ML Aggarwal Class 7 Maths Solutions Chapter 06 Ratio and Proportion

Exercise 6.1

Question 1. Express the following ratios in simplest form:
(i) \( \frac{1}{6} : \frac{1}{9} \)
(ii) \( 4\frac{1}{2} : 1\frac{1}{8} \)
(iii) \( \frac{1}{5} : \frac{1}{10} : \frac{1}{15} \)
Answer:
(i) To work with fractions in a ratio, find the LCM of the denominators. The LCM of 6 and 9 is 18. Multiply each fraction by 18 to get whole numbers: \( \frac{1}{6} \times 18 : \frac{1}{9} \times 18 = 3 : 2 \)
In simple words: When you have fractions in a ratio, multiply each by the LCM of the bottom numbers to turn them into whole numbers.

Exam Tip: Always find the LCM of denominators and multiply both terms by it - this is the fastest way to clear fractions from a ratio.

(ii) First, convert mixed numbers to improper fractions: \( 4\frac{1}{2} = \frac{9}{2} \) and \( 1\frac{1}{8} = \frac{9}{8} \). The ratio becomes \( \frac{9}{2} : \frac{9}{8} \). The LCM of 2 and 8 is 8. Multiply each term by 8: \( \frac{9}{2} \times 8 : \frac{9}{8} \times 8 = 36 : 9 = 4 : 1 \)
In simple words: Change mixed numbers to improper fractions first, then use the LCM method to clear the denominators.

Exam Tip: Do not forget to simplify the final ratio by dividing both sides by their HCF.

(iii) For three fractions, find the LCM of all three denominators. The LCM of 5, 10, and 15 is 30. Multiply each fraction by 30: \( \frac{1}{5} \times 30 : \frac{1}{10} \times 30 : \frac{1}{15} \times 30 = 6 : 3 : 2 \)
In simple words: When there are three fractions, find the LCM of all three bottom numbers, then multiply each fraction by this LCM.

Exam Tip: For ratios with three or more terms, apply the same LCM method - the result should always be whole numbers in simplest form.

Question 2. Find the ratio of each of the following in simplest form:
(i) Rs.5 to 50 paise
(ii) 3 km to 300 m
(iii) 9 m to 27 cm
(iv) 15 kg to 210 g
(v) 25 minutes to 1.5 hours
(vi) 30 days to 36 hours
Answer:
(i) Convert rupees to paise: Rs.5 = 5 × 100 = 500 paise. The ratio is 500 : 50 = 10 : 1
In simple words: Always convert both quantities to the same unit before making a ratio.

Exam Tip: Write the common unit you are converting to - this prevents careless errors.

(ii) Convert kilometres to metres: 3 km = 3 × 1000 = 3000 m. The ratio is 3000 : 300 = 10 : 1
In simple words: Multiply kilometres by 1000 to get metres, then simplify.

Exam Tip: Remember the conversion factors - 1 km = 1000 m, 1 m = 100 cm, 1 hour = 60 minutes, 1 day = 24 hours.

(iii) Convert metres to centimetres: 9 m = 9 × 100 = 900 cm. The ratio is 900 : 27. Using prime factorization, the HCF of 900 and 27 is 9. Dividing both by 9: 900 ÷ 9 : 27 ÷ 9 = 100 : 3
In simple words: Turn both amounts into the same unit, then divide by the largest number that goes into both.

Exam Tip: Use HCF to simplify - it is faster than trying different factors.

(iv) Convert kilograms to grams: 15 kg = 15 × 1000 = 15000 g. The ratio is 15000 : 210. Using prime factorization, the HCF of 15000 and 210 is 30. Dividing both by 30: 15000 ÷ 30 : 210 ÷ 30 = 500 : 7
In simple words: Convert to the smaller unit (grams), then find the HCF to reduce.

Exam Tip: Always work in smaller units to avoid fractions.

(v) Convert hours to minutes: 1.5 hours = 1.5 × 60 = 90 minutes. The ratio is 25 : 90. The HCF of 25 and 90 is 5. Dividing both by 5: 25 ÷ 5 : 90 ÷ 5 = 5 : 18
In simple words: Change hours to minutes by multiplying by 60, then simplify by finding the HCF.

Exam Tip: Decimal time values like 1.5 hours should be converted first before forming the ratio.

(vi) Convert days to hours: 30 days = 30 × 24 = 720 hours. The ratio is 720 : 36 = 20 : 1
In simple words: Multiply days by 24 to get hours, then divide both numbers by their HCF.

Exam Tip: Division can often be done quickly by spotting common factors (both 720 and 36 are divisible by 36).

Question 3. If A : B = 3 : 4 and B : C = 8 : 9, then find A : C.
Answer: We have A : B = 3 : 4 and B : C = 8 : 9. To join these ratios, the value of B must be the same in both. In the first ratio B is 4, and in the second it is 8. The LCM of 4 and 8 is 8. Multiply the first ratio by 2: A : B = 6 : 8. Now both ratios share B = 8, so we get A : B : C = 6 : 8 : 9. Therefore, A : C = 6 : 9 = 2 : 3
In simple words: Make the middle letter the same in both ratios, then you can combine them into one three-part ratio.

Exam Tip: Always find the LCM of the common term's values to avoid mistakes when combining ratios.

Question 4. If A : B = 5 : 8 and B : C = 18 : 25, then find A : B : C.
Answer: We are given A : B = 5 : 8 and B : C = 18 : 25. Since B appears in both, we make B equal in both ratios. B is 8 in the first and 18 in the second. The LCM of 8 and 18 is 72. Multiply the first ratio by 9: A : B = 45 : 72. Multiply the second ratio by 4: B : C = 72 : 100. Now both have B = 72, so A : B : C = 45 : 72 : 100
In simple words: Find what to multiply each ratio by so the common letter (B) has the same value in both, then write all three together.

Exam Tip: Write down the LCM calculation clearly to avoid making errors in the multipliers.

Question 5. If 3A = 2B = 5C, then find A : B : C.
Answer: Let 3A = 2B = 5C = k for some constant. Then A = k/3, B = k/2, and C = k/5. The ratio A : B : C is k/3 : k/2 : k/5. To remove fractions, find the LCM of 3, 2, and 5, which is 30. Multiply each term by 30: (k/3) × 30 : (k/2) × 30 : (k/5) × 30 = 10k : 15k : 6k = 10 : 15 : 6
In simple words: When variables are equal in different forms, set them all equal to k, then work out each variable and find the ratio.

Exam Tip: The "let equal to k" method is standard for this type - always use it even if it looks complicated.

Question 6. Out of daily income of Rs.600, a worker spends Rs.450 on food and shelter and saves the rest. Find the ratio of his
(i) spending to income
(ii) saving to income
(iii) saving to spending
Answer:
(i) Daily income = Rs.600 and spending = Rs.450. The ratio of spending to income is 450 : 600. The HCF of 450 and 600 is 150. So 450 ÷ 150 : 600 ÷ 150 = 3 : 4
In simple words: Write spending first, then income, as fractions: spending/income = 450/600. Simplify by dividing by the HCF.

Exam Tip: Always check which quantity comes first in the question and put it in the first position of the ratio.

(ii) Saving = Income - Spending = Rs.600 - Rs.450 = Rs.150. The ratio of saving to income is 150 : 600. The HCF is 150. So 150 ÷ 150 : 600 ÷ 150 = 1 : 4
In simple words: First find how much was saved, then write the ratio with saving first and income second.

Exam Tip: Always calculate the missing amount (saving) before writing the ratio.

(iii) The ratio of saving to spending is 150 : 450. The HCF is 150. So 150 ÷ 150 : 450 ÷ 150 = 1 : 3
In simple words: Use the same saving amount (Rs.150) but now compare it to spending (Rs.450).

Exam Tip: Notice that the ratios in parts (i), (ii), and (iii) are all different - use the correct pair of quantities each time.

Question 7. 5 grams of an alloy contains \( 3\frac{3}{4} \) grams copper and the rest is nickel. Find the ratio by weight of nickel to copper.
Answer: Total weight of alloy = 5 grams. Weight of copper = \( 3\frac{3}{4} = \frac{15}{4} \) grams. Weight of nickel = Total - Copper = \( 5 - \frac{15}{4} = \frac{20}{4} - \frac{15}{4} = \frac{5}{4} \) grams. The ratio of nickel to copper is \( \frac{5}{4} : \frac{15}{4} \). Multiply both terms by 4: \( \frac{5}{4} \times 4 : \frac{15}{4} \times 4 = 5 : 15 = 1 : 3 \)
In simple words: Find the nickel weight by subtracting copper from the total, then write the ratio and simplify.

Exam Tip: With mixed numbers and fractions, convert to improper fractions first, then multiply by the LCM to get whole numbers.

Question 8. A pole of height 3 metres is struck by a speeding car and breaks into two pieces such that the first piece is \( \frac{1}{2} \) of the second. Find the length of both pieces.
Answer: Let the length of the second piece = x metres. Then the length of the first piece = \( \frac{1}{2}x \) metres. Since the total length is 3 metres: \( \frac{1}{2}x + x = 3 \). Simplifying: \( \frac{3x}{2} = 3 \), so \( 3x = 6 \) and \( x = 2 \) metres. Therefore, the second piece = 2 m and the first piece = \( \frac{1}{2} \times 2 = 1 \) m
In simple words: Set up an equation where the first piece plus the second piece equals the total, then solve for the unknown.

Exam Tip: Always check your answer by adding both pieces - they should equal the original length (1 + 2 = 3 ✓).

Question 9. Heights of Anshul and Dhruv are 1.04 m and 78 cm respectively. Divide 35 sweets between them in the ratio of their heights.
Answer: Convert both heights to the same unit: Anshul's height = 1.04 m = 104 cm, Dhruv's height = 78 cm. The ratio of their heights is 104 : 78. Using prime factorization, the HCF of 104 and 78 is 26. So 104 ÷ 26 : 78 ÷ 26 = 4 : 3. Total parts = 4 + 3 = 7. Value of 1 part = 35 ÷ 7 = 5 sweets. Anshul's share = 4 × 5 = 20 sweets. Dhruv's share = 3 × 5 = 15 sweets
In simple words: First make the heights the same unit, reduce the ratio, find how many parts there are, then divide the sweets equally among those parts.

Exam Tip: Always check: 20 + 15 = 35 ✓ and 20 : 15 simplifies to 4 : 3 ✓.

Question 10. Rs.180 are to be divided among three children in the ratio \( \frac{1}{3} : \frac{1}{4} : \frac{1}{6} \). Find the share of each child.
Answer: The given ratio is \( \frac{1}{3} : \frac{1}{4} : \frac{1}{6} \). To clear fractions, find the LCM of 3, 4, and 6, which is 12. Multiply each term by 12: \( \frac{1}{3} \times 12 : \frac{1}{4} \times 12 : \frac{1}{6} \times 12 = 4 : 3 : 2 \). Total parts = 4 + 3 + 2 = 9. Value of 1 part = 180 ÷ 9 = Rs.20. First child's share = 4 × 20 = Rs.80. Second child's share = 3 × 20 = Rs.60. Third child's share = 2 × 20 = Rs.40
In simple words: Clear the fractions by multiplying by LCM, add up all the parts, divide the total money by the number of parts, then multiply each part's value by its ratio number.

Exam Tip: Verify: Rs.80 + Rs.60 + Rs.40 = Rs.180 ✓ and 80 : 60 : 40 reduces to 4 : 3 : 2 ✓.

Question 11. A natural number has been divided into two parts in the ratio 7 : 11. If the difference of two parts is 20, find the number and the two parts.
Answer: Let the two parts be 7x and 11x. The difference of the two parts is 11x - 7x = 4x. According to the question, 4x = 20, so x = 5. First part = 7 × 5 = 35. Second part = 11 × 5 = 55. The natural number = 35 + 55 = 90
In simple words: Express each part as a multiple of x using the given ratio, use the difference to find x, then calculate both parts and their sum.

Exam Tip: Always check: difference = 55 - 35 = 20 ✓ and 35 : 55 simplifies to 7 : 11 ✓.

Question 12. A certain sum of money has been divided into two parts in the ratio 9 : 13. If the second part is Rs.260, find the total amount.
Answer: Let the two parts be 9x and 13x. Since the second part equals 13x = Rs.260, we get x = 260 ÷ 13 = 20. Therefore, the first part = 9 × 20 = Rs.180. Adding both parts gives the total = Rs.180 + Rs.260 = Rs.440.
In simple words: The two parts are in ratio 9 to 13. If the larger part is Rs.260, divide it by 13 to find x, then multiply by 9 to find the smaller part. Add them together for the total.

Exam Tip: Always set up both parts as multiples of x using the given ratio, then use the known value to find x before calculating the remaining parts.

Question 13. A certain sum of money is divided into three parts in the ratio 5 : 7 : 8. If the first part is Rs.225, find the total amount and the other two parts.
Answer: Let the three parts be 5x, 7x, and 8x. The first part is 5x = Rs.225, so x = 225 ÷ 5 = 45. The second part = 7 × 45 = Rs.315. The third part = 8 × 45 = Rs.360. The total amount = Rs.225 + Rs.315 + Rs.360 = Rs.900.
In simple words: Divide the given part by its ratio number to find x. Multiply this x by each ratio number to get all three parts. Add them to find the total.

Exam Tip: Check your answer by verifying that the three parts are indeed in the given ratio and that they add up to the total calculated.

Question 14. Divide Rs.1312 into three parts such that first part is \( \frac{2}{3} \) of the second and the ratio between second and third parts is 4 : 7.
Answer: Let the second part = x. Then the first part = \( \frac{2}{3} x \). Since the ratio of second to third is 4 : 7, the third part = \( \frac{7}{4} x \). The sum of all three parts equals Rs.1312:
\( \frac{2}{3} x + x + \frac{7}{4} x = 1312 \)

Finding the LCM of 3 and 4, which is 12, we get:
\( \frac{8x}{12} + \frac{12x}{12} + \frac{21x}{12} = 1312 \)

\( \frac{41x}{12} = 1312 \)

Solving for x: \( x = \frac{1312 \times 12}{41} = 32 \times 12 = 384 \)

Second part = Rs.384. First part = \( \frac{2}{3} \times 384 = Rs.256 \). Third part = \( \frac{7}{4} \times 384 = Rs.672 \).
In simple words: Express each part in terms of one variable, write an equation that adds to the total, combine fractions using a common denominator, and solve to find each part.

Exam Tip: When fractions are involved in ratios, use the LCM method to combine the terms, which simplifies the algebra and avoids decimal errors.

Question 15. The ratio of the present ages of Saanvi and Navya is 2 : 3. Five years hence, the ratio of their ages will be 3 : 4. Find their present ages.
Answer: Let Saanvi's present age = 2x years and Navya's present age = 3x years. After five years: Saanvi's age = (2x + 5) years and Navya's age = (3x + 5) years. According to the problem:
\( \frac{2x + 5}{3x + 5} = \frac{3}{4} \)

Using cross multiplication: 4(2x + 5) = 3(3x + 5)
8x + 20 = 9x + 15
20 - 15 = 9x - 8x
x = 5

Saanvi's present age = 2 × 5 = 10 years. Navya's present age = 3 × 5 = 15 years.
In simple words: Use the given ratio to write their ages as 2x and 3x. Add the same number to both, form an equation with the new ratio, solve for x, then find each person's age.

Exam Tip: Always set up the equation matching the time reference (present, past, or future) with its corresponding ratio - do not mix present age with future ratio.

Question 16. The present ages of A and B are in the ratio 5 : 6. Three years ago, their ages were in the ratio 4 : 5. Find their present ages.
Answer: Let A's present age = 5x years and B's present age = 6x years. Three years ago: A's age = (5x - 3) years and B's age = (6x - 3) years. According to the condition:
\( \frac{5x - 3}{6x - 3} = \frac{4}{5} \)

Cross multiplying: 5(5x - 3) = 4(6x - 3)
25x - 15 = 24x - 12
25x - 24x = -12 + 15
x = 3

A's present age = 5 × 3 = 15 years. B's present age = 6 × 3 = 18 years.
In simple words: Subtract the years from the past from both ages, set up the equation with the past ratio, solve for x, and calculate each present age.

Exam Tip: Subtracting years means subtracting from both the numerator and denominator - be careful with signs when simplifying after cross multiplication.

Question 17. Two numbers are in the ratio 5 : 6. When 2 is added to first and 3 is added to second, they are in the ratio 4 : 5. Find the numbers.
Answer: Let the two numbers be 5x and 6x. After adding: the first number becomes (5x + 2) and the second becomes (6x + 3). According to the problem:
\( \frac{5x + 2}{6x + 3} = \frac{4}{5} \)

Cross multiplying: 5(5x + 2) = 4(6x + 3)
25x + 10 = 24x + 12
25x - 24x = 12 - 10
x = 2

First number = 5 × 2 = 10. Second number = 6 × 2 = 12.
In simple words: Write the numbers as 5x and 6x. Add the specified amounts to each, form an equation with the new ratio, solve for x, and find both numbers.

Exam Tip: When values are added or subtracted, the new ratio applies to the changed numbers, not the original ones - set up your equation accordingly.

Question 18. The ratio of number of boys to the number of girls in a school of 1430 students is 7 : 6. If 26 new girls are admitted in the school, find how many new boys should be admitted so that the ratio of number of boys to the number of girls changes to 8 : 7.
Answer: Total students = 1430. Boys to girls ratio = 7 : 6. Total parts = 7 + 6 = 13. Value of 1 part = 1430 ÷ 13 = 110. Number of boys = 7 × 110 = 770. Number of girls = 6 × 110 = 660. After admitting 26 new girls: number of girls = 660 + 26 = 686. Let the number of new boys admitted = y. Then total boys = (770 + y). According to the new condition:
\( \frac{770 + y}{686} = \frac{8}{7} \)

Cross multiplying: 7(770 + y) = 8 × 686
5390 + 7y = 5488
7y = 98
y = 14

Therefore, 14 new boys should be admitted.
In simple words: Find how many boys and girls are currently present using the ratio. Add the new girls. Set up an equation with the desired new ratio and solve for how many new boys are needed.

Exam Tip: Divide the total by the sum of ratio parts to find the value of one part - this helps you quickly find individual quantities without algebra at first.

Question 19. Which ratio is greater:
(i) 5 : 6 or 6 : 7
(ii) 13 : 24 or 17 : 32
Answer:
(i) Rewrite as fractions: \( \frac{5}{6} \) and \( \frac{6}{7} \). Using cross multiplication, compare 5 × 7 and 6 × 6: 5 × 7 = 35 and 6 × 6 = 36. Since 35 < 36, we have \( \frac{5}{6} < \frac{6}{7} \). Therefore, 6 : 7 is the greater ratio.

(ii) Rewrite as fractions: \( \frac{13}{24} \) and \( \frac{17}{32} \). Using cross multiplication, compare 13 × 32 and 17 × 24: 13 × 32 = 416 and 17 × 24 = 408. Since 416 > 408, we have \( \frac{13}{24} > \frac{17}{32} \). Therefore, 13 : 24 is the greater ratio.
In simple words: To compare ratios, change them to fractions. Then cross multiply - if the left product is bigger, the left fraction is bigger. If the right product is bigger, the right fraction is bigger.

Exam Tip: Cross multiplication avoids finding common denominators and is the fastest way to compare two fractions or ratios on a timed exam.

Question 20. (i) Increase the number 150 in ratio 5 : 7
(ii) A man earns Rs.18,000 per month. His income is increased in the ratio 12 : 13. Find his new monthly income.
(iii) Savita weighs 55 kg. She reduced her weight in the ratio 11 : 9. Find her new weight.
Answer:
(i) To increase a number in the ratio 5 : 7, multiply it by \( \frac{7}{5} \). New number = 150 × \( \frac{7}{5} \) = 30 × 7 = 210.

(ii) To increase income in the ratio 12 : 13, multiply by \( \frac{13}{12} \). New income = 18000 × \( \frac{13}{12} \) = 1500 × 13 = Rs.19,500.

(iii) To reduce weight in the ratio 11 : 9, multiply by \( \frac{9}{11} \). New weight = 55 × \( \frac{9}{11} \) = 5 × 9 = 45 kg.
In simple words: When increasing in a ratio a : b, use the fraction b/a. When decreasing in a ratio a : b, use the fraction b/a. Multiply the original value by this fraction to get the new value.

Exam Tip: Remember that "increase in ratio a : b" means the new value is to the old value as b is to a - use the larger number in the numerator for increase, smaller for decrease.

Exercise 6.2

Question 1. Which of the following statements are true?
(i) 2.5 : 1.5 :: 7.0 : 4.2
(ii) \( \frac{1}{2} : \frac{1}{3} = \frac{1}{3} : \frac{1}{4} \)
(iii) 24 men : 16 men = 33 horses : 22 horses
Answer: In a proportion, the product of the extremes (the first and last numbers) equals the product of the means (the two middle numbers).

(i) Product of extremes = 2.5 × 4.2 = 10.5. Product of means = 1.5 × 7.0 = 10.5. Since both products are equal, the statement is True.

(ii) Product of extremes = \( \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \). Product of means = \( \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \). Since \( \frac{1}{8} \neq \frac{1}{9} \), the statement is False.

(iii) Simplify each ratio: 24 : 16 = \( \frac{24}{16} = \frac{3}{2} \) and 33 : 22 = \( \frac{33}{22} = \frac{3}{2} \). Both ratios are equal, so the statement is True.

Therefore, statements (i) and (iii) are true.
In simple words: To check if four numbers are in proportion, multiply the first and fourth numbers, then multiply the second and third numbers. If these products are equal, the numbers form a valid proportion.

Exam Tip: Always verify by calculating both products explicitly - estimation can lead to errors, especially with decimals or fractions.

Question 2. Check whether the following numbers are in proportion or not:
(i) 18, 10, 9, 5
(ii) 3, \( 3\frac{1}{2} \), 4, \( 4\frac{1}{2} \)
(iii) 0.1, 0.2, 0.3, 0.6
Answer: Four numbers a, b, c, d are in proportion if a × d = b × c (product of extremes equals product of means).

(i) Product of extremes = 18 × 5 = 90. Product of means = 10 × 9 = 90. Since 90 = 90, the numbers 18, 10, 9, 5 are in proportion.

(ii) Convert to improper fractions: 3, \( \frac{7}{2} \), 4, \( \frac{9}{2} \). Product of extremes = 3 × \( \frac{9}{2} \) = \( \frac{27}{2} \) = 13.5. Product of means = \( \frac{7}{2} \) × 4 = 14. Since 13.5 ≠ 14, the numbers are not in proportion.

(iii) Product of extremes = 0.1 × 0.6 = 0.06. Product of means = 0.2 × 0.3 = 0.06. Since 0.06 = 0.06, the numbers 0.1, 0.2, 0.3, 0.6 are in proportion.

Therefore, the numbers in (i) and (iii) are in proportion.
In simple words: Multiply the first and last numbers. Also multiply the second and third numbers. If both products match, the four numbers form a proportion; if not, they do not.

Exam Tip: When dealing with mixed numbers, convert them to improper fractions first to avoid arithmetic mistakes during multiplication.

Question 3. Find x in the following proportions:
(i) x : 4 = 9 : 12
(ii) \( \frac{1}{13} : x :: \frac{1}{2} : \frac{1}{5} \)
(iii) 3.6 : 0.4 = x : 0.5
Answer:
(i) Using the property product of extremes = product of means:
x × 12 = 4 × 9
12x = 36
x = 3

(ii) \( \frac{1}{13} \times \frac{1}{5} = x \times \frac{1}{2} \)
\( \frac{1}{65} = \frac{x}{2} \)
x = \( \frac{2}{65} \)

(iii) 3.6 × 0.5 = 0.4 × x
1.8 = 0.4x
x = \( \frac{1.8}{0.4} \) = 4.5
In simple words: In a proportion, the product of the outer numbers equals the product of the inner numbers. Use this to set up an equation and solve for x.

Exam Tip: Always cross multiply first before trying to simplify - this reduces computational errors and keeps the algebra clear and organized.

Question 4. Find the fourth proportional to
(i) 42, 12, 7
(ii) \( \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \)
(iii) 3 kg, 12 kg, 15 kg
Answer: If a, b, c, x are in proportion, then x is the fourth proportional and a × x = b × c.

(i) Set up the proportion: 42 : 12 :: 7 : x. Cross multiplying: 42 × x = 12 × 7. This gives 42x = 84, so x = 2. The fourth proportional is 2.

(ii) Set up the proportion: \( \frac{1}{3} : \frac{1}{4} :: \frac{1}{5} : x \). Using the property: \( \frac{1}{3} \times x = \frac{1}{4} \times \frac{1}{5} \). This gives \( \frac{x}{3} = \frac{1}{20} \), so x = \( \frac{3}{20} \). The fourth proportional is \( \frac{3}{20} \).

(iii) Set up the proportion: 3 : 12 :: 15 : x. Cross multiplying: 3 × x = 12 × 15. This gives 3x = 180, so x = 60 kg. The fourth proportional is 60 kg.
In simple words: The fourth proportional is found by using the rule: first × fourth = second × third. Set up the equation and solve for the unknown fourth number.

Exam Tip: Write down the proportion clearly with all four numbers before attempting to solve - this visual arrangement prevents mixing up which numbers multiply together.

Question 4. (iii) 3 kg, 12 kg, 15 kg
Answer: Set up the proportion: 3 : 12 :: 15 : x

Multiply the extremes and means: 3 × x = 12 × 15

3x = 180

x = 180 ÷ 3 = 60 kg

The fourth proportional is 60 kg.
In simple words: Use cross multiplication. Multiply 12 by 15, then divide by 3 to get the missing number.

Exam Tip: Always check your answer by verifying the proportion holds: 3 : 12 :: 15 : 60 simplifies to 1 : 4 on both sides.

Question 5. Check whether 7, 49, 343 are in continued proportion or not.
Answer: For three numbers a, b, c to be in continued proportion, the condition a : b :: b : c must hold, which means a × c = b × b.

Here, a = 7, b = 49, c = 343.

Product of extremes = 7 × 343 = 2401

Product of means = 49 × 49 = 2401

Since 2401 = 2401, the numbers follow the continued proportion rule.

Therefore, yes, 7, 49, 343 are in continued proportion.
In simple words: When you multiply the first and last numbers, you get the same answer as multiplying the middle number by itself. This means they are in continued proportion.

Exam Tip: Always check that the product of the first and third terms equals the product of the middle term with itself - this is the key test for continued proportion.

Question 6. Find the third proportional to
(i) 36, 18
(ii) \( 5\frac{1}{4} \), 7
(iii) 3.2, 0.8
Answer: When a, b, x are in continued proportion, x is the third proportional and we use a × x = b × b.

(i) 36, 18

Set up: 36 : 18 :: 18 : x

36 × x = 18 × 18

36x = 324

x = 324 ÷ 36 = 9

The third proportional is 9.

(ii) \( 5\frac{1}{4} \), 7

Convert: \( 5\frac{1}{4} = \frac{21}{4} \)

Set up: \( \frac{21}{4} \) : 7 :: 7 : x

\( \frac{21}{4} \) × x = 7 × 7

\( \frac{21x}{4} \) = 49

x = \( \frac{49 × 4}{21} = \frac{196}{21} = \frac{28}{3} = 9\frac{1}{3} \)

The third proportional is \( 9\frac{1}{3} \).

(iii) 3.2, 0.8

Set up: 3.2 : 0.8 :: 0.8 : x

3.2 × x = 0.8 × 0.8

3.2x = 0.64

x = 0.64 ÷ 3.2 = 0.2

The third proportional is 0.2.
In simple words: To find the third proportional, multiply the two middle numbers, then divide by the first number.

Exam Tip: Remember the formula a × x = b × b. Always convert mixed numbers and decimals to fractions when needed to make calculation clearer.

Question 7. The ratio between the length and width of a rectangular sheet of paper is 7 : 5. If the width of the sheet is 20.5 cm, find its length.
Answer: Given information:

Length : Width = 7 : 5

Width = 20.5 cm

Let the length = x cm.

Set up the proportion: \( \frac{x}{20.5} = \frac{7}{5} \)

Using cross multiplication: 5 × x = 7 × 20.5

5x = 143.5

x = 143.5 ÷ 5 = 28.7 cm

The length of the sheet is 28.7 cm.
In simple words: The length and width are in the ratio 7 : 5. Since the width is 20.5 cm, multiply it by 7 ÷ 5 to get the length.

Exam Tip: Set up the ratio correctly with the unknown length in the numerator and known width in the denominator, matching the given ratio order.

Question 8. The ages of Advik and Anaya are in the ratio 4 : 5. If Advik is 4 years 8 months old, find the age of Anaya.
Answer: Given information:

Advik : Anaya = 4 : 5

Advik's age = 4 years 8 months = (4 × 12 + 8) months = 56 months

Let Anaya's age = x months.

Set up the proportion: \( \frac{56}{x} = \frac{4}{5} \)

Using cross multiplication: 4 × x = 56 × 5

4x = 280

x = 280 ÷ 4 = 70 months

Converting back: 70 months = 5 years 10 months

The age of Anaya is 5 years 10 months.
In simple words: Convert ages to months to work with whole numbers, then use cross multiplication to find the unknown age, and convert back to years and months.

Exam Tip: Always convert time to a single unit (months) before using the proportion formula to avoid errors with mixed units.

Exercise 6.3

Question 1. 6 bowls cost Rs.90. What would be cost of 10 such bowls?
Answer: Given information:

Cost of 6 bowls = Rs.90

Cost of 1 bowl = 90 ÷ 6 = Rs.15

Cost of 10 bowls = 10 × Rs.15 = Rs.150

The cost of 10 bowls is Rs.150.
In simple words: Find the cost of one bowl first, then multiply by 10 to get the total cost.

Exam Tip: Breaking the problem into unit cost first (cost of 1 item) makes it easier to find the cost of any quantity.

Question 2. Ten pencils cost Rs.15. How many pencils can be bought with Rs.72?
Answer: Given information:

Cost of 10 pencils = Rs.15

Cost of 1 pencil = 15 ÷ 10 = Rs.1.50

Number of pencils that can be bought with Rs.72 = 72 ÷ 1.50 = 48

48 pencils can be bought with Rs.72.
In simple words: Find the price of one pencil first, then divide the total money by the unit price to see how many you can buy.

Exam Tip: Always find the unit cost (cost per item) first, then use it to solve for any other quantity.

Question 3. 400 grams cake costs 800 rupees. How much would a 1.5 kg cake cost?
Answer: Given information:

Cost of 400 g cake = Rs.800

Cost of 1 g cake = 800 ÷ 400 = Rs.2

Convert 1.5 kg to grams: 1.5 kg = 1.5 × 1000 g = 1500 g

Cost of 1500 g cake = 1500 × Rs.2 = Rs.3000

A 1.5 kg cake would cost Rs.3000.
In simple words: Find the cost per gram, change kilograms to grams, then multiply to get the total cost.

Exam Tip: Convert all measurements to the same unit before calculating, as mixing units can lead to incorrect answers.

Question 4. A man earns Rs.18000 in 3 months.
(i) How much time would he take to earn Rs.30000?
(ii) How much money will he earn in 7 months?
Answer: Given information:

Earning in 3 months = Rs.18000

Earning in 1 month = 18000 ÷ 3 = Rs.6000

(i) Time taken to earn Rs.30000 = 30000 ÷ 6000 = 5 months

He would take 5 months to earn Rs.30000.

(ii) Money earned in 7 months = 7 × Rs.6000 = Rs.42000

He will earn Rs.42000 in 7 months.
In simple words: Calculate monthly earnings first. Then use it to find how long for a target amount, or how much in a given time.

Exam Tip: Finding the rate per unit (monthly earnings) is the key to solving both parts of this problem quickly.

Question 5. 12 mangoes weigh 2.4 kg. What is the weight of 8 mangoes?
Answer: Given information:

Weight of 12 mangoes = 2.4 kg

Weight of 1 mango = 2.4 ÷ 12 = 0.2 kg

Weight of 8 mangoes = 8 × 0.2 kg = 1.6 kg

The weight of 8 mangoes is 1.6 kg.
In simple words: Find the weight of one mango, then multiply by 8 to get the total weight.

Exam Tip: Always calculate the unit measure first (weight or cost of one item) before solving for larger quantities.

Question 6. If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh \( 2\frac{1}{2} \) kilograms?
Answer: Given information:

Weight of 12 sheets = 40 grams

Convert: \( 2\frac{1}{2} \) kg = \( \frac{5}{2} \) × 1000 g = 2500 g

Number of sheets weighing 40 g = 12

Number of sheets weighing 1 g = 12 ÷ 40 = \( \frac{12}{40} \)

Number of sheets weighing 2500 g = 2500 × \( \frac{12}{40} \) = \( \frac{30000}{40} \) = 750

750 sheets would weigh \( 2\frac{1}{2} \) kilograms.
In simple words: Convert kilograms to grams. Find how many sheets per gram, then multiply by the total grams.

Exam Tip: Convert all units to the same measurement system before starting calculations to avoid confusion.

Question 7. A bus consumes 25 litres of diesel in covering a distance of 90 kilometres. How much diesel is needed to cover 288 kilometres?
Answer: Given information:

Diesel needed for 90 km = 25 litres

Diesel needed for 1 km = 25 ÷ 90 litre = \( \frac{5}{18} \) litre

Diesel needed for 288 km = 288 × \( \frac{5}{18} \) = 16 × 5 = 80 litres

80 litres of diesel is needed to cover 288 kilometres.
In simple words: Find how much diesel is used per kilometre, then multiply by the total distance to get the amount needed.

Exam Tip: Simplify the fraction early (90 and 288 share common factors) to make the multiplication easier.

Question 8. If \( \frac{4}{5} \) metre cloth costs Rs.36, find the cost of \( 2\frac{1}{5} \) metres of cloth.
Answer: Given information:

Cost of \( \frac{4}{5} \) metre cloth = Rs.36

Cost of 1 metre cloth = 36 ÷ \( \frac{4}{5} \) = 36 × \( \frac{5}{4} \) = Rs.45

Convert: \( 2\frac{1}{5} \) metres = \( \frac{11}{5} \) metres

Cost of \( \frac{11}{5} \) metres = \( \frac{11}{5} \) × Rs.45 = 11 × 9 = Rs.99

The cost of \( 2\frac{1}{5} \) metres of cloth is Rs.99.
In simple words: Find the cost per metre by dividing 36 by the fraction of the metre given. Then multiply by the total metres needed.

Exam Tip: When dividing by a fraction, flip it and multiply - this avoids errors when working with fractional quantities.

Question 9. If 15 men can pack 540 parcels per day, how many men are needed to pack 396 parcels per day?
Answer: Given information:

Number of men to pack 540 parcels = 15

Number of parcels packed by 1 man = 540 ÷ 15 = 36

Number of men needed to pack 396 parcels = 396 ÷ 36 = 11

11 men are needed to pack 396 parcels per day.
In simple words: Find how many parcels one man can pack, then divide the target number of parcels by that amount to find how many men you need.

Exam Tip: Always find the rate per unit person or thing first, then use it to calculate the number of units needed.

Question 10. Which is a better buy : 12 kg potatoes for Rs.132 or 16 kg potatoes for Rs.168?
Answer: For 12 kg potatoes:

Cost of 12 kg = Rs.132

Cost of 1 kg = 132 ÷ 12 = Rs.11

For 16 kg potatoes:

Cost of 16 kg = Rs.168

Cost of 1 kg = 168 ÷ 16 = Rs.10.50

Since Rs.10.50 is less than Rs.11, the second option costs less per kilogram.

16 kg potatoes for Rs.168 is the better buy.
In simple words: Calculate the price per kilogram for each option and pick the one with the lower cost per kilogram.

Exam Tip: To compare value between offers, always calculate the unit cost (cost per one item) and compare those numbers, not just the total prices.

Exercise 6.4

Question 1. Convert the following speeds into m/sec :
(i) 72 km/h
(ii) 9 km/h
(iii) 1.2 km/minute
(iv) 600 m/hour
Answer: To convert km/h into m/sec, multiply by \( \frac{5}{18} \).

(i) 72 km/h = 72 × \( \frac{5}{18} \) = 4 × 5 = 20 m/sec

Hence, 72 km/h = 20 m/sec.

(ii) 9 km/h = 9 × \( \frac{5}{18} \) = \( \frac{45}{18} \) = 2.5 m/sec

Hence, 9 km/h = 2.5 m/sec.

(iii) 1.2 km/minute

1.2 km = 1.2 × 1000 m = 1200 m and 1 minute = 60 seconds.

\( \frac{1200}{60} \) = 20 m/sec

Hence, 1.2 km/minute = 20 m/sec.

(iv) 600 m/hour

1 hour = 3600 seconds.

\( \frac{600}{3600} = \frac{1}{6} \) m/sec

Hence, 600 m/hour = \( \frac{1}{6} \) m/sec.
In simple words: To change km/h to m/sec, multiply by 5/18. To change distance/time pairs directly, divide metres by seconds.

Exam Tip: Memorise the conversion factor 5/18 for km/h to m/sec, as it appears in many speed problems and saves time.

Question 2. Convert the following speeds into km/h :
(i) 15 m/sec
(ii) 1.5 m/sec
Answer: To convert m/sec into km/h, multiply by \( \frac{18}{5} \).

(i) 15 m/sec = 15 × \( \frac{18}{5} \) = 3 × 18 = 54 km/h

Hence, 15 m/sec = 54 km/h.

(ii) 1.5 m/sec = 1.5 × \( \frac{18}{5} \) = \( \frac{27}{5} \) = 5.4 km/h

Hence, 1.5 m/sec = 5.4 km/h.
In simple words: To change m/sec to km/h, multiply by 18/5, which is the reverse of the conversion from km/h to m/sec.

Exam Tip: Notice that 18/5 is the reciprocal of 5/18 - knowing one conversion makes the other easy to remember.

Question 3. Which is greater - a speed of 30 m/sec or 30 km/h?
Answer: Convert 30 m/sec into km/h by multiplying by \( \frac{18}{5} \).

30 m/sec = 30 × \( \frac{18}{5} \) = 6 × 18 = 108 km/h

Now compare 108 km/h and 30 km/h.

Since 108 km/h > 30 km/h, a speed of 30 m/sec is greater.

Hence, a speed of 30 m/sec is greater.
In simple words: Convert both speeds to the same unit, then compare the numbers to see which is larger.

Exam Tip: Always convert speeds to the same unit before comparing them - never compare m/sec directly with km/h.

Question 4. An aeroplane is flying at a speed of 720 km/h
(i) If the aerial distance between two cities is 1800 km, how much time will the aeroplane take in crossing these cities?
(ii) How much distance does the aeroplane cover in 40 minutes?
(iii) How far will it fly in 15 seconds?
Answer: Given information:

Speed of aeroplane = 720 km/h

(i) Time = \( \frac{\text{Distance}}{\text{Speed}} = \frac{1800}{720} = \frac{5}{2} = 2\frac{1}{2} \) hours

Hence, the aeroplane will take \( 2\frac{1}{2} \) hours to cross the two cities.

(ii) 40 minutes = \( \frac{40}{60} \) hour = \( \frac{2}{3} \) hour

Distance = Speed × Time = 720 × \( \frac{2}{3} \) = 240 × 2 = 480 km

Hence, the aeroplane covers 480 km in 40 minutes.

(iii) Convert the speed into m/sec: 720 × \( \frac{5}{18} \) = 40 × 5 = 200 m/sec

Distance covered in 15 seconds = 200 × 15 = 3000 m = 3 km

Hence, the aeroplane will fly 3 km in 15 seconds.
In simple words: Use the formulas: time = distance ÷ speed, distance = speed × time. Convert units when needed to match the given values.

Exam Tip: Always ensure time and speed units match (if speed is in km/h, convert time to hours; if speed is in m/sec, convert time to seconds) before multiplying or dividing.

Question 5. A dog is walking at a speed of 6 km/h.
Answer: [Question text incomplete in source; answer section not provided. Complete answer cannot be generated without the specific question being asked about the dog's walking speed.]
In simple words: [Awaiting question completion to provide answer.]

Exam Tip: Ensure the complete question is stated before solving problems involving rates and measurements.

Question. (i) How much distance does it cover in 5 minutes?
Answer: The dog travels at 6 km/h. In 5 minutes (which is \( \frac{1}{12} \) of an hour), it covers \( 6 \times \frac{1}{12} = \frac{1}{2} \) km, or 500 metres.
In simple words: The dog moves 500 metres in 5 minutes.

Exam Tip: Always convert time to the same unit as speed — here, 5 minutes becomes a fraction of an hour, then multiply by the speed to get distance.

Question. (ii) How much time would it take to cover 200 metres?
Answer: First, convert 6 km/h to metres per hour: \( 6 \times 1000 = 6000 \) m/h. Now use the formula: Time = \( \frac{\text{Distance}}{\text{Speed}} = \frac{200}{6000} = \frac{1}{30} \) hour. Converting to minutes: \( \frac{1}{30} \times 60 = 2 \) minutes. So the dog takes 2 minutes to cover 200 metres.
In simple words: First change speed to metres per hour, then divide distance by speed to get time in hours, then change to minutes.

Exam Tip: Make sure your units match throughout — convert everything to the same system before doing calculations.

Question 6. A swimming pool is 50 metres long. A boy can swim across the length and then return to his starting position in 5 minutes. What is his swimming speed in km/h?
Answer: The boy swims from one end to the other and back, covering a total distance of \( 50 \times 2 = 100 \) metres, which equals \( \frac{100}{1000} = 0.1 \) km. The time taken is 5 minutes = \( \frac{5}{60} = \frac{1}{12} \) hour. Using Speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{0.1}{\frac{1}{12}} = 0.1 \times 12 = 1.2 \) km/h. Therefore, his swimming speed is 1.2 km/h.
In simple words: The boy swims a round trip of 100 metres in 5 minutes, which works out to 1.2 km/h.

Exam Tip: Remember to count the total distance (both ways) and convert time to hours when finding speed in km/h.

Question 7. A bus takes 48 minutes to cover a certain distance when travelling at a speed of 50 km/h. How much time will it take to cover the same distance when travelling at a speed of 30 km/h?
Answer: First, find the distance. Time = 48 minutes = \( \frac{48}{60} = \frac{4}{5} \) hour. Distance = Speed \( \times \) Time = \( 50 \times \frac{4}{5} = 40 \) km. Now, at 30 km/h: Time = \( \frac{\text{Distance}}{\text{Speed}} = \frac{40}{30} = \frac{4}{3} \) hours. Converting: \( \frac{4}{3} \times 60 = 80 \) minutes = 1 hour 20 minutes. Thus, the bus will take 1 hour 20 minutes to cover 40 km at 30 km/h.
In simple words: Find how far it goes in the first case, then use that distance to find time at the slower speed.

Exam Tip: When distance is unknown, calculate it first from the first set of conditions, then apply it to the second set.

Question 1. Fill in the blanks: (i) The simplest form of the ratio \( \frac{1}{6} : \frac{1}{4} \) is ............... .
Answer: (i) 2 : 3. The LCM of 6 and 4 is 12. Multiplying both parts of the ratio by 12: \( \frac{1}{6} \times 12 : \frac{1}{4} \times 12 = 2 : 3 \).
In simple words: To simplify a ratio with fractions, multiply both by their LCM to make whole numbers.

Exam Tip: Always find the LCM of the denominators when working with fractional ratios.

Question 1. (ii) 75 cm : 1.25 m = ............... .
Answer: (ii) 3 : 5. Convert both to the same unit: 1.25 m = 125 cm. So the ratio is 75 : 125. Dividing both by their GCD of 25 gives 3 : 5.
In simple words: Change both quantities to the same unit first, then simplify by dividing by the common factor.

Exam Tip: Never compare measurements in different units — always convert to the same unit before simplifying.

Question 1. (iii) If two ratios are equivalent, then the four quantities are said to be in ............... .
Answer: (iii) proportion. When two ratios are equal, the four numbers involved are said to be in proportion with each other.
In simple words: If two ratios match exactly, their four numbers form a proportion.

Exam Tip: The key difference: a ratio compares two quantities, while proportion states that two ratios are equal.

Question 1. (iv) If 8, x, 48 and 18 are in proportion then the value of x is ............... .
Answer: (iv) 3. Using the property of proportion: product of extremes = product of means. So \( 8 \times 18 = x \times 48 \), which gives \( 144 = 48x \), therefore \( x = 3 \).
In simple words: In a proportion, the first and last numbers multiplied equal the middle two numbers multiplied together.

Exam Tip: Always use the cross-multiplication rule (extremes and means) when solving for an unknown in a proportion.

Question 1. (v) If the cost of 10 pencils is Rs 15, then the cost of 6 pencils is ............... .
Answer: (v) Rs 9. Cost of one pencil = \( \frac{15}{10} = \text{Rs } 1.50 \). Cost of 6 pencils = \( 6 \times 1.50 = \text{Rs } 9 \).
In simple words: Divide the total cost by the quantity to find the cost per unit, then multiply by the new quantity.

Exam Tip: For problems involving unit rates, always find the cost (or amount) per single item first.

Question 1. (vi) If a cyclist is travelling at a speed of 15 km/h, then the distance covered by him in 20 minutes is ............... .
Answer: (vi) 5 km. First, convert 20 minutes to hours: 20 minutes = \( \frac{20}{60} = \frac{1}{3} \) hour. Then, Distance = Speed \( \times \) Time = \( 15 \times \frac{1}{3} = 5 \) km.
In simple words: Convert time to hours, then multiply speed by time to find distance.

Exam Tip: Always express time in the same unit as the speed (hours for km/h, seconds for m/sec).

Question 2. State whether the following statements are true (T) or false (F): (i) A ratio is always greater than 1.
Answer: (i) False. A ratio can be less than 1. For instance, 2 : 3 equals \( \frac{2}{3} \), which is smaller than 1.
In simple words: Some ratios give numbers less than 1, some give numbers greater than 1.

Exam Tip: Remember that a ratio compares two quantities and can result in any positive number.

Question 2. (ii) Ratio of half an hour to 20 seconds is 30 : 20.
Answer: (ii) False. Half an hour = 30 minutes = \( 30 \times 60 = 1800 \) seconds. So the actual ratio is 1800 : 20, which simplifies to 90 : 1, not 30 : 20.
In simple words: First convert both measurements to the same unit (seconds), then form the ratio.

Exam Tip: Always ensure both quantities are in the same unit before writing the ratio.

Question 2. (iii) The ratio 5 : 7 is greater than the ratio 5 : 6.
Answer: (iii) False. Converting to compare: \( \frac{5}{7} \) and \( \frac{5}{6} \). Cross-multiplying: \( 5 \times 6 = 30 \) and \( 5 \times 7 = 35 \). Since 30 < 35, we have \( \frac{5}{7} < \frac{5}{6} \). Therefore, 5 : 7 is smaller than 5 : 6.
In simple words: When the first number stays the same, a smaller second number makes the ratio larger.

Exam Tip: To compare two ratios, cross-multiply and compare the products.

Question 2. (iv) If the numbers 3, 5, 12 and x are in proportion then the value of x is 20.
Answer: (iv) True. Using the property: \( 3 \times x = 5 \times 12 \), so \( 3x = 60 \), giving \( x = 20 \). The statement is true.
In simple words: Apply the cross-multiplication rule: multiply the outer numbers and set them equal to the product of the middle numbers.

Exam Tip: For proportions, product of extremes always equals product of means.

Question 2. (v) The ratios 3 : 2 and 4 : 5 are equivalent.
Answer: (v) False. When written as fractions, 3 : 2 = 1.5 and 4 : 5 = 0.8, which are not equal. Therefore, these ratios are not equivalent.
In simple words: Two ratios are equivalent only if they simplify to the same value.

Exam Tip: Always convert ratios to decimal or simplified form to check if they are equivalent.

Question 3. A ratio equivalent to 6 : 10 is
(a) 3 : 4
(b) 18 : 30
(c) 12 : 40
(d) 5 : 3
Answer: (b) 18 : 30
In simple words: A ratio stays equivalent when you multiply (or divide) both numbers by the same value. Since \( 6 \times 3 = 18 \) and \( 10 \times 3 = 30 \), the ratio 18 : 30 is equivalent to 6 : 10.

Exam Tip: To find equivalent ratios, multiply or divide both terms by the same non-zero number.

Question 4. A ratio equivalent to the ratio \( \frac{2}{3} : \frac{3}{4} \) is
(a) 4 : 6
(b) 8 : 9
(c) 6 : 8
(d) 9 : 8
Answer: (b) 8 : 9
In simple words: To remove fractions from a ratio, multiply both parts by their LCM. The LCM of 3 and 4 is 12, so \( \frac{2}{3} \times 12 = 8 \) and \( \frac{3}{4} \times 12 = 9 \), giving 8 : 9.

Exam Tip: When a ratio contains fractions, multiply both parts by the LCM of the denominators to get a whole-number ratio.

Question 5. The ratio of 75 mL to 3 litres is
(a) 25 : 1
(b) 40 : 1
(c) 1 : 40
(d) 3 : 200
Answer: (c) 1 : 40
In simple words: Convert 3 litres to millilitres: \( 3 \times 1000 = 3000 \) mL. The ratio becomes 75 : 3000. Divide both by 75 to get 1 : 40.

Exam Tip: Always convert quantities to the same unit before forming a ratio, then simplify by dividing by the GCD.

Question 6. The ratio of the number of sides of a rectangle to the number of edges of a cuboid is
(a) 1 : 2
(b) 1 : 3
(c) 2 : 3
(d) none of these
Answer: (b) 1 : 3
In simple words: A rectangle has 4 sides and a cuboid has 12 edges. The ratio is 4 : 12, which simplifies to 1 : 3.

Exam Tip: Make sure you identify the correct geometric property (sides, edges, vertices) before setting up the ratio.

Question 7. In a class of 35 students, the number of girls is 20. The ratio of number of boys to the number of girls in the class is
(a) 3 : 4
(b) 4 : 3
(c) 7 : 4
(d) 7 : 3
Answer: (a) 3 : 4
In simple words: Number of boys = 35 - 20 = 15. The ratio of boys to girls = 15 : 20. Simplify by dividing by 5 to get 3 : 4.

Exam Tip: Always subtract to find the missing quantity before forming the ratio.

Question 8. The ratio of number of girls to the number of boys in a class is 6 : 7. If there are 21 boys in the class, then the number of girls in the class is
(a) 39
(b) 24
(c) 18
(d) 13
Answer: (c) 18
In simple words: In the ratio 6 : 7, the part for boys is 7. Since 7 parts = 21 boys, 1 part = 3. The number of girls = 6 parts = 6 × 3 = 18.

Exam Tip: Break down the ratio into equal parts, find the value of one part, then multiply to find the unknown quantity.

Question 9. Two numbers are in the ratio 3 : 5. If the sum of the numbers is 144, then the larger number is
(a) 48
(b) 54
(c) 72
(d) 90
Answer: (d) 90
In simple words: Total parts = 3 + 5 = 8. Each part = 144 ÷ 8 = 18. The larger number = 5 × 18 = 90.

Exam Tip: When given a ratio and a sum, add the ratio numbers to find total parts, then divide the sum by that total.

Question 10. If x, 12, 8 and 32 are in proportion, then x is
(a) 6
(b) 4
(c) 3
(d) 2
Answer: (c) 3
In simple words: Using \( x : 12 :: 8 : 32 \), cross-multiply: \( x \times 32 = 12 \times 8 \), so \( 32x = 96 \), giving \( x = 3 \).

Exam Tip: In a proportion, always apply the rule: product of extremes = product of means.

Question 11. If 3, 12 and x are in continued proportion, then x is
(a) 4
(b) 6
(c) 16
(d) 48
Answer: (d) 48
In simple words: In continued proportion \( 3 : 12 :: 12 : x \), the middle term is used twice. Cross-multiplying: \( 3 \times x = 12 \times 12 \), so \( 3x = 144 \), giving \( x = 48 \).

Exam Tip: For continued proportion, the middle term appears in both ratios, and you apply the same cross-multiplication rule.

Question 12. If the weight of 5 bags of sugar is 27 kg, then the weight of one bag of sugar is
(a) 5.04 kg
(b) 5.2 kg
(c) 5.4 kg
(d) 5.6 kg
Answer: (c) 5.4 kg
In simple words: Divide the total weight by the number of bags: \( \frac{27}{5} = 5.4 \) kg per bag.

Exam Tip: To find unit weight (or unit cost), always divide the total by the quantity.

Question 13. Sonali bought one dozen notebooks for Rs 66. What did she pay for one notebook?
(a) Rs 6.50
(b) Rs 6.60
(c) Rs 5.60
(d) Rs 5.50
Answer: (d) Rs 5.50
In simple words: One dozen = 12 notebooks. Cost per notebook = \( \frac{66}{12} = \text{Rs } 5.50 \).

Exam Tip: Always find the unit rate by dividing total cost by total quantity.

Question 14. The speed of 90 km/h is equal to
(a) 10 m/sec
(b) 18 m/sec
(c) 25 m/sec
(d) none of these
Answer: (c) 25 m/sec
In simple words: To convert km/h to m/sec, multiply by \( \frac{5}{18} \). So \( 90 \times \frac{5}{18} = 5 \times 5 = 25 \) m/sec.

Exam Tip: Remember the conversion factor: multiply by \( \frac{5}{18} \) to change km/h to m/sec, or multiply by \( \frac{18}{5} \) to go the other way.

Question 15. Statement I: Two numbers are in the ratio 4 : 5. If the sum of the numbers is 27, then the smaller number is 12. Statement II: We can multiply or divide both the terms of a ratio by the same non-zero integer.
(a) Statement I is true but Statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (c) Both Statement I and Statement II are true
In simple words: For Statement I, total parts = 4 + 5 = 9, so 1 part = 3, and the smaller number = 4 × 3 = 12 (true). For Statement II, multiplying or dividing both parts of a ratio by the same non-zero number always gives an equivalent ratio (true).

Exam Tip: Verify both statements independently before selecting your answer on two-statement questions.

Question. If 7 bananas cost Rs.49, then the cost of a dozen bananas is Rs.70
Answer: The cost of one banana works out to Rs.7 (49 ÷ 7). A dozen bananas would therefore cost Rs.84 (12 × 7), not Rs.70. This statement is false.
In simple words: If 7 bananas cost Rs.49, one banana costs Rs.7. So 12 bananas cost Rs.84, not Rs.70.

Exam Tip: Always divide the total cost by the number of items first to find the unit cost, then multiply by the required quantity to check if the claim is correct.

Question. If 7 pens cost Rs.77 and 5 notebooks cost Rs.60, then the cost of one dozen pens is higher than the cost of one dozen notebooks.
Answer: A single pen costs Rs.11 (77 ÷ 7), so 12 pens would cost Rs.132 (12 × 11). A single notebook costs Rs.12 (60 ÷ 5), so 12 notebooks would cost Rs.144 (12 × 12). Since Rs.132 is less than Rs.144, a dozen pens actually cost less than a dozen notebooks. This statement is false.
In simple words: One pen costs Rs.11, so 12 pens cost Rs.132. One notebook costs Rs.12, so 12 notebooks cost Rs.144. Pens are cheaper.

Exam Tip: When comparing two items, always calculate the unit cost first, then find the cost for the same quantity (like a dozen) before making the comparison.

Question 17. If the speed of car A is 57 km/h and the speed of car B is 17.5 m/s, then car B is faster than car A.
Answer: First, change car A's speed to m/sec: 57 × 5/18 = 285/18 = 15.83 m/sec. Since 17.5 m/sec is greater than 15.83 m/sec, car B is indeed faster than car A. This statement is true.
In simple words: Car A travels at 15.83 m/sec (when converted from 57 km/h). Car B travels at 17.5 m/sec. So car B is faster.

Exam Tip: Always convert speeds to the same unit (usually m/sec) before comparing. Use the conversion factor 5/18 to change km/h to m/sec.

Question 17. Speed = Distance / Time
Answer: This is the correct formula for calculating speed. This statement is true.
In simple words: Speed tells you how far something travels in a fixed amount of time. You find it by dividing the total distance by the total time.

Exam Tip: Memorize the three speed-related formulas: Speed = Distance ÷ Time, Distance = Speed × Time, and Time = Distance ÷ Speed. They are rearrangements of each other.

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Question 1. A rectangular park is 120 m long and 75 m wide. Find the ratio of:
(i) breadth to its length
(ii) length to its perimeter
Answer:
(i) The ratio of breadth to length is 75 : 120. To simplify, find the H.C.F. of 75 and 120 by prime factorization:
75 = 3 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
H.C.F. = 3 × 5 = 15
Therefore: (75 ÷ 15) : (120 ÷ 15) = 5 : 8

(ii) The perimeter of the park is 2 × (120 + 75) = 2 × 195 = 390 m. The ratio of length to perimeter is 120 : 390. To simplify, find the H.C.F. of 120 and 390:
120 = 2 × 2 × 2 × 3 × 5
390 = 2 × 3 × 5 × 13
H.C.F. = 2 × 3 × 5 = 30
Therefore: (120 ÷ 30) : (390 ÷ 30) = 4 : 13
In simple words: To simplify a ratio, divide both numbers by their highest common factor. This gives you the simplest form of the ratio.

Exam Tip: Always reduce ratios to their simplest form by dividing by the H.C.F. Show your prime factorization work to earn full marks.

Question 2. Divide the angles of a triangle in the ratio 2 : 3 : 4
Answer: The sum of all angles in a triangle is always 180°. The total number of parts in the ratio is 2 + 3 + 4 = 9. Each part represents 180° ÷ 9 = 20°. Therefore:
First angle = 2 × 20° = 40°
Second angle = 3 × 20° = 60°
Third angle = 4 × 20° = 80°
Verification: 40° + 60° + 80° = 180° ✓
In simple words: Add up all the ratio parts to find how many equal pieces the 180° is divided into. Then multiply each ratio number by the value of one piece.

Exam Tip: Always verify your answer by adding the three angles to confirm they equal 180°. This shows your work is correct.

Question 3. Heights of Anshul, Ankita and Dhruv are 1.04 m, 1.30 m and 91 cm respectively. Divide 100 sweets among them in the same ratio as their heights.
Answer: First, convert all heights to the same unit (centimetres): Anshul = 104 cm, Ankita = 130 cm, Dhruv = 91 cm. The ratio of their heights is 104 : 130 : 91. To simplify, find the H.C.F. by prime factorization:
104 = 2 × 2 × 2 × 13
130 = 2 × 5 × 13
91 = 7 × 13
H.C.F. = 13
Simplified ratio: (104 ÷ 13) : (130 ÷ 13) : (91 ÷ 13) = 8 : 10 : 7
Total parts = 8 + 10 + 7 = 25. Each part represents 100 ÷ 25 = 4 sweets.
Anshul's share = 8 × 4 = 32 sweets
Ankita's share = 10 × 4 = 40 sweets
Dhruv's share = 7 × 4 = 28 sweets
Verification: 32 + 40 + 28 = 100 ✓
In simple words: Convert all measurements to the same unit first. Then simplify the ratio by dividing by the H.C.F. Finally, divide the total by the number of parts and multiply by each ratio number.

Exam Tip: Always convert all quantities to the same unit before creating a ratio. Verify that your shares add up to the total you started with.

Question 4. The weights of Divya and Himanshu are in the ratio 5 : 7. If Himanshu weighs 28 kg, find the weight of Divya.
Answer: Let Divya's weight be 5x kg and Himanshu's weight be 7x kg. Since Himanshu weighs 28 kg, we have 7x = 28, so x = 4 kg. Therefore, Divya's weight = 5 × 4 = 20 kg.
In simple words: The ratio 5 : 7 means Himanshu's weight (7 parts) equals 28 kg. So one part equals 4 kg. Divya has 5 parts, which is 20 kg.

Exam Tip: When you know one actual value and a ratio, find the value of one part by dividing the known value by its ratio number. Then multiply by the other ratio number.

Question 5. The areas of three flats are in the ratio 5 : 6 : 8. If the difference in the areas of flat C and flat A is 180 square metres, find the area of flat B.
Answer: Let the areas be 5x, 6x, and 8x square metres for flats A, B, and C respectively. The difference between flat C and flat A is 8x - 5x = 3x. According to the problem, 3x = 180, so x = 60. Therefore, the area of flat B = 6x = 6 × 60 = 360 square metres.
In simple words: Use the ratio numbers as coefficients (with an unknown x). Find what 3x equals from the given difference, then calculate the value of x and use it to find flat B's area.

Exam Tip: When given the difference between two quantities in a ratio, use subtraction to find the difference in ratio parts, then solve for x.

Question 6. The income of a man is increased in the ratio 7 : 8. If the increase in his income is Rs.4500 per month, find his new income.
Answer: Let the original income be 7x and the new income be 8x. The increase is 8x - 7x = x. Since the increase is Rs.4500, we have x = 4500. Therefore, the new income = 8 × 4500 = Rs.36000 per month.
In simple words: When income increases in a ratio, the difference between the two ratio numbers tells you how many "parts" the increase represents. Divide the increase by that number of parts to find the value of one part.

Exam Tip: The increase in a ratio is found by subtracting the two ratio numbers. For example, in 7 : 8, the increase is 8 - 7 = 1 part.

Question 7. If 3A = 5B and 4B = 6C, then find A : C.
Answer: From 3A = 5B, divide both sides by 3 and 5 to get A/B = 5/3, or A : B = 5 : 3. From 4B = 6C, divide both sides by 4 and 6 to get B/C = 6/4 = 3/2, or B : C = 3 : 2. Since B has a value of 3 in both ratios, we can write A : B : C = 5 : 3 : 2. Therefore, A : C = 5 : 2.
In simple words: Find the ratio between each pair first. Then make sure the middle quantity (B) has the same value in both ratios so you can combine them into a single three-part ratio.

Exam Tip: When combining ratios, the middle term must have the same numerical value in both ratios. If it doesn't, multiply one or both ratios to make it match.

Question 8. Which ratio is smaller - 9 : 13 or 7 : 11?
Answer: Convert both ratios to fractions: 9/13 and 7/11. Use cross multiplication to compare: 9 × 11 = 99 and 7 × 13 = 91. Since 99 > 91, we have 9/13 > 7/11. Therefore, 7 : 11 is the smaller ratio.
In simple words: To compare two ratios, multiply the numerator of one by the denominator of the other and vice versa. The one that gives the smaller product is the smaller ratio.

Exam Tip: Cross multiplication is the fastest way to compare two fractions or ratios without calculating their decimal values.

Question 9. Find the fourth proportional to
(i) 4, 7, 20
(ii) 2 1/2, 1 1/4, 2.2
Answer:
(i) Set up the proportion as 4 : 7 :: 20 : x. Using the property that the product of extremes equals the product of means: 4 × x = 7 × 20, so 4x = 140, and x = 35. The fourth proportional is 35.

(ii) Convert the mixed numbers to improper fractions: 2 1/2 = 5/2 and 1 1/4 = 5/4. Set up the proportion as 5/2 : 5/4 :: 2.2 : x. Using the property: (5/2) × x = (5/4) × 2.2. Solving: 5x/2 = 11/4, so 5x = 11/2, and x = 11/10 = 1.1. The fourth proportional is 1.1.
In simple words: To find the fourth proportional, use the rule: first number times fourth number equals second number times third number. Then solve for the unknown fourth number.

Exam Tip: Always convert mixed numbers and decimals to fractions before working with proportions to keep calculations precise.

Question 10. A typist types 70 pages in 3 hours 30 minutes. How long will she take to type 300 pages?
Answer: Convert 3 hours 30 minutes to 210 minutes (or 3.5 hours). The time to type one page is 210 ÷ 70 = 3 minutes. Therefore, the time to type 300 pages is 300 × 3 = 900 minutes. Convert back to hours: 900 ÷ 60 = 15 hours. She will take 15 hours to type 300 pages.
In simple words: Find how long it takes to type one page. Then multiply that by 300 to find the total time for 300 pages.

Exam Tip: Always find the unit rate first (time per page or pages per hour), then use it to calculate the answer for a different quantity.

Question 11. 12 looms weave 210 m cloth per day. How many metres of cloth will 8 looms weave per day?
Answer: If 12 looms weave 210 m per day, then 1 loom weaves 210 ÷ 12 = 17.5 m per day. Therefore, 8 looms will weave 8 × 17.5 = 140 m per day.
In simple words: Find how much cloth one loom makes per day. Then multiply by the number of looms to find the total production.

Exam Tip: When changing the quantity of workers (or machines), always calculate the output per unit worker first, then scale to the new number.

Question 12. A journey takes 4 hours 30 minutes at a speed of 60 km/h. How long will the same journey take at a speed of 15 m/sec?
Answer: First, convert 4 hours 30 minutes to 9/2 hours (or 4.5 hours). Calculate the distance: 60 × 9/2 = 30 × 9 = 270 km. Next, convert 15 m/sec to km/h: 15 × 18/5 = 3 × 18 = 54 km/h. Finally, find the time at this speed: 270 ÷ 54 = 5 hours. The journey will take 5 hours at 15 m/sec.
In simple words: Find the distance first (speed times time). Then convert the new speed to the same unit as the first speed. Finally, divide the distance by the new speed to get the new time.

Exam Tip: Always ensure both speeds are in the same units before comparing or calculating. Use 5/18 to convert km/h to m/sec, or 18/5 to convert m/sec to km/h.

Question 13. Present ages of Rohit and Mayank are in the ratio 11 : 8. 8 years hence the ratio of their ages will be 5 : 4. Find their present ages.
Answer: Let Rohit's present age be 11x years and Mayank's present age be 8x years. After 8 years, Rohit's age will be (11x + 8) years and Mayank's age will be (8x + 8) years. According to the condition, (11x + 8) : (8x + 8) = 5 : 4. Cross multiply: 4(11x + 8) = 5(8x + 8), which gives 44x + 32 = 40x + 40. Solving: 4x = 8, so x = 2. Therefore, Rohit's present age = 11 × 2 = 22 years and Mayank's present age = 8 × 2 = 16 years.
In simple words: Use a variable for each ratio part. Write equations for the present and future conditions. Solve the equation by cross multiplication to find the value of the variable.

Exam Tip: Always set up two equations - one for the present ratio and one for the future ratio - then solve by cross multiplication to find the unknown multiplier.

Question 14. Ratio of length and breadth of a rectangle is 3 : 2. If the length of rectangle is 5 m more than the breadth, find the perimeter of the rectangle.
Answer: Let the length and breadth be 3x m and 2x m respectively. According to the condition, 3x - 2x = 5, so x = 5. Therefore, length = 3 × 5 = 15 m and breadth = 2 × 5 = 10 m. The perimeter = 2 × (15 + 10) = 2 × 25 = 50 m.
In simple words: Use the ratio numbers with a variable. Write an equation using the given difference. Solve for the variable, then calculate the dimensions and perimeter.

Exam Tip: When a difference is given between two quantities in a ratio, subtract the ratio numbers to set up your equation. This directly gives you the difference in terms of x.