ssc mathematice

7/29/2019 SSC Mathematice

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Salient Features:

9 Written as per the new textbook.

9 Exhaustive coverage of entire syllabus.

9 Facilitates complete and thorough preparation of HOTS section.

9 Variety of questions provided in order of importance.

9 Complex and intellectually challenging questions in the form of activities.

9 Constructions drawn with accurate measurements.

9 Attractive layout of the content.9 Self evaluative in nature.

Target PUBLICATI ONS PVT. LTD.

Mumbai, Maharashtra

Tel: 022 6551 6551

Website: www.targetpublications.in

www.targetpublications.org

email : [email protected]

Written according to the New Text book (2011-2012) published by the Maharashtra State

Board of Secondary and Higher Secondary Education, Pune.


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Target Publications Pvt Ltd.

First Edition : May 2012

Price : 110/-

Printed at:

Vijaya EnterprisesSion,Mumbai

Published by

Target PUBLICATIONS PVT. LTD.Shiv Mandir Sabagriha,Mhatre Nagar, Near LIC Colony,Mithagar Road,Mulund (E),Mumbai - 400 081Off.Tel: 022 6551 6551email: [email protected]


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PREFACEHOTS stands for Higher Order Thinking Skills. As the name implies, this book is filled to the brim with

questions that challenge your Thinking Skills. Unlike the traditional methods that focus on drill and

repetition as a mode of education, HOTS brings out the problem solving ability of a child.Inclusion of HOTS in a Question Paper is to attest if the students factually have an in depth understanding

of the said topic. It is not just how much but how well you understand the subject.

These twisted and brain challenging questions (HOTS) that form a part of the Question Paper demand an

out of the box thinking. Students are assessed through this format of Questions in terms of skills

pertaining to their analyzing, reasoning, comprehending, application and evaluation of a subject.

This book is full of such thought provoking and curiosity crunching questions across the subjects of

Algebra, Geometry and Science. The Questions throughout are framed in an innovative format including

Maps, Crosswords and Tree Diagrams. The emphasis in this courseware lies not just on the subject

knowledge but also its applications in real life.

With a progressive thought we have thoroughly followed the new syllabus pattern while composing this

book. There is no doubt that in the long run this book would be a proven aid for students to encounter

questions pertaining to Higher Order Thinking Skills.

We wish all the aspirants all the best and hope this book turns out to be a formidable aid in your tryst with

success.

Best of luck to all the aspirants!

Yours faithfully

Publisher


5/18

ContentsSubject No. Topic Name Page No.

Algebra

1 Arithmetic Progression and Geometric Progression 12 Quadratic Equations 103 Linear Equations in Two Variables 224 Probability 325 Statistics I 406 Statistics II 52

Geometry

1 Similarity 692 Circle 823 Geometric Constructions 954 Trigonometry 1045 Co-ordinate Geometry 1206 Mensuration 134

ScienceandTechnology

1 School of Elements 1492 The Magic of Chemical Reactions 1543 The Acid Base Chemistry 1584 The Electric Spark 1635 All about Electromagnetism 1686 Wonders of Light Part I 1717 Wonders of Light Part II 1758 Understanding Metals and Non-metals 1799 Amazing World of Carbon Compounds 18310

Lifes Internal Secrets 188

11 The Regulators of Life 19312 The Life Cycle 19913 Mapping our Genes 204


6/18

TARGET Publications Std. X : HOTS (Algebra)

1Arithmetic Progression and Geometric Progression

01 Arithmetic Progression and Geometric Progression1 Mark Questions

1. Find t5, if a = 5 and r = 2.

Solution:

Here a = 5, r = 2

tn = arn 1

t5 = 5 (2)5 1

= 5 (2)4 = 5 16

t5 = 80

2. If Sn = np +1

2n(n 1)Q where Sn denotes

the sum of first n terms of an A.P. Find the

common difference of the A.P.

Solution:

Sum of n terms of an A.P is given by

Sn =n

2[2at(n 1)d] = na +

1

2(n 1)d

Compare with Sn = np +1

2n(n 1)Q

d = Q

3. Find tn of the following A.P.

5

6, 1 , 1

1

6, tn

Solution:

The given A.P is5

6, 1 ,

11

6.

where a =5

6, d =

1

6

tn = a + (n 1)d

=5

6+ (n 1)

1

6

=5

6+

n 1

6

tn =n + 4

6

4. Find the first two terms of the following

sequence.

Sn =( ) n4n 3 3 + 3

2

Solution:

Sn=( ) n4n 3 3 3

2

+

S1=( ) 14 1 3 3 3

2

+

=( )4 3 3 3

2

+=

1 3 3

2

+=

6

2

S1 = 3

S2 =( ) 24 2 3 3 3

2

+

=( )8 3 9 3

2

+

=5 9 3

2

+=

45 3

2

+=

48

2

S2 = 24

5. Ifx 1 is the geometric mean between

(x + 5) and (x 3) then find the value ofgeometric mean.

Solution:

We know that, G = ( )( )5 3+ x x

(x 1)2 = (x + 5)(x 3)x

2 2x + 1 =x2 + 2x 154x = 16x = 4

Geometric mean =x 1 = 3

6. For a G.P 3, 3, 3, 3, ..... find Sn.

Solution:

3, 3, 3, 3,......... are in G.P.

Here, a = 3, r =

3

3

=

1

Sn = an1 r

1 r

= 3

( )

( )

n1 1

1 1

= 3( )n1 1

1 1

+

Sn =3

2[1 ( 1)

n]


7/18

TARGET PublicationsStd. X : HOTS (Algebra)

Arithmetic Progression and Geometric Progression2

2 Marks Questions

7. A person has 2 parents, 4 grandparents, 8

great grandparents and so on. Find the

number of his ancestors during the ten

generations preceeding his own.

Solution:Here, a = 2, r = 2, n = 10

Sn =( )na r 1

r 1

S10 =( )102 2 1

2 1

= 211

2 S10 = 2046

The number of ancestors preceeding the

person is 2046

8. Find the sum of all natural numbers lying

between 50 and 500, which are multiple of 5.Solution:

Required numbers are 55, 60, 65, 70, ......., 495

This is an A.P where a = 55, d = 60 55 = 5Now, tn = a + (n 1)d = 495 55 + (n 1) 5 = 495 5(n 1) = 495 55

n 1 =440

5

n 1 = 88 n = 89Here a = 55, l = 495, n = 89

Sn =n

2(a + l) =

89

2(55 + 495) =

89

2 550

Sn = 24475

9. The first term of a G.P is 1. The sum of its

third and fifth terms is 20. Find the

common ratio of the G.P

Solution:

First term of the given G.P is 1.

Let r be the common ratio of this G.P .

t3 = 1r3 1 = r

2and t5 = 1r

(5 1) = r4

t3 + t5 = 20

r2 + r4 = 20 r4 + r2 20 = 0

(r2 + 5)(r2 4) = 0

r25and r

2= 4

r = 2

Common ratio = 2 or 2

10. The first, second and last terms of an A.Pare a, b and 2a. Find the number of terms

in A.P.

Solution:

a, b, ......., 2a are in A.P

Since, tn = a + (n 1)d 2a = a + (n 1)(b a)

2a a = (n 1)(b a) a = (n 1)(b a)

n 1 =a

b a

n =a

b a+ 1 =

a b a

b a

+

n =b

b a

11. Find the sum of all three digit numbers

which leave the remainder 3 when divided

by 5.

Solution:

The sequence is 103, 108, 113, ., 998

The sequence is an A.P with a = 103, d = 5, tn = 998

tn = a + (n 1)d 998 = 103 + (n 1)5 998 = 103 + 5n 5 5n = 900 n = 180

Sn =n

2[2a + (n 1)d]

=180

2[2 103 + (180 1)5]

= 90[206 + 179 5] = 90 1101

Sn = 99090

12. First term of a G.P isx2. If 10

thterm of the

sequence isx20

then find 2nd

term.

Solution:

Here a =x2, t10 =x

20

We know that,

tn = arn 1

t10 =x2 r10 1

t10 =x2 r9 =x20

r9 =20

2

x

x

r9 =x18

r =18

9x =x2

t2 = ar2 1 =x2 (x2)1

t2 =x4


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13. If for on A.P, Sn = 0.02 (2n 1), find tn.

Solution:

Sn = 0.02 (2n 1)

We know that,

tn = Sn Sn 1tn = 0.02(2

n 1) [0.02 (2n 11)]

= 0.02 (2n 1) [0.02 (2n 21 1)]

= 0.02 2n 0.02 n.02 2

20 + 0.02

= 0.02 2n 0.01 2n

tn = 2n

(0.01)

14. The sum of the first ten terms of an A.P is

three times the sum of the first five terms,

then find ratio of the first term to the

common difference.

Solution:

We know that,

Sn = n2

[2a + (n 1)d]

S10 =10

2[2a + (10 1)d]

S10 = 5(2a + 9d)

S5 =5

2[2a + (5 1)d]

S5 =5

2[2a + 4d]

According to given condition,

S10 = 3 S5

5(2a + 9d) = 3 52

(2a + 4d)

10(2a + 9d) = 15(2a + 4d)

2(2a + 9d) = 3(2a + 4d)

4a + 18d = 6a + 12d

2a = 6d

a

d= 3

3 Marks Questions

15. For a G.P if a = 5 and t7 =

1

3125 ,

find r and t9.

Solution:

a = 5, t7 =1

3125

tn = arn 1

t7 = ar6

1

3125= 5r6 r6 =

1

3125 5

r6 =5

1

5 5 r6 =

61

5

r =1

5

t9 = ar8= 5

81

5

=7

1

5

t9=

1

78125

16. How many terms of an A.P 6,11

2, 5,

are needed to give the sum 25? Explain the

reason for getting two answer.

Solution:

Here, a = 6, d =1

2

Let 25 be the sum of n terms of this A.P

Where n N Sn = 25

Sn =n

2[2a + (n 1)d]

25 = ( )n 1

2 6 n 12 2

+

25 =n n 25

2 2

100 = n2 25n n2 25n + 100 = 0

(n

5)(n

20) = 0 n 5 = 0 or n 20 = 0

n = 5 or n = 20

Both the values of n are natural numbers

and hence admissible.

17. Find the geometric mean of ( )9 0 + 3

and ( )90 3 .Solution:

Let G be the geometric mean.

Letx = 90 + 3 andy = 90 3

G = xy

= ( )( )90 3 90 3+ = ( ) ( )2 2

90 3

= 90 9 = 81

G = 9

90 + 3 and 90 3 are positive.

The geometric mean of 90 + 3 and

90 3 is 9


9/18



18. Insert three numbers between l and 625 so

that the resulting sequence is a G.P.

Solution:

Let the required numbers be t2, t.3, t4...

Then 1, t2, t3, t4, 625 are in G.P.

Here, a = 1, t5 = 625

t5 = 625 = ar5 1 [ tn = ar

n 1]

1 r4 = 625 r4 = 625 r = 5When r = 5

t2 = r = 5, t3 = r2 = 25, t4 = r

3 = 125

When r = 5 t2 = r = 5, t3 = r

2 = 25, t4 = r3 = 125

The required numbers are (5, 25, 125) or

( 5, 25, 125)

19. Divide 45 into three terms which are in A.P

in such a way that the product of the lasttwo terms is 256.

Solution:

Let the three terms in A.P be a d, a, a + d. a d + a + a + d = 45 3a = 45 a = 15 The three terms are 15 d, 15, 15 + d

It is given that the product of last two terms is

255.

a (a + d) = 255

a + d =255

a =

255

15

a + d = 17 15 + d = 17 d = 2

Three terms are 13, 15 and 17

20. Find two positive number x and y whose

A.M and G.M are 25 and 15 respectively.

Solution:

SinceA.M =2

+x y, G.M = xy


2

+x y= 25 xy = 15

x +y = 50 xy = 225

(xy) = ( )2

4+ x y xy

= ( )2

50 4 225

= 2500 900 = 1600

xy = 40

x +y = 50 xy = 40Case 1: x +y = 50,xy = 40

x = 45,y = 5

Case 2: x +y = 50,xy = 40x = 5,y = 45

The numbers are (x = 45,y = 5) or

(x = 5,y = 45)

21. If the pth

term of an A.P is q and the qth

term is p, prove that nth

term is (p + q n).

Solution:

Let a be the first term and d be the common

difference of the given A.P.

tp = q .(given)

a + (p 1)d = q .(i)

and tq = p .(given)

a + (q 1)d = p .(ii)

Subtracting (ii) from (i), we get

(a a) + (p 1)d d(q 1) = q p

d(p 1 q + 1) = q p

d(p q) = (p q)

d = 1

Substituting value of d in equation (i), we get

a + (p 1) ( 1) = q

a p + 1 = q

a = p + q 1

tn = a + (n 1)d

= (p + q 1) + (n 1) (1)

= p + q 1 n + 1

tn = p + q n

4 Marks Questions

22. A man repays a loan of 3250 by paying

305 in the first month and then decreases

the payment by 15 every month. How

long will it take to clear his loan?

Solution:

Here, a = 305, d = 15, Sn = 3250

Let time required to clear the loan be n months.

Sn =n

2[2a + (n 1)d]

3250 =n

2[2 305 + (n 1)(15)]

6500 = n[610 15n + 15]

6500 = n[625 15n]


10/18



6500 = 625n 15n2

15n2 625n + 6500 = 0

3n2 125n + 1300 = 0

3n2 60n 65n + 1300 = 0

3n(n 20) 65(n 20) = 0 n 20 = 0 or 3n 65 = 0

n = 20 or n = 653

Since n is a natural number.

n 65

3

n = 20

The time required to clear the loan is

20 months

23. If1

x y

,1

2y

,1

zy

are the three

consecutive terms of an A.P, prove that

x,y and z are the three consecutive terms of

a G.P.

Solution:

1

+x y,

1

2y,

1

z+yare in A.P.

1

2y

1

+x y=

1

z+y

1

2y

( )

2

2

+

+

x y y

y x y=

( )

2 z

2

y y

y y + z

( )2

x y

y x y

+=

( )

z

2 z

y

y y

+

+

x y

x y=

z

z

y

y

+

By componendo-dividendo

( ) ( )

( ) ( )

+ +

+

x y x y

x y x y=

( ) ( )

( ) ( )

z z

z z

y y

y y

+ +

+

2

2

x

y=

2

2z

y

x

y =

z

y

x

y=

z

y

By Invertendo

y

x=

z

y

x, y, z are in G.P

24. If in an A.P the sum of m terms is equal to

n and the sum of n terms is equal to m, then

show that sum of (m + n) terms is (m + n).

Solution:

Let a be the first term and d be the common

difference of A.P.

According to the given condition,

Sm = n

n =m

2[2a + (m 1)d]

2n = m[2a + md d] 2n = 2am + m2d md .(i)Also, Sn = m

m =n

2[2a + (n 1)d]

n[2a + nd d] = 2m 2an + n2d nd = 2m .(ii)

Subtracting (ii) from (i), we get

2am 2an + m2

d n2

d md + nd = 2n 2m 2a(m n) + d(m2 n2) d(m n) = 2(n m) (m n)[2a + (m + n)d d] = 2(m n) (m n)[2a + (m + n 1)d] = 2(m n) [2a + (m + n 1)d] = 2 .(iii)

Sm + n =m n

2

+[2a + (m + n 1)d]

=m n

2

+ (2) .[From (iii)]

Sm+ n = (m + n)

25. A contract on construction job specifies a

penalty for delay of completion beyond a

certain limit as follows:

200 for first day,

250 for second day,

300 for third day, etc.

If the contractor pays 27,750 as penalty,

find the numbers of days for which the

construction work is delayed.

Solution:

Here, a = 200, d = 50, Sn = 27750

Sn = n2

[2a + (n 1)d]

27750 =n

2[2 200 + (n 1)50]

27750 =n

2[350 + 50n]

27750 = 175n + 25n2 25n2 + 175n 27750 = 0 n2 + 7n 1110 = 0


11/18



Comparing it with ax2

+ bx + c = 0, we get

a = 1, b = 7, c = 1110

= b2 4ac

= (7)2 4 1 (1110)

= 4489

n =2

b b 4ac

2a

=7 4489

2

=7 67

2

n = 30 or n = 37

But n 37 as number of days cannot be

negative.

Number of days are 30

26. The interior angles of a polygon are in

arithmetic progression. The smallest angle

is 52 and the common difference is 8 . Find

the number of sides of the polygon.

Solution:

Let n be the sides of the polygon.

Sum of all interior angles of polygon =

(n 2) 180

Here, a = 52, d = 8

Sn =n

2

[2a + (n 1)d]

(n 2) 180 =n

2[2 52 + (n 1)8]

180n 360 =n

2[104 + 8n 8]

180n 360 =n

2[96 + 8n]

360n 720 = 96n + 8n2

8n2 + 96n 360n + 720 = 0

8n2 264n + 720 = 0

n2 33n + 90 = 0

(n 30)(n 3) = 0

n 30 = 0 or n 3 = 0

n = 30 or n = 3

But when n = 30, the last angle is

52 + (30 1)8 = 284, which is not possible asinterior angle of a polygon cannot be more than

180.

Number of sides of given polygon is 3

27. A man set out on a cycle ride of 50 km. He

covers 5 km in the first hour and during

each successive hour his speed falls by

1

4km/hr. How many hours will he take to

finish his ride?

Solution:

Here, a = 5, Sn = 50, d =1

4

Let number of hours required to finish the ride be

n.

Sn =n

2[2a + (n 1)d]

50 =n

2

12 5 (n 1)

4

+

50 = n21 n104 4 +

100 = n41 n

4 4

100 = n 41 n

4

400 = 41n n2

n2 41n + 400 = 0

(n 25)(n 16) = 0

n 25 = 0 or n 16 = 0 n = 25 or n = 16

If n = 25, speed would be negative.

n = 16

16 hours are required to finish the ride.

5 Marks Questions

28. The A.M of two positive numbers a and b

(a > b) is twice their G.M prove that

a : b = 2 3 : 2 3 Solution:

A.M = 2 G.M

a + b

2= 2 ab

a + b = 4 ab

(a + b)2 = 16ab a2 + 2ab + b2 = 16ab a214ab + b2 = 0


12/18



Dividing throughout by b2, we get

2

a

b

14a

b+ 1 = 0

Let a : b =a

b= x

x 2 14x + 1 = 0

x =( ) ( )14 196 4 1 1

2

=14 192

2

=

14 64 3

2

=14 8 3

2

= 7 4 3

a

b= 7 4 3

Since a > b,a

b 7 4 3

ab

= 7 + 4 3

Now, (2 + 3 ) : ( 2 3 )

=2 3

2 3

+

=(2 3)(2 3)

(2 3)(2 3)

+ +

+

=2 2

(2) 2(2) 3 ( 3)

4 3

+ +

= 4 4 3 31

+ +

= 7 + 4 3

a : b =( )2 3 : ( )2 3

29. Prove that the sequence Sn = 2n2

+ 5n is in

A.P. Hence find tn.

Solution:

Sn = 2n2 + 5n

For n = 1,

S1 = 2(1)

2

+ 5(1) = 2 + 5 = 7S1 = 7 = t1

For n = 2,

S2 = t1 + t2 = 2(2)2 + 5(2)

= 2 4 + 10

= 18

S2 = 18

t2 = S2 S1 = 18 7 = 11

For n = 3,

S3 = t1 + t2 + t3 = 2(3)2 + 5(3)

= 2 9 + 15

= 33

S3 = 33

t3 = S3 S2 = 33 18 = 15

Now,

t2 t1 = 11 7 = 4

t3 t2 = 15 11 = 4

The sequence is an A.P

Here a = t1 = S1 = 7 and d = 4

tn = a + (n 1)d

= 7 + (n 1)4

= 7 + 4n 4

tn = 4n + 3

30. If the ratio of the sum of m terms and n

terms of an A.P be m2 : n2, prove that the

ratio of mth

and nth

terms is (2m 1) :

(2n 1).

Solution:

It is given that m

n

S

S=

2

2

m

n

( )[ ]

( )[ ]

m2a m 1 d

2

n2a n 1 d

2

+

+

=

2

2

m

n

( )

( )

2a m 1 d

2a n 1 d

+

+ =

m

n

2a + (m 1)d = km .(i)

2a + (n 1)d = kn .(ii)

Subtracting equation (ii) from (i),

(m 1)d d(n 1) = km kn

md d nd + d = k(m n)

d(m n) = k(m n) d = k

Substituting value of d in equation (i),

2a + (m 1)k = km

2a + mk k = km

2a = k

a =k

2

[k is constant]


13/18



Now, tm = a + (m 1)d and tn = a + (n 1)d

m

n

t

t=

( )

( )

a m 1 d

a n 1 d

+

+

=

k(m 1)k

2

k(n 1)k

2

+

+

=

kmk k

2

knk k

2

+

+

=

1k m 1

2

1k n 1

2

+

+

=

2m 1

2

2n 1

2

m

n

t

t=

2m 1

2n 1

31. Find the number of terms common to the

two A.Ps 3, 7, 11, ., 407 and 2, 9, 16, .,

709.

Solution:

Let the number of terms in the two A.Ps be m and n

respectively.

tm = 407 and tn = 709

As tm = a + (m 1)d

407 = 3 + (m 1)4

407 = 3 + 4m 4

407 = 4m 1

408 = 4m m = 102

tn = a + (n 1)d

709 = 2 + (n 1)7

709 = 2 + 7n 7

709 = 7n 5

714 = 7n

n = 102

Each A.P consists of 102 terms.

Let pth

term of 1st

A.P be equal to qth

term of the 2nd

A.P.

3 + (p 1)4 = 2 + (q 1)7

3 + 4p 4 = 2 + 7q 7

4p 1 = 7q 5

4p + 4 = 7q

4(p + 1) = 7q

p 1

7

+=

q

4

Letp 1

7

+=

q

4= k (k 0)

p = 7k 1 and q = 4kSince each A.P consists of 102 terms.

p 102 and q 102 7k 1 102 and 4k 102

k5

14 7 and k1

25 2

k 14

k = 1, 2, 3, 4, , 14

For each value of k, we get a pair of

identical terms. Hence there are 14

identical terms in two A.Ps .

___________________________________________

32. The sum of three numbers in G.P is 56. If

we subtract 1, 7, 21 from these numbers in

the same order, we obtain an arithmeticprogression. Find the numbers.

Solution:

Let the three numbers be a, ar, ar2.

a + ar + ar2 = 56 .(i)


a 1, ar 7, ar2 21 are in A.P.

ar 7 =2a 1 ar 21

2

+

2(ar 7) = a + ar2 22 2ar 14 = a + ar2 22

2ar = a + ar2 8 a + ar2 = 2ar + 8Since a + ar

2= 56 ar .[From (i)]

56 ar = 2ar + 8

3ar = 48

ar = 16

r =16

a

Substituting the value of r in equation (i), we get

a + 16 +256

a

= 56

a2 + 16a + 256 = 56a a2 40a + 256 = 0 (a 32)(a 8) = 0 a = 32 or a = 8When a = 8, r = 2

When a = 32, r =16

32=

1

2

The numbers are 8, 16, 32 or 32, 16, 8


14/18

TARGET Publications Std. X : HOTS (Science & Technology)

149School of Elements

01 School of Elements1 Marks Questions

1. Check whether following set of elements

follow Dobereiners triad.Calcium (40.1), Strontium (87.6), Barium

(137.3)

Ans: Condition for Dobereiners triad

Atomic mass of strontium = Arithmetic means

of atomic masses of calcium and barium

=40.1 137.3

2

+

87.6 88.7

The given set follows Dobereiners triad.

2. On what basis can metallic and non-

metallic properties of elements judged?Ans: Electronegativity

3. Which period is considered as an

incomplete period?

Ans: The seventh period is considered as an

incomplete period because there are still

vacant spaces for the elements to be

discovered or artificially prepared.

4. In the following set of elements, one element

does not belong to the set. Select this

element and explain reason for it.

Oxygen, Nitrogen, Carbon, Chlorine,Fluorine, Boron

Ans: Chlorine does not belong to the set. This isbecause Oxygen, Nitrogen, Carbon, Fluorine

and Boron are the 2nd

period elements whereas

Chlorine belongs to the 3rd

period.

5. Out of the following which element has

strongest metallic character?

Al, P, S, Cl

Ans: Al has strongest metallic character.

6. Give three examples of lanthanide elements.

Ans: Lanthanum, cerium, lutetium

2 Marks Questions

7. State with example the limitations of

Newlands law of octaves.

Ans: Newland arranged the element only upto

calcium. After calcium every eighth element

did not possess properties similar to that of the

first.

In order to fit the existing elements Newland

placed two elements in the same position

which differed in the properties.8. In the modern periodic table:

i. How many periods and groups are

present there?

ii. Identify the shortest and longest

period.

Ans: i. The horizontal rows in the modern

periodic table are called as periods and

the vertical columns are called as

groups. The modern periodic table

consists of seven periods and eighteen

groups.

ii. The first period is the shortest period

containing only 2 elements. The sixth

period is the longest and it contains 32

elements in it.

9. On moving from left to right in a period of

the periodic table how does the following

properties of elements change?

i. Atomic size

ii. Metallic character

iii. Non-metallic character

iv. No. of valence electrons

Ans: On moving from left to right in a period of theperiodic table

i. Atomic size decreases

ii. Metallic character decreases

iii. Non-metallic character increases

iv. Number of valence electrons increases

10. On moving from top to bottom in a group

in the periodic table how does the following

properties change?

i. Metallic character

ii. Non-metallic characteriii. Atomic size

iv. No. of valence electrons

Ans: On moving from top to bottom in a group of

the periodic table

i. Metallic character of elements increases.

ii. Non-metallic character decreases.

iii. Size of atoms increases

iv. All elements will have the same number

of valence electrons.


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TARGET PublicationsStd. X : HOTS (Science & Technology)

School of Elements150

11. Why is group 18 of modern periodic table

also known as group zero?

Ans: The outermost shells of the elements belonging

to group 18 contain eight electrons (except

helium), so they are already completely filled

with electrons. They have no tendency to lose

or gain electrons and hence have a stable

electronic configuration with zero valency.Therefore, group 18 of modern periodic table is

also known as group zero.

12. The atomic numbers of five elements A, B,

C and D are given below.

Element Atomic number

A 3

B 5

C 7

D 9

E 10

i. Which element belongs to group 18?

ii. Which element belongs to group 15?iii. Which element belongs to halogen

group?

iv. Which element belongs to alkali metals?

Ans: i. E ii. C

iii. D iv. A

13. From the given electronic configuration,

identify the element and give its position in

periodic modern table.

i. (2, 7)

ii. (2, 8, 8)

Ans: i. (2, 7)It is flourine

It is placed at 17th

group and period 2.

ii. (2, 8, 8)

It is Argon

It is placed at 18th group and period 3.

3 Marks Questions

14. P and Q are the two elements having

similar properties which obey Newlands

law of octaves.

i. How many elements are there in

between P and Q?ii. State Newlands law.

iii. Why is it known as Newlands law of

Octaves?

Ans: i. There are six elements between P and Q.

ii. Newland law states that When the

elements are arranged in an increasing

order of their atomic masses, the

properties of the 8th

element are similar

to the properties of the 1st element.

iii. The repetition in the properties of the

element is just like the repetition of

eighth note in an octave of music, so it

is known as the Newlands law of

octaves.

15. The Physical and chemical properties of

elements are a periodic function of their

atomic masses.

i. Name the law.

ii. What do you understand by this

statement?

iii. How does it differ from the modern

periodic law?

Ans: i. This law is known as Mendeleevsperiodic law.

ii. Mendeleevs periodic law means that if

elements are arranged in the order of

increasing atomic masses, then the

properties of the elements are repeatedafter certain intervals or periods.

iii. It differs in the following way - The

Modern periodic law states that The

chemical and physical properties of

elements are a periodic function of their

atomic numbers.

16. The three elements predicted by Mendeleev

from the gaps in his periodic table were

known as eka-boron, eka-aluminium and

eka-silicon.

What names were given to these elementswhen they were discovered later?

Ans: Eka-boron was named as Scandium

Eka-aluminium was named as Gallium

Eka-silicon was named as Germanium

17. In Period 2 of the modern periodic table,

name the following:

i. an alkali metal

ii. a noble gas with 2 shells

iii. a non-metal with valency 4

iv. a halogenv. an alkaline earth metal

vi. a metalloid

Ans: i. Lithium

ii. Neon

iii. Carbon

iv. Fluorine

v. Beryllium

vi. Boron


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18. Element A belongs to the 3rd period and group 15 of periodic table. State the following:

i. Name of the element.

ii. Atomic number

iii. Number of valence electrons

iv. Valency

v. Nature of the element (Metal/ Non-metal)

vi. Electronic configuration

Ans: i. Phosphorousii. 15

iii. 5

iv. 3

v. Non-metal

vi. Electronic configuration (2, 8, 5)

5 Marks Question

19. The following table shows the position of nine elements in the periodic table. The elements are shown

by the letters which are not their usual symbols. Using the following table answer the following

questions:

Groups

Periods

1 2 3 to 12 13 14 15 16 17 18

Period 1A

Period 2 B C D E F

Period 3 G H I J

Period 4 K L M

i. Which of these elements is a halogen?

ii. Which of these elements are metals?

iii. Which element is a metal with valency 2?

iv. Element D is a (metal/ metalloid/ non-metal)

v. Which of these elements are metalloids?

vi. Which element is a non-metal with valency 3?

vii. Write a common group name of elements A and F.

viii. Out of G and J which one has a larger atomic radius and why?

Ans: i. Mii. B, G, H, K, L

iii. K

iv. Non-metal

v. C and I

vi. J

vii. Inert gases or Noble gases

viii. G has a larger radius because on moving from left to right in a period, the size of the atoms

decreases. Hence atomic radius will also decrease.


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TARGET PublicationsStd. X : HOTS (Science & Technology)

School of Elements152

20. In the periodic table given below, Hydrogen, Calcium, Boron, Nitrogen, Oxygen and Neon are placed

in their correct positions and the positions of thirteen other elements are represented by letters. These

letters are not their usual symbols for the elements.

Groups

Periods

1 2 13 14 15 16 17 18

Period 1 Hydrogen

Period 2 A Boron X Nitrogen Oxygen P Neon

Period 3 B Y L Q

Period 4 C Calcium Z M R

Period 5 D S

With the help of the table, answer the following questions:

i. Give the letter which has maximum metallic character.

ii. Give the letter which has maximum non-metallic character.

iii. Name the family of elements represented by P, Q, R, and S.

iv. Name the elements A, X, Q and S.

v. Write the valency of letters (a) X, Y and Z (b) L and M (c) P, Q, R and S

vi. Which element has a larger atomic radius out of (a) A and D (b) A and P?

vii. Give the common name of the group 1 and group 2.

Ans: i. D

ii. P

iii. Halogen

iv. A - Lithium, X - Carbon, Q - Chlorine, S - Iodine

v. a. Valency of X,Y and Z is 4.

b. Valency of L and M is 2.

c. Valency of P, Q, R and S is 1.

vi. a. D has larger radius (out of A and D).

b. A has larger radius (out of A and P).

vii. Group 1 is known as alkali metals.

Group 2 is known as alkaline earth metals.

21. The electronic configuration of an element X is 2, 8, 1 and element Y is 2, 8, 18, 8, 1.

i. Identify the group of X and Y.

ii. What is the period number of Y?

iii. How many valence electrons are there in the atoms of X and Y?

iv. What is the valency of X and Y?

v. State whether X and Y are metals or non-metals.

Ans: i. Group of X and Y is IA.

ii. X belongs to 3rd

period and Y belongs to 5th

period.

iii. Valence electrons of X and Y is 1.

iv. Valency of X and Y is 1.

v. X and Y are alkali metals.


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22. Fill in the following table:

GroupGroup

IA

Group

IIA

Group

IIIA

Group

IVA

Group

VA

Group

VIA

Group

VIIA

Group

0

Element Na Mg Al Si P S Cl Ar

Atomic number

Electronic

configuration

Metallic property

Valency

Ans:

GroupGroup

IA

Group

IIA

Group

IIIA

Group

IVA

Group

VA

Group

VIA

Group

VIIAGroup 0

Element Na Mg Al Si P S Cl Ar

Atomic

number11 12 13 14 15 16 17 18

Electronic

configuration2, 8, 1 2, 8, 2 2, 8, 3 2, 8, 4 2, 8, 5 2, 8, 6 2, 8, 7 2, 8, 8

Metallic

propertyMetal Metal Metal Metalloid Non-Metal Non-Metal Non- Metal Noble gases

Valency 1 2 3 4 3 2 1 0

23. Identify the main characteristic of the elements.

i. Element A has completely filled outermost shellii. Element B is in 4th group and has period 4

iii. Element C is a s-block element.

iv. Element D is placed in 3rd group and having period 7.

v. Element E is in group 1.

Ans: i. It is an inert gas.

ii. It is titanium with 4 electrons in the outermost shell.

iii. It is a metal.

iv. It is an actinides.

v. It is alkali metal.

ssc mathematice

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