Three water pipes A, B and C can fill a tank to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three water pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes
| Part filled by (A + B + C) in 3 minutes = 3 |  |
1 | + | 1 | + | 1 |  |
= | |
3 x | 11 | |
= | 11 | . |
| 30 | 20 | 10 | 60 | 20 |
| Part filled by C in 3 minutes = | 3 | . |
| 10 |
 Required ratio = |
|
3 | x | 20 | |
= | 6 | . |
| 10 | 11 | 11 |
A pump can fill a drum with acidic water in 2 hours. Because of a leak, it took 2
hours to fill the tank. The leak can drain all the water of the tank in:
| Work done by the leak in 1 hour = | |
1 | - | 3 | |
= | 1 | . |
| 2 | 7 | 14 |
Leak will empty the tank in 14 hrs.
A well is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
Suppose pipe A alone takes y hours to fill the tank.
| Then, pipes B and C will take | y | and | y | hours respectively to fill the tank. |
| 2 | 4 |
|
1 | + | 2 | + | 4 | = | 1 |
| y | y | y | 5 |
 |
7 | = | 1 |
| y | 5 |
y = 35 hrs.
A pipe can fill a tank in 6 hours. After half the tank is filled, three more similar pipes are opened. What is the total time taken to fill the tank completely
Time taken by one tap to fill half of the tank = 3 hrs.
| Part filled by the four taps in 1 hour = | |
4 x | 1 | |
= | 2 | . |
| 6 | 3 |
| Remaining part = | |
1 - | 1 | |
= | 1 | . |
| 2 | 2 |
|
2 | : | 1 | :: 1 : x |
| 3 | 2 |
| x = |
|
1 | x 1 x | 3 | |
= | 3 | hours. |
| 2 | 2 | 4 |
So, total time taken = 3 hrs. 45 mins
A hollow steel pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:
External radius = 4 cm,
Internal radius = 3 cm.
| Volume of iron |
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| = 462 cm3. |
Weight of iron = (462 x 8) gm = 3696 gm = 3.696 k
A piece of wood having a length 3 m and breadth 2 m is floating on an ocean. The piece of wood sinks by 1 cm when a man gets on it. The mass of the man is:
| Volume of water displaced | = (3 x 2 x 0.01) m3 |
| = 0.06 m3. |
| Mass of man |
= Volume of water displaced x Density of water |
| = (0.06 x 1000) kg | |
|
= 60 kg.
|
A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm.
| Required ratio = |
|
6 x (32 + 42 + 52) | |
= | 50 | = 25 : 18. |
| 6 x 62 | 36 |
A can contains a mixture of two petrol solutions A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initiall
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
| Quantity of A in mixture left = | |
7x - | 7 | x 9 | |
litres = | |
7x - | 21 | litres. |
| 12 | 4 |
| Quantity of B in mixture left = | |
5x - | 5 | x 9 | |
litres = | |
5x - | 15 | litres. |
| 12 | 4 |
|
|
= | 7 | |||||
|
9 |
|
28x - 21 | = | 7 |
| 20x + 21 | 9 |
252x - 189 = 140x + 147
112x = 336
x = 3.
So, the can contained 21 litres of A.
A container contains 40 litres of alcoholic solution. From this container 4 litres of the solution was taken out and replaced by water. This process was repeated further two times. How many litres of the solution is now contained by the container
| Amount of solution left after 3 operations = |  |
40 | |
1 - | 4 | |
3 |  litres |
| 40 |
| = | |
40 x | 9 | x | 9 | x | 9 | |
= 29.16 litres. |
| 10 | 10 | 10 |
|
8 litres are drawn from a cask full of alcohol and is then filled with water. This operation is performed three more times. The ratio of the quantity of alcohol now left in cask to that of water is 16 : 81. How much alcohol did the cask hold originally?
Let the quantity of the alcohol in the cask originally be x litres.
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The average age of mine workers is 15.8 years. The average age of men in the mine is 16.4 years while that of women is 15.4 years. What is the ratio of men to women in the class?
Let the ratio be k:1.
Then k*16.4+1*15.4 + (k+1)*15.8 (16.4-15.8)k = 15.8 - 15.4 k=0.4/0.6 = 2/3 so required ratio = 2:3
Then k*16.4+1*15.4 = (k+1)*15.8 (16.4-15.8)k = 15.8 - 15.4 k=0.4/0.6 = 2/3 so required ratio = 2:3
Diamond is 19 times as heavy as sulphur and Tin is 9 times as sulphur as water. In what ratio should these be mixed to get an alloy 15 times as heavy as sulphur?
Let 1gm of diamond be mixed with x gm of Tin to give (1+x)gm of the alloy.
1G=19W, 1C = 9W and alloy = 15W 1gm gold + xgm Copper = (1+x)gm alloy 19W+9Wx = (1+x)*15W x = 4W/6W = 2/3
So ratio of diamond and tin is 1:2/3 or 3:2
