New product is $24times27$. Using the distributive property, $(a+1)b = ab+b$, with $a=23, b=27$:

New product is $23times28$. Using $a(b+1) = ab + a$, with $a=23, b=27$:

New product is $24times28 = (23+1)(27+1)$. Expanding using distributivity:

Pattern noticed: increasing one number by 1 increases the product by the value of the other number; increasing both by 1 increases the product by the sum of both numbers plus 1.

No — the product does not always increase. Recall the identity:

The increase is $b – a – 1$. This is negative (i.e. the product decreases) whenever $b lt a+1$, i.e. when $a ge b$.

The distributive property holds for all integers, positive or negative, since $x(y+z)=xy+xz$ is true for any integers $x,y,z$. Let’s verify with the identity $(a+1)(b-1)=ab+b-a-1$.

Treat $(b+1)$ as one block and distribute over $(a+1)$:

• $(a+1)(b+1) = a(b+1) + 1(b+1)$
• $= ab + a + b + 1$

Write as $(a + (-2))(b+3)$, so $m=-2, n=3$:

Write as $(a+(-3))(b+(-4))$, so $m=-3, n=-4$:

Both results match what you’d get by direct distribution, confirming Identity 1 works for decreases too — just treat the decrease as adding a negative number.

We know $(a+b)^2 = a^2+2ab+b^2$. So:

• If $ab gt 0$ (both same sign, both positive or both negative) → $(a+b)^2 gt a^2+b^2$.
• If $ab lt 0$ (opposite signs) → $(a+b)^2 lt a^2+b^2$.
• If $a=0$ or $b=0$ → $(a+b)^2 = a^2+b^2$.

Decompose $104 = 100+4$:

Decompose $37 = 30+7$:

Two terms can only be combined (added/subtracted into one term) if they are like terms — meaning they have exactly the same letter-numbers (variables) raised to the same powers.

To multiply any number by 11: write down the last (units) digit as it is. Then, moving leftward, add each pair of neighbouring digits and write down that sum (carrying over if the sum is 10 or more). Finally, write down the first digit (plus any carry).

For a 4-digit number $dcba$: $ dcba times 11 = d (c+d) (b+c) (a+b) a$  (handling carries as needed).

Using $dcba times 101 = dcbatimes100 + dcba$, for digits $d{=}3, c{=}8, b{=}7, a{=}4$:

For 101: leave the last two digits as-is, then for each pair moving left, add digits two places apart, carrying as needed, finishing with the leading digit(s). This generalises: multiplying by $10^k+1$ means “shift and add” — write the number, then add the same number shifted left by $k$ places.

• By $101 = 100+1$: shift left by 2 places and add.
• By $1001 = 1000+1$: shift left by 3 places and add.
• By $10001 = 10000+1$: shift left by 4 places and add.

Write $99 = 100-1$:

Write $999 = 1000-1$:

The identity $(a+b)^2 = a^2+2ab+b^2$ works for any split of 65 into two parts — the total area of the big square stays the same regardless of how we partition the side length.

Using $(a+b)^2=a^2+2ab+b^2$ with $a=m, b=3$:

With $a=6, b=p$:

With $a=3j, b=2k$:

Draw a square of side $a$, and remove a square of side $b$ from one corner (where $b lt a$), leaving an L-shaped region with side $a-b$.

The square of side $(a-b)$ is obtained by taking the big square ($a^2$), and removing the two rectangle strips of dimensions $atimes b$ — but this removes the small corner square of side $b$ twice, so we add it back once.

Write $99 = 100-1$, so $a=100, b=1$:

Write $58 = 60-2$, so $a=60, b=2$:

Take $a’=-2a, b’=3$ (or equivalently use $(3-2a)^2$):

Take $a=4, b=7$. Sum of squares: $4^2+7^2 = 16+49=65$. Twice this sum: $2times65=130$.

Add the two identities:

Write as $(100-2)(100+2)$, so $a=100, b=2$:

Write as $(50-5)(50+5)$, so $a=50, b=5$:

Start with a large square of side $a$, with a smaller square of side $b$ cut from one corner. The remaining L-shaped region has area $a^2-b^2$.

Cut this L-shape into two rectangles as shown, then re-arrange (slide the smaller rectangle) to form a single rectangle of dimensions $(a+b)$ by $(a-b)$.

This is just Identity 1C, $(a+b)(a-b)=a^2-b^2$, rearranged by adding $b^2$ to both sides:

The pattern shows hollow square frames: Step 1 is a $3times3$ frame with a $1times1$ hole, Step 2 is a $4times4$ frame with a $2times2$ hole, Step 3 is a $5times5$ frame with a $3times3$ hole.

Number of tiles = (area of outer square) − (area of inner hole) = $(n+2)^2 – n^2$.

Expanding $(n+2)^2-n^2$ using the difference-of-squares idea:

The full bounding rectangle has dimensions $s times p$. The missing piece (top-left, not part of the L) is $(s-r)times r$ — wait, let’s set it up carefully using the figure: the overall width is $s$, overall height is $p$; the small square cut out (bottom-left corner notch) has side $r$.

This comes from: take rectangle $p times s$, subtract the two rectangle strips of width $r$ along two adjoining sides ($ptimes r$ and $stimes r$), then add back the doubly-subtracted corner square $r^2$ — exactly like the $(a-b)^2$ identity logic.

The grid is a multiplication table where the entry at row $i$, column $j$ is $itimes j$. If the centre cell of a $3times3$ block is $ptimes q = pq$ (row $p$, column $q$), then the row above is $(p-1)$, row below is $(p+1)$; column to the left is $(q-1)$, column to the right is $(q+1)$.

Let the numbers be $a$ and $b$. We need $(a+2)(b-4) = ab$.

$(a-b)$ multiplied by a sum of $n$ terms (powers of $a$ decreasing, powers of $b$ increasing) always telescopes down to a simple difference of powers: $a^2-b^2, a^3-b^3, a^4-b^4, ldots$

Verification by expansion: $a^5+a^4b+a^3b^2+a^2b^3+ab^4 -a^4b-a^3b^2-a^2b^3-ab^4-b^5 = a^5-b^5$ ✓ (all middle terms cancel out).

Note that $(b-a) = -(a-b)$. So:

Verification with numbers: Let $a=7, b=3$. Then $(a-b)^2=4^2=16$ and $(b-a)^2=(-4)^2=16$.

We want $a^2-b^2 = 100$, i.e. $(a+b)(a-b)=100$. Try $a-b=4$ and $a+b=25$… but those need to give integer $a,b$. Easier: try $a-b=2, a+b=50 Rightarrow a=26, b=24$.

Both patterns are derived purely from the algebraic identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, which are true for any real numbers $a$ and $b$ — there is no restriction requiring them to be counting numbers.

Let $a=-3, b=2$:

Let $a=tfrac32, b=tfrac12$:

The student treated $-3p$ as if it were just $-3$, forgetting to multiply the variable $p$ into each term. Multiplication, not addition, must be distributed across the bracket.

In the first step, $2$ was multiplied only into the $x$ term of $(x-1)$, but not into $-1$. (It should be $2 times x = 2x$ and $2times(-1)=-2$, not just $-1$.)

This treats $y+2(y+2)$ as if it were $(y+2)times(y+2)$ — but the expression has $y$ added to $2(y+2)$, not multiplied. The mistake is converting an addition into a squaring operation.

The middle “cross term” $2ab$ from the identity $(a+b)^2=a^2+2ab+b^2$ was forgotten entirely. Squaring a sum is not the same as squaring each term separately.

The sign of the cross term was computed incorrectly. With $a=-q, b=2$: the middle term is $2ab = 2(-q)(2) = -4q$ — that part is actually right — but $a^2=(-q)^2=q^2$ is correct too. Let’s verify carefully: actually $(-q+2)^2 = (-q)^2+2(-q)(2)+2^2 = q^2-4q+4$. This is correct!

The expression $3a(2btimes3c)$ has only one multiplication needed — first simplify inside the bracket, then multiply by $3a$ once. Instead, the student incorrectly distributed $3a$ separately into $2b$ and $3c$ as if it were $3a(2b+3c)$, creating two separate products and multiplying them together — a double mistake.

Actually, this one is computed correctly! $tfrac12(10s-6) = 5s-3$, and adding $3$ gives $5s-3+3=5s$.

$5w^2$ and $6w$ are not like terms — one has $w^2$, the other has $w^1$. Only like terms (same variable, same power) can be combined into a single term. The student incorrectly added the coefficients while ignoring that the powers of $w$ are different.

Since the terms are unlike, the expression is already in its simplest form:

Two separate mistakes: (1) $2a^3+3a^3=5a^3$ is correct, since these are like terms. (2) But $6a^2b$ and $6ab^2$ are not like terms (different powers of $a$ and $b$ — $a^2b$ vs $ab^2$), so they cannot be combined into $12a^2b^2$.

Nothing — this expansion is done correctly, step by step, using the distributive property properly.

Only the “outer” terms ($atimes b$) and the product of the two constants ($2times4=8$) were multiplied; the two “cross” terms ($atimes4$ and $2times b$) were completely skipped. All four pairwise products are needed when expanding $(a+2)(b+4)$.

Let’s verify by factoring out the common factor $ab$ from each term: $ab^2 = abtimes b$; $a^2b = abtimes a$; $a^2b^2 = abtimes ab$. So:

“A square number” means some number squared, say $s^2$. “Two more than” it means add 2:

If one number is $m$, the next consecutive number is $m+1$. Sum of their squares:

Label the 2×2 block as: top-left $=a$, top-right $=a+1$; bottom-left $=a+7$ (one row below, same weekday column), bottom-right $=a+8$.

This depends on $k$ (e.g. $k=1$ gives $1+2-1=2$, but $k=2$ gives $4+4-1=7ne2$).

$4q^2-4q$ is a multiple of 4, but subtracting 3 means the whole expression leaves remainder $-3 equiv 1 pmod 4$ — not a multiple of 4. Check $q=1$: $(3)(-1)=-3$, not a multiple of 4.

Even number: $2n$. $(2n)^2=4n^2$ — clearly a multiple of 4. ✓

Odd number: $2n+1$. $(2n+1)^2 = 4n^2+4n+1 = 4n(n+1)+1$. Since $n(n+1)$ is always even (product of consecutive integers), $4n(n+1)$ is a multiple of 8, so $(2n+1)^2$ is 1 more than a multiple of 8. ✓

For this to be “5 less than a square number,” we’d need $20n^2-5 = (text{some square})-5$, i.e. $20n^2$ itself a perfect square for all $n$ — but $20n^2$ is a perfect square only when $20$ is a perfect square times something, which fails generally (e.g., $n=1$: $20$ is not a perfect square).

Let $n_1 = 7a+3$ and $n_2=7b+5$ for some integers $a, b$.

Remainder when divided by 7: 1

Remainder when divided by 7: 5  (equivalently, $n_2-n_1$ leaves remainder 2)

Remainder when divided by 7: 1

Take $4, 5, 6$: middle squared $= 5^2=25$; product of outer two $=4times6=24$. Difference $=25-24=1$.

Try $9,10,11$: $10^2=100$; $9times11=99$; difference $=1$.

Let consecutive numbers be $(n-1), n, (n+1)$:

Let the two numbers be $a$ and $b$.

• Add the two numbers: $a+b$
• Multiply by half of the sum: $(a+b) times tfrac12(a+b)$

Write $14times26 = (16-2)(24+2)$:

Write $25times75 = (26-1)(74+1)$:

General insight: for a fixed sum, the product of two numbers is larger when the numbers are closer together — moving them further apart (while keeping the sum fixed) always decreases the product.

Looking at the layout: the path of width $w$ surrounds and separates the two square plots. The full park’s length (left-to-right) $= w + g + 2w + g + w = 2g+4w$. The park’s breadth (top-to-bottom) $= w+g+w = g+2w$.

Counting units: Step 1 has $9=3^2$ units, Step 2 has $16=4^2$, Step 3 has $25=5^2$ — each step is a perfect square of (Step number + 2).

Step 1 has $5$ squares (a $2times2$ block of 4, plus 1 extra) $=2^2+1$. Step 2 has $11 = 3^2+2$. Step 3 has $19=4^2+3$. Pattern: Step $y$ has $(y+1)^2+y$.

For a triangle with $n$ rows, the total number of coins is the triangular number $T_n = tfrac{n(n+1)}{2}$.

The minimum number of moves needed follows a known result for “reversing a triangular array”: for the cases above, the minimums $1, 2, 3, 5$ for $n=2,3,4,5$ rows match a quadratic-style growth pattern related to $n^2$.

More simply, by direct construction: for the 15-coin (5-row) triangle, the minimum is 5 moves — found by identifying which corner coins lie outside the “flipped” footprint and relocating just those.

Images A, C, and D all show the tiger in the same proportions — just scaled up or down uniformly. Image B looks stretched (elongated), and Image E looks squashed (fatter/compressed).

Width of C is half of A’s width ($30 = \tfrac12\times60$), and height of C is also half of A’s height ($20=\tfrac12\times40$). Both dimensions change by the same factor ($\tfrac12$) — so the shapes look the same.

Width of B is 20 mm less than A ($40 = 60-20$), and height of B is also 20 mm less than A ($20=40-20$). The difference is the same (subtraction), but checking the factor (multiplication/division): height of B is half of A’s height, but width of B ($40$) is not half of A’s width ($60$) — it would need to be 30, not 40.

Image A: width $60$, height $40$. Image D: width $90$, height $60$.

We need the factor $f$ such that $60\times f = 90$:

Checking the second term: $40 \times \tfrac32 = 60$ ✓ — matches image D’s height.

$3:4$ is already in simplest form. To simplify $72:96$, we find $\text{HCF}(72,96)$.

Dividing both terms of $72:96$ by 24:

To keep the same sweetness, the ratio of glasses of lemonade to spoons of sugar must stay proportional. We model this as:

First simplify the example ratio: $5:170 = 1:34$ (dividing by HCF $=5$).

Suppose a school has 8 teachers and 240 students. Ratio is $8:240$. HCF of 8 and 240 is 8, so simplest form is $1:30$.

Suppose a blackboard measures $200\,\text{cm}$ wide and $120\,\text{cm}$ tall. The ratio of width to height is $200:120$, which simplifies (HCF $=40$) to $5:3$.

To draw a smaller proportional rectangle, pick any common factor, e.g. multiply $5:3$ by $2$ to get $10:6$ (cm) as a notebook-sized rectangle.

Mother’s age $=10\times3=30$. Ratio of Neelima’s age to mother’s age:

Mother’s age then $=30+9=39$. Ratio:

$42$ is $2$ times $21$ (the second term of $14:21$), so the factor is $2$. The first term should also be multiplied by $2$:

We need the factor $y$ such that $14y=6$:

Dividing $14$ by its HCF with $21$ (which is $7$) gives $2$. Dividing $21$ by $7$ also:

Regular ratio $15:35$ simplifies to $3:7$ (dividing by HCF 5). The “stronger” ratio $20:30$ simplifies to $2:3$ (dividing by HCF 10). Comparing decoction’s share: $\tfrac{3}{10}$ (regular) vs $\tfrac{2}{5}=\tfrac{4}{10}$ (stronger) — a larger fraction of decoction, hence stronger.

$10:40$ simplifies to $1:4$. Decoction’s share is $\tfrac{1}{5} = \tfrac{2}{10}$ — a smaller fraction of decoction than the regular $\tfrac{3}{10}$, hence lighter.

No, this formulation is incorrect as written, because the units don’t match: $150$ is in minutes, but $4$ is in hours. Both terms representing time must use the same unit for the ratio to make sense.

Convert 4 hours to minutes: $4 \times 60 = 240$ minutes.

It would take less time at the higher speed — speed and time are inversely related for a fixed distance (faster speed means the same distance is covered sooner).

The Rule of Three only works when two quantities increase or decrease together in the same direction (direct proportion) — e.g., more distance needs more time at constant speed. But here, as speed increases, time decreases — this is an inverse relationship, not a direct one.

Since $1:30 \ne 1:77$, the ratios are not proportional.

Larger containers are usually relatively cheaper per mL because manufacturing, packaging, and transport costs don’t scale linearly with volume — bigger packaging is more cost-efficient per unit. Small sachets carry a higher “convenience premium” per mL, since fixed costs (packaging material, machinery) are spread over a smaller volume.

Sharing 12 counters equally means each person gets $12\div2=6$ counters.

If the partner gets 5 counters out of the total 12, you get the remaining:

Multiply both terms of $4:9$ by the same factor (2, 3, 4, …):

Simplify $18:24$ first: HCF is 6, so simplest form is $3:4$. Every proportional ratio must equal $3:4$ when simplified.

$12 = 3\times4$, so multiply 4 by the same factor: $4\times4=16$.

$20 = 3\times\tfrac{20}{3}$, so the second term is $4\times\tfrac{20}{3} = \tfrac{80}{3}$.

$27 = 3\times9$, so the second term is $4\times9=36$.

Two rectangles are similar if the ratio of their width to height (in simplest form) is the same, regardless of orientation (rotation doesn’t affect the ratio).

• Measure the width and height of each rectangle (A, B, C, D, E) using a ruler/scale.
• Compute the ratio width:height for each, and reduce to simplest form.
• Group together the rectangles that share the same simplified ratio.

Measure the given rectangle’s width and height, find the ratio in simplest form, then draw new rectangles using any common multiple of that ratio (smaller using a smaller multiplier, bigger using a larger one).

Counting one repeating block: grey bricks $=2+3+4=9$; coloured (red) bricks $=3+2+1=6$.

Counting one repeating block: grey bricks $=16$; coloured (orange) bricks $=12$.

Suppose: head $=20$ cm, torso $=45$ cm, arms $=60$ cm, legs $=75$ cm.

There are 52 weeks in a year (approximately). We set up the proportion (year : distance) :: (week : distance):

Total groups $= 2+3=5$. Size of each group $=4500\div5=900$.

Total groups $=1+5=6$. Size of each group $=240\div6=40$ mL.

Total groups $=3+5=8$. Size of each group $=40\div8=5$ mL.

Total groups $=2+1=3$. Size of each group $=6\div3=2$ cups.

In the original bucket, red is $\tfrac{3}{8}$ of the total and yellow is $\tfrac58$ of the total (treating the bucket as 1 whole unit).

Red remains $\tfrac38$. New ratio of red to yellow:

HCF of 600 and 900 is 300.

Since we can’t have a fraction of a bus, we round up to 4 buses.

The ratio $4:6$ means the neck is $\tfrac{4}{4+6}=\tfrac{4}{10}$ of the total height.

Example: if your height is 150 cm, your neck (by this ratio) would be $\tfrac{4}{10}\times150=60$ cm.

Set up the proportion: $2\tfrac12 : \tfrac37 :: x : 9$, i.e., $\tfrac52 : \tfrac37 :: x:9$.

Let $x$ be the number of years from now. Harmain’s age becomes $x+1$, brother’s age becomes $x+5$.

Convert 10 litres to mL: $10\times1000=10{,}000$ mL.

$1$ acre $=43{,}560$ sq.ft.

Total groups $=3+1=4$. Size of each group $=7.74\div4=1.935$ g.

• Look for any two adjacent identical symbols (e.g., two horizontal lines in a row) — the cell next to them must be the opposite symbol, since three in a row is forbidden.
• Count filled symbols in each row/column. If a row already has, say, half its cells as one symbol (for a 6×6 grid, that’s 3), every remaining empty cell in that row must be the other symbol.
• After filling using the above two rules, check for duplicate rows or columns. If completing a row a certain way would duplicate an existing row, flip the remaining empty cells appropriately.
• Repeat these steps, alternating between rows and columns, until the entire grid is filled.