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Questions:

Generic 5

25 x 5 = 125

Draw concentric circles for the following measurement of radii / diameters. Find out the width of circulating.
r = 7.1 cm and d = 12 cm

Draw concentric circles for the following measurement of radii / diameters. Find out the width of circular ring.
r = 5 cm and r = 7.5 cm.

Draw circles for the following measurements of radius (r)/ diameters(d).
d = 6cm

Draw circles for the following measurements of radius (r)/ diameters(d).
r = 3.5 cm

Draw concentric circles for the following measurements of radii / diameters. Find out the width of each circular ring.
r = 3 cm and r = 5 cm.

Find the mode of the following data. 14, 15, 12, 14, 16, 15, 17, 13, 16, 16, 15, 12, 16, 15, 13, 14, 15, 13, 15, 17, 15, 14, 18, 19, 12, 14, 15, 16, 15, 16, 13, 12.

The marks obtained by 11 students of a class in a test are 23, 2, 15, 38, 21, 19, 23, 23, 26, 34, 23. Find the mode of the marks.

Solve the inequation: −2 ≤ z + 3 ≤ 5, where z is an integer. Also, represent the solution, graphically.

Represent the solutions −8 < 2x < 10 in a number line, where x is a natural number.

10.

Solve: 5 − 7x ≥ 33, where x is an integer.

11.

A principal becomes Rs.10,050 at the rate of 10% in 5 years. Find the principal.

12.

Sathish kumar borrowed Rs. 52,000 from a money lender at a particular rate of simple interest. After 4 years, he paid Rs. 79,040 to settle his debt. At what rate of interest he borrowed the money?

13.

Kumaravel has paid simple interest on a certain sum for 2 years at 10% per annum is Rs. 750. Find the sum.

14.

How many per cent in the Fig. is shaded as blue?

15.

Convert the given percentage to decimals
(i) 58%
(ii) 8%
(iii) 30%
(iv) 120%
(v) 1.25%

16.

Convert the given decimals to percentage.
0.025

17.

Convert the given decimals to percentage.
0.05

18.

The perimeter of a regular polygon is 17.5 cm. Each side measures 2.5 cm. Find the number of sides of a polygon.

19.

Sugarcane juice of 1.51 has to be shared equally among five members. What is the share of each member.

20.

Naren bought 7.4 kg of mangoes. On the way home, he gave 4.650 kg of mangoes to his sister’s family. Find the weight of the remaining mangoes.

21.

Subtract 3.09 from 12.7.

22.

Subtract 2.85 from 4.97.

23.

Add the following : (i) 30.9 + 52.73 (ii) 25.67 + 33.856

24.

Add 3.2 + 6.4.

25.

Find 0.52 – 0.08 using decimal grid models.

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Answers Key:

Generic 5

25 x 5 = 125

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 = 6 cm
Step 4: With O as centre and draw another circle of radius OB = 7.1 cm.
Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB - OA = 7.1 - 6 = 1.1cm

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 = 5 cm.
Step 4: With O as centre and draw another circle of radius OB = 7.5 cm.
Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB - OA = 7.5 - 5 = 2.5cm.

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 3 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.

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 3.5cm.
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.

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 cm
Step 4: With O as centre and draw another circle of radius OB = 5 cm.
Thus the concentric circles C₁ and C₂ are drawn.
Width of the circular ring = OB - OA = 5 - 3 = 2cm

We tabulate the data as follows.
The whole data ranges from 12 to 19.

Observation	Tally marks	Frequency
12	\|\|\|\|	4
13	\|\|\|\|	4
14		5
15		9
16		6
17	II	2
18	I	1
19	I	1
	Total	32

The highest frequency is 9 which corresponds to the value 15. Hence the mode of this data is 15.

Arranging the given marks in ascending order, we get,
2, 15, 19, 21, 23, 23, 23, 23, 26, 34, 38.
Clearly, 23 occurs maximum number of times. Hence mode of marks = 23.

−2 ≤ z + 3 ≤ 5
−2 −3 ≤ z + 3 −3 ≤ 5 −3 [Subtracting −3 from the inequation]
−5 ≤ z ≤ 2
Since the solution belongs to integers, the solutions are –5, –4 , –3, –2, –1, 0, 1 and 2. It’s graph on the number line is shown below:

−8 < 2x < 10
\(\cfrac { -8 }{ 2 } <\cfrac { 2x }{ 2 } <\cfrac { 10 }{ 2 } \) [Dividing the inequation by 2]
−4 < x < 5

10.

5 − 7x ≥ 33
5 − 5 − 7x ≥ 33 − 5 [Subtracting 5 from both sides]
−7x ≥ 28
\(\cfrac { -7x }{ -7 } \ge \cfrac { 28 }{ 7 } \) [Dividing both sides by –7]
\(x\le -4\) [since, it is divided by a negative number, the inequality is reversed]
Since, solution belongs to the set of integers, that are less than −4, we take the values of x as –4, –5, –6, ...
Therefore, the solutions are –4, –5, –6,...

11.

A = Rs. 10,050
n = 5 years
r = 10%
P = ?
For calculating principal with the given data, we proceed as follows.
We know that,
\(I=\cfrac { Pnr }{ 100 } \)
A = P + I
\(A=P\left( 1+\cfrac { nr }{ 100 } \right) \)
Therefore, \(10,050=P\left( 1+\cfrac { 10\times 5 }{ 100 } \right) \)
= \(P(1+\cfrac { 50 }{ 100 } )\)
= \(P(\cfrac { 150 }{ 100 } )\)
= \(P(\cfrac { 3 }{ 2 } )\)
Therefore, \(P=10,050\times \cfrac { 2 }{ 3 } =3350\)
Hence, Principal = Rs.3,350.

12.

Principal (P) = Rs. 52,000
Time (n) = 4 years
Interest = Amount – Principal
= 79,040 − 52,000 = 27,040
We find the Simple Interest \((I)=\cfrac { Pnr }{ 100 } \)
Therefore,\(27040=\cfrac { 52000\times r\times 4 }{ 100 } \)
Rate of interest he borrowed \((r)=\cfrac { 27040\times 100 }{ 52000\times 4 } =13\%\)

13.

Given the rate of interest (r) = 10% per annum
Time (n) = 2 years
We know that Simple Interest (I) = \(\cfrac { Pnr }{ 100 } \)
\(750=\cfrac { P\times 2\times 10 }{ 100 } \)
Therefore, Principal \((P)=\cfrac { 750\times 100 }{ 2\times 10 } =3750\)
Therefore, Kumaravel has borrowed a sum of Rs.3,750.

14.

The fraction in the Fig. 2.3, that is shaded = \(\cfrac { 2 }{ 4 } =\cfrac { 1 }{ 2 } \)
That is half of the Fig. 2.3 is shaded blue.
So, the percentage of shaded portion = \(\cfrac { 1 }{ 2 } \times 100\%=50\%\)
Thus, 50% of the Fig. s shaded

15.

(i) \(58\%=\cfrac { 58 }{ 100 } =0.58\)
(ii) \(8\%=\cfrac { 8 }{ 100 } =0.08\)
(iii) \(30\%=\cfrac { 30 }{ 100 } =0.3\)
(iv) \(120\%=\cfrac { 120 }{ 100 } =1.2\)
(v) \(1.25\%=\cfrac { 1.25 }{ 100 } =0.0125\)
From the above examples we see that to convert percentage to decimals we first convert it into fraction and get the solution.

16.

\(0.025=\cfrac { 25 }{ 1000 } \times 100\%=\cfrac { 25 }{ 10 }\% =\cfrac { 5 }{ 2 }\%\) or 2.5%

17.

\(0.05=\cfrac { 5 }{ 100 } \times 100\%=5\%\)

18.

Since it is given that the polygon is a regular polygon, all sides will be equal.
Therefore, the number of sides of a polygon = \(\cfrac { perimeter }{ oneside } \)
= \(\cfrac { 17.5 }{ 2.5 } =\cfrac { 175 }{ 25 } =7\) (Equal number of decimal digits in both dividend and divisor)
Hence, the polygon has 7 sides.

19.

Quantity of sugarcane juice = 1.51
Number of persons to be shared = 5.
Individual share = \(\cfrac { 1.5 }{ 5 } 0.3l\)
Therefore, the individual share is 0.3 litres.

20.

Mangoes bought by Naren = 7.4 kg
Mangoes given to Naren’s sister = 4.650 kg
Mangoes left for Naren’s family = 7.400 − 4.650
Weight of remaining mangoes = 2.750 kg
Therefore, the weight of the remaining mangoes is 2.750 kg.

21.

12.7 – 3.09 = ?
Let us use the place value grid

Decimal No.	Tens	Ones	Tenth	Hundredth
12.7	1	2	7	0
3.09		3	0	9
9.61		9	6	1

Therefore, 12.7 – 3.09 = 9.61

22.

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

Decimal No	Ones	Tenth	Hundredth
4.97	4	9	7
2.85	2	8	5
2.12	2	1	2

Therefore, 4.97 – 2.85 = 2.12.

23.

(i) 30.9 + 52.73
Let us use the place value grid.
(Since the digits in the decimal place of 52.73 is 2 and 30.9 is 1, we should add 0 at the hundredth place of 30.9 to equalize the digits in the decimal place)

Decimal No.	Tens	Ones	Tenth	Hundredth
30.9	3	0	9	0
52.73	5	2	7	3
83.63	8	3	6	3

Therefore, 30.9 + 52.73 = 83.63.
(ii) 25.67 + 33.856
Let us use
the place value grid.

Decimal No.	Tens	Ones	Tenth	Hundredth	Thousandth
25.67	2	5	6	7	0
33.856	3	3	8	5	6
59.526	5	9	5	2	6

Therefore, 25.67 + 33.856 = 59.526.

24.

Here 3.2 is represented in Blue colour and 6.4 is represented in Green colour. Hence, the sum of 3.2 and 6.4 is 9.6.

25.

Here \(0.52=\cfrac { 52 }{ 100 } \) and \(0.08=\cfrac { 8 }{ 100 } \). First shade the region 0.52 then cross out 0.08, which is \(\cfrac { 8 }{ 100 } \) from the shaded area. The left out shaded region without cross marks is the difference. So, 0.52 − 0.08 = 0.44 .