If two quantities are directly proportional then their quotient is an invariant(it does not change, it is constant). Further, if we have another pair of two quantities that are directly proportional then their quotient is also an invariant. Moreover, if you multiple these two quotients, you end up with a third invariant. But, consider the following example.
Given that x is directly proportional to y and to z and is inversely propotional to w, and that x=4 when (w,y,z)=(6,8,5), what is x when (w,y,z)=(4,10,9)?
xw is a contant, x/z is a constant, and x/y is a constant.
Thus, xw/yz is constant. (They just combined the constant terms)
so, xw/yz=(4)(6)/(8)(5)=3/5.
Thus, when (w, y, z)=(4,10,9), we find x=3yz / 5w=27/2.
question 1. Jane ran 100 meters in 15 seconds. How long did she take to run 1 meter?
Solution: Step 1: Think of the word problem as:
If 100 then 15. If 1 then how many?
Step 2: Write the proportional relationship:
100 → 15
She took 0.15 seconds
question 2. If of a tank can be filled in 2 minutes, how many minutes will it take to fill the whole tank?
Solution: Step 1: Think of the word problem as:
If then 2. If 1 then how many? (Whole tank is )
Step 2: Write the proportional relationship:
It took 3.5 minutes
question 3. A car travels 125 miles in 3 hours. How far would it travel in 5 hours?
Solution:Step 1: Think of the word problem as:
If 3 then 125. If 5 then how many?
Step 2: Write the proportional relationship:
3 → 125
He traveled miles.
e. g. Let the four quantities 3, 4, 9 and 12 be in proportion.
or 3 : 4 :: 9 : 12.
These quantities of the same kind are said to be in ‘continued’ proportion when the ratio of the first to the second is equal to the ratio of the second to the third.
The second quantity is called the mean proportional between the first and the third, and the third quantity is called the third proportional to the first and second.
Then 9, 6 and 4 are in continued proportion for 9 : 6 :: 6 : 4.
Hence, 6 is the mean proportional between 9 and 4, and 4 is the third proportional to 9 and 6.
Mean proportion
a : b :: b : c ⇒ b2 = ac
question 4. Find the fourth proportional to the number 6, 8 and 15.
Solution: Let x be fourth proportional
∴ 6 : 8 :: 15 : x
∴ 6 × x = 8 × 15
∴
question 5. Find the third proportional to 15 and 20.
Solution: Here, we have to find a fourth proportional to 15, 20 and 20.
( it is continued proportion).
Let x be the fourth proportional.
We have 15 : 20 :: 20 : x
15 × x = 20 × 20
∴ x =
question 6. Find the mean proportional between 3 and 75.
Solution: Let x be the required mean proportion.
3 : x :: x : 75
∴ x2 = 3 × 75
x2 = 225
∴
x = 15
Rule: Two quantities a and b are in direct variation, the ratio of any two values of a is equal to the ratio of the corresponding values of b.
∴ a1b2 = a2b1 ⇒
∴ a1 : a2 :: b1 : b2
question 1. If one score oranges cost Rs. 45, how many oranges can be brought for Rs. 72?
Solution: For Rs. 45, number of oranges brought = 20
For Rs. 1 number of oranges brought =
For Rs. 72 number of oranges brought =
Number of oranges = 32
question 2. A taxi charges a fare of Rs. 260 for a journey of 200 km. How much would it travel for Rs.279.50.
Solution: But taxi travels x km for Rs. 279.50.
Fare (Rs.) |
260 |
279.50 |
Distance (km) |
200 |
x |
∴ Ratio of Rs = Ratio of distance
⇒ x = 215
The taxi will come 215 km
question 3. If 3 men and 4 women earn Rs. 480 in a day. Find how much will 7 men and 11 women earn in a day?
Solution: Let men will earn x Rs.
Women will earn y Rs.
For men
Men |
3 |
7 |
Rs. |
480 |
x |
x = 1120
Men will earn Rs. 1120 (1)
For women
Women |
4 |
11 |
Rs. |
480 |
y |
⇒
∴
y = 1320
Women will earn Rs. 1320.
Total money for 7 men and 11 women in day
= 1120 + 1320 = 2440
i.e. Rs. 2440
question 4. If the wages of 15 workers for 6 days are Rs. 9450. Find the wages of 19 workers for 5 days.
Solution:Let the wages be x.
First find relation between wages and workers.
Workers |
15 |
19 |
Wages |
9450 |
x |
∴
∴
x = Rs. 11970
Now ratio between days wages
Days |
6 |
5 |
Wages |
11970 |
y |
⇒
⇒
y = Rs. 9975
Hence the wages of 19 workers for 5 day = Rs. 9975.
5. The cost of 16 packets of salt, each weighing 900 grams is Rs. 84. What will be the cost of 27 packets of salt, each weighing 1 kg?
Solution: Cost of 16 packets each weighing kg = Rs. 84
Cost of 16 packets, each weighing 1 kg = Rs.
Cost of 1 packet, each weighing 1 kg = Rs.
Cost of 27 packets, each weighing 1 kg = Rs.
= Rs. = Rs. 157.50
Hence the cost of 27 packets, each weighing 1 kg is Rs. 157.50
Inversely Proportional questions are similar to directly proportional problems, but the difference is that when x increase y will decrease and vice versa – which is the inverse proportion relationship. The most common example of inverse proportion problems would be “the more men on a job the less time taken for the job to complete”
Again, the technique is to change the proportion problems into the form:
If x then y. If x is changed to a then what will be the value of y?
and then write the inverse relationship (take note of the "inverse" form):
x → y
(i) The time taken to finish a piece of work varies inversely as the number of men at work.
(ii) The speed varies inversely as the time taken to cover a distance.
Rule: If two quantities a and b vary inversely as each other, and b1, b2 are the values if b corresponding to the values a1, a2 of a respectively, then
a1b1 = k and a2b2 = k
∴ a1b1 = a2b2
⇒ a1 : a2 = b1 : b2 or a1 : a2 :: b1 : b2
class 8 maths NCERT solutions Chapter 1: Rational Numbers
class 8 maths NCERT solutions Chapter 2: Linear Equations in One Variable
class 8 maths NCERT solutions Chapter 3: Understanding Quadrilaterals
class 8 maths NCERT solutions Chapter 4: Practical Geometry
class 8 maths NCERT solutions Chapter 5: Data Handling
class 8 maths NCERT solutions Chapter 6: Square and Square Roots
class 8 maths NCERT solutions Chapter 7: Cube and Cube Roots
class 8 maths NCERT solutions Chapter 8: Comparing Quantities
NCERT Class 8 Maths solution Chapter 9: Algebraic Expressions and Identities
class 8 maths NCERT solutions Chapter 10: Visualizing Solid Shapes
class 8 maths NCERT solutions Chapter 11: Mensuration
class 8 maths NCERT solutions Chapter 12: Exponents and Powers
class 8 maths NCERT solutions Chapter 13: Direct and Inverse Proportions
class 8 maths NCERT solutions Chapter 14: Factorization
class 8 maths NCERT solutions Chapter 15: Introduction to Graphs
class 8 maths NCERT solutions Chapter 16: Playing with Numbers
Notes,worksheet and solved question for Maths class 8
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