1.Following are the car parking charges near a railway station upto:
4 hours – Rs.60
8 hours – Rs.100
12 hours – Rs.140
24 hours – Rs.180
Check if the parking charges are in direct proportion to the parking time.
Solution:
We have the ratio of time period and parking charges per hour:
C1 = 60/4 = Rs. 15
C2 = 100/8 = Rs. 12.50
C3 = 140/12 = Rs. 11.67
C4 = 180/24 = Rs.7.50
Here, the charges per hour are not same, i.e., C1 ≠ C2 ≠ C3 ≠ C4
Hence, the parking charges are not in directly proportion.
2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
| Parts of base | 8 | …… | …….. | ……… | ……. |
Solution:
Let the required field be a, b, c and d respectively.
1/8 = 4/a
So, a = 32
1/8 = 7/b
So, b = 56
1/8 = 12/c
So, c = 96
1/8 = 20/d
So, d = 160
| Parts of red pigment | 1 | 4 | 7 | 12 | 20 |
| Parts of base | 8 | 32 | 56 | 96 | 160 |
3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Solution:
Let the parts of red pigment mix with 1800 mL base be x.
1/x = 75 ml/ 1800 ml
So, x = 1800/75
= 24
Hence, 24 parts red pigment should be mixed with base 1800 mL
4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the required number of bottles filled in five hours be x.
| No. of bottles | Time in hours |
| 840 | 6 |
| X | 5 |
Now,
6x = 5×840
x = 5×840/6
= 700
Hence, machine will fill 700 bottles in five hours.
5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Solution:
Let enlarged length of bacteria be x cm.
| Enlargement | Length |
| 50,000 | 5 cm |
| 20,000 | X cm |
Now,
50,000/20,000 = 5/x
X = (5 x 20,000)/ 50,000
= 2 cm
Now,
Original length of bacteria = 5/50,000
= 1/10,000
= 10-4
6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length if the ship is 28 m, how long is the model ship?
Solution:
Let the length of the model ship be X m.
| Height of mast | Length of ship |
| 9 cm | 12 m |
| X cm | 28 |
Here, length of mast and actual length of ship are in directly proportion.
Now,
9/X = 12/28
12 X = 9 x 12
X = 21
Hence, required length of the model ship is 21 cm.
7. Suppose 2 kg of sugar contains 9×106 crystals. How many sugar crystals are there in
(i)5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution:
(i) Let the required number of sugar crystals be x.
| No. of sugar crystals | Weight of sugar |
| 9 x 106 | 2 kg |
| x | 5 kg |
Here, weight of sugar and number of crystals are in direct proportion.
Now,
(9 x 106) /x = 2/5
2x = 5 x 9 x 106
x = 2.25 x 107
(ii) Let sugar crystals be y.
| No. of sugar crystals | Weight of sugar |
| 9 x 106 | 2 kg |
| y | 1.2 kg |
Here weight of sugar and number of crystals are in directly proportion.
(9 x 106)/y = 1/1.2
2y = 1.2 x 9 x 106
Y = 5.4 x 9 x 106
8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:
Let required distance covered in the map be x.
| Scale in cm | Distance in km |
| 1 | 18 |
| x | 72 |
Here actual distance and distance covered in the map are in directly proportion.
1/x = 18/72
18x = 72
x = 4 cm
Hence, distance covered in the map is 4 cm.
9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long.
Solution:
(i)
Let the required length of shadow be x.
| Height of Pole | Length of Shadow |
| 5 m 60 cm = 5.60 m | 3 m 20 cm = 3.20 m |
| 10 m 50 cm = 10.50 m | X m |
Here height of the pole and length of the shadow are in directly proportion.
Now,
5.60/10.50 = 3.20/x
x = (3.20 x 10.50)/5.60
= [(320 x 1050)/560] x [1/100]
= 6 m
(ii) Let the height of the pole be y.
| Height of Pole | Length of shadow |
| 5.60 | 3.20 |
| y | 5 |
5.60/y = 3.20/5
Y = (5.60 x 5)/3.20
= 35/4
= 8 m 75 cm
Hence, required height of the pole is 8 m 75 cm.
10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Let the required distance covered in 5 hours be x km.
| Distance | Time |
| 14 km | 25 min |
| X km | 5 hours = 300 min |
1 hour = 60 minutes
5 hours = 5×60 = 300 minutes
Here distance covered and time in directly proportion.
14/x = 25/300
X = (14 x 300)/25
x = 168 km
Hence, the required distance travelled by truck in 5 hours = 168 km
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