Simple Interest
In the case of simple interest, we calculate interest paid by borrower over certain period of time without changing its principle.
Calculate the simple interest on a sum of $1200 at the rate of 5 % per annum or two years.
Solution:
The following steps are to be followed
Step 1: The interest for first year when
P =$ 1200, R = 5 % and T = 1 year
S.I. \[=\frac{P\times R\times T}{100}=\frac{1200\times 5\times 1}{100}=\$60\]
Therefore, interest for the first year is $ 60.
Step 2: Interest for the second year
P = $ 1200, R = 5 % and T = 1 year
S.I. \[=\frac{P\times R\times T}{100}=\frac{1200\times 5\times 1}{100}=\$60\]
Therefore, interest for the second year is $60.
Total interest paid by borrower after 2 years =$ 60 + $ 60 = $ 120. 
Compound Interest
In this case we observe that the interest paid by borrower is same for every year.
There are different methods to calculate the interest in case of bank transaction loan etc. In these methods, interest is calculated either quarterly, half yearly or yearly, what so ever may be the case. i.e. The agreement between the lender and borrower on the principle. The amount after that first fixed interval of time will be principle for second interval of time, the amount after second interval of time will be the principle for the third interval and so on. The interest paid by borrower under the above conditions is called compound interest.
Abbreviation used in Interest
Find the compound interest on a sum of $ 1500 for two years at the rate opf 5% per annum compounded annually.
Solution:
The following steps are followed
Step 1: Principle for the first year = $ 1500, Time (T) =1 year R = 5 % p.a.
\[S.I.=\frac{P\times R\times T}{100}=\frac{1500\times 5\times 1}{100}\$75\]
Amount after 1st year \[=\$1500+\$75=\$1575\]
Step 2: Principle for second year = $ 1575, Time = 1 year, R = 5 %
\[S.I.=\frac{P\times R\times T}{100}=\frac{1575\times 5\times 1}{100}\$78.75\]
The amount after second year \[=\$1575+\$\] \[78.75=\$1653.75\]
Therefore, the compound interest \[=A-P\]
\[=\$1653.75-\$1500=\$153.75\]
Form the above discussion we conclude that
Find the compound interest on $15,000 for 2 year at 8 % per annum.
(a) Rs 2496
(b) Rs 2393
(c) Rs 2293
(d) Rs 2593
(e) None of these
Answer: (a)
Explanation
Step 1: Principle for 1st year =$ 15,000, time (T) = 1 years, R = 8 % p.a.
\[S.I.=\frac{P\times R\times T}{100}=\frac{15000\times 8\times 1}{100}\$1200\]
The amount after 1st year =$15,000 +$ 1,200 = $16,200
Step 2: Principle for the second year = Amount after 1st year =$ 16,200
Time (T) = 1 year, R = 8 % P.a.
\[S.I.=\frac{P\times R\times T}{100}=\frac{15000\times 8\times 1}{100}\$1296\]
Amount after 2nd year =$16200 +$ 1296 = $17496
Compound interest (C.I.) =Amount - Principle
=$ 17496-$15000 =$ 2496
more... 
Introduction
The word triangle is derived from Greek word, triangle means three and hence it to a shape consisting three internal angles. Obviously the shape consists of sides. Hence, a triangle can be defined as a polygon having three sides. 
Basic Concepts of Triangles
The general shape of a triangle is shown below:
The vertices of a triangle are denoted by the Capital letters of English alphabets.
In the above figure \[\Delta ABC,\]the sides are AB, BC and CA.
Altitude
A Perpendicular drawn from a vertex to the opposite side is called the altitude of the triangle and denoted as h.
Some Basic Facts Related to Triangle
Proof: In \[\Delta ABC,a+b+c=180{}^\circ \]
but,\[c+x=180{}^\circ \] (Linear pair) \[c={{180}^{0}}-x\]
Putting "c" in the above equation, we get
\[a+b+180{}^\circ -x=180{}^\circ \Rightarrow a+b=x\]
Therefore,\[~\Delta ACD=\Delta A+\Delta B\]
Types of Triangle
Classification based on angles
Equilateral Triangle
A triangle in which all sides are equal is known as equilateral triangle.
Isosceles Triangle
A triangle in which any two sides are equal is said to be and isosceles triangle.
The angles opposite to the equal sides are equal.
In the triangle given below, sides AB and AC are equal as well as \[\angle B\]and \[\angle C\]are also equal.
Scalene Triangle
A triangle in which all side are unequal is said to be scalene triangle.
In scalene triangle all angles are different.
In the above given triangle all the sides of triangle denoted by a, b and c are unequal and angles x, y and z are also unequal.
Pythagoras Theorem more... 
Factorization of the Polynomials
Let us recall that an algebraic expression that is expressed as the product of two or more expressions & each of these is a factor of the given algebraic expression. The process of writing a given algebraic expression as the product of two or more factors is called factorization.
Some of the common methods of factorization of the algebraic expressions are as follows:
Factorization of the Algebraic Expression \[a{{x}^{2}}+bx+c\]
Step 1: Find the product of constant term and coefficient of \[{{x}^{2}}\] i.e. ac.
Step 2: Find factors of "ac".
Step 3: Select the factors of 'ac' in such a way that addition or subtraction of factors must be equal to the coefficient of \[x\] i. e "b".
Step 4: If the product 'ac' is positive then both the factors are either positive or negative.
Step 5: If the product 'ac' is negative then the two factors of 'ac' will have different sign.
Find the factors of\[~{{x}^{2}}+9x+18\]
Solution:
Step 1: The product of constant term and coefficient of \[{{x}^{2}}\] is 18
Step 2: The product is positive. Therefore, both the factors of 18 will be either be positive or negative.
Step 3: But the coefficient of \[x\] is positive therefore, both the factors of 18 will be positive.
Step 4: Factors of 18 are \[1\times 18,2\times 9\] and \[3\times 6.\]One pair whose sum is 9 should betaken.
Step 5: The required number are 3 & 6.
\[\therefore {{x}^{2}}+9x+18={{x}^{2}}+(6+3)x+18\]
\[=(x+6)(x+3)\][By using identity \[(x+a)(x+b)=\]\[{{x}^{2}}+(a+b)x+ab]\]
or \[{{x}^{2}}+9x+18={{x}^{2}}+6x+3x\text{ }+18\] \[=x(x+6)+3(x+6)=(x+6)(x+3).\]
Taking out x from first two terms and 3 common from last two terms.
Factorize: \[{{x}^{2}}-11x+30.\]
(a)\[(x-6)(x-5)\]
(b)\[(x-6)(x+5)\]
(c)\[~(7{{x}^{2}}-6)(8x-9)\]
(d) \[\left( 9{{x}^{2}}-62 \right)\left( 2x+3 \right)\]
(e) None of these
Answer: (a)
Explanation
Let us find two numbers whose product is 30 & sum is \[(-11)\] Since, the product is positive, therefore, either both the numbers will be positive or negative.
But the sum is negative hence, both the numbers will also be negative.
Required factors of 30 are \[(-6)\] and \[(-5).\]
\[\therefore {{x}^{2}}-11x+30={{x}^{2}}+\left\{ \text{(}-6\text{)}+(-5) \right\}\text{ }x+30\]
\[={{x}^{2}}-6x-5x+30=x(x-6)-5(x-6)=(x-6)(x-5)\]
Factorize: \[{{x}^{2}}+x-12\]
(a)\[(x+5)(x-3)\]
(b) \[(x-9)(x-52)\]
(c)\[(x+4)(x-3)\]
(d) \[(7x+{{5}^{2}})(6x-7)\]
(e) None of these
Answer: (c)
Explanation
To find two numbers whose product is (- 12) & sum is 1.
Since the product is negative, one number will be positive & the other number will be negative.
Since the sum is positive, the numerically greater of two numbers will be positive.
So, the required factors more... 
Algebraic identities
Abatement of equality which holds, for all values of the variable is called algebraic identities.
Now we recall some important algebraic identities:
Simplify: \[{{(2p+3q-+4r)}^{2}}+{{(2p-3q-4r)}^{2}}\]
(a) \[2(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-16rp)\]
(b) \[-2(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-16rp)\]
(c) \[2(-4{{p}^{2}}+9{{q}^{2}}-16{{r}^{2}}+16rp)\]
(d)\[~2(5{{p}^{2}}+9{{q}^{2}}-16{{r}^{2}}-98rp)\]
(e) None of these
Answer: (a)
Explanation
Let us first solve,
\[{{\left[ 2p+3q+(-4r) \right]}^{2}}={{(2p)}^{2}}+{{(3q)}^{2}}\]\[+{{(-4r)}^{2}}+2(2p)(3q)+2(3q)(-4r)\] \[+2(-4r)(2p)=4{{p}^{2}}+4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}\]\[+12pq-24qr-16rp..........(i)\]
Now solve, \[~{{(2p-3q-4r)}^{2}}={{\left[ 2p+(-3q)+(-4r) \right]}^{2}}\]
\[={{(2p)}^{2}}+{{(-3q)}^{2}}+{{(-4r)}^{2}}+2(2p)(-3q)+2(-3q)\]\[(-4r)+2(-4r)(2p)\]
\[=4{{p}^{2}}+9{{q}^{2}}+16r2-12pq+24qr-16rp\text{ }.......\left( ii \right)\]
Adding, (i) & (ii) we get,
\[{{(2p+3a-4r)}^{2}}+{{(2p-3q-4r)}^{2}}\]
\[=4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}+12pq-24qr-16rp\]+ \[(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-12pq+24qr-16rp)\]
\[=4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}+12pq-24qr-16rp+4{{p}^{2}}+9{{q}^{2}}\] \[+16{{r}^{2}}-12pq+24qr-16rp\]
\[=8{{p}^{2}}+18{{q}^{2}}+32{{r}^{2}}-32rp\]
\[=2(4{{p}^{2}}+9{{q}^{2+}}16r2-16rp)\]
The expanded form of \[{{(2x+3y-5z)}^{2}}\] is:
(a) \[4{{x}^{2}}+9{{y}^{2}}+25{{z}^{2}}+12xy-30yz-20zx\]
(b) \[5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36{{y}^{6}}+21x{{y}^{2}}\]
(c) \[8{{a}^{2}}+{{a}^{3}}-{{c}^{2}}+7{{x}^{2}}+2{{c}^{2}}+9{{c}^{2}}y\]
(d) \[1{{z}^{2}}-{{z}^{4}}-{{8}^{c}}-{{75}^{2}}+2{{c}^{2}}-98{{c}^{2}}\]
(e) None of these
Answer: (a)
Explanation
\[{{(2x+3y-5z)}^{2}}={{(2x)}^{2}}+{{(3y)}^{2}}+{{(-5z)}^{2}}\]\[+2(2x)(3y)+2(3y)(-5z)+2(5z)(2x)\text{ }\]
Expand: \[{{(3a-b+4c)}^{2}}\]
(a) \[~5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36y+21xy\]
(b) \[9{{a}^{2}}+{{b}^{2}}+16{{c}^{2}}-6ab-8bc+24ac\]
(c) \[2{{x}^{2}}+7{{y}^{2}}+2{{z}^{2}}+xy+3yz+2zx\]
(d) \[15a2+22{{b}^{2}}+2{{c}^{4}}+ab+4bc+102ca\]
(e) \[5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36y+21xy\]
Answer: (b)
Explanation
\[{{(3a-b+4c)}^{2}}={{(3a(-b)+4c)}^{2}}={{(3a)}^{2}}\]\[+{{(-b)}^{2}}+{{(4c)}^{2}}+2(3a)(-b)+2(-b)\] \[(4c)+2(4c)(3a)=9{{a}^{2}}+{{b}^{2}}+16{{c}^{2}}-6ab-8bc+24ac\]
Find the value of \[{{(3a+5b)}^{3}}.\]
(a) \[27{{a}^{3}}+125{{b}^{3}}+135{{a}^{2}}b+225a{{b}^{2}}\]
(b) \[27{{a}^{3}}-125{{b}^{3}}-135{{a}^{2}}b+2225a{{b}^{2}}\]
(c) \[27{{a}^{2}}+155{{b}^{2}}-135{{a}^{2}}b-225{{a}^{2}}b\]
(d) \[29{{a}^{2}}-156{{b}^{2}}-156{{a}^{2}}b-225{{a}^{4}}c\]
(e) None of these
Answer: (a)
Find the Cube of \[x-2y.\]
(a)\[~{{x}^{3}}+8{{y}^{2}}+6{{x}^{2}}y-12x{{y}^{3}}\]
(b) \[{{x}^{3}}-8{{y}^{3}}-6{{x}^{2}}y+12x{{y}^{2}}\]v
(c)\[{{x}^{2}}+87y+7xy-7xy\]
(d) \[7x{{y}^{3}}-6x-74{{x}^{2}}y+2xy\]
(e) None of these
Answer: (b)
Evaluate: \[{{\left( -6p+\frac{1}{3}q-r \right)}^{2}}\]
(a) \[36{{p}^{2}}\frac{9}{1}{{p}^{2}}+{{5}^{2}}-4pq-\frac{2}{3}gr+12rp\]
(b) \[36{{p}^{3}}\frac{9}{1}{{q}^{2}}+{{p}^{2}}+4qp-\frac{3}{2}qr+13rp\]
(c)\[~36{{p}^{2}}+\frac{1}{9}{{q}^{2}}+{{r}^{2}}-4qp-\frac{2}{3}qr+12rp\]
(d) \[36{{x}^{2}}-\frac{99}{8}{{s}^{2}}-{{s}^{2}}-4pq-\frac{2}{3}qr+14rp\]
(e) None of these
Answer: (c) 
Operations on Algebraic Expression
Addition and subtraction of an algebraic expressions mean addition and subtraction of like terms.
Addition of an Algebraic Expression
For addition of algebraic expression we may follow any one of the following methods:
Row Method
In this method, write all expression in a single row then arrange the terms to collect all like terms together and add it.
Column Method
In this method, arrange each expression in such a way that each like term is placed one below to other in a column.
Add \[3x+2y+3z\]and \[2x-3y+4z\]
Solution:
\[(3x+2y+3z)+(2x-3y+4z)\]
\[=(3x+2x)+(2y-3y)+(3z+4z)=5x-y+7z\]
Column method
\[3x+2y+3z\]
\[\frac{+2x-3y+4z}{5x-y+7z}\]
Subtraction of an Algebraic Expression
For subtraction also you may follow any one of the following method
Row Method
We arrange algebraic expression in a row and change the sign (from + to ?, ? or ? to +) of all terms which is to be subtracted. The two expression then added as above.
Column Method
Arrange two expression in such a way that like terms are placed one below the other and change the sign (from + to ?,or ? to +) of algebraic expression which is to be subtracted.
Subtract \[5{{a}^{2}}{{b}^{2}}+6{{a}^{2}}{{b}^{2}}+4\] from \[7{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}+5\]
Solution:
By row method,
\[(7{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}+5)-(5{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}+4)\]
\[=7{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}+5-5{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}-4\]
Arrange like terms and add, we get \[=2{{a}^{2}}b-12{{a}^{2}}{{b}^{2}}+1\]
By column method,
\[7{{a}^{2}}b-6{{a}^{2}}{{b}^{2}}+5\] \[\frac{{{\underline{5a}}^{2}}b+6\underline{{{a}^{2}}}{{b}^{2}}+\underline{4}}{2{{a}^{2}}b-12{{a}^{2}}{{b}^{2}}+1}\] 
The value of an Algebraic Expression
Step 1: If possible simplify the given algebraic expression.
Step 2: Replace variable with given numerical value.
Step 3: Simplify it.
Find the value of \[\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx},\] lf \[x=1,y=2\] and \[z=-1.\]
(a) 24
(b) 14
(c) 7
(d) 2
(e) None of these
Answer: (d)
Explanation
\[\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx}=\frac{(x+y+z+)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}\]
\[=(x+y+z).\] Now putting the values of \[x,\text{ }y\]and \[z\] we get = 2
Find the value of \[4\text{ }xy\text{(}x-y\text{)}-6{{x}^{2}}\text{(}y-{{y}^{2}}\text{)}-3{{y}^{2}}(2{{x}^{2}}-x)+2xy(x-y)\] for \[x=5\]and \[y=13.\]
(a) - 195
(b) 2535
(c) - 2535
(d) 215
(e) None of these
Answer: (c)
Explanation
\[4xy(x-y)-6{{x}^{2}}(y-{{y}^{2}})-3{{y}^{2}}(2{{x}^{2}}-x)+2xy(x-y)\]\[4{{x}^{2}}y-4x{{y}^{2}}-6{{x}^{2}}y+6{{x}^{2}}{{y}^{2}}-6{{y}^{2}}{{x}^{2}}+3x{{y}^{2}}+2{{x}^{2}}y-2x{{y}^{2}}\]
After simplification, we get \[-\text{ }3x{{y}^{2}}=-3\times 5\times 13\times 13=-2535\]
The value of A, B, C, D, E, F, G, H and I for which \[(ABC)\times (DEF)=GHI\]
(a) (6, 5, 7, 8, 2, 6, 1, 8, 4, 0)
(b) (7, 9, 5, 2, 6, 4, 8, 4, 0)
(c) (4, 6, 7, 8, 5, 7, 0, 8, 9)
(d) (1, 0, 3, 7, 8, 4, 2, 1, 0)
(e) None of these
Answer: (d)
Explanation
If we put the value of A, B, C, D, E, F, G, H and I from option D in the left hand side and right hand side of given expression, then it becomes zero.
Find the value of \[3{{x}^{3}}y+4{{x}^{y}}+2{{y}^{x}},\] if \[x=2\]\[x=2\]and \[y=-2.\]
(a) - 39
(b) 39
(c) 13
(d) Cannot be determined
(e) None of these
Answer: (a)
Find the value of \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca,\]if \[a=2b=-2\]and \[c=1.\]
(a) 12
(b) 1
(c) 13
(d) 7
(e) None of these
Answer: (d)
Which one of the following options is the correct value or the expression \[\frac{axy+byz+cxy}{ax+by+cz+1},\] if \[a=1,b=2,c=-1,x=-1,y=2,z=3?\]
(a) 12
(b) 10
(c) 13
(d) 11
(e) None of these
Answer: (a) 
Introduction
We have discussed about the addition, subtraction, multiplication and division of the arithmetic expression into previous chapter. In this chapter, we will discuss about the operation on algebraic expression.
Algebraic Expression
It is the combination of constants and variables along with the fundamental operations \[(+,\,-,\,\,\times ,\,\,\div )\]
Terms: It is the part of an algebraic expression which is separated by the sign of addition and subtraction.
\[5{{x}^{4}}{{y}^{2}},35{{x}^{4}}{{y}^{2}}-13{{x}^{2}}y,6xy,-3\] is an algebraic expression having \[8{{x}^{3}}{{y}^{2}},-4{{x}^{2}}y,6xy,-3\] as its term.
Like and Unlike Terms
The terms having similar variable(s) are called like terms otherwise it is unlike. In an algebraic expression \[5{{x}^{4}}{{y}^{2}},-13{{x}^{2}}y+6xy-3-35{{x}^{4}}{{y}^{2}};5{{x}^{4}}{{y}^{2}},35{{x}^{4}}{{y}^{2}}\]are like terms and are unlike terms.
Types of Algebraic Expression
Monomials
An algebraic expression which contains one term is called monomial. i.e. \[3x,4x,-xy,{{b}^{2}}a\] etc. are monomials.
Binomials
An algebraic expression which contains two terms is called binomial. i.e. \[x+y,a-b,{{b}^{2}}a+{{a}^{2}}b\] etc. are binomials.
Trinomials
An algebraic expression which contains three terms is called trinomial.
i.e. \[\left( x+y+z \right),\text{ }\left( a+b+c \right),\text{ }\left( {{a}^{2}}+{{b}^{2}}+{{a}^{2}}{{b}^{2}} \right)\]etc. are trinomials.
i.e. \[\left( a-b-c+d \right),\left( x+y+z-c \right)\]etc. are quadrimonials.
Polynomials
An algebraic expression in which the variables have only non-negative integral power is called polynomial.
Degree of Polynomial
In case of one variable, the highest power of variable is the degree of polynomial e.g \[5{{x}^{4}}-4{{x}^{2}}+6x-3\] is a polynomial of degree 4. If polynomial is in the more than one variable then the highest sum of degree of the variables in each term is the degree of polynomial. e.g. \[8{{x}^{3}}{{y}^{2}}-4{{x}^{2}}y+6xy-3\]is a polynomial of degree 5.
Linear Polynomial
A polynomial of one degree is called linear polynomial.
\[p\left( x \right)=6x-3\] is a linear polynomial.
Quadratic Polynomial
A polynomial of two degree is called quadratic polynomial.
\[q(x)=4x2+3x+45\] is a quadratic polynomial. 
Negative Rational Number as Exponent
Let us consider 'a' be a positive rational number and \[x\left( i.e{{.}^{\frac{-m}{n}}} \right)\] be a negative rational exponent then it is defined as \[{{a}^{x}}\left( i.e.{{a}^{\frac{-m}{n}}} \right).\]
Find the value of \[{{4}^{\frac{-3}{2}}}\] and \[{{\left( 512 \right)}^{\frac{-2}{9}}}\]
Solution:
\[{{4}^{\frac{-3}{2}}}=\frac{1}{^{{{4}^{\frac{3}{2}}}}}=\frac{1}{^{\left( {{4}^{3}} \right)\frac{1}{2}}}=\frac{1}{{{\left( 64 \right)}^{\frac{1}{2}}}}=\frac{1}{8}\]and \[\frac{1}{{{\left( 512 \right)}^{\frac{-2}{9}}}}=\frac{1}{{{\left( {{512}^{2}} \right)}^{\frac{-2}{9}}}}\]\[=\frac{2}{\left( {{\left\{ {{\left( {{2}^{9}} \right)}^{2}} \right\}}^{\frac{1}{9}}} \right)}=\frac{1}{^{_{2}\cancel{9}\times 2\times \frac{1}{\cancel{9}}}}=\frac{1}{4}\]
Evaluate : \[{{\left( \frac{4}{9} \right)}^{\frac{3}{2}}}\times {{\left( \frac{4}{9} \right)}^{\frac{1}{2}}}\]
(a) \[{{\left( \frac{4}{9} \right)}^{2}}\]
(b) \[{{\left( \frac{81}{16} \right)}^{4}}\]
(c) \[{{\left( \frac{4}{9} \right)}^{4}}\]
(d) \[\frac{4}{27}\]
(e) None of these
Answer: (a)
Solution:
\[{{\left( \frac{4}{9} \right)}^{\frac{3}{2}}}\times {{\left( \frac{4}{9} \right)}^{\frac{1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{3}{2}+\frac{1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{3+1}{2}}}={{\left( \frac{4}{9} \right)}^{\frac{\cancel{4}}{\cancel{2}}}}={{\left( \frac{4}{9} \right)}^{2}}\]
Evaluate: \[{{(27)}^{\frac{6}{5}}}\div {{(27)}^{\frac{1}{5}}}\]
(a) \[5\times {{3}^{3}}\]
(b) \[1\times {{3}^{3}}\]
(c) \[2\times {{3}^{3}}\]
(d) \[7\times {{3}^{3}}\]
(e) None of these
Answer: (b)
Solution:
\[{{(27)}^{\frac{6}{5}}}\div {{(27)}^{\frac{1}{5}}}={{(27)}^{\frac{6}{5}-\frac{1}{5}}}={{(27)}^{\frac{5}{5}}}=27={{3}^{3}}\]
Evaluate: \[\frac{{{\left( 27 \right)}^{\frac{-2}{3}}}\times {{\left( 81 \right)}^{\frac{5}{4}}}}{{{\left( \frac{1}{3} \right)}^{-3}}}\]
(a) 1
(b) 2
(c) 3
(d) 4
(e) None of these
Answer: (a)
Solution:
\[\frac{{{\left( 27 \right)}^{\frac{-2}{3}}}\times {{\left( 81 \right)}^{\frac{5}{4}}}}{{{\left( \frac{1}{3} \right)}^{-3}}}=\frac{{{3}^{\cancel{3}\times \left( \frac{2}{\cancel{3}} \right)}}\times {{3}^{\cancel{4}\times \frac{5}{\cancel{4}}}}}{{{3}^{3}}}=\frac{{{3}^{-2}}\times {{3}^{5}}}{{{3}^{3}}}=\frac{{{3}^{-2+5}}}{{{3}^{3}}}=\frac{{{3}^{3}}}{{{3}^{3}}}=1\]
Evaluate: \[{{\left[ {{\left( \frac{36}{25} \right)}^{\frac{3}{2}}} \right]}^{\frac{5}{3}}}\]
(a) \[\frac{7776}{3125}\]
(b) 1
(c) \[\frac{75}{31}\]
(d) 2
(e) None of these
Answer: (a)
Solution:
\[{{\left( \frac{36}{25} \right)}^{\frac{\cancel{3}}{2}\times \frac{5}{\cancel{3}}}}{{\left( \frac{36}{25} \right)}^{\frac{5}{2}}}={{\left( \frac{6}{2} \right)}^{\cancel{2}\times \frac{5}{\cancel{2}}}}{{\left( \frac{6}{2} \right)}^{5}}=\frac{7776}{3125}\]
Evaluate: \[{{\left[ {{\left\{ {{\left( 625 \right)}^{-\frac{1}{2}}} \right\}}^{\frac{1}{4}}} \right]}^{2}}\]
(a) 2
(b) 3
(c) 4
(d) 5
(e) None of these
Answer: (d)
Solution
\[{{\left( 625 \right)}^{-\frac{1}{\cancel{2}}\times \left( -\frac{1}{4} \right)\times \cancel{2}}}={{\left( 625 \right)}^{\frac{1}{4}}}={{\left( 5 \right)}^{\cancel{4}\times \frac{1}{4}=5}}\]
Evaluate: \[{{64}^{\frac{2}{3}}}\times {{27}^{\frac{2}{3}}}\]
(a) 144
(b) 12
(c) 4
(d) 3
(e) None of these
Answer: (a)
Solution:
\[{{\left( 64\times 27 \right)}^{\frac{2}{3}}}={{\left( {{4}^{3}}\times {{3}^{3}} \right)}^{\frac{2}{3}}}={{\left( 4\times 3 \right)}^{\cancel{3}\times \frac{2}{\cancel{3}}}}=16\times 9=144\]
Evaluate: \[{{\left[ {{\left\{ {{\left( \frac{1}{x} \right)}^{-12}} \right\}}^{\frac{1}{4}}} \right]}^{-\frac{2}{3}}}\]
(a) \[\frac{1}{x}\]
(b) \[\frac{1}{{{x}^{2}}}\]
(c) \[\frac{1}{{{x}^{3}}}\]
(d)\[\frac{1}{{{x}^{4}}}\]
(e) None of these
Answer: (b)
Evaluate: \[{{\left[ {{\left( 729 \right)}^{\frac{-5}{3}}} \right]}^{-\frac{1}{2}}}\]
(a) 243
(b) 81
(c) 27
(d) 9
(e) None of these
Answer: (a)
Solution:
\[{{\left( 729 \right)}^{\frac{-5}{3}\times \left( \frac{-1}{2} \right)}}={{\left( 9 \right)}^{^{\cancel{3}\times \left( \frac{-5}{\cancel{3}} \right)\left( \frac{-1}{2} \right)}}}={{9}^{\frac{5}{2}}}={{3}^{\cancel{2}{{\times }^{\frac{5}{\cancel{2}}}}}}={{3}^{5}}=243\]
Find the value of\[x\], if \[{{\left( \sqrt{6} \right)}^{x-2}}=1.\]
(a) 1
(b) 2
(c) 3
(d) 4
(e) None of these
Answer: (b)
Solution:
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