RD Sharma Solutions Class 7 Chapter 3 Decimals

Read RD Sharma Solutions Class 7 Chapter 3 Decimals below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations

Exercise 3.1
 
Question 1: Write each of the following as decimals:
(i) (8/100)
(ii) 20 + (9/10) + (4/100)
(iii) 23 + (2/10) + (6/1000)
 
Solution 1:
 
(i) (8/100)
Put the decimal point two places from right to left
(8/100) = 0.08
 
(ii) 20 + (9/10) + (4/100)
Firstly, change the fractions to decimals
=(9/10)
Put the decimal point one place from right to left
(9/10) = 0.9
=(4/100)
Put the decimal point two places from right to left
(4/100) = 0.04
= 20 + 0.9 + 0.04
= 20.94
 
(iii) 23 + (2/10)+ (6/1000)
Firstly, change the fractions to decimals
=(2/10)
Put the decimal point one place from right to left
(2/10) = 0.2
= (6/1000)
Put the decimal point three places from right to left
(6/1000) = 0.006
= 23 + 0.2 + 0.006
= 23.206
 
Question 2: Convert each of the following decimals as fractions:
(i) 0.04
(ii) 2.34
(iii) 0.342
(iv) 17.38
 
Solution 2:
 
(i) 0.04
Change the given decimals into fractions
We written 0.04 as (0.04/1)
Now multiply the fraction by 100:
(0.04/1) = (((0.04 × 100))/((1 × 100))) 
= (4/100)
= (1/25)
 
(ii) 2.34
Change the given decimals into fractions
We written 2.34 as (2.34/1)
Now multiply the fraction by 100:
(2.34/1) = (((2.34 × 100))/((1 × 100)))
= (234/100)
= (117/50)
 
(iii) 0.342
Change the given decimals into fractions
We written 0.342 as (0.342/1)
Now multiply the fraction by 100:
(0.342/1) = (((0.342 × 1000))/((1 × 1000)))
= (342/1000)
= (171/500)
 
(iv) 17.38
Change the given decimals into fractions
We written 17.38 as (17.38/1)
Now multiply the fraction by 100:
(17.38/1) = (((17.38 × 100))/((1 × 100)))
= (1738/100)
= (869/50)
 
Question :3. Express the following fractions as decimals:
(i) (23/10)
(ii) 251/8
(iii) 397/35
(iv) 151/25
 
Solution 3:
 
(i) (23/10)
Divide 23 by 10 
(23/10) = 2.3
 
(ii) 251/8
251/8 we written as 25 + (1/8
=(1/8) Now multiply the fraction by 125 to get 1000 as denominator
251/8  = 25 + (1/8) = 25 +(((1 × 125))/((8 × 125)))
= 25 + (125/1000)
= 25 + 0.125
= 25.125
 
(iii) 39 7/35
First convert given mixed fraction 397/35 into improper fraction
397/35 = (35 ×39+7)=(1372/35)
By dividing we get
397/35 = 39.2
 
(iv) 15 1/25
151/25 we written as
151/25 = 15 + (1/25)
= (1/25),
Now multiply the fraction by 4 to get 100 as denominator
151/25 = 15 + (1/25) = 15 +(((1 × 4 ))/((25 × 4)))
= 15 + (4/100)
= 15 + 0.04
= 15.04
 
 

Question 4: Add the following:

(i) 41.8, 39.24, 5.01 and 62.6

(ii) 18.03, 146.3, .829 and 5.324 

Solution 4:

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

 

Question :5. Find the value of:

(i) 9.756 – 6.28

(ii) 48.1 – 0.37

(iii) 108.032 – 86.8

(iv) 100 – 26.32 

Solution 5:

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question :6. Take out of 3.547 from 7.2 

Solution 6:

 RD Sharma Solutions Class 7 Chapter 3 Decimals

 

 
Question :7. What is to be added to 36.85 to get 59.41?
 
Solution 7:
 
Let x  be the unknown number:
x + 36.85 = 59.41
x = 59.41 – 36.85
x = 22.56
Therefore, 22.56 added to 36.85 to get 59.41
 
 
Question :8. What is to be subtracted from 17.1 to get 2.051?
 
Solution 8:
 
Let x be  the unknown number:
Given that x subtracted from 17.1 to get 2.051
17.1 – x = 2.051
17.1 – 2.051 = x
x = 17.1 – 2.051
x = 15.049
 
 

Question 9: By how much should 34.79 be increased to get 70.15? 

Solution 9:

Let  be the unknown number

 + 34.79 = 70.15

 = 70.15 – 34.79

 = 35.36

35.36 should be increased to 34.79 to get 70.15

 

Question 10: By how much should 59.71 be decreased to get 34.58? 

Solution 10:

Let  be the unknown number

59.71 –  =34.58

59.71 – 34.58 =

 = 59.71 – 34.58

 = 25.13

25.13 should be decreased by 59.71 to get 34.58

 

Exercise 3.2

 

Question :1. Find the product:

(i) 4.74 × 10

(ii) 0.45 × 10

(iii) 0.0215 × 10

(iv) 0.0054 × 10 

Solution 1:

(i) 4.74 × 10

Now we’ve to do normal multiplication by shifting the decimal point by one place to the right

Therefore, 4.74 × 10 = 47.4

 

(ii) 0.45 × 10

Now we’ve to do normal multiplication by shifting the decimal point by one place to the right

Therefore, 0.45 × 10 = 4.5

 

(iii) 0.0215 × 10

Now we’ve to do normal multiplication by shifting the decimal point by one place to the right

Therefore, 0.0215 × 10 = 0.215

 

(iv) 0.0054 × 10

Now we’ve to do normal multiplication by shifting the decimal point by one place to the right

Therefore, 0.0054 × 10 = 0.054

 

Question :2. Find the product:

(i) 35.853 × 100

(ii) 42.5 × 100

(iii) 12.075 × 100

(iv) 100 × 0.005 

Solution 2:

(i) 35.853 × 100

Now we’ve to do normal multiplication by shifting the decimal point by two places to the right

Therefore, 35.853 × 100 = 3585.3

 

(ii) 42.5 × 100

Now we’ve to do normal multiplication by shifting the decimal point by two places to the right

Therefore, 42.5 × 100 = 4250

 

(iii) 12.075 × 100

Now we’ve to do normal multiplication by shifting the decimal point by two places to the right

Therefore, 12.075 × 100 = 1207.50

 

(iv) 100 × 0.005

Now we’ve to do normal multiplication by shifting the decimal point by two places to the right

Therefore, 100 × 0.005 = 0.5

 

Question :3. Find the product:

(i) 2.506 × 1000

(ii) 20.708 × 1000

(iii) 0.0529 × 1000

(iv) 1000 × 0.1 

Solution 3:

(i) 2.506 × 1000

Now we’ve to do normal multiplication by shifting the decimal point by three places to the right

Therefore 2.506 × 1000 = 2506

 

(ii) 20.708 × 1000

Now we’ve to do normal multiplication by shifting the decimal point by three places to the right

Therefore 20.708 × 1000 = 20708

 

(iii) 0.0529 × 1000

Now we’ve to do normal multiplication by shifting the decimal point by three places to the right

Therefore 0.0529 × 1000 = 52.9

 

(iv) 1000 × 0.1

Now we’ve to do normal multiplication by shifting the decimal point by three places to the right

Therefore 1000 × 0.1 = 100

 

Question :4. Find the product:

(i) 3.14 × 17

(ii) 0.745 × 12

(iii) 28.73 × 47

(iv) 0.0415 × 59 

Solution 4:

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question :5. Find:

(i) 1.07 × 0.02

(ii) 211.9 × 1.13

(iii) 10.05 × 1.05

(iv) 13.01 × 5.01 

Solution 5:

(i) 1.07 × 0.02

Multiply normally without observing at the decimal point

1.07 × 0.02 = 00214

Now put the decimal point in the product to have four places of decimal as there in the given decimal

1.07 × 0.02 = 0.0214

 

(ii) 211.9 × 1.13

Multiply normally without observing at the decimal point

211.9 × 1.13 = 239447

Now put the decimal point in the product to have three places of decimal as there in the given decimal

211.9 × 1.13 = 239.447

 

(iii) 10.05 × 1.05

Multiply normally without observing at the decimal point

10.05 × 1.05 = 105525

Now put the decimal point in the product to have four places of decimal as there in the given decimal

10.05 × 1.05 = 10.5525

 

(iv) 13.01 × 5.01

Multiply normally without observing at the decimal point

13.01 × 5.01 = 651801

Now put the decimal point in the product to have four places of decimal as there in the given decimal

13.01 × 5.01 = 65.1801

 

Question :6. Find the area of a rectangle whose length is 5.5m and breadth is 3.4m. 

Solution 6:

According to the given information Length of rectangle = 5.5m

Breadth of rectangle = 3.4 m

Area of rectangle = length × breadth

= 5.5 × 3.4

= 18.7 m2

 

Question :7. If the cost of a book is Rs 25.57, find the cost 0f 24 such books. 

Solution 7:

According to the given information the cost of a book is Rs 25.57

The Cost of 24 books are = 25.57 × 24

= Rs 618.00 

 

Question 8: A car covers a distance of 14.75km in one liter of petrol. How much distance it will cover in 15.5 liters of petrol?
 
Solution 8:
 
According to the given information that distance covered by car in 1 liter of petrol = 14.75 km
Distance covered by car in 15.5 liters of petrol = 14.75 × 15.5
= 228.625 km
 
 
Question 9: One kg of rice costs Rs 42.65. What will be the cost of 18.25 kg of rice?
 
Solution 9:
 
According to the given information that the cost of 1kg of rice = 42.65
Cost of 18.25kg of rice = 42.65 × 18.25
= Rs 778.3625
 
 
Question 10: One meter of cloth costs Rs 152.50. What is the cost of 10.75 meters of cloth?
 
Solution 10:
 
According to the given information that cost of 1m cloth = Rs 152.50
Cost of 10.75 m of cloth = 152.50 × 10.75
= Rs 1639.375
 
Exercise 3.3 
 
Question 1: Divide:
(i) 142.45 by 10
(ii) 54.25 by 10
(iii) 3.45 by 10
(iv) 0.57 by 10
(v) 0.0043 by 10
(vi) 0.004 by 10
 
Solution 1:
 
(i) 142.45 by 10
Arranging the decimal point by one place to the left we can get the result
142.45/10 = 14.245
 
(ii) 54.25 by 10
Arranging the decimal point by one place to the left we can get the result
54.25/10 = 5.425
 
(iii) 3.45 by 10
Arranging the decimal point by one place to the left we can get the result
3.45/10 = 0.345
 
(iv) 0.57 by 10
Arranging the decimal point by one place to the left we can get the result
0,57/10 = 0.057
 
(v) 0.0043 by 10
Arranging the decimal point by one place to the left we can get the result
0.0043/10 = 0.00043
 
(vi) 0.004 by 10
Arranging the decimal point by one place to the left we can get the result
0.004/10 = 0.0004
 
 
Question 2: Divide:
(i) 459.5 by 100
(ii) 74.3 by 100
(iii) 5.8 by 100
(iv) 0.7 by 100
(v) 0.48 by 100
(vi) 0.03 by 100
 
Solution 2:
 
(i) 459.5 by 100
Arranging the decimal point by two places to the left we can get the result
459.5/100 = 4.595
 
(ii) 74.3 by 100
Arranging the decimal point by two places to the left we can get the result
74.3/100 = 0.743
 
(iii) 5.8 by 100
Arranging the decimal point by two places to the left we can get the result
5.8/100 = 0.058
 
(iv) 0.7 by 100
Arranging the decimal point by two places to the left we can get the result
0.7/100 = 0.007
 
(v)  0.48 by 100
Arranging the decimal point by two places to the left we can get the result
0.48/100 = 0.0048
 
(vi) 0.03 by 100
Arranging the decimal point by two places to the left we can get the result
0.03/100 = 0.0003


Question :3. Divide:
(i) 235. 41 by 1000
(ii) 29.5 by 1000
(iii) 3.8 by 1000
(iv) 0.7 by 1000
 
Solution 3:
 
(i) 235. 41 by 1000
Arranging the decimal point by three places to the left we can get the result
235.41/1000 = 0.23541
 
(ii) 29.5 by 1000
Arranging the decimal point by three places to the left we can get the result
29.5/1000 = 0.0295
 
(iii) 3.8 by 1000
Arranging the decimal point by three places to the left we can get the result
3.8/1000 = 0.0038
 
(iv) 0.7 by 1000
Arranging the decimal point by three places to the left we can get the result
0.7/1000 = 0.0007
 

Question :4. Divide:

(i) 0.45 by 9

(ii) 217.44 by 18

(iii) 319.2 by 2.28

(iv) 40.32 by 9.6

(v) 0.765 by 0.9

(vi) 0.768 by 1.6 

Solution 4:

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question 5: Divide:

(i) 16.64 by 20

(ii) 0.192 by 12

(iii) 163.44 by 24

(iv) 403.2 by 96

(v) 16.344 by 12

(vi) 31.92 by 228 

Solution 5:

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question :6. Divide:

(i) 15.68 by 20

(ii) 164.6 by 200

(iii) 403.80 by 30 

Solution 6:

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question :7. Divide:

(i) 76 by 0.019

(ii) 88 by 0.08

(iii) 148 by 0.074

(iv) 7 by 0.014 

Solution 7:

(i) 76 by 0.019 = 76/0.019

Multiply the fraction by 1000 then divide

= 76000/19

 RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question 8: Divide:
(i) 20 by 50
(ii) 8 by 100
(iii) 72 by 576
(iv) 144 by 15
 
Solution 8:

 RD Sharma Solutions Class 7 Chapter 3 Decimals

RD Sharma Solutions Class 7 Chapter 3 Decimals

 

Question 9: A vehicle covers a distance 0f 43.2 km in 2.4 liters of petrol. How much distance will travel in 1 liter of petrol?
 
Solution 9:
 
Vehicle covered distance in 2.4 liters of petrol is = 43.2 km
Vehicle distance travel in 1 liter of petrol = 43.2/2.4
Multiply both by 10 then we get 432/24
On dividing
 RD Sharma Solutions Class 7 Chapter 3 Decimals
Hence, 18km can travel in 1litre of petrol.
 
 
Question 10: The total weight of some bags of wheat is 1743 kg. If each bag weighs 49.8 kg, how many bags are there?
 
Solution 10:
 
Total weight of bags are 1743 kg
Each bag weighs are 49.8 kg
Number of bags are 1743/49.8
Multiply both by 10 then we get 17430/498
On dividing
RD Sharma Solutions Class 7 Chapter 3 Decimals
 
 
Question 11: Shikha cuts 50 m of cloth into pieces 0f 1.25 m each. How many pieces does she get?
 
Solution 11:

That total length of cloth is 50 m
Length of each piece is 1.25 m
Number of cloth piece are 50/1.25
Multiply both by 100 then we get 5000/125
On dividing
 
RD Sharma Solutions Class 7 Chapter 3 Decimals
 
Question :12. Each side of a rectangular polygon is 2.5cm in length. The perimeter of the polygon is 12.5 cm. How many sides does the polygon have?
 
Solution 12:
 
The length of each side of polygon is 2.5 cm
Perimeter of polygon is 12.5 cm
Number of sides is 12.5/2.5
Multiply both by 10 then we get 12.5/25
On dividing
 
RD Sharma Solutions Class 7 Chapter 3 Decimals
 
 
Question :13. The product of two decimals is 42.987. If one of them is 12.46, find the other.
 
Solution 13:

The product of two decimals  42.987
One of the numbers is 12.46
Another number is  42.987/12.46
Multiply both by 1000 then we get 42987/12460 
On dividing
 
RD Sharma Solutions Class 7 Chapter 3 Decimals
Another number is 3.45
 
 
Question :14. The weight of 34 bags of sugar is 3483.3 kg. If all bags weigh equally, find the weight of each bag.
 
Solution 14:

The weight of 34 bags of sugar is = 3483.3 kg
Weight of each bag is 3483.3/34
Multiply both by 10 then we get 34833/34
On dividing
 
RD Sharma Solutions Class 7 Chapter 3 Decimals
 
Each bag’s weight 102.45 kg
 
 
Question :15. How many buckets of equal capacity can be filled from 586.5 liters of water, if each has capacity of 8.5 liters?
 
Solution 15:
 
The capacity of each bucket = 8.5 liters
Total water available = 586.5 liters
Number of buckets = 586.5/8.5
Multiply both by 10 then we get 5865/85
On dividing

 RD Sharma Solutions Class 7 Chapter 3 Decimals

 

 

NCERT Exemplar Solutions Class 7 Maths Algebraic Expressions
NCERT Exemplar Solutions Class 7 Maths Comparing Quantities
NCERT Exemplar Solutions Class 7 Maths Data Handling
NCERT Exemplar Solutions Class 7 Maths Exponents and Powers
NCERT Exemplar Solutions Class 7 Maths Fractions and Decimals
NCERT Exemplar Solutions Class 7 Maths Integers
NCERT Exemplar Solutions Class 7 Maths Lines and Angles
NCERT Exemplar Solutions Class 7 Maths Perimeter and Area
NCERT Exemplar Solutions Class 7 Maths Practical Geometry
NCERT Exemplar Solutions Class 7 Maths Rational Numbers
NCERT Exemplar Solutions Class 7 Maths Simple Equation
NCERT Exemplar Solutions Class 7 Maths Triangles and Its Properties
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability
RD Sharma Solutions Class 7 Maths
RD Sharma Solutions Class 7 Chapter 1 Integers
RD Sharma Solutions Class 7 Chapter 2 Fraction
RD Sharma Solutions Class 7 Chapter 3 Decimals
RD Sharma Solutions Class 7 Chapter 4 Rational Numbers
RD Sharma Solutions Class 7 Chapter 5 Operation on Decimal Numbers
RD Sharma Solutions Class 7 Chapter 6 Exponents
RD Sharma Solutions Class 7 Chapter 7 Algebraic Expressions
RD Sharma Solutions Class 7 Chapter 8 Linear Equations in One Variable
RD Sharma Solutions Class 7 Chapter 9 Ratio and Proportion
RD Sharma Solutions Class 7 Chapter 10 Unitary Method
RD Sharma Solutions Class 7 Chapter 11 Percentage
RD Sharma Solutions Class 7 Chapter 12 Profit and Loss
RD Sharma Solutions Class 7 Chapter 13 Simple Interest
RD Sharma Solutions Class 7 Chapter 14 Line and Angles
RD Sharma Solutions Class 7 Chapter 15 Properties of Triangle
RD Sharma Solutions Class 7 Chapter 16 Congruence
RD Sharma Solutions Class 7 Chapter 17 Constructions
RD Sharma Solutions Class 7 Chapter 18 Symmetry
RD Sharma Solutions Class 7 Chapter 19 Visualising Solid Shapes
RD Sharma Solutions Class 7 Chapter 20 Mensuration Perimeter and Area of Rectilinear Figures
RD Sharma Solutions Class 7 Chapter 21 Mensuration Area of Circle
RD Sharma Solutions Class 7 Chapter 22 Data Handling Collection and Organisation of Data
RD Sharma Solutions Class 7 Chapter 23 Data Handling Central Values
RD Sharma Solutions Class 7 Chapter 24 Data Handling Constructions of Bar Graph
RD Sharma Solutions Class 7 Chapter 25 Data Handling Probability
RS Aggarwal Class 7 Mathematics Solutions
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability