HEGP106 — We Distribute, Yet Things Multiply
Complete Answers to the Mathematical Problems
This guide answers the mathematical problems, “Figure it Out”
questions, algebraic corrections, pattern activities, and applications in the
attached Class 8 chapter.
1. Increments in products
Problem 1. How much does (23\times27) increase if the first
number is increased by 1?
[ (23+1)\times27-23\times27=27. ]
The product increases by 27.
Problem 2. How much does (23\times27) increase if the second
number is increased by 1?
[ 23\times(27+1)-23\times27=23. ]
The product increases by 23.
Problem 3. How much does the product increase if both numbers
are increased by 1?
[ (23+1)(27+1)-23\times27=24\times28-23\times27. ]
Using the distributive property,
[ (a+1)(b+1)=ab+a+b+1. ]
Therefore, the increase is
[ \boxed{a+b+1}. ]
For 23 and 27, the increase is
[ 23+27+1=\boxed{51}. ]
Problem 4. What happens when one factor is increased by 1 and
the other is decreased by 1?
[ (a+1)(b-1)=ab+b-a-1. ]
The change is
[ \boxed{b-a-1}. ]
It can be positive, zero, or negative. For example, with
(a=10,b=3),
[10\times3=30,
\qquad(10+1)(3-1)=22,]
so the product decreases.
2. General product-change identity
If the original product is (ab), and the factors become (a+m)
and (b+n), then
[ (a+m)(b+n)=ab+mb+an+mn. ]
Therefore, the change in the product is
[ \boxed{mb+an+mn}. ]
Important special cases are:
[ (a+u)(b-v)=ab+ub-av-uv, ]
[ (a-u)(b+v)=ab+av-bu-uv, ]
and
[ (a-u)(b-v)=ab-av-bu+uv. ]
Problem 5. Use the identity for the two specified changes.
(i) One factor decreases by 2 and the other increases by 3
[ (a-2)(b+3)=ab+3a-2b-6. ]
The change is
[ \boxed{3a-2b-6}. ]
(ii) One factor decreases by 3 and the other by 4
[ (a-3)(b-4)=ab-4a-3b+12. ]
The change is
[ \boxed{-4a-3b+12}. ]
3. Multiplication-grid exercise
If the centre of a 3×3 multiplication frame is (pq), the
factors at the centre are (p) and (q). The surrounding entries are:
[\begin{array}{ccc}
(p-1)(q-1)&(p-1)q&(p-1)(q+1)\
p(q-1)&pq&p(q+1)\(p+1)(q-1)&(p+1)q&(p+1)(q+1)
\end{array}]
Expand the following products
|
Expression |
Expansion |
|
((3+u)(v-3)) |
(3v-9+uv-3u) |
|
(3(15+6a)) |
(45+18a) |
|
((10a+b)(10c+d)) |
(100ac+10ad+10bc+bd) |
|
((3-x)(x-6)) |
(-x^2+9x-18) |
|
((-5a+b)(c+d)) |
(-5ac-5ad+bc+bd) |
|
((5+z)(y+9)) |
(5y+45+yz+9z) |
Problem 6. Give three examples where the product remains
unchanged when one factor increases by 2 and the other decreases by 4.
We require
[ (a+2)(b-4)=ab. ]
Expanding gives
[ab-4a+2b-8=ab
\Rightarrow 2b=4a+8
\Rightarrow b=2a+4.]
Examples are
[ (1,6):\quad1\times6=3\times2, ]
[ (2,8):\quad2\times8=4\times4, ]
[ (3,10):\quad3\times10=5\times6. ]
4. More expansions and identities
Expand the following expressions
[(a+ab-3b^2)(4+b)
=4a+5ab+ab^2-12b^2-3b^3.]
[(4y+7)(y+11z-3)
=4y^2+44yz-5y+77z-21.]
Difference-of-powers pattern
[ (a-b)(a+b)=a^2-b^2, ]
[ (a-b)(a^2+ab+b^2)=a^3-b^3, ]
[ (a-b)(a^3+a^2b+ab^2+b^3)=a^4-b^4. ]
The next identity is
[ \boxed{(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5}. ]
In general,
[ (a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})=a^n-b^n. ]
5. Multiplication by 11, 101, 1001, 99, and 999
Evaluate products by 11
[ 94\times11=\boxed{1034}, ]
[ 495\times11=\boxed{5445}, ]
[ 3279\times11=\boxed{36069}, ]
[ 4791256\times11=\boxed{52703816}. ]
General rules
[ N\times101=N\times(100+1)=100N+N, ]
[ N\times1001=N\times(1000+1)=1000N+N, ]
[ N\times10001=N\times(10000+1)=10000N+N. ]
Also,
[ N\times99=N(100-1)=100N-N, ]
and
[ N\times999=N(1000-1)=1000N-N. ]
Evaluate the requested products
[ 89\times101=\boxed{8989}, ]
[ 949\times101=\boxed{95849}, ]
[ 265831\times1001=\boxed{266096831}, ]
[ 1111\times1001=\boxed{1112111}, ]
[ 9734\times99=\boxed{963666}, ]
[ 23478\times999=\boxed{23454522}. ]
6. Square identities
The three important identities are
[ (a+b)^2=a^2+2ab+b^2, ]
[ (a-b)^2=a^2-2ab+b^2, ]
and
[ (a+b)(a-b)=a^2-b^2. ]
Useful consequences are
[ 2(a^2+b^2)=(a+b)^2+(a-b)^2, ]
and
[ a^2=(a+b)(a-b)+b^2. ]
Use the identities to calculate squares
[ 104^2=(100+4)^2=10000+800+16=\boxed{10816}, ]
[ 37^2=(40-3)^2=1600-240+9=\boxed{1369}, ]
[ 99^2=(100-1)^2=10000-200+1=\boxed{9801}, ]
[ 58^2=(60-2)^2=3600-240+4=\boxed{3364}. ]
Also,
[ 98\times102=(100-2)(100+2)=10000-4=\boxed{9996}, ]
[ 45\times55=(50-5)(50+5)=2500-25=\boxed{2475}. ]
7. Figure it Out: square identities
Question 1. Which is greater: ((a-b)^2) or ((b-a)^2)?
They are equal because
[ b-a=-(a-b), ]
and the square of a number and its opposite are equal:
[ \boxed{(a-b)^2=(b-a)^2}. ]
Question 2. Express 100 as a difference of two squares.
One answer is
[ \boxed{26^2-24^2=676-576=100}. ]
There are infinitely many integer solutions because
[ (a-b)(a+b)=100. ]
Question 3. Find the following using identities.
[ 406^2=(400+6)^2=\boxed{164836}, ]
[ 72^2=\boxed{5184}, ]
[ 145^2=\boxed{21025}, ]
[ 1097^2=(1100-3)^2=\boxed{1203409}, ]
[ 124^2=\boxed{15376}. ]
Question 4. Do the patterns hold for negative integers and
fractions?
Yes. The identities follow from the distributive property,
which is valid for integers, rational numbers, and real numbers. For example,
[ (a+b)^2=a^2+2ab+b^2 ]
holds regardless of whether (a) and (b) are positive,
negative, or fractional.
8. Correct the algebraic mistakes
|
No. |
Correct result |
|
1 |
(-3p(-5p+2q)=15p^2-6pq) |
|
2 |
(2(x-1)+3(x+4)=5x+10) |
|
3 |
(y+2(y+2)=3y+4) |
|
4 |
((5m+6n)^2=25m^2+60mn+36n^2) |
|
5 |
((-q+2)^2=q^2-4q+4), so this one is correct |
|
6 |
(3a(2b\times3c)=18abc) |
|
7 |
(\frac12(10s-6)+3=5s) |
|
8 |
(5w^2+6w) cannot be combined further |
|
9 |
(2a^3+3a^3+6a^2b+6ab^2=5a^3+6a^2b+6ab^2) |
|
10 |
((x+2)(x+5)=x^2+7x+10), so this one is correct |
|
11 |
((a+2)(b+4)=ab+4a+2b+8) |
|
12 |
(ab^2+a^2b+a^2b^2=ab(a+b+ab)), so this one is correct |
9. Circle and tile patterns
Circle pattern
The number of circles at Step (k) is
[ k^2+2k=k(k+2)=(k+1)^2-1. ]
Thus, the next figure after Step 3 is Step 4, containing
[ 4^2+2(4)=\boxed{24} ]
circles. Step 10 contains
[ 10^2+2(10)=\boxed{120} ]
circles, and Step 15 contains
[ 15^2+2(15)=\boxed{255} ]
circles.
Square-tile pattern
The number of tiles at Step (n) is
[ (n+2)^2-n^2=4n+4. ]
Therefore:
[ \text{Step 4}: (4+2)^2-4^2=36-16=\boxed{20}, ]
[ \text{Step 10}: (10+2)^2-10^2=144-100=\boxed{44}. ]
The next figure is the next border of tiles, and the
algebraic expression is
[ \boxed{(n+2)^2-n^2=4n+4}. ]
10. Shaded-area problems
Slanting-line region
For the first shaded arrangement, the area can be written in
more than one equivalent way as
[ \boxed{x^2-xy=x(x-y)}. ]
For the second arrangement, the area is
[ \boxed{x(x+2y)-3xy=x^2-xy=x(x-y)}. ]
The different methods agree because they simplify to the same
expression.
Dashed region
The area of the dashed region is
[ \boxed{ps-pr-sr+r^2}. ]
Substituting (p=6), (r=3.5), and (s=9):
[ps-pr-sr+r^2
=6\cdot9-6\cdot3.5-3.5\cdot9+3.5^2]
[=54-21-31.5+12.25
=\boxed{13.75\text{ square units}}.]
11. Figure it Out: identity products
Question 1. Compute the products.
[ 46^2=(40+6)^2=\boxed{2116}, ]
[ 397\times403=(400-3)(400+3)=400^2-3^2=\boxed{159991}, ]
[ 91^2=(100-9)^2=\boxed{8281}, ]
[ 43\times45=(44-1)(44+1)=44^2-1=\boxed{1935}. ]
Question 2. Expand the products.
[ (p-1)(p+11)=\boxed{p^2+10p-11}, ]
[ (3a-9b)(3a+9b)=\boxed{9a^2-81b^2}, ]
[ -(2y+5)(3y+4)=\boxed{-6y^2-23y-20}, ]
[ (6x+5y)^2=\boxed{36x^2+60xy+25y^2}, ]
[ \left(2x-\frac12\right)^2=\boxed{4x^2-2x+\frac14}, ]
[ (7p)(3r)(p+2)=\boxed{21p^2r+42pr}. ]
Question 3. Identify the correct expressions.
“Two more than a square number” is
[ \boxed{s^2+2}. ]
The sum of the squares of two consecutive numbers is
[ \boxed{m^2+(m+1)^2}. ]
Equivalent forms include ((m-1)^2+m^2) when the consecutive
numbers are (m-1,m).
Question 4. Calendar diagonal products
Label a 2×2 calendar block as
[\begin{matrix}
a&a+1\a+7&a+8\end{matrix}. ]
The difference between the diagonal products is
[(a+7)(a+1)-a(a+8)
=a^2+8a+7-(a^2+8a)=\boxed{7}.]
Thus, the two diagonal products always differ by 7,
regardless of the starting date.
Question 5. Verify the statements.
(i) ((k+1)(k+2)-(k+3)) is always 2
False. It simplifies to
[ k^2+3k+2-k-3=k^2+2k-1, ]
which is not always 2.
(ii) ((2q+1)(2q-3)) is a multiple of 4
False. Both factors are odd, so their
product is odd, not divisible by 4.
(iii) Squares of even numbers are multiples of 4, and squares
of odd numbers are 1 more than a multiple of 8
True. An even number is (2n), whose square
is (4n^2). An odd number is (2n+1), and
[ (2n+1)^2=4n(n+1)+1. ]
Since one of (n,n+1) is even, (4n(n+1)) is divisible by 8.
(iv) ((6n+2)^2-(4n+3)^2) is 5 less than a square number
False as stated. The expression is
[ 36n^2+24n+4-(16n^2+24n+9)=20n^2-5. ]
It is 5 less than (20n^2), but (20n^2) is not generally a
square number.
Question 6. Remainders modulo 7
Let the numbers be (7a+3) and (7b+5).
Their sum has remainder
[ 3+5=8\equiv\boxed{1}\pmod7. ]
Their difference has remainder
[ 3-5=-2\equiv\boxed{5}\pmod7, ]
if the first number is subtracted from the second; if the
second is subtracted from the first, the remainder is (3-5\equiv5). For the
order used in the chapter’s answer, (n_1-n_2), the remainder is 5.
Their product has remainder
[ 3\times5=15\equiv\boxed{1}\pmod7. ]
Question 7. Three consecutive numbers
For (n-1,n,n+1),
[ n^2-(n-1)(n+1)=n^2-(n^2-1)=\boxed{1}. ]
The result is always 1.
Question 8. Add two numbers and multiply by half their sum
For numbers (a,b), the required expression is
[(a+b)\cdot\frac{a+b}{2}
=\boxed{\frac{(a+b)^2}{2}}.]
Thus, the result is half the square of their sum.
Question 9. Compare the products.
[ 14\times26=(16-2)(24+2)=16\times24-20, ]
so
[ \boxed{16\times24>14\times26}. ]
Similarly,
[ 25\times75=(26-1)(74+1)=26\times74-49, ]
so
[ \boxed{26\times74>25\times75}. ]
12. Park-area problem
The outer rectangular park has dimensions
[ 2g+4w\quad\text{and}\quad g+2w. ]
Its area is
[ (2g+4w)(g+2w)=2g^2+8gw+8w^2. ]
The two green square plots have total area (2g^2). Therefore,
the tiled walking-path area is
[(2g+4w)(g+2w)-2g^2
=\boxed{8w(g+w)}.]
13. Coin Conjoin
The minimum number of moves required to invert the 3-coin
triangle is 1, and for the 6-coin triangle it is 2. The 10-coin triangle can be
inverted in 3 moves.
For the next triangular arrangement of 15 coins, the minimum
is 4 moves. In general, a triangular arrangement
with (n) rows can be inverted in (n-1) moves by moving the appropriate corner
coins. Thus, the minimum number of moves for a triangle of (T_n=\frac{n(n+1)}2)
coins is
[ \boxed{n-1}. ]
0 Comments