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About Math Homework: At Tutorhelpdesk we offer efficient

and cost effective math homework help service for students

looking for quality math study support for completing their

math homework projects. Students need to send their

homework task to us and once the initial payment formality

etc. will be completed we assign one of our best math tutors to

undertake the project for due completion. We only appoint

tutors with comprehensive math background for doing math

homework services. We also offer free of cost modification

service and online solution clarification session between the

related students and the working tutor. At Tutorhelpdesk.com we value your time line and

your privacy while offering you math assignment help.

Sample of Math Homework Illustrations and Solutions:

Question 1: The side of a square exceeds the side of the another square by 4 cm and the

sum of the areas of the two squares is 400 sq. cm. Find the dimensions of the squares.

Solution. Set 𝑆1 and 𝑆2 be two squares. Let the side of the square 𝑆2 be x cm in length.

Then, the side of square S1 is (x + 4) cm.

∴ Area of square 𝑆1 = 𝑥 + 4 2

And, Area of square 𝑆2 = 𝑥2

It is given that

Area of square 𝑆1 + Area of square 𝑆2 = 400 cm2

= 𝑥 + 4 2 + 𝑥2 = 400 = (𝑥2 + 8x + 16) + x = 400 = 2𝑥2 + 8x – 384 = 0 = 𝑥2 + 4x – 192 =

0

= 𝑥2 + 16x – 12x – 192 = 0 =x(x + 16) -12 (x + 16) = 0 = (x + 16) (x – 12) = 0 x = 12 or

x = -16

Since the length of the side of a square cannot be negative: x = 12

∴ Side the length of the side of a square cannot be negative: x = 12

∴ Side of square 𝑆1 = x + 4 = 12 + 4 = 16 cm and Side of square 𝑆2 = 12 cm.

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Question 2:

There is square field whose side is 44 m. A square flower bed is prepared in its centre

leaving gravel path all round the flower bed and gravelling the path at $ 2.75 and $ 1.50

per square meter respectively is $ 4904. Find the width of the gravel path.

Solution. Let the width of the gravel path be x meters. Then, Each side of the square

flower bed is (44 – 2x) meters.

Now area of the square field = 44 × 44 = 1936 m2

Area of the flower bed = (44 - 2𝑥)2 m2

∴ Area of the gravel path

= Area of the field – Area of the flower bed

= 1936 – (44 - 2𝑥)2

= 1936 – (1936 – 176x + 4𝑥2) m2

Cost of laying the flower bed

= (Area of the flower bed) (Rate per sq. m)

= (44 - 2𝑥)2 × 275

100 =

11

4 (44 - 2𝑥)2 = 11 ( 22 – 𝑥)2

Cost of gravelling the path = (Area of the path) × (Rate per sq. m) = (176x - 4𝑥2) 150

100 =

6(44x - 𝑥2)

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It is given that the total cost of laying the flower bed and gravelling the path is $ 4904.

∴ 11 (22 – 𝑥)2 + 6 (44x - 𝑥2) = 4904 = 11(484 – 44x + 𝑥2) + (364x - 6𝑥2) = 4904.

= 5𝑥2 - 220x + 5324 = 4908 = 5𝑥2 – 220x + 420 = 0 = 𝑥2 -44x + 84 = 0

= x2 – 42x – 2x + 84 = 0 = x(x – 42) -2(x – 42) = 0

= (x -2) (x – 42) = 0 = x = 2 or x = 42.

But, x ≠ 42, as the side of the square is 44m. Therefore, x = 2.

Hence, the width of the gravel path is 2 meters.

Question 3:

A chess board contains 64 equal squares and the area of each square is 6.25 cm2. A border

round the board is 2 cm wide. Find the length of the side of the chess board.

Solution. Let length of the side of the chess board be x cm. Then Area of 64 squares =

𝑥 − 4 2

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∴ 𝑥 − 4 2 = 64 × 6.25 = 𝑥2 - 8x + 16 = 400

= 𝑥2 – 8x – 384 = 0 = 𝑥2 - 24x + 16x – 384 = 0

= (x – 24) (x + 16) = 0 = x = 24 cm

Question 4:

A swimming pool is filled with three pipes with uniforms flow. The first tow pipes operating

simultaneously, fill the pool in the same time during which the pool is filled by the third pipe

alone. The second pipe fills the pool five hours faster than the first pipe and four hours

slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Solution. Let v be the volume of the pool and x the number of hours required by the

second pipe alone to fill the pool. Then, first pipe takes (x + 5) hours, while the third pipe

takes (x -4) hours to fill the pool. So, the parts of the pool filled by the first, second and

third pipes in one hour are respectively

V

x+5, V

x and

V

x−4

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Let the time taken by the first and second pipes fill the pool simultaneously be 6 hours.

Then, the third pipe also takes the same time to fill the pool.

∴ V

x+5,V

x t =

𝑉

𝑥−4t =

1

𝑥+5 +

1

𝑥 =

1

𝑥−4

= (2x + 5) (x – 4) = 𝑥2 + 5x = 𝑥2 - 8x- 20 = 0

= 𝑥2 - 10x + 2x – 20 = 0 = (x – 10) (x + 2) = 0 = 10 or x = -2

But, x cannot be negative, So, x = 10.

Hence, the timings required by first, second and third pipes to fill the pool individually are

15 hours, 10 hours respectively.