CBSE Class 4 Mathematics NCERT Solutions: Unit 11 Tables And Shares

This question asks for different ways to arrange 18 plants in rows. We need to find pairs of numbers that multiply to give 18.

One way is to have 1 row with all 18 plants. This can be written as:

$$18\; =\; 1\; \backslash times\; 18$$

So, there is 1 row with 18 plants.

Another way is to have 2 rows. To find out how many plants are in each row, we divide 18 by 2:

$$18\; \backslash div\; 2\; =\; 9$$

This can be written as:

$$18\; =\; 2\; \backslash times\; 9$$

So, there are 2 rows with 9 plants each.

We can also think of 3 rows. To find the number of plants in each row, we divide 18 by 3:

$$18\; \backslash div\; 3\; =\; 6$$

This can be written as:

$$18\; =\; 3\; \backslash times\; 6$$

So, there are 3 rows with 6 plants each.

Similarly, we can have 6 rows with 3 plants each ($$18\; =\; 6\; \backslash times\; 3$$), 9 rows with 2 plants each ($$18\; =\; 9\; \backslash times\; 2$$), and 18 rows with 1 plant each ($$18\; =\; 18\; \backslash times\; 1$$).

We need to draw a garden with 48 plants, ensuring each row has the same number of plants. This means we need to find pairs of numbers that multiply to 48.

Let's consider some possibilities:

• If we have 4 rows, we need to find how many plants are in each row: $$48\; \backslash div\; 4\; =\; 12$$. So, we can have 4 rows with 12 plants each ($$4\; \backslash times\; 12\; =\; 48$$).
• If we have 6 rows, we need to find how many plants are in each row: $$48\; \backslash div\; 6\; =\; 8$$. So, we can have 6 rows with 8 plants each ($$6\; \backslash times\; 8\; =\; 48$$).
• If we have 8 rows, we need to find how many plants are in each row: $$48\; \backslash div\; 8\; =\; 6$$. So, we can have 8 rows with 6 plants each ($$8\; \backslash times\; 6\; =\; 48$$).
• If we have 3 rows, we need to find how many plants are in each row: $$48\; \backslash div\; 3\; =\; 16$$. So, we can have 3 rows with 16 plants each ($$3\; \backslash times\; 16\; =\; 48$$).
• If we have 12 rows, we need to find how many plants are in each row: $$48\; \backslash div\; 12\; =\; 4$$. So, we can have 12 rows with 4 plants each ($$12\; \backslash times\; 4\; =\; 48$$).

The answer provided in the source suggests an arrangement of 6 rows with 8 flowers in each row, which is a valid solution ($$6\; \backslash times\; 8\; =\; 48$$). Another example given is 12 plants x 4 rows = 48 plants ($$12\; \backslash times\; 4\; =\; 48$$), and 16 plants x 3 rows = 48 plants ($$16\; \backslash times\; 3\; =\; 48$$).

A drawing would show these arrangements visually, with dots or symbols representing plants arranged in the specified number of rows and plants per row.

We need to find different ways to arrange 30 jars on a shelf such that each row has the same number of jars. This means finding pairs of numbers that multiply to 30.

The source mentions one way: three rows with 10 jars in each row. This can be written as:

$$3\; \backslash times\; 10\; =\; 30$$

Other possible arrangements include:

• 5 rows with 6 jars in each row: $$5\; \backslash times\; 6\; =\; 30$$
• 6 rows with 5 jars in each row: $$6\; \backslash times\; 5\; =\; 30$$
• 10 rows with 3 jars in each row: $$10\; \backslash times\; 3\; =\; 30$$
• 15 rows with 2 jars in each row: $$15\; \backslash times\; 2\; =\; 30$$
• 30 rows with 1 jar in each row: $$30\; \backslash times\; 1\; =\; 30$$

Each of these represents a different way to arrange 30 jars on a shelf with an equal number of jars in each row.

This question asks us to draw a shelf arrangement for a certain number of jars. The source provides examples for arranging 60 jars. We can choose one of these arrangements or create our own for 60 jars.

Let's consider the arrangement of 5 rows with 12 jars in each row. This means:

$$5\; \backslash times\; 12\; =\; 60$$

To draw this, we would sketch a shelf and divide it into 5 horizontal sections (rows). In each section, we would draw or represent 12 jars. The drawing would visually show 5 rows, and each row would contain 12 jars.

Other arrangements mentioned in the source for 60 jars include:

• 10 rows with 6 jars in each row ($$10\; \backslash times\; 6\; =\; 60$$)
• 6 rows with 10 jars in each row ($$6\; \backslash times\; 10\; =\; 60$$)
• 12 rows with 5 jars in each row ($$12\; \backslash times\; 5\; =\; 60$$)
• 20 rows with 3 jars in each row ($$20\; \backslash times\; 3\; =\; 60$$)

We can choose any of these valid arrangements to draw. For example, drawing 10 rows, with each row containing 6 jars, would fulfill the requirement.

To help Bunty make the table of 7 using the tables of 4 and 3, we can add the corresponding results from the table of 4 and the table of 3. This works because $$7\; =\; 4\; +\; 3$$.

First, let's write down the multiplication tables of 4 and 3:

Multiplication table of 4:

$4\; \backslash times\; 1\; =\; 4$ $4\; \backslash times\; 2\; =\; 8$ $4\; \backslash times\; 3\; =\; 12$ $4\; \backslash times\; 4\; =\; 16$ $4\; \backslash times\; 5\; =\; 20$ $4\; \backslash times\; 6\; =\; 24$ $4\; \backslash times\; 7\; =\; 28$ $4\; \backslash times\; 8\; =\; 32$ $4\; \backslash times\; 9\; =\; 36$ $4\; \backslash times\; 10\; =\; 40$

Multiplication table of 3:

$3\; \backslash times\; 1\; =\; 3$ $3\; \backslash times\; 2\; =\; 6$ $3\; \backslash times\; 3\; =\; 9$ $3\; \backslash times\; 4\; =\; 12$ $3\; \backslash times\; 5\; =\; 15$ $3\; \backslash times\; 6\; =\; 18$ $3\; \backslash times\; 7\; =\; 21$ $3\; \backslash times\; 8\; =\; 24$ $3\; \backslash times\; 9\; =\; 27$ $3\; \backslash times\; 10\; =\; 30$

Now, we can create the table of 7 by adding the results from the table of 4 and the table of 3 for each corresponding multiplication:

• $7\; \backslash times\; 1\; =\; (4\; \backslash times\; 1)\; +\; (3\; \backslash times\; 1)\; =\; 4\; +\; 3\; =\; 7$
• $7\; \backslash times\; 2\; =\; (4\; \backslash times\; 2)\; +\; (3\; \backslash times\; 2)\; =\; 8\; +\; 6\; =\; 14$
• $7\; \backslash times\; 3\; =\; (4\; \backslash times\; 3)\; +\; (3\; \backslash times\; 3)\; =\; 12\; +\; 9\; =\; 21$
• $7\; \backslash times\; 4\; =\; (4\; \backslash times\; 4)\; +\; (3\; \backslash times\; 4)\; =\; 16\; +\; 12\; =\; 28$
• $7\; \backslash times\; 5\; =\; (4\; \backslash times\; 5)\; +\; (3\; \backslash times\; 5)\; =\; 20\; +\; 15\; =\; 35$
• $7\; \backslash times\; 6\; =\; (4\; \backslash times\; 6)\; +\; (3\; \backslash times\; 6)\; =\; 24\; +\; 18\; =\; 42$
• $7\; \backslash times\; 7\; =\; (4\; \backslash times\; 7)\; +\; (3\; \backslash times\; 7)\; =\; 28\; +\; 21\; =\; 49$
• $7\; \backslash times\; 8\; =\; (4\; \backslash times\; 8)\; +\; (3\; \backslash times\; 8)\; =\; 32\; +\; 24\; =\; 56$
• $7\; \backslash times\; 9\; =\; (4\; \backslash times\; 9)\; +\; (3\; \backslash times\; 9)\; =\; 36\; +\; 27\; =\; 63$
• $7\; \backslash times\; 10\; =\; (4\; \backslash times\; 10)\; +\; (3\; \backslash times\; 10)\; =\; 40\; +\; 30\; =\; 70$

So, the table of 7 is: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70.