NCERT Solutions for Class 5 Maths - Be My Multiply

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1. The hungry cat is trying to catch Kunjan the mouse. Kunjan is now on the 14th step and it can jump 2 steps at a time. The cat is on the third step. She can jump 3 steps at a time. If the mouse reaches 28 it can hide in the hole. Find out whether the mouse can get away safely!
(a) The steps on which the mouse jumps.
(b) The steps on which the cat jumps.
(c) The steps on which both the cat and the mouse jump.
(d) Can the mouse get away?

 Ans.
(a) The mouse jumps from the 14th step to the 16, 18, 20, 22, 24, 26, and 28th steps.
(b) The cat jumps from the 3rd step to the 6, 9, 12, 15, 18, 21, 24, and 27th steps.
(c) The steps on which both the cat and the mouse jump are 18 and 24.
(d) The mouse can get away safely as they do not come together at steps 18 and 24.

2. If the cat starts from the 5th step and jumps five steps at a time and the mouse starts from the 8th step and jumps four steps at a time, can the mouse get away?

Ans. 
If the cat starts from the 5^th step and jumps five steps at a time, the cat jumps from the 5^th step to the 10, 15, 20, and 25^th steps,
Then, the mouse starts from the 8th step and jumps four steps at a time, the mouse jumps from the 8th step to 12, 16, 20, and 24^th steps.
Then, the mouse cannot get away. The cat will easily catch the mouse at the 20^th step.

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3. Who is Monto waiting for?
 Monto cat is waiting for somebody. Do you know for whom he is waiting? There is a   trick to find out.

Mark with a red dot all the numbers which can be divided by 2. Mark a yellow dot on the numbers which can be divided by 3 and a blue dot on the numbers which can be divided by 4. Which are the boxes which have dots of all three colours? What are the letters on top of those boxes? Write those letters below in order.

Ans. 
Let us mark the coloured dots are desired on the above number:
Red dot will appear on the numbers: 2, 4, 6,8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32,      34, 36, 38, 40, 42, 44, 46,48, 50, 52, 54, 56, 58 and 60.
Yellow dot will appear on the numbers: 3,6, 9, 12, 15, 18, 21, 24, 27, 30,33, 36, 39, 42, 45, 48, 51, 54, 57 and 60.
Blue dot will appear on the numbers: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56 and 60.
All the three coloured dots will appear on the boxes: 122, 24, 36, 48 and 60.
The respective letters on the top of these boxes are: M, O, U, S and E.
Writing these letters in order: MOUSE.

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4.  Meow Game: To play this game, everyone stands in a circle. One player calls out ‘one’. The next player says ‘two’ and so on. A player who has to call out 3 or a number which can be divided by 3 has to say ‘Meow’ instead of the number. One who forgets to say ‘Meow’ is out of the game. The last player left is the winner. 

(a) Which numbers did you replace with ‘Meow’?        
Ans. The number which will be replaced with ‘Meow’ are:  
3,6, 9, 12, 15, 18, 21, 24, 27, 30…

(b) Now, which numbers did you replace with ‘Meow’?
Ans. The numbers which will be replaced with ‘Meow’ are:
 4, 8, 12, 16, 20, 24, 28, 32…

(c) Write any ten multiple of 5.
Ans. Ten multiple of 5 are: 5, 10, 15, 220, 25, 30, 35, 40, 45 and 50.

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5. Common Multiples: 
(i) Think of a number. If it is a multiple of 3, write it in the red circle. If it is a multiple of 5, write it in the blue circle.

Ans. 

(ii) Some numbers are multiples of both 3 and 5. So we can say that they are both 3 and 5. Think! If you write the multiples common to 3 and 5 in the purple part, then will they still be in both the red and the blue circles?
Ans. 15 and 30 are the numbers which are multiples of both 3 and 5.

6. Repeat the game using the numbers 2 and 7. Write the common multiples of 2 and 7.
Ans. LCM of 2 and 7 is 14. If you observe the common multiples of 2 and 7, there are 14, 28, 42, 56 and so on. Out of these common multiples, 14 is the least number or the smallest number that is common

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7. Repeat the game by putting the multiples of 4, 6 and 5 in the circles.

Ans. 

(i) What common multiples of 5 and 6 did you write in the green part?
Ans. 30 and 60 are the common multiples of 5 and 6.

(ii) What common multiples of 4 and 6 are written in the orange part?
Ans. 12 and 24 are the common multiples of 4 and 6.

(iii) In which coloured part did you write the common multiples of 4, 6 and 5?
Ans. The common multiples of 4, 6 and 5 are written in the grey part.

(iv) What is the smallest common multiple of 4, 6 and 5?
Ans. The smallest common multiple of 4, 6 and 5 is 60.

8. Tamarind seeds: Sunita took some tamarind (imli) seeds. She made groups of five with them, and found that one seed was left over. She tried making groups of six and groups of four. Each time one seed was left over. What is the smallest number of seeds that Sunita had? 
Ans. We know that the smallest number divisible by 5,6 and 4 is their smallest common multiple. Therefore, the required number be 1 more than this smaller common multiple.
The smallest common multiple of 5,6 and 4 =60
Therefore, the required number =60+1=61
Hence, the smallest number of seeds Sunita had 61.

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9. More tamarind seeds:  
Ammini arranges 12 tamarind seeds in the form of different rectangles. Try to make more rectangles like this using 12 tamarind seeds. How many different rectangles can you make?
Ans.

I got four different rectangles, as shown in the figure below,

If there are 15 tamarind seeds, how many rectangles can you make?
Ans.

10. Colouring the Grid

In the grid here, a rectangle made of 20 boxes is drawn. The width of this rectangle is 2 boxes.
(i) What is its length?
Ans. By observing the rectangle coloured in the grid, we can say that length of the rectangle is 10 boxes.(ii) Colour a rectangle made of 20 boxes in some other way.
Ans.

(iii) What is the length and width of the rectangle you coloured?
Ans. The length of the rectangle is 5 boxes, and the width of the rectangle is 4 boxes.

(iv) In how many ways can you colour a rectangle of 20 boxes? Colour them all in the grid, and write the length and width of each rectangle you have coloured.
Ans. 

The length and width of rectangle 1 are 5 and 4, respectively.
The length and width of rectangle 2 are 4 and 5, respectively.
The length and width of rectangle 3 are 20 and 1, respectively.

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11. Bangles: There are 18 bangles on the rod. Meena trying to group them. She can put them in groups of 2, 3,6, 9 and 18 – without any bangle being left.

How many bangles will she have if she makes groups of 1 bangle each? Now, complete the table, for different numbers of bangles. For each number see what different groups can be made.

Ans. Meena will have 18 groups of 1 bangle each. The complete table is on the given here:    

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12.  Complete the multiplication chart given here:

Ans. 

13. Look at the green boxes in the chart. These show how we can get 12 by multiplying different numbers. 12 = 4 × 3, so 12 is a multiple of both 4 and 3. 12 is also a multiple of 6 and 2, as well as 12 and 1. We say 1, 2, 3, 4, 6, 12 are factors of 12.
(a) What are the factors of 10? Can you do this from the chart?
(b) What are the factors of 36?
(c) Find out all the factors of 36 from the multiplication chart.
(e)  What can you do for number bigger than chart?
Ans.
(a) The factors of 10 are 1, 2, 5, 10. Yes, we can find these factors from the above chart.
(b) The factors of 36 are 1, 2,3,4, 6, 9, 12, 18, 36.
(c) From the multiplication chart, we find that
36=3 12, 36=49, 36=66, 36=94
 3,4,6, 9, 12 are factors of 36 as found from the above chart.
(d) The biggest number for which we can find the factors from this chart is 144.
(e) We can find the factors of bigger numbers just by reducing to smaller numbers by division and then using and then using the above chart. 

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14. Common Factors

(i) Write the factors of 25 in the red circle and the factors of 35 in the blue circle.

Ans.

(ii) Which are the factors of 25 and 35 you have written in the common part (purple) of both circles? 
Ans. 5 and 1 are the common factors of 25 and 35 written in the common part (purple).

(iii) Now write the factors of 40 in the red circle and 60 in the blue circle.

Ans.

(iv) What are the factors written in the common (purple) part of the circle? Which is the biggest common factor of 40 and 60?
Ans. The factors written in the common (purple) part of the circle are 1, 2, 4, 5, 10 and 20. 20 is the biggest common factor of 40 and 60.

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15. Factor Tree

(i) Look at the factor tree. Now can you make another tree like this?

Ans.

(ii) In how many ways can you draw a factor tree for 24? Draw three of them below.

Ans.

Tiling Problems

16. There is a garden in Anu’s house. In the middle of the garden there is a path. They decided to tile the path using tiles of length 2 feet, 3 feet and 5 feet. The mason tiled the first row with 2 feet tiles, the second row with 3 feet tiles and the third row with 5 feet tiles. The mason has not cut any of the tiles. Then what is the shortest length of the path?
Ans. From the question, it is given that the mason tiled the first row with 2 feet tiles, the second row with 3 feet tiles and the third row with 5 feet tiles.
So, now we have to find the LCM of 2, 3 and 5.
The LCM of 2, 3, and 5 is 30.
Therefore, the shortest length of the path is 30 m. 

17.  Manoj has made a new house. He wants to lay tiles on the floor. The size of the room is 9 feet x  12 feet. In the market, there are three kinds of square tiles: 1 foot × 1 foot, 2 feet x 2 feet and 3 feet x 3 feet. Which size of tile should he buy for his room, so that he can lay it without cutting?
Ans. From the question, it is given that the size of the room is 9 feet × 12 feet.
Size of the tiles available in the market are: 1 foot × 1 foot, 2 feet × 2 feet and 3 feet × 3 feet.
Here, 2 is not a factor of the width of the room, i.e. 9 feet, so Manoj cannot lay 2 feet × 2 feet tiles.
1 and 3 are the factors of 3 and 9.
Therefore, Monoj can buy 1 foot × 1 foot and 3 feet × 3 feet so that he can lay it without cutting.

18. Rani, Geetha and Naseema live near each other. The distance from their houses to the road is 90 feet. They decided to tile the path to the road. They all bought tiles of different designs and length. Rani bought the shortest tile, Geetha bought the middle sized one and Naseema bought the longest one. If they could tile the path without cutting any of the tiles, what is the size of the tiles each has bought? Suggest 3 different solutions. Explain how you get this answer.
Ans. From the question, the distance from their house to the road is 90 feet.
Factors of 90 are 1, 2, 3, 5, 6, 9 etc.
Then, possible sizes of tiles are 1 foot × 1 foot, 2 feet × 2 feet, 3 feet × 3 feet, 5 feet × 5 feet, 6 feet × 6 feet and 9 feet × 9 feet etc.
Then the size of the tiles each has bought are,
(i) Rani 1 foot × 1 foot
Geetha 2 feet × 2 feet
Naseema 3 feet × 3 feet
(ii) Rani 2 feet × 2 feet
Geetha 3 feet × 3 feet
Naseema 5 feet × 5 feet
(iii) Rani 3 feet × 3 feet
Geetha 5 feet × 5 feet
Naseema 6 feet × 6 feet