RD Sharma solution class 8 chapter 9 Linear Equation In One Variable Exercise 9.4

Exercise 9.4

Page-9.29

Question 1:

Four-fifth of a number is more than three-fourth of the number by 4. Find the number.

Answer 1:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{4}{5}\mathrm{x}-\frac{3}{4}\mathrm{x}=4 \\ \mathrm{or} \frac{16\mathrm{x}-15\mathrm{x}}{20}=4 \\ \mathrm{or} \mathrm{x}=80 [\text{After} \text{c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{Thus}, \mathrm{the} \mathrm{required} \mathrm{number} \mathrm{is} 80. \end{gathered}$$

Question 2:

The difference between the squares of two consecutive numbers is 31. Find the numbers.

Answer 2:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{numbers} \mathrm{be} \mathrm{x} \mathrm{and} \mathrm{x}+1. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ (\mathrm{x} + 1{)}^2 - {\mathrm{x}}^2 = 31 \\ \mathrm{or} {\mathrm{x}}^2 + 2\mathrm{x} + 1 - {\mathrm{x}}^2 = 31 \\ \mathrm{or} 2\mathrm{x} = 31 - 1 \\ \mathrm{or} \mathrm{x} = \frac{30}{2} \\ \mathrm{or} \mathrm{x} = 15 \\ \mathrm{Thus}, \mathrm{the} \mathrm{numbers} \mathrm{are} 15 \mathrm{and} 16. \end{gathered}$$

Question 3:

Find a number whose double is 45 greater than its half.

Answer 3:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 2\mathrm{x} = \frac{1}{2}\mathrm{x} + 45 \\ or 2\mathrm{x} - \frac{1}{2}\mathrm{x} = 45 \\ \mathrm{or} \frac{4\mathrm{x}-\mathrm{x}}{2} = 45 \\ \mathrm{or} 3\mathrm{x} = 90 [\text{After} \text{c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} \mathrm{x} = \frac{90}{3} \\ \mathrm{or} \mathrm{x} = 30 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 30. \end{gathered}$$

Question 4:

Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 5\mathrm{x} - 5 = 2\mathrm{x} + 4 \\ \mathrm{or} 5\mathrm{x} - 2\mathrm{x} = 4 + 5 \\ \mathrm{or} 3\mathrm{x} = 9 \\ \mathrm{or} \mathrm{x} = \frac{9}{3} \\ \mathrm{or} \mathrm{x} = 3 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 3. \end{gathered}$$

Question 5:

A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.

Answer 5:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{\mathrm{x}}{5} + 5 = \frac{\mathrm{x}}{4} - 5 \\ \mathrm{or} \frac{\mathrm{x}}{5} - \frac{\mathrm{x}}{4} = -5 - 5 \\ \mathrm{or} \frac{4\mathrm{x}-5\mathrm{x}}{20} = -10 \\ \mathrm{or} -\mathrm{x} = -200 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} \mathrm{x} = 200 \\ \mathrm{Thus}, \mathrm{the} \mathrm{number} \mathrm{is} 200. \end{gathered}$$

Question 6:

A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number.

Answer 6:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{units} \mathrm{digit} \mathrm{be} \mathrm{x}. \\ \because \mathrm{Sum} \mathrm{of} \mathrm{two} \mathrm{digits} = 9 \\ \therefore \mathrm{Tens} \mathrm{digit} = (9-\mathrm{x}) \\ \therefore \mathrm{Original} \mathrm{number} =10\times (9-\mathrm{x})+\mathrm{x} \\ \mathrm{Reversed} \mathrm{number} = 10\mathrm{x}+(9-\mathrm{x}) \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 10\times (9-\mathrm{x})+\mathrm{x}-27 = 10\mathrm{x}+(9-\mathrm{x}) \\ \mathrm{or} 90-10\mathrm{x}+\mathrm{x}-27 = 10\mathrm{x}+9-\mathrm{x} \\ \mathrm{or} 9\mathrm{x}+9\mathrm{x} = 90-27-9 \\ \mathrm{or} 18\mathrm{x} = 54 \\ \mathrm{or} \mathrm{x} = \frac{54}{18} = 3 \\ \therefore \text{The n}\mathrm{umber} =10\times (9-3)+3 = 63 \end{gathered}$$

Question 7:

Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.

Answer 7:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{first} \mathrm{part} \mathrm{of} 184 \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{other} \mathrm{part} \mathrm{will} \mathrm{be} (184-\mathrm{x}). \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{1}{3}\mathrm{x}-\frac{1}{7}(184-\mathrm{x}) = 8 \\ \mathrm{or} \frac{7\mathrm{x}-552+3\mathrm{x}}{21} = 8 \\ \mathrm{or} 10\mathrm{x}-552 = 168 [\text{After c}\mathrm{ross} \mathrm{multipl}\text{ication}] \\ \mathrm{or} 10\mathrm{x} = 168+552 \\ \mathrm{or} \mathrm{x} = \frac{720}{10} = 72 \\ \mathrm{Thus}, \mathrm{the} \mathrm{parts} \mathrm{of} 184 \mathrm{are} 72 \mathrm{and} 112 (184-72 = 112). \end{gathered}$$

Question 8:

The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to $\frac{2}{3}$. What is the original fraction equal to?

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{denominator} \mathrm{of} \mathrm{the} \mathrm{fraction} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{numerator} \mathrm{will} \mathrm{be} ( \mathrm{x}-6). \\ \therefore \mathrm{Fraction} = \frac{\mathrm{x}-6}{\mathrm{x}} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{\mathrm{x}-6+3}{\mathrm{x}} = \frac{2}{3} \\ \mathrm{or} \frac{\mathrm{x}-3}{\mathrm{x}} = \frac{2}{3} \\ \mathrm{or} 3\mathrm{x}-9 = 2\mathrm{x} \left[\text{After cross multiplication}\right] \\ \mathrm{or} 3\mathrm{x}-2\mathrm{x} = 9 \\ \mathrm{or} \mathrm{x} = 9 \\ \mathrm{Thus}, \mathrm{the} \mathrm{original} \mathrm{fraction} = \frac{9-6}{9}=\frac{1}{3} \end{gathered}$$

Question 9:

A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50, find the number of notes of each type.

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{Rs}. 10 \mathrm{notes} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{Rs}. 20 \mathrm{notes} \mathrm{will} \mathrm{be} (50-\mathrm{x}). \\ \mathrm{Value} \mathrm{of} \mathrm{Rs}. 10 \mathrm{notes} = 10\mathrm{x} \\ \mathrm{Value} \mathrm{of} \mathrm{Rs}. 20 \mathrm{notes} = 20(50-\mathrm{x}) \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 10\mathrm{x}+20(50-\mathrm{x}) = 800 \\ \mathrm{or} 10\mathrm{x}+1000-20\mathrm{x} = 800 \\ \mathrm{or} 10\mathrm{x} = 1000-800 \\ \mathrm{or} \mathrm{x} = \frac{200}{10} = 20 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{Rs}. 10 \mathrm{notes} = 20 \\ \mathrm{Number} \mathrm{of} \mathrm{Rs}. 20 \mathrm{notes} = (50-20) = 30. \end{gathered}$$ 

Question 10:

Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have?

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} 50 \mathrm{paise} \mathrm{coins} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} 25 \mathrm{paise} \mathrm{coins} \mathrm{will} \mathrm{be} 2\mathrm{x}. \\ \mathrm{Value} \mathrm{of} 50 \mathrm{paise} \mathrm{coins} = \mathrm{Rs}. 0.5\mathrm{x} \\ \mathrm{Value} \mathrm{of} 25 \mathrm{paise} \mathrm{coins} = \mathrm{Rs}. 0.25\times 2\mathrm{x} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 0.5\mathrm{x}+0.25\times 2\mathrm{x} = 9 \\ \mathrm{or} \mathrm{x} = 9 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{fifty} \mathrm{paise} \mathrm{coins} = 9 \\ \mathrm{Number} \mathrm{of} \mathrm{twenty} \mathrm{five} \mathrm{paise} \mathrm{coins} = 2\times 9 = 18 \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{coins} = 9+18 = 27. \end{gathered}$$

Page-9.30

Question 11:

Sunita is twice as old as Ashima. If six years is subtracted from Ashima's age and four years added to Sunita's age, then Sunita will be four times Ashima's age. How old were they two years ago?

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{age} \mathrm{of} \mathrm{Ashima} \text{be} \mathrm{x} \mathrm{years}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{age} \mathrm{of} \mathrm{Sunita} \mathrm{will} \mathrm{be} 2\mathrm{x} \mathrm{years}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 4(\mathrm{x}-6) = 2\mathrm{x}+4 \\ \mathrm{or} 4\mathrm{x}-24 = 2\mathrm{x}+4 \\ \mathrm{or} 4\mathrm{x}-2\mathrm{x} = 4+24 \\ \mathrm{or} 2\mathrm{x} = 28 \\ \mathrm{or} \mathrm{x} = 14 \\ \therefore \mathrm{Age} \mathrm{of} \mathrm{Ashima} = 14 \mathrm{years}. \\ \mathrm{Age} \mathrm{of} \mathrm{Sunita} = 2\times 14 = 28 \mathrm{years}. \end{gathered}$$

Question 12:

The ages of sonu and Monu are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.

Answer 12:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{ratio} \mathrm{of} \mathrm{the} \mathrm{ages} \mathrm{of} \mathrm{Sonu} \mathrm{and} \mathrm{Monu} \mathrm{is} 7:5. \\ \mathrm{Let} \mathrm{the} \mathrm{present} \mathrm{ages} \mathrm{of} \mathrm{Sonu} \mathrm{and} \mathrm{Monu} \mathrm{be} 7x \mathrm{and} 5x \mathrm{years}. \\ \mathrm{After} \mathrm{ten} \mathrm{years}: \\ \mathrm{Age} \mathrm{of} \mathrm{Sonu} = 7\mathrm{x} + 10 \mathrm{years} \\ \mathrm{Age} \mathrm{of} \mathrm{Monu} = 5\mathrm{x} + 10 \mathrm{years} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{7\mathrm{x} + 10}{5\mathrm{x} + 10} = \frac{9}{7} \\ \mathrm{or} 49\mathrm{x} + 70 = 45\mathrm{x} + 90 \\ \mathrm{or} 49\mathrm{x} - 45\mathrm{x} = 90 - 70 \\ o\text{r} 4\mathrm{x} = 20 \\ \mathrm{or} \mathrm{x} = 5 \\ \therefore \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{Sonu} = 7 \times 5 = 35 \mathrm{years}. \\ \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{Monu} = 5 \times 5 = 25 \mathrm{years}. \end{gathered}$$

Question 13:

Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

Answer 13:

$$\begin{gathered} \mathrm{Five} \mathrm{years} \mathrm{ago}: \\ \mathrm{Let} \mathrm{the} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{son} \mathrm{be} x \mathrm{years}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{father} \mathrm{will} \mathrm{be} 7\mathrm{x} \mathrm{years}. \\ \therefore \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{son} = (\mathrm{x} + 5) \mathrm{years} \\ \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{father} = (7\mathrm{x} + 5) \mathrm{years} \\ \mathrm{After} \mathrm{five} \mathrm{years}: \\ \mathrm{Age} \mathrm{of} \mathrm{the} \mathrm{son} = (\mathrm{x} + 5 + 5) = (\mathrm{x} + 10) \mathrm{years} \\ \mathrm{Age} \mathrm{of} \mathrm{the} \mathrm{father} = (7\mathrm{x} + 5 + 5) = (7\mathrm{x} + 10) \mathrm{years} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 7\mathrm{x} + 10 = 3(\mathrm{x} + 10) \\ \mathrm{or} 7\mathrm{x} - 3\mathrm{x} = 30 - 10 \\ \mathrm{or} 4\mathrm{x} = 20 \\ \mathrm{or} \mathrm{x} = 5 \\ \therefore \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{son} = (5 + 5) = 10 \mathrm{years}. \\ \mathrm{Present} \mathrm{age} \mathrm{of} \mathrm{the} \mathrm{father} = (7 \times 5 + 5) = 40 \mathrm{years}. \end{gathered}$$

Question 14:

I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now?

Answer 14:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{age} \mathrm{of} \mathrm{my} \mathrm{son} \mathrm{be} \mathrm{x} \mathrm{years}. \\ \mathrm{Therefore}, \mathrm{my} \mathrm{age} \mathrm{will} \mathrm{be} 5\mathrm{x} \mathrm{years}. \\ \mathrm{After} 6 \mathrm{years}: \\ \mathrm{Age} \mathrm{of} \mathrm{my} \mathrm{son} = (\mathrm{x} + 6) \mathrm{years} \\ \mathrm{My} \mathrm{age} = (5\mathrm{x} + 6) \mathrm{years} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 5\mathrm{x} + 6 = 3(\mathrm{x} + 6) \\ \mathrm{or} 5\mathrm{x} - 3\mathrm{x} = 18 - 6 \\ \mathrm{or} 2\mathrm{x} = 12 \\ \text{or} \mathrm{x} = 6 \\ \therefore \mathrm{Age} \mathrm{of} \mathrm{my} \mathrm{son} = 6 \mathrm{years}. \\ \mathrm{My} \mathrm{age} = 5 \times 6 = 30 \mathrm{years}. \end{gathered}$$

Question 15:

I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

Answer 15:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{five}-\mathrm{rupee} \mathrm{notes} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{ten}-\mathrm{rupee} \mathrm{notes} \mathrm{will} \mathrm{be} (\mathrm{x}+10). \\ \mathrm{Now}, \\ \mathrm{Value} \mathrm{of} \mathrm{five}-\mathrm{rupee} \mathrm{notes} = \mathrm{Rs}. 5\mathrm{x} \\ \mathrm{Value} \mathrm{of} \mathrm{ten}-\mathrm{rupee} \mathrm{notes} = \mathrm{Rs}. 10(\mathrm{x} + 10) \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 5\mathrm{x} + 10(\mathrm{x} + 10) = 1000 \\ \mathrm{or} 15\mathrm{x} = 1000 - 100 \\ \mathrm{or} \mathrm{x} = \frac{900}{15} = 60 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{five}-\mathrm{rupee} \mathrm{notes} = 60. \\ \mathrm{Number} \mathrm{of} \mathrm{ten}-\mathrm{rupee} \mathrm{notes} = 60 + 10 = 70. \end{gathered}$$

Question 16:

At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink any thing. How many guests were in all?

Answer 16:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{guests} \mathrm{be} \mathrm{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{guests}, \mathrm{who} \mathrm{drank} \mathrm{colas}, \mathrm{would} \mathrm{be} \frac{1}{4}\mathrm{x}. \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{guests}, \mathrm{who} \mathrm{drank} \mathrm{squash}, \mathrm{would} \mathrm{be} \frac{1}{3}\mathrm{x}. \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{guests}, \mathrm{who} \mathrm{drank} \mathrm{fruit} \mathrm{juice}, \mathrm{would} \mathrm{be} \frac{2}{5}\mathrm{x}. \\ \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{guests}, \mathrm{who} \mathrm{did} \mathrm{not} \mathrm{drink}, \mathrm{would} \mathrm{be} 3. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \mathrm{x} - (\frac{\mathrm{x}}{4} + \frac{\mathrm{x}}{3} + \frac{2\mathrm{x}}{5}) = 3 \\ \mathrm{or} \frac{60\mathrm{x} - 15\mathrm{x} - 20\mathrm{x} - 24\mathrm{x}}{60} = 3 \\ \mathrm{or} \mathrm{x} = 180 \\ \mathrm{Thus}, \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{guests} =180. \end{gathered}$$

Question 17:

There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

Answer 17:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{correctly} \mathrm{answered} \mathrm{questions} \mathrm{be} x. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{unattempted} \mathrm{or} \mathrm{wrongly} \mathrm{answered} \mathrm{questions} \mathrm{will} \mathrm{be} (180 - \mathrm{x}). \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 4\mathrm{x} - 1(180 - \mathrm{x}) = 450 \\ \mathrm{or} 5\mathrm{x} = 450 + 180 \\ \mathrm{or} \mathrm{x} = \frac{630}{5} = 126 \\ \mathrm{Thus}, \mathrm{number} \mathrm{of} \mathrm{correctly} \mathrm{answered} \mathrm{questions} =126. \\ \mathrm{Number} \mathrm{of} \mathrm{unattempted} \mathrm{or} \mathrm{wrongly} \mathrm{answered} \mathrm{questions} =180 - 126 = 54. \end{gathered}$$

Question 18:

A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day, he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all, for how many days he remained absent?

Answer 18:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{days} \text{for which} \mathrm{the} \mathrm{labour}\text{er} \text{is absent} \mathrm{be} \text{x}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} \text{days for which he is} \mathrm{present} \mathrm{will} \mathrm{be} (20-\mathrm{x}). \\ \therefore \mathrm{Earnings} = \mathrm{Rs}. 60(20 - \mathrm{x}) \\ \mathrm{Fine} = \mathrm{Rs}. 5\mathrm{x} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ 60(20 - \mathrm{x}) - 5\mathrm{x} = 745 \\ \mathrm{or} 1200 - 60\mathrm{x} - 5\mathrm{x} = 745 \\ \mathrm{or} 65\mathrm{x} = 1200 - 745 \\ \mathrm{or} \mathrm{x} = \frac{455}{65} = 7 \\ \mathrm{Thus}, \mathrm{the} \mathrm{labourer} \text{was} \mathrm{absent} \mathrm{for} 7 \mathrm{days}. \end{gathered}$$

Question 19:

Ravish has three boxes whose total weight is $60\frac{1}{2}$ kg. Box B weighs $3\frac{1}{2}$ kg more than box A and box C weighs $5\frac{1}{3}$ kg more than box B. Find the weight of box A.

Answer 19:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{weight} \mathrm{of} \mathrm{box} \mathrm{A} \mathrm{be} \mathrm{x} \mathrm{kg}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{weights} \mathrm{of} \mathrm{box} \mathrm{B} \mathrm{and} \mathrm{box} \mathrm{C} \mathrm{will} \mathrm{be} (\mathrm{x} + 3\frac{1}{2}) \mathrm{kg} \mathrm{and} (\mathrm{x} + 3\frac{1}{2} + 5\frac{1}{3}) \mathrm{kg}, \mathrm{respectively}. \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \mathrm{x} + (\mathrm{x} + 3\frac{1}{2}) + (\mathrm{x} + 3\frac{1}{2} + 5\frac{1}{3}) = 60\frac{1}{2} \\ \mathrm{or} 3\mathrm{x} = \frac{121}{2} - \frac{7}{2} - \frac{7}{2} - \frac{16}{3} \\ \mathrm{or} 3\mathrm{x} = \frac{363 - 21 - 21 - 32}{6} \\ \mathrm{or} 3\mathrm{x} = \frac{289}{6} \\ \mathrm{or} \mathrm{x} = \frac{289}{18} \\ \mathrm{Thus}, \mathrm{weight} \mathrm{of} \mathrm{box} \mathrm{A} = \frac{289}{18} \mathrm{kg} \end{gathered}$$

Question 20:

The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2.  Find the rational number.

Answer 20:

$$\begin{gathered} \mathrm{Let}, \mathrm{the} \mathrm{denominator} \mathrm{of} \mathrm{the} \mathrm{rational} \mathrm{number} \mathrm{be} \mathrm{x}. \\ \therefore \text{T}\mathrm{he} \mathrm{numerator} \mathrm{of} \mathrm{the} \mathrm{rational} \mathrm{number} \mathrm{will} \mathrm{be} \mathrm{x}-3. \\ \therefore \text{T}\mathrm{he} \mathrm{rational} \mathrm{number} =\frac{\mathrm{x}-3}{\mathrm{x}} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{\mathrm{x} - 3 + 2}{\mathrm{x} + 5} = \frac{1}{2} \\ \mathrm{or} \frac{\mathrm{x} - 1}{\mathrm{x} + 5} = \frac{1}{2} \\ \mathrm{or} 2\mathrm{x} - 2 = \mathrm{x} + 5 \\ \mathrm{or} 2\mathrm{x} - \mathrm{x} = 5 + 2 \\ \mathrm{or} \mathrm{x} = 7 \\ \therefore \text{T}\mathrm{he} \mathrm{rational} \mathrm{number} = \frac{7 - 3}{7} = \frac{4}{7} \end{gathered}$$

Question 21:

In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to each, the numerator and the denominator, the new fraction is 2/3. Find the original number.

Answer 21:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{denominator} \mathrm{be} \mathrm{x}. \\ \therefore \text{T}\mathrm{he} \mathrm{numerator} = \frac{x + 2}{2} \\ \therefore \text{T}\mathrm{he} \mathrm{rational} \mathrm{number} = \frac{\mathrm{x} + 2}{2\mathrm{x}} \\ \mathrm{According} \mathrm{to} \mathrm{the} \mathrm{question}, \\ \frac{{\displaystyle \frac{\mathrm{x} + 2}{2} + 3}}{\mathrm{x} + 3} = \frac{2}{3} \\ or \frac{x + 2 + 6}{2(x + 3)} = \frac{2}{3} \\ \mathrm{or} \frac{\mathrm{x} + 8}{2\mathrm{x} + 6} = \frac{2}{3} \\ \mathrm{or} 3\mathrm{x} + 24 = 4\mathrm{x} + 12 \\ \mathrm{or} \mathrm{x} = 24 - 12 \\ \mathrm{or} \mathrm{x} = 12 \\ \therefore \text{T}\mathrm{he} \mathrm{rational} \mathrm{number}=\frac{12 + 2}{2 \times 12} = \frac{14}{24} = \frac{7}{12} \end{gathered}$$

Question 22:

The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.

Answer 22:

$$\begin{gathered} \mathrm{Let}, \mathrm{the} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{train} \mathrm{be} \mathrm{x} \mathrm{km}/\mathrm{h}. \\ \mathrm{Then}, \mathrm{the} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{train} \mathrm{will} \mathrm{be} (\mathrm{x} + 5) \mathrm{km}/\mathrm{h}. \\ 2 \mathrm{hours} \text{after they} \mathrm{start}\text{ed:} \\ \mathrm{Distance} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{train} \mathrm{from} \mathrm{the} \mathrm{starting} \mathrm{point} = 2\mathrm{x} \mathrm{km} \\ \mathrm{Distance} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{train} \mathrm{from} \mathrm{the} \mathrm{starting} \mathrm{point} = 2(\mathrm{x} + 5) \mathrm{km} \\ \mathrm{Now}, \\ 2(\mathrm{x} + 5) + 2\mathrm{x} + 30 = 340 \\ \mathrm{or} 4\mathrm{x} + 10 + 30 = 340 \\ \mathrm{or} 4\mathrm{x} = 340 - 40 \\ \mathrm{or} \mathrm{x} = \frac{300}{4} = 75 \\ \therefore \mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{train} = 75 \mathrm{km}/\mathrm{h}. \\ \mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{other} \mathrm{train} = (75 + 5)=80 \mathrm{km}/\mathrm{h}. \end{gathered}$$

Question 23:

A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports.

Answer 23:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{stream} \mathrm{is} 1 \mathrm{km}/\mathrm{h}. \\ \mathrm{Let} \mathrm{the} \mathrm{speed} \mathrm{of} \mathrm{the} \mathrm{steamer} \mathrm{in} \mathrm{still} \mathrm{water} \mathrm{be} \mathrm{x} \mathrm{km}/\mathrm{h}. \\ \therefore \mathrm{Downstream} \mathrm{speed} = (\mathrm{x} + 1) \mathrm{km}/\mathrm{h} \\ \mathrm{Upstream} \mathrm{speed} = (\mathrm{x} - 1) \mathrm{km}/\mathrm{h} \\ \mathrm{The} \mathrm{downstream} \mathrm{and} \mathrm{upstream} \mathrm{distances} \mathrm{are} \mathrm{same}; \mathrm{therefore}, \mathrm{we} \mathrm{have}: \\ 9(\mathrm{x} + 1) = 10(\mathrm{x} - 1) \\ \mathrm{or} 9\mathrm{x} + 9 = 10\mathrm{x} - 10 \\ \mathrm{or} \mathrm{x} = 19 \\ \therefore \mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{steamer} \mathrm{in} \mathrm{still} \mathrm{water} = 19 \mathrm{km}/\mathrm{h}. \\ \mathrm{Distance} \mathrm{between} \mathrm{the} \mathrm{ports} = 9(19 + 1) = 180 \mathrm{km}. \end{gathered}$$

Question 24:

Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00. How much did she invest at each rate?

Answer 24:

$$\begin{gathered} \mathrm{At} \text{the rate of} 10\%, \mathrm{let} \mathrm{the} \mathrm{investment} \mathrm{by} \mathrm{Bhagwanti} \mathrm{be} \mathrm{Rs}. x. \\ \mathrm{Therefore}, \mathrm{at} \text{the rate of} 12\%, \mathrm{the} \mathrm{investment} \mathrm{will} \mathrm{be} \mathrm{Rs}. (12000-\mathrm{x}). \\ \mathrm{At} \text{the rate of} 10\%, \mathrm{her} \mathrm{annual} \mathrm{income} = \mathrm{x} \times 10\% \\ \mathrm{At}\text{ the rate of} 12\%, \mathrm{her} \mathrm{annual} \mathrm{income} = (12000 - \mathrm{x}) \times 12\% \\ \mathrm{So}, \\ \mathrm{x} \times 0.1 + 0.12(12000 - \mathrm{x}) = 1280 \\ \mathrm{or} 0.1\mathrm{x} - 0.12\mathrm{x} = 1280 - 1440 \\ \mathrm{or} 0.02\mathrm{x} = 160 \\ \mathrm{or} \mathrm{x} = 8000 \\ \mathrm{Thus}, \mathrm{at} \text{the rate of} 10\%, \mathrm{she} \mathrm{invested} \mathrm{Rs}. 8000 \mathrm{and} \mathrm{at} \text{the rate of} 12\%, \mathrm{she} \mathrm{invested} \mathrm{Rs}. 4000 (12000-8000). \end{gathered}$$

Question 25:

The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm² more than that of the given rectangle. Find the length and breath of the given rectangle.

Answer 25:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{be} \mathrm{x} \mathrm{cm}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{will} \mathrm{be} (\mathrm{x} + 9) \mathrm{cm}. \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{rectangle} = \mathrm{x}(\mathrm{x} + 9) {\mathrm{cm}}^2. \\ \mathrm{If} \mathrm{the} \mathrm{length} \mathrm{and} \mathrm{breadth} \mathrm{are} \mathrm{increased} \mathrm{by} 3 \mathrm{cm} \mathrm{each}, \\ \text{a}\mathrm{rea} = (\mathrm{x} + 3)(\mathrm{x} + 9 + 3) {\mathrm{cm}}^2. \\ \mathrm{Now}, \\ (\mathrm{x} + 3)(\mathrm{x} + 12) - \mathrm{x}(\mathrm{x} + 9) = 84 \\ \mathrm{or} {\mathrm{x}}^2 + 15\mathrm{x} + 36 - {\mathrm{x}}^2 - 9\mathrm{x} = 84 \\ \mathrm{or} 6\mathrm{x} = 84 - 36 \\ \mathrm{or} \mathrm{x} = \frac{48}{6} = 8. \\ \mathrm{Thus}, \mathrm{brea}\text{d}\mathrm{th} \mathrm{of} \mathrm{the} \mathrm{rectangle} = 8 \mathrm{cm}. \\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{rectangle} =(8+9)=17 \mathrm{cm}. \end{gathered}$$

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Question 26:

The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?

Answer 26:

$$\begin{gathered} \mathrm{Let} \mathrm{Anup}\text{'}\mathrm{s} \mathrm{age} \mathrm{be} \text{x} \mathrm{years}. \\ \mathrm{Therefore}, \mathrm{his} \mathrm{father}\text{'}\mathrm{s} \mathrm{age} \mathrm{will} \mathrm{be} (100 - \mathrm{x}) \mathrm{years}. \\ \mathrm{When} \mathrm{Anup}\text{ i}s \mathrm{as} \mathrm{old} \mathrm{as} \mathrm{his} \mathrm{father} \mathrm{after} (100 - 2\mathrm{x}) \mathrm{years}, \\ \mathrm{Anuj}\text{'}\mathrm{s} \mathrm{age} = \left(\frac{100 - \mathrm{x}}{5} + 100 - 2\mathrm{x}\right) \mathrm{years}=\frac{600 - 11\mathrm{x}}{5} \mathrm{years}. \\ \mathrm{Again}, \mathrm{when} \mathrm{Anup} \mathrm{is} \mathrm{as} \mathrm{old} \mathrm{as} \mathrm{his} \mathrm{father}, \\ \mathrm{Anuj}\text{'}\mathrm{s} \mathrm{age} = \mathrm{x} + 8. \\ \mathrm{Now}, \\ \frac{600 - 11\mathrm{x}}{5} = \mathrm{x} + 8 \\ \mathrm{or} 600 - 11\mathrm{x} = 5\mathrm{x} + 40 \\ \mathrm{or} 16\mathrm{x} = 560 \\ \mathrm{or} \mathrm{x} = 35. \\ \mathrm{Thus}, \mathrm{Anup}\text{'}\mathrm{s} \mathrm{age} = 35 \mathrm{years} \\ \mathrm{Anup}\text{'}\mathrm{s} \text{f}\mathrm{ather}\text{'}s \mathrm{age}=100 - \mathrm{x} = 100 - 35 = 65 \mathrm{years} \\ \mathrm{Anuj}\text{'}\mathrm{s} \mathrm{age} = \mathrm{x} + 8 = 35 + 8 = 43 \mathrm{years} \end{gathered}$$

Question 27:

A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?

Answer 27:

$$\begin{gathered} \text{Suppose}, \text{the} \mathrm{lady} \mathrm{start}\text{ed} \mathrm{with} \mathrm{x} \mathrm{rupee}s. \\ \text{Money} \mathrm{spent} \mathrm{on} \mathrm{shopping}=\frac{\mathrm{x}}{2} \mathrm{rupee}\text{s} \\ \text{Remaining amount}=\mathrm{x}-\frac{\mathrm{x}}{2}=\frac{\mathrm{x}}{2} \mathrm{rupee}\text{s} \\ \text{A}\mathrm{fter} \mathrm{giving} \mathrm{a} \mathrm{rupee} \mathrm{she} \mathrm{had}=(\frac{\mathrm{x}}{2}-1) \mathrm{rupee}\text{s} \\ \text{Money} \mathrm{spent} \mathrm{on} \mathrm{lunch}=\frac{1}{2}(\frac{\mathrm{x}}{2}-1) \mathrm{rupee}\text{s} \\ \text{A}\mathrm{fter} \mathrm{giving} a \mathrm{two}-\mathrm{rup}ee \mathrm{tip} \mathrm{she} \mathrm{had}=\frac{1}{2}(\frac{\mathrm{x}}{2}-1)-2=\frac{\mathrm{x}-2-8}{4}=\frac{\mathrm{x}-10}{4} \mathrm{rupee}\text{s} \\ \text{Money} \mathrm{spent} \mathrm{on} \mathrm{a} \mathrm{book}=\frac{1}{2}\left(\frac{\mathrm{x}-10}{4}\right) \mathrm{rupee}\text{s} \\ \text{A}\mathrm{fter} \mathrm{spending} \mathrm{three} \mathrm{rupees} \mathrm{on} \mathrm{bus} \mathrm{fare} \mathrm{she} \mathrm{had}=\frac{1}{2}\left(\frac{\mathrm{x}-10}{4}\right)-3=\frac{\mathrm{x}-10-24}{8}=\frac{\mathrm{x}-34}{8} \mathrm{rupee}\text{s} \\ \mathrm{Now}, \\ \frac{\mathrm{x}-34}{8}=1 \\ \mathrm{or} \mathrm{x}-34=8 \\ \mathrm{or} \mathrm{x}=42 \\ \mathrm{Therefore}, \mathrm{she} \mathrm{start}\text{ed} \mathrm{with} 42 \mathrm{rupees}. \end{gathered}$$