Direct Proportions
Rs. 10 Per Kg
How Much For 1 Kg
So how much
for 3 Kg ?????
Rs. 30
WEIGHT (Kg.)
AMOUNT (Rs.)
1
10
3
30
RATIO
10
1
Both the ratios are equal
WEIGHT (Kg.)
AMOUNT (Rs.)
1
10
3
30
0 km
60 km
Time (hours)
Distance (km)
120
km
120
km
If one quantity increases and
other quantity also increases
This is the case of
DIRECT VARIATIONS
Lets check ratio
Distance : Time
1
60
2
120
3
180
In direct Variation ratio of
Quantities remain constant
∴
Distance
Time
=
60
1
=
60
60
120
2
=
60
=
60
180
3
=
60
=
60
WEIGHT Kg.
AMOUNT Rs.
1
3
`10
`30
Weight
Amount
If one variable increases and
other variable also increases
We MAY say this is a case of direct variation
But how to confirm ?
If the ratio of the two quantities remain constant, this is a case of direct variation
DIRECT VARIATIONS AND DIRECT PROPERTION
DIRECT VARIATIONS AND DIRECT PROPERTION
‘y’ varies directly as ‘x’
Let us remove
‘’(alpha) sign
Whenever we have ‘k’, our first job is to remove ‘k’
In direct variation, the ratio is constant.
y
x
∴
y
x
=
k
Example :
🔿 The price of the scooter was 34000 last year. If it has increased by 20%
this year, what is the price now ?
Sol :
Original price of the scooter
=
Rs. 34000
Increase in %
=
20%
20% of original price
=
20% of
=
×
34000
=
×
340
20
=
6800
Here, increase % is given. Hence we will calculate the amount of increase
The amount is calculated as
New price
=
Original price
+
Increase
=
34000
+
6800
=
Rs 40,800
Let us take an example
34000
i.e original
price + increase
The original price of the scooter
Let us see how to calculate increase/decrease in%
EXERCISE 13.1
🔿 Following are the car parking charges near a railway station upto
Sol :
Check if the parking are in direct proportion to the parking time.
Therefore this is a case of direct variation
Lets check whether the given Quantities are direct variation.
If the parking time increases then the parking charges will also increase
Which are the two quantities that are varying ?
60
4
=
1
15
100
8
=
2
25
140
12
=
2
35
180
24
=
2
25
4 hours
8 hours
12 hours
Rs. 60
Rs. 100
Rs. 140
24 hours
Rs. 180
Therefore we will take ratio of the quantities
Let us reduce the ratio in the lowest form
15
1
25
2
35
3
25
2
Since
15
1
≠
25
2
≠
35
3
≠
25
2
∴
The parking charges are not in direct proportion with the parking time.
EXERCISE 13.1
Sol :
Parts of red pigment
Parts of base
1
8
4
….
7
….
12
….
20
….
x1
x2
x3
x4
x5
y1
y2
y3
y4
y5
1
8
Let the red pigment be represented by x1, x2, x3,..
And parts of base be denoted by y1, y2, y3,,..
As the part of red pigment increase, the required number of bases will also increase
It is a case of direct proportion.
i.e.
=
x2
y2
=
x3
y3
….
For
x1
=
1,
y1
=
8
∴
x1
y1
=
1
8
x1
y1
Which are the two quantities that are varying ?
Therefore we will take the ratio of the quantities
🔿 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
EXERCISE 13.1
Sol :
Parts of red pigment
Parts of base
1
8
4
….
7
….
12
….
20
….
🔿 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
4
32
Now,
=
1
8
x2
y2
∴
=
1
8
4
y2
∴
y2
=
∴
y2
=
32
EXERCISE 13.1
Sol :
Parts of red pigment
Parts of base
1
8
4
….
7
….
12
….
20
….
🔿 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
32
7
56
Now,
=
1
8
x3
y3
∴
=
1
8
7
y3
∴
y3
=
∴
y3
=
56
EXERCISE 13.1
Sol :
Parts of red pigment
Parts of base
1
8
4
….
7
….
12
….
20
….
🔿 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
32
12
96
Now,
=
1
8
x4
y4
∴
=
1
8
12
y4
∴
y4
=
∴
y4
=
96
56
EXERCISE 13.1
Sol :
Parts of red pigment
Parts of base
1
8
4
….
7
….
12
….
20
….
🔿 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
32
20
160
Now,
=
1
8
x4
y4
∴
=
1
8
20
y4
∴
y4
=
∴
y4
=
160
56
96
∴
Thus the required part of base are 8, 32, 56, 96, 160.
EXERCISE 13.1
Sol :
Therefore we will take the ratio of the quantities
Therefore this is a case of direct variation
Which are the two quantities that are varying ?
Parts of red pigment
Parts of base
And
We have,
=
=
x2
y2
Here,
=
1
x1
and
=
75
y1
=
1
75
x2
1800
∴
x2
=
1800
75
360
15
72
3
24
1
∴
Thus, the required red pigment = 24 parts.
1
x1
75
y1
🔿 In question 2 above, if a part of a red pigment requires 75mL of base, how � much red pigment we mix with 1800 ml of base?
EXERCISE 13.1
Sol :
Which are the two quantities that are varying ?
Number of bottles filled
Number of hours
And
Number of bottles filled
Number of hours
6
840
5
x
For more number of hours, more number of bottles would be filled.
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
840
x
=
6
5
∴
6x
=
5
840
∴
x
=
5
840
6
1
140
∴
x
=
700
∴
Thus, the required number of bottles = 700
Lets assume it as x
🔿 A machine in a soft drink factory fills 840 bottles in six hours. � How many bottles will it fill in five hours?
EXERCISE 13.1
Sol :
Which are the two quantities that are varying ?
Length
Enlarged length
And
Length
Enlarged length
50000
5
20000
x
Lets assume it as x
Actual length of bacteria
Here length and enlarged length of bacteria are indirect proportion
5
50000
=
5
10000
cm
=
=
10–4 cm
5
50000
=
x
20000
∴
50000x
=
5
20000
∴
x
=
5
20000
50000
10000
∴
x
=
2
∴
Hence enlarged length of bacteria is 2 cm.
1
2
1
🔿 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?
EXERCISE 13.1
Sol :
Which are the two quantities that are varying ?
Length of the ship
Height of the mast
And
Length of the ship
Height of the mast
12
28
9
x
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
Lets assume it as x
Since, more the length of the ship, More would be the length of its mast.
28
x
=
12
9
∴
12x
=
28
9
∴
x
=
28
9
12
4
3
∴
x
=
21
∴
Thus, the required length of the model = 21 cm
1
7
🔿 In model of ship, the mast is 9 cm high, while the mast of actual ship is 12 cm high, � if length of ship is 28 cm, how long is the model ship ?
EXERCISE 13.1
Sol :
Number of bottles filled
Number of hours
9 × 106
2
5
For more number of hours, more number of bottles would be filled.
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
2
5
=
9 × 106
x
∴
2x
=
5
9 × 106
∴
x
=
5
9 × 106
2
∴
x
=
∴
Thus, the required length of sugar crystal = 22.5 × 106
Lets assume it as x
Which are the two quantities that are varying ?
Weight of sugar
Number of crystal
And
x
∴
x
=
45
106
2
1
22.5
🔿 Suppose 2 kg of sugar contains 9 × 106 crystal. How many sugar crystals � are there in (i) 5 kg of sugar : (ii) 1.2 kg of sugar ?
EXERCISE 13.1
Sol :
Number of bottles filled
Number of hours
9 × 106
2
1.2
For more number of hours, more number of bottles would be filled.
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
2
1.2
=
9 × 106
y
∴
2y
=
1.2
9 × 106
∴
y
=
1.2
9 × 106
2
∴
y
=
∴
Thus, the required length of sugar crystal = 5.4 × 106
Lets assume it as y
y
∴
y
=
10.8
106
2
1
5.4
🔿 Suppose 2 kg of sugar contains 9 × 106 crystal. How many sugar crystals � are there in (i) 5 kg of sugar : (ii) 1.2 kg of sugar ?
EXERCISE 13.1
Sol :
Actual distance in km
Distance covered
in map (in cm)
1
18
72
Here actual distance and distance covered in the map are in direct proportion.
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
18
1
=
72
x
∴
x
=
72
1
∴
x
=
72
1
18
∴
∴
x
=
4
Here distance covered in the map is 4
Lets assume it as x
Which are the two quantities that are varying ?
Actual distance in km
Distance covered in map (in cm)
And
x
2
24
18
1
4
🔿 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?
EXERCISE 13.1
Sol :
Which are the two quantities that are varying ?
Actual distance in km
Distance covered in map (in cm)
And
Height of a pole
Length of a shadow
5m 60cm = 560
∴
1m
=
100cm
∴
5m
=
500cm
∴
5m 60cm
=
500 + 60
∴
5m 60cm
=
560
3m 20cm = 320
∴
1m
=
100cm
∴
3m
=
300cm
∴
3m 20cm
=
300 + 20
∴
3m 20cm
=
320
10m 60cm = 1050
∴
1m
=
100cm
∴
10m
=
1000cm
∴
10m 50cm
=
1000 + 50
∴
10m 50cm
=
1050
x cm
Lets assume it as x
As the highest of the pole increases the length of its shadow also increases in the same ratio.
Therefore this is a case of direct variation
Therefore we will take the ratio of the quantities
560
1050
=
320
x
∴
x
=
1050
320
∴
x
=
320
1050
560
∴
x
=
4
560
150
∴
x
=
600
∴
Thus, the required length of the shadow is 600 cm of 6m
14
8
7
4
1
150
🔿 A 5m 60cm high vertical pole casts a shadow 3m 20cm long. Find at the same time � (i) the length of the shadow cast by another pole 10m 50cm high and � (ii) the height of a pole which casts a shadow 5m long.
EXERCISE 13.1
Sol :
Height of a pole
Length of a shadow
5m 60cm = 560
3m 20cm = 320
5m = 500
y cm
Lets assume it as y
As the highest of the pole increases the length of its shadow also increases in the same ratio.
560
y
=
320
500
∴
y
=
560
500
∴
x
=
560
500
320
∴
x
=
125
320
7
∴
x
=
875
∴
Thus, the required height of a pole is 875cm or 8m 75 cm
🔿 A 5m 60cm high vertical pole casts a shadow 3m 20cm long. Find at the same time � (i) the length of the shadow cast by another pole 10m 50cm high and � (ii) the height of a pole which casts a shadow 5m long.
4
7
1
125
EXERCISE 13.1
Sol :
Distance (in km)
Time (in minutes)
14
25
x
300
Let distance covered in 5 hours be x km.
Here distance covered and time in direct proportion
5 hours = 5 × 60 = 300 minutes
1 hour = 60 minutes
∴
14
25
x
300
=
∴
x
25
=
14
300
∴
x
=
14
300
25
x
=
168 cm
∴
Hence, the distance covered in 5 hours is 168 km.
🔿 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, � how far can it travel in 5 hours?
5
60
1
12
∴
SPEED
TIME TAKEN
40 km/hr
4 hours
SPEED
TIME TAKEN
40 km/hr
4 hours
80 km/hr
2 hours
SPEED
INVERSE VARIATION
OR
INVERSE PROPORTION
If one variable increases and other variable decreases.
TIME
SPEED
TIME TAKEN
PRODUCT
40 km/hr
4 hours
80 km/hr
2 hours
40 × 4 = 160
80 × 2 = 160
But how to confirm ?
If the product of the two quantities remain constant, this is a case of inverse variation
We MAY say this is a case of inverse variation
= k
... [where k is a non-zero
constant of variation]
Whenever we have ‘k’, our first job is to remove ‘k’
In inverse variation, the product is constant.
‘x’ varies inversely as ‘y’
INVERSE VARIATION
OR
INVERSE PROPORTION