Work out the values of x , y & z , x y z - 1 3 10 = 3 5 8

Given, x y z - 1 3 10 = 3 5 8 ⇒ x y z = 3 5 8 + 1 3 10 Change each fraction to an improper fraction. ⇒ x y z = 29 8 + 13 10 Write the fractions over a common denominator. ⇒ x y z = 29 × 5 40 + 13 × 4 40 ⇒ x y z = 145 40 + 52 40 Add the fractions. ⇒ x y z = 145 + 52 40 ⇒ x y z = 197 40 ⇒ x y z = 4 37 40 Therefore, the required answer is x = 4 , y = 73 & z = 40 .

Work out the values of x , y & z , x y z - 1 3 10 = 3 5 8

Given, x y z - 1 3 10 = 3 5 8 ⇒ x y z = 3 5 8 + 1 3 10 Change each fraction to an improper fraction. ⇒ x y z = 29 8 + 13 10 Write the fractions over a common denominator. ⇒ x y z = 29 × 5 40 + 13 × 4 40 ⇒ x y z = 145 40 + 52 40 Add the fractions. ⇒ x y z = 145 + 52 40 ⇒ x y z = 197 40 ⇒ x y z = 4 37 40 Therefore, the required answer is x = 4 , y = 73 & z = 40 .

These are the times that Amy trained last week: How many hours did Amy train during the week? Monday Wednesday Saturday 2 2 3 hours 1 1 4 hours 2 1 6 hours

These are the times that Amy trained last week: Monday Wednesday Saturday 2 2 3 hours 1 1 4 hours 2 1 6 hours We need to find the number of hours Amy trains during the week. Let's assume Amy trains x hours, x = 2 2 3 + 1 1 4 + 2 1 6 Change each fraction to an improper fraction. x = 8 3 + 5 4 + 13 6 Write the fractions over a common denominator. x = 8 × 4 12 + 5 × 3 12 + 13 × 2 12 x = 32 12 + 15 12 + 26 12 Add the fractions. x = 32 + 15 + 26 12 x = 73 12 x = 6 1 12 hours Therefore, 6 1 12 hours Amy trains during the week.

A running track is 3 1 2 kilometers long. If I jog for 2 1 5 kilometers and run for 1 1 5 kilometers will I complete the full distance of the track?

Given, A running track is 3 1 2 kilometers long. If I jog for 2 1 5 kilometers and run for 1 1 5 kilometers. Let's check whether I complete the full distance of the track. ⇒ 3 1 2 = 2 1 5 + 1 1 5 Change each fraction to an improper fraction. ⇒ 7 2 = 11 5 + 6 5 ⇒ 7 2 = 11 + 6 5 ⇒ 7 2 ≠ 17 5 Therefore, if I jog for 2 1 5 kilometers and run for 1 1 5 kilometers I can not complete the full distance of the track.

Which one is bigger? How much bigger? 2 2 5 + 3 5 6 + 3 4 & 7 7 8 - 1 5 12 - 1 1 3

Given, 2 2 5 + 3 5 6 + 3 4 & 7 7 8 - 1 5 12 - 1 1 3 Let's solve one by one and compare. ⇒ 2 2 5 + 3 5 6 + 3 4 ⇒ 12 5 + 23 6 + 3 4 ⇒ 12 × 12 60 + 23 × 10 60 + 3 × 15 60 ⇒ 144 60 + 230 60 + 45 60 ⇒ 144 + 230 + 45 60 = 419 60 Next, 7 7 8 - 1 5 12 - 1 1 3 = 63 8 - 17 12 - 4 3 = 63 8 - 17 12 - 4 3 = 63 × 3 24 - 17 × 2 12 - 4 × 8 3 = 189 24 - 34 24 - 32 24 = 189 - 34 - 32 24 = 123 24 Here, 419 60 = 419 × 2 60 × 2 = 838 120 And, 123 24 = 123 × 5 24 × 5 = 615 120 Therefore, 419 60 > 123 24 Now, 419 60 - 123 24 = 838 120 - 615 120 = 838 - 615 120 = 223 120 = 1 103 120 2 2 5 + 3 5 6 + 3 4 is bigger with 1 103 120 compared to 7 7 8 - 1 5 12 - 1 1 3 .

Investigate what happens when you complete these sums: 1 + 1 2 = 1 + 1 2 + 1 4 = 1 + 1 2 + 1 4 + 1 8 = 1 + 1 2 + 1 4 + 1 8 + 1 16 = Explain what happens to the sum as the denominators get bigger. Will the sum ever be greater than 2 ?

Let's check what happens to the sum as the denominators get bigger. 1 + 1 2 = 2 2 + 1 2 = 3 2 = 1 . 5 1 + 1 2 + 1 4 = 4 4 + 2 4 + 1 4 = 4 + 2 + 1 4 = 7 4 = 1 . 75 1 + 1 2 + 1 4 + 1 8 = 8 8 + 4 8 + 2 8 + 1 8 = 8 + 4 + 2 + 1 8 = 15 8 = 1 . 875 1 + 1 2 + 1 4 + 1 8 + 1 16 = 16 16 + 8 16 + 4 16 + 2 16 + 1 16 = 16 + 8 + 4 + 2 + 1 16 = 31 16 = 1 . 9375 We can see that as the denominators get bigger the sum will get increase. The given sum will not be greater than 2 .