Fractions

Q1: A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is:

i. the length of AB to that of CD?

Step 1: The length of AB is 6 cm and the length of CD is 15 cm. To find the fraction of the length of AB to that of CD, we write it as: \[ \frac{\text{Length of AB}}{\text{Length of CD}} = \frac{6}{15} \] Step 2: Now, simplify the fraction: \[ \frac{6}{15} = \frac{6 \div 3}{15 \div 3} = \frac{2}{5} \] Answer: \(\frac{2}{5}\)

ii. \(\frac{1}{2}\) the length of AB to that of \(\frac{1}{3}\) of CD?

Step 1: The length of AB is 6 cm and the length of CD is 15 cm. To find \(\frac{1}{2}\) the length of AB to \(\frac{1}{3}\) of CD, we calculate: \[ \frac{\frac{1}{2} \times 6}{\frac{1}{3} \times 15} = \frac{3}{5} \] Answer: \(\frac{3}{5}\)

iii. \(\frac{1}{5}\) of CD to that of AB?

Step 1: The length of AB is 6 cm and the length of CD is 15 cm. To find \(\frac{1}{5}\) of CD to AB, we calculate: \[ \frac{\frac{1}{5} \times 15}{6} = \frac{3}{6} = \frac{1}{2} \]Answer: \(\frac{1}{2}\)

Q2: Subtract \((\frac{2}{7}-\frac{5}{21}\) from the sum of \(\frac{3}{4},\ \frac{5}{7}\ and\ \frac{7}{12}\).

Step 1: Calculate the sum of \(\frac{3}{4}, \frac{5}{7}\) and \(\frac{7}{12}\).
First, we need to find a common denominator for \(\frac{3}{4}, \frac{5}{7}\) and \(\frac{7}{12}\). The least common denominator (LCD) of 4, 7, and 12 is 84.
Convert each fraction to have the denominator 84: \[ \frac{3}{4} = \frac{3 \times 21}{4 \times 21} = \frac{63}{84}, \quad \frac{5}{7} = \frac{5 \times 12}{7 \times 12} = \frac{60}{84}, \quad \frac{7}{12} = \frac{7 \times 7}{12 \times 7} = \frac{49}{84} \]Now add the fractions: \[ \frac{63}{84} + \frac{60}{84} + \frac{49}{84} = \frac{63 + 60 + 49}{84} = \frac{172}{84} \] Simplify the fraction: \[ \frac{172}{84} = \frac{172 \div 4}{84 \div 4} = \frac{43}{21} \]Step 2: Subtract \(\left(\frac{2}{7} – \frac{5}{21}\right)\) from the sum \(\frac{43}{21}\).
First, simplify the expression inside the parentheses: \[ \frac{2}{7} – \frac{5}{21} \] To subtract, find a common denominator. The LCD of 7 and 21 is 21, so: \[ \frac{2}{7} = \frac{2 \times 3}{7 \times 3} = \frac{6}{21} \] Now subtract: \[ \frac{6}{21} – \frac{5}{21} = \frac{6 – 5}{21} = \frac{1}{21} \]Now subtract \(\frac{1}{21}\) from \(\frac{43}{21}\): \[ \frac{43}{21} – \frac{1}{21} = \frac{43 – 1}{21} = \frac{42}{21} \]Simplify the result: \[ \frac{42}{21} = 2 \]Answer: 2

Q3: From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, \(5\frac{1}{4}\) kg, \(9\frac{1}{2}\) kg and \(9\frac{3}{4}\) kg respectively.

i. How many kg did he sell?

To find the total weight sold, we need to add the weights of all the portions sold.
Step 1: The weights sold are: \[ 6\ \text{kg}, \ 5\frac{1}{4}\ \text{kg}, \ 9\frac{1}{2}\ \text{kg}, \ 9\frac{3}{4}\ \text{kg} \]Step 2: Convert the mixed fractions into improper fractions: \[ 5\frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{21}{4}, \quad 9\frac{1}{2} = \frac{9 \times 2 + 1}{2} = \frac{19}{2}, \quad 9\frac{3}{4} = \frac{9 \times 4 + 3}{4} = \frac{39}{4} \]Step 3: Now, we need to find a common denominator. The least common denominator (LCD) of 1, 4, 2, and 4 is 4. We convert the fractions: \[ 6 = \frac{6 \times 4}{1 \times 4} = \frac{24}{4}, \quad \frac{21}{4} \text{ is already in the correct form}, \quad \frac{19}{2} = \frac{19 \times 2}{2 \times 2} = \frac{38}{4}, \quad \frac{39}{4} \text{ is already in the correct form} \]Step 4: Add the fractions: \[ \frac{24}{4} + \frac{21}{4} + \frac{38}{4} + \frac{39}{4} = \frac{24 + 21 + 38 + 39}{4} = \frac{122}{4} \]Step 5: Simplify the fraction: \[ \frac{122}{4} = 30\frac{2}{4} = 30\frac{1}{2} \]Answer: The merchant sold 30\(\frac{1}{2}\) kg of potatoes.

ii. How many kg are still left in the sack?

To find the remaining potatoes in the sack, we subtract the total sold weight from the original weight of the sack. Step 1: The original weight of the sack is 120 kg, and the weight sold is \(30\frac{1}{2}\) kg.
Step 2: Convert the mixed fraction into an improper fraction: \[ 30\frac{1}{2} = \frac{30 \times 2 + 1}{2} = \frac{61}{2} \] Step 3: To subtract, we need a common denominator. The LCD of 120 and 2 is 2. Convert 120 kg to a fraction with denominator 2: \[ 120 = \frac{120 \times 2}{2} = \frac{240}{2} \] Step 4: Now subtract: \[ \frac{240}{2} – \frac{61}{2} = \frac{240 – 61}{2} = \frac{179}{2} \] Step 5: Convert the improper fraction back to a mixed number: \[ \frac{179}{2} = 89\frac{1}{2} \] Answer: The remaining potatoes in the sack are 89\(\frac{1}{2}\) kg.

Q4: If a boy works for six days for 8 hours, \(7\frac{1}{2}\) hours, \(8\frac{1}{4}\) hours, \(6\frac{1}{4}\) hours, \(6\frac{3}{4}\) hours and 7 hours respectively, how much money will he earn at the rate of ₹36 per hour.

Step 1: First, let’s add up the hours worked by the boy over the six days.
The hours worked on each day are:
– 8 hours
– \(7\frac{1}{2}\) hours = \(\frac{15}{2}\) hours
– \(8\frac{1}{4}\) hours = \(\frac{33}{4}\) hours
– \(6\frac{1}{4}\) hours = \(\frac{25}{4}\) hours
– \(6\frac{3}{4}\) hours = \(\frac{27}{4}\) hours
– 7 hours
Step 2: To add these fractions, we need a common denominator. The least common denominator (LCD) of 1, 2, 4 is 4. Now, convert all fractions to have the denominator 4:
– \(8 = \frac{32}{4}\)
– \(7\frac{1}{2} = \frac{15}{2} = \frac{30}{4}\)
– \(8\frac{1}{4} = \frac{33}{4}\)
– \(6\frac{1}{4} = \frac{25}{4}\)
– \(6\frac{3}{4} = \frac{27}{4}\)
– \(7 = \frac{28}{4}\)
Now, add them all together: \[ \frac{32}{4} + \frac{30}{4} + \frac{33}{4} + \frac{25}{4} + \frac{27}{4} + \frac{28}{4} = \frac{32 + 30 + 33 + 25 + 27 + 28}{4} = \frac{175}{4} \] Simplifying this: \[ \frac{175}{4} = 43\frac{3}{4} \text{ hours} \]Step 3: Now that we know the boy worked for \(43\frac{3}{4}\) hours, we need to calculate how much he will earn. At the rate of ₹36 per hour: \[ 43\frac{3}{4} \times 36 \] Convert \(43\frac{3}{4}\) to an improper fraction: \[ 43\frac{3}{4} = \frac{175}{4} \] Now multiply by 36: \[ \frac{175}{4} \times 36 = \frac{175 \times 36}{4} = \frac{6300}{4} = 1575 \]Answer: ₹1575

Q5: A student bought \(4\frac{1}{3}\) m of yellow ribbon, \(6\frac{1}{6}\) m of red ribbon and \(3\frac{2}{9}\) m of blue ribbon for decorating a room. How many metres of ribbon did he buy?

Step 1: To find the total length of ribbon the student bought, we need to add the lengths of all three ribbons.
The lengths are:
– \(4\frac{1}{3}\) m = \(\frac{13}{3}\) m
– \(6\frac{1}{6}\) m = \(\frac{37}{6}\) m
– \(3\frac{2}{9}\) m = \(\frac{29}{9}\) m
Step 2: To add these fractions, we need a common denominator. The least common denominator (LCD) of 3, 6, and 9 is 18. Now, convert each fraction to have the denominator 18:
– \(\frac{13}{3} = \frac{78}{18}\)
– \(\frac{37}{6} = \frac{111}{18}\)
– \(\frac{29}{9} = \frac{58}{18}\)
Step 3: Now, we add the fractions: \[ \frac{78}{18} + \frac{111}{18} + \frac{58}{18} = \frac{78 + 111 + 58}{18} = \frac{247}{18} \] Simplify: \[ \frac{247}{18} = 13\frac{13}{18} \]Answer: The student bought \(13\frac{13}{18}\) metres of ribbon.

Q6: In a business, Ram and Deepak invest \(\frac{3}{5}\) and \(\frac{2}{5}\) of the total investment. If ₹ 40,000 is the total investment, calculate the amount invested by each.

Step 1: We are given the total investment of ₹ 40,000 and the fractions of the total investment made by Ram and Deepak.
Ram’s share is \(\frac{3}{5}\) and Deepak’s share is \(\frac{2}{5}\).
Step 2: To find Ram’s investment, we multiply the total investment by \(\frac{3}{5}\): \[ \text{Ram’s investment} = \frac{3}{5} \times 40,000 = \frac{3 \times 40,000}{5} = \frac{120,000}{5} = 24,000 \]Step 3: To find Deepak’s investment, we multiply the total investment by \(\frac{2}{5}\): \[ \text{Deepak’s investment} = \frac{2}{5} \times 40,000 = \frac{2 \times 40,000}{5} = \frac{80,000}{5} = 16,000 \]Answer: Ram’s investment = ₹ 24,000, Deepak’s investment = ₹ 16,000

Q7: Geeta had 30 problems for home work. She worked out \(\frac{2}{3}\) of them. How many problems were still left to be worked out by her?

Step 1: We are given that Geeta had a total of 30 problems and worked out \(\frac{2}{3}\) of them.
Step 2: To find out how many problems Geeta worked out, we multiply the total number of problems by \(\frac{2}{3}\): \[ \text{Problems worked out} = \frac{2}{3} \times 30 = \frac{2 \times 30}{3} = \frac{60}{3} = 20 \]Step 3: To find how many problems are still left, we subtract the number of problems worked out from the total number of problems: \[ \text{Problems left} = 30 – 20 = 10 \]Answer: Problems left = 10

Q8: A picture was marked at ₹ 90. It was sold at \(\frac{3}{4}\) of its marked price. What was the sale price?

Step 1: We are given that the marked price of the picture is ₹ 90, and it was sold at \(\frac{3}{4}\) of the marked price.
Step 2: To find the sale price, we multiply the marked price by \(\frac{3}{4}\): \[ \text{Sale Price} = \frac{3}{4} \times 90 = \frac{3 \times 90}{4} = \frac{270}{4} = 67.5 \]Answer: Sale Price = ₹ 67.50

Q9: Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, each weighed \(10\frac{1}{2}\) kg?

Step 1: We are given that each parcel weighs \(10\frac{1}{2}\) kg, and there are 15 parcels.
Step 2: First, we convert the mixed fraction \(10\frac{1}{2}\) into an improper fraction: \[ 10\frac{1}{2} = \frac{21}{2} \]Step 3: To find the total weight, we multiply the weight of one parcel by the number of parcels: \[ \text{Total Weight} = 15 \times \frac{21}{2} = \frac{15 \times 21}{2} = \frac{315}{2} = 157\frac{1}{2} \]Answer: Total Weight = \(157\frac{1}{2} = 157.5\) kg

Q10: A rope is \(25\frac{1}{2}\) m long. How many pieces each of \(1\frac{1}{2}\) m length can be cutout from it?

Step 1: We are given the total length of the rope as \(25\frac{1}{2}\) m and the length of each piece as \(1\frac{1}{2}\) m.
Step 2: Convert both mixed fractions into improper fractions.
– The total length of the rope: \[ 25\frac{1}{2} = \frac{51}{2} \] – The length of each piece: \[ 1\frac{1}{2} = \frac{3}{2} \]Step 3: To find how many pieces can be cut from the rope, divide the total length of the rope by the length of each piece: \[ \text{Number of Pieces} = \frac{51}{2} \div \frac{3}{2} \]Step 4: Dividing by a fraction is the same as multiplying by its reciprocal: \[ \frac{51}{2} \div \frac{3}{2} = \frac{51}{2} \times \frac{2}{3} = \frac{51 \times 2}{2 \times 3} = \frac{102}{6} = 17 \]Answer: Number of Pieces = 17

Q11: The heights of two vertical poles, above the earth’s surface, are \(14\frac{1}{4}\) m and \(22\frac{1}{3}\) m respectively. How much higher ig the Second pole ag compared with the height of first pole?

Step 1: We are given the heights of two poles:
– Height of the first pole = \(14\frac{1}{4}\) m
– Height of the second pole = \(22\frac{1}{3}\) m
Step 2: Convert both mixed fractions into improper fractions.
– The height of the first pole: \[ 14\frac{1}{4} = \frac{57}{4} \] – The height of the second pole: \[ 22\frac{1}{3} = \frac{67}{3} \]Step 3: To find how much higher the second pole is compared to the first, subtract the height of the first pole from the height of the second pole: \[ \text{Difference in Heights} = \frac{67}{3} – \frac{57}{4} \]Step 4: To subtract these fractions, we need a common denominator. The LCM of 3 and 4 is 12, so we convert both fractions: \[ \frac{67}{3} = \frac{67 \times 4}{3 \times 4} = \frac{268}{12} \] \[ \frac{57}{4} = \frac{57 \times 3}{4 \times 3} = \frac{171}{12} \] Now subtract the fractions: \[ \frac{268}{12} – \frac{171}{12} = \frac{268 – 171}{12} = \frac{97}{12} \]Step 5: Convert the improper fraction back into a mixed fraction: \[ \frac{97}{12} = 8\frac{1}{12} \]Answer: Difference in Heights = \(8\frac{1}{12}\) m

Q12: Vijay weighed \(65\frac{1}{2}\) kg. He gained \(1\frac{2}{5}\) kg during the first week, \(1\frac{1}{4}\) kg during the second week, but lost \(\frac{5}{16}\) kg during the third week. What was his weight after third week?

Step 1: Vijay’s initial weight is \(65\frac{1}{2}\) kg.
Step 2: Convert the mixed fraction into an improper fraction: \[ 65\frac{1}{2} = \frac{131}{2} \]Step 3: He gained \(1\frac{2}{5}\) kg in the first week and \(1\frac{1}{4}\) kg in the second week. Convert these mixed fractions into improper fractions: \[ 1\frac{2}{5} = \frac{7}{5}, \quad 1\frac{1}{4} = \frac{5}{4} \]Step 4: He lost \(\frac{5}{16}\) kg in the third week. This is already in improper fraction form.
Step 5: To find his weight after the third week, add the weight gained during the first and second weeks and subtract the weight lost during the third week from his initial weight: \[ \text{Final weight} = \frac{131}{2} + \frac{7}{5} + \frac{5}{4} – \frac{5}{16} \]Step 6: To add and subtract these fractions, we need a common denominator. The LCM of 2, 5, 4, and 16 is 80. Convert each fraction to have a denominator of 80: \[ \frac{131}{2} = \frac{131 \times 40}{2 \times 40} = \frac{5240}{80} \] \[ \frac{7}{5} = \frac{7 \times 16}{5 \times 16} = \frac{112}{80} \] \[ \frac{5}{4} = \frac{5 \times 20}{4 \times 20} = \frac{100}{80} \] \[ \frac{5}{16} = \frac{5 \times 5}{16 \times 5} = \frac{25}{80} \]Now, substitute the fractions into the equation: \[ \text{Final weight} = \frac{5240}{80} + \frac{112}{80} + \frac{100}{80} – \frac{25}{80} \]Step 7: Combine the fractions: \[ \text{Final weight} = \frac{5240 + 112 + 100 – 25}{80} = \frac{5427}{80} \]Step 8: Convert the improper fraction back into a mixed fraction: \[ \frac{5427}{80} = 67\frac{67}{80} \]Answer: Final weight = \(67\frac{67}{80}\) kg

Q13: A man spends \(\frac{2}{5}\) of his salary on food and \(\frac{3}{10}\) on house rent, electricity, etc. What fraction of his salary is still left with him ?

Step 1: Let the total salary of the man be represented as 1 (i.e., his entire salary).
Step 2: The man spends \(\frac{2}{5}\) of his salary on food and \(\frac{3}{10}\) on house rent, electricity, etc. To find the total amount spent, we add the fractions: \[ \text{Total spent} = \frac{2}{5} + \frac{3}{10} \]Step 3: To add these fractions, we need a common denominator. The LCM of 5 and 10 is 10. Convert \(\frac{2}{5}\) to have a denominator of 10: \[ \frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} \] Now, add the fractions: \[ \text{Total spent} = \frac{4}{10} + \frac{3}{10} = \frac{7}{10} \]Step 4: The total amount spent is \(\frac{7}{10}\) of the salary.
Step 5: To find the fraction of salary still left with the man, subtract the total spent from the total salary: \[ \text{Remaining salary} = 1 – \frac{7}{10} \]Step 6: Convert 1 to \(\frac{10}{10}\): \[ \text{Remaining salary} = \frac{10}{10} – \frac{7}{10} = \frac{3}{10} \]Answer: Remaining salary = \(\frac{3}{10}\)

Q14: A man spends \(\frac{2}{5}\) of his salary on food and \(\frac{3}{10}\) bf the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?

Step 1: Let the total salary of the man be represented as 1 (i.e., his entire salary).
Step 2: The man spends \(\frac{2}{5}\) of his salary on food. So, the fraction of the salary remaining after food expenditure is: \[ \text{Remaining after food} = 1 – \frac{2}{5} = \frac{5}{5} – \frac{2}{5} = \frac{3}{5} \]Step 3: Now, he spends \(\frac{3}{10}\) of the remaining salary on house rent, electricity, etc. The fraction of the remaining salary spent on these expenses is: \[ \text{Amount spent on rent} = \frac{3}{10} \times \frac{3}{5} = \frac{9}{50} \]Step 4: To find how much salary is still left, subtract the amount spent on rent from the remaining salary after food: \[ \text{Remaining salary} = \frac{3}{5} – \frac{9}{50} \]Step 5: To subtract these fractions, we need a common denominator. The LCM of 5 and 50 is 50. Convert \(\frac{3}{5}\) to have a denominator of 50: \[ \frac{3}{5} = \frac{30}{50} \] Now, subtract: \[ \text{Remaining salary} = \frac{30}{50} – \frac{9}{50} = \frac{21}{50} \]Answer: Remaining salary = \(\frac{21}{50}\)

Q15: Shyam bought a refrigerator for ₹ 5,000. He paid \(\frac{1}{10}\) of the price in cash and the rest in 12 equal monthly instalments. How much did he have to pay each month?

Step 1: The total price of the refrigerator is ₹ 5,000.
Step 2: Shyam paid \(\frac{1}{10}\) of the price in cash. So, the amount he paid in cash is: \[ \text{Amount paid in cash} = \frac{1}{10} \times 5000 = 500 \]Step 3: The remaining amount to be paid through instalments is: \[ \text{Remaining amount} = 5000 – 500 = 4500 \]Step 4: The remaining amount of ₹ 4500 is to be paid in 12 equal monthly instalments. So, the amount Shyam has to pay each month is: \[ \text{Monthly instalment} = \frac{4500}{12} = 375 \]Answer: Monthly instalment = ₹ 375

Q16: A lamp post has half of its length in mud and \(\frac{1}{3}\) of its length in water.

i. What fraction of its length is above the water?

Step 1: The total length of the lamp post is divided into three parts:
– Half of it is in mud.
– \(\frac{1}{3}\) of it is in water.
– The remaining part is above the water.
Step 2: Let the total length of the lamp post be \(L\).
– Length in mud = \(\frac{1}{2}L\)
– Length in water = \(\frac{1}{3}L\)
Step 3: The remaining part, above the water, is: \[ \text{Length above the water} = L – \left(\frac{1}{2}L + \frac{1}{3}L\right) \]Step 4: To subtract the fractions, we need a common denominator. The LCM of 2 and 3 is 6: \[ \frac{1}{2}L = \frac{3}{6}L \quad \text{and} \quad \frac{1}{3}L = \frac{2}{6}L \] Now subtracting: \[ \text{Length above the water} = L – \left(\frac{3}{6}L + \frac{2}{6}L\right) = L – \frac{5}{6}L = \frac{1}{6}L \]Answer: Fraction above the water = \(\frac{1}{6}\)

ii. If \(3\frac{1}{3}\) m of the lamp post is above the water, find the whole length of the lamp post.

Step 1: Let the total length of the lamp post be \(L\). From part i, we know that the length above the water is \(\frac{1}{6}\) of the total length.
Step 2: We are given that the length above the water is \(3\frac{1}{3}\) meters, or \(\frac{10}{3}\) meters. \[ \frac{1}{6}L = \frac{10}{3} \]Step 3: To find \(L\), multiply both sides of the equation by 6: \[ L = 6 \times \frac{10}{3} = 20 \]Answer: Total length of the lamp post = 20 meters

Q17: I spent \(\frac{3}{5}\) of my savings and still have ₹ 2,000 left. What were my savings?

Step 1: Let the total savings be \(S\).
Step 2: According to the problem, I spent \(\frac{3}{5}\) of my savings. Therefore, the amount I spent is: \[ \text{Amount spent} = \frac{3}{5}S \]Step 3: The amount left after spending is ₹ 2,000, which is the remaining part of my savings. The remaining savings are: \[ \text{Remaining savings} = S – \frac{3}{5}S \]Step 4: Simplifying the remaining savings: \[ \text{Remaining savings} = \frac{2}{5}S \] We are told that the remaining savings are ₹ 2,000. Therefore, we have the equation: \[ \frac{2}{5}S = 2000 \]Step 5: To find \(S\), multiply both sides of the equation by \(\frac{5}{2}\): \[ S = 2000 \times \frac{5}{2} = 5000 \]Answer: Total savings = ₹ 5,000

Q18: In a school \(\frac{4}{5}\) of the children are boys. If the number of girls is 200, find the number of boys.

Step 1: Let the total number of children in the school be \(T\).
Step 2: According to the problem, \(\frac{4}{5}\) of the children are boys, and the remaining \(\frac{1}{5}\) of the children are girls. We are told that the number of girls is 200. So, we have: \[ \frac{1}{5}T = 200 \]Step 3: To find \(T\), multiply both sides of the equation by 5: \[ T = 200 \times 5 = 1000 \]Step 4: Now that we know the total number of children is 1000, we can find the number of boys by using the fact that \(\frac{4}{5}\) of the children are boys: \[ \text{Number of boys} = \frac{4}{5} \times 1000 = 800 \]Answer: Number of boys = 800

Q19: If \(\frac{4}{5}\) of an estate is worth ₹ 42,000, find the worth of the whole estate. Also, find the value of \(\frac{3}{7}\) of it.

Step 1: Let the worth of the whole estate be \(x\).
Step 2: We are told that \(\frac{4}{5}\) of the estate is worth ₹ 42,000. Therefore, we can set up the equation: \[ \frac{4}{5} \times x = 42000 \]Step 3: To find \(x\), multiply both sides of the equation by \(\frac{5}{4}\): \[ x = \frac{42000 \times 5}{4} = 52500 \] So, the worth of the whole estate is ₹ 52,500.
Step 4: Now, we need to find the value of \(\frac{3}{7}\) of the estate. To do this, we multiply \(\frac{3}{7}\) by the total worth of the estate: \[ \frac{3}{7} \times 52500 = \frac{3 \times 52500}{7} = 22500 \] Thus, the value of \(\frac{3}{7}\) of the estate is ₹ 22,500.
Answer: Worth of the whole estate = ₹ 52,500 Answer: Value of \(\frac{3}{7}\) of the estate = ₹ 22,500

Q20: After going \(\frac{3}{4}\) of my journey, I find that I have covered 16 km. How much journey is still left?

Step 1: Let the total distance of the journey be \(x\) km.
Step 2: We are told that \(\frac{3}{4}\) of the journey is 16 km. Therefore, we can set up the equation: \[ \frac{3}{4} \times x = 16 \]Step 3: To find \(X\), multiply both sides of the equation by \(\frac{4}{3}\): \[ x = \frac{16 \times 4}{3} = \frac{64}{3} \text{ km} \] So, the total distance of the journey is \(\frac{64}{3}\) km.Step 4: To find how much of the journey is still left, we subtract the distance covered (\(\frac{3}{4}\) of the journey) from the total journey: \[ \text{Distance left} = x – \frac{3}{4}X = \frac{1}{4} \times x = \frac{1}{4} \times \frac{64}{3} = \frac{64}{12} = \frac{16}{3} \text{ km} \]Answer: Distance left = \(\frac{16}{3} = 5\frac{1}{3}\) km

Q21: When Krishna travelled 25 km, he found that \(\frac{3}{5}\) of his journey was still left. What was the length of the whole journey?

Step 1: Let the total length of the journey be \(x\) km.
Step 2: We are told that \(\frac{3}{5}\) of the journey is still left after Krishna travelled 25 km. This means that the remaining distance is \(\frac{3}{5}x\), and the distance Krishna has already covered is \(x – \frac{3}{5}x = \frac{2}{5}x\).
Step 3: The distance Krishna has already covered is 25 km, so: \[ \frac{2}{5}x = 25 \]Step 4: To find \(X\), multiply both sides of the equation by \(\frac{5}{2}\): \[ x = 25 \times \frac{5}{2} = \frac{125}{2} = 62.5 \text{ km} \]Answer: Total journey = 62.5 \(= 62\frac{1}{2}\) km

Q22: From a piece of land, one-third is bought by Rajesh and one-third Of remaining is bought by Manoj. If 600 m2 lend is still left unsold, find the total area of the piece of land.

Step 1: Let the total area of the land be \(A\) m².
Step 2: Rajesh buys \(\frac{1}{3}\) of the land, so the area Rajesh bought is: \[ \text{Rajesh’s area} = \frac{1}{3}A \]Step 3: After Rajesh’s purchase, the remaining land is: \[ \text{Remaining land after Rajesh’s purchase} = A – \frac{1}{3}A = \frac{2}{3}A \]Step 4: Manoj buys \(\frac{1}{3}\) of the remaining land, so the area Manoj bought is: \[ \text{Manoj’s area} = \frac{1}{3} \times \frac{2}{3}A = \frac{2}{9}A \]Step 5: The remaining land after Manoj’s purchase is: \[ \text{Remaining land after Manoj’s purchase} = \frac{2}{3}A – \frac{2}{9}A = \frac{6}{9}A – \frac{2}{9}A = \frac{4}{9}A \]Step 6: We are told that 600 m² of land is still left unsold, so: \[ \frac{4}{9}A = 600 \]Step 7: To find \(A\), multiply both sides of the equation by \(\frac{9}{4}\): \[ A = 600 \times \frac{9}{4} = 1350 \text{ m²} \]Answer: Total area of land = 1350 m²

Q23: A boy spent \(\frac{3}{5}\) of his money on buying clothes and \(\frac{1}{4}\) of the remaining money on buying shoes. If he initially had ₹ 2,400 how much did he spend on shoes?

Step 1: Let the total money the boy initially had be ₹ 2400.
Step 2: The boy spent \(\frac{3}{5}\) of his money on buying clothes. Therefore, the amount spent on clothes is: \[ \text{Amount spent on clothes} = \frac{3}{5} \times 2400 = 1440 \text{ rupees} \]Step 3: The remaining money after buying clothes is: \[ \text{Remaining money} = 2400 – 1440 = 960 \text{ rupees} \]Step 4: The boy spent \(\frac{1}{4}\) of the remaining money on buying shoes. Therefore, the amount spent on shoes is: \[ \text{Amount spent on shoes} = \frac{1}{4} \times 960 = 240 \text{ rupees} \]Answer: Amount spent on shoes = ₹ 240

Q24: A boy spent \(\frac{3}{5}\) of his money on buying clothes and \(\frac{1}{4}\) of his money on buying shoes. If he initially had ₹ 2,400, how much did he spend on shoes?

Step 1: Let the total money the boy initially had be ₹ 2400.
Step 2: The boy spent \(\frac{3}{5}\) of his money on buying clothes. Therefore, the amount spent on clothes is: \[ \text{Amount spent on clothes} = \frac{3}{5} \times 2400 = 1440 \text{ rupees} \]Step 3: The boy spent \(\frac{1}{4}\) of his total money on buying shoes. Therefore, the amount spent on shoes is: \[ \text{Amount spent on shoes} = \frac{1}{4} \times 2400 = 600 \text{ rupees} \]Answer: Amount spent on shoes = ₹ 600