GSEB Class 6 Maths Solutions Chapter 12 Ratio and Proportion Exercise 12.1

Question 1. There are 20 girls and 15 boys in a class.
(a) What is the ratio of the number of girls to the number of boys?
(b) What is the ratio of the number of girls to the total number of students in the class?
Answer:
Number of girls in the class = 20
Number of boys in the class = 15
Total number of students in the class = \( 20 + 15 = 35 \)
Now,
(a) Ratio = \( \frac{\text{Number of girls}}{\text{Number of boys}} = \frac{20}{15} = \frac{20 \div 5}{15 \div 5} = \frac{4}{3} = 4:3 \)
[HCF of 15 and 20 is 5]
(b) Ratio = \( \frac{\text{Number of girls}}{\text{Total Number of students}} = \frac{20}{35} = \frac{20 \div 5}{35 \div 5} = \frac{4}{7} = 4:7 \)
[HCF of 20 and 35 is 5]
In simple words: First, count the total number of students. Then, for part (a), you compare girls to boys. For part (b), you compare girls to all students. Always simplify the ratio to its simplest form.

Exam Tip: Remember to reduce ratios to their simplest form by dividing both numbers by their Highest Common Factor (HCF). Always label what each number in your ratio represents clearly.

Question 2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of
(a) Number of students liking football to a number of students liking tennis.
(b) Number of students liking cricket to the total number of students.
Answer:
Total number of students = 30
Number of students who like football = 6
Number of students who like cricket = 12
Remaining number of students liking tennis = \( 30 - (12 + 6) = 30 - 18 = 12 \)
Now,
(a) Ratio = \( \frac{\text{Number of students liking football}}{\text{Number of students who like tennis}} = \frac{6}{12} = \frac{6 \div 6}{12 \div 6} = \frac{1}{2} = 1:2 \)
[HCF of 6 and 12 is 6]
(b) Ratio = \( \frac{\text{Number of students liking cricket}}{\text{Total Number of students}} = \frac{12}{30} = \frac{12 \div 6}{30 \div 6} = \frac{2}{5} = 2:5 \)
[HCF of 12 and 30 is 6]
In simple words: First, work out how many students like tennis. After that, create the ratios as requested, making sure to simplify them by dividing by the largest common factor.

Exam Tip: When dealing with multiple categories, always find the quantity of the "remaining" category first before calculating any ratios involving it. This prevents errors in your calculations.

Question 3. See the figure and find the ratio of
(a) Number of triangles to the number of circles inside the rectangle.
(b) Number of squares to all the figures inside the rectangle.
(c) Number of circles to all the figures inside the rectangle.
Answer:
Number of triangles = 3
Number of squares = 2
Number of circles = 2
Total number of figures = \( 3 + 2 + 2 = 7 \)
(a) Ratio of triangles to circles = \( \frac{\text{Number of triangles}}{\text{Number of circles}} = \frac{3}{2} = 3:2 \)
(b) Ratio of squares to all figures = \( \frac{\text{Number of squares}}{\text{Number of all figures}} = \frac{2}{7} = 2:7 \)
(c) Ratio of circle to all figures = \( \frac{\text{Number of circles}}{\text{Number of all figures}} = \frac{2}{7} = 2:7 \)
In simple words: Count each shape type first. Then, for each question, make a comparison of the numbers. The total count includes all the shapes together.

Exam Tip: Be careful when counting the different shapes. Double-check your counts before calculating the ratios, especially when the figures are similar.

Question 4. Distances traveled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar.
Answer:
Speed of Hamid = 9 km/hr
Speed of Akhtar = 12 km/hr
Ratio = \( \frac{\text{Speed of Hamid}}{\text{Speed of Akhtar}} = \frac{9}{12} = \frac{9 \div 3}{12 \div 3} = \frac{3}{4} = 3:4 \)
[HCF of 9 and 12 is 3]
In simple words: To find this ratio, you just compare Hamid's speed with Akhtar's speed directly. Then, simplify the numbers by dividing them by their greatest common factor.

Exam Tip: Make sure the units for both quantities being compared are the same. In this case, both are in km/hr, so direct comparison is possible.

Question 5. Fill in the following blanks:
\( \frac{15}{18} = \frac{\text{_}}{6} = \frac{10}{\text{_}} = \frac{\text{_}}{30} \)
[Are these equivalent ratios?]
Answer:
By cross products, we have:
For \( \frac{15}{18} = \frac{\text{_}}{6} \):
\( 15 \times 6 = 18 \times \text{_} \)
\( 90 = 18 \times \text{_} \)
\( \text{_} = \frac{90}{18} = 5 \)
So, \( \frac{15}{18} = \frac{5}{6} \)

For \( \frac{15}{18} = \frac{10}{\text{_}} \):
\( 15 \times \text{_} = 18 \times 10 \)
\( 15 \times \text{_} = 180 \)
\( \text{_} = \frac{180}{15} = 12 \)
So, \( \frac{15}{18} = \frac{10}{12} \)

For \( \frac{15}{18} = \frac{\text{_}}{30} \):
\( 15 \times 30 = 18 \times \text{_} \)
\( 450 = 18 \times \text{_} \)
\( \text{_} = \frac{450}{18} = 25 \)
So, \( \frac{15}{18} = \frac{25}{30} \)

Now, we have:
\( \frac{15}{18} = \frac{5}{6} = \frac{10}{12} = \frac{25}{30} \)
Again, simplifying each ratio:
\( \frac{15}{18} = \frac{15 \div 3}{18 \div 3} = \frac{5}{6} \)
[HCF of 15 and 18 is 3]
\( \frac{25}{30} = \frac{25 \div 5}{30 \div 5} = \frac{5}{6} \)
[HCF of 25 and 30 is 5]
Also, \( \frac{10}{12} = \frac{10 \div 2}{12 \div 2} = \frac{5}{6} \)
[HCF of 10 and 12 is 2]
Thus, all the ratios are equivalent to \( \frac{5}{6} \).
In simple words: To fill in the blanks, use cross-multiplication. For instance, if \( \frac{A}{B} = \frac{C}{D} \), then \( A \times D = B \times C \). Once all blanks are filled, simplify each fraction to check if they are all equal.

Exam Tip: Equivalent ratios are crucial. Always remember that multiplying or dividing both terms of a ratio by the same non-zero number does not change its value. Cross-multiplication is an effective way to find missing terms in equivalent ratios.

Question 6. Find the ratio of the following:
(a) 81 to 108
(b) 98 to 63
(c) 33 km to 121 km
(d) 30 minutes to 45 minutes
Answer:
(a) \( \frac{81}{108} = \frac{81 \div 27}{108 \div 27} = \frac{3}{4} = 3:4 \)
[HCF of 81 and 108 is 27]
(b) \( \frac{98}{63} = \frac{98 \div 7}{63 \div 7} = \frac{14}{9} = 14:9 \)
[HCF of 98 and 63 is 7]
(c) \( \frac{33 \text{ km}}{121 \text{ km}} = \frac{33 \div 11}{121 \div 11} = \frac{3}{11} = 3:11 \)
[HCF of 33 and 121 is 11]
(d) \( \frac{30 \text{ minutes}}{45 \text{ minutes}} = \frac{30 \div 15}{45 \div 15} = \frac{2}{3} = 2:3 \)
[HCF of 30 and 45 is 15]
In simple words: Write each comparison as a fraction. Then, divide both the top and bottom numbers by their largest shared factor to make the fraction as simple as possible. This gives you the simplified ratio.

Exam Tip: Always make sure to simplify the ratio to its lowest terms. Identifying the HCF of the two numbers is key for quick and accurate simplification.

Question 7. Find the ratio of the following:
(a) 30 minutes to 1.5 hours
(b) 40 cm to 1.5 m
(c) 55 paise to Rs 1
(d) 500 ml to 2 litres
Answer:
(a) To compare 30 minutes to 1.5 hours, convert hours to minutes:
\( 1.5 \text{ hours} = 1.5 \times 60 \text{ minutes} = 90 \text{ minutes} \)
Ratio = \( \frac{30 \text{ minutes}}{90 \text{ minutes}} = \frac{30 \div 30}{90 \div 30} = \frac{1}{3} = 1:3 \)
[HCF of 30 and 90 is 30]

(b) To compare 40 cm to 1.5 m, convert meters to centimeters:
\( 1 \text{ m} = 100 \text{ cm} \)
\( 1.5 \text{ m} = 1.5 \times 100 \text{ cm} = 150 \text{ cm} \)
Ratio = \( \frac{40 \text{ cm}}{150 \text{ cm}} = \frac{40 \div 10}{150 \div 10} = \frac{4}{15} = 4:15 \)
[HCF of 40 and 150 is 10]

(c) To compare 55 paise to Rs 1, convert rupees to paise:
\( \text{Rs } 1 = 100 \text{ paise} \)
Ratio = \( \frac{55 \text{ paise}}{100 \text{ paise}} = \frac{55 \div 5}{100 \div 5} = \frac{11}{20} = 11:20 \)
[HCF of 55 and 100 is 5]

(d) To compare 500 ml to 2 litres, convert liters to milliliters:
\( 1 \text{ litre} = 1000 \text{ ml} \)
\( 2 \text{ litres} = 2 \times 1000 \text{ ml} = 2000 \text{ ml} \)
Ratio = \( \frac{500 \text{ ml}}{2000 \text{ ml}} = \frac{500 \div 500}{2000 \div 500} = \frac{1}{4} = 1:4 \)
[HCF of 500 and 2000 is 500]
In simple words: Before comparing numbers, make sure their units are the same. Convert hours to minutes, meters to centimeters, rupees to paise, and litres to milliliters. Then, simplify each ratio.

Exam Tip: Unit conversion is a common step in ratio problems. Always convert both quantities to the smallest common unit before calculating the ratio to avoid mistakes. Write down the conversion factor explicitly.

Question 8. In a year, Seema earns Rs 1,50,000 and saves Rs 50,000. Find the ratio of
(a) Money that Seema earns to the money she saves.
(b) Money that she saves to the money she spends.
Answer:
Total earnings = Rs 1,50,000
Savings = Rs 50,000
(a) Ratio of earnings to savings:
\( \frac{\text{Earnings}}{\text{Savings}} = \frac{\text{Rs } 1,50,000}{\text{Rs } 50,000} = \frac{1,50,000 \div 50,000}{50,000 \div 50,000} = \frac{3}{1} = 3:1 \)
[HCF of 1,50,000 and 50,000 is 50,000]

(b) Money she spends = Total earnings - Savings
Money she spends = \( \text{Rs } 1,50,000 - \text{Rs } 50,000 = \text{Rs } 1,00,000 \)
Ratio of savings to expenditure:
\( \frac{\text{Savings}}{\text{Expenditure}} = \frac{\text{Rs } 50,000}{\text{Rs } 1,00,000} = \frac{50,000 \div 50,000}{1,00,000 \div 50,000} = \frac{1}{2} = 1:2 \)
[HCF of 50,000 and 1,00,000 is 50,000]
In simple words: First, figure out how much money Seema spends. Then, set up the comparisons requested in each part, making sure to simplify the ratios.

Exam Tip: In financial ratio questions, always calculate all relevant values (earnings, savings, spending) first. This ensures you have all the necessary information before forming the ratios, reducing calculation errors.

Question 9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.
Answer:
Number of teachers = 102
Number of students = 3300
Ratio of teachers to students = \( \frac{102}{3300} = \frac{102 \div 6}{3300 \div 6} = \frac{17}{550} = 17:550 \)
[HCF of 102 and 3300 is 6]
In simple words: To find the ratio, simply divide the number of teachers by the number of students. Then, reduce the fraction to its most basic form.

Exam Tip: When forming a ratio, the order of the items matters. "Ratio of A to B" means A:B or A/B. Always simplify the ratio by finding the greatest common divisor.

Question 10. In a college, out of 4320 students, 2300 are girls. Find the ratio of
(a) Number of girls to the total number of students.
(b) Number of boys to the number of girls.
(c) Number of boys to the total number of students.
Answer:
Total number of students = 4320
Number of girls = 2300
Number of boys = \( 4320 - 2300 = 2020 \)

(a) Ratio of number of girls to total number of students:
\( \frac{\text{Number of girls}}{\text{Total number of students}} = \frac{2300}{4320} = \frac{2300 \div 20}{4320 \div 20} = \frac{115}{216} = 115:216 \)
[HCF of 2300 and 4320 is 20]

(b) Ratio of number of boys to the number of girls:
\( \frac{\text{Number of boys}}{\text{Number of girls}} = \frac{2020}{2300} = \frac{2020 \div 20}{2300 \div 20} = \frac{101}{115} = 101:115 \)
[HCF of 2020 and 2300 is 20]

(c) Ratio of number of boys to the total number of students:
\( \frac{\text{Number of boys}}{\text{Total number of students}} = \frac{2020}{4320} = \frac{2020 \div 20}{4320 \div 20} = \frac{101}{216} = 101:216 \)
[HCF of 2020 and 4320 is 20]
In simple words: First, calculate the number of boys by subtracting the girls from the total. Then, form each requested ratio and reduce it to its lowest terms.

Exam Tip: When given a total and one part, remember to subtract to find the other part. Always clarify what each part of the ratio represents to avoid mixing them up.

Question 11. Out of 1800 students in a school, 750 opted basketball, 800 opted cricket, and the remaining opted table tennis. If a student can opt for only one game, find the ratio of
(a) Number of students who opted for basketball to the number of students who opted for table tennis.
(b) Number of students who opted for cricket to the number of students opting for basketball.
(c) Number of students who opted for basketball to the total number of students.
Answer:
Total number of students = 1800
Number of students who opted for basketball = 750
Number of students who opted for cricket = 800
Number of students who opted for table tennis = \( 1800 - [750 + 800] = 1800 - 1550 = 250 \)
Now,
(a) Ratio of number of students opting for basketball to table tennis:
\( \frac{750}{250} = \frac{750 \div 250}{250 \div 250} = \frac{3}{1} = 3:1 \)
[HCF of 750 and 250 is 250]

(b) Ratio of number of students opting for cricket to basketball:
\( \frac{800}{750} = \frac{800 \div 50}{750 \div 50} = \frac{16}{15} = 16:15 \)
[HCF of 800 and 750 is 50]

(c) Ratio of number of students opting for basketball to total number of students:
\( \frac{750}{1800} = \frac{750 \div 150}{1800 \div 150} = \frac{5}{12} = 5:12 \)
[HCF of 750 and 1800 is 150]
In simple words: First, determine the number of students who chose table tennis by subtracting the other groups from the total. Then, set up each ratio as asked and simplify them.

Exam Tip: Pay close attention to the order of the quantities when forming ratios (e.g., A to B vs. B to A). Always calculate the "remaining" category by subtracting known quantities from the total.

Question 12. Cost of a dozen pens is Rs 180 and cost of 8 ball pens are Rs 56. Find the ratio of the cost of a pen to the cost of a ball pen.
Answer:
Cost of 12 pens = Rs 180
Cost of 1 pen = \( \frac{\text{Rs } 180}{12} = \text{Rs } 15 \)
Cost of 8 ball pens = Rs 56
Cost of 1 ball pen = \( \frac{\text{Rs } 56}{8} = \text{Rs } 7 \)
Now, ratio of the cost of a pen to the cost of a ball pen = \( \frac{\text{Rs } 15}{\text{Rs } 7} = 15:7 \)
In simple words: First, find the price of a single pen and a single ball pen. Then, compare these two individual prices to get your final ratio.

Exam Tip: When comparing items that are sold in different quantities, always find the unit price (cost per single item) for each before forming the ratio. This ensures a fair and accurate comparison.

Question 13. Consider the statement: Ratio of breadth and length of a hall is 2:5. Complete the following table that shows some possible breadths, and lengths of the hall.
Answer:
The ratio of breadth to length is 2:5, or \( \frac{\text{Breadth}}{\text{Length}} = \frac{2}{5} \).
Let's fill the blanks using this ratio.

For the first blank (Breadth when Length is 50 m):
Let the unknown breadth be 'x'.
\( \frac{x}{50} = \frac{2}{5} \)
\( 5x = 2 \times 50 \)
\( 5x = 100 \)
\( x = \frac{100}{5} = 20 \text{ m} \)
So, the breadth is 20 m.

For the second blank (Length when Breadth is 40 m):
Let the unknown length be 'y'.
\( \frac{40}{y} = \frac{2}{5} \)
\( 2y = 40 \times 5 \)
\( 2y = 200 \)
\( y = \frac{200}{2} = 100 \text{ m} \)
So, the length is 100 m.

Now, the table is completed as given below:

Exam Tip: For problems involving tables and ratios, remember that the ratio between corresponding values in the rows/columns must remain constant. Use cross-multiplication for efficiency to find unknown values.

Question 14. Divide 20 pens between Sheela and Sangeeta in the ratio of 3 : 2.
Answer:
Total number of pens = 20
Sheela's share : Sangeeta's share = 3 : 2
Sum of ratios = \( 3 + 2 = 5 \)
Sheela's share in 20 pens = \( \frac{3}{5} \times 20 = 12 \)
Sangeeta's share in 20 pens = \( \frac{2}{5} \times 20 = 8 \)
In simple words: Add the parts of the ratio together to get the total parts. Then, divide the total number of pens by this sum. Multiply that result by each part of the ratio to find their individual shares.

Exam Tip: When dividing a quantity in a given ratio, first find the sum of the ratio terms. Then, determine each share by multiplying the total quantity by the fraction of each ratio term over the sum of terms.

Question 15. Does the mother want to divide Rs 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.
Answer:
Age of Shreya = 15 years
Age of Bhoomika = 12 years
Ratio of their ages = \( \frac{15 \text{ years}}{12 \text{ years}} = \frac{15 \div 3}{12 \div 3} = \frac{5}{4} \)
[HCF of 15 and 12 is 3]
Sum of the ratios = \( 5 + 4 = 9 \)
Total amount to be divided = Rs 36
Shreya's share = \( \frac{5}{9} \times \text{Rs } 36 = \text{Rs } 20 \)
Bhoomika's share = \( \frac{4}{9} \times \text{Rs } 36 = \text{Rs } 16 \)
In simple words: First, find the simplest ratio of their ages. Add the parts of this ratio. Then, divide the total money by this sum and multiply by each girl's ratio part to find her share.

Exam Tip: Always simplify the age ratio first before calculating shares. This makes the numbers smaller and easier to work with, minimizing calculation mistakes.

Question 16. The present age of the father is 42 years and that of his son is 14 years. Find the ratio of
(a) Present age of father to the present age of the son.
(b) Age of the father to the age of the son, when the son was 12 years old.
(c) Age of father after 10 years to the age of son after 10 years.
(d) Age of father to the age of son when father was 30 years old.
Answer:
(a) Present age of father = 42 years
Present age of son = 14 years
Ratio of their ages = \( \frac{\text{Age of father}}{\text{Age of son}} = \frac{42 \text{ years}}{14 \text{ years}} = \frac{42 \div 14}{14 \div 14} = \frac{3}{1} = 3:1 \)
[HCF of 14 and 42 is 14]

(b) When son's age was 12 years, this means 2 years ago (since \( 14 - 12 = 2 \)).
Father's age then = \( 42 \text{ years} - 2 \text{ years} = 40 \text{ years} \)
Son's age then = 12 years
Ratio of father's age to son's age = \( \frac{40 \text{ years}}{12 \text{ years}} = \frac{40 \div 4}{12 \div 4} = \frac{10}{3} = 10:3 \)
[HCF of 40 and 12 is 4]

(c) After 10 years:
Age of father = \( 42 \text{ years} + 10 \text{ years} = 52 \text{ years} \)
Age of son = \( 14 \text{ years} + 10 \text{ years} = 24 \text{ years} \)
Ratio of father's age to son's age = \( \frac{52 \text{ years}}{24 \text{ years}} = \frac{52 \div 4}{24 \div 4} = \frac{13}{6} = 13:6 \)
[HCF of 52 and 24 is 4]

(d) When father was 30 years old, this means 12 years ago (since \( 42 - 30 = 12 \)).
Son's age then = \( 14 \text{ years} - 12 \text{ years} = 2 \text{ years} \)
Father's age then = 30 years
Ratio of father's age to son's age = \( \frac{30 \text{ years}}{2 \text{ years}} = \frac{30 \div 2}{2 \div 2} = \frac{15}{1} = 15:1 \)
[HCF of 30 and 2 is 2]
In simple words: For each part, first calculate the actual ages of the father and son at that specific time. Then, create the ratio using those ages and simplify it. Remember that 'ago' means subtract, and 'after' means add.

Exam Tip: Age-related ratio problems require careful calculation of ages at different points in time (past, present, future). Always determine the exact ages for both individuals for the specific scenario before forming the ratio, and clearly state the time reference.