Pioneer Education {The Best Way To Success} IIT JEE … · 2014-09-12 · Pioneer Education {The...

Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes

www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3

Revision Question Bank

Number System

1. Prove that:1/2

3/2 0 19 3 5 15

81

Solution :

L . H. S. : – 1/2

3/2 0 19 3 5

81

= 1

23/22

2

13 3 1

9

( a0 = 1)

=

123 22

21

3 39

(

nm mna a )

= 1

3 13 3

9

= 27 – 3 – 9 1 1a

a

= 24 – 9

= 15 = R. H. S.

2. Find the value of 23 .

Sol :

We have, 1

2 2 23 3

= 1

2123 3

[

nm m na a ]

= m

m

1 1a

3 a

3. Find the value of 1

10, when 10 3.162 .

Sol : (c)

Let x = 1

10



= 1 10 10

10010 10 [by rationalization]

= 3.162

100 [given, 10 3.162]

= 0.03162

4. Write the rational number 329

400 in decimal form. Also, find the kind of decimal expansion.

Sol ;

Given rational number is 329

400.

Now, we divided 329 by 400, using long division method.

400 3290 0.8225

3200

900

800

1000

800

2000

2000

0

Hence, decimal number of given rational number is 0.8225.

Here, we see that remainder is zero, so it is a terminating decimal.

5. Find the value of 213

1/4 3

64 1 25

125 64256

625

.

Sol :

We have, 213

1/4 3

64 1 25

125 64256

625



=

12

23

1 1

4 3

5 54 4 4 1

5 5 54 4 44 4 4 4

5 5 5 5

=

12

23 223

1 14 3

4 2

54 1 4 1 4

45 5 54 4

55

[ n

m mna a ]

=

124 5 5

5 4 4

= 1

16 5 5

25 4 4

= 25 5 5

16 4 4 1 1

a4

= 25 20 20 65

16 16

6. If x = 3+2 2 , then find whether x + 1

x is rational or irrational.

Sol :

Given that, x = 3 + 2 2

= 2(3 2 2) 1

3 2 2

= 9 8 12 2 1

3 2 2

[

2 2 2a b a b 2ab]

= 18 12 2 6(3 2 2)

3 2 2 (3 2 2)

= 6, which is a rational number.

7. Represent 7.3 on the number line.

Sol :

Firstly we draw a line segment AB = 53 units and extend it to C such that BC = 1 unit. Let O

be the mid-point of AC.



Now, draw a semi-circle with centre O and radius OA. Let us draw BD perpendicular to AC

passing through point B intersecting the semi-circle at point D.

The distance (BD) is 53 units.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = 5.3

units.

8. Express with rational denominator 1

2 3 5 .

Sol :

We have,

1 1 {( 2 3) 5}

( 2 3) 5 {( 2 3) 5} {( 2 3) 5}

[multiplying numerator and denominator by {( 2 3) 5)}]

= 2 2

( 2 3) 5

( 2 3) ( 5)

[ (a – b) (a + b) = a2 – b2]

= 2 3 5

2 3 2 6 5

[ (a + b)2 = a2 + b2 + 2ab]

= 2 3 5

2 6

= 2 3 5 6

2 6 6

[multiplying numerator and denominator by 6 ]

= 2 3 3 2 30

12

9. If 3 2 3 2

a and b3 2 3 2

, then find the value of a2 + b2 – 5ab.

Sol :

Given, a = 3 2 3 2

and b3 2 3 2



Now, 3 2 3 2

a3 2 3 2

[multiplying numerator and denominator by 3 2 ]

= 2( 3 2)

3 2

[ (a – b) (a + b) = a2 – b2]

= 3 2 6

(5 6)1

[ (a – b)2 = a2 + b2 – 2ab]

and b = 3 2 3 2

3 2 3 2


= 2( 3 2)

3 2

[ (a – b) (a + b) = a2 – b2]

= 3 2 6

1

[ (a – b)2 = a2 + b2 + 2ab]

= 5 6

Now, a2 + b2 – 5ab

= 2 2(5 6) (5 6) 5(5 6)(5 6)

= 25 + 6 – 10 6 + 25 + 6 + 10 6 – 5(52 – 6)

[ (a – b)2 = a2 + b2 – 2ab, (a + b)2

= a2 + b2 + 2ab and (a – b) (a + b) = a2b2]

= 31 10 6 31 10 6 5 25 6

= 62 – 95 = – 33

10. Simplify 1 1 1 1

.2 5 5 6 6 7 7 8

Sol :

We have,

1 1 1 1

2 5 5 6 6 7 7 8

Taking Ist term,

1 1or

2 5 5 2

1 5 2

5 2 5 2

[multiplying numerator and denominator by



( 5 2)]

Chapter Test {Number System}

M:M: 40 M: Time: 40

1. Show that :

3/4 3/2 381 25 5

116 9 2

[4]

Sol :

LHS =

3/4 3/2 381 25 5

16 9 2

=

3/23/4 2 381 5 5

16 3 2

=

3/4 3 381 5 2

16 3 5

[ m n mn(a ) a ]

=

3/44 3 33 3 2

2 5 5

m

m1a

a

=

3 3 33 3 2

2 5 5

=

3 3 3

3 3 3

2 3 5

3 5 2



= 3 3 3 3 3 3

2 3 5

[m

m n

n

aa

a ]

= 0 0 02 3 5

= 1 × 1 × 1 = 1 = RHS Hence proved.

2. Represents 1 3 on the number line. [4]

Sol : We can write,

2 21 3 1 1 2 1 1 ( 2)

Draw a number line and mark all the points on it at equal distance of l unit, such that OA = AB = BC = 1 unit.

Leave 1 unit (OA) and start from A, draw BL l and cut-off BL=1 unit. Now, in ABL, by Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 (AL)2 =(AB)2+(BL)2 = 12 + 12 = 2

AL 2 [taking positive square root] Again, at L, draw ML AL and cut-of LM = 1 unit. Now, in ALM, by Pythagoras theorem, we have (AM)2 =(AL)2 +(LM)2

= ( 2 )2+(1)2 = 2 + 1=3

AM = 3 [taking positive square root]

Now, take A as centre and AM = 3 as a radius. Draw an arc which intersects the number line at point .R.

Hence, the point R represents 1 + 3 .

3. Express in a rational denominator3

3 2 5 . [4]

Sol : 3

3 2 5



= 3 [( 3 2) 5]

[( 3 2) 5] [( 3 2) 5]

[multiplying numerator and denominator by ( 3 2) 5]

= 2 2

2 2

3 3 3 2 3 5[ a b a b a b ]

( 3 2) ( 5)

= 2 2 23 3 3 2 3 5

[ a b a b 2ab]3 2 2 6 5

= 3 3 3 2 3 5 6

2 6 6

[multiplying numerator and denominator by 6 ]

= 1

[3 3 6 3 2 6 3 5 6]12

= 1

[3 18 3 12 3 30]12

= 1

[9 2 6 3 3 30]12

[ 18 9 2 3 2, 12 4 3 2 3 ]

= 1

[3 30 6 3 9 2]12

= 1

3[ 30 2 3 3 2]12

= 30 2 3 3 2

4

4. If 3 = 1.73205, then what is the value of 3 1

23 1

? [4]

Sol :

Now,

3 1 3 1 3 12 2

3 1 3 1 3 1


=

2

22

( 3 1)2

( 3) 1 1



= 2( 3 1)

23 1

[ 2 2a b a b a b ]

= 2( 3 1)

22

= ( 3 1)

= 1.73205 – 1 = 0.73205 [put 3 1.73205 ]

5. If 5 3

a 155 3

b, then find the values of a and b. [4]

Sol : We have,

5 3 5 3 5 3

5 3 5 3 5 3

[multiplying numerator and denominator]

= 2 2( 5) ( 3) 2 5 3

5 3

[using (a + b)2 = a2 + b2 + 2ab and a2 – b2 = (a – b) (a + b)]

= 5 3 2 15

2

= 8 2 15 2(4 15)

2 2

= 4 15

But it is given 5 3

a 15 b5 3

4 15 a 15 b

On comparing both sides of coefficients a and b, we get a = 4 and b = – 1

6. Simplify:

7 51 2 2 32 2

2 4 3 5

5 7 5 7

5 7 5 7

. [4]

Sol :

We have,

7 51 2 2 32 2

2 4 3 5

5 7 5 7

5 7 5 7



= 1 2 2 4 7/2 2 3 3 5 5/2(5 7 ) (5 7 ) m

m n

n

aa

a

= 3 6 7/2 5 8 5/2(5 7 ) (5 7 )

= 3 7/2 6 7/2 5 5/2 8 5/2(5 ) (7 ) (5 ) (7 ) [ m m ma b a b ]

= 7 7 5 5

3 6 5 82 2 2 2(5) (7) (5) (7)

[ m n mn(a ) a ]

= 21 25

21 202 2(5) (7) (5) (7)

= 21 25 4

21 20 1 22 2 25 7 5 7 5 7

= 25 × 7 = 175

7. Express 1.27 as a vulgar fraction. [4]

Express 1.27 as a vulgar fraction.

SoL ;

Let x = 1.27

x = 1.272727...

On multiplying both sides by 100 in Eq. (i), we get

100x =100 × 1.2727...

100x = 127.2727...

100x = 126 + 1.2727...

100 x = 126 + x [ from Eq. (i)]

99x = 126

126

x99

14

x11

Hence, 14

1.2711

8. Find five rational numbers between 3

5 and

4

5. [4]

Sol :

We have, 3 3 6 18

5 5 6 30

and 4 4 6 24

5 5 6 30

know that,



18< 19< 20< 2I< 22< 23< 24 On dividing all the numbers by 30, we get 18 19 20 21 22 23 24

30 30 30 30 30 30 30

3 19 20 21 22 23 4

5 30 30 30 30 30 5

Hence, five rational numbers between 3

5 and

4

5 are

19 20 21 22 23

, , , ,30 30 30 30 30

or 19 2 7 11 23

, , , ,30 3 10 15 30

.

9. Simplify:7 3 2 5 3 2

10 3 6 5 15 3 2

. [4]

Sol :

7 3 2 5 3 2

10 3 6 5 15 3 2

= 7 3 ( 10 3)

( 10 3) ( 10 3)

2 5 ( 6 5) 3 2 ( 15 3 2)

( 6 5) ( 6 5) 15 3 2 ( 15 3 2)

[by rationalising]

= 2 2 2 2

7 3( 10 3) 2 5 6 2 5 5

( 10) ( 3) ( 6) ( 5)

2 2

3 2( 15 3 2)

( 15) (3 2)

[ 2 2a b a b a b ]

= 7 3 ( 10 3) 2 30 10

10 3 6 5

3 2( 15 3 2)

15 18

= 7 3 ( 10 3) 3 2 ( 15 3 2

(2 30 10)7 3

= 3 ( 10 3) (2 30 10) 2( 15 3 2)

= ( 30 3) (2 30 10) ( 30 6)

= 30 3 2 30 10 30 6 = 1



10. Find the value of x, if 24 × 25 =(23)x. [4] Solution :

x

4 5 32 2 2

4 5 3x2 2 ( am × an = am + n and n

m mna a )

29 = 23x 9 = 3x x = 3

Pioneer Education {The Best Way To Success} IIT JEE … · 2014-09-12 · Pioneer Education {The...

Documents

Transcript of Pioneer Education {The Best Way To Success} IIT JEE … · 2014-09-12 · Pioneer Education {The...