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,
, ,
and
So ,
Therefore, the parking charges are not in direct proportion to the parking time .
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 and are the parts of base respectively , then
Parts of red pigment |
1 |
4 |
7 |
12 |
20 |
Parts of base |
8 |
|
|
|
|
So,
We have ,
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 part of a red pigment requires 1800 mL of base .
A/Q,
Therefore, 24 parts of a red pigment requires 1800 mL of base .
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 be the number of bottles .
A/Q,
Therefore, the number of bottles is 700 .
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: The actual length of the bacteria
cm
Let be the length of a photograph of a bacteria .
A/Q ,
Therefore, the length of a photograph of the bacteria is 2 cm .
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 of the ship is 28 m, how long is the model ship?
Solution: let be the length of the model ship (in metres) .
We have,
High of mast (in cm) |
Length of ship (in m) |
9
|
12 28 |
A/Q,
Therefore, the length of the model ship is 21 m
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 be the number of sugar crystals .
We have ,
Weight of sugar (In kg) |
No. of sugar crystals |
2 |
|
5 |
|
A/Q,
Therefore, the number of sugar crystals is .
(ii) let be the number of sugar crystals .
We have ,
Weight of sugar (In kg) |
No. of sugar crystals |
2 |
|
1.2 |
|
A/Q,
Therefore, the number of sugar crystals is .
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 be the distance covered in the map (in cm) .
We have,
Scale (in cm) |
Distance (in km) |
1 |
18 |
|
72 |
A/Q,
Therefore, the 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 5m long.
Solution: (i) let be the length of the shadow cast by vertical pole (in cm) .
We have,
5m + 60 cm = 5×100+60 = 500+60 = 560 cm
10m + 50 cm = 10×100+50 = 1000+50 = 1050 cm
3m + 20 cm = 3×100+20 = 300+20 = 320 cm
Height of pole |
Length of shadow |
560 |
320 |
1050 |
|
A/Q,
Therefore, the length of the shadow cast by vertical pole is 6 m .
(ii) let be the length of the vertical pole (in metres) .
We have,
5m + 60 cm = 5.60 m = 5.6 m
3m + 20 cm = 3.20 m = 3.2 m
Height of pole |
Length of shadow |
5.6 |
3.2 |
|
5 |
A/Q,
m = 875 cm = 8m 75cm
Therefore , the length of the vertical pole is 8m75 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? Photo
Solution: let be the distance travels by the truck in 5 hours .
We have,
Distance (in km) |
Time (in minutes) |
14 |
25 |
|
5 × 60 = 300 |
A/Q,
Therefore, the distance travels by the truck is 168 km .
Solution: (i) The number of workers on a job and the time to complete the job.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Number of winners
|
1
|
2
|
4
|
5
|
8
|
10
|
20
|
Prize for each winner (in Rs)
|
1,00,000
|
50,000
|
...
|
...
|
...
|
....
|
…
|
Solution: Firstly ,we observe , 1×100000 = 2×50000 = 100000 . So, it is inversely proportional .
Let ,
Number of winners |
1 |
2 |
4 |
5 |
8 |
10 |
20 |
Prize for each winner (in Rs) |
1,00,000 |
50,000 |
|
|
|
p |
|
Now,
Again,
Again ,
Again,
And
Solution: let
Number of spokes |
Angle between a pair of consecutive spokes |
4 |
90° |
6 |
60° |
8 |
|
10 |
|
12 |
|
Now,
Again,
and
We have,
Number of spokes |
Angle between a pair of consecutive spokes |
4 |
90° |
6 |
60° |
8 |
45° |
10 |
36° |
12 |
30° |
(i) Firstly , we observe 4 × 90° = 6 × 60° = 360°
So, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion .
(ii) let be the angle between a pair of consecutive spokes on a wheel .
A/Q,
Therefore, the angle between a pair of consecutive spokes on a wheel is 24° .
(iii) let be the number of spokes .
A/Q,
Therefore, the number of spokes is 9 .
Solution: let be the number of sweets box .
No. of children |
No. of sweets box |
24 |
5 |
24 – 4 = 20 |
|
A/Q,
Therefore, the number of sweets box is 6 .
Solution: let be the number of days .
We have ,
No. of cattle |
No. of days |
20 |
6 |
20+10 = 30 |
|
A/Q,
Therefore, the number of days is 4 .
Solution: let be the days take to complete the job .
We have,
No. of persons |
No. of days |
3 |
4 |
4 |
|
A/Q,
Therefore, the days take to complete the job is 3 .
Solution: let be the number of boxes .
We have,
No. of boxes |
No. of bottles |
25 |
12 |
|
20 |
A/Q,
Therefore, the number of boxes is 15
Solution: let be the number of machines .
We have,
No. of machine |
No. of days |
42 |
63 |
|
54 |
A/Q,
Therefore, the number of machines is 49 .
Solution: let be the time taken travel by the car .
We have ,
Speed (in km) |
Times (in hours) |
60 |
2 |
80 |
|
A/Q,
Therefore ,the time taken travel by the car is hours .
Solution: (i) let be the number of days .
We have ,
No. of persons |
No. of days |
2 |
3 |
(2 – 1 = ) 1 |
|
A/Q,
Therefore, the number of days is 6 .
(ii) Let be the number of persons .
We have,
No. of persons |
No. of days |
2 |
3 |
|
1 |
A/Q,
Therefore, the number of persons is 6 .
Solution: let be the number of period duration (in minute) .
We have ,
Periods |
Duration (minutes) |
8 |
45 |
9 |
|
A/Q,
Therefore , the number of period duration is 40 minute .