Equations

Question 1

Justin is 2 years older than his friend, Sam. Justin’s father is 3 times Justin’s age. Their accumulative age is 86. How old is Justin, Sam and Justins father?

Answer 1

Justin = J
Sam = J - 2
Father = 3J
= J + J - 2 + 3J = 86
= 5J - 2 = 86
= 5J - 2 + 2 = 86 + 2
=5J = 88
=5J/5 = 88/5
J = 17.6.
S = 17.6 - 2 = 15.6
F = 17.6 x 3 = 52.8

Question 2

Mary is 4 years older than her sister, Stacey. Mary’s mother is 2 times Mary’s age. Their accumulative age is 72. How old is Mary, Stacey and Mary’s mother?

Answer 2

Mary = Ma
Stacey = Ma - 4
Mother = 2Ma
= Ma + Ma - 4 + 2Ma = 72
= 4Ma - 4  = 72
= 4Ma - 4 + 4 = 72 + 4
= 4Ma = 76
= 4Ma/4 = 76/4
Ma = 19
S = 19 - 4 = 15
Mo = 19 x 2 = 38

Question 3

Jack is 10 years older than his little brother, William. Jack’s father is 3 times ‘s age. Their accumulative age is 94. How old is Jack, William and Jack’s father?

Answer 3

Jack = J
William = J - 10
Father = 3J
= J + J - 10 + 3J = 94
= 5J - 10 = 94
= 5J - 10 + 10 = 94 + 10
= 5J = 104
=5J/5 = 104/5
J = 20.8
W = 20.8 - 10 = 10.8
F = 20.8 x 3 = 6.24

Question 4

Convert 3km to m.

Answer 4

3 x 1000 = 3000

= 3000m

Question 5

Convert 740L to kL

Answer 5

740/1000 = 0.74

= 0.74kL

Question 6

Convert 200cm to m

Answer 6

200/100 = 2

= 2m

Question 7

Convert 7400km to mm

Answer 7

= 7400 x 1000 = 7,400,000m

= 7,400,000 x 1000 = 74,000,000mm

Question 8

Convert 18L to mL

Answer 8

= 18 x 1000 = 18,000mL

Question 9

State the limit of accuracy.

Answer 9

The limit of accuracy is ±0.5 of an instruments scale.

Question 10

Find the limit of accuracy of a 15cm ruler with 1ml increments.

Answer 10

1 x ±0.5

= 0.5ml.

Question 11

Find the limit of accuracy of a clock with 1 min increments.

Answer 11

1 x ±0.5.

= 30 seconds.

Question 12

Find the limit of accuracy of a scale with 25kl increments..

Answer 12

25 x±0.5

= 12.5kl.