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This is very helpful for those students who want to know more about Science and Math, because it has word problems....

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WORD PROBLEM With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind? Solution Let the speed of the plane be ‘a’ and that of the wind be ‘w’. Our table looks like this:

Distance

Speed

Time

With the wind

2400kms

s+w

4

Against the wind

2400kms

s-w

6

(with wind) 4h(s+w) = 2400km

speed AGAINST the wind 6h (s-w) = 2400km

4w= 2400 – s

6s = 2400 + w

4w= 2400/4 – s

6s = 2400/ 6 + w

w=600-s

s= 400 + w

s-w= 400

s+w= 600

s-(600-s)=400

(400 +w) + w = 600

s-600-s = 400

400 + w + w = 600

s+s = 400 + 600

2w = 600 -400

2s = 1000

2w = 200

s= 1000/2

w= 200/2

s=500 kmph

w= 100kmph

The speed of the plane is 500 kmph and that of the wind is 100 kmph.

2. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.

A car takes 4 hours to cover a distance, if it travels at a speed of 40 kph. What should be its speed to cover the same distance in 1.5 hours?

3.

D 160 km 160 km

S 40kph 106.66kph

Solution Distance covered d= s x t =40 * 4 = 160 km Speed required to cover the same distance in 1.5 hours s = d/t = 160/1.5 = 106.66 kph

T 4 hours 1.5 hours

4. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.

5. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°

6. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.

7. In a class of 42 students, the number of boys is 2/5 of the girls. Find the number of boys and girls in the class. Let x= girls x+ 2/5 = boys total = 42 x+2/5 = 42

x+ 2/5

10x=42

= 30 + 2/5

x= 42-10= 30 girls

30/5= 6(2) = 12 boys...