?In-text Question — Convert 60,00,000 cm to kilometres. It is 60 km. Verify this.
1 m \(=100\) cm and 1 km \(=1000\) m.
So 1 km \(=1000\times100\) cm \(=1{,}00{,}000\) cm.
60,00,000 cm \(=\dfrac{60{,}00{,}000}{1{,}00{,}000}\) km
\(=60\) km
?In-text Question — Using the map above, can you find the geographical distance between Bengaluru and Chennai? Also, find the geographical distance between Mangaluru and Chennai. [Hint: Use a ruler to find the distance between the cities on the map. Then, use the ratio given on the map to find the actual geographical distance.]
RF \(=1:60{,}00{,}000\).
So 1 cm on the map \(=60{,}00{,}000\) cm on the ground \(=60\) km.
Bengaluru to Chennai
Distance on the map, measured with a ruler \(\approx2.4\) cm
Actual distance \(=2.4\times60\) km
\(=144\) km
Mangaluru to Chennai
Distance on the map, measured with a ruler \(\approx4.9\) cm
Actual distance \(=4.9\times60\) km
\(=294\) km
Mangaluru to Chennai ≈ 294 km
?In-text Question — Try to find the distances between the same two pairs of cities with different maps that have different scales (ratios). Do they all give the same geographical distance, approximately?
Take any other map, for example one whose RF is \(1:30{,}00{,}000\).
So 1 cm on this map \(=30{,}00{,}000\) cm \(=30\) km.
On this map, the cities are twice as far apart on paper.
Suppose the Bengaluru–Chennai distance on this map measures \(d\) cm.
Actual distance \(=d\times30\) km
If the same distance were measured on the 1:60,00,000 map, it would be \(\dfrac d2\) cm.
Actual distance \(=\dfrac d2\times60\) km \(=d\times30\) km
The two answers are equal. The same works for Mangaluru–Chennai.
The distance on paper changes with the scale, but (distance on the map) \(\times\) (what 1 cm stands for) gives the same actual distance.
AActivity — Guide students to make a sketch of their classroom with an accurate scale (ratio of 1: 50). They should mark the location of various objects in the classroom like the teacher’s desk, blackboard, fans and lights, according to scale. Students can use appropriate symbols to represent different objects like fans, lights, tables, chairs, and so on.
The scale is \(1:50\).
So 1 cm on the sketch \(=50\) cm in the classroom.
Length on the sketch \(=\) actual length \(\div\,50\).
Steps
Step 1: Measure each length in the classroom in cm.
Step 2: Divide each length by 50.
Step 3: Draw the lengths with a ruler and mark the objects with symbols.
Example (suppose the classroom is 6 m by 5 m)
| Object | Actual length | Length on the sketch |
|---|---|---|
| Classroom length | 600 cm | \(600\div50=12\) cm |
| Classroom width | 500 cm | \(500\div50=10\) cm |
| Blackboard | 300 cm | \(300\div50=6\) cm |
| Teacher's desk | 120 cm by 60 cm | \(2.4\) cm by \(1.2\) cm |
Figure it Out — Set 1
Five questions from page 60 of the textbook.
1A cricket coach schedules practice sessions that include different activities in a specific ratio — time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5. If each session is 150 minutes long, how much time is spent on each activity?
Total parts \(=3+4+3+5=15\)
Value of 1 part \(=150\div15=10\) minutes
Warm-up/cool-down \(=3\times10=30\) minutes
Batting \(=4\times10=40\) minutes
Bowling \(=3\times10=30\) minutes
Fielding \(=5\times10=50\) minutes
Check: \(30+40+30+50=150\) minutes
Batting: 40 minutes
Bowling: 30 minutes
Fielding: 50 minutes
2A school library has books in different languages in the following ratio — no. of Odiya books : no. of Hindi books : no. of English books :: 3 : 2 : 1. If the library has 288 Odiya books, how many Hindi and English books does it have?
Odiya books are 3 parts.
3 parts \(=288\)
Value of 1 part \(=288\div3=96\)
Hindi books \(=2\times96=192\)
English books \(=1\times96=96\)
English books: 96
3I have 100 coins in the ratio — no. of ₹10 coins : no. of ₹5 coins : no. of ₹2 coins : no. of ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in coins?
Total parts \(=4+3+2+1=10\)
Value of 1 part \(=100\div10=10\) coins
Number of ₹10 coins \(=4\times10=40\)
Number of ₹5 coins \(=3\times10=30\)
Number of ₹2 coins \(=2\times10=20\)
Number of ₹1 coins \(=1\times10=10\)
Money in ₹10 coins \(=40\times10=\) ₹400
Money in ₹5 coins \(=30\times5=\) ₹150
Money in ₹2 coins \(=20\times2=\) ₹40
Money in ₹1 coins \(=10\times1=\) ₹10
Total money \(=400+150+40+10=\) ₹600
4Math Talk — Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?
Take 1 unit \(=1\) cm.
The sides are 3 cm, 4 cm and 5 cm.
Construction
Step 1: Draw BC \(=4\) cm.
Step 2: With B as centre and radius 3 cm, draw an arc.
Step 3: With C as centre and radius 5 cm, draw an arc cutting the first arc at A.
Step 4: Join AB and AC.
Are all such triangles congruent?
Take 1 unit \(=2\) cm instead.
The sides are 6 cm, 8 cm and 10 cm, which are also in the ratio 3 : 4 : 5.
This triangle has the same shape as the first, but it is bigger.
Two triangles are congruent only when their corresponding sides are equal.
Sides \(3k, 4k, 5k\) are equal to \(3m, 4m, 5m\) only when \(k=m\).
5Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?
Let the sides be \(k, 3k, 5k\).
In a triangle, the sum of any two sides must be greater than the third side.
The two shorter sides: \(k+3k=4k\)
The longest side: \(5k\)
\(4k<5k\), so the two shorter sides together are shorter than the longest side.
Figure it Out — Set 2
Three questions from pages 62–63 of the textbook.
1A group of 360 people were asked to vote for their favourite season from the three seasons — rainy, winter and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information.
Number who liked winter \(=360-90-120=150\)
Total people \(=360\) and total angle \(=360^\circ\).
So 1 person is shown by \(1^\circ\).
Summer \(=\dfrac{90}{360}\times360^\circ=90^\circ\)
Rainy \(=\dfrac{120}{360}\times360^\circ=120^\circ\)
Winter \(=\dfrac{150}{360}\times360^\circ=150^\circ\)
Check: \(90^\circ+120^\circ+150^\circ=360^\circ\)
Construction
Step 1: Draw a circle and mark a radius OA.
Step 2: Measure \(90^\circ\) from OA anticlockwise and draw the radius OB.
The slice AOB is summer.
Step 3: Measure \(120^\circ\) from OB and draw the radius OC.
The slice BOC is rainy.
Step 4: The remaining slice COA (\(150^\circ\)) is winter.
Rainy: 120°
Winter: 150°
2Draw a pie chart based on the following information about viewers’ favourite type of TV channel: Entertainment — 50%, Sports — 25%, News — 15%, Information — 10%.
The whole circle \(=360^\circ\) stands for \(100\%\).
So \(1\%\) of the viewers \(=\dfrac{360^\circ}{100}=3.6^\circ\).
Entertainment \(=50\times3.6^\circ=180^\circ\)
Sports \(=25\times3.6^\circ=90^\circ\)
News \(=15\times3.6^\circ=54^\circ\)
Information \(=10\times3.6^\circ=36^\circ\)
Check: \(180^\circ+90^\circ+54^\circ+36^\circ=360^\circ\)
Construction
Step 1: Draw a circle with centre O and mark a radius OA.
Step 2: With a protractor at O, measure \(180^\circ\) from OA. This sector is Entertainment.
Step 3: From the end of that sector, measure \(90^\circ\). This sector is Sports.
Step 4: From there, measure \(54^\circ\). This sector is News.
Step 5: The remaining sector is \(36^\circ\). This is Information.
Step 6: Label each sector with its name and percentage.
Sports: 90°
News: 54°
Information: 36°
3Prepare a pie chart that shows the favourite subjects of the students in your class. You can collect the data of the number of students for each subject shown in the table (each student should choose only one subject). Then write these numbers in the table and construct a pie chart:Subject Language Arts Education Vocational Education Social Science Physical Education Maths Science Number of Students
| Subject | Language | Arts Education | Vocational Education | Social Science | Physical Education | Maths | Science |
|---|---|---|---|---|---|---|---|
| Number of Students |
This question uses the data of your own class.
The method is shown below with sample data for a class of 40 students.
Method
Step 1: Ask each student for one favourite subject and write the numbers in the table.
Step 2: Add the numbers to get the total number of students.
Step 3: Angle for a subject \(=\dfrac{\text{number of students}}{\text{total students}}\times360^\circ\)
Step 4: Draw the slices one after the other with a protractor.
Sample data (total \(=40\) students)
1 student is shown by \(\dfrac{360^\circ}{40}=9^\circ\).
Angle for a subject \(=\) number of students \(\times\,9^\circ\)
| Subject | Language | Arts Education | Vocational Education | Social Science | Physical Education | Maths | Science |
|---|---|---|---|---|---|---|---|
| Number of Students | 6 | 3 | 2 | 5 | 4 | 8 | 12 |
| Angle | \(54^\circ\) | \(27^\circ\) | \(18^\circ\) | \(45^\circ\) | \(36^\circ\) | \(72^\circ\) | \(108^\circ\) |
Check: \(54+27+18+45+36+72+108=360\)
MTMath Talk — Puneeth’s father went from Lucknow to Kanpur in 3 hours by riding his motorcycle at a speed of 30 km/h. If he takes a car instead and drives at 60 km/h, how long will it take him to reach Kanpur? Can we represent this problem with the following statement of proportionality — 30 : 60 :: 3 : x ? Will the travel time increase or decrease as the speed of the motorcycle increases?
The speed increases from 30 km/h to 60 km/h, which is 2 times.
If \(30:60::3:x\) were true, then \(x=\dfrac{60\times3}{30}=6\) hours.
That would mean the time becomes 2 times as long when the speed becomes 2 times.
But a faster vehicle takes less time to cover the same distance.
So \(30:60::3:x\) is not correct.
Distance from Lucknow to Kanpur \(=30\times3=90\) km
Time by car \(=\dfrac{90}{60}=1.5\) hours
| Walk | Bicycle | Motorcycle | Car | |
|---|---|---|---|---|
| Speed (km/h) | 5 | 15 | 30 | 60 |
| Time (in hours) | 18 | 6 | 3 | 1.5 |
In the table, as the speed increases, the time decreases.
?In-text Question — Does it decrease by the same rate (or factor)? Going by bicycle is 3 times faster than walking (15 ÷ 5). The speed has increased 3 times. The travel time has decreased 3 times too (18 ÷ 6). The speed has increased by the same factor by which the travel time has decreased. Check if this is the case for the other modes of transport.
Walking: speed \(=5\) km/h and time \(=18\) hours.
Motorcycle
Speed increases by \(30\div5=6\) times.
Time decreases by \(18\div3=6\) times.
Car
Speed increases by \(60\div5=12\) times.
Time decreases by \(18\div1.5=12\) times.
In each case the speed increases by the same factor by which the time decreases.
Product of speed and time: \(5\times18=90\), \(15\times6=90\), \(30\times3=90\), \(60\times1.5=90\).
Figure it Out — Set 3
Two questions from page 65 of the textbook.
1Which of these are in inverse proportion?
(i)x 40 80 25 16 y 20 10 32 50
(ii)x 40 80 25 16 y 20 10 12.5 8
(iii)x 30 90 150 10 y 15 5 3 45
(i)
| x | 40 | 80 | 25 | 16 |
| y | 20 | 10 | 32 | 50 |
(ii)
| x | 40 | 80 | 25 | 16 |
| y | 20 | 10 | 12.5 | 8 |
(iii)
| x | 30 | 90 | 150 | 10 |
| y | 15 | 5 | 3 | 45 |
x and y are in inverse proportion if \(xy\) is the same for every pair.
(i)
\(40\times20=800\), \(80\times10=800\), \(25\times32=800\), \(16\times50=800\)
\(xy=800\) every time, so x and y are in inverse proportion.
(ii)
\(40\times20=800\), \(80\times10=800\), \(25\times12.5=312.5\), \(16\times8=128\)
\(xy\) is not the same every time, so x and y are not in inverse proportion.
(iii)
\(30\times15=450\), \(90\times5=450\), \(150\times3=450\), \(10\times45=450\)
\(xy=450\) every time, so x and y are in inverse proportion.
(ii) Not in inverse proportion
(iii) Inverse proportion (xy = 450)
2Fill in the empty cells if x and y are in inverse proportion.x 16 12 36 y 9 48
| x | 16 | 12 | 36 | |
| y | 9 | 48 |
x and y are in inverse proportion, so \(xy=k\) is constant.
From the first column, \(k=16\times9=144\).
When \(x=12\): \(12\times y=144\), so \(y=\dfrac{144}{12}=12\).
When \(y=48\): \(x\times48=144\), so \(x=\dfrac{144}{48}=3\).
When \(x=36\): \(36\times y=144\), so \(y=\dfrac{144}{36}=4\).
| x | 16 | 12 | 3 | 36 |
| y | 9 | 12 | 48 | 4 |
MTMath Talk — If Ram takes 1 hour to cut a given quantity of vegetables and Shyam takes 1.5 hours to cut the same quantity of vegetables, how much time will they take to cut the vegetables if they do it together? How much time will it take them to complete 1 unit of work? Is the quantity of work and time taken to complete it directly or inversely proportional?
Take the work of cutting the vegetables as 1 unit.
Ram does 1 unit in 1 hour.
So in 1 hour Ram does 1 unit of work.
Shyam does 1 unit in 1.5 hours.
So in 1 hour Shyam does \(\dfrac{1}{1.5}=\dfrac23\) unit of work.
Work done by both in 1 hour \(=1+\dfrac23=\dfrac53\) units
More work needs more time at the same rate, so the quantity of work and the time taken are directly proportional.
\(\dfrac53\) units of work take 1 hour, and 1 unit of work takes \(x\) hours.
\(\dfrac53:1::1:x\)
\(\dfrac53\times x=1\times1\)
\(x=\dfrac{1}{5/3}=\dfrac35\)
Figure it Out — Set 4
Twelve questions from pages 67–68 of the textbook.
1Which of the following pairs of quantities are in inverse proportion?
(i) The number of taps filling a water tank and the time taken to fill it.
(ii) The number of painters hired and the days needed to paint a wall of fixed size.
(iii) The distance a car can travel and the amount of petrol in the tank.
(iv) The speed of a cyclist and the time taken to cover a fixed route.
(v) The length of cloth bought and the price paid at a fixed rate per metre.
(vi) The number of pages in a book and the time required to read it at a fixed reading speed.
(i) More taps fill the tank in less time.
So they are in inverse proportion.
(ii) More painters finish the wall in fewer days.
So they are in inverse proportion.
(iii) More petrol lets the car travel a greater distance.
Both increase together, so this is direct proportion.
(iv) A faster cyclist takes less time for the same route.
So they are in inverse proportion.
(v) More cloth costs more money.
Both increase together, so this is direct proportion.
(vi) More pages need more time to read.
Both increase together, so this is direct proportion.
(ii) Inverse proportion
(iii) Not inverse (direct proportion)
(iv) Inverse proportion
(v) Not inverse (direct proportion)
(vi) Not inverse (direct proportion)
2If 24 pencils cost ₹120, how much will 20 such pencils cost?
More pencils cost more money, so this is direct proportion.
Cost of 24 pencils \(=\) ₹120
Cost of 1 pencil \(=\dfrac{120}{24}=\) ₹5
Cost of 20 pencils \(=20\times5=\) ₹100
3Math Talk — A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem?
More families use the same water in fewer days, so this is inverse proportion.
Number of families now \(=20+10=30\)
\(20\times6=30\times x\)
\(x=\dfrac{20\times6}{30}=\dfrac{120}{30}=4\)
Assumptions
Every family uses the same amount of water each day.
The water in the tank is the only supply and no water is added to it.
The daily use of each family does not change after the new families move in.
4Fill in the average number of hours each living being sleeps in a day by looking at the charts. Select the appropriate hours from this list : 15, 2.5, 20, 8, 3.5, 13, 10.5, 18.
The whole ring stands for 24 hours.
The blue part shows the time asleep.
A bigger blue part means more hours of sleep.
Order of the blue parts, smallest to largest: giraffe, elephant, human, dog, cat, squirrel, python, bat.
Order of the given hours, smallest to largest: 2.5, 3.5, 8, 10.5, 13, 15, 18, 20.
Matching them in this order gives the answer.
The blue shares below are read from the picture, so they are estimates.
Giraffe: blue \(\approx\dfrac1{10}\) of the ring, and \(24\times\dfrac1{10}\approx2.5\) hours.
Elephant: blue \(\approx\dfrac17\) of the ring, and \(24\times\dfrac17\approx3.5\) hours.
Dog: blue \(\approx0.44\) of the ring, and \(24\times0.44\approx10.5\) hours.
Cat: blue \(\approx0.54\) of the ring, and \(24\times0.54\approx13\) hours.
Check
Human: the blue part is \(\dfrac13\) of the ring, and \(24\times\dfrac13=8\) hours.
Squirrel: the blue part is \(\dfrac58\) of the ring, and \(24\times\dfrac58=15\) hours.
Python: the blue part is \(\dfrac34\) of the ring, and \(24\times\dfrac34=18\) hours.
Bat: the blue part is \(\dfrac56\) of the ring, and \(24\times\dfrac56=20\) hours.
Elephant: 3.5 hours
Human: 8 hours
Dog: 10.5 hours
Cat: 13 hours
Squirrel: 15 hours
Python: 18 hours
Bat: 20 hours
5The pie chart on the right shows the result of a survey carried out to find the modes of transport used by children to go to school. Study the pie chart and answer the following questions.
(i) What is the most common mode of transport?
(ii) What fraction of children travel by car?
(iii) If 18 children travel by car, how many children took part in the survey? How many children use taxis to travel to school?
(iv) By which two modes of transport are equal numbers of children travelling?
Angle for Car \(=360^\circ-(90^\circ+120^\circ+60^\circ+60^\circ)\)
\(=360^\circ-330^\circ=30^\circ\)
(i)
The largest slice is Bus (\(120^\circ\)).
(ii)
Fraction by car \(=\dfrac{30^\circ}{360^\circ}=\dfrac1{12}\)
(iii)
\(30^\circ\) stands for 18 children.
\(1^\circ\) stands for \(\dfrac{18}{30}=0.6\) child.
\(360^\circ\) stands for \(360\times0.6=216\) children.
There is no slice for taxis in the pie chart, so no child uses a taxi.
(iv)
Cycle and Two-wheeler both have \(60^\circ\).
(ii) 1/12
(iii) 216 children took part; 0 children use taxis
(iv) Cycle and Two-wheeler
6Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make?
More workers finish the work in fewer days, so this is inverse proportion.
Number of workers now \(=3+1=4\)
\(3\times4=4\times x\)
\(x=\dfrac{3\times4}{4}=3\)
Assumptions
All the workers paint at the same speed.
All of them work the same number of hours each day.
The workers do not get in each other’s way.
7It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump?
More tanks need more time with the same pump, so this is direct proportion.
Time for 2 tanks \(=6\) hours
Time for 1 tank \(=6\div2=3\) hours
Time for 5 tanks \(=5\times3=15\) hours
8A given set of chairs are arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have?
The number of chairs stays the same.
More chairs in each row means fewer rows, so this is inverse proportion.
\(25\times12=x\times20\)
\(300=20x\)
\(x=\dfrac{300}{20}=15\)
9A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same?
The total school time stays the same.
More periods means each period is shorter, so this is inverse proportion.
\(8\times45=9\times x\)
\(360=9x\)
\(x=\dfrac{360}{9}=40\)
10A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill?
Take filling the whole tank as 1 unit of work.
The small pump fills the tank in 3 hours.
So in 1 hour it fills \(\dfrac13\) of the tank.
The large pump fills the tank in 2 hours.
So in 1 hour it fills \(\dfrac12\) of the tank.
Work done by both in 1 hour \(=\dfrac13+\dfrac12=\dfrac26+\dfrac36=\dfrac56\) of the tank
\(\dfrac56:1::1:x\)
\(\dfrac56\times x=1\)
\(x=\dfrac65=1\dfrac15\) hours
11A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days?
Fewer days need more machines, so this is inverse proportion.
\(42\times63=x\times54\)
\(2646=54x\)
\(x=\dfrac{2646}{54}=49\)
12A car takes 2 hours to reach a destination, travelling at a speed of 60 km/h. How long will the car take if it travels at a speed of 80 km/h?
The distance is the same.
A greater speed takes less time, so this is inverse proportion.
\(60\times2=80\times x\)
\(120=80x\)
\(x=\dfrac{120}{80}=1.5\)
Continue with Part 2, Chapter 4
Move on to the next Part 2 chapter next.