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NCERT Solutions for Class 5th Maths Chapter 4 Quick Sums and Differences — Quick Sums and Differences

Book page 51–52 Updated on2026-09-19

Q1.
Sukanta likes the numbers 10, 100, 1,000, and 10,000. He wants to figure out what number he should add to a given number such that the sum is 100 or 1,000. Help him fill in the blanks with an appropriate number. 32 + ______ = 100
Answer

32 + 68 = 100

Piku's way, in three steps.

  1. First aim for 99 instead of 100. Take each digit of 32 away from 9: 9 − 3 = 6 and 9 − 2 = 7. So 32 + 67 = 99.
  2. 100 is 1 more than 99, so add 1 to the 67: 67 + 1 = 68.
  3. So 32 + 68 = 100.
3 2
+ 6 7
———
  9 9     →     32 + 68 = 100
Why aiming for 99 first is easier: Every digit of 99 is a 9, and no digit can be bigger than 9. So you never have to borrow — you just take each digit away from 9 in your head.
Check it yourself: 32 + 68 → ones 2 + 8 = 10, write 0 carry 1; tens 3 + 6 + 1 = 10. That is 100 ✓
Q2.
Do you think this method will always work?
Answer

Yes — for any whole number less than 100 when you are making 100, and in the same way for 1,000 and 10,000.

Why it always works: 100 is always 99 + 1. Reaching 99 needs no borrowing, because 9 is the biggest a digit can be, so every digit of your number is 9 or less. Once you are at 99 you are just one short of 100, so adding 1 finishes the job. The same reasoning works for 1,000 (which is 999 + 1) and for 10,000 (which is 9,999 + 1).

The rule to remember

  1. To make 100, subtract each digit from 9 to reach 99, then add 1.
  2. To make 1,000, subtract each digit from 9 to reach 999, then add 1.
  3. To make 10,000, subtract each digit from 9 to reach 9,999, then add 1.
Try it on 46: 9 − 4 = 5, 9 − 6 = 3, so 46 + 53 = 99
Then 53 + 1 = 54, so 46 + 54 = 100 ✓
Q3.
59 + ______ = 100. Try this method for the number 59.
Answer

59 + 41 = 100

  1. Aim for 99 first. 9 − 5 = 4 and 9 − 9 = 0. So 59 + 40 = 99.
  2. Add 1 to reach 100: 40 + 1 = 41.
  3. So the missing number is 41.
5 9
+ 4 0
———
  9 9     →     59 + 41 = 100
Check it yourself: 59 + 41 → ones 9 + 1 = 10, write 0 carry 1; tens 5 + 4 + 1 = 10. That is 100 ✓
Q4.
Now, use this method to solve the following. 877 + ______ = 1,000 and 666 + ______ = 1,000. 4,103 + ______ = 10,000 and 5,555 + ______ = 10,000
Answer
QuestionStep 1 — reach all 9sStep 2 — add 1Answer
877 + ___ = 1,000999 − 877 = 122122 + 1 = 123123
666 + ___ = 1,000999 − 666 = 333333 + 1 = 334334
4,103 + ___ = 10,0009,999 − 4,103 = 5,8965,896 + 1 = 5,8975,897
5,555 + ___ = 10,0009,999 − 5,555 = 4,4444,444 + 1 = 4,4454,445

877 + 123 = 1,000

  1. Take each digit of 877 away from 9: 9 − 8 = 1, 9 − 7 = 2, 9 − 7 = 2. That gives 122, and 877 + 122 = 999.
  2. Add 1 more to reach 1,000: 122 + 1 = 123.

666 + 334 = 1,000

  1. 9 − 6 = 3 three times, so 666 + 333 = 999.
  2. 333 + 1 = 334.

4,103 + 5,897 = 10,000

  1. Use 9,999 now, because 10,000 has one more digit. 9 − 4 = 5, 9 − 1 = 8, 9 − 0 = 9, 9 − 3 = 6. That gives 5,896.
  2. 5,896 + 1 = 5,897.

5,555 + 4,445 = 10,000

  1. 9 − 5 = 4 four times, so 5,555 + 4,444 = 9,999.
  2. 4,444 + 1 = 4,445.
Check it yourself: 5,555 + 4,445 → ones 5 + 5 = 10, and each column after it becomes 5 + 4 + 1 = 10. You end with 10,000 ✓
Q5.
Will this method work if the units digit is 0? What do you think? What other methods can you use to find the missing number to fill in the blanks? Share your thoughts in the class. (a) 180 + ______ = 1,000 (b) 760 + ______ = 1,000 (c) 400 + ______ = 1,000
Answer

Yes, it still works. A units digit of 0 changes nothing, because 9 − 0 = 9 and there is still no borrowing.

QuestionReach 999Add 1Answer
(a) 180 + ___ = 1,000999 − 180 = 819819 + 1 = 820820
(b) 760 + ___ = 1,000999 − 760 = 239239 + 1 = 240240
(c) 400 + ___ = 1,000999 − 400 = 599599 + 1 = 600600

Other methods you can use

  1. Count on in easy jumps. For 180: go up to the next hundred, 180 + 20 = 200. Then 200 + 800 = 1,000. Add the jumps: 20 + 800 = 820.
  2. Use the zeros. 760 ends in 0, so ask instead: 76 tens + how many tens = 100 tens? 100 − 76 = 24 tens, which is 240.
  3. Straight subtraction. 1,000 − 400 = 600. This is easy when the number is a round hundred.

Sample answer for the class discussion: "The 99s trick still works when the last digit is 0, because 9 − 0 is just 9. But when a number ends in one or two zeros, counting on in jumps is even faster — for 400 I can see straight away that 600 more makes 1,000."

Check it yourself: 180 + 820 = 1,000 ✓   760 + 240 = 1,000 ✓   400 + 600 = 1,000 ✓
Q6.
Namita likes the number 9. She wants to subtract 9 or 99 from any number. Find a way to quickly subtract 9 or 99 from any number. (a) 67 – 9 (b) 83 – 9 (c) 144 – 9 (d) 187 – 99 (e) 247 – 99 (f) 763 – 99
Answer

The trick: to take away 9, take away 10 and then put 1 back. To take away 99, take away 100 and then put 1 back.

QuestionWorkingAnswer
(a) 67 − 967 − 10 = 57, then 57 + 158
(b) 83 − 983 − 10 = 73, then 73 + 174
(c) 144 − 9144 − 10 = 134, then 134 + 1135
(d) 187 − 99187 − 100 = 87, then 87 + 188
(e) 247 − 99247 − 100 = 147, then 147 + 1148
(f) 763 − 99763 − 100 = 663, then 663 + 1664
Why you put 1 back: 9 is one less than 10. If you take 10 away you have taken one rupee too much, so you hand one rupee back. In the same way 99 is one less than 100, so after taking 100 away you put 1 back.

Taking away 10 or 100 is easy because only one digit changes — the tens digit for 10, the hundreds digit for 100.

67 − 9
Step 1: 67 − 10 = 57
Step 2: 57 + 1 = 58
Check: 58 + 9 = 67 ✓
Check them all: 58 + 9 = 67 ✓, 74 + 9 = 83 ✓, 135 + 9 = 144 ✓, 88 + 99 = 187 ✓, 148 + 99 = 247 ✓, 664 + 99 = 763 ✓
Q7.
Now, use the above solutions to find answers to the following problems. Do not calculate again. Namita wonders if she can get 9 or 99 as the answer to any subtraction problem. Find a way to get the desired answer. (a) 32 – ______ = 9 (b) 56 – ______ = 9 (c) 877 – ______ = 99 (d) 666 – ______ = 99
Answer

Yes, she can — from any number. The number to take away is simply the starting number minus 9 (or minus 99), and that is exactly the quick trick from the last question.

QuestionWorking with the trickAnswer
(a) 32 − ___ = 932 − 10 = 22, then 22 + 1 = 2323
(b) 56 − ___ = 956 − 10 = 46, then 46 + 1 = 4747
(c) 877 − ___ = 99877 − 100 = 777, then 777 + 1 = 778778
(d) 666 − ___ = 99666 − 100 = 566, then 566 + 1 = 567567

How to see it

  1. The three numbers in a subtraction always make one family: whole − part = other part.
  2. Here 9 is one part and the starting number is the whole. So the missing part is starting number − 9.
  3. And "take away 9" is the quick trick — take away 10 and put 1 back.
(a) 32 − 23 = 9 ✓
(b) 56 − 47 = 9 ✓
(c) 877 − 778 = 99 ✓
(d) 666 − 567 = 99 ✓
Try This: Pick any number, say 415. Take away 9 the quick way: 415 − 10 = 405, then + 1 = 406. So 415 − 406 = 9. It works every time.
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