10. Solve the following
a. 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5
pipes of the same type are used?
b. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. how long
will it take when the car travels at the speed of 80 km/h?
Answers
Answer:
a)1hr 20 min=80 minutes
6 pipes fill tank in 80 minutes
1 pipe fill tank in 80×6 minutes
5 pipes fill tank in 80×6/5=96 minutes
b) Distance=60×2=120 km
Time taken=120/80=3/2hr=1 hr 30 minutes
➺6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? That 5 pepes take 1 hour and 36 minutes by same type used.
6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? That 5 pepes take 1 hour and 36 minutes by same type used....
6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? That 5 pepes take 1 hour and 36 minutes by same type used....Direct and Inverse Proportions.
6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? That 5 pepes take 1 hour and 36 minutes by same type used....Direct and Inverse Proportions.Weight of sugar Number of sugar crystals
6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? That 5 pepes take 1 hour and 36 minutes by same type used....Direct and Inverse Proportions.Weight of sugar Number of sugar crystals5 X
➺∴
∴ 80
∴ 8060
∴ 8060
∴ 8060 =
∴ 8060 = 2
∴ 8060 = 2x
∴ 8060 = 2x
∴ 8060 = 2x
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x=
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 80
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x=
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21 The car will take 1\frac{1}{2}1
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21 The car will take 1\frac{1}{2}1 2
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21 The car will take 1\frac{1}{2}1 21
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21 The car will take 1\frac{1}{2}1 21
∴ 8060 = 2x \Rightarrow\ x=\frac{60\times2}{80}⇒ x= 8060×2 x=\frac{3}{2}x= 23 x=1\frac{1}{2}x=1 21 The car will take 1\frac{1}{2}1 21 hours to reach the destination.