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A car requires 108 litres of petrol for covering a distance pf 594 km. How much petrol will be required by the car to cover 1650 km? |
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Answer» Step-by-step EXPLANATION: Given, diesel required for 594km = 108ldiesel required for 1km=\FRAC{108}{594}L\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 l Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 ldiesel required for 1650km=\frac{2}{11}\times1650l=2\times150=300l Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 ldiesel required for 1650km=\frac{2}{11}\times1650l=2\times150=300l 11 Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 ldiesel required for 1650km=\frac{2}{11}\times1650l=2\times150=300l 112 Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 ldiesel required for 1650km=\frac{2}{11}\times1650l=2\times150=300l 112 Given, diesel required for 594km = 108ldiesel required for 1km=\frac{108}{594}l\ =\frac{108\div54}{594\div54}=\frac{2}{11}l 594108 l = 594÷54108÷54 = 112 ldiesel required for 1650km=\frac{2}{11}\times1650l=2\times150=300l 112 ×1650l=2×150=300l |
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