Notes - Algebra
Category : 8th Class
Algebra
Learning Objectives
Linear Equations in one Variable
Example
1. Solve the equation: \[\frac{\mathbf{0}\mathbf{.5}\left( \mathbf{z-0}\mathbf{.4} \right)}{\mathbf{3}\mathbf{.5}}\mathbf{-}\frac{\mathbf{0}\mathbf{.6}\left( \mathbf{z-2}\mathbf{.7} \right)}{\mathbf{4}\mathbf{.2}}\mathbf{=z+6}\mathbf{.1}\]
(a) \[-\frac{202}{35}\] (b) \[\frac{202}{35}\]
(c) \[\frac{35}{202}\] (d) \[-\frac{35}{202}\]
(e) None of these
Answer: (a)
Explanation: \[\frac{5\left( z-0.4 \right)}{35}-\frac{6\left( z-2.7 \right)}{42}=z+6.1\Rightarrow \frac{30z-12-30z+81}{210}=z+6.1\Rightarrow \frac{69}{210}=z+6.1\Rightarrow z=-\frac{202}{35}\]
2. David cuts a bread into two equal pieces and cuts one half into smaller pieces of equal size. Each of the small pieces is twenty gram in weight. If he has seven pieces of the bread all with him, how heavy is the original cake.
(a) 120 gm (b) 180 gm
(c) 300 gm (d) 240 gm
(e) None of these
Answer: (d)
Explanation: There are total of seven pieces, so number of smaller pieces is six.
Weight of each smaller piece is 20 gm
Therefore, weight of six such pieces is \[6\times ~20=120\text{ }gm\]
Hence the total weight of original cake \[=2\times ~120=240\text{ }gm\]
Algebraic expressions and identities
For example, \[3{{x}^{2}},8xy,\,-6z,\,9x{{y}^{2}},\,2x,\,-3,\,22qrs,\,\,\] etc. are the monomials.
For example, \[4a+5b,\,3l-8m,\,2m+7,3-7{{x}^{2}}y,\,4{{x}^{2}}-{{z}^{2}},\] etc. are the binomials.
For example, \[a+b+c,\text{ }2x+3y-5z,{{x}^{2}}{{y}^{2}}z-{{x}^{3}}{{y}^{2}}z+1,\] etc. are the trinomials.
For example, \[6xy,\,\,8{{x}^{2}}yz-7,\,\,5x+9y+8z,\] etc. are the polynomials.
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
\[\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab\]
Example
1. The product of \[\left( \mathbf{-3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-x+7} \right)\] and \[\left( \mathbf{3-2x+}{{\mathbf{x}}^{\mathbf{2}}} \right)\] is____.
(a) \[-3{{x}^{4}}+5{{x}^{3}}-17x+21\]
(b) \[{{x}^{5}}+24{{x}^{4}}+5{{x}^{2}}+x+21\]
(c) \[8{{x}^{5}}+{{x}^{4}}-12+7x+1\]
(d) \[3{{x}^{5}}-4{{x}^{4}}+1{{x}^{3}}-5{{x}^{2}}-7x+2\]
(e) None of these
Answer: (a)
Explanation: \[\left( 2{{x}^{2}}-5{{x}^{2}}-x+7 \right)\left( 3-2x+{{x}^{2}} \right)\]
= \[-9{{x}^{2}}+6{{x}^{3}}-3{{x}^{4}}-3x+2{{x}^{2}}-{{x}^{3}}+21-14x+7{{x}^{2}}\]
= \[3{{x}^{4}}+5{{x}^{3}}-17x+21\]
Factorisation
\[{{x}^{2}}~+\left( a+b \right)x+ab\]
\[={{x}^{2}}~+\text{ }ax+bx+ab\]
\[=x\left( x+a \right)+b\left( x+a \right)\]
\[=\left( x+a \right)\left( x+b \right)\]
Here \[\left( x+a \right)\] and \[\left( x+b \right)\] are the required factors.
Example
1. Factorise the given polynomial: \[\mathbf{36}{{\mathbf{u}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{25}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{+25}{{\mathbf{w}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{12}}{\mathbf{5}}\mathbf{uv-2vw+60wu}\]
(a) \[{{\left( 6u-\frac{v}{5}-5w \right)}^{2}}\]
(b) \[{{\left( 6u+\frac{v}{5}-5w \right)}^{2}}\]
(c) \[{{\left( 6u-\frac{v}{5}+5w \right)}^{2}}\]
(d) \[{{\left( 6u+\frac{v}{5}+5w \right)}^{2}}\]
(e) None of these
Answer: (c)
2. Simplify the expression \[\mathbf{39}{{\mathbf{y}}^{\mathbf{2}}}\left( \mathbf{50}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{-98} \right)\mathbf{\div 26}{{\mathbf{y}}^{\mathbf{2}}}\left( \mathbf{5y+7} \right)\].
(a) \[3\left( 5y+8 \right)\]
(b) \[15y+21\]
(c) \[15y-21\]
(d) \[18y-21\]
(e) None of these
Answer: (c)
3. Simplify: \[\left( \mathbf{2x}+\mathbf{3y} \right)\left( \mathbf{2x}-\mathbf{3y} \right).\]
(a) \[{{x}^{2}}+3{{y}^{2}}\]
(b) \[2{{x}^{2}}+3{{y}^{2}}\]
(c) \[4{{x}^{2}}-9{{y}^{2}}\]
(d) \[{{x}^{2}}-9{{y}^{2}}\]
(e) None of these
Answer: (c)
Explanation: \[\left( 2x+3y \right)\left( 2x-3y \right)\left( 2x-3y \right)={{\left( 2x \right)}^{2}}-{{\left( 3y \right)}^{2}}=4{{x}^{2}}-9\,{{y}^{2}}\]
Exponents and Powers
(a) \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
(b) \[{{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}\]
(c) \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
(d) \[{{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}\]
(e) \[{{a}^{0}}=1\]
(f) \[\frac{{{a}^{m}}}{{{b}^{m}}}={{\left( \frac{a}{b} \right)}^{m}}\]
Example
1. Find the value of x such that \[{{\left( \frac{\mathbf{64}}{\mathbf{125}} \right)}^{\mathbf{2}}}{{\left( \frac{\mathbf{4}}{\mathbf{5}} \right)}^{\mathbf{4}}}{{\left( \frac{\mathbf{16}}{\mathbf{25}} \right)}^{\mathbf{2x+1}}}\mathbf{=}{{\left( \frac{\mathbf{256}}{\mathbf{625}} \right)}^{\mathbf{3x}}}\mathbf{.}\]
(a) \[\frac{3}{2}\] (b) \[\frac{2}{3}\]
(c) \[\frac{1}{3}\] (d) \[\frac{1}{2}\]
(e) None of these
Answer: (a)
Direct and Indirect Proportion
Example
1. If women or 3 men earn Rs. 960 in a day, then the earning of 11 women and 7 men in a day will be:
(a) Rs. 4880 (b) Rs. 2200
(c) Rs. 1860 (d) Rs. 1480
(e) None of these
Answer: (a)
Explanation: One day earning of 4 women or 3 men = Rs. 960
Therefore, one day earning of 1 women or 1 men =\[\frac{960}{4}\] or Rs. \[\frac{960}{3}\]
One day earning of 11 women or 7 men = Rs. \[\frac{960}{4}\times 11\] or Rs. \[\frac{960}{3}\times 7\]
= Rs. 2640 or Rs. 2240
Therefore, total earning of 11 women and 7 men for one day is Rs 4880.
2. A university has its own hostel for its students. 8t provides fooding and lodging to the students. Due to festive season some of the students are left for their home and 100 students stays in the hostel. There is a food provision for 20 days for these students. How long the food will last if 25 more students decide to stay back in the hostel?
(a) 12 days (b) 13 days
(c) 16 days (d) 14 days
(e) None of these
Answer: (c)
Explanation: Initially the number of students = 100.
Provision for food = 20 days.
Finally number of students = 125.
100x20
Number of days food will last \[=\frac{100\times 20}{125}~=16\text{ }days.\]
Comparing Quantities
Discount = Marked Price - Sale Price (When M.P. and S.P. is given)
Discount = Discount % of Marked Price (When discount percentage is given)
Cost Price (C.P.) = Buying price + Overhead expenses
Sales tax = Tax% of Bill Amount
Profit = S.P. - C.P.
Loss = C.P. - S.P
SI = PRT/100 where P is principal, R is rate of interest and T is time,
\[C.I.=A-P\]
Here, \[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\]
(P = principal, r = rate of interest and n = time period)
Example
1. A landlord has a large piece of agricultural land which he wants to sell. James wants to buy the land and buys it for Rs. 400000. After some time he was in need of money and wanted to sell that land piece. He sells one third of the land at the loss of 20% and two fifth at the gain of 25%. At what price he must sell the remaining land so that he can make the overall profit of 10% on the whole transaction?
(a) \[Rs\frac{80000}{3}\]
(b) \[Rs\frac{400000}{3}\]
(c) \[Rs\frac{320000}{3}\]
(d) \[Rs\frac{920000}{3}\]
(e) None of these
Answer: (b)
Explanation: Cost price of entire land = Rs. 400000
Proposed profit on the whole land = 10% of 400000 = Rs. 40000
Proposed selling price of whole land = Rs. 440000
Cost price of one third land = Rs. \[\frac{400000}{3}\]
Loss on that land = 20% of \[Rs\frac{400000}{3}=Rs\frac{80000}{3}\]
Selling price of that land = \[\frac{400000}{3}-\frac{80000}{3}=Rs\frac{320000}{3}\]
Cost price of two fifth of the land = \[\frac{2}{5}\times 400000=Rs160000\]
Gain on the transaction \[=25%\]of \[160000=Rs.\text{ }40000\]
Selling price of that land \[=Rs.\left( 160000+40000 \right)=Rs.\text{ }200000\]
Total selling price of the land = \[\frac{320000}{3}+200000=Rs\frac{920000}{3}\]
Selling price of the remaining land = Rs. \[(440000-\frac{920000}{3})=Rs\frac{400000}{3}\]
2. If the cost price of 20 greeting cards is equal to the selling price of 16 greeting cards, find the gain or loss percent.
(a) 20% (b) 25%
(c) 30% (d) 40%
(e) None of these
Answer: (b)
Explanation: Let cost price of each card be\[Rs\text{ }x.\].
Then C. P. of 16 greeting cards \[=Rs.\text{ }16x\]
S. P. of 16 greeting cards = C.P. of 20 cards \[=Rs.\text{ }20x\]
Therefore, gain \[=S.P.-C.P.=\text{ }Rs\left( 20-16 \right)x=Rs.\text{ }4x\]
% gain = \[\frac{4x}{16x}\times 100=25%\]
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