Multiply the distance per hour by the number of hours:

$$3 \times \tfrac{1}{4} = \tfrac{1}{4}+\tfrac{1}{4}+\tfrac{1}{4} = \tfrac{3}{4}\text{ km}$$

In $\tfrac15$ hour, the distance covered is what we get by splitting the 1-hour distance (3 km) into 5 equal parts:

$$\tfrac{1}{5} \times 3 = \tfrac{3}{5}\text{ km}$$

$$5 \times \tfrac{2}{3} = \tfrac23+\tfrac23+\tfrac23+\tfrac23+\tfrac23 = \tfrac{10}{3}\text{ acre}$$

Convert the mixed fraction: $1\tfrac14 = \tfrac54$ hours.

$$\tfrac{5}{4} \times 8 = \tfrac{5\times 8}{4} = \tfrac{40}{4}=10$$

We need $\tfrac12 \times \tfrac14$. Take a unit square as the “whole”, shade $\tfrac14$ of it (one row out of four), then split that shaded strip into 2 equal parts. The whole square is now divided into 8 equal parts, and exactly 1 of them is the shaded-then-halved region.

Since the whole is divided into 8 equal parts and 1 part is shaded:

$$\tfrac{1}{2} \times \tfrac{1}{4} = \tfrac{1}{8}$$

$$\tfrac{3}{4}\times\tfrac{2}{5} = \tfrac{6}{20} = \tfrac{3}{10}\text{ km}$$

First represent $\tfrac32$ on a unit square (one whole and a half). Dividing this into 4 equal parts: the whole is split into 2 rows × 4 columns = 8 equal parts, and 3 of them are shaded, so $\tfrac32 \div 4 = \tfrac38$.

Now multiply this by 5 (the numerator of $\tfrac54$):

$$\tfrac{5}{4}\times\tfrac{3}{2} = 5\times\tfrac38 = \tfrac{15}{8}$$

Instead of multiplying first and simplifying later, cancel the common factor of 12 and 24 before multiplying:

$$\tfrac{12}{7}\times\tfrac{5}{24} = \tfrac{\cancel{12}^{\,1}\times 5}{7\times \cancel{24}^{\,2}} = \tfrac{1\times5}{7\times2} = \tfrac{5}{14}$$

Compare: $6\div3=2$ (quotient < dividend, since divisor > 1); but $6\div\tfrac14=24$ and $\tfrac18\div\tfrac14=\tfrac12$ (quotient > dividend in both, since divisor < 1).

We need $\tfrac14 \div 5$. Rewrite as multiplication by the reciprocal of 5 (which is $\tfrac15$):

$$\tfrac{1}{4}\div5 = \tfrac{1}{5}\times\tfrac{1}{4} = \tfrac{1}{20}$$

Area of one brick $= \tfrac15\times\tfrac15=\tfrac{1}{25}$ sq. units. Total area $=7\tfrac12=\tfrac{15}{2}$ sq. units.

Number of bricks $=$ Total area $\div$ area of one brick:

$$\tfrac{15}{2}\div\tfrac{1}{25} = \tfrac{15}{2}\times25 = \tfrac{375}{2} = 187.5$$

Find how many times each fountain would fill the cistern in one day:

• Fountain 1: $1\div1=1$ time
• Fountain 2: $1\div\tfrac12=2$ times
• Fountain 3: $1\div\tfrac14=4$ times
• Fountain 4: $1\div\tfrac15=5$ times

Together, in one day they fill the cistern $1+2+4+5=12$ times over.

In both cases the method is the same: identify what fraction each nested region is of the one containing it, then multiply those fractions together (exactly like the area-model for fraction multiplication).

For unit fractions, the unit square is split into (denominator 1) rows × (denominator 2) columns, and exactly one small rectangle is shaded:

So: $$\tfrac{1}{12}\times\tfrac{1}{18} = \tfrac{1}{12\times18} = \tfrac{1}{216}$$

Using $\tfrac{a}{b}\times\tfrac{c}{d}=\tfrac{a\times c}{b\times d}$:

Land remaining with Somu after the road $= 1-\tfrac16=\tfrac56$.

$$3\tfrac34\times9\tfrac35 = \tfrac{15}{4}\times\tfrac{48}{5} = \tfrac{15\times48}{4\times5}=\tfrac{720}{20}=36\text{ sq ft}$$

4 saplings in a row have 3 gaps between the first and last:

$$3\times\tfrac34 = \tfrac94 = 2\tfrac14\text{ m}$$

$\tfrac{12}{15}$ of 500 g $=\tfrac{12}{15}\times500=400$ g $=0.4$ kg.
$\tfrac{3}{20}$ of 4 kg $=\tfrac{3}{20}\times4=\tfrac{12}{20}=0.6$ kg.

¼ kg makes 12 rotis, so flour per roti $=\tfrac14\div12=\tfrac{1}{48}$ kg. For 6 rotis: $6\times\tfrac{1}{48}=\tfrac{6}{48}=\tfrac18$ kg.

$1\div\tfrac16=6,\; 1\div\tfrac{1}{10}=10,\; 1\div\tfrac{1}{13}=13,\; 1\div\tfrac19=9,\; 1\div\tfrac12=2$

$$6+10+13+9+2 = 40$$

Yesterday: $\tfrac15\times400=80$ pages. Today: $\tfrac{3}{10}\times400=120$ pages. Total read $=200$ pages.

$$16\times2\tfrac34 = 16\times\tfrac{11}{4}=\tfrac{176}{4}=44\text{ km}$$

Train: $5\tfrac16=\tfrac{31}{6}$ hr. Time saved by plane $=\tfrac{31}{6}-\tfrac12=\tfrac{31}{6}-\tfrac36=\tfrac{28}{6}=\tfrac{14}{3}=4\tfrac23$ hours.

Cake remaining after cousins ate $\tfrac45$: $1-\tfrac45=\tfrac15$. Shared equally among 3 friends: $\tfrac15\div3=\tfrac{1}{15}$.

Both fractions are greater than 1 (since $565>465$ and $707>676$). The product of two numbers each greater than 1 is greater than both individual numbers, and certainly greater than 1.

At each branch point the group of ants splits into two equal halves. Tracking the fraction of the original colony that survives down each branch to the two food sources, and checking that all the fractional paths add up to the whole colony (1):

Each factor $\left(1-\tfrac1k\right)=\tfrac{k-1}{k}$, so consecutive terms cancel in a chain: $\tfrac12\times\tfrac23\times\tfrac34\times\cdots\times\tfrac{n-1}{n}=\tfrac{1}{n}$ (every numerator cancels with the denominator before it, leaving only the very first numerator, 1, and the very last denominator, $n$).

A queen attacks along its row, column, and both diagonals, so no two queens may share any of these. This is the classic “Eight Queens Puzzle” — it has 92 total solutions (12 of them genuinely distinct once you remove rotations/reflections). Here is one valid placement, given as (column, row) pairs:

(1,1) (2,5) (3,8) (4,6) (5,3) (6,7) (7,2) (8,4)

Check: no two queens share a row, a column, or a diagonal — this is one of the 92 valid arrangements. (For the warm-up 4×4 puzzle, one valid placement is rows 2, 4, 1, 3 for columns 1–4.)