Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 3 Ratio and Proportion Exercise 3.5

Miscellaneous Practice Problems

 

Question 1. The maximum speed of some of the animals are given below: the Elephant = 20 km/h; the Lion = 80 km/h; the Cheetah = 100 km/h. Find the following ratios of their speeds in simplified form and find which ratio is the least?
(i) the Elephant and the Lion
(ii) the Lion and the Cheetah
(iii) the Elephant and the Cheetah
Answer:
(i) The Elephant : the Lion
\( = 20 : 80 = \frac{20}{80} = \frac{1}{4} = 1:4 \)
(ii) The Lion : the Cheetah
\( = 80 : 100 = \frac{80}{100} = \frac{4}{5} = 4 : 5 \)
(iii) The Elephant : the Cheetah
\( = 20 : 100 = \frac{20}{100} = \frac{1}{5} = 1 : 5 \)
Comparing the simplified ratios: \( 1:4 \), \( 4:5 \), and \( 1:5 \). The ratio of Elephant to Cheetah, which is \( 1:5 \), is the smallest among them. The speed of a cheetah is much higher than that of an elephant, making its ratio the least when compared to the elephant.
In simple words: We find the speed ratio for each pair of animals and simplify them. The smallest ratio is between the Elephant and the Cheetah.

🎯 Exam Tip: Always simplify ratios to their lowest terms to accurately compare them and identify the smallest or largest ratio.

 

Question 2. A particular high school has 1500 students, 50 teachers, and 5 administrators. If the school grows to 1800 students and the ratios are maintained, then find the number of teachers and administrators.
Answer: The initial ratio of Administrators : teachers : students is \( 5 : 50 : 1500 \).
We can simplify this ratio by dividing all numbers by 5: \( 1 : 10 : 300 \).
So, 1 part represents administrators, 10 parts represent teachers, and 300 parts represent students.
If the school now has 1800 students, then 300 parts \( = 1800 \).
To find the value of 1 part, we divide the total number of students by 300:
1 part \( = \frac{1800}{300} = 6 \).
Now we can find the number of teachers and administrators:
Number of administrators \( = 1 \) part \( = 6 \).
Number of teachers \( = 10 \) parts \( = 10 \times 6 = 60 \).
So, for 1800 students, there will be 6 administrators and 60 teachers, keeping the same proportion. Maintaining these ratios ensures a balanced school environment.
In simple words: First, simplify the ratio of administrators, teachers, and students. Then, use the new total number of students to find out how many administrators and teachers there should be to keep the same balance.

🎯 Exam Tip: Always simplify ratios first to make calculations easier when dealing with proportional increases or decreases.

 

Question 3. I have a box which has 3 green, 9 blue, 4 yellow, 8 orange coloured cubes in it.
(a) What is the ratio of orange to yellow cubes?
(b) What is the ratio of green to blue cubes?
(c) How many different ratios can be formed, when you compare each colour to any one of the other colours?
Answer:
Number of green cubes = 3
Number of blue cubes = 9
Number of yellow cubes = 4
Number of orange cubes = 8

(a) Ratio of orange to yellow cubes:
\( = \frac{\text{Number of orange cubes}}{\text{Number of yellow cubes}} = \frac{8}{4} = \frac{2}{1} = 2:1 \)

(b) Ratio of green to blue cubes:
\( = \frac{\text{Number of green cubes}}{\text{Number of blue cubes}} = \frac{3}{9} = \frac{1}{3} = 1:3 \)

(c) To find how many different ratios can be formed when comparing each color to any one of the other colors, we list all possible unique pairs of colors:
Orange : Yellow, Orange : Blue, Orange : Green
Yellow : Orange, Yellow : Blue, Yellow : Green
Blue : Green, Blue : Orange, Blue : Yellow
Green : Orange, Green : Yellow, Green : Blue
Each comparison involves a pair of colors, and since the order matters (Orange:Yellow is different from Yellow:Orange in its numerical ratio), we count each pair. There are 4 colors. For each color, it can be compared with the other 3 colors. So, \( 4 \times 3 = 12 \) different ratios can be formed. These comparisons help in understanding the relative quantities of different items.
In simple words: First, write down how many cubes of each color there are. Then, for parts (a) and (b), make a fraction of the two colors and simplify it. For part (c), count all the possible ways to compare one color to another color.

🎯 Exam Tip: When simplifying ratios, always ensure you divide both numbers by their greatest common factor to get the simplest form. For counting combinations, list them systematically to avoid missing any.

 

Question 4. A gets double of what B gets and B gets double of what C gets. Find A : B and B : C and verify whether the result is in proportion or not.
Answer: Let C get \( x \) amount.
Since B gets double of what C gets, B gets \( 2x \).
Since A gets double of what B gets, A gets \( 2 \times (2x) = 4x \).
So, A : B : C \( = 4x : 2x : x \).
The ratio A : B \( = 4x : 2x = 2 : 1 \).
The ratio B : C \( = 2x : x = 2 : 1 \).
To check if the result is in proportion, we compare A : B and B : C.
Here, A : B \( = 2 : 1 \) and B : C \( = 2 : 1 \).
Since A : B \( = \) B : C, the ratios are equal, which means they are in proportion.
This demonstrates a geometric progression in terms of quantities received.
In simple words: We find out how much A, B, and C get in relation to each other. Then we write down the ratio A to B and B to C. If these two ratios are the same, then they are in proportion.

🎯 Exam Tip: To check for proportion, compare the simplified ratios. If the first ratio equals the second ratio, they are in proportion.

 

Question 5. The ingredients required for the preparation of Ragi Kali, a healthy dish of Tamilnadu is given below.

(a) If one cup of ragi flour is used then, what would be the amount of raw rice required?
(b) If 16 cups of water are used, then how much ragi flour should be used?
(c) Which of these ingredients cannot be expressed as a ratio? Why?
Answer:
(a) From the table, for 4 cups of Ragi flour, 1 cup of Raw rice broken is required.
So, the ratio of Ragi flour to Raw rice is \( 4:1 \).
If 1 cup of Ragi flour is used, then the amount of raw rice required will be \( \frac{1}{4} \) cup. This maintains the same proportion of ingredients.

(b) From the table, for 8 cups of Water, 4 cups of Ragi flour are required.
So, the ratio of Water to Ragi flour is \( 8:4 \), which simplifies to \( 2:1 \).
If 16 cups of water are used, then the amount of ragi flour required will be \( \frac{16}{2} = 8 \) cups.

(c) Ragi flour, Raw rice, and Water are all measured in 'cups' (units of volume). Sesame oil is measured in 'ml' (milliliters, also a unit of volume). Salt is measured in 'mg' (milligrams, a unit of mass).
Ratios can only be formed between quantities that have the same units. We can compare quantities measured in cups with other quantities measured in cups, or milliliters with milliliters, and so on.
Therefore, we can form ratios between Ragi flour, Raw rice, and Water (all in cups). We can also form a ratio between Sesame oil and other ingredients if they were also in 'ml'.
However, Salt (in milligrams) cannot be directly compared in a ratio with Ragi flour, Raw rice, Water, or Sesame oil, because its unit of measurement is different (mass vs. volume). Ratios are useful for comparing similar types of quantities.
In simple words: (a) If you use 1 cup of ragi flour, you will need one-fourth cup of raw rice. (b) If you use 16 cups of water, you will need 8 cups of ragi flour. (c) Salt cannot be used in a ratio with the other ingredients because its measurement unit (milligrams) is different from the others (cups or milliliters).

🎯 Exam Tip: Remember that a ratio can only be formed between two quantities that have the same unit of measurement. If units are different, conversion is needed before forming a ratio, or the ratio simply cannot be made.

 

Question 6. Antony brushes his teeth in the morning and night on all days of the week. Shabeen brushes her teeth only in the morning. What is the ratio of the number of times Antony brushes his teeth to the number of times Shabeen brushes her teeth in a week?
Answer: Antony brushes his teeth twice a day (morning and night).
In a week (7 days), Antony brushes his teeth: \( 2 \times 7 = 14 \) times.
Shabeen brushes her teeth once a day (only in the morning).
In a week (7 days), Shabeen brushes her teeth: \( 1 \times 7 = 7 \) times.
The ratio of Antony's brushing to Shabeen's brushing in a week is:
Antony : Shabeen \( = 14 : 7 \).
To simplify the ratio, divide both numbers by 7:
\( 14 \div 7 : 7 \div 7 = 2 : 1 \).
So, the required ratio is \( 2:1 \). This shows Antony brushes twice as often as Shabeen.
In simple words: Antony brushes his teeth 14 times a week and Shabeen brushes hers 7 times a week. So the ratio of how often they brush is 2 to 1.

🎯 Exam Tip: Carefully read the question to count the total number of actions for each person over the specified time period before forming the ratio.

 

Question 7. Thirumagal's mother wears a bracelet made of 35 red beads and 30 blue beads. Thirumagal wants to make smaller bracelets using the same two coloured beads in the same ratio. In how many different ways can she make the bracelets?
Answer: The original bracelet has 35 red beads and 30 blue beads.
The ratio of Red : Blue beads \( = 35 : 30 \).
To simplify this ratio, we find the greatest common divisor (GCD) of 35 and 30, which is 5.
Dividing both numbers by 5, we get: \( 35 \div 5 : 30 \div 5 = 7 : 6 \).
This means for every 7 red beads, there are 6 blue beads in the smallest possible ratio.
Thirumagal can make smaller bracelets by using multiples of this simplified ratio, as long as the total number of beads used for each color does not exceed the available beads (35 red, 30 blue).
The possible ways are the common factors of 35 and 30, which are also divisors of the total beads:
(i) Using the simplest ratio: 7 red : 6 blue (This uses 7 red and 6 blue beads)
(ii) Using 2 times the simplest ratio: \( (7 \times 2) \) red : \( (6 \times 2) \) blue \( = 14 \) red : \( 12 \) blue
(iii) Using 3 times the simplest ratio: \( (7 \times 3) \) red : \( (6 \times 3) \) blue \( = 21 \) red : \( 18 \) blue
(iv) Using 4 times the simplest ratio: \( (7 \times 4) \) red : \( (6 \times 4) \) blue \( = 28 \) red : \( 24 \) blue
We cannot use 5 times the ratio because that would require \( 7 \times 5 = 35 \) red beads and \( 6 \times 5 = 30 \) blue beads. This uses up all the beads. The question implies "smaller" bracelets, suggesting partial use of the total beads for each bracelet. If she makes one bracelet of this size, she cannot make another one. If "smaller bracelets" means breaking the whole set into smaller, identical bracelets, then the common divisors of 35 and 30 that allow for more than one bracelet are 1 and 5. So, she could make 5 bracelets of 7:6 beads each. But if it means making *a* smaller bracelet, these are the possible combinations. The common factors of 35 and 30 are 1 and 5. Since we are looking for ways to create smaller bracelets while maintaining the ratio, we can consider multiples of the basic ratio (7:6) that are less than or equal to the original bead counts. Therefore, the possible ways are 4. Each of these combinations represents a valid smaller bracelet.
In simple words: First, find the simplest ratio of red to blue beads. Then, find how many times you can multiply this simple ratio to make different smaller bracelets, without using more beads than available. There are 4 different ways to make smaller bracelets while keeping the bead ratio the same.

🎯 Exam Tip: When forming smaller sets from a larger one while maintaining a ratio, look for multiples of the simplified ratio that do not exceed the original quantities. The number of ways usually corresponds to how many times the simplified ratio can be multiplied without surpassing the initial counts.

 

Question 8. Team A wins 26 matches out of 52 matches. Team B wins three fourth of 52 matches played. Which team has a better winning record?
Answer: To compare the winning records, we need to find the winning fraction or percentage for each team.
For Team A:
Number of matches won \( = 26 \)
Total matches played \( = 52 \)
Winning fraction for Team A \( = \frac{26}{52} = \frac{1}{2} \).

For Team B:
Team B wins three fourth of 52 matches.
Number of matches won by Team B \( = \frac{3}{4} \times 52 \)
\( = 3 \times \frac{52}{4} = 3 \times 13 = 39 \).
Winning fraction for Team B \( = \frac{39}{52} \).

Now we compare the winning fractions: \( \frac{1}{2} \) for Team A and \( \frac{39}{52} \) for Team B.
To compare them easily, we can convert \( \frac{1}{2} \) to a fraction with a denominator of 52:
\( \frac{1}{2} = \frac{1 \times 26}{2 \times 26} = \frac{26}{52} \).
Comparing \( \frac{26}{52} \) (Team A) and \( \frac{39}{52} \) (Team B), we see that \( \frac{39}{52} > \frac{26}{52} \).
Therefore, Team B has a better winning record. This shows that converting fractions to a common denominator helps in easy comparison.
In simple words: Team A won half of its matches. Team B won three-fourths of its matches. Since three-fourths is more than half, Team B has a better winning record.

🎯 Exam Tip: To compare fractions, convert them to a common denominator or convert them to decimals or percentages for a clear comparison.

 

Question 9. In a school excursion, 6 teachers and 12 students from 6th standard and 9 teachers and 27 students from 7th standard, 4 teachers and 16 students from 8th standard took part. Which class has the least teacher to student ratio?
Answer: We need to calculate the teacher to student ratio for each standard and then compare them.
For Standard VI:
Teachers : Students \( = 6 : 12 \)
Simplify the ratio by dividing both numbers by 6: \( 1 : 2 \).

For Standard VII:
Teachers : Students \( = 9 : 27 \)
Simplify the ratio by dividing both numbers by 9: \( 1 : 3 \).

For Standard VIII:
Teachers : Students \( = 4 : 16 \)
Simplify the ratio by dividing both numbers by 4: \( 1 : 4 \).

Comparing the simplified ratios: \( 1:2 \), \( 1:3 \), and \( 1:4 \).
A smaller number on the student side of the ratio (when the teacher side is 1) means fewer students per teacher, indicating a higher teacher presence. Conversely, a larger number on the student side means more students per teacher, indicating a lower teacher presence or a smaller teacher to student ratio.
The ratio \( 1:4 \) means 1 teacher for every 4 students. This is the smallest teacher to student ratio (meaning the teachers are fewer per student). Therefore, Standard VIII has the least teacher to student ratio. This helps manage student-teacher interactions effectively.
In simple words: We find the teacher-to-student ratio for each class and simplify it. The class with the highest student number for each teacher (meaning 1:4) has the lowest teacher-to-student ratio. So, Standard VIII has the least ratio.

🎯 Exam Tip: Always simplify ratios to their unit form (e.g., 1:X) to make direct comparisons straightforward and identify which ratio is the least or greatest.

 

Question 10. Fill the boxes using any set of suitable numbers \( 6 : \boxed{\phantom{X}} = \boxed{\phantom{Y}} : 15 \).
Answer: For a proportion, the product of the extremes (first and last terms) must be equal to the product of the means (second and third terms).
So, in \( 6 : A = B : 15 \), we have \( 6 \times 15 = A \times B \).
Product of the extremes \( = 6 \times 15 = 90 \).
Therefore, the product of the means (A and B) must also be 90. We need to find pairs of numbers whose product is 90.
Set of suitable numbers (A and B pairs):
\( (A, B) \):
(1, 90) \( \implies 6 : 1 = 90 : 15 \)
(2, 45) \( \implies 6 : 2 = 45 : 15 \)
(3, 30) \( \implies 6 : 3 = 30 : 15 \)
(5, 18) \( \implies 6 : 5 = 18 : 15 \)
(6, 15) \( \implies 6 : 6 = 15 : 15 \)
(9, 10) \( \implies 6 : 9 = 10 : 15 \)
Any pair of factors of 90 can fill the boxes. For instance, if we choose A=2 and B=45, the proportion would be \( 6 : 2 = 45 : 15 \). This property of cross-multiplication is fundamental in proportions.
In simple words: In a proportion, the number at the start times the number at the end must equal the two middle numbers multiplied together. Here, 6 times 15 is 90. So, the two numbers in the boxes must also multiply to 90. We can pick any two numbers that multiply to 90.

🎯 Exam Tip: Remember the property of proportions: "product of extremes equals product of means" (ad = bc for a:b = c:d). This is key to solving questions involving missing numbers in a proportion.

 

Question 11. From your school diary, write the ratio of the number of holidays to the number of working days in the current academic year.
Answer: Let's assume typical values for an academic year for an example. Actual values should be taken from a real school diary.
Number of holidays \( = 145 \)
Number of working days \( = 220 \)
Ratio of Holidays : Working days \( = 145 : 220 \).
To simplify the ratio, we find the greatest common divisor (GCD) of 145 and 220. Both numbers are divisible by 5.
\( \frac{145}{220} = \frac{145 \div 5}{220 \div 5} = \frac{29}{44} \).
So, the simplified ratio is \( 29:44 \). This indicates that for every 29 holidays, there are 44 working days, showing a balance between study and rest.
In simple words: We take the number of holidays and the number of working days from a school diary. Then, we write them as a ratio and simplify it by dividing both numbers by the biggest number that can divide them both evenly.

🎯 Exam Tip: When presenting a ratio, always simplify it to its lowest terms to make it easier to understand and compare.

 

Question 12. If the ratio of Green, Yellow and Black balls in a bag is 4 : 3 : 5, then
(a) Which is the most likely ball that you can choose from the bag?
(b) How many balls in total are there in the bag if you have 40 black balls in it?
(c) Find the number of green and yellow balls in the bag.
Answer: The ratio of Green : Yellow : Black balls is \( 4 : 3 : 5 \).

(a) The likelihood of choosing a ball depends on its proportion in the bag. The higher the number in the ratio, the more likely that ball is to be chosen. Since 5 is the largest number in the ratio \( 4 : 3 : 5 \), the Black balls have the highest proportion.
Therefore, the most likely ball that you can choose from the bag is a Black ball.

(b) The ratio of black balls is 5 parts. If there are 40 black balls, then 5 parts \( = 40 \) balls.
To find the value of 1 part, we divide 40 by 5: \( 1 \) part \( = \frac{40}{5} = 8 \) balls.
The total number of parts in the ratio is \( 4 + 3 + 5 = 12 \) parts.
Total number of balls in the bag \( = 12 \) parts \( \times 8 \) balls/part \( = 96 \) balls.
So, there are 96 balls in total in the bag.

(c) Using the value of 1 part (which is 8 balls):
Number of green balls \( = 4 \) parts \( = 4 \times 8 = 32 \) green balls.
Number of yellow balls \( = 3 \) parts \( = 3 \times 8 = 24 \) yellow balls.
We can verify the total: \( 32 \) (green) \( + 24 \) (yellow) \( + 40 \) (black) \( = 96 \) balls. Understanding ratios helps predict quantities and probabilities.
In simple words: (a) The black balls are the most likely to be picked because there are more of them (ratio of 5 is the largest). (b) If 5 parts of the ratio are 40 black balls, then one part is 8 balls. The total parts are 12, so there are 96 balls in total. (c) There are 32 green balls (4 parts) and 24 yellow balls (3 parts).

🎯 Exam Tip: In ratio problems, identify the value of one 'part' in the ratio. This value can then be used to calculate the quantity of each component and the total quantity.