NCERT Solutions for Class 6 Maths Exercise 3.7

NCERT Solutions for Class 6 Maths Exercise 3.7 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 6 Maths chapter wise NCERT solution for Maths Book all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Playing with numbers Download as PDF

NCERT Solutions for Class 6 Maths Playing with numbers

Class –VI

Mathematics

(Ex. 3.7)

Question 1.Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.

Answer: For finding maximum weight, we have to find H.C.F. of 75 and 69.

Factors of 75 = 3 x 5 x 5

Factors of 69 = 3 x 69

H.C.F. = 3

Therefore the required weight is 3 kg.

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Question 2.Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the maximum distance each should cover so that all can cover the distance in complete steps?

Answer: For finding minimum distance, we have to find L.C.M of 63, 70, 77.

L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm.

Therefore, the minimum distance is 6930 cm.

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NCERT Solutions for Class 6 Maths Exercise 3.7

Question 3.The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Answer: The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm.

Factors of 825 = 3 x 5 x 5 x 11

Factors of 675 = 3 x 5 x 5 x 3 x 3

Factors of 450 = 2 x 3 x 3 x 5 x 5

H.C.F. = 3 x 5 x 5 = 75 cm

Therefore, the longest tape is 75 cm.

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Question 4.Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Answer: L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24

The smallest 3-digit number = 100

To find the number, we have to divide 100 by 24

24{tex}\mathop{\left){\vphantom{1\begin{gathered}

{\text{ }}24 \\\\\\

\underline { – 24} \\\\\\

{\text{ }}4 \\\\\\

\end{gathered} }}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered}

{\text{ }}24 \\\\\\

\underline { – 24} \\\\\\

{\text{ }}4 \\\\\\

\end{gathered} }}}

\limits^{\displaystyle\,\,\, 4}{/tex}

Therefore, the required number = 100 + (24 – 4) = 120.

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NCERT Solutions for Class 6 Maths Exercise 3.7

Question 5.Determine the largest 3-digit number which is exactly divisible by 8, 10 and 12.

Answer: L.C.M. of 8, 10, 12 = 2 x 2 x 2 x 3 x 5 = 120

The largest three digit number = 999

Now,

120{tex}\mathop{\left){\vphantom{1\begin{gathered}

{\text{ }}999 \\\\\\

\underline { – 960} \\\\\\

{\text{ }}39 \\\\\\

\end{gathered} }}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered}

{\text{ }}999 \\\\\\

\underline { – 960} \\\\\\

{\text{ }}39 \\\\\\

\end{gathered} }}}

\limits^{\displaystyle\,\,\, 8}{/tex}

Therefore, the required number = 999 – 39 = 960

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Question 6.The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?

Answer: L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec.

After 432 seconds, the lights change simultaneously.

432 second = 7 minutes 12 seconds

Therefore the time = 7 a.m. + 7 minutes 12 seconds

= 7 : 07 : 12 a.m.

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NCERT Solutions for Class 6 Maths Exercise 3.7

Question 7.Three tankers contain 403 liters and 465 liters of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.

Answer: The maximum capacity of container = H.C.F. (403, 434, 465)

Factors of 403 = 13 x 31

Factors of 434 = 2 x 7 x 31

Factors of 465 = 3 x 5 x 31

H.C.F. = 31

Therefore, 31 liters of container is required to measure the quantity.

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Question 8.Find the least number which when divided by 6, 15 and 18, leave remainder 5 in each case.

Answer: L.C.M. of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90

Therefore the required number = 90 + 5 = 95

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NCERT Solutions for Class 6 Maths Exercise 3.7

Question 9.Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Answer: L.C.M. of 18, 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

The smallest four-digit number = 1000

Now,

288{tex}\mathop{\left){\vphantom{1\begin{gathered}

{\text{ 1000}} \\\\\\

\underline {{\text{ }} – 864} \\\\\\

{\text{ 136}} \\\\\\

\end{gathered} }}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered}

{\text{ 1000}} \\\\\\

\underline {{\text{ }} – 864} \\\\\\

{\text{ 136}} \\\\\\

\end{gathered} }}}

\limits^{\displaystyle\,\,\, 3}{/tex}

Therefore, the required number is 1000 + (288 – 136) = 1152.

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Question 10.Find the L.C.M. of the following numbers:

(a) 9 and 4

(b) 12 and 5

(c) 6 and 5

(d) 15 and 4

Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?

Answer: (a) L.C.M. of 9 and 4

= 2 x 2 x 3 x 3 = 36

(b) L.C.M. of 12 and 5

= 2 x 2 x 3 x 5 = 60

(c) L.C.M. of 6 and 5

= 2 x 3 x 5 = 30

(d) L.C.M. of 15 and 4

= 2 x 2 x 3 x 5 = 60

Yes, the L.C.M. is equal to the product of two numbers in each case.

And L.C.M. is also the multiple of 3.

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NCERT Solutions for Class 6 Maths Exercise 3.7

Question 11.Find the L.C.M. of the following numbers in which one number is the factor of other:

(a) 5, 20

(b) 6, 18

(c) 12, 48

(d) 9, 45

What do you observe in the result obtained?

Answer: (a) L.C.M. of 5 and 20

= 2 x 2 x 5 = 20

(b) L.C.M. of 6 and 18

2 x 3 x 3 = 18

(c) L.C.M. of 12 and 48

2 x 2 x 2 x 2 x 3 = 48

(d) L.C.M. of 9 and 45

= 3 x 3 x 5 = 45

From these all cases, we can conclude that if the smallest number if the factor of largest number, then the L.C.M. of these two numbers is equal to that of larger number.

NCERT Solutions for Class 6 Maths Exercise 3.7

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