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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
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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
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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]
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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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TARGET Publications Std. X : HOTS (Algebra)
3Arithmetic Progression and Geometric Progression
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
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TARGET PublicationsStd. X : HOTS (Algebra)
Arithmetic Progression and Geometric Progression4
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
According to given condition,
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]
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TARGET Publications Std. X : HOTS (Algebra)
5Arithmetic Progression and Geometric Progression
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
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TARGET PublicationsStd. X : HOTS (Algebra)
Arithmetic Progression and Geometric Progression6
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
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TARGET Publications Std. X : HOTS (Algebra)
7Arithmetic Progression and Geometric Progression
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]
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TARGET PublicationsStd. X : HOTS (Algebra)
Arithmetic Progression and Geometric Progression8
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)
According to given condition,
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
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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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TARGET Publications Std. X : HOTS (Science & Technology)
151School of Elements
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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TARGET Publications Std. X : HOTS (Science & Technology)
153School of Elements
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.