Number system 1surds & indices

NUMBER SYSTEMSURDS & INDICES

If 32x = 81, what is the value of x?

1

2

3

4

Apr 15, 2023C.S.VEERARAGAVAN 98948 34264 [email protected] 2

I

Apr 15, 2023 C.S.VEERARAGAVAN 98948 34264 [email protected] 3

34 = 81.2x = 4X = 2.

If 32x = 81, what is the value of x?I

+ 2

– + 2

Either 1 or 2

None of these.


What is the conjugate of –2? II


What is the conjugate of –2?

1. + 22. – + 23. Either 1 or 24. None of these.

Conjugate is – – 2

II

2 –

2 –

3 –

2 – 2


III


Simplify:

x

III


Simplify:

x = √3 −√23−2

III


Simplify:

x = √3 −√23−2 =

III


Simplify:

x = √3 −√23−2 =

12+√3

𝑋 2−√32−√3

III


Simplify:

x = √3 −√23−2 =

12+√3

𝑋 2−√32−√3

¿ 2−√34−3

III


Simplify:

x = √3 −√23−2 =

12+√3

𝑋 2−√32−√3

¿ 2−√34−3 =2−√3

III


Simplify:

x = √3 −√23−2 =

12+√3

𝑋 2−√32−√3

¿ 2−√34−3 =2−√3

Adding we get 2 –

III

What is the positive square root of ?

312

318

314

None of these


IV



IV



√√3

IV

What is the positive square root of 7 + 4

√3+12 + 2

√2+√32 +


V


√3+12 + 2

√2+√32 +


√7+4 √3=√7+2√3(4)

V


√3+12 + 2

√2+√32 +


√7+4 √3=√7+2√3(4)

√3+4+2√3(4)=√ (√3+√4 )2

V

What is the positive square root of

√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

VI


√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

√√72−√40

VI


√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

√√72−√40

√√36 (2)−√20(2)

VI


√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

√√72−√40

√√36 (2)−√20(2)

√√2 (√36−√20 )

VI


√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

√√72−√40

√√36 (2)−√20(2)

√√2 (√36−√20 )

√√2 (6−2√(5 ) )

VI


√2 (√5−1 )4√2 (√5−1 )4√2 (√3−√2 )


4√2 (√5−√3 )

√√72−√40

√√36 (2)−√20(2)

√√2 (√36−√20 )

√√2 (6−2√(5 ) )

√√2 (5+1−2√ (5 ) )

VI

5

323 −6

13 +2

23

=

31/3+21/3

31/3−21/3

21 /3−31/3

32/3+22 /3


VII

5

323 −6

13 +2

23

=

31/3+21/3

31/3−21/3

21 /3−31/3

32/3+22 /3


5

323 −6

13 +2

23

5

(313 )2

−(2)13 (3)

13 + (21

3 )2

VII

5

323 −6

13 +2

23

=

31/3+21/3

31/3−21/3

21 /3−31/3

32/3+22 /3


5

323 −6

13 +2

23

5

(313 )2

−(2)13 (3)

13 + (21

3 )2

5

(313 )2

−(2)13 (3)

13 + (21

3 )2 𝑥3

13 +2

13

313 +2

13

VII

5

323 −6

13 +2

23

=

32/3+21 /3

31/3−21/3

21 /3−31/3

32/3+22 /3


5

323 −6

13 +2

23

5

(313 )2

−(2)13 (3)

13 + (21

3 )2

5

(313 )2

−(2)13 (3)

13 + (21

3 )2 𝑥3

13 +2

13

313 +2

13

5(313 +2

13 )

3+ 2 VII

What is the rationalizing factor of

423 +12+3

23

423−12+3

23

423 +12

13+3

23

443 +12

23 +3

43


VIII


423 +12+3

23

423−12+3

23

423 +12

13+3

23

443 +12

23 +3

43


is of the form a – b

VIII


423 +12+3

23

423−12+3

23

423 +12

13+3

23

443 +12

23 +3

43


is of the form a – b

should be multiplied to get

423 +4

13 3

13 +3

23

VIII

Simplify: .

1

x

0

xabc


IX

Which is greater?

1

√2+√5

Cannot be determined

None of these.


X

Which is greater?

1

√5+√2


None of these.


X

8 > 5

Which is greater?

1

√5+√2


None of these.


X

8 > 58 + > 5 +

Which is greater? ,

222

322


Both are equal


XI

–6 – 3

√5−2–

None of these


What is the rationalizing factor of 6 + 3

XII

–6 – 3

√5−2–

None of these



6 + 3

XII

–6 – 3

√5−2–

None of these



6 + 3 3(2) + 3

XII

–6 – 3

√5−2–

None of these



6 + 3 3(2) + 3

3(2 +

XII

–6 – 3

√5−2–

None of these



6 + 3 3(2) + 3

3(2 +

is the factor req.

XII

What is the value of x if

0

1

3

2


XIII

If = 216 and x and y are integers then x,y?

3,2

2,2

3,4

3,3


If , what is the value of x?

4

3

2

1


(17)3.5 x (17)? = 178

2.29

2.75

4.25

4.5


11

If then the value of x is

2

1

1272


03


2

1

1272


(𝑎𝑏 )𝑥−1

=(𝑏𝑎 )𝑥−3

(𝑎𝑏 )𝑥−1

=(𝑎𝑏 )−(𝑥− 3)

03


2

1

1272


(𝑎𝑏 )𝑥−1

=(𝑏𝑎 )𝑥−3

(𝑎𝑏 )𝑥−1

=(𝑎𝑏 )−(𝑥− 3)

x – 1 = 3 – xSolving x = 2

03

Given 100.48 = x, 100.7 = y and xz = y2, value of z = ?

1.45

1.88

2.9

3.7


12


1.45

1.88

2.9

3.7


xz = y2

100.48z = 102*0.7

12


1.45

1.88

2.9

3.7


xz = y2

100.48z = 102*0.7

0.48z = 1.4

12


1.45

1.88

2.9

3.7


xz = y2

100.48z = 102*0.7

0.48z = 1.4Z = 2.9

12

If 5a = 3125, then the value of 5a – 3 is?

25

125

625

1625


13

If 3x – y = 27 and 3x + y = 243, then x =?

0

2

4

6


14

If 3x – y = 27 and 3x + y = 243, then x = ?

0

2

4

6


3x – y = 27 = 33

3x + y = 243 = 35

14

If 3x – y = 27 and 3x + y = 243, then x = ?

0

2

4

6


3x – y = 27 = 33

3x + y = 243 = 35

x – y = 3

14

If 3x – y = 27 and 3x + y = 243, then x = ?

0

2

4

6


3x – y = 27 = 33

3x + y = 243 = 35

x – y = 3x + y = 5

14

If 3x – y = 27 and 3x + y = 243, then x = ?

0

2

4

6


3x – y = 27 = 33

3x + y = 243 = 35

x – y = 3x + y = 5Solving x = 4

14

2560.16 x 2560.09 = ?

4

16

64

256.25


01

2560.16 x 2560.09 = ?

4

16

64

256.25


2560.16 x 2560.09 = 2560.25

01

2560.16 x 2560.09 = ?

4

16

64

256.25


2560.16 x 2560.09 = 2560.25

256 = 44

01

1

1+ x( b − a ) +x ( c− a ) +1

1+ x (a − b )+ x (c − b) +1

1+x ( b − c )+ x (a −c ) =?

0

1

Xa – b – c

None of these


15

1

1+ x( b − a ) +x ( c− a ) +1

1+ x (a − b ) + x (c − b) +1

1+x ( b − c ) + x (a −c ) =?

0

1

Xa – b – c

None of these


1

1+xb

xa +xc

xa

+1

1+xa

xb +xc

x b

+1

1+xb

xc +xa

xc

15

1

1+ x( b − a ) +x ( c− a ) +1

1+ x (a − b ) + x (c − b) +1

1+x ( b − c ) + x (a −c ) =?

0

1

Xa – b – c

None of these


1

1+xb

xa +xc

xa

+1

1+xa

xb +xc

x b

+1

1+xb

xc +xa

xc

xa

xa + xb+ xc +xb

xa + xb + xc +xc

xa + xb + xc

15

257.5 x 52.5 1251.5 = 5?

8.5

13

16

17.5


06

(243 )n5 X 32 n +1

9n X 3n −1

1

2

9

3n


04

(243 )n5 X 32 n +1

9n X 3n −1

1

2

9

3n


(35 )n5 X 32 n +1

( 32 )n X 3n −1

04

(243 )n5 X 32 n +1

9n X 3n −1

1

2

9

3n


(35 )n5 X 32 n +1

( 32 )n X 3n −1

=3𝑛+2𝑛+ 1

32𝑛+𝑛−1

04

(243 )n5 X 32 n +1

9n X 3n −1

1

2

9

3n


= 33n + 1 – 3n + 1

04

+

0

12

1

am+

n


07

+

0

12

1

am+

n


+

07

If mn = 121, the value of (m – 1)n+1 is

1

10

121

1000


08

(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

xabc

1

xab+bc+c

a

xa+b+c


09

(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

xabc

1

xab+bc+c

a

xa+b+c


(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

09

(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

xabc

1

xab+bc+c

a

xa+b+c


(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

= x(b - c)(b + c - a) . x(c - a)(c + a - b) . x(a - b)(a + b - c)

09

(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

xabc

1

xab+bc+c

a

xa+b+c


(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )


= x(b - c)(b + c) – (b-c)(a) . x(c - a)(c + a) – (c-a)(b) . x(a - b)(a + b – (a-b)(c)

09

(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )

xabc

1

xab+bc+c

a

xa+b+c


(xb

xc )( b + c − a )

(xc

xa )( c + a − b)

(xa

xb )(a + b − c )


= x(b - c)(b + c) – (b-c)(a) . x(c - a)(c + a) – (c-a)(b) . x(a - b)(a + b – (a-b)(c)

=

09

If x = 3 + 2, then is

1

2

2

3


10


1

2

2

3


X = 3 + 2

10


1

2

2

3


X = 3 + 2

= 3 + 2 +

10


1

2

2

3


X = 3 + 2

= 3 + 2 + = 3 + 2 +

10


1

2

2

3


X = 3 + 2

= 3 + 2 + = 3 + 2 + = 3 + 2 + (3 - 2

10


1

2

2

3


X = 3 + 2

= 3 + 2 + = 3 + 2 + = 3 + 2 + (3 - 2= 4

10

Number system 1surds & indices

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