I. Binary Numbers: A Discussion In order to understand what a binary number is we should first understand what a decimal number is. Decimal Numbers (i) In our everyday life we represent numbers by using ten

digits (which are 0, 1,2, 3, 4, 5,6, 7,8,9) and therefore these numbers are called decimal numbers (deci means ten in Latin).

(ii) Any number can be represented using these ten digits. (iii) Consider for example a sequence of digits 197. Here we

have three digits 1,9 and 7 written in that order. All of us know that when 1, 9 and 7 are written in that order this sequence of digits is equal to the number: "One hun-dred and ninety-seven". How do we arrive at this value for the given sequence? We do it in the following man-ners:

197= l x l O 2 +9x10' +7x10 =100+90 + 7=197 Note: io = 1. In fact, the value is always equal to 1 i f any number is raised to the power zero.

This means that to get the value of any decimal num-ber we follow the following rules:

The 1st digit from the right is multiplied by io

The 2nd digit from right is multiplied by J O 1

The 3rd digit from right is multiplied by i o 2 55 51 55 55

51 51 55 55

The nth digit from right is multiplied by io"" 1 And finally, all these are added.

Ex. 1: What is the value i f 3, 5, 7 and 9 are written in this order: 5793.

Soln: 5793= 5 x l 0 3 + 7 x l 0 2 +9x10' +3x10 = 5 x 1000 + 7 x 100 + 9x 10 + 3 x 1=5793

(Five thousand seven hundred and ninety-three). (iv) We get the value of numbers in these cases by multiply-

ing every digit by power of 10. Here this 10 is called the

Miscellaneous

base or the radix. Therefore, under a decimal system our base is 10 and we use a total of ten digits to repre-sent any number.

Binary Numbers (i) Just as we use ten digits to represent a decimal number,

we may as well use only two digits (which are: 0, 1) to represent any number. This will be called a binary sys-tem as bi means two in Latin.

(ii) Any number can be represented using these two digits: 0 and 1.

(iii) Consider, for example, a sequence of the digits: 1010. Here we have a sequence of digits: 1, 0, 1 and 0, in that order. What is the value of this number? We get the value in the following manner:

1010 = l x 2 3 + 0 x 2 2 +1x2 ' + 0 x 2 = 1 x 8 + 0 x 4 + l x 2 + 0 x 1 = 8 + 0 + 2 + 0=10(ten).

Therefore, 1010 in the binary system represents the num-ber: ten. (Which is represented as 10 in our usual decimal system.).

This means that to get the value of any binary number we follow the following rules:

The 1 st digit from right is multiplied by 2 (= 1). The 2nd digit from right is multiplied by 2' (= 2). The 3rd digit from right is multiplied by 2 2 (= 4).

u M

7 4 2 P R A C T I C E B O O K ON Q U I C K E R MATHS

decimal system we represent numbers by ten digits (0,1,2,3, 4,5,6,7, 8 and 9) and the value of the number is obtained by multiplying different digits of the sequence by powers of 10 and adding; while in case of binary system we represent numbers by two digits (0 and 1) and the value of the number is obtained by multiplying different digits of the sequence by powers of 2 and adding.

Converting binary numbers to decimal numbers We have already seen how to do it. A binary number is

converted to a decimal number by (1) multiplying the nth digit from right by 2"', where n = 1,

2,... (2) adding all these. Ex. 3: Convert the following binary numbers into decimal numbers: (a) 1010 (b) 1111 (c)100 (d) 10000 (e) 1110010

Soln: (a) 1010= l x 2 3 + 0 x 2 2 +1x2' + 0x2 = 8 + 0 + 2 + 0=10(Ten)

(b) l l l l = l x 2 3 + l x 2 2 + l x 2 I +1x2 = 8+4 + 2 + 1 = 15 (Fifteen)

(c) 100= l x 2 2 + 0 x 2 ' + 0 x 2 = 4 + 0 + 0 = 4(Four)

(d) 10000= l x 2 4 + 0 x 2 3 + 0 x 2 2 + 0 x 2 ' + 0 x 2 = 16 + 0 + 0 + 0 + 0= 16(Sixteen)

(e) 1110010 = l x 2 6 + l x 2 5 + l x 2 4 + 0 x 2 3 +

0 x 2 2 +1x2' + 0 x 2 = 64 + 32+ 16 + 0 + 0 + 2 + 0 = 114 (Hundred Fourteen)

Table 1: List of powers of 2

Power Value

2 1

2' 2 = 2

2 2 4 = 2 x 2

2 3 8 = 2 x 2 x 2

2 4 16 = 2 x 2 x 2 x 2

Power Value

2 5 32 = 2 x2 x2 x2 x 2

2 6 64 = 2 x2 x2 x2 x 2x 2

2 7 128 = 2 x 2... 7 times

2 s 256 = 2 x 2... 8 times

29 512 = 2 x 2... 9 times

For a quicker conversion of binary numbers to decimal numbers we must remember the above-mentioned table by heart. Thus, we can save time by directly writing the value to be multiplied (for example instead of writing 1 x 2 3 we can directly write 1 x 8). Now, since anything multiplied by 1 gives the same number and since anything multiplied by 0 gives zero, we can further save time by writing (i) only the value of the power of 2 wherever it has to be multiplied by 1, and (ii) zero, wherever it has to be multiplied by zero. Thus, in Ex 2, to convert 1010 we may directly write as 1 * 2 3 + 1 * 2' or 8 + 2 (because other terms will be multiplied by 0 and give 0 in any case). Thus we can develop the following quicker method for conversion of binary into decimal numbers:

Quicker method for converting binary num-bers to decimal Step I : Starting from the rightmost digit of the given binary

number, write 1, 2, 4, 8, 16, 32... and so on below each digit as you proceed towards the left.

Step I I : Ignore the numbers below the 0s (zeroes). Add all the remaining numbers below the Is.

Ex. 4: Solve Ex. 3 by quicker method. Soln: (a) 1010 Step I : Starting from right we write 1,2,4 and 8 below the digits. We get

1 0 1 0 8 4 2 1

Step I I : 4 and 1 fall below the zeros. We ignore them and add the remaining. We get 8 + 2 = 10 (Ten).

O) 1111 Step I : Starting from right, we write 1,2,4, and 8 below the

digits. We get

1 1 1 1 8 4 2 1

Step I I : Al l numbers fall below Is. So we add all of them to get8 + 4 + 2+ 1 = 15 (fifteen)

(c) 100. Step I : Starting from right, we write 1, 2 and 4 below the digits. We get:

1 0 0 4 2 1

Step I I : 1 and 2 fall below the zeros, we ignore them. This leaves 4 (four), (d) 10000. Step I : Starting from right, we write 1,2,4,8, and 16 below the digits. We get

1 0 0 0 0 16 8 4 2 1

Step I I : 1, 2, 4, 8 fall below the zeros. Ignore them. That leaves 16. (Sixteen)

Miscellaneous 743

(e) 1110010 Step I : Starting from right, we write 1,2,4, 8,16,32 and 64 below the digits. We get

1 1 1 0 0 1 0 64 32 16 8 4 2 1

Step I I : We ignore 8, 4 and 1 as they fall below the zeros. Adding the rest, 64 + 32 + 16 + 2 we get 114 (One hundred and fourteen)

Converting decimal numbers into binary A decimal number is converted into binary by the method

of successive divisions. Each time the dividend is divided by 2. The remainder is noted and the quotient becomes the next dividend, which is again divided by 2. This process is re-peated until no more division is possible. We will explain it by the following example. Ex. 5: Convert 17 into a binary number: Soln: Step I : We divide 17 by 2. The remainder is 1 and the

dividend is 8. (See below)

17 8 1

Step I I : In the previous step the dividend was 8. That is, our new quotient. We divide it again by 2. Now the re-mainder is 0 and dividend is 4. (See below)

17 8 1 4 0

Step I I I : Last step's dividend is our new quotient. I f we di-vide it again by 2, our dividend is 2 and remainder 0. (See below)

17 1 8 0 4 0 2

Step IV: Last step's dividend is now our quotient, i.e. 2. I f we divide it by 2 our dividend is 1 and remainder 0. (See below)

17 1 8 0 4 0 2 0 1

Note that any more divisions are not possible be-cause 1 is not divisible by 2. Now, we write all our remainders from left to right in the order shown be-low by the arrows

17 1 8 0 4 0 2 0 1

.-. Our binary number for 17 is 10001.

Ex. 6: Convert (a) 11 and (b) 14 into binary: Soln: (a)

11 1 5 1 2 0 1

.-. 11= 1011 in binary (b)

11 1 5 1 2 0 1

.-. 14 = 1110 in binary [Note: We can use the method of 18.3.1 to check i f

our answer is right.

Forexample, 1011 = l x 2 3 + 0 x 2 2 + 1 x 2 ' + 1 x 2 = 8 + 2 + 1=11 (Eleven)

1110 = l x 2 3 + l x 2 2 + 1 x 2 ' + 0 x 2 = 8 + 4 + 2 = 14 (Fourteen)]

Table 2: Some decimal numbers and their binary repre-sentation

(The following table could be used for read\-ence in case you require a quick solution. Yes ma\o like to memorise the binary representation of first 16 numbers: it will save you a lot of time.)

Decimal number

Binary form

Decimal number

Brian form

Decimal number

Brian form

1 1 12 1100 23 10111

2 10 13 1101 24 11000

3 11 14 1110 25 11001

4 100 15 1111 26 11010

5 101 16 10000 27 11011

6 110 17 10001 28 11100

7 111 18 10010 29 11101

8 1000 19 10011 30 11110

9 1001 20 10100 31 i n n

10 1010 21 10101 32 100000

11 1011 22 10110

Some tips for quick answers Tip 1: (A) The binary form of an odd number will always

have a 1 in the end and the binary form of an even

7 4 4 P R A C T I C E B O O K ON Q U I C K E R MATHS

number will always have a 0 in the end. (B) Conversely, if the binary form has a 0 in the end it

must be an even number and if it has a 1 in the end it must be an odd number

Tip 2: (A) The binary form of 4, 5, 6 and 7 has three digits; that of8, 9,15; has four digits and that of 16, 17,

31 has five digits. (B) Conversely, if the binary form has three digits, it

must be one of 4, 5, 6 or 7; if it has four digits it must be one of 8, 9,15; if it has five digits it must be one of 16, 17..., 31.

Tip 3: Just as a zero at the leftmost place has no value for a decimal number, it has no value for a binary num-ber also. (For example, 010 is the same as 10 in binary system and it equals 2 in the decimal. So, effectively, 010 is not a 3- digit number but a 2-digit number).

I I . Questions Based on Equations In recent years, there have been some changes in

the pattern of test papers of various exams. Some new chap-ters and some new types of question have been introduced in Quantitative Aptitude paper. We are going to discuss one of those newly introduced patterns of question. These ques-tions are a part of Algebra. They are mainly related to Linear Equation, Quadratic equation, Inequality and Exponential chapters.

Type of question: See the following direction and questions.

Directions: In each of the following questions one or more equations is/are given. On the basis of the given equation(s) find the relationship between p and q. Mark an-swer:

l ) i f p = q 4) if p > q

Questions 1. I.4p + 8q = 3

4p 8

2) i f p > q

5) i f q > P

3 ) i f q > p

I . 15 = 0

I I . 12p + 4q = 4

I I . 9q 2 =12q-4

3. I . q 2 15q + 56 = 0 I I . 2p 2 -10p + 12 = 0

4. I . pq+30 = 6p + 5q

5. I . p ( p - ' ) = p - ' n. q 2 = 4 q - 1

6. I . 2p 2 = 23p-63 H. 2q (q - 8 )=q- 3 6

7. I . 2p(p + 4) = 8(p + 5) I I . q + 4 = 7q"'

Note: 1. All the above questions have been asked in PO Exams

during 1999-2000 and 2000-01. 2. Q 1 is based on linear equations. In Q 2 equation I gives

direct value of P, whereas equation II is quadratic. In Q 3 both the equations are quadratic. In Q 4 two linear equa-tions are combined together. In Q 5 both the equations give direct values of p and q. In Q 6 equation I is qua-dratic and equation II gives direct value of q. In Q 7 both the equations are quadratic.

3. First we will learn to solve the questions based on linear equation and then on quadratic equation. After that the questions based on combined equations and other types of equation will be discussed,

(i) Questions based on Linear Equations In such questions we are given two linear equa-

tions. To find the value of p & q, we are required to solve these two equations. Now, the problem is to solve the two linear equations. There are some well-known methods to solve them. See the following example (Q No. 1).

I . 4p + 8q = 3 I I . 12p + 4q = 4 We can't say which is greater (p or q) in the first

glance. We will have to solve these equations and find the values of p and q. These two equations can be solved through graph. But it is not useful for us. The second method, which is most in vogue, is to equate the coefficient of p in the two equations and subtract one equation from the other to get the value of q. See.

4p + 8q = 3 ....(1) 12p + 4q = 4 ....(2) We multiply eqn(l) by (3) and eqn (2) by 1 to equate

the coefficients of p. Now, the two equations become 12p + 24q = 9 ....(3) 12p + 4q = 4 ....(4) Now we subtract (4) from (3) and we have

20q = 5 .-. q = \_

20 4 Now, substitute q in either (1) or (2) and get the

value of p.

Miscellaneous 745

P = b ]C 2 - b?c 2H

and q 3.1 C _ a 2 H l*-2

a,b 2 - a 2 b ,

Note:

a,b 2 - a 2 b ,

Note that the denominators of both the values are the same (a,b 2 - a 2 b ] ) . It is very systematic and easy to remember. Once we find the value of p, we can get q by putting p in either of the equations. So you don't need to remember both the formulae. In the above example: I . 4 p + 8 q - 3 = 0 II . 12p + 4 q - 4 = 0

8 ( - 4 ) - 4 ( - 3 ) _ - 3 2 + 12 _ 1

4 x 4 - 1 2 x 8 * - 8 0 ~ 4 P =

Putting p = - in I , q p = q

If you have good practice of multiplication and addi-tion you can write the values of p and q direct. Thus it minimises writing work at the cost of mental work, which ultimately saves our time.

See the following two examples: E x l .

Ex 2.

Soln

Soln

(I) p + 2q-95 = 0 (II) 2p + 3q-151=0 (I) 3p + 4q-25 = 0 (II) 2p + 3q-18=0

-302 + 285 ,_ ( i ) P = - 7 T T - = 1 7

Puttingp= 17 i n l , q = 3

72 + 75

q > p

(2)P = 3 9 - 8

Puttingp = 3 in I , q = 4 => q > p Exercise: Solve the following questions which are

based on the direction given earlier. First try yourself. I f you find any problem only then see the given solution.

1.1.p + 4q = 6 II. 5p + 8q= 18 2.1.3p + 4q+ 1 =0 II .p + 6q-9 = 0

. 3 3.1. 6p + q= 4 - II.2p + 3 q = 3 ^

4.1.6p + 3q = 3 ,2

II.3p + 2q= l j

5.1. 3p + 2q = 2.3 II. 4p + q=1.9 6.1.p + q+ 1=0 I I . p - q - 5 = 0 7.1.2p + 2q = 7 II .4p + q = 5 8.1.p + 7q = 6 II . 3p + 5q = 2 9.1.2p-q=16 II.3p + 2q = 66

P 7q , 10.1. T - + - T - = 1

2 4 H.3p + ^ = 2

Mark the following points 1. Your answer would be choice (1), (2) or (3) only. Why?

Because, both the equations together give single val-ues of p and q. So in linear equation cases, one value (p) can't be greater than or equal to other value (q). So our answer can't be choices (4) or (5).

2. From the given formula

P b , c 2 - b 2 C i

3 2 31C l*-2

> , => p > q

p = q

p < q

Can we apply the above rule to find our answer directly? No. Our conclusion may be wrong for -ve values of p and q. Check question (6).

p + q + 1 = 0 p - q - 5 = 0

p _ -5 + 1 _ -4 _ - 2 q ~ 1+5 ~ T ~ q > p 1 + 5 6 3 q

But if we solve, the correct value of p = 2 and q = -3, which implies that p > q. So the above method does not give the correct answer in all the cases. Solution

(1)2; I. p + 4q-6 = 0 II.5p + 8q-18 = 0

-72 + 48 -24 r P 8-20 -12 Putting p = 2 in I , q = 1

(2)3;I.3p + 4 q + l = 0

- 3 6 - 6 p = = -3

18-4 Putting p in I , q = 2 =;

p > q I I .p + 6q-9 = 0

q > p

(3 )3 ; 6p + q - 4 - = 0

2p + 3 q - 3 - = 0

43 57 12 + 4

1 9 6p + q - = 0

43 . > 2p + 3 q - = 0

-43 + 171 128

18-2 12x16

3 Putting P = y i n l , q = 4

(4) l ; 6p + 3 q - 3 = 0

12x16 12

q > p

7 4 6 P R A C T I C E B O O K ON Q U I C K E R MATHS

3p + 2 q - y = 0

P = -5 + 6 12-9

Putting P 3 i n l , q = - p = q

(5) 3;3p + 2q-2.3=0 4p + q-1.9 = 0

-3.8 + 2.3 -1.5 P = =

3-8 - 5 Let p = 0.3 in I , q = 0.7

(6) 2;p + q+ 1 =0 p - q - 5 = 0

= 0.3

q > p

P = -5 + 1

= 2

Putting p = 2 in 1, q = -3 (7)3;2p + 2q-7 = 0

4p + q - 5 = 0

-10 + 7 -6 ~ 2

p > q

P = 2 - 8

1 Putting p = in I , q = 3

(8)3;p + 7q-6 = 0 3p + 5q-2 = 0

q > p

-14 + 30 -1

5 -21

Putting p = -1 in I , q = 1 (9)2;2p-q-16 = 0

3p + 2q-66 = 0

98 7

q > p

66 + 32 P = - 14 4 + 3 Putting p = 14 in I , q = 12 p > q

P 7q

(10)3; J + ^

5q

1 => 2p + 7q-4 = 0

3p + -^- = 2 r > 6p + 5q-4 = 0

_ -28 + 20 -8 _ 1 P ~ 10-42 ~ ^ 3 2 ~ 4

1 I Putting p = in I , q = q > p

(ii) Questions Related to Quadratic Equations The general form of quadratic equation is

ax2 + bx + c = 0 . The two roots or the two values of x are

b V b 2 -4ac -b . Sum of the two values of x - and mul-

2a a

tiplication of the two values = . a

Now, with the help of the above information we will try to solve the problems.

Take question (3). I. q 2 15q + 56 = 0

II. 2p 2 -10p + 12 = 0

I. q 2 -15q + 56 = 0

+ 151^225-4x56 15 + 1 => q = ~ = = 7 ' 8

2 2

II. 2p 2 -10p + 12 = 0 + 10 + V100-96 10 + 2 . ,

=> p = = _ ^ = 2,3 4 4

Thus q > p

Other Method: (By factorisation) I. q 2 - 1 5 q + 56 = 0

or, q 2 - 7 q - 8 q + 56 = 0 or,q(q-7)-8(q-7) = 0 or , (q-8)(q-7) = 0=> q = 7,8

SimilarlyII. = (p-2)(p-3) = 0 => p = 2,3 Therefore, q > p.

Suggested Method

q > p means both the values of q are more than both the values of p. This further implies that sum of both the values of q is more than sum of both the values of p. In-versely, i f sum of two values of q is greater than sum of two values of p then q > p.

See the same in above case:

Inequation I. q 2 15q +56 = 0

Sum of two roots (or two values of q)

_ - ( - 1 5 ) . 1

= 15

In equation II . 2p 2 - 1 Op +12 = 0

Sum of two values of p = - ^ = 5 q>p .

Now our work becomes so easy. We don't need to find the roots of the quadratic equations. With the help of the above explanation we can find our answer quickly. But what happens when one value of q is equal to one value of p,

which requires the choice (5) q > p ?

Miscellaneous

What happens when one value of q is more and the other value of q is less than the respective values of p?

To get the solution of the above questions mark the following points. (1) Suppose the quadratic equations give the value of p

and q like: p = 3,7 ancfq =1,8

In such case, we can't say p > q or p < q because 3 is less than 8 but more than 1; similarly, 7 is more than 1 but less than 8. Then what should be our answer? We have no choice to mark !! Don't worry. Such a case will never

i come i f you have no option among given choices. (2) Our method suggests only about q > p or p > q but what

happens when one value of p is equal to one value of q,

which subsequently changes our answer as q > p or

p > q ? To know the^efrdttfOfTofequality of one root in two quadratic equations. See the following explanation. Suppose the two given quadratic equations are

I. a ,p 2 +b,p + c, = 0 II. a 2 q 2 + b 2 q t-c2 = 0 Suppose one value of p and one value of q are equal

ie, P| =q , = a (say). Then

a,a 2 + b,a + c, = 0 and a2cc2 + b 2 a + c, =0

Then o r

or,

b ] C 2 b-)C| a 2 C j a | C 2 a j b i a 2 b |

b | C 2 b-jCj a 2 C | a | C 2

a 2 C ] ajC-> a j b 2 a->b|

of,^a2c, - a , c 2 ) 2 = (a,b 2 - a 2 b ,Xb 1 c 2 - b 2 c , )

or, (a,c2 - a 2 c , ) 2 = (a,b 2 - a 2 b ,Xb ,c 2 - b 2 c , ) **

Now, we may conclude that i f the relation given in** is true then one of the values of p is equal to one of the values of q. The above relationship is very systematic. Mark and remember it.

Take the example:

I. p 2 -10p + 24 = 0 II. q 2 - 9 q + 20 = 0

Sum of roots (p, + q 2 ) = ^ p ^ = 10

Sum of roots fai + q 2 ) = ^ ^ = 9 p > q

Now, check the equality of root.

(20-24) 2 = ( - 9 + 10X-200 + 216)

=> 16=16, which is true. Hence one root of p is equal to one root of q. Thus our required answer should be p > q

Note: The roots (or values of p&q) of the quadratic tions in the above two examples can be found easiK by factor method. This does not imply that the suggesk : method (discussed above) is useless. It is useful when the equations are difficult to factorise and roots come in fractional value.

Take some more examples: E x : l . 1.18p2 + 3p - 3 = 0 Soln: By factor Method:

I. 18p2 + 9p -6p -3=0 9 p ( 2 p + l ) - 3 ( 2 p + l ) = 0

=>(9p i -3) (2p+l ) = 0= i > p = X . - J * II. 14q: + 9q + 1 = 0 => 14q2 + 7q - 2q - 1 = 0

=> 7q (2q+ l ) + ( 2 q + l ) = 0

II. 14q : -9q- 1 =0

S> (7q+ l ) ( 2 q + l ) = 0rr> q =

Therefore p > q . By Solution Method:

1

:

i .p =

II. q =

- 3 V 9 + 12xl8 -3 + 15 1 1 36 36.

9 V81-56 - 9 + 5 1 1 28

Therefore p > q By Suggested Method:

-3 " 18

28

I . Sum of roots :

I I . Sum of roots : 14 p > q

For equality of roots, (18 + 42)2 = (162-42)(3 + 27)

3600 = 3600 = > p > q Ex:2. I .p 2 -12p + 36 = 0 Il .q 2 -14q + 48 = 0 Soln: By Factor Method:

I. p 2 - 12p + 36 = 0 => p 2 - 6p - 6p + 36 = 0 =>(p-6 ) (p -6 ) = 0 s p = 6

II. q 2 - 14q+ 48 = 0 => q 2 -8q-6q + 48 = 0 => q (q -8 ) -6 (q -8 ) = 0 => (q-6)(q-8) = 0 .-. q = 6,8 Therefore q > p By Solution Method:

j p = 12+Vl2 2 -144 = 6

14 + V196-192. 14 + 2 . n II. q= - = = 6,8

7 4 8

Therefore q > p By Suggested Method: I. Sum of roots = 12 II . Sum of roots = 14 => q > p For equality of roots, (48-36)2 = (-14 + 12) (-12*48+ 14x36) (12)2 = (-2)(72)(7-8) ^

(12)2 = (12) 2=> q > p Note: We call the third method as suggested method and not

the quicker method, because for students whose basic calculation is not fast this method is not quicker.

Ex: 3. 1.2p2+ 12p+ 16 = 0 => p 2 + 6p + 8 = 0 II . 2q : +14q + 24 = 0"=> q 2 + 7q+12 = 0

Soln: By factor Method: I. p 2 + 4p + 2p + 8 = 0 => p(p + 4) + 2(p-

=>(p + 2){p + 4) = 0 > p = -2,-4 II. q 2 + 4q + 3q+12 = 0

=> q(q + 4)+3(q + 4) = 0 => (q + 3)(q + 4) = 0 => q = -3,-4

Therefore p > q By Solution Method:

- 6 V 3 6 - 3 2 - 6 2

4) = 0

I.P =

II. q =

2 7 + V49^ 48

2 -71

=-2,-4

= -3,-4 2 2

Therefore p > q By Suggested Method: I . Sum of roots = -6 I I . Sum of roots = -7 => p > q For equality:

(12-8) 2 = (72 - 56)(7 - 6)

16=16 = > p > q

I I I . To find the part of the equations which is not equal to the other given equations. First see the format of the question given below.

Direction: Four of the following five parts numbered (1), (2), (3), (4) and (5) in the following equation are exactly equal. Which of the parts is not equal to the other four? The number of that part is the answer.

xy2 -x2y + 2x2y2 = xy2(\ + x2y(2y-\) =

1) 2)

xy2 (l + x) - x2y{l - y) = xy\y{\ x) - x(l - y)] =

3) 4)

xy\x + y)2 + y{] - y)-x(\ x)]

5) (SBI Bank PO Exam, 1999)

P R A C T I C E B O O K ON Q U I C K E R MATHS

Suggested Method: Put the value ofx = - l andy= 1.Check the equality of the given equations as suggested below.

We have,

xy2 -x2y + 2x2y2 = xy2(\2y{2y-\) =

1) . 2)

xy2 {\+x)-x2y{\-y) = xy\y{l + x)-x(l - y)] =

" 3) 4)

xy^x + y)2+y(l-y)-x{\+x)\

Now, put x = - 1 and y = 1 and check the equality of all given equations.

> ( - l X l ) 2 - ( - l ) 2 0 ) + 2 ( - l ) 2 ( l ) 2

> - 1 - 1 + 2 => 0

> ( - i X r ) 2 [ i - 2 ( - i ) ] + ( - i ) 2 ( i ) [ 2 ( i ) - i ]

> l [ l + 2 ] + l [ 2 - l ] => 4

> ( - i X i ) 2 [ i + ( - i ) ] - ( - i ) 2 x i [ i - i ] ^ o - o = > o

> ( - l ) ( l ) [ l ( l + ( - l ) ] _ ( - l ) [ l - ( + ! ) ] => 0 -0 ^ 0

( - I X I ) [ ( - I + I ) 2 ] + I ( I - I ) - ( - I X I + ( - I ) ]

=^ 0 + 0 => 0. From the above calculation we can conclude that the all

parts are equal except the equation (2). Hence (2) is the cor-rect answer. Note: The above suggested method is not true for each and

every case. But aspirants are advised to try this method. Here your luck has to play its role. I f you are lucky enough, you may save atleast 1 minute. Now, take the case given below carefully.

Direction: Four of the five parts numbered (1), (2), (3), (4) and (5) in the following equation are exactly equal. Which part is not equal to the other four? The number of that part is the answer.

0)

(2)

(3)

(4)

(5)

{\)x{x + y)2 -2x2y =

(3) x(x2+y2) =

(5) x\x + y)2-2xy2\

(2) x{x-y)2+2x2y =

^)x\x + y)2 -2xy] =

(SBI Associates PO 1999) Now, we try to solve by the suggested method. By put-

ting the value of x = - 1 and y = 1, we find that every part of equation is equal to (-2). Hence, the given method doesnot hold true in this case. Therefore, in such cases traditional method will be applied. After simplification of the equations given in the above question, we find that all equations are

equal to x3 +xy2 except equation (5). Hence (5) is the cor-

rect answer.

M

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it. 1

I i the ten Ex:

Ex-

Miscellaneous 749

IV. Questions Based on Inequality "Equation" means A statement of equality which has

L.H.S (Left Hand Side) and R.H.S (Right Hand Side) and con-nected by a equal sign (=).

Now see the following statements. 6 is greater than 5 (6 > 5) x is less than y (x < y)

a is greater than or equal to b (a > b)

-3 is less than or equal to x ( - 3 < x) Here the signs are >, > , < and < so these are called

inequalities or inequation and not equations. Now go through the following rules and try to remember

it. * I f a >b then

(i) a + c> b + c (ii) a - c > b - c

(iii) ac>bc (cis+ve) a b

(iv) > (c is +ve) b c

Ex:

* In an inequality i f one term goes from one side to the other the sign of the inequality remains the same but the sign of term changes from +ve to -ve, from x to * and vice versa.

I f a - c > b then a > c + b Ifa + b > c t h e n a > b * c. * I f the signs of all terms of an inequality changes then the sign of the inequality is reversed. I f a > b t h e n - a < - b Ex:

* I fa>b then a " > D " and 1 1

~ 7 < 7 7 (Ifnis+ve) a b

(a,b,c,x,y,z>0)

How to solve an inequality We should divide it into two categories 1. (i) I f (x -a ) (x -b) i ie. x e R-(a,b)

Some Solved Examples Ex.1: So lve (3x - l ) (x -2 ) < 0 Soln: Divide 3 on both side (because the term 3 x is there

so to get x we have to divide it by 3)

Soln:

then

- 0

-+-2 5

X5 ie xeR-[2, 5J

Ex.3: Solve 2x2 -7x-6>0

3xz - 9 x + 2 x - 6 > 0

=> 3X(X -3 )+2(JC -3 )>0

=> (x-3X3x + 2 )>0

(x- 3^x +1 j > 0 (dividing both sides by 3).

( x - 3 >0 . x3 3JI 3

Ex.4: Solve Sx2 +6x + l < 0

Soln: 5x2+6x + \ 5x2 +5x + x + \0

=> ( x + l ) ( 5 x + l ) < 0

' -l