Mathsma

[ Solved By Asma Hossain Sarah]

Mathematics: A; Edexcel International GCSE ( 9 - 1 ); Student Book 1; Unit : 2; Shape and Space : 2; Q)6; Exercise : 1*.

Q) Calculate the area of a square whose diagonals
are 20 cm.

Solve :

Let 'x' m be the length of each side of the square.

By the Pythagoras' Theorem :

20^2 = x^2 + x^2

=> x^2 + x^2 = 20^2

=> 2x^2 = 400

=> x^2 = 400/2

=> x^2 = 200

=> x = √200

=> x = 14.14 ( to 2 d. p )

Thus, the length of each side of the square is 14.14 cm.

Answer : 14.14 cm.

Mathematics: A; Edexcel International GCSE ( 9 - 1 ); Student Book 1; Unit : 2; Shape and Space : 2; Q)6; Exercise : 1*.

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition, Purple Book.

Q) Express 24 : 6 : 15 in it's simplest form.

Answer :

24 : 6 : 15

= 24/3 : 6/3 : 15/3

= 8 : 2 : 5.

Answer : 8 : 2 : 5

[ Question 9 - Primary Mathematics 6B - 3rd Edition Purple Book - Chaprer 4; Triangles and 4 Sided Figures - Review C ]

[ Solved By Asma Hossain Sarah ]

Mathematics: A; Edexcel International GCSE ( 9 - 1 ); Student Book : 1; Question 7; Exercise: 1; Unit : 2; Shape and Space: 2.

Q) Find the diagonal length of a square field of
side 50 m.

Answer :

Let 'x' m be the length of the diagonal of the square field.

x^2 = 50^2 + 50^2

=> x^2 = 50^2 + 50^2

=> x^2 = 2500+ 2500

=> x^2 = 5000

=> x = √5000

=> x = 70.71. ( to 2 d.p )

Thus, the diagonal length of the square field is 70.71 m

Answer : 70.71 m.

Mathematics: A; Edexcel International GCSE ( 9 - 1 ); Student Book : 1; Question 7; Exercise: 1; Unit : 2; Shape and Space: 2.

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition; Purple Book.

Q) Write the following in figures :

(a) Thirty thousand and thirty

(b) Three million and forty thousand.

Answer :

(a) 30,030

(b) 3,040,000.

[ Question 1 - Primary Mathematics 6B - 3rd Edition; Purple Book. - Chapter 4; Triangles and 4 - Sided Figures - Review C].

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition, Purple Book.

Q) Write 100 ÷ 3 as a mixed number.

Answer :

100 ÷ 3

= 100/3

= 33 1/3 .

Answer : 33 1/3.

[ Question 5 - Primary Mathematics 6B - 3rd Edition, Purple Book - Chapter 4; Triangles and 4 - Sided Figures - Review C].

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition; Purple Book.

Q) 9t - 6t + t - 2

Answer :

9t - 6t + t - 2

= 9t + t - 6t - 2

= 10t - 6t - 2

= 4t - 2

= 2( 2t - 1 ).

Answer : 2( 2t - 1 ).

[ Question 24 - Primary Mathematics 6B - 3rd Edition; Purple Book - Chapter 4; Triangles and 4 - Sided Figures - Review C].

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition; Purple Book.

Q) Find the value of 20% of $14.50.

Answer :

20% of $14.50

= 20/100 of $14.50

= 20/100 × $14.50

= 2/10 × $14.50

= 1/5 × $14.50

= 14.50/5

= $2.90.

Answer : $2.90.

[ Question 16 - Primary Mathematics 6B - 3rd Edition; Purple Book - Chapter 4; Triangles and 4 Sided Figures - Review C ].

[ Solved By Asma Hossain Sarah ]

Primary Mathematics 6B - 3rd Edition; Purple Book.

Q) The ratio of John's money to Peter's money is 3:5. Express Peter's money as a fraction of John's money.

Answer :

Because,

ratio of John's money to Peter's money = 3:5 .

Let John's money = 3x

and Peter's money = 5x.

Therefore,

the fraction of Peter's money to John's money

= 5x/3x .

= 5/3 .

Answer : 5/3.

[ Question 11 - Primary Mathematics 6B - 3rd Edition; Purple Book - Chapter 4; Triangles and 4 - Sided Figures - Review C].