Important Questions for CBSE Class 6 Maths Chapter 3 - Playing with Numbers

Very Short Answer Questions                                                  1 Mark

1. \[2\] is a composite number. True or False

Ans: A number having more than two factors is a Composite number. 

The number \[2\] has factors \[2\], $1$.

\[2\] is not a composite number.

Therefore, the statement is False.

2. $1$ is neither prime nor composite. Say True or False.

Ans: Given: A prime number is a number having exactly two positive factors $1$ and the number itself. $1$ is neither prime nor composite. 

So, the given statement is true.

3. Prime numbers have more than two factors. Say True or False.

Ans: Given: We know that A prime number is a number having exactly two positive factors $1$ and the number itself. 

Therefore, the given statement is False.

4. If a number is a factor of each of the two given numbers, then it must be a factor of their sum. Say True or False.

Ans: Given: If a number is a factor of each of the two given numbers, then it must be a factor of their sum.

For example, $3$ is a factor of $6{\text{ and }}9$.

Their sum is,

 $6 + 9 $

 $ = 15 $

It is also divisible by $3$. 

Therefore, we can see that the statement given is True.

5. The greatest number which is not a common factor of two or more given numbers is called HCF. Say True or False.

Ans: Given: We know that the greatest common factor (GCF) is the greatest common divisor that is common to two or more numbers. However, it is the product of all prime factors common to the numbers.

Therefore, ${\text{GCF = HCF}}$.

Therefore, the given statement is true.

6. Lowest of the common multiple of two or more numbers is called the ______ of the given numbers.

Ans: Given: We know that Lowest of the common multiple of two or more numbers is called the LCM of the given numbers.

Short Answer Questions                                                       2 Mark

1. Write all the factors of \[78.\]

Ans: Given: \[78\]

We need to find all the factors of the given number.

Therefore, factors of \[78\] will be

$1 \times 78 $

$  2 \times 39 $

$  3 \times 26 $

$  6 \times 13 $

So, factors are $1,2,3,6,13,26,39,78$

2. Write the first five multiples of \[7.\]

Ans: Given: $7$

We need to find the first five multiples of the given numbers.

A Multiple of a number is any number that is the product of that number and any other integer.

Therefore, the multiples of $7$ are

$ 7 \times 1 = 7 $

$  7 \times 2 = 14 $

$ 7 \times 3 = 21 $

 $ 7 \times 4 = 28 $

 $ 7 \times 5 = 35 $

3. Determine whether the number \[5555\] is divisible by \[5\] and \[10.\]

Ans: Given: We know that a number is divisible by 5 if the unit digit of the number is divisible by \[5\] or it is zero.

Therefore, \[5555\] has unit place \[5\] and it is divisible by \[5\].

A number is divisible by \[10\]if the unit place of the number is zero.

Therefore, the number \[5555\] has a unit place \[5\]. Hence, it is not divisible by \[10.\]

4. Determine whether the number \[1890164\] is divisible by \[2,{\text{ }}4,{\text{ }}6\] and \[8.\]

Ans: Given: \[1890164\]

We need to find if the given number is divisible by \[2,{\text{ }}4,{\text{ }}6\] and \[8.\]

A number is divisible by $2,$ if the unit place of the number is divisible by $2.$

The number \[1890164\] has unit place $4$, so the number is divisible by $2.$

A number is divisible by $4$, if the number formed by tens and one's place is divisible by $4$.

The number \[1890164\] has $64$, so it is divisible by $4$.

A number is divisible by $6$, if it is divisible by $2{\text{ and }}3.$ We can see that the number is divisible by $2.$ It is not divisible by $3$as the sum of digits $1 + 8 + 9 + 0 + 1 + 6 + 4 = 29$ is not divisible by $3.$

Therefore, the number \[1890164\] is not divisible by $6$.

A number is divisible by $8,$ if the number formed by hundreds, tens, and ones digit is divisible by $8.$

Therefore, the number \[1890164\] is not divisible by $8.$

5. Determine whether the number 70169308 is divisible by 11?

Ans: Given: \[70169308\]

We need to find if the given number is divisible by \[11.\]

A number is divisible by $11$if the difference of sum of digits in odd places and sum of digits in even places is divisible by \[11.\]

The digits at odd places are

 $  = 8,3,6,0 $

$  {\text{Sum}} = 8 + 3 + 6 + 0 $

$   = 17 $

The digits at even places are 

$ = 0,9,1,7 $

$  {\text{Sum}} = 0 + 9 + 1 + 7 $

$   = 17 $

The difference will be

$ 17 - 17 $

$   = 0 $

Therefore, the number \[70169308\] is divisible by \[11.\]

6. The HCF of two numbers is 18 and their product is 3072. Find the LCM.

Ans: Given: HCF of two numbers \[ = 18\]

Product of the numbers \[ = 3072\]

We need to find the LCM of the numbers.

We know that, 

${\text{LCM}} = \dfrac{{{\text{product of given numbers}}}}{{{\text{HCF}}}}$

Therefore, the LCM will be

$ = \dfrac{{3072}}{{18}} $

$   = 170.6 $

$   \approx 171 $

Short Answer Questions                                                       3 Mark

1. Give the prime factorization of \[30580.\]

Ans: Given: \[30580\]

We need to find factors using Prime Factorization.

$\underline{2 \left| \,\,{30580\,\,}\right.} $

$\underline{2 \left|\,\, {15290\quad}\right.} $

$\underline{5 \left| \,\,{7645\quad}\right.} $

$\underline{11 \left| \,\,{1529\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{139\quad}\right.} $

Therefore, the prime factors will be

$2,2,5,11,139$

2. Reduce \[\mathbf{\dfrac{{279}}{{381}}}\] to lowest forms.

Ans: Given: \[\dfrac{{279}}{{381}}\]

We need to reduce the given fraction to its lowest form.

The lowest fraction will be when the numbers do not have common factors other than $1.$

Therefore, the reduced form of the given number will be

$  \dfrac{{279}}{{381}} \div \dfrac{3}{3} $

$   = \dfrac{{93}}{{127}}  $

3. Find the LCM of \[\mathbf{48,{\text{ }}96}\] and \[\mathbf{108.}\] by prime factorization method.

Ans: Given: $48,96,108$

We need to find the LCM of the given numbers.

LCM stands for least common multiple.

So, LCM of the given numbers will be

$\underline{2 \left| \,\,{48,96,108\,\,}\right.} $

$\underline{2 \left|\,\, {24,48,54\quad}\right.} $

$\underline{2 \left| \,\,{12,24,27\quad}\right.} $

$\underline{2 \left| \,\,{6,12,27\quad}\right.} $

$\underline{3 \left| \,\,{3,6,27\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{1,2,9\quad}\right.} $

$\therefore {\text{LCM}} = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 $

$   = 864  $

4. Find LCM of \[\mathbf{8,{\text{ }}18,{\text{ }}27,{\text{ }}36}\] by division method.

Ans: Given: \[8,{\text{ }}18,{\text{ }}27,{\text{ }}36\]

We need to find the LCM of the given numbers using the division method.

LCM stands for least common multiple.

So, LCM of the given numbers will be

$\underline{2 \left| \,\,{8,18,27,36\,\,}\right.} $

$\underline{2 \left|\,\, {4,9,27,18\quad}\right.} $

$\underline{3 \left| \,\,{2,9,27,9\quad}\right.} $

$\underline{3 \left| \,\,{2,3,9,3\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{2,1,3,1\quad}\right.} $

 $ \therefore LCM = 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$   = 216 $

Long Answer Questions                                                    4 Mark

1. Determine whether 64275 and 65403are divisible by 3 and 9 respectively.

Ans: Given: \[64275\], \[65403\]

We need to find if \[64275\] is divisible by \[3\] and \[65403\] is divisible by \[9\].

A number is divisible by \[3\] if the sum of its digits is divisible by\[3\].

Sum of digits of \[64275\] is 

$  6 + 4 + 2 + 7 + 5 $

$   = 24 $

This is divisible by \[3\]. Therefore, the number is divisible by \[3\].

A number is divisible by \[9\] if the sum of its digits is divisible by \[9\].

Sum of digits of \[65403\] is

$  6 + 5 + 4 + 0 + 3 $

$   = 18 $ 

This is divisible by \[9\]. Therefore, the number is divisible by \[9\].

2. Find the HCF and LCM of 496 and 2080 by prime factorization or division method.

Ans: Given: \[496\], \[2080\]

We need to find the LCM and HCF of the given numbers by prime factorization or division method.

We know that 

${\text{LCM}} = \dfrac{{{\text{product of the numbers}}}}{{{\text{HCF}}}}$

The HCF of the given numbers will be

$\quad \,\,\,\,\,\, 4$

$496{\overline{)\,\,{2080}\quad}}$

$\quad\underline{\,1984\,\,} $

$\,\,\,\,\, 96{\overline{)\,\,{496}{(5}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,480\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 16{\overline{)\,\,{96}{(6}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,96\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,0\,\,} $

Therefore, ${\text{HCF}} = 16$

The LCM of the given numbers using the formula given above will be

   $= \dfrac{{496 \times 2080}}{{16}} $

$   = 31 \times 2080 $

$   = 64,480 $

3. Find HCF by division method for 613and 893.

Ans: Given: The numbers \[613\], \[893\]

We need to find the HCF of the given numbers using the division method.

Therefore, HCF by division method will be

$\quad \,\,\,\,\,\, 1$

$613{\overline{)\,\,{893}\quad}}$

$\quad\underline{\,613\,\,} $

$\,\,\,\,\, 280{\overline{)\,\,{613}{(2}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,560\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 53{\overline{)\,\,{280}{(5}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,265\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 15{\overline{)\,\,{53}{(3}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,45\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 8{\overline{)\,\,{15}{(1}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,8\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 7{\overline{)\,\,{8}{(1}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,7\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 1{\overline{)\,\,{7}{(7}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,7\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,0\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,0\,\,} $

Therefore, the HCF of the given numbers will be $1.$

4. Find HCF of \[\mathbf{138,{\text{ }}180,{\text{ }}260.}\]

Ans: Given: \[138,{\text{ }}180,{\text{ }}260\]

We need to find the HCF of the given numbers.

First, find the factors and then HCF.

Factors of $138$ will be

$\underline{2 \left| \,\,{138\,\,}\right.} $

$\underline{3 \left|\,\, {69\quad}\right.} $

$\underline{23 \left| \,\,{23\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{1\quad}\right.} $

$ 138 = 2 \times 3 \times 23 $

Factors of $180$ will be

$\underline{2 \left| \,\,{180\,\,}\right.} $

$\underline{2 \left|\,\, {90\quad}\right.} $

$\underline{3 \left| \,\,{45\quad}\right.} $

$\underline{3 \left| \,\,{15\quad}\right.} $

$\underline{5 \left| \,\,{5\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{1\quad}\right.} $

$180$ = $2 \times 2 \times 5 \times 3 \times 3$

= ${2^2} \times {3^2} \times 5$

The factors of $260$ will be

$\underline{2 \left| \,\,{260\,\,}\right.} $

$\underline{2 \left|\,\, {130\quad}\right.} $

$\underline{5 \left| \,\,{65\quad}\right.} $

$\underline{13 \left| \,\,{13\quad}\right.} $

$\,\,\,\, \underline{\left| \,\,{1\quad}\right.} $

$260$ = $2 \times 2 \times 5 \times 13$

= ${2^2} \times 5 \times 13$
Therefore, HCF of the given numbers will be the factors common in each number.

HCF $ = 2$

5. The circumferences of three tyres are \[\mathbf{40,{\text{ }}50}\] and \[\mathbf{70{\text{ cm}}{\text{.}}}\] If they moving simultaneously, then what is the least distance they should cover before they make one revolution?

Ans: Given: circumferences of three tyres are \[40,{\text{ }}50\] and \[70{\text{ cm}}\]

We need to find the least distance covered by the tyres before they make one revolution.

We will find the LCM of the circumference given and that will be the distance covered by the tyres.

Therefore, the least distance covered by the tyres before they make one revolution will be

$\underline{2 \left| \,\,{260\,\,}\right.} $

$\underline{2 \left|\,\, {40,50,70\quad}\right.} $

$\underline{2 \left| \,\,{20,25,35\quad}\right.} $

$\underline{5 \left| \,\,{10,25,35\quad}\right.} $

$\underline{5 \left| \,\,{2,5,7\quad}\right.} $

$\text{LCM}$ = $2 \times 2 \times 2 \times 5 \times 5 \times 7$

   = $1400 {\text{cm}}$ 

Very Long Answer Questions                                                5 Mark

1. The length, breadth and height of the cuboid is \[45,{\text{ }}85\] and \[115{\text{ cm}}\] respectively. Find the length of the longest tape which can measure the three dimensions exactly.

Ans: Given: length of cuboid $ = 45$

Breadth of cuboid $ = 85$

Height of cuboid $ = 115$

We need to find the length of the longest tape which can measure the three dimensions exactly.

For this, find the HCF and the HCF found will be the length of the Tape.

HCF of $45{\text{ and }}85$

$\quad \,\,\,\,\,\, 1$

$45{\overline{)\,\,{85}\quad}}$

$\quad\underline{\,45\,\,} $

$\,\,\,\,\, 40{\overline{)\,\,{45}{(1}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,40\,\,} $

$\,\,\,\,\,\,\,\,\,\,\,\, 5{\overline{)\,\,{40}{(8}\quad}}$

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,40\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,\,\,\,\,0\,\,} $

 $ {\text{HCF}} = 5 $

Now, we will find HCF of $5,115$

$\quad \,\,\,\,\,\, 23$

$5{\overline{)\,\,{115}\quad}}$

$\quad\underline{\,10\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,15\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,15\,\,} $

$\,\,\,\,\,\,\, \underline{\,\,\,\,\,\,0\,\,} $

Therefore, the length of the tape which can measure the three dimensions exactly $ = d$

2. What is the least number to be subtracted from 1045 to get a number exactly divisible by 22?

Ans: Given: Number \[1045\]

To find: the least number to be subtracted from \[1045\] to get a number exactly divisible by \[22.\]

Divide the given number with \[22.\] The remainder will be the required number.

Therefore, 

$\quad \,\,\,\,\,\, 47$

$22{\overline{)\,\,{1045}\quad}}$

$\quad\underline{\,88\,\,} $

$\,\,\,\,\, \underline{\,\,\,\,\,\,165\,\,} $

$\,\,\,\,\,\,\, \underline{\,\,\,\,\,\,154\,\,} $

$\,\,\,\,\,\,\, \underline{\,\,\,\,\,\,11\,\,} $

So, the least number to be subtracted from \[1045\] to get a number exactly divisible by \[22\] is $11.$

Important Questions for Class 6 Maths Chapter 3 

Question 1. What will be the sum of any two (a) Odd numbers? (b) Even numbers?

Solution:

1. Let’s take any two numbers. The sum of any two given odd numbers is even numbers.

Examples: 5 + 3 = 8

15 + 13 = 28

2. The sum of any two given even numbers is an even number

Examples: 2 + 8 = 10

12 + 28 = 40

Question 2. The numbers 13 and 31 are prime numbers. Both these numbers have the same digits that are 1 and 3. Find such pairs of prime numbers upto 100.

Solution: 

The prime numbers with the same digits upto 100 are as follows:

17 and 71

37 and 73

79 and 97

These are some important questions for class 6 maths chapter 3.

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Important Related Links for CBSE Class 6 Maths 

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