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A conveyor belt delivers baggage at the rate of 3 tons in 5 minutes and second conveyor belt delivers baggage at the rate of 1 ton in 2 minutes. How much time will it take to get 33 tons of baggage delivered using both the conveyor belts together ?
Answer & Solution
Correct Answer:
Option
B
Baggage delivered by first belt in 1 minute
$$ = \left( {\frac{3}{5}} \right){\text{tons}}$$
Baggage delivered by second belt in 1 minute
$$ = \left( {\frac{1}{2}} \right){\text{tons}}$$
Baggage delivered by both belt in 1 minute
$$\eqalign{ & = \left( {\frac{3}{5} + \frac{1}{2}} \right){\text{tons}} \cr & = \frac{{11}}{{10}}{\text{ tons}} \cr & \therefore {\text{Required time}} \cr & = \left( {33 \div \frac{{11}}{{10}}} \right){\text{ minutes}} \cr & = \left( {33 \times \frac{{10}}{{11}}} \right){\text{minutes}} \cr & = {\text{30 minutes}} \cr} $$
$$ = \left( {\frac{3}{5}} \right){\text{tons}}$$
Baggage delivered by second belt in 1 minute
$$ = \left( {\frac{1}{2}} \right){\text{tons}}$$
Baggage delivered by both belt in 1 minute
$$\eqalign{ & = \left( {\frac{3}{5} + \frac{1}{2}} \right){\text{tons}} \cr & = \frac{{11}}{{10}}{\text{ tons}} \cr & \therefore {\text{Required time}} \cr & = \left( {33 \div \frac{{11}}{{10}}} \right){\text{ minutes}} \cr & = \left( {33 \times \frac{{10}}{{11}}} \right){\text{minutes}} \cr & = {\text{30 minutes}} \cr} $$
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