Question

The ratio of the time taken by a caterpillar to walk past a 50 cm and a metre scale is 109:209. What is the
length (in cm) of the caterpillar?​

Answer 1

Given info : The ratio of the time taken by a caterpillar to walk past 50 cm and a metre scale is 109 : 209.

To find : the length of the caterpillar is ..

solution : let the length of caterpillar is l.

case 1 : to walk 50 cm caterpillar takes 109x seconds.

i.e., speed of caterpillar = total distance covered/time taken

= (50cm + length of caterpillar)/time taken

= (50 + l)/109x ...(1)

case 2 : To walk 100 cm (a metre scale) , caterpillar takes 209x seconds.

so, speed = (100 + l)/209x ...(2)

from equations (1) and (2) we get,

(50 + l)/(100 + l) = 109/209

⇒209(50 + l) = 109(100 + l)

⇒10450 + 209l = 10900 + 109l

⇒100l = 10900 - 10450

⇒l = 450/100 = 4.5 cm

Therefore the length of caterpillar is 4.5 cm

Answer 2

Given info: The ratio of the time taken by a caterpillar to walk past 50 cm and a metre scale is 109: 209.

by a caterpillar to walk past 50 cm and a metre scale is 109: 209.To find the length of the caterpillar is...

by a caterpillar to walk past 50 cm and a metre scale is 109: 209.To find the length of the caterpillar is...solution : let the length of caterpillar is 1. case 1: to walk 50 cm caterpillar takes 109x seconds.

by a caterpillar to walk past 50 cm and a metre scale is 109: 209.To find the length of the caterpillar is...solution : let the length of caterpillar is 1. case 1: to walk 50 cm caterpillar takes 109x seconds.i.e., speed of caterpillar = total distance covered/time taken

by a caterpillar to walk past 50 cm and a metre scale is 109: 209.To find the length of the caterpillar is...solution : let the length of caterpillar is 1. case 1: to walk 50 cm caterpillar takes 109x seconds.i.e., speed of caterpillar = total distance covered/time taken= (50cm + length of caterpillar)/time taken

= (50 + 1)/109x...(1)

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209→ 209(50 + 1) = 109(100 + 1)

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209→ 209(50 + 1) = 109(100 + 1)→ 10450 + 2091 10900 + 1091

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209→ 209(50 + 1) = 109(100 + 1)→ 10450 + 2091 10900 + 1091 → 1001 = 10900 - 10450

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209→ 209(50 + 1) = 109(100 + 1)→ 10450 + 2091 10900 + 1091 → 1001 = 10900 - 10450→1=450/100 = 4.5 cm

= (50 + 1)/109x...(1)case 2: To walk 100 cm (a metre scale), caterpillar takes 209x seconds.so, speed = (100 + 1)/209x...(2)from equations (1) and (2) we get,(50 + 1)/(100 + 1) = 109/209→ 209(50 + 1) = 109(100 + 1)→ 10450 + 2091 10900 + 1091 → 1001 = 10900 - 10450→1=450/100 = 4.5 cmTherefore the length of caterpillar is 4.5 cm