Chapter 8 Working with Fractions (Class 7 - Latest Maths NCERT (Ganita Prakash I) NCERT Solutions)

Given:

Amount of milk consumed daily = $\frac{1}{2}$ glass.

To Find:

1. Total glasses of milk in a week.

2. Total glasses of milk in January.

Solution:

1. Milk consumed in a week (7 days):

Total milk = Number of days $\times$ Daily consumption

$\text{Total} = 7 \times \frac{1}{2} = \frac{7}{2}$

$\text{Total} = \mathbf{3 \frac{1}{2}}$ glasses.

2. Milk consumed in January:

The month of January has $31$ days.

Total milk = $31 \times \frac{1}{2} = \frac{31}{2}$

$\text{Total} = \mathbf{15 \frac{1}{2}}$ glasses.

Final Answer:

Tenzin drinks $3 \frac{1}{2}$ glasses of milk in a week and $15 \frac{1}{2}$ glasses in January.

To Find: The daily and weekly progress of the work.

Solution:

Since the team completes $1\text{ km}$ in $8$ days, the work done in a single day is the total length divided by the total number of days.

$\text{Daily progress} = \frac{1}{8} \text{ km}$

Now, to find the weekly progress (working 5 days a week):

$\text{Weekly progress} = \text{Daily progress} \times 5$

$\text{Weekly progress} = \frac{1}{8} \times 5 = \frac{5}{8} \text{ km}$

Final Answer:

In one day, the team can make $\frac{1}{8}$ km and in a week they can make $\frac{5}{8}$ km.

Given:

Total oil bought per week = $5$ litres.

Total number of families = $3$ (Manju + 2 neighbours).

Solution:

1. Weekly share per family:

To share equally, we divide the total quantity by the number of families.

$\text{Share} = 5 \div 3 = \frac{5}{3} = \mathbf{1 \frac{2}{3}}$ litres.

2. Share for one family in 4 weeks:

We multiply the weekly share by $4$.

$\text{Total in 4 weeks} = 4 \times \frac{5}{3} = \frac{20}{3}$

$\text{Total in 4 weeks} = \mathbf{6 \frac{2}{3}}$ litres.

Final Answer:

Each family gets $1 \frac{2}{3}$ litres per week, and in 4 weeks, a family receives $6 \frac{2}{3}$ litres.

Given:

Base setting time on Monday = $10\text{ pm}$.

Delay per day = $\frac{5}{6}$ hour.

To Find:

The total delay from $10\text{ pm}$ on Thursday.

Solution:

First, we determine the number of days passed from Monday to Thursday:

1. Monday to Tuesday (1 day)

2. Tuesday to Wednesday (2 days)

3. Wednesday to Thursday (3 days)

The total number of days is 3.

Now, calculate the total delay:

$\text{Total Delay} = 3 \times \frac{5}{6}$

$\text{Total Delay} = \frac{15}{6} = \frac{5}{2}$ hours.

Convert to a mixed number:

$\text{Total Delay} = \mathbf{2 \frac{1}{2}}$ hours.

Final Answer:

The moon will set $2 \frac{1}{2}$ hours (or 2 hours 30 minutes) after $10\text{ pm}$ on Thursday.

This would mean the moon sets at 12:30 am on Friday morning.

To Find: The product of the given numbers and their representation as mixed fractions.

(a) $7 \times \frac{3}{5}$

First, we multiply the whole number by the numerator of the fraction:

$7 \times \frac{3}{5} = \frac{7 \times 3}{5} = \frac{21}{5}$

Now, we convert the improper fraction $\frac{21}{5}$ into a mixed fraction by dividing the numerator by the denominator. $21$ divided by $5$ gives a quotient of $4$ and a remainder of $1$.

Result: $4 \frac{1}{5}$

(b) $4 \times \frac{1}{3}$

Multiply the whole number and the numerator:

$4 \times \frac{1}{3} = \frac{4 \times 1}{3} = \frac{4}{3}$

Converting $\frac{4}{3}$ to a mixed fraction: $4$ divided by $3$ gives a quotient of $1$ and a remainder of $1$.

Result: $1 \frac{1}{3}$

(c) $\frac{9}{7} \times 6$

Multiply the numerator by the whole number:

$\frac{9}{7} \times 6 = \frac{9 \times 6}{7} = \frac{54}{7}$

Converting $\frac{54}{7}$ to a mixed fraction: $54$ divided by $7$ gives a quotient of $7$ (since $7 \times 7 = 49$) and a remainder of $5$.

Result: $7 \frac{5}{7}$

(d) $\frac{13}{11} \times 6$

Multiply the numerator by the whole number:

$\frac{13}{11} \times 6 = \frac{13 \times 6}{11} = \frac{78}{11}$

Converting $\frac{78}{11}$ to a mixed fraction: $78$ divided by $11$ gives a quotient of $7$ (since $11 \times 7 = 77$) and a remainder of $1$.

Result: $7 \frac{1}{11}$

These steps are essential for solving word problems involving quantities, ensuring the final answer is in the most readable form for practical use.

(a) Solution for $\frac{1}{3} \times \frac{1}{5}$:

To represent the product, we take a unit square. First, we divide the square vertically into 5 equal parts and shade 1 part to represent $\frac{1}{5}$. Next, we divide the square horizontally into 3 equal parts and shade 1 part to represent $\frac{1}{3}$.

The total number of small equal rectangles formed is $3 \times 5 = 15$. The overlapping shaded region is 1 unit.

Therefore, $\frac{1}{3} \times \frac{1}{5} = \frac{1 \times 1}{3 \times 5} = \frac{1}{15}$.

(b) Solution for $\frac{1}{4} \times \frac{1}{3}$:

We divide a unit square vertically into 3 equal parts and shade 1 part for $\frac{1}{3}$. Then, we divide it horizontally into 4 equal parts and shade 1 part for $\frac{1}{4}$.

The total number of small equal parts is $4 \times 3 = 12$. The common shaded region is 1 unit.

Therefore, $\frac{1}{4} \times \frac{1}{3} = \frac{1 \times 1}{4 \times 3} = \frac{1}{12}$.

(c) Solution for $\frac{1}{5} \times \frac{1}{2}$:

We divide a unit square vertically into 2 equal parts and shade 1 part for $\frac{1}{2}$. Then, we divide it horizontally into 5 equal parts and shade 1 part for $\frac{1}{5}$.

The total number of small equal parts is $5 \times 2 = 10$. The common shaded region is 1 unit.

Therefore, $\frac{1}{5} \times \frac{1}{2} = \frac{1 \times 1}{5 \times 2} = \frac{1}{10}$.

(d) Solution for $\frac{1}{6} \times \frac{1}{5}$:

We divide a unit square vertically into 5 equal parts and shade 1 part for $\frac{1}{5}$. Then, we divide it horizontally into 6 equal parts and shade 1 part for $\frac{1}{6}$.

The total number of small equal parts is $6 \times 5 = 30$. The common shaded region is 1 unit.

Therefore, $\frac{1}{6} \times \frac{1}{5} = \frac{1 \times 1}{6 \times 5} = \frac{1}{30}$.

(a) Solution for $\frac{2}{3} \times \frac{4}{5}$:

First, we divide the unit square vertically into 5 equal parts and shade 4 parts to represent $\frac{4}{5}$. Next, we divide it horizontally into 3 equal parts and shade 2 parts to represent $\frac{2}{3}$.

The total number of small parts is $3 \times 5 = 15$. The overlapping region contains $2 \times 4 = 8$ small rectangles.

$\text{Product} = \frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5} = \frac{8}{15}$.

(b) Solution for $\frac{1}{4} \times \frac{2}{3}$:

Divide the unit square vertically into 3 equal parts and shade 2 parts for $\frac{2}{3}$. Then, divide it horizontally into 4 equal parts and shade 1 part for $\frac{1}{4}$.

The total number of small parts is $4 \times 3 = 12$. The overlapping region consists of $1 \times 2 = 2$ parts.

$\text{Product} = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12}$

On simplifying, we get: $\frac{\cancel{2}^{1}}{\cancel{12}_{6}} = \frac{1}{6}$.

(c) Solution for $\frac{3}{5} \times \frac{1}{2}$:

Divide the unit square vertically into 2 equal parts and shade 1 part for $\frac{1}{2}$. Then, divide it horizontally into 5 equal parts and shade 3 parts for $\frac{3}{5}$.

The total number of small parts is $5 \times 2 = 10$. The overlapping region consists of $3 \times 1 = 3$ parts.

$\text{Product} = \frac{3}{5} \times \frac{1}{2} = \frac{3 \times 1}{5 \times 2} = \frac{3}{10}$.

(d) Solution for $\frac{4}{6} \times \frac{3}{5}$:

Divide the unit square vertically into 5 equal parts and shade 3 parts for $\frac{3}{5}$. Then, divide it horizontally into 6 equal parts and shade 4 parts for $\frac{4}{6}$.

The total number of small parts is $6 \times 5 = 30$. The overlapping region consists of $4 \times 3 = 12$ parts.

$\text{Product} = \frac{4}{6} \times \frac{3}{5} = \frac{12}{30}$

On simplifying by dividing by 6, we get: $\frac{\cancel{12}^{2}}{\cancel{30}_{5}} = \frac{2}{5}$.

Given:

Part of the tank filled in $1$ hour = $\frac{7}{10}$

(a) Part of the tank filled in $\frac{1}{3}$ hour:

$\text{Portion filled} = \text{Time} \times \text{Rate per hour}$

$\text{Portion filled} = \frac{1}{3} \times \frac{7}{10} = \frac{1 \times 7}{3 \times 10} = \frac{7}{30}$

So, $\frac{7}{30}$ of the tank is filled in $\frac{1}{3}$ hour.

(b) Part of the tank filled in $\frac{2}{3}$ hour:

$\text{Portion filled} = \frac{2}{3} \times \frac{7}{10}$

$\text{Portion filled} = \frac{2 \times 7}{3 \times 10} = \frac{\cancel{14}^{7}}{\cancel{30}_{15}} = \frac{7}{15}$

So, $\frac{7}{15}$ of the tank is filled in $\frac{2}{3}$ hour.

(c) Part of the tank filled in $\frac{3}{4}$ hour:

$\text{Portion filled} = \frac{3}{4} \times \frac{7}{10} = \frac{3 \times 7}{4 \times 10} = \frac{21}{40}$

So, $\frac{21}{40}$ of the tank is filled in $\frac{3}{4}$ hour.

(d) Part of the tank filled in $\frac{7}{10}$ hour:

$\text{Portion filled} = \frac{7}{10} \times \frac{7}{10} = \frac{7 \times 7}{10 \times 10} = \frac{49}{100}$

So, $\frac{49}{100}$ of the tank is filled in $\frac{7}{10}$ hour.

(e) Time required for the tank to be full:

To find the time for a full tank (1 whole), we divide the whole by the rate of filling per hour.

$\text{Time} = 1 \div \frac{7}{10}$

$\text{Time} = 1 \times \frac{10}{7} = \frac{10}{7}$ hours

Converting into mixed fraction: $\frac{10}{7} = 1 \frac{3}{7}$ hours.

Therefore, the tap should be running for $1 \frac{3}{7}$ hours to fill the tank completely.

Given:

Let the total original land be $1$ unit.

Land taken by government = $\frac{1}{6}$

To Find: Remaining land and shares of Krishna, Bora, and Somu.

Solution:

First, we find the land remaining after the government took its share:

$\text{Remaining land} = 1 - \frac{1}{6} = \frac{6 - 1}{6} = \frac{5}{6}$

(a) Part of the original land Krishna got:

Krishna gets half ($\frac{1}{2}$) of the remaining land ($\frac{5}{6}$).

$\text{Krishna's share} = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$

So, Krishna got $\frac{5}{12}$ of the original land.

(b) Part of the original land Bora got:

Bora gets $\frac{1}{3}$ of the remaining land ($\frac{5}{6}$).

$\text{Bora's share} = \frac{1}{3} \times \frac{5}{6} = \frac{5}{18}$

So, Bora got $\frac{5}{18}$ of the original land.

(c) Part of the original land Somu kept for herself:

Somu kept the land remaining after giving shares to Krishna and Bora.

$\text{Total share given to children} = \frac{5}{12} + \frac{5}{18}$

To add these, we find the LCM of 12 and 18:

$\begin{array}{c|cc} 2 & 12 \;, & 18 \\ \hline 2 & 6 \; , & 9 \\ \hline 3 & 3 \; , & 9 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$

$\text{LCM} = 2 \times 2 \times 3 \times 3 = 36$

$\text{Total given} = \frac{5 \times 3}{12 \times 3} + \frac{5 \times 2}{18 \times 2} = \frac{15}{36} + \frac{10}{36} = \frac{25}{36}$

Now, Somu's remaining share from the original land is:

$\text{Somu's land} = \text{Remaining land after road} $$ - \text{Total given to children}$

$\text{Somu's land} = \frac{5}{6} - \frac{25}{36}$

$\text{Somu's land} = \frac{5 \times 6}{6 \times 6} - \frac{25}{36} = \frac{30 - 25}{36} = \frac{5}{36}$

So, Somu kept $\frac{5}{36}$ of the original land for herself.

Given:

Length of the rectangle ($l$) = $9 \frac{3}{5}\text{ ft} = \frac{(9 \times 5) + 3}{5} = \frac{48}{5}\text{ ft}$

Breadth of the rectangle ($b$) = $3 \frac{3}{4}\text{ ft} = \frac{(3 \times 4) + 3}{4} = \frac{15}{4}\text{ ft}$

To Find: Area of the rectangle.

Solution:

We know that,

$\text{Area of Rectangle} = \text{Length} \times \text{Breadth}$

$\text{Area} = \frac{48}{5} \times \frac{15}{4}$

$\text{Area} = \frac{\cancel{48}^{12}}{\cancel{5}_{1}} \times \frac{\cancel{15}^{3}}{\cancel{4}_{1}}$

$\text{Area} = 12 \times 3 = 36\text{ sq. ft}$

Therefore, the area of the rectangle is $36\text{ sq. ft}$.

Given:

Number of saplings = $4$

Distance between two adjacent saplings = $\frac{3}{4}\text{ m}$

Construction Required:

A rough diagram representing 4 saplings in a row:

From the diagram, we can see that there are 3 gaps between 4 saplings.

Solution:

Number of gaps between the first and the last sapling = $3$

$\text{Total distance} = \text{Number of gaps} \times \text{Distance of one gap}$

$\text{Total distance} = 3 \times \frac{3}{4}$

$\text{Total distance} = \frac{9}{4}\text{ m}$

Converting into mixed fraction: $\frac{9}{4} = 2 \frac{1}{4}\text{ m}$

Therefore, the distance between the first and the last sapling is $2 \frac{1}{4}\text{ m}$ (or $2.25\text{ m}$).

To Find: Which of the two given quantities is heavier.

Step 1: Calculate $\frac{12}{15}$ of $500$ grams

$\text{Weight 1} = \frac{12}{15} \times 500$

First, simplify the fraction $\frac{12}{15}$ by dividing by 3:

$\frac{\cancel{12}^{4}}{\cancel{15}_{5}} \times 500 = \frac{4}{5} \times 500$

$\text{Weight 1} = 4 \times \frac{\cancel{500}^{100}}{\cancel{5}_{1}} = 4 \times 100 = 400\text{ g}$

Step 2: Calculate $\frac{3}{20}$ of $4\text{ kg}$

First, convert kg to grams ($1\text{ kg} = 1000\text{ g}$):

$4\text{ kg} = 4 \times 1000 = 4000\text{ g}$

Now, find $\frac{3}{20}$ of $4000\text{ g}$:

$\text{Weight 2} = \frac{3}{20} \times 4000$

$\text{Weight 2} = 3 \times \frac{\cancel{4000}^{200}}{\cancel{20}_{1}} = 3 \times 200 = 600\text{ g}$

Comparison:

Weight 1 = $400\text{ g}$

Weight 2 = $600\text{ g}$

Since $600\text{ g} > 400\text{ g}$, Weight 2 is heavier.

Therefore, $\frac{3}{20}$ of $4\text{ kg}$ is heavier.

Given:

Time taken by the first fountain to fill the cistern = $1$ day

Time taken by the second fountain = $\frac{1}{2}$ day

Time taken by the third fountain = $\frac{1}{4}$ day

Time taken by the fourth fountain = $\frac{1}{5}$ day

To Find:

The total time taken by all four fountains flowing together to fill the cistern.

Solution:

To find how many times each fountain fills the cistern in one full day, we divide the total time ($1$ day) by the time taken for one fill.

Number of times the first fountain fills the cistern in a day = $1 \div 1 = 1$

Number of times the second fountain fills the cistern in a day = $1 \div \frac{1}{2} = 1 \times 2 = 2$

Number of times the third fountain fills the cistern in a day = $1 \div \frac{1}{4} = 1 \times 4 = 4$

Number of times the fourth fountain fills the cistern in a day = $1 \div \frac{1}{5} = 1 \times 5 = 5$

Now, let us find the total number of times the cistern is filled in one day when all fountains work together:

$\text{Total fills in a day} = 1 + 2 + 4 + 5 = 12$

This means that in $1$ day, the fountains together can fill the cistern $12$ times.

To find the time taken for one fill, we take the reciprocal of the total fills per day:

$\text{Time taken together} = \frac{1}{12}\text{ of a day}$

Since $1$ day = $24$ hours, we convert this into hours:

$\text{Time in hours} = \frac{1}{12} \times 24$

$\text{Time} = \frac{1}{\cancel{12}_{1}} \times \cancel{24}^{2} = 2\text{ hours}$

Therefore, if all the fountains flow together, they will fill the cistern in $2$ hours.

Alternate Solution:

We can use the combined work formula. Let $T$ be the total time taken:

$\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + \frac{1}{T_4}$

$\frac{1}{T} = \frac{1}{1} + \frac{1}{1/2} + \frac{1}{1/4} + \frac{1}{1/5}$

$\frac{1}{T} = 1 + 2 + 4 + 5 = 12$

$T = \frac{1}{12}\text{ day}$

$T = \frac{1}{12} \times 24\text{ hours} = 2\text{ hours}$

To Find: The fraction of the total area of the big square that is covered by the shaded regions in both geometric figures.

Solution for Figure 1:

We consider the area of the big square to be $1$ unit. To find the shaded fraction, we first observe the division of the square by its diagonals. The diagonals divide the big square into $4$ equal large triangles.

The shaded area is contained within the left triangle and the bottom triangle. Let us analyze the left triangle first. If we divide this triangle into $2$ smaller triangles by drawing an altitude from the center to the side, each small triangle has an area of $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

$\bullet$ The top part of this left triangle is completely shaded. Shaded area $= \frac{1}{8}$.

$\bullet$ The bottom part of this left triangle is half shaded. Shaded area $= \frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$.

Total shaded area in the left triangle is calculated as:

Similarly, the bottom triangle follows the exact same pattern of division and shading:

Now, we find the total shaded area of the whole square by adding the fractions from both triangles:

Solution for Figure 2:

From the visual arrangement, it is clear that the top-left square (indicated by the bold lines) occupies a specific portion of the whole square.

Now, we observe the internal division of this top-left square. By drawing the diagonals and mid-lines of this smaller square, it is divided into $8$ identical triangles. Out of these $8$ triangles, exactly $2$ are shaded.

To find the shaded region's fraction relative to the whole square, we multiply the fraction of the small square by its area relative to the whole:

Final Answer:

The shaded region occupies $\frac{3}{8}$ of the area of the whole square in the first figure and $\frac{1}{16}$ in the second figure.

Solution:

1. $3 \div \frac{7}{9} = 3 \times \frac{9}{7} = \frac{27}{7} = \mathbf{3 \frac{6}{7}}$

2. $\frac{14}{4} \div 2 = \frac{14}{4} \times \frac{1}{2} = \frac{\cancel{14}^{7}}{4} \times \frac{1}{\cancel{2}_{1}} = \frac{7}{4} = \mathbf{1 \frac{3}{4}}$

3. $\frac{2}{3} \div \frac{2}{3} = \frac{2}{3} \times \frac{3}{2} = \frac{\cancel{2}^{1}}{\cancel{3}_{1}} \times \frac{\cancel{3}^{1}}{\cancel{2}_{1}} = \mathbf{1}$

4. $\frac{14}{6} \div \frac{7}{3} = \frac{14}{6} \times \frac{3}{7} = \frac{\cancel{14}^{2}}{\cancel{6}_{2}} \times \frac{\cancel{3}^{1}}{\cancel{7}_{1}} = \frac{2}{2} = \mathbf{1}$

5. $\frac{4}{3} \div \frac{3}{4} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9} = \mathbf{1 \frac{7}{9}}$

6. $\frac{7}{4} \div \frac{1}{7} = \frac{7}{4} \times \frac{7}{1} = \frac{49}{4} = \mathbf{12 \frac{1}{4}}$

7. $\frac{8}{2} \div \frac{4}{15} = \frac{\cancel{8}^{4}}{\cancel{2}_1} \times \frac{15}{4} = 4 \times \frac{15}{4} = \frac{\cancel{4}^{1} \times 15}{\cancel{4}_{1}} = \mathbf{15}$

8. $\frac{1}{5} \div \frac{1}{9} = \frac{1}{5} \times \frac{9}{1} = \frac{9}{5} = \mathbf{1 \frac{4}{5}}$

9. $\frac{1}{6} \div \frac{11}{12} = \frac{1}{6} \times \frac{12}{11} = \frac{1}{\cancel{6}_{1}} \times \frac{\cancel{12}^{2}}{11} = \mathbf{\frac{2}{11}}$

10. $3 \frac{2}{3} \div 1 \frac{3}{8} = \frac{11}{3} \div \frac{11}{8} = \frac{11}{3} \times \frac{8}{11} = \frac{\cancel{11}^{1}}{3} \times \frac{8}{\cancel{11}_{1}} = \frac{8}{3} = \mathbf{2 \frac{2}{3}}$

(a) Solution:

Total lace bought = $8\text{ m}$

Lace used for each bag = $\frac{1}{4}\text{ m}$

To find the number of bags, we divide the total length by the length per bag.

The correct expression is (iii) $8 \div \frac{1}{4}$.

$\text{Simplification: } 8 \div \frac{1}{4} = 8 \times 4 = 32$

Maria decorated 32 bags.

(b) Solution:

Total length of ribbon = $\frac{1}{2}\text{ m}$

Number of badges made = $8$

To find the length for each badge, we divide the total length by the number of badges.

The correct expression is (iv) $\frac{1}{2} \div 8$.

$\text{Simplification: } \frac{1}{2} \div 8 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$

The length of ribbon used for each badge is $\frac{1}{16}\text{ m}$.

(c) Solution:

Total flour available = $5\text{ kg}$

Flour required for one loaf = $\frac{1}{6}\text{ kg}$

To find the number of loaves, we divide the total flour by the flour per loaf.

The correct expression is (iii) $5 \div \frac{1}{6}$.

$\text{Simplification: } 5 \div \frac{1}{6} = 5 \times 6 = 30$

The baker can make 30 loaves of bread.

Given:

Flour used for 12 rotis = $\frac{1}{4}\text{ kg}$

To Find:

Flour used for 6 rotis.

Solution:

First, we find the flour used for 1 roti:

$\text{Flour for 1 roti} = \frac{1}{4} \div 12$

$\text{Flour for 1 roti} = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48}\text{ kg}$

Now, we find the flour used for 6 rotis:

$\text{Flour for 6 rotis} = 6 \times \frac{1}{48}$

$\text{Flour for 6 rotis} = \frac{\cancel{6}^{1}}{\cancel{48}_{8}} = \frac{1}{8}\text{ kg}$

Therefore, $\frac{1}{8}\text{ kg}$ of flour is used to make 6 rotis.

Alternate Solution:

Since 6 rotis is exactly half of 12 rotis, the amount of flour required will also be half of the flour used for 12 rotis.

$\text{Flour for 6 rotis} = \frac{1}{2} \times \text{Flour for 12 rotis}$

$\text{Flour for 6 rotis} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\text{ kg}$

Given:

The expression to be summed is: $1 \div \frac{1}{6} + 1 \div \frac{1}{10} + 1 \div \frac{1}{13} + 1 \div \frac{1}{9} + 1 \div \frac{1}{2}$

To Find:

The total sum of the expression.

Solution:

To divide 1 by a fraction, we multiply 1 by the reciprocal of that fraction.

Now, we add all these values together:

$\text{Sum} = 6 + 10 + 13 + 9 + 2$

$\text{Sum} = 40$

Therefore, the friend should say that the sum is 40.

Given:

Total number of pages in the novel = $400$

Fraction of pages read yesterday = $\frac{1}{5}$

Fraction of pages read today = $\frac{3}{10}$

To Find:

The number of pages remaining to be read.

Solution:

First, we calculate the number of pages read on each day:

$\text{Pages read yesterday} = \frac{1}{5} \times 400 = \frac{\cancel{400}^{80}}{\cancel{5}_{1}} = 80\text{ pages}$

$\text{Pages read today} = \frac{3}{10} \times 400 = 3 \times \frac{\cancel{400}^{40}}{\cancel{10}_{1}} = 120\text{ pages}$

Now, we find the total pages read so far:

$\text{Total pages read} = 80 + 120 = 200\text{ pages}$

Finally, we subtract the read pages from the total number of pages:

$\text{Remaining pages} = 400 - 200 = 200\text{ pages}$

Therefore, Mira needs to read 200 more pages to finish the novel.

Alternate Solution:

We can first find the total fraction of pages read:

$\text{Total fraction read} = \frac{1}{5} + \frac{3}{10} = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2}$

$\text{Fraction of pages remaining} = 1 - \frac{1}{2} = \frac{1}{2}$

$\text{Number of remaining pages} = \frac{1}{2} \times 400 = 200\text{ pages}$

Given:

Distance covered with $1\text{ litre}$ of petrol = $16\text{ km}$

Quantity of petrol available = $2 \frac{3}{4}\text{ litres}$

To Find:

Total distance covered with $2 \frac{3}{4}\text{ litres}$ of petrol.

Solution:

First, we convert the mixed fraction into an improper fraction:

$2 \frac{3}{4}\text{ litres} = \frac{(2 \times 4) + 3}{4} = \frac{11}{4}\text{ litres}$

Now, we find the total distance:

$\text{Total distance} = \text{Petrol quantity} \times \text{Distance per litre}$

$\text{Total distance} = \frac{11}{4} \times 16$

$\text{Total distance} = 11 \times \frac{\cancel{16}^{4}}{\cancel{4}_{1}}$

$\text{Total distance} = 11 \times 4 = 44\text{ km}$

Therefore, the car will go 44 km using $2 \frac{3}{4}$ litres of petrol.

Given:

Time taken by train = $5 \frac{1}{6}\text{ hours}$

Time taken by plane = $\frac{1}{2}\text{ hour}$

To Find:

The time saved by taking the plane.

Solution:

To find the time saved, we subtract the time taken by the plane from the time taken by the train.

$\text{Time saved} = 5 \frac{1}{6} - \frac{1}{2}$

Convert the mixed fraction to an improper fraction:

$5 \frac{1}{6} = \frac{(5 \times 6) + 1}{6} = \frac{31}{6}$

Now, subtract the fractions by finding a common denominator (LCM of 6 and 2 is 6):

$\text{Time saved} = \frac{31}{6} - \frac{1 \times 3}{2 \times 3}$

$\text{Time saved} = \frac{31}{6} - \frac{3}{6}$

$\text{Time saved} = \frac{31 - 3}{6} = \frac{28}{6}$

Simplify the fraction:

$\text{Time saved} = \frac{\cancel{28}^{14}}{\cancel{6}_{3}} = \frac{14}{3}\text{ hours}$

Convert back to a mixed fraction:

$\text{Time saved} = 4 \frac{2}{3}\text{ hours}$

To express this in hours and minutes:

$\text{Minutes} = \frac{2}{3} \times 60 = 2 \times 20 = 40\text{ minutes}$

Therefore, the plane saves $4 \frac{2}{3}$ hours (or 4 hours and 40 minutes).

Given:

Portion of cake eaten by Mariam and cousins = $\frac{4}{5}$

Number of friends sharing the remaining cake = $3$

To Find:

The portion of the cake received by each friend.

Solution:

First, we find the remaining portion of the cake:

$\text{Remaining cake} = 1 - \frac{4}{5}$

$\text{Remaining cake} = \frac{5 - 4}{5} = \frac{1}{5}$

Now, this $\frac{1}{5}$ portion is divided equally among 3 friends:

$\text{Share of each friend} = \frac{1}{5} \div 3$

$\text{Share of each friend} = \frac{1}{5} \times \frac{1}{3}$

$\text{Share of each friend} = \frac{1 \times 1}{5 \times 3} = \frac{1}{15}$

Therefore, each friend got $\frac{1}{15}$ of the cake.

Given:

The product to be evaluated is $\frac{565}{465} \times \frac{707}{676}$.

To Find:

The relationship between the product and its factors or $1$.

Solution:

First, let us analyze the nature of the two fractions:

We know that if we multiply a number by a factor greater than 1, the product will be greater than the original number.

1. Since we are multiplying $\frac{565}{465}$ by a number greater than 1 ($\frac{707}{676}$), the product will be greater than $\frac{565}{465}$. Thus, option (a) is correct.

2. Since we are multiplying $\frac{707}{676}$ by a number greater than 1 ($\frac{565}{465}$), the product will be greater than $\frac{707}{676}$. Thus, option (c) is correct.

3. Since the product of two numbers greater than 1 is always greater than 1, the product will be greater than 1. Thus, option (e) is correct.

Therefore, the correct options are (a), (c), and (e).

To Find: The fraction of the total area of the big square that is covered by the shaded region.

Solution:

In the given figure, the big square is first divided into $4$ identical squares (quadrants). One small square occupies exactly one-fourth of the area of the big square.

Now, let us consider the bottom-right small square where the shading is present. To find the exact fraction, we need to subdivide this small square into equal parts.

Construction Required:

By drawing the diagonals and the horizontal/vertical midlines within this small square, it is divided into $8$ identical triangles. The image below illustrates this subdivision:

From the subdivided small square, we can see that $3$ triangles are shaded out of the total $8$ identical triangles.

To find the shaded part relative to the whole big square, we multiply the fraction of the small square by its area relative to the whole:

Final Answer:

Hence, $\frac{3}{32}$ of the whole square is shaded.

To Find: The fraction of the original colony of ants that reaches the Mango tree and the Sugarcane field by splitting equally at each junction.

Solution:

We assume the total group of ants starting at the bottom is $1$. At each red circular point (node), the ants split into equal fractions based on the number of outgoing paths.

1. At the first point: The ants split into $2$ ways. The fraction of ants in each way is $1 \div 2 = \frac{1}{2}$. One path goes directly towards the Mango tree.

2. At the second point: The remaining $\frac{1}{2}$ group reaches here and splits into $2$ ways. The fraction of ants in each way is $\frac{1}{2} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. One path goes towards the Mango tree.

3. At the third point: The remaining $\frac{1}{4}$ group reaches here and splits into $4$ ways. The fraction of ants in each way is $\frac{1}{4} \div 4 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$. Out of these four paths, two go to the Mango tree, one goes to the fourth point, and one goes to the Sugarcane field.

4. At the fourth point: The remaining $\frac{1}{16}$ group reaches here and splits into $2$ ways. The fraction of ants in each way is $\frac{1}{16} \div 2 = \frac{1}{16} \times \frac{1}{2} = \frac{1}{32}$. One path goes to the Mango tree and the other to the Sugarcane field.

Now, we calculate the total fraction for each destination by adding the fractions from all paths leading to it:

$\bullet$ Fraction of ants at the Mango tree:

$\bullet$ Fraction of ants near Sugarcane field:

Final Answer:

The fraction of the original group that reaches the Mango tree is $\frac{29}{32}$ and the fraction that reaches the Sugarcane field is $\frac{3}{32}$.

Solution:

1. $1 - \frac{1}{2} = \frac{2-1}{2} = \mathbf{\frac{1}{2}}$

2. $(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) = \frac{1}{2} \times \frac{2}{3} = \frac{1 \times \cancel{2}^{1}}{\cancel{2}_{1} \times 3} = \mathbf{\frac{1}{3}}$

3. $(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{4}) \times (1 - \frac{1}{5})$

$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{1 \times \cancel{2} \times \cancel{3} \times \cancel{4}}{\cancel{2} \times \cancel{3} \times \cancel{4} \times 5} = \mathbf{\frac{1}{5}}$

4. $(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times ... \times (1 - \frac{1}{10})$

$= \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} \times \frac{6}{7} \times \frac{7}{8} \times \frac{8}{9} \times \frac{9}{10}$

Notice that the numerator of each fraction cancels out with the denominator of the preceding fraction. Only the first numerator ($1$) and the last denominator ($10$) remain.

$= \mathbf{\frac{1}{10}}$

General Statement:

The product of the sequence $(1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times \dots \times (1 - \frac{1}{n})$ is always equal to $\frac{1}{n}$.

Explanation:

When we simplify each term $(1 - \frac{1}{k})$, we get $\frac{k-1}{k}$. The full expression looks like:

$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \dots \times \frac{n-1}{n}$

In this product, every intermediate term (like 2, 3, 4...) appears once as a numerator and once as a denominator, causing them to cancel out. This is known as a telescoping product. The only terms left behind are the numerator of the first fraction ($1$) and the denominator of the last fraction ($n$).