﻿ Inverse and Joint Variation Worksheet | Problems & Solutions
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Inverse and Joint Variation Worksheet
• Page 1
1.
According to Boyle's law PV = constant. Which of the following ordered pair fail to satisfy the Boyle's law equation PV = 800? a. (540, 1.30) b. (500, 1.60) c. (200, 4) d. (400, 2)

#### Solution:

The ordered pair (540, 1.30) fails to satisfy the equation PV = 800
[540 × 1.30 = 702.]

Correct answer : (1)
2.
If $x$ and $y$ are varying so that their product is 162, which of the following is the graph of this variation? a. Ellipse b. Parabola c. Circle d. Rectangular hyperbola

#### Solution:

x and y are varying so that their product is 162, which is a constant.

The graph of xy = 162 is a rectangular hyperbola.
[As x increases then y decreases, and x decreases then y increases.]

Correct answer : (4)
3.
If the strength of a beam is inversely proportional to its length ($s$ $\propto$ $\frac{1}{l}$), then how much weight does a 10 ft long beam support ? [A beam of length 2 yard supports atmost 300 lb.] a. 180 lb b. 1790 lb c. 190 lb d. 1800 lb

#### Solution:

Strength 1length of the beam
[Inverse proportion.]

Strength × length = k
[k is constant of variation.]

s × l = k

300 lb × 2 yard = k
[Replace s = 300 lb and l = 2 yard.]

300 lb × 6 feet = k
[1 yard = 3 feet.]

k = 1800

s × l = k
[Step 3.]

s × 10 = 1800
[Replace k = 1800 and l = 10.]

s = 180 lb
[Divide both sides by 10.]

Correct answer : (1)
4.
The area of a rectangular plot is 5600 $m$2. Into how many sub-plots can it be divided such that each sub-plot is a square of 800 $m$2 area. a. 8 b. 28 c. 4800 d. 7

#### Solution:

The area of a rectangular plot is 5600 m2.

The area of a sub-plot square is 800 m2.

Number of sub-plots × area of each square plot = 5600

The number of sub-plots = 5600 / 800 = 7.

Correct answer : (4)
5.
A sprinter completes a 200 $m$ race in 5.5 seconds and wins a silver medal. If the winner of the gold medal completes the race in 0.5 seconds earlier than silver medallist, then find the average speed of gold medallist. a. 1000 m/s b. 1100 m/s c. 36.36 m/s d. 40 m/s

#### Solution:

Time taken by the gold medallist to complete the race is 5.5 - 0.5 = 5 seconds.

Average speed × time = distance
[Use the formula.]

Average speed = distance traveledtime taken

The average speed of gold medallist = 200 / 5 = 40 m/s.

Correct answer : (4)
6.
A sprinter completes a 100 $m$ race in 15.7 seconds and wins a silver medal. If the winner of the gold medal completes the race 0.7 seconds earlier than silver medallist, then what is the difference between the average speeds of the gold medallist and silver medallist? a. 0.6 m/s b. 6.6 m/s c. 0.3 m/s d. 136.5 m/s

#### Solution:

Average speed of silver medallist = distancetime taken = 100 / 15.7 = 6.3 m/s
[Simplify.]

Time taken by the gold medallist to complete the race is 15.7 - 0.7 = 15 seconds.

The average speed of gold medallist = distancetime taken = 100 / 15 = 6.6 m/s

Difference in their average speed = 6.6 - 6.3 = 0.3 m/s

Correct answer : (3)
7.
The intensity(I) of light varies inversely as the square of the distance($d$) between the light source and the observer. Suppose the intensity is measured as 25 units at a distance of 5 $m$ from the source of the light. What is the intensity at a distance of 2.5 $m$? a. 75 units b. 625 units c. 100 units d. 25 units

#### Solution:

I 1d2
[Inverse proortion.]

I d2 = k
[Constant of variation.]

k = 25 × (5)2 = 625
[Replace I with 25 and d with 5.]

Intensity of light at a distance of 2.5 m is I = kd2 = 6252.52 = 100 units.
[Replace k with 625 and d with 2.5.]

Correct answer : (3)
8.
The intensity(I) of light varies inversely as the square of the distance($d$) between the light source and the observer. Suppose the intensity is measured as 42 units at a distance of 2 $m$, at what distance would be the intensity of the light source, half of what it was? a. 1.4 $m$ b. 2.8 $m$ c. 8 $m$ d. 1.4 $m$

#### Solution:

I 1d2
[Intensity(I) of light varies inversely as the square of the distance(d).]

I d2 = k
[Constant of variation.]

k = 42 × (2)2 = 168
[Replace I with 42 and d with 2.]

d ² = kI = 168422 = 8
[Intensity is half of the original.]

d = 8 = 2.8
[Simplify.]

So, the distance between the light source and the observer when intensity of the light source is half of the original is 2.8 m.

Correct answer : (2)
9.
A graph shown below is plotted with pressure in atmosphere on $y$-axis and volume in $m$3 on $x$-axis. Which of the following holds well?  a. $\frac{P}{V}$ = constant b. PV = constant c. PV2 = $k$ d. P2 V3 = constant

#### Solution:

The graph shown is the graph of an inverse variation.

So, the pressure P and the volume V are inversely proportional to each other and hence PV = constant holds good.

Correct answer : (2)
10.
Gay-lussac's law states that at fixed volume, for a given amount of ideal gas the pressure(P) is directly proportional to the absolute temperature(T). Which of the following curves represents the relationship between P and T.  a. Graph B b. Graph C c. Graph A

#### Solution:

P T
[Direct proportion.]

P / T = constant (k)

P = k T which is in the form y = m x and hence it represents a line passing through origin as shown. Correct answer : (1)

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