\(\mathrm{r}_{1}=\frac{6.2}{2}=3.1 \mathrm{~cm}\)

Step 1: Draw a rough diagram and mark the given measurements.
Step 2: Take a point O and mark it as the centre.
Step 3: With O as centre and draw a circle of radius OA.= 3.1 cm
Step 4: With O as centre and draw another circle of radius OB = 3.1 cm.
Thus the concentric circles C₁, and C₂ are drawn.
Width of the circular ring = OB - OA = 6.2 - 3.1 = 3.1 cm

Step 1:Draw a rough diagram and mark the given measurements
Step 2: Take a point O and mark it as the centre.
Step 3: With O as centre and draw a circle of radius OA = 3.5 cm
Step 4: With O as centre and draw another circle of radius OB = 6.5 cm.
Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB -OA
= 6.5 - 3.5 = 3cm

Step 1 : Draw a rough diagram and mark the given measurements.
Step 2: Mark a point O on the paper.
Step 3: Extend the compass distance equal to the radius 6.5cm.
Step 4: At center O, Hold the compass firmly and place the pointed end of the compass.
Step 5: Slowly rotate the compass around to get the circle.

\(r=\frac{12}{2}=6 \mathrm{~cm}\)


Step 1 : Draw a rough diagram and mark,the given measurements.
Step 2: Mark a point O on the paper.
Step 3: Extend the compass distance equal to the radius 6 cm.
Step 4: At center O, Hold the compass firmly and place the pointed end of the compass.
Step 5: Slowly rotate the compass around to get the circle.

Since the one hour study time is spent by maximum number of students, the mode of the data is 1 hour.

From the above data, we can see that there is no repetition of values in the given data. Each observation occurs only once, so there is no mode.

Arranging the numbers in ascending order without leaving any value, we get,
3, 3, 4, 4, 5, 5, 5, 5, 5, 7, 7, 9, 10, 10, 10, 12, 12, 16, 20
Mode of this data is 5, because it occurs more number of times than the other values.

6y − 5 ≤ 2y + 7
6y − 2y − 5 ≤ 2y − 2y + 7 [Subtracting 2y from both sides]
4y − 5 ≤ 7
4y − 5 + 5 ≤ 7 + 5 [Adding 5 on both sides]
4y ≤ 12
\(\cfrac { 4y }{ 4 } \le \cfrac { 12 }{ 4 } \) [Dividing by 4 on both sides]
y ≤ 3
Since the solution belongs to integers, the solutions are 3, 2, 1, 0, –1, –2,..... It’s graph on the number line is shown below.

3x + 9 ≤ 12
\(\cfrac { 3x }{ 3 } +\cfrac { 9 }{ 3 } \le \cfrac { 12 }{ 3 } \)  [Dividing the inequation by 3 on both sides]
x + 3 ≤ 4
x + 3 – 3 ≤ 4–3 [Subtracting 3 from both sides]
x ≤ 1
Since the solution belongs to integers, the solutions are 1, 0, −1, −2, …. It’s graph on the number line is shown below:

Let the number of workers be x.
Then, the wages that x workers will earn per day = Rs.200x
The wages that x workers will earn per month = Rs.(200x x 30) = Rs.6000x
Given that, this amount cannot exceed Rs.300000.
Otherwise, it can be written as 6000 x ≤ 300000
\(\frac {6000x}{6000 } \le \frac {300000 }{6000 } \) 
\(x\le 50\) 
Thus, up to 50 workers can be employed on a monthly budget of Rs.300000.

2x + 4 < 18
2x + 4 – 4 < 18 − 4 [Subtracting 4 from both sides]
2x < 14 [Divide by 2 on both sides]
x < 7
Since the solution belongs to natural numbers, that are less than 7, we take the values of the x as 1, 2, 3, 4, 5 and 6.
Therefore, the solutions are 1, 2, 3, 4, 5 and 6.

A = P + I
I = A – P
= 52440 – 46000
= Rs. 6,440
r = ? 
1 Year and 9 months = \(1\cfrac { 9 }{ 12 } \) 
= \(1\cfrac { 3 }{ 4 } \)  = \(\cfrac { 7 }{ 4 } \) 
we know that, \(I=\cfrac { Pnr }{ 100 } \) 
Therefore, \(6440=\cfrac { 46000\times r\times \frac { 7 }{ 4 } }{ 100 } \) 
\(r=\cfrac { 6440\times 4\times 100 }{ 46000\times 7 } \) 
= 8%

Principal (P) = Rs. 5,600
Rate (r) = 6% per annum
Amount = Rs. 6,720
Amount = principal + interest
Interest = Amount – Principal
= 6720 − 5600 = 1120
We know that Simple Interest \(\left( I \right) =\cfrac { Pnr }{ 100 } \) 
\(1120=\cfrac { 6720\times 6\times n }{ 100 } \) 
Therefore, \(n=\cfrac { 1120\times 100 }{ 5600\times 6 } =3\cfrac { 1 }{ 3 } years\) 

Here, the Principal (P) = Rs. 25,000
Rate of interest (r) = 8% per annum
Time (n) = 3 years
Simple Interest (I) = \(\cfrac { Pnr }{ 100 } \) 
= \(\cfrac { 2500\times 3\times 8 }{ 100 } =6000\) 
Hence, Simple Interest (I) is Rs. 6,000.

Total length of the fabric = 25 m
Length of the fabric bought = 1.75 m
Percentage of the fabric bought = \(\cfrac { 1.75 }{ 25 } \times \cfrac { 100 }{ 100 } =\cfrac { 175 }{ 25\times 100 } =\cfrac { 7 }{ 100 } =7\%\)

\(2.25=\cfrac { 225 }{ 100 } \times 100\%=225\%\)

\(0.3=\cfrac { 3 }{ 10 } \times 100\%=30%\)

\(0.85=0.85\times 100\%=\cfrac { 85 }{ 100 } \times 100\%=85\%\)

The distance covered by the car in one hour = \(\cfrac { 16.8 }{ 0.21 } \) 
Multiply the divisor by 100 to make it a whole number. Hence multiply the dividend also by 100.
Therefore,\(\cfrac { 16.8\times 100 }{ 0.21\times 100 } =\cfrac { 1680 }{ 21 } =80\) 

The decimal number 6.3 is shown in Figure
Since 6.3 is divided by 3, we seperate the areas into 3 equal groups using three different colours as in Figure


Each group to represents 2.1, which is the quotient.
Therefore, 6.3 ÷3 = 2.1.
(ii) Division by 10, 100 and 1000
Let us learn how to divide decimal numbers by 10, 100 and 1000.
For example, consider 51.7 ÷ 10
Now, \(51.7=\cfrac { 517 }{ 10 } \) 
Therefore, \(\cfrac { 51.7 }{ 10 } =\cfrac { 517 }{ 10 } \times \cfrac { 1 }{ 10 } =\cfrac { 517 }{ 100 } =5.17\) 
Similarly, \(51.7\div 100=\cfrac { 517 }{ 10 } \times \cfrac { 1 }{ 100 } =\cfrac { 517 }{ 10000 } =0.0517\)
Also,\(\\ \\ \\ 51.7\div 1000=\cfrac { 517 }{ 10 } \times \cfrac { 1 }{ 10000 } =\cfrac { 517 }{ 10000 } =0.0517\) 
When decimal numbers are divided in powers of 10, it can be noted that the decimal number so obtained contains the same number of decimal digits as that of zeros in powers of 10.
Let us see whether there is a pattern for dividing numbers by 10, 100 and 1000.
Observe the following and complete it.

Therefore, 86.9 – 32.042 = 54.858.

4.97 – 2.85 = ?
Let us use the place value grid.

Therefore, 4.97 – 2.85 = 2.12.

100 m = 1 km, 1m = \(\cfrac { 1 }{ 1000 } km\) 
Hence, 295 m = \(\\ \\ \\ \cfrac { 295 }{ 1000 } \) m
= 0.295km
Distance travelled by bus  = 1.820 km
Distance covered by walk = 0.295 km
Total distance = 1.820 + 0.295
= 1.820 + 0.295
= 2.115 km
Therefore, the school is situated at the distance of 2.115 km from her house.

First represent the decimal number 7.5 using 7 squares and 5 rectangular strips. Cross out 3 squares from 7 squares and 4 rectangular strips from 5 rectangular strips to get the difference (see Figure). Hence, 7.5 – 3.4 = 4.1.

Take a square of 100 boxes. Shade 72 boxes to represent 0.72.
Then strike out 51 boxes out of 72 shaded boxes to subtract 0.51 from 0.72.
The left over shaded boxes represent the required value.
Therefore, 0.72 – 0.51 = 0.21.