Exercise 3.12 in Class 6 Maths Chapter 3-Playing with Numbers implements all the number concepts learnt in Chapter 3. It involves various word problems and reasoning questions that will lead you to use factors, multiples, prime numbers and rules of divisibility.
This exercise enables you to put those concepts that you have learnt to real-life scenarios, while developing your thinking skills. It serves as a reasonable way to revise the whole chapter and prepare you for your CBSE exams. The more you practice these worded problems, the increase of ability you will have to work at speed and accuracy with different Maths questions. All the questions in this exercise follow current NCERT syllabus so are useful in the way it can be used for school tests and class fomentations.
We have provided NCERT Solutions for Exercise 3.12 in detailed, free PDF format as to assist you in your revision. The answers are written in a clear way and simple steps and explanations can help for easier understanding.
This exercise covers word problems using all number concepts learned in this chapter. The NCERT Solutions from Class 6 Maths Chapter 3 exercise 3.12 answers each answer step by step. You can download a free PDF to practice well and prepare for the exams with full confidence.
1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.
Sol. Here, 62,871 is the only one supercell.
If we swap first and last digit of central number 62,871, we get desired result.
Now, 4 supercells are ⇒(39,344),(50,319),(45,306),(23,609).
2. How many rounds does your year of birth take to reach the Kaprekar constant?
Sol. My year of birth is 1995.
Round 1
Greatest number : 9951
Smallest number : 1599
Subtract: 9951-1599=8352
Round 2
Greatest number : 8532
Smallest number : 2358
Subtract: 8532-2358 = 6174
After 2 rounds, I reach the Kaprekar constant 6174.
3. We are the group of 5 -digit numbers between 35,000 and 75,000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group?
Who among us is the closest to 50,000 ?
Sol. Possible odd digits
Digits: 1, 3, 5, 7, 9
Largest Number: 73,999
Smallest Number: 35,111
Closest To 50,000: 51,111
4. Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then try to get an exact number and see how close your estimate is.
Sol. Holidays in a Year
Weekends: 104 Days
Summer Vacation: 55 Days
Winter Vacation: 10 Days
Festival Holidays: 10 Days
Public Holidays: 5 Days
Total Estimated Holidays =104+55+10+10+5=184 Days.
5. Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Sol. As per my house
Mug: 0.10-0.25 Litres
Bucket: 30 Litres
Overhead Tank: 3000 Litres
6. Write one 5 -digit number and two 3 -digit numbers such that their sum is 18,670 .
Sol. 5-digit number: 17,355
3-digit number - 1: 784
3-digit number - 2: 531
Sum: 17,355+784+531=18,670
7. Choose a number between 210 and 390. Create a number pattern similar to those shown in from NCERT will sum up to this number.
Sol. Simple Addition Pattern
We can create a simple addition pattern using a series of repeated numbers:
30+30+30+30+60+60+60=300
Pattern: 30 Repeated 4 Times, 60 Repeated 3 Times,
30+30+30+30=120
60+60+60=180
120+180=300
8. Recall the sequence of powers of 2 how student will relate. Why is the collatz conjecture correct for all the starting numbers in this sequence?
Sol. The square of power of 2 is:
1,2,4,8,16,32,64
Let's take the number 64 as per collatz conjecture.
64 is even, divide by 2=32
32 is even, divide by 2=16
16 is even, divide by 2=8
8 is even, divide by 2=4
4 is even, divide by 2=2.
2 is even, divide by 2=1
Hence collatz conjecture is correct in all numbers in the power of 2 sequence. As it is power of 2 , and in collatz conjecture even number is divided by 2 in each step.
9. Check if the Collatz conjecture holds for the starting number 100.
Sol. 100 is even, divide by 2=50.
50 is even, divide by 2=25
25 is odd, so multiply by 3 and add 1→76 and so on ⋯
The sequence formed with starting number 100 is as follows: 100,50,25,76,38,19,58,29, 88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1.
Yes, the Collatz conjecture holds for the starting number 100.
10. Starting with 0 , players alternate by adding numbers between 1 and 3 . The first person to reach 22 wins. What is the winning strategy now?
Sol. Do it yourself.
The exercise includes word problems that connect factors, multiples and divisibility rules.
It helps review all concepts from the chapter in an organized and simple manner.
All questions follow NCERT guidelines and fit the CBSE pattern of examination.
Practicing using NCERT solutions increases speed and accuracy in school tests and final exams.
The solutions provide support for students that prepare for Math olympiads by building number skills and reasoning skills.
The step by step answers will help students understand the method of working out each problem correctly and will help them complete all problems in confidence.
(Session 2025 - 26)