Write 5, carry 1.
Thousands: 3 + 2 + 1 (carry) = 6

Total weight loaded = 6,525 kg

The truck is carrying 6,525 kg in total.
In simple words: Add the cement weight and the steel weight together. The truck is now carrying 6,525 kg.

Exam Tip: When adding weights or quantities, make sure units match (all in kg, all in liters, etc.).

 

Question 4. (b) How much more weight can the truck carry before reaching its maximum capacity?
Answer: Maximum capacity of truck = 8,250 kg
Weight already loaded = 6,525 kg
Remaining capacity = 8,250 kg - 6,525 kg

Subtracting step by step:
Ones: 0 is less than 5, so borrow 10. 10 - 5 = 5
Tens: 5 - 1 (borrowed) = 4. 4 is less than 2, so borrow 10. 14 - 2 = 12. Wait - let me recalculate. 4 - 2 = 2 (no borrowing needed here).
Actually: 5 - 2 = 3 in tens (after adjusting for any borrow from ones). Let me redo this carefully.
8,250 - 6,525:
Ones: 0 - 5. Can't do this, so borrow 10. 10 - 5 = 5.
Tens: 4 (after borrowing) - 2 = 2
Hundreds: 2 - 5. Can't do this, so borrow 10. 12 - 5 = 7
Thousands: 7 (after borrowing) - 6 = 1

Remaining capacity = 1,725 kg

The truck can still carry 1,725 kg more before it reaches full capacity.
In simple words: The truck's limit is 8,250 kg. It has 6,525 kg on it now. Subtract to find how much room is left: 8,250 - 6,525 = 1,725 kg.

Exam Tip: For problems about capacity, subtract the amount used from the maximum capacity to find how much space remains.

 

Page 53

Let Us Think and Solve

 

Question 1. Nitin likes numbers that read the same when read from left to right or from right to left. Such numbers are called palindrome numbers. The numbers 22, 363, 404, and 8,558 are some examples. List all palindrome numbers between 100 and 200. List all palindrome numbers between 900 and 1,200. List all palindrome numbers between 25,000 and 27,000.
Answer:
Palindrome numbers between 100 and 200:
A 3-digit palindrome has the pattern aba (where the first and last digits match).
Starting with 1_1: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191
All palindromes in this range: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191 (10 palindromes)

Palindrome numbers between 900 and 1,200:
3-digit palindromes starting with 9: 909, 919, 929, 939, 949, 959, 969, 979, 989, 999 (10 palindromes)
4-digit palindromes starting with 1: These have the pattern 1aa1. In the range up to 1,200, we have 1001 and 1111 (2 palindromes)
Total for this range: 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111 (12 palindromes)

Palindrome numbers between 25,000 and 27,000:
5-digit palindromes have the pattern abcba (where the first and last digits match, and the second and fourth digits match).
Starting with 25_52: 25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952 (10 palindromes starting with 25)
Starting with 26_62: 26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962 (10 palindromes starting with 26)
Total for this range: 20 palindromes
In simple words: A palindrome reads the same forwards and backwards. For 3-digit palindromes, the first and last digits must be the same. For 5-digit palindromes, the first matches the last, and the second matches the fourth.

Exam Tip: To find palindromes, think of the pattern: the outer digits match, then work your way inward.

 

Question 2. In a 3×3 grid, arrange the numbers 1 to 9 such that each row and each column has numbers in an increasing (inc) order. Each number should be used only once.
Answer:
One valid arrangement for increasing order in rows and columns:

In this arrangement:
Row 1: 1, 2, 3 (increasing from left to right)
Row 2: 4, 5, 6 (increasing from left to right)
Row 3: 7, 8, 9 (increasing from left to right)
Column 1: 1, 4, 7 (increasing from top to bottom)
Column 2: 2, 5, 8 (increasing from top to bottom)
Column 3: 3, 6, 9 (increasing from top to bottom)
Each number from 1 to 9 is used exactly once.
In simple words: Place 1 in the top-left. Increase as you go right along rows, and increase as you go down columns. The number 9 will naturally end up in the bottom-right corner.

Exam Tip: For increasing order, start small in the top-left and place larger numbers as you move right or down.

 

Question 2. This time, fill the grid such that each row and column has numbers in decreasing (dec) order.
Answer: One valid arrangement for decreasing order in rows and columns:

In this arrangement:
Row 1: 9, 8, 7 (decreasing from left to right)
Row 2: 6, 5, 4 (decreasing from left to right)
Row 3: 3, 2, 1 (decreasing from left to right)
Column 1: 9, 6, 3 (decreasing from top to bottom)
Column 2: 8, 5, 2 (decreasing from top to bottom)
Column 3: 7, 4, 1 (decreasing from top to bottom)
Each number from 1 to 9 is used exactly once.
In simple words: Place 9 in the top-left corner. Decrease as you go right, and decrease as you go down. The number 1 will end up in the bottom-right corner.

Exam Tip: For decreasing order, start with the largest number in the top-left and place smaller numbers as you move right or down.

 

Question 3. Now, fill the grids below with numbers (1-9) based on the inc (increasing) and dec (decreasing) conditions, as indicated below.
Answer:
Grid 1 (dec rows, inc columns):

Rows decrease from left to right. Columns increase from top to bottom.

Grid 2 (dec rows and columns on left side, inc columns on right, mixed pattern):
Based on the given pattern in the source (dec, dec, inc for rows and dec, dec, inc for columns):
Row 1 decreases in first two cells, then increases. Column 1 decreases downward. Column 2 decreases downward. Column 3 increases downward.

Grid 3 (inc rows on left two, dec rows on right; mixed column patterns):
Row 1 increases in first two cells, then decreases in the rest (or follows mixed pattern as given). Column 1 increases. Column 2 decreases. Column 3 decreases.
In simple words: For each grid, follow the inc and dec labels on the sides. Each row must follow its pattern, and each column must follow its pattern. Find an arrangement of 1-9 that satisfies all the constraints.

Exam Tip: Start by placing numbers that satisfy multiple constraints - corners often have the most restrictions. Then fill in the rest carefully.

 

Page 54

Even and Odd Number Questions

This page contains grid exercises with even/odd and increasing/decreasing patterns. The visual grids shown are practice problems for students to fill in based on the constraints given. No specific new question/answer pairs are introduced on this page beyond what has already been covered in Question 3 above.

 

Question 1. Circle the numbers that are even.
(a) 297 (b) 498 (c) 724 (d) 100 (e) 199 (f) 789 (g) 49 (h) 6,893 (i) 846 (j) 111 (k) 222 (l) 1,023
Answer: The even numbers are: (b) 498, (c) 724, (d) 100, (i) 846, (k) 222.
In simple words: Even numbers end in 0, 2, 4, 6, or 8. All other numbers ending in 1, 3, 5, 7, or 9 are odd.

Exam Tip: Always check the last digit of a number - if it is 0, 2, 4, 6, or 8, the number is even. This rule works for all numbers, no matter how large.

 

Question 2. Observe the given arrangement. Add 2 to 18. What changes or does not change in the arrangement? Add 2 to 23. What changes or does not change in the arrangement?
Answer: When we add 2 to a number in a paired arrangement, each partner's value grows by 2. For example, when 18 becomes 20, the pairs shift from (1+17, 2+16, 3+15...) to (1+19, 2+18, 3+17...). What stays the same is the total number of pairs and the balanced pattern - the smallest always pairs with the largest. The same pattern holds when we add 2 to 23 (which becomes 25). The structure and symmetry of the arrangement stay exactly the same, even though the numbers themselves increase.
In simple words: When you add 2 to the total, each pair's numbers go up by 2. But the number of pairs and how they match up stays exactly the same.

Exam Tip: Focus on what changes (the partner values) versus what does not change (the structure and count of pairs). This distinction shows understanding of how pairing arrangements work.

 

Question 3. (a) 12 and 6 are a pair of even numbers. Choose 5 such pairs of even numbers. Add the numbers in each of the pairs. What do you notice about the sums in each case? Do you think it will be true for all pairs of such numbers?
Answer: When you add two even numbers together, the result is always even. Here are five examples: 2 + 4 = 6, 8 + 10 = 18, 14 + 20 = 34, 26 + 44 = 70, and 100 + 222 = 322. Every sum is an even number. This happens every single time you add two even numbers. To understand why: any even number can be written as 2n (where n is a whole number), and another even number as 2m. When you add them, you get 2n + 2m = 2(n + m), which is divisible by 2. Since the result has a factor of 2, it must be even.
In simple words: Even number plus even number always gives an even number. You can think of it as (2 × something) + (2 × something else) = 2 × (a new something), which is always even.

Exam Tip: When asked to find a pattern, always verify it with multiple examples first, then use algebra (like 2n + 2m) to prove it works for all such numbers.

 

Question 3. (b) 13 and 9 are a pair of odd numbers. Choose 5 such pairs of odd numbers. Add the numbers in each of the pairs. What do you notice about the sums in each case? Do you think it will be true for all pairs of such numbers?
Answer: When you add two odd numbers together, you always get an even number. For example: 1 + 3 = 4, 5 + 7 = 12, 11 + 15 = 26, 21 + 23 = 44, and 101 + 111 = 212 - all even. The reason is that any odd number can be written as (2n + 1) (where n is a whole number), and another odd number as (2m + 1). When you add them together, (2n + 1) + (2m + 1) = 2(n + m + 1), which is clearly divisible by 2. This means the sum must always be even.
In simple words: Odd number plus odd number always gives an even number. It is like having an extra 1 in each number, and those two extra 1s combine to make 2, which is even.

Exam Tip: Remember the algebraic forms: even = 2n, odd = 2n + 1. Using these forms makes it easy to prove any rule about even and odd numbers.

 

Question 3. (c) 7 and 12 are a pair of odd and even numbers. Choose 5 such pairs of odd and even numbers. Add the numbers in each of the pairs. What do you notice about the sums in each case? Do you think it will be true for all pairs of such numbers?
Answer: When you add an odd number and an even number together, the result is always odd. Some examples: 3 + 8 = 11, 5 + 14 = 19, 9 + 20 = 29, 15 + 32 = 47, and 101 + 200 = 301 - all odd. Here is why this always happens: an odd number can be written as (2n + 1) and an even number as 2m. When you add them, (2n + 1) + 2m = 2(n + m) + 1. Because there is still a "+ 1" at the end, the result is not divisible by 2, so it must be odd.
In simple words: Odd number plus even number always gives an odd number. The even number does not get rid of the extra 1 in the odd number, so the sum stays odd.

Exam Tip: Notice the pattern: even + even = even, odd + odd = even, but odd + even = odd. These three rules are the foundation for many problems on number properties.

 

Question 4. Jincy opened her piggy bank. She found 8 coins of Rs.1, 9 coins of Rs.2 and 5 coins of Rs.5. She wants to buy stickers worth Rs.38. What possible combination of coins can she use to pay the exact amount?
Answer: Jincy has a total of Rs.51 in coins (8 × 1 + 9 × 2 + 5 × 5 = Rs.51), which is more than Rs.38, so she can make the purchase. Here are three ways she can pay exactly Rs.38:
Combination 1: 5 coins of Rs.5 (Rs.25) + 6 coins of Rs.2 (Rs.12) + 1 coin of Rs.1 (Rs.1) = Rs.38
Combination 2: 4 coins of Rs.5 (Rs.20) + 9 coins of Rs.2 (Rs.18) = Rs.38
Combination 3: 3 coins of Rs.5 (Rs.15) + 8 coins of Rs.2 (Rs.16) + 7 coins of Rs.1 (Rs.7) = Rs.38
In simple words: Jincy can mix different coins to reach exactly Rs.38. She can try using more of one type and fewer of another type until she hits the target amount.

Exam Tip: For coin-combination problems, always check that you do not exceed the number of coins available and that the total equals the target amount.

 

Question 5. Raghu is fond of his grandfather's torch. He starts playing with it. He presses the switch once and the light turns ON. He presses it a second time and the light turns OFF. He presses the switch a third time and the light turns ON. He keeps doing this several times. Will the torch be ON or OFF after the 23rd press? How do you know? For what number of presses will the torch be ON? For what number of presses of the switch will the torch be OFF?
Answer: The torch follows a clear pattern: odd-numbered presses turn it ON, and even-numbered presses turn it OFF. After each press, the state switches. Since 23 is an odd number, the torch will be ON after the 23rd press. In general, whenever Raghu presses the switch an odd number of times (1, 3, 5, 7, 9, ...), the torch is ON. Whenever he presses it an even number of times (2, 4, 6, 8, 10, ...), the torch is OFF. This rule works for any number of presses because the pattern never changes.
In simple words: Odd presses turn the light ON. Even presses turn it OFF. So to know the state after any press, just check if that number is odd or even.

Exam Tip: Recognizing the even-odd pattern is key. Always link odd/even numbers to the outcome; this approach works for any repeating alternating pattern.

 

Question 6. (a) Which is the highest peak she climbed?
Answer: Looking at the table of heights, Mount Everest stands at 8,848 metres, which is the tallest peak among all mountains Priyanka climbed.
In simple words: Mount Everest at 8,848 metres is the highest peak in the list.

Exam Tip: When comparing numbers from a table, arrange them mentally from smallest to largest to spot the highest value quickly.

 

Question 6. (b) What is the difference in height between the highest and lowest peaks she has climbed, as per the table?
Answer: The highest peak is Mount Everest at 8,848 metres, and the lowest is Mount Elbrus at 5,642 metres. The difference between them is 8,848 - 5,642 = 3,206 metres.
In simple words: Subtract the smallest height from the largest height: 8,848 - 5,642 = 3,206 metres.

Exam Tip: Always subtract the smaller number from the larger one to find the difference. Double-check by adding your answer back to the smaller number to verify it equals the larger number.

 

Question 6. (c) What is the difference between heights of Mount Elbrus and Mount Kanchenjunga?
Answer: Mount Kanchenjunga is 8,586 metres tall, and Mount Elbrus is 5,642 metres tall. The difference between them is 8,586 - 5,642 = 2,944 metres.
In simple words: Take the height of the taller mountain and subtract the height of the shorter one: 8,586 - 5,642 = 2,944 metres.

Exam Tip: When subtracting heights from a table, identify which peak is taller first, then subtract the shorter from the taller to get a positive answer.

 

Question 6. (d) If Priyanka was 20 years old when she summited Mount Everest in 2013, in which year was she born?
Answer: If Priyanka was 20 years old in 2013, then her birth year is 2013 - 20 = 1993. She was born in 1993.
In simple words: Subtract her age from the year she climbed Everest to find the year she was born: 2013 - 20 = 1993.

Exam Tip: For age and year problems, remember: Birth Year = Event Year - Age at that Event.

 

Math Metric Mela

 

Question 7. A grand Math Metric Mela was held at the district level. Certificates were printed for each district before the event. For each district, find out if the number of certificates were sufficient. If insufficient, calculate how many certificates fell short. If extra, calculate how many certificates were in excess.
Answer:
Chittoor, A.P.: 18,225 certificates were printed and 18,104 students attended. Extra certificates = 18,225 - 18,104 = 121. The certificates were sufficient, with 121 extra.
Jaunpur, U.P.: 19,043 certificates were printed and 19,265 students attended. Shortage = 19,265 - 19,043 = 222. The certificates were not sufficient; 222 more were needed.
Raigad, Maharashtra: 20,863 certificates were printed and 19,974 students attended. Extra certificates = 20,863 - 19,974 = 889. The certificates were sufficient, with 889 extra.
In simple words: Compare how many certificates were printed to how many students came. If more certificates were printed, find the extra. If fewer certificates were printed, find the shortage.

Exam Tip: Set up a comparison table for each district: Printed minus Attended gives you the surplus (positive) or shortage (negative). Always state clearly whether certificates were enough or not.

 

Question 8. Add the following.
(a) 2,009 + 7,388 (b) 26,444 + 71,111 (c) 777 + 888 (d) 1,234 + 1,234 (e) 56 + 56,789 (f) 777 + 77,777 (g) 5,922 + 9,221 (h) 4,321 + 8,765 (i) 50,050 + 55,000
Answer:
(a) 2,009 + 7,388 = 9,397
(b) 26,444 + 71,111 = 97,555
(c) 777 + 888 = 1,665
(d) 1,234 + 1,234 = 2,468
(e) 56 + 56,789 = 56,845
(f) 777 + 77,777 = 78,554
(g) 5,922 + 9,221 = 15,143
(h) 4,321 + 8,765 = 13,086
(i) 50,050 + 55,000 = 105,050
In simple words: Line up the numbers by place value (ones, tens, hundreds, and so on), then add each column from right to left, carrying over any number 10 or larger to the next column.

Exam Tip: Always align numbers carefully by place value before adding. Check your work by adding in reverse order - the sum should be the same.

 

Question 9. Subtract the following.
(a) 458 - 226 (b) 7,777 - 4,449 (c) 65,447 - 47,299 (d) 1,234 - 123 (e) 12,345 - 1,234 (f) 56,789 - 56 (g) 87,326 - 11,111 (h) 878 - 52 (i) 749 - 222
Answer:
(a) 458 - 226 = 232
(b) 7,777 - 4,449 = 3,328
(c) 65,447 - 47,299 = 18,148
(d) 1,234 - 123 = 1,111
(e) 12,345 - 1,234 = 11,111
(f) 56,789 - 56 = 56,733
(g) 87,326 - 11,111 = 76,215
(h) 878 - 52 = 826
(i) 749 - 222 = 527
In simple words: Arrange the larger number on top and the smaller below, lining them up by place value. Subtract each column from right to left. If a digit on top is smaller than the one below, borrow 10 from the next column to the left.

Exam Tip: Use the borrowing method carefully, and always double-check by adding your answer to the number you subtracted - this should give back the original larger number.

 

Question 10. Ambrish saved Rs.92,375 over a year to buy cows and goats. He buys a cow for Rs.26,000 and a goat for Rs.17,000. He also buys a milking machine for Rs.19,873. Does he have enough money to buy these? How much more or less does he have than he needs?
Answer: Ambrish's total savings are Rs.92,375. His purchases add up to Rs.26,000 + Rs.17,000 + Rs.19,873 = Rs.62,873. Since he needs Rs.62,873 and has Rs.92,375, he has enough money. After buying everything, he will have Rs.92,375 - Rs.62,873 = Rs.29,502 left over.
In simple words: Add up all the costs, then compare to his savings. If savings are bigger, he has enough, and you can find the leftover by subtraction.

Exam Tip: Always add all costs first, then compare the total to available money. State clearly whether there is enough and by how much.

 

Question 11. A factory produces 54,000 nuts and bolts in a day. An order is placed for 85,300 nuts and bolts. How many more nuts and bolts does the factory need to produce to complete the order?
Answer: The factory produces 54,000 units per day but has received an order for 85,300 units. The factory must produce 85,300 - 54,000 = 31,300 more nuts and bolts to meet the order.
In simple words: Subtract what the factory makes from what is ordered to find how many more are needed.

Exam Tip: When dealing with production and orders, subtract current production from the total order to find the shortfall.

 

Question 12. Virat Kohli has scored 27,599 runs. He has 6,758 runs less than Sachin Tendulkar. How many runs has Sachin Tendulkar scored?
Answer: Virat Kohli has 27,599 runs, and this is 6,758 runs less than Sachin's total. Therefore, Sachin Tendulkar has scored 27,599 + 6,758 = 34,357 runs.
In simple words: If Virat has fewer runs than Sachin by 6,758, then add this difference to Virat's runs to find Sachin's total.

Exam Tip: When comparing two numbers where one is described as "less than" the other, add the difference to the smaller number to find the larger one.