sample question paper for class x – mathematics &...

Ganga Academy [Sample Question Paper for Class X – Mathematics & Science ]

CBSE PATTERN SPECIALIST SINCE 2002 CBSE BOARD SAMPLE PAPERS 2012- 13 2012:

20012 RESULTS X & XII CBSECLASS X1) ASMITA A GAIND ENGLISH COMM.: 10 , MATHEMATICS : 09 SCIENCE: 10 CGPA 9.82)RAJ HANMANT KATKAR ENGLISH COMM.: 10 MATHEMATICS : 09 SCIENCE :10 CGPA 9.8 3)Hemang PancholiMATHEMATICS: 10

CLASS XII (Sci)1) KIRAN ASHOK SALUNKE PHYSICS : 088 ; CHEMISTRY :086 ; BIOLOGY: 090 Total:85.2%2) PRABODHINI .B. BHUMKAR MATHEMATICS : 077 ; PHYSICS : 094 ;CHEMISTRY : 084 Total:84.8%

3)INDERJEET SINGH PHYSICS: 84 ;Total:83.2%

OVER ALL 100% Result ! Thax. to all dedicated teacherSYLLABUS MATHEMATICS X SAI I. NUMBER SYSTEMS 11 ;II. ALGEBRA 23 ; III. G EOMETRY 17 ;IV TRIGONOMETRY 22 ; V STATISTICS 17 TOTAL 90

BY;CBSE EXPERT AJAY KUMAR MISHRA

Note : This booklet contains mathematics Important formulae on next page.

Algebra

SOLUTIONS MATHEMATICS 2012-131 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104

Sides )

SUMMATIVE ASESSMENT- I CLASS X


SAMPLE PAPER –I

Solutions:1]

One decimal place2]

An odd integer3]

Since -3 is the root of quadratic polynomial, we have

(K - 1)(-3)2 + 1 = 0 4]

5]Median

6]x = 3 sec2 - 1, y = tan2 - 2x - 3y = 3sec2 - 1 - 3tan2 + 6= 3(sec2 - tan2 ) + 5= 3 + 5= 8

7]cos + cos2 = 1 1 - cos2 = cos

sin2 + sin4 = + (1 - cos2 )2

= cos + cos2

= 18]

2 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir : 9766607104


9]Since, 870 = 225 3 + 195225 = 195 1 + 30195 = 30 6 + 1530 = 15 2 + 0

HCF (870,225) = 1510]

x + y = 3 -x + y = 1

OR

Subtracting equation (1) from 3x - = 5, we get

From equation (1), x = 311]

are roots of x2 - (k + 6)x + 2(2k - 1)

12]

From



1 + 3 = 1 + 2 3 = 2

13]

Cot = (given)

= =

= 14]

C.I f c.f.

135 - 140 4 4

140 - 145 7 11

145 - 150 11 22

150 - 155 6 28

155 - 160 7 35

160 - 165 5 40

Here, n = 40 Median class is 145 - 150



Also, since highest frequency is 11, Modal class is 145 - 150.15]

To prove 5 + is irrational, let us assume 5 + is rational.

We can find integers a and b where a, b are co-prime, b

Such that,

Now a, b are integers, is rational.

is rational.

Which is a contradiction. So 5 + is irrational.OR

Let us assume to the contrary, that is a rational number.

is rational.

(n-1)+(n+1)-2 is rational

2n+2 is rational

But we know that is an irrational number

So 2n+2 is also an irrational numberSo our basic assumption that the given number is rational is wrong.

Hence, is an irrational number.16]

f(x) = x2 - 2x + 1Zeroes of f(x) are , Sum of zeroes + = 2 and . = 1



Required polynomial = k(x2 - 4x + 4), where k is any integer.17]

Let the length and breadth of the rectangle be x and y respectively.So the original area of the rectangle=xyAccording to question,(x+2)(y-2)=xy-28i.e. xy-2x+2y-4=xy-282x-2y=24 ...(i)Next, (x-1)(y+2)=xy+33i.e. xy+2x-y-2=xy+332x-y=35 ..(ii)Now we need to solve (i) and (ii)From (ii) we get,y=2x-35substituting this value in (i) we get,2x-4x+70=24-2x=-46x=23substituting this value in (ii)we get,y=11So the length and breadth of the rectangle are 23 metres and 11 metres respectively.OR



Let 40 % acids in the solution be x litresLet 60 % of other solution be y litresTotal Volume in the mixture = x + yGiven volume is 10 litresx + y = 10 ---(i)

Also, So, 40 x + 60 y =500 or 2x+3y =25 …(ii)Solving (i) and (ii) we get x = y = 5 litres

18]The system has infinitely many solution

Equating (1) and (2), we get a = 5bEquating (2) and (3), we get 2a - 4b = 6 On solving, we get b = 1 and a = 5.

19]By BPT



20]

To prove AB² = AC² = 2AD² +

Draw AE BC

In ?ABD since D > 90°

AB² = AD² + BD² + 2BD x DE …(1) (using Obtuse angle property) ?ACD = since D < 90°

AC² = AD² + DC² - 2DC x DE …(2) (using acute angle property)

Adding (1) and (2)

AB² + AC² = 2(AD² + BD²)

=2(AD² + )

Or AB² + AC² = 2(AD² + BD²)

Hence proved.21]

Given:

L.H.S. = (m² + n²) cos²



OR

22], we have

23]CI 50-60 60-

7070-80

80-90 90-100

100-110

Total

fi 5 3 4 p 2 13 27+pxi 55 65 75 85 95 105fi xi 275 195 300 85p 190 1365 2325+85

p

Mean = Substituting the values we get9 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


86 = 86p + 2322 = 2325 + 85pp = 3

24]Age in yrs. ( more than or equal to)

0-10

10-20

20-30

30-40

40-50

50-60

60-70

No. of persons (fi)

10 15 25 22 13 10 5

Since the maximum frequency is 25 and it lies in the class interval 20-30.Therefore, modal class = 20 - 30??= 20, h = 10, f0 = 15, f1 = 25, f2 = 22

mode = ??+

= 20 + = 20+7.69 = 27.69 years (approx.)

25]If the number 15n where n N, were to end with a zero, then its prime factorisation must have 2 and 5 as its factors.But 15=5 315n = (5 3)n = 5n 3n

So Prime factors of 15n includes only 5 but not 2Also from the Fundamental theorem of Arithmetic, the prime factorisation of a number is unique.Hence a number of the form 15n where n N, will never end with a zero.

26]To solve the equations, make the table corresponding to each equation.



x ?1 ?2 ?3y 4 2 0

x 4 ?1y 0 4

Now plot the points and draw the graph.

Since the lines intersect at the point (?1, 4), so x = ?1 and y = 4 be the solution.Also by observation vertices of triangle formed by lines and x-axis are A (?1, 4),B (?3, 0) and C (4, 0).

27]Let p(x) = x3 - 6x2 - 15x + 80Let say that we subtracted ax + b so that it is exactly divisible by x2 + x - 12s(x) = x3 - 6x2 - 15x + 80 - (ax + b)= x3 - 6x2 - (15 + a)x + (80 - b)



Dividend = Divisor x Quotient + RemainderBut remainder = 0 Dividend = Divisor x Quotients(x) = (x2 + x -12) x quotients(x) = x3 - 6x2 - (15 + a)x + (80 - b)x (x2 + x - 12) - 7(x2 + x - 12)= x3 + x2 - 7x2 - 12x - 7x + 84= x3 - 6x2 - 19x + 84Hence, x3 - 6x2 - 19x + 84 = x3 - 6x2 - (15 + a)x + (80 - b)-15 - a = - 19 a = +4and 80 - b = 84 b = -4Hence if in p(x) we subtracted 4x - 4 then it is exactly divisible byx2 + x -12.

28]AD is the median of ABC since D is mid-point of BC

BD = DC= ....(i)

In right triangle AEB,

...Pythagoras theorem

=( )+

Using Pythagoras theorem for right triangle AED and BE=BD-DE

= +( ....from (i)



AB2= + + -2

-BC x DE +

Hence proved.OR

Given: A right triangle ABC right angled at B.To prove: that AC2 = AB2 + BC2

Construction:Let us draw BD AC (See fig.)

Proof :Now, ADB ABC (Using Theorem:If a perpendicular is drawn from

the vertex of the right angle of a right triangle to the hypotenuse ,then triangles on both sides of the perpendicular are similar to the whole triangle and to each other)

So, (Sides are proportional)Or, AD.AC = AB2

Also, BDC ABC (Theorem)

So, Or, CD. AC = BC2

Adding (1) and (2),



AD. AC + CD. AC = AB2 + BC2

OR, AC (AD + CD) = AB2 + BC2

OR, AC.AC = AB2 + BC2

OR AC2 = AB2 + BC2

Hence Proved.29]

ABC ADE (by AA Similarity)

In right AB2 = BC2 + AC2 (by PT)

AB2 = 52 + 122 = 25 + 144 = 169

AB = 13Subsisting AB = 13 cm, BC = 12 cm and AC = 5 cm in (1) and

Getting DE = and AE = 30]

To prove: Using the identity cosec2 A = 1 + cot2A

L.H.S =



= cosecA + cotA= R.H.S

OR



=RHS31]

LHS =

= sin = RHS32]

Hence, LHS = RHS.33]

We can find frequency distribution table of less than type as following -Daily income(in Rs)(upper class limits)

Cumulative frequency

Less than 120

12

Less than 140

12 + 14 = 26

Less than 160

26 + 8 = 34

Less than 180

34 + 6 = 40



Less than 200

40 + 10 = 50

Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we can draw its ogive as following -

34]Class Interval 0-6 6-12 12-18 18-24 24-30Frequency 4 x 5 y 1Cumulative frequency

4 4+x 9+x 9+x+y

10+x+y

It is given that total frequency N is 20So, 10+x+y = 20 i.e. x + y = 10 ….(i)Given 50% of the observations are greater than 14.4.So median = 14.4, which lies in the class interval 12-18.= 12, cf = 4 + x, h = 6, f = 5, N = 20

Median =

14.4 = 12 + x 6

14.4 - 12 = x6

= 6 - xx =4



Now using equation, 10+x+y = 20, we get y = 6.Hence x =4 and y = 6.

SAMPLE PAPER –IISolutions:1]

tan B tan c = =12]

Number of zeros is one as the graph touches the x-axis at one point.3]

3Median = Mode + 2 mean

Median = = = 4]

Since

Therefore, lines are parallel.5]

BC =



6]

Therefore, PN = 4.8 cm

7]

8]

Let Then, x=0.7777…… … (1) Here ,the number of digits recurring is only 1,so we multiply both sides of the equation by 10.

… (2)Subtracting(1) from(2),we get

9x=7

9]

-3 and -4 are zeroes of the polynomial



Sum of zeros = -3 - 4 = -7=

Product of zeros = (-3) (-4) = = 10]

Let it possible 6n ends with digit 0 6n = 10 q

(2 3)n = 2 5 q2n 3n = 2 5 q5 is a prime factor of 2n 3n

Which is not possible 2n 3n can have only 2 and 3 are prime factors. Hence, it is not possible the number ends with digit 5.

11]Marks f cf

0 - 10 5 5

10 - 30 15 20

30 - 60 30 50

60 - 80 8 58

80 - 100 2 6

N=Here, N=60 So, N/2=30The cumulative frequency is just greater than N/2=30 is 50 and the corresponding class is 30-60.Hence,30-60 is the median class.Therefore, l=30,f=30 ,F=20,h=30

Now, Median=Median = 40

12]Condition for infinitely many solution



(½)

similarly, k + 2 = 4 3k + 2 = 8k = 2 k = 2

k = 2 is the common solution.13]

Sin(45 + 30) = sin 45o cos 30o + cos 45o sin 30o

Sin 75o =

=

= OR

Cosec =

AB =



Cot + tan =

=

= 14]

In ABD and ECF

D = F = 90o

B = C

By AA similarity, ABD ECF15]

Let Rekha's Age be 'x'yearsAnd her mother's age be 'y' yearsy = 5x as per given data … (1)After 5 yearsy + 5 = 3(x+5)y - 3x = 10 … (2)Solving (1) and (2) equation.Rekha's age = 5 yearsMother's age = 25 years

ORLet the two number be 5x, 6x

Two numbers are 40,48.16]

Sin = tan =



17]a=12 a+b=25 b=13 c=25+10=35 c+d=43 d=43-35=8 43+e=48 e=5f=48+2=50

18]To calculate the mean, first obtain the column of mid value and then multiply the corresponding values of frequency and mid value.

C.I. f Mid value

(x)

fx

0-100 2 50 100100-200

3 150 450

200-300

5 250 1250

300-400

2 350 700

400-500

3 450 1350

15 3850

Here and , so the mean is given as

.



19]

=

=

=

Simplifying we get, 1 + sec . cosec OR

20]ADE ABC by AA similarity

…(1)

21]Given p(x) = 6x3 + 3x2 - 5x + 1



a = 6, b = 3, c = -5 d = 1

,r are zero.

= = 22]

Let be rational and equal to

then where a and b are co primes,

here a,b are integers

is rational .Therefore, is rational is rational which is a contradiction

6 + is an irrational numberOR

Let 5 - be rational equal to

Then 5 - =

a,b are integers

is rational which is a contradiction



Hence 5 - is an irrational number23]

In fig, PQD,

Using Pythagoras thm.PD2 = a2 - c2 …(1)

Similarly in PD2 = b2 - d2 …(2)From (1) and (2) a2 - c2 = b2 - d2

a2 - b2 = c2 - d2

(a + b) (a - b) = (c + d) (c - d)24]

Multiply equation (1) by 3 and add in equation (2), we get

Using equation (1),



y - 2 = 3 y = 5Hence x = 4, y = 5.

25]Let 5q + 2, 5q + 3 be any positive integers(5q + 2)2 = 25q2 + 20q + 4= 5q (5q + 4) + 4 is not of the form 5q + 2Similarly for 2nd

(5q + 3)2 = 25q2 + 30q + 9=5q(5q+6)+ 9 is not of the form 5q+3So, the square of any positive integer cannot be of the form5q+2 or 5q+3For any integer q

26]Statement If a line is drawn parallel of one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ration.

Given:A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see fig.)

To prove that Construction:Let us join BE and CD and then draw DM AC and EN

AB.

Proof:Now, area of



Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.So, ar(BDE) = ar(DEG)Therefore, from (1), (2) and (3), we have:

OR

Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ?ABC ~ ?PQR To Prove: Construction: Draw AD?

BC and PS?QR Proof: ?ADB ~ ?

PSQ (AA) Therefore, … (iii) But ?ABC ~ ?

PQR Therefore, … (iv) Therefore,

Therefore, From (iii)27]

2 sec2 - sec4 - 2sec2 + cosec4



= 2 (1 + tan2 ) - (1 + tan2 )2 - 2(1 + cot2 ) + (1 + cot2 )2

= (1 + tan2 ) (1 - tan2 ) - (1 + cot2 ) (1 - cot2 )= 1 - tan4 -1+cot4

= cot4 - tan4

28]We have,x-y=12x+y=8Graph of the equation x-y=1:We have,x-y=1 =>y=x-1 and x=y+1Putting x=0,we get y=-1Putting y=0,we get x=1Thus, we have the following table for the points on the line x-y=1:

x 0 1

y -1 0

Plotting points A(0,-1),B(1,0) on the graph paper and drawing a line passing through them,we obtain the graph of the line represented by the equation x-y=1 as shown in fig.Graph of eqn 2x+y=8:We have,2x+y=8 =>y=8-2xPutting x=0,we get y=8Putting y=0,we get x=4Thus,we have the following table giving two points on the line represented by the equation 2x+y=8.

x 0 4

y 8 0

Plotting points C(0,8) and D(4,0) on the same graph paper and rawing a line passing through them, we obtain the graph of the line represented by the equation 2x+y=8 as shown in fig.



Clearly ,the 2 line intersect at P(3,2).The area bounded by these 2 lines and y-axis is shaded in the given fig.

29]

=

= OR

30]

Marks Frequency



25 - 35 5

35 - 45 10

45 - 55 20

55 - 65 9

65 - 75 6

75 - 85 2

Total 52

Here the maximum frequency is 20 and the corresponding class is 45-55.So,45-55 is the modal class.

We have,l=45,h=10,f=20,

Mode = + Mode=49.7

31]Let p(x) = x3 + 2x2 + kx + 3Then using Remainder theoremp(3) = 33 + 2 32 + 3k + 3 = 21

k = -9

Quotient of p(x) is x2 + 5x + 6Hence, x3 + 2x2 - 9x + 3 = (x2 + 5x + 6) (x - 3) + 21

x3 + 2x2 -9x - 18 = (x - 3) (x + 2) (x + 3)



All the zeros of p(x) are 3,-2,-3.32]

Given: ABC is a right angled triangle, B = 900

To prove: AB2 + BC2 = AC2

Construction: Drop a perpendicular BD on the side AC.

Proof: From triangle ADB and triangle ABC, We can re-write as, AC × AD = AB2

Also, triangle BDC is similar to triangle ABC.Equating the proportional sides of the similar triangles BDC and ABC,

AC × CD = BC2

Now adding this to the equation that we had obtained,AC × AD + AC × CD = AB2 + BC2

AC × (AD + CD) = AB2 + BC2

AC × AC = AB2 + BC2

AC2 = AB2 + BC2

33]

LHS=



= Cos A + sin A = RHS34]

We first prepare the cumulative frequency distribution table by less than method as given below:

Marks no. of students marks less than cumulative frequency

0-10 7 10 7

10-20 10 20 17

20-30 23 30 40

30-40 51 40 91

40-50 6 50 97

50-60 2 60 100

Other than the given class intervals ,we assume a class-10-0 before the first class interval 0-10 with zero frequency.Now, we mark the upper class limits along X-axis on a suitable scale and the cumulative frequencies along Y-axis on a suitable scale.Thus, we plot the points(0,0),(10,7),(20,17),(30,40),(40,91),(50,97)and(60,100).Now, we join the plotted points by a free hand curve to obtain the required ogive.

SAMPLE PAPER –IIISolutions:



1]

2]Zeroes of a polynomial are the x- coordinates of the points where its graph crosses or touches the X- axis. Graph of y = p(x) intersects the X-axis at 4 points. Therefore, the polynomial p(x) has 4 zeroes

3]Range is not a measure of central tendency.

4]7×11×13+13 +13 x 2 = 13(7×11 + 1+2) = 13(80)

5]Smallest composite number= 4 and the smallest prime number=2. HCF of 2 and 4 = 2.

6]For the system of linear equations: a1x + b1y + c1 = 0, and

a2x + b2y + c2 = 0 to have no solution

For the system of equations 3x + y = 1,(2K - 1) x + (K - 1) y = 2K + 1 a1 =3, b1 = 1, c1 = 1, anda2 = 2K-1 b2 = K-1 c2 = 2K+1

This gives K=2.7]

Given that 3cot A=4, we have, cot A=

Then, tan A=

Now, (1-tan2A)= 1- =

And, ( 1+tan2A) = 1+ =



THerefore, 8]

tan A=cot Btan A=tan(900-B)A=900-B or A+B=900

9]In ADC, AD2 = AC2 - CD2

In ABD, AD2 = AB2 - DB2

AB2 - B2 = AC2 - CD2

AB2 + CD2 = AC2 + BD2

10]120 = 23 3 5

144 = 24 32

LCM = 24 32 5 = 720HCF = 23 3 = 24

11]3x + 4y = 104x - 4y = 4

x = 2

x - y = 1 y = x - 1 = 2 - 1 = 1The solution x = 2, y = 1

12]We know that sin2 + cos2 =1 Taking cube of both sides (sin2 + cos2 )3 =13

(sin2 )3 + (cos2)3 + 3 sin2 .cos2 (sin2 + cos2)= 1Therefore, sin6 + cos6 + 3sin2cos2=1

13]Marks No. of students

0 - 1010 - 2020 - 30

276



30 - 4040 - 50

510

Total 3014]

Let p(x) = x3 + ax2 + b x + 16g(x) = x - 2 is a factor of p(x) p(2) = 0so, p(2) = 8 + 4a + 2b + 16 = 0= 4a + 2b = -24= 2a + b = -12 (i)Also a -b = 16 (ii)Solving (i) and (ii)

3a = 4 a = ORGiven polynomial p(x) = ax2 - 5x + c

Sum of zeroes m + n =

Product of zeroes mn = Given: m + n = mn = 10

15]Modal Class = 20-30 ? = 20, f0 = 12, f1 = 32, f2 = 20, h = 10 Mode

=



Mode = = 26.2516]

Construction - draw AE BCIn right triangle AEB and AECAB2 + AC2 = BC2 + AE2 + EC2 + AC2 = 2AE2 + (BD - ED)2 + (ED + DC)2

= 2AE2 + 2ED2 + BD2 + DC2

AB2 + AC2 = 2AE2 + 2ED2 + 2BD2

= 2 [AE2 + ED2] + 2BD2

= 2 (AD2 + BD2)17]

LHS =

= = 2 cosec A = RHS18]

Let the fraction be: According to the questionx + y = 8



On solving we get x = 3, y = 5

The fraction is OR

Let the tens and units digits of the number be x and y respectively then7(10x + y) = 4 (10y + x)

70x + 7y = 40y + 4x

66x = 33y

y = 2xAlso, y - x = 3On solving, x = 3, y = 6Number = 36.

19]

Let be a rational number

Let =

= p11q2 = p2

11 divides p2 hence 11 divides P.Let p = 11c

11q2 = 121 c2

Or q2 = 11c2

11 divides q2 Hence 11 divides qFrom (1) and (2) p and q have a common factor 11 which contradicts our assumption. is irrational

OR

Let 2 - 7 is rational



=

Since p and q are integer is rational

is rational

But we have that is irrational

our assumption is wrong. Hence 2 - 7 is irrational.20]

Class Interval

Frequency

0 - 20 7 7

20 - 40 8 15

40 - 60 12 27

60 - 80 10 37

80 - 100 8 45

100 - 120 5 50

Total 50

Median class 40 - 60

= 40, f = 12 CF = 15 h = 20

Median = +

= 40 +

= 40 +



= 40 + = 56.7OR

Age (in Years)

No. of people fi

Value xi

fixi

0 - 20 15 10 150

20 - 40 f1 30 30f1

40 - 60 21 50 1050

60 - 80 f2 70 70f2

80 - 100 17 90 1530

Total 100 2730 + 30f1 + 70 f2

53 + f1 + f2 = 100f1 + f2 = 47 …(1)

53 = 5300 - 2730 = 30f1 + 70f230f1 + 70f2 = 2570

Or 3f1 + 7f2 = 257 …(2) f1 + f2 = 47

f2 = 47 - 29 = 1821]

In ABC and DACBAC = ADC

C = C



22]

LHS = = 1 + cos A

=

= OR

23]

p(x) = x3 - 3x2 + x + 2g(x) = ?q(x) = x - 2 r(x) = -2x + 4p(x) = g(x) + r(x)x3 - 3x2 + x + 2 = g(x) (x - 2) - 2x + 4

g(x) =

=

= x2 - x + 124]

The system has infinitely many solutions:

Equating (1) and (2), we get a = 5bEquating (2) and (3), we get 2a - 4b = 6On solving, we get b = 1 and a = 5.



25]If a and b are one two positive integers. Then a = bq + r, 0 r b Let b = 3 Therefore, r = 0, 1, 2 Therefore, a = 3q or a = 3q + 1 or a = 3q + 2 If a = 3q a2 = 9q2 = 3(3q2) = 3m or where m = 3q2 a = 3q + 1 a2 = 9q2 + 6q + 1 = 3(3q2+ 2q) + 1 = 3m + 1 where m = 3q2 + 2q or a = 3q + 2 a2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1= 3m + 1, where m 3q2 + 4q + 1Therefore, the squares of any positive integer is either of the form 3m or 3m + 1.

26]CI 1-4 4-7 7-10 10-13 13-16 16-19fi 6 30 40 16 4 4xi 2.5 5.5 8.5 11.5 14.5 17.5

Average most suitable here is the Mode because we are interested in knowing the length of surname for maximum no. of peopleSince the maximum frequency is 40 and it lies in the class interval 7-10.Therefore, modal class = 7-10?= 7, h=3, f0=30 , f1=40 , f2=16

Mode =

= 7+= 7 +.88 = 7.88 years(approx.)

27]

LHS =

= tan? + cot? + 1 = RHSORLHS=(cosec A - sin A) (sec A - cos A) (tan A + cot A)



=

=1= RHS

28]Basic proportionality theoremStatement If a line is drawn parallel of one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given:A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively (see fig.)

To prove that Construction: Let us join BE and CD and then draw DM AC and EN AB.

Proof:

Now, area of Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.So, ar(BDE) = ar(DEG)Therefore, from (1), (2) and (3), we have :

Hence proved.



ORPythagoras Theorem : Statement:In a right angled triangle,the square of the hypotenuse is equal to the sum of squares of the other two sides.Given: A right triangle ABC right angled at B.To prove: that AC2 = AB2 + BC2

Construction:Let us draw BD AC (See fig.)

Proof :Now, ADB ABC (Using Theorem: If a perpendicular is drawn from

the vertex of the right angle of a right triangle to the hypotenuse ,then triangles on both sides of the perpendicular are similar to the whole triangle and to each other)

So, (Sides are proportional)Or, AD.AC = AB2 (1)Also, BDC ABC (Theorem)

So, Or, CD. AC = BC2

Adding (1) and (2),AD. AC + CD. AC = AB2 + BC2



OR, AC2 = AB2 + BC2

Hence proved.



29]m2 - n2 = (m + n) (m - n)

=

= 2 tan . 2 sin

= 4 tan . sin

= 4

= 4

= 4

= 4

= 4 30]

Let p(x) = x3 - 6x2 - 15x + 80Let say that we subtracted ax + b so that it is exactly divisible byx2 + x - 12s(x) = x3 - 6x2 - 15x + 80 - (ax + b)= x3 - 6x2 - (15 + a)x + (80 - b)Dividend = Divisor x Quotient + RemainderBut remainder = 0 Dividend = Divisor x Quotients(x) = (x2 + x -12) x quotients(x) = x3 - 6x2 - (15 + a)x + (80 - b)x (x2 + x - 12) - 7(x2 + x - 12)= x3 + x2 - 7x2 - 12x - 7x + 84= x3 - 6x2 - 19x + 84Hence, x3 - 6x2 - 19x + 84 = x3 - 6x2 - (15 + a)x + (80 - b)-15 - a = - 19 a = +4and 80 - b = 84 b = -4



Hence if in p(x) we subtracted 4x - 4 then it is exactly divisible byx2 + x -12.

31]We have to solve the pair of equations graphically2x + 3y = 11 … (1)2x - 4y = -24 … (2)For (1)

X 1 4 -2y 3 1 5

For (2)

X -12 0 -10y 0 6 1

point of intersection x = -2, = 5The triangle formed is shaded as ABC coordinates areA (-2,5) B (-12,0) C(5.5,0).

32]Class Interval Frequency

0 - 15 6

15 - 30 7

30 - 45 f1



45 - 60 15

60 - 75 10

75 - 90 f2

Total 51

Mode = 55 (Given) Modal Class 45 - 60

= 45, fo = f1 and f1 = 15f2 = 10 h = 1538 + f1 + f2 = 51f1 + f2 = 51 - 38f1 + f2 = 13 …(1)

55 = 45 +

10 = 200 - f1 = 225 - 15f15f1 = 25f1 = 5f1+f2=13

f2=13-5=8

The missing frequencies are 5 and 8.33]




Given: ?ABC ~ ?PQR To Prove: Construction: Draw AD?

BC and PS?QR Proof: ?ADB ~ ?

PSQ (AA) Therefore, … (iii) But ?ABC ~ ?

PQR Therefore, … (iv) Therefore,

Therefore, From (iii)34]

LHS =

RHS =

= tan

Hence, LHS = RHS.SAMPLE PAPER –IV

olutions:1]

Mode = 3 median - 2 mean

2]

3]



4]A rational number can be expressed as a terminating decimal if the denominator has factors 2 or 5.

5]9 cm 5 cm 7cm cannot form the sides of a right triangle as the Pythagoras theorem is not satisfied in this case.

6]For the system of equations: 2x + 3y = 7

4x + 6y = 5we have

Since is the condition for no solution and henceinconsistent system of equations.

7]

Since is a root of the equation (p + q)2 x2 - 2 ( p + q) x + k =0 So, (p +

q)2 - 2 ( p + q ) . + k = 025 - 10 + k = 0k = -15

8]

9]HCF LCM = Product of the number

9 90 = 18 x

x =

10]

Monthly income range(In Rs.)

No. offamilies

10000-13000 5



13000-16000 16

16000-19000 19

19000-22000 17

22000-25000 18

25000-28000 15

No. of families having income range (in Rs.) 16000-19000 is 19. From the graph it is clear that median is 4.

11]

LHS = 1 +

= 1 +

= 1 + cosec - 1 = cosec = = RHS

12]Since A = B, AC = BC … (1)Also AD = BE … (2)Subtracting (2) from (1),

AC - AD = BC - BE

DC = EC … (3)From (2) and (3), we have

Therefore, DE || AB by converse of BPT.

13]One of the zero =- 15

Sum of the zeroes = 42 Other zero = 42 + 15 = 57 Product of the zeroes = 57 -15 = 855

The quadratic polynomial is x2 - 42x - 85550 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


OR

Let p(x) = 2x2 - 4x + 5

?² + ?²= (? + ?)² - 2??

Substituting the values, we get = ?² + ?²= -1

14]For no solution:

3p - 3 = 2p - 1

p = 215]

Modal class - 30 - 40

= 30 fo = 12 fi = 20 f2 = 11 h = 10

Mode = +

= 30 +

= 30 +

16]Construction: Draw PA QR and SB GR



(As one angle is 90 degrees and one is vertically opposite

angles)

From (1) and (2), we get

17]

Let 5 + 7 is rational number

5 + 7 =

7 =

= Since p and q are integers

a rational number

is rational

But we know that is rational Out assumption is wrong

5 + is irrational. OR

Let be a rational number

Let = p/q where q 0, p and q are integers and coprime.

7 q2 = p252 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


7 divides pLet p = 7m7q2 = 49 m2Q2 = 7m2

7 divides q2

7 divides q

7 divides p and q both.

Which is a contradiction for the that p and q are co-prime.18]

LHS =

=

= sin A cos A

Hence, LHS = RHS.

19]

The polynomial p(x) can be divided by the polynomial g(x) as follows:

Quotient = x - 3

Remainder = 7x - 9

20]

Class Interval Fi frequency Mid value xi Fixi



30 - 40 2 35 70

40 - 50 3 45 135

50 - 60 8 55 440

60 - 70 6 65 390

70 - 80 6 75 450

80 - 90 3 85 255

90 - 100 2 95 190

Total 30 1930

Mean = OR

Daily expenses (in Rs) No, of families C.F

20 - 40 6 6

40 - 60 9 15

60 - 80 11 26

80 - 100 14 40

100 - 120 20 60

120 - 140 15 75

140 - 160 10 85

160 - 180 8 93

180 - 200 7 100

Total 100

Median class - 100 - 120f = 20 cf = 40 h = 20 l = 100



Median =

= 100 + = 100 + 10 = 110

21]

Construction: - Draw AE BCIn right triangle AEBAB2 = AE2 + BE2= AE2 + (BD + DE)2= AC2 - DE2 + BD2 + DE2 + 2BD.DE

= AD2 + BC2 + 2 BC BC

= AD2 + BC29AB2 = 9AD2 + 2BC29AB2 - 2AB2 = 9AD2 AB = BCOr 9AD2 = 7AB2

22]Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing upstream = km/h

Speed of Ritu while rowing downstream = km/h



According to the question,

Adding equations (1) and (2), we obtain:

Putting the value of x in equation (1), we obtain:

y = 4

Thus, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

ORLet the speed of train and bus be u km/h and v km/h respectively.

According to the question,

Let and

The given equations reduce to:

Multiplying equation (3) by 10, we obtain:

Subtracting equation (4) from equation (5), we obtain:



Substituting the value of q in equation (3), we obtain:

Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.23]

Given: sin + cos = m and sec +cosec = n

Consider,

L.H.S. = n

=

=

=

24]Let the cost of one T-shirt be Rs x and that of one jacket be Rs y.

According to given condition

2x+y=625 …(i)

3x+2y=1125 …(ii)

Multiplying (i) by 2 we get

4x+2y=1250 …(iii)

Subtracting (ii) from (iii) we get,

x=125

Substituting this value of x in (i) we get

250+y=625 y=375

Therefore cost of one T-shirt is Rs125 and the cost of one jacket is Rs 375.57 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


25]Let a and b be any positive Integers

a = b + r, 0 r <bLet b = 6 Thes r = 0,1,2,3,4,5Where r = 0, a = 6m + 0 = 6m. which is evenWhere r = 1 a = 6m + 1 oddWhere r = 2 a = 6m + 2 evenWhere = 3 a - 6m + 3 oddWhere r = 4 a = 6m + 4 evenWhere = 5 a - 6m + 5 odd

All positive even integers are of the from 6m, 6m + 2 or 6m + 4.26]

We have5 + f1 + 10 + f2 + 7 + 8 = 50f1 + f2 = 20f1 = 20 - f2

C.I fi Xi fixi

0 - 20 5 10 50

20 - 40 fi 30 30fi

40 - 60 10 50 500

60 - 80 20 - fi 70 1400 - 70fi

80 - 100 7 90 630

100 - 120 8 110 882

Mean =



Therefore, f2 = 20 - 8 =12.

27]

In

…(1)

Similarly

…(2)From (1) and (2)

xy = xz + yzDividing by xyz

28]

q = Consider,q (p2 - 1)

= = 2p = RHS



29]

Since = Zeroes of the given polynomial

(x + ) (x - ) will be a factorOr x2 = 2 will be a factorLong division.

2x2 - 3x+ 1 = 2x2 - 2x - 2x + 1= 2x (x- 1) - 1(x - 1)= (2x - 1) (x- 1 )

The other zeroes are and 1.30]

On dividing the numerator and denominator of lts by cos , we get

OR



= 31]

Pythagoras Theorem: Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Given: A right triangle ABC right angled at B.

To prove: that AC2 = AB2 + BC2

Construction: Let us draw BD AC (See fig.)

Proof :

Now, ADB ABC (Using Theorem:If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse ,then triangles on both sides of the perpendicular are similar to the whole triangle and to each other)

So, (Sides are proportional)

Or, AD.AC = AB2

Also, BDC ABC (Theorem)

So,

Or, CD. AC = BC2

Adding (1) and (2),






OR AC2 = AB2 + BC2

ORStatement:Ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: Two triangles ABC and PQR such that

To prove :

Proof For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles.

Now, ar

And ar (PQR) =



So,

Now, in

B = Q (As

And m = n (Each is of 90o)

So, (AA similarity criterion)

Therefore,

Also,

So,

Therefore, [from (1) and (3)]

= [From (2)]

=

Now using (3), we get

32]The given cumulative frequency distributions of less than type is -

Weight

(in kg)

upper class limits

Number of students

(cumulative frequency)

Less than 38 0



Less than 40 3

Less than 42 5

Less than 44 9

Less than 46 14

Less than 48 28

Less than 50 32

Less than 52 35

Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following -

Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data is 46.5.

33]x - y + 1 = 0 x = y - 1

Three solutions of this equation can be written in a table as follows:

x 0 1 2y 1 2 3

3x + 2y - 12 = 0



Three solutions of this equation can be written in a table as follows:

x 4 2 0y 0 3 6

Now, these equations can be drawn on a graph. The triangle formed by the two lines and the x-axis can be shown by the shaded part as:

34]We have

= Cos (90o - 50o - ) - cos (40o - ) +



SAMPLE PAPER –V

Solutions:1] Let x1,x2,x3……..,x12 be the 12 values of the given data. Let the 13th observation

be x13.x1+x2+x3……..+x12 = 12x19.25 = 231x1+x2+x3……..,x12+x13= 13x20=260(x1+x2+x3……..+x12)+x13= 260x13=260-231 = 29

2] Given, triangle ABC is right angled at C. Therefore,A+B=90o or A=90o-B1+cot2A = 1 + cot2(90o-B) = 1+tan2B = sec2B

3]A real number is an irrational number when it has a non terminating non repeating decimal representation.

4] x2 +2x+1= (x+1)2

x = -1? = ?= -11/ ? and 1/? are also -1. 1/ ? + 1/? = -2

5]Since the x-axis y=0 does not intersect y=-7 at any point.

6]

Since .7]

Because cosec 90°=1, others are not defined.8]

9] If a and b are one two positive integers. Then a = bq + r, 0 r b Let b = 3 Therefore, r = 0, 1, 2 Therefore, a = 3q or a = 3q + 1 or a = 3q + 2 If a = 3q a2 = 9q2 = 3(3q2) = 3m or where m = 3q2 a = 3q + 1 a2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q or a = 3q + 2 a2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1 = 3m + 1 where m 3q2 + 4q + 1 Therefore, the squares of any positive integer is either of the form 3m or 3m + 1.

10] Given polynomial P(x) = x4 + 2x3 - 2x2 + x -1

Let g(x) must be added to it.

So, number to be added=-(-x+ 2) = x - 2



11] For infinite number of solution,

Consider

Again,

12]

AC2 = AD2 + CD2 … (2)[By Pythagoras theorem](1) - (2) gives,

Hence proved.13]

ORConsider,7 sin2 + 3 cos2 = 4

7Sin2 + 3 (1 - sin2 ) = 4 7Sin2 + 3 - 3sin2 = 44Sin2 = 1

Sin =

Thus, Sec 30o + Cosec30o = 14] Class Interval Cumulative Frequency

More then 50More then 60More then 70More then 80More then 90More then 102

1089580634219

15] Let 3 - be a rational number.

3 - = [ p,q are integers, 2 0]



Here,LHS = Rational No.RHS = irrational No.

But, Irrational no Rational no

our assumption is wrong is an irrational.

ORLet us assume to the contrary, that is a rational number.

is rational.(n-1)+(n+1)-2 is rational2n+2 is rational

But we know that is an irrational numberSo 2n+2 is also an irrational numberSo our basic assumption that the given number is rational is wrong.

Hence, is an irrational number.16] … (1)

… (2)Multiplying (1) with a and (2) with b, we get

From (1), bx + ab = 2abbx = ab x = a

Hence, x = a and y = b.17] C.I Fi Xi Fi. .Xi

10 - 20 5 15 7520 - 30 3 25 7530 - 40 4 35 14040 - 50 F 45 45f50 - 60 2 55 11060 - 70 6 65 39070 - 80 13 75 975

33+f

1765+45f

Mean =



ORC.I fi xi di fidi100 - 150 4 125 -2 -8150 - 200 5 175 -1 -5200 - 250 12 225 0 0250 - 300 2 275 1 2300 - 350 2 325 2 4

-7

Where:

18] Let us assume that Prema invests Rs x @10% and Rs y @8% in the first year.We know that

Interest = ATQ,

+ =1640 10x + 8y = 164000 …(i)After interchanging,

+ =1600we get 10y+8x=1600008x+10y=160000 ...(ii)Adding (i) and (ii)18x+18y=324000 x + y = 18000 ... (iii)Subtracting (ii) from (i),2x-2y=4000 x - y = 2000 ...(iv)Adding (iii) and (iv)2x=20000 x = 10000.Substituting this value of x in (iii)y=8000So the sums invested in the first year at the rate 10% and 8% are Rs 10000 and Rs 8000 respectively.

OR



Solving (1) and (2), we getx = 30 y = 6.

19] Let is one zero. = 7 is another zero then

+ 7 =

8 = = and = Now,

20] The two angles and being the acute angles of a right triangle, must be complementary angles.So,

Substituting, in above equation

21]

Now, AP + AQ = 120 + 60 = 180 cm



22]

23]

Let the area of triangle =x sq units

Area of trapezium = 3x sq units

Area triangle ABC = x + 3x = 4x sq units

Now,

Consider triangles AED and ABC,

ED ll BC...given

AED = ABC Corresponding angles

A = A Common

?AED ~ ?ABC [By AA rule]

= (since Ratio of areas of two similar triangles is equal to ratio of square of corresponding sides)



So 24] C.I F Cf

9.5 - 19.5 2 219.5 - 29.5 4 629.5 - 39.5 8 1432.5 - 49.5 9 2349.5 - 59.5 4 2759.5 - 69.5 2 2969.5 - 79.5 1 30

Here, l = 39.5 c.f = 14 f = 9 h = 10

M = 39.5 + 25] Given 2 and 3 are the zeroes of the polynomial.

Thus(x - 2) (x - 3) are factors of this polynomial.

4x4 - 20x3 + 23x2 =5x - 6 = (x2 - 5x + 6) (4x2 - 1)Thus, 4x4 - 20x3 + 23x2 +5x-6=(x - 2) (x - 3) (2x - 1) (2x + 1)

Therefore, 2,3, 26]

Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively

To prove that

Construction: Let us join BE and CD and then draw DM AC and EN AB.

Proof: Now, area of



Note that BDE and DEC are on the same base DE and between the same parallels BC and DE.

So, ar(BDE) = ar(DEG)

Therefore, from (1), (2) and (3), we have :

OR

Given: A right triangle ABC right angled at B.

To prove: that AC2 = AB2 + BC2

Construction: Let us draw BD AC (See fig.)

Proof:

Now, ADB ABC (Using Theorem: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse ,then triangles on both sides of the perpendicular are similar to the whole triangle and to each other)

So, (Sides are proportional)

Or, AD.AC = AB2 … (1)

Also, BDC ABC (By Theorem)

So,

Or, CD. AC = BC2 … (2)



Adding (1) and (2),




OR AC2 = AB2 + BC2

Hence proved.

27]

OR

Dividing numerator and denominator of LHS by sin , we get



28]

29] Let the number of girls and boys in the class be x and y respectively.According to the given conditions, we have:x + y = 10x - y = 4x + y = 10 x = 10 - yThree solutions of this equation can be written in a table as follows:

x 5 4 6y 5 6 4

x - y = 4 x = 4 + yThree solutions of this equation can be written in a table as follows:

x 5 4 3y 1 0 -1

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).So, x = 7 and y = 3.

30] We can obtain cumulative frequency distribution of more than type as following:Production yield

(lower class limits)Cumulative frequency

more than or equal to 50 100more than or equal to 55 100 - 2 = 98more than or equal to 60 98 - 8 = 90



more than or equal to 65 90 - 12 = 78more than or equal to 70 78 - 24 = 54more than or equal to 75 54 - 38 = 16

Now taking lower class limits on x-axis and their respective cumulative frequencies on y-axis we can obtain its ogive as following.

31]


Given: ABC ~ PQR To Prove: Construction: Draw ADBC and

PSQR Proof: ADB

~ PSQ (AA) Therefore, … (iii) But ABC ~ PQR Therefore,

… (iv) Therefore, Therefore, From (iii)



32]

Hence, L.H.S = R.H.S33] Let 5q + 2, 5q + 3 be any positive integers

(5q + 2)2 = 25q2 + 20q + 4= 5q (5q + 4) + 4 is not of the form 5q + 2Similarly for 2nd

(5q + 3)2 = 25q2 + 30q + 9=5q(5q+6)+ 9 is not of the form 5q+3So, the square of any positive integer cannot be of the form5q+2 or 5q+3For any integer q

34] Marks Frequency25 - 35 535 - 45 1045 - 55 2055 - 65 965 - 75 675 - 85 2Total 52

Here the maximum frequency is 20 and the corresponding class is 45-55.So,45-55 is the modal class.We have,l=45,h=10,f=20,

Mode = +

Mode=49.7

SOLUTIONS SCIENCE 2012-13



SAMPLE PAPER –I

Solutions:1]

Y – Diamond and Z – Graphite2]

1/253]

Sometimes there is a lack of oxygen in our muscle cells hence the pyruvate breaks down to form lactic acid which builds up in our muscles during sudden activity and causes cramps.

4]Plaster of Paris is calcium sulphate hemihydrates. It is obtained by heating gypsum

Uses:- (1) Plastering fractured bones

(2) statues, toys etc.5]

It is a double displacement and precipitation reaction.6]

Plant kept in continuous light will live longer, because it will be able to produce oxygen required for its respiration by the process of photosynthesis.

7]

Maximum current through resistor A = 3A

Thus the maximum current through resistors B and C each 8]

The following events occur during photosynthesis:

(i) Absorption of light energy by chlorophyll.

(ii) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.

(iii) Reduction of carbon dioxide to carbohydrates9]

The steps involved in reflex action are:

(i) Receptors receive the stimulus

(ii) Nerve cells takes the stimulus, converts it into electrical signal and passes it through sensory nerve.

(iii) The impulse passes through spinal cord where it gets transmitted through relay neuron and it passes further through motor neuron to the effector organ, i.e. muscles.



10]

11](a) Silver chloride on exposure to sunlight may decompose as per the following reaction.

Therefore, it is stored in dark coloured bottles.

(b) Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) releasing a large amount of heat

12]

(a)

(b) 13]

In absence of litmus paper a reagent phenolphthalein can be used. The solution which gets pink colour with phenolphthalein is base other is acid.

14]A stimulus received by a neuron travels through it in the form of an electrochemical disturbance. When an electrical signal reaches the axonal end of one neuron, it releases chemical substance called neurotransmitter that cross the synapse and move towards the dendritic end of next neuron generating another electrical singal. A narrow fluid filled space, called synaptic cleft occurs between the two neurons. An impulse travels through the neurons only in one direction because the neurotransmitter releases only on one side of the synapse.

15](a)

(b) 16]

The various plant hormones are auxin, gibberellin, cytokinin and abscisic acid. Functions of plant hormones:

(a) Auxin - It helps the cells to grow longer at shoot tips.

(b) Gibberellin – Growth of stem.

(c) Cytokinin – Promotes cell division.

(d) Abscisic acid – Inhibits growth in plants17]

The thumb in right hand rule indicates the direction of current in the straight



conductor held by curled fingers, whereas the Fleming's left-hand rule gives the direction of force experienced by current carrying conductor placed in an external magnetic field.

18]Hints – Biomass: Plant and animal wastes. Give description of biogas plant with the

following diagram.

19](a)A current carrying conductor produces a magnetic field around it. When it interacts

with magnetic field it experiences a force. Direction of the force depends upon direction of current and that of magnetic field.

Fleming's left hand rule. State the rule.

(b)The deflection will increase. As the magnetic field produced around the conductor is proportional to the current flowing through the wire.

OR

a) The earth wire, which has insulation of green colour, is

usually connected to a metal plate deep in the earth near the house.

This is used as a safety measure, especially for those appliances that

have a metallic body, for example, electric press, toaster, table fan,

refrigerator, etc. The metallic body is connected to the earth wire, which

provides a low-resistance conducting path for the current. Thus, it

ensures that any leakage of current to the metallic body of the appliance

keeps its potential to that of the earth, and the user may not get a severe

electric shock.

b) due to electromagnetic induction, there would be change in the magnetic field around the coil A and so will be around B . So current will be induced in coil B.

20](a) Baking powder (NaHCO3), salt A is commonly used in bakery products. On heating



it forms sodium carbonate (Na2CO3), B and CO2 gas, C is evolved. When CO2 gas is passed through lime water it forms calcium carbonate (CaCO3), which is slightly soluble in water making it milky.

A – NaHCO3

B – Na2CO3

C – CO2 gas

(b) Bleaching powder is used –

(i) for bleaching wood pulp in paper factories.

(ii) for disinfecting drinking water to make it free of germs.

OR

(a) (i) Sulphuric acid + Zinc Zinc sulphate + Hydrogen H2SO4 (aq) + Zn(s) ZnSO4 (aq) + H2 (g) (ii) Hydrochloric acid + Magnesium Magnesium chloride + Hydrogen 2HCl (aq) + Mg (s) MgCl2(aq) + H2 (g)

(b) A reaction in which an acid and base react with each other to give a salt and water is termed as neutralisation reaction. In this reaction, energy is evolved in the form of heat.

(c) Uses of washing soda:-

(i) Sodium carbonate (washing soda) is used in glass, soap and paper industries.

(ii) It is used in the manufacture of sodium compounds such as borax.21]

(a)

(b)

(c) Na+ cation, Mg+2 cation O-2 anion

(d) Strong inter ionic attraction.

OR

(a) Noble, highly malleable, ductile shinglustre

(b) Forms a protective layer of aluminium oxide



(c) Strong oxidizing agent oxides H2 to H2O

(d) Easy to reduce metallic oxide to metal

(e) Good conductor of heat, Non corrosive.22]

(a)

(b)

(c) 4V. Hint – V = IR = = 4V

(d) No difference.

Hint – Same current flows through each element in a series circuit.

OR

Current carrying loops behave like bar magnets and both have their associated lines of field. This modifies the already existing earth's magnetic field and a deflection result. Magnetic field has both direction and magnitude. Magnetic field lines emerge from N-pole and enter Spole. The magnetic field strength is represented diagrammatically by the degree of closeness of the field lines. Field lines cannot cross each other as two values of net field at a single point cannot exist. Only one value a unique net value, can exist. If in a given region, lines of field are shown to be parallel and equispaced, the field is understood to be uniform.

23](a)



(b) The digestion of carbohydrates (starch) starts in the mouth. The saliva has salivary amylase or ptyalin enzyme which acts on the starch and converts it into maltose sugar. The partially digested food reaches the stomach where the food gets mixed with gastric juice secreted by gastric glands. The gastric juice contains large quantities of mucus, HCl and pepsin. The enzyme pepsin acts on proteins and coverts them into peptones and proteoses. The gastric lipase emulsifies fats.

Bile juice secreted from liver lipids into small droplets.

Pancreas secretes pancreatic juice which contains the pancreatic amylase, trypsin and lipase. Pancreatic amylase digests the leftover starches into maltose. Trypsin converts peptones and proteoses into small peptides. Lipase acts on small droplets and converts them into triglycerides.

The food reaches the small intestine where complete digestion of food takes place in the presence of an enzyme called succus entericus. The maltose converts into glucose, peptide emulsifies into amino acids and the triglycerides emulsifies into fatty acids and glycerol. [

OR

(a) In four chambered heart, left half is completely separated from right half by septa.This prevents oxygenated and deoxygenated blood from mixing. This allows a highly efficient supply of oxygenated blood to all parts of the body. This is useful in animals that have energy needs, such as birds and mammals.

(b)



24](d) in set up A and (a) in set up B

25]Sodium bicarbonate > Water > Fruit juice

26]I

27]A, C, D, B

28]A

29]An antacid

30]The final solution becomes light green

31]Metal Al2(SO4)3 CuSO4 FeSO4

(a) Al x ? ?32]

all the three cases33]

(ii)34]

(-4mA, +0.2V) and (2mA, 0.1V) respectively35]

Student A will determine the equivalent resistance of series combination while student B will determine the equivalent resistance of parallel combination of the two resistors

36]epidermal cells, stomata, guard cells each with one nucleus and many chloroplasts

37]I Cytoplasm II nucleus

III stoma IV chloroplast



38]

39]Absorbs CO2 released by germinating seeds

40]Thin layer of fleshy leaf of onion

41]The glow of the bulbs B2 and B3 will remain the same

SAMPLE PAPER –II

solutions:1]

Gallium or Cesium (any one).2]

Cerebellum3]

Since ventricles have to pump blood into various organs, therefore, they have thicker muscular walls than the auricles.

4]Commercial unit of energy is Kilowatt hour or KWh.

5]Chemical reaction in which one reactant gets oxidized while the other gets reduced is called a redox reaction.

Mg gets oxidized on burning in air.

6](i) Fresh milk is acidic and gets spoiled easily. In presence of baking soda, milk becomes alkaline and can be stored for a longer time.

(ii) When milk sets to curd, pH decreases. The alkali does not allow it to become more acidic easily. So it takes more time.

7]Name of gland Secretion

Salivary gland Saliva

Gastric gland Gastric juice

Liver Bile

Pancreas Pancreatic Juice

(any two)8]

(a) When Iron nails are kept into a copper sulphate solution, Iron displaces copper



from CuSO4 solution and blue colour of CuSO4 solution becomes light green.

(b)

9](a) Ionic compounds are formed by the transfer of electrons from metal to non metal.

(b)

10](a) Respiration is an exothermic process since energy is released during the

process.

(b) Energy in the form of heat, light or electricity is required for decomposition Reactions hence these are endothermic.

(c)

The solution becomes colourless because of copper sulphate formed which is colourless.11]

(i) Test the three solutions with blue litmus paper, the one solution will change blue litmus into red it is Acidic solution

(ii) Test the remaining two solution with the red litmus (Changed in activity (i)) one solution will change it again to blue It is basic solution. (iii) Remaining third solutions is distilled water



12]

13]In series (i) R1 = 5+5 =10

(ii) R2 = 6+4 = 10

In pararell

(i)

14](i) Glottis gets covered by a small cartilaginous flap of skin called epiglottis which prevents the entry of food particles into wind pipe while swallowing.

(ii) The lung alveoli are richly supplied with blood capillaries for the exchange of gases. From the thin walls of alveoli, oxygen diffuses into blood and is supplied to the tissues while carbon dioxide is absorbed by blood from the tissues and is carried to the alveoli of lungs for exhalation.

(iii) The walls of trachea is supported by C – shaped cartilage rings which does allow the trachea to collapse in the presence of less air in it.

15]Hind brain has three parts:

(i) Cerebellum: It is the regulating centre for swallowing, coughing, sneezing and vomiting and helps to maintain the balance of the body and coordinate the muscular activities.

(ii) Pons: It takes part in the regulation of respiration.

(iii) Medulla oblongata: It is responsible for the coordination and adjustment of movement and posture.

16](a) Diagram of Neuron:



Functions of Neuron:

(i) To carry information from receptors to brain and spinal cord.

(ii) To transfer respose from brain and spinal cord to effectors.17]

(a) Auxin.

When light is coming from one side of the plant auxin diffuses towards shady sides.

This concentration stimulates the cells to grow longer on the side of the shoot which is away from the light and shoot bends.

18](i) Magnetic field will increase

(ii) Magnetic field will decrease

(iii) Magnetic field will increase

19](i) Wave energy be a viable proposition only when waves are very strong. [1x3=3]

(ii) A tidal power station at the mouth of a river blocks the flow of polluted water into the sea. It creates health and pollution hazards.

(iii) Efficient commercial exploitation is difficult or any other.

20](a)

(b) Functions of two chambers of human heart:

Left Atrium – Receive oxygenated blood from pulmonary vein



Right Atrium – Receive deoxygenated blood from vena cava

Left Ventricle – Pumps oxygenated blood to all parts of body

Right Ventricle – Pumps deoxygenated blood to lungs. (Any two)

OR

(a) Diagram of stomata:

(b) Functions of stomata:

(a) Exchange of gases

(b) Transpiration

Raw materials for photosynthesis CO2, H2O

21](a) The chemical name of washing soda is Sodium carbonate Decahydrate.

The formula is Na2CO3.10H2O

It is obtained by heating baking Soda and then recrystallization.

(b) Distilled water is a pure form of water and is devoid of any ionic species. Therefore, it does not conduct electricity. Rain water, being an impure form of water, contains many ionic species such as acids and therefore it conducts electricity.

OR

(a)

(i) Neutral- Solution D with pH 7 (ii) Strongly alkaline- Solution C with pH 11 (iii) Strongly acidic- Solution B with pH 1 (iv) Weakly acidic- Solution A with pH 4 (v) Weakly alkaline-



Solution E with pH 9

The pH can be arranged in the increasing order of the concentration of hydrogen ions as: 11 < 9 < 7 < 4 < 1

(b) The fizzing will occur strongly in test tube A, in which hydrochloric acid (HCl) is added.

This is because HCl is a stronger acid than CH3COOH and therefore produces hydrogen gas at a faster speed due to which fizzing occurs.

22](i) Metal – sodium

(ii)

(iii)Electrolysis of molten chloride (NaCl) is the process.

i.e. we can write:

OR

Definition: Homogeneous mixture of two or more metals or a metal and a non metal is called an alloy.

Prepared: (i) By melting the primary metal.

(ii) By dissolving the other elements in a definite proportion and cooled to room temperature.

23]A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder.



The magnetic field is uniform inside the solenoid

Used to magnetise a piece of magnetic material.

OR

(a)

(i) Right hand Thumb rule

(ii) Fleming's Left hand rule

(iii) Fleming's Right hand rule

(b)

(a) To avoid the risk of electric shocks

(b)To save the electrical appliances from damage.

24](a)

(i) Momentary deflection in the galvanometer to one side.

(ii) Momentary deflection in the galvanometer now in opposite direction.

(iii) No deflection in the galvanometer. Phenomenon involved is electromagnetic induction.

(b) An energy source that can be replenished in a short period of time and state one use each.

Eg. Solar Energy, Wind energy, Ocean energy etc. Any use.

OR

(a)Argon or Neon gas is filled in the electric bulbs.

These are used since these are inactive or inert gases to prolong life of the filament. (b)

(i) Length of the conductor: proportional91 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


(ii) Area of cross section: inversely proportional

(iii) Material of conductor25]

26]II

27]1A in Circuit I and 0A in Circuit II

28]9

29]Parallel in circuit I and in series in circuit II

30]A=Digestor,B=Gas outlet

31]Safrainin

32]Alcohol

33]Alcohol is inflammable

34]Germinating seeds

35]To absorb CO2 released by germinating seeds

36]Covered portion of any of the above

37]Green


39]A

40]Light green

41]NaOH

42](i) and (ii)

SAMPLE PAPER –III

Solutions:1]

2]The receptors present in the specific regions of the tongue are called gustatory receptors.



3]A large number solar cell combined in an arrangement.

4](i) Human kidney performs the excretion by cleaning the blood of metabolic wastes.

(ii) It performs the function of osmoregulation by maintaining normal levels of water and mineral ions in the body fluids.

5]'Zn' being more reactive displaces 'Cu' from CuSO4 solution CuSO4 solution becomes Colourless.

(blue) (colourless)6]

The pH of milk decreases from '6' as it turns into curd. That is curd is more acidic than milk.

7]Resistance depends upon:

(i) Length of the conductor

(ii) Area of cross section

(iii) Material of conductor

Resistivity depends on:

(i) Material

(ii) Temperature

SI Unit of resistance is 'Ohm' or 8]

(i) Each appliance have equal potential differenc

(ii) Each appliance have a separate switch to on/off

(iii) Each appliances can be operated on different current9]

(a) Momentary deflection in the galvanometer to one side.

(b)Momentary deflection in the galvanometer now in opposite direction.

(c)No deflection in the galvanometer.

(d)Phenomenon involved is electromagnetic induction.

(e) Momentary deflection in the galvanometer now in opposite to the direction of the first case.

(f) No deflection in the galvanometer.

10](i) Xylem – water and minerals in plants

(ii) Pulmonary Artery – Deoxygenated blood from heart to lungs.



(iii) Pulmonary Vein – Oxygenated blood from lungs to heart.

Phloem – Synthesised food in plants

11]1)Methane or CH4

2) Hydrogen

3) Hydrogen sulphide

4) CO2

Uses Any two:

. For the production of electricity ii. As a fuel to produce heat and light. iii. Biogas generation is an effective method of the disposal of bio wastes like, animal wastes and sewage.

12](i) Ionic Compounds which are held together by strong ionic bonds so high amount of energy is required to break these bonds.

(ii)Ionic compounds are very hard solids due to strong force of attraction between positive and negative ions.

(iii)Hydrogen gas is evolved when dilute hydrochloric acid is added to a reactive metal.

13](a) Chemical reaction in which one reactant gets oxidized while the other get reduced is known as a redox reaction.

When Magnesium ribbon burns in air, magnesium is oxidized and the white residue formed is of magnesium oxide.

(b) Chemical equations must always be balanced to follow the law of conservation of mass, according to which "Mass can neither be created nor destroyed in a chemical reaction."

14](i) This is an example of decomposition reaction and it is endothermic in nature.

(ii) This is an example of combination reaction and it is also a type of endothermic reaction.

(iii) This is an example of displacement reaction and it is exothermic reaction.

15](i) The chemical name of 'Plaster of Paris' is 'Calcium sulphate Hemi hydrate'

(ii) It is prepared on heating gypsum at 373 K

(iii)



16]

(i)

(ii) In series R2 =7 + 3 =10

R3 =6 + 4 = 10

In Parallal

OR

17]The gastric glands in the stomach secrete:

(i) HCl – It kills the bacteria ingested with food and makes the medium of food acidic so as to facilitate the action of the enzyme pepsin.

(ii) Pepsin – It helps to digest proteins.

(iii) Mucus – It protects the inner lining of the stomach from the action of HCl



18]The functions of fore brain are:

(i) It has sensory area where information is received from sense organs.

(ii) It has motor area where impulses are sent to muscles or effector organs.

(iii) It has centres for visual reception, touch, smell, temperature and muscular activities.

19](a) Diagram of Neuron:

Functions:

(i) To carry information from receptors to brain and spinal cord.

(ii) To transfer respose from brain and spinal cord to effectors.20]

(a) The glands which does not have duct and secretes it's product directly into blood stream is called an endocrine gland.

Two glands are: Thyroid and pancreas.

Thyroid secretes thyroxin and pancreas secretes insulin.21]

(a)

(i) Test the three solutions with blue litmus paper, the solution that changes l blue litmus into red is Acidic solution.

(ii) Test the remaining two solution with the red litmus one solution will change it again to blue –then that solution is basic solution.

(iii) Remaining third solutions is distilled water which being neutral does not cause any colour change of either the blue or red litmus paper.

(b) Plaster of Paris should be stored in a moisture-proof container because Plaster

of Paris, a powdery mass absorbs water (moisture) to form a hard solid known as gypsum.



OR

(a) Two important used of washing soda and baking soda are as follows: Washing soda:

(i) It is used in glass, soap, and paper industries. (ii) It is used to remove permanent hardness of water.

Baking soda:- (i) It is used as baking powder. Baking powder is a mixture of baking soda and a mild acid known as tartaric acid. When it is heated or mixed in water, it releases CO2 gas that makes bread or cake fluffy. (ii) It is used in soda-acid fire extinguishers.

(b) Acids dissociate in the presence of water to give free hydrogen ions. It is the hydrogen ions that are responsible for the acidic behaviour.

22](i) Metal – sodium

(ii)

(ii) Electrolysis of molten chloride (NaCl)

OR

Definition: Homogeneous mixture of two or more metals or a metal and a non metal

Prepared:

(i) Melting the primary metal.

(ii) Dissolving the other elements in a definite proportion and then cooling them to room temperature .

23]A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder.



The magnetic field is uniform inside the solenoid

Used to magnetise a piece of magnetic material.

OR

(a) (i) Right hand Thumb rule

(ii) Fleming's Left hand rule

(iii) Fleming's Right hand rule

(b) (a) To avoid the risk of electric shocks (b) To save the electrical appliances from damage.

24](a)

(b) Functions of chambers of human heart:



Left Atrium – Receive oxygenated blood from pulmonary vein

Right Atrium – Receive deoxygenated blood from vena cava

Left Ventricle – Pumps oxygenated blood to all parts of body

Right Ventricle – Pumps deoxygenated blood to lungs. (Any two)

OR

(a) Diagram of Stomata:

Functions

(a) Exchange of gases

(b) Transpiration

Raw materials for photosynthesis are CO2, H2O

25]Green


27]A

28]Hydrogen

29]Red

30]Calcium

31]0.25

32]0A in Circuit I and 1A in Circuit II

33]9



34]3

35]Parallel in circuit I and in series in circuit II

36]

37]Glycerine

38]Alcohol

39](I) Guard cell

(II) Nucleus

(III) Stoma

(IV) Chloroplast

40]Germinating seeds

41]To absorb CO2 released by germinating seeds


SAMPLE PAPER –IV

Solutions:1]

Metal: Mercury: Non metal: Bromine2]

The brain box called cranium protects the brain and has shock absorbing fluid in it which prevents it from shock and injuries.

3](i) Volmeter (ii) Ammeter

4]It is a reaction in which there is an exchange of ions between the reactants

Na2SO4(aq) + BaCl2(aq) ? BaSO4(s) +2NaCl(aq) or any other5]

HCl and HNO3 produce H+ ions in aqueous solution. Hence, they show acidic character.

Alcohol and Glucose do not produce H+ ions in aqueous solution, hence, they do not show acidic characters.

6]Glucose that enters the nephron along with the filtrate after passing through the



glomerulus, passes from the tubule of nephron where it is selectively reabsorbed and sent back to blood.

7]

8](a)

Let Resistance of a wire = R

When cut into 3 equal parts

Resistance of each part =

When connected in Parallel, net resistance Rp:

(b) Resistivity will remain same. (c) Current through each part will be triple the previous and total current in the circuit will be 9 times.

9](a) Strength of magnetic field (B) is inversely proportional to radius of the coil (r)

(b) Directly proportional to the Number of turns in the coil (N)

( c) The magnetic field lines will be like:



10]At the roots, cells in contact with the soil actively take up ions. It creates a difference in the concentration of these ions between roots of soil. To eliminate this difference, water moves in to the roots from the soil. Thus water of minerals are absorbed by the plants.

11](i) Buring of coal or Petroleum Products lead to air pollution

(ii) The oxides of C, N and S are released on buring fossil fuels as acidic oxides it leads to acidic rain and affect water soil.

(iii) CO2, a green house gas causes global warming (or any other relevant point)

i. Source of electricity in artificial satellites ii. Used as source of electricity in radio and wireless transmissions, at TV relay stations, traffic lights and research centers.

12](a) Oxidation: gain of oxygen by a substance

or any other example

Reduction: loss oxygen by a substance or any example

(b) Sodium (Na) is oxidised to sodium oxide as it gains oxygen and oxygen gets reduced.

13](a) The compounds formed by transfer of electrons from metal to non metal are called ionic compounds

In ionic compounds, there are strong electrostatic forces of attraction between the oppositely charged ions. Due to this, movement of ions is not possible due to their rigid structure.

In molten state, the electrostatic forces of attraction between the oppositively charged ions are overcome due to heat.

(b) Ionic compounds have strong electrostatic forces of attraction between the ions. Therefore, it requires a lot of energy to overcome these forces. That is why ionic compounds have high melting points.

14]Exothermic: Reactions in which heat is released.

Endothermic: Reactions in which heat is absorbed.

or any other

15]When the pH of mouth of a person is lower than 5.5 tooth decay starts

Tooth enamel made up of calcium phosphate is corroded.

Measures:

(i) Clean the mouth after eating food



(ii) Use of toothpaste (generally basic in nature)16]

The functions of fore brain are:

(i) It has sensory area where information is received from sense organs.

(ii) It has motor area where impulses are sent to muscles or effector organs.

(iii) It has centres for visual reception, touch, smell, temperature and muscular activities.

17]The following events occur during photosynthesis:

(i) Absorption of light energy by chlorophyll.

(ii) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.

(iii) Reduction of carbon dioxide to carbohydrates.

18]Gland Harmone Function

Thyroid Thyroxin Regulates fat protein and carbohydrate metabolisms

Pancrease Insulin Regular blood sugar level

Adrenal gland Adrenaline Prepage the body to face the emergency situation and blood pressure

19]Auxin diffuses to the other part which is away from the support. Part of the tendril in contact from the support does not grow rapidly

Part of the tendril away from the support grow rapidly and cause the tendril to grow around a support.

20](a) Bleaching Powder

(b) CaOCl2

(c)

(d) Bleaching powder is used:

(i) as an oxidising agent in many chemical industries; and

(ii) for disinfecting drinking water to make it free of germs.

OR

(a) The chemical equation for the reaction of Plaster of Paris and water can be



represented as:

(b) Calcium hydroxide [Ca(OH)2], on treatment with chlorine, yields bleaching

powder.

(c) Washing soda (Na2CO3.10 H2O) is used for softening hard water.

(d) When a solution of sodium hydrogencarbonate is heated, sodium carbonate and water are formed with the evolution of carbon dioxide gas.

21]Corrosion: Process of slowly eating up of metals due to action of atmospheric gases

Metal does not corrode ? Gold

Which corrode ? Iron or any other

Prevention:

(i) By painting

(ii) By greasing or oiling

(iii) Galvinization

(iv) Tin plating/ Chromium plating etc. (any three)

OR

(i)

(ii) [1]

(iii)

(iv)

(v)



22]

(b) Exchange of gases take place in the alveoli

OR

(a)

It receive secretion from liver and pancreas in the form of juice and trypsin

Trysin digest proteins.

Lipase digest fats

Bile juice makes basic medium and emulsifies fats

Intestinal Juice converts proteins to amino acids Carbohydrates to glucose and fats to 105 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


fatty acids and glycerol.23]

(b)

(i) Over loading is caused when too many short circuit appliances are connected to a singl socket

(ii) When live wire and neutral wire come into direct contact

OR

(i) North Pole to South Pole

(ii) Tesla

(iii) Strong magnetic field

(iv) 50 Hz

(v) Heart and brain24]

(i) Separately: Current

(ii) Series R=R1+R2=44

(iii) In Parallel

Advantages of solar cooker:(any two) [1/2x4=2]



(1) Free of cost

(2) Does not cause any pollution

(3) Nutritive value of food is retained

Disadvantages:

(i) Can not used in night

(ii) Can not be used in cloudy day

(iii) Direction of sunlight is need and to be adjusted frequently or any other

OR

(a) Deflection increases

(b) Direction of deflection changes

(c) Deflection decreases. [1]

(d)

Looking at a crossection lying in the plane of the

paper. At a distance d from the wire, the magnetic field direction is as shown and the magnitude is constant for all points shown.

25]sodium bicarbonate > water > lemon juice

26]HCl

27]II and IV

28]Set HCl+Zn HCl+Na2CO3 NaOH+Zn NaOH+Na2CO3

(d) ? ? ? X29]

R > P > Q



30]Displacement reaction

31]B

32]A

33]

34]0.5

35]Ammeter and rheostat

36]1


38]Entry of air bubbles

39]

40]Dark room for 03 days

41]I cytoplasm II nucleus III stoma IV chloroplast

42]

SAMPLE PAPER –V1]

Adrenaline controls the rate of respiration. Pons is the part of the brain which controls the rate of respiration.

2]Acid present in pickles or curd will react with metals like copper and poisonous salts are produced.108 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


3]

60 W bulb because ie resistance is inversely proportional to the power.4]

When aluminium powder is heated with manganese dioxide the following reaction takes place:

When Mn powder is heated with aluminum dioxide no reaction takes place because Mn is less reactive than Al and hence there will be no displacement reaction.

5]Baking powder is added to make breads soft and fluffy. Baking soda and an edible acid like tartaric acid are its main ingredients.

Baking soda (Sodium hydrogen carbonate) is added to release CO2 gas when heated.

Tartaric acid is added to avoid the bitter taste by reacting with the Na2CO3 which is formed by the heating of NaHCO3.

6](i) Kidney infection or injury to kidneys.

(ii) Restricted blood flow to kidneys.

7]The lining of canal has muscles that contract rhythmically in order to push the food forward.

Peristaltic movement is necessary to move the food in a regulated manner along the digestive tube so that the food can be processed properly in each part.

8]Solar cooker with concave mirror reflector is more efficient.

Because concave mirror can focus the heat radiations to the material kept inside to increase the temperature.

Concave mirror is used in headlight of vehicles or as shaving mirror.

9]Ohm's law relates current and potential difference.

According to this law:

V/I = constant = R



10]Ammeter, A1 will show higher reading.

Because, as wire A is thicker than B, A has lesser resistance. So higher current will be drawn by A from the battery and hence ammeter A1 will show higher reading.

11]Observation a is correct.

Because force experienced by a current carrying conductor in a magnetic field is proportional to the strength of the current. Resistance of a conductor is due to the obstruction to the flow of electrons due to the collisions with atoms and other electrons.

12](i) The substance 'X' is calcium oxide. Its chemical formula is CaO. (ii) Calcium oxide reacts vigorously with water to form calcium hydroxide (slaked lime).

(iii)

13]Main ore is cinnarbar, HgS.

When it is heated in air, it is first converted into mercuric oxide (HgO). Mercuric oxide is then reduced to mercury on further heating.

14](a)

Water of crystallisation is the number of water molecules that combine chemically in definite molecular proportion, with the concerned salt in the crystalline state.

(b) Two correct examples are:

Copper sulphate ,chemical formula = CuSO4.5H2O

Washing soda, chemical name = Na2CO3.10 H2O.

15](a) we can show it by an experiment of displacement reaction in which Mg rod is dipped in ZnSO4 solution. Magnesium displaces zinc from zinc sulphate solution.

(b)Cu is less reactive then hydrogen and thus place below hydrogen in the reactive series of metals, hence cannot displace hydrogen in dilute HCl.

16](a)



Functions:

(i) Cerebrum:It controls the conscious sensations.

(ii) Mid brain: It controls the auditory impulses.17]

Chemotropism – Movement of plant parts due to chemical stimulus is called chemotropism.

Example - Growth of pollen tube.

Plant Hormones:

(i) Auxin - It helps the cells to grow longer at shoot tips.

(ii) Gibberellin – Growth of stem.18]

(i) Glycogen and starch – Both are carbohydrates and stored food products.

(ii) Chlorophyll and haemoglobin – Both are pigments.

(iii) Arteries and veins – Both are blood vessels.19]

Involuntary actions are the actions which cannot be controlled by us if we want to do so. There is no external stimulus involved. The action takes place on its own like digestion, heart beat, etc.

Reflex action is a kind of involuntary action controlled by and takes due to external stimulus. Its respond to the stimulus is quick. Example – sneezing, blinking of eyes, etc.

20]Series:

OR

For maximum resistance, resistors are to be connected in series.

For maximum current, resistance has to be minimum and so resistors are to be



connected in parallel.

I series = V/R series = 6V/9 = 0.67 A

I parallel = V/R parallel = 6V/2 = 3A

Parallel connection is preferred over series for house hold circuits.21]

(a) Chloralkali process is used to prepare sodium hydroxide using sodium chloride.

When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide. The process is called the chlor-alkali process because of the products formed– chlor for chlorine and alkali for sodium hydroxide.

(b) (i) 7 (ii) less than 7

OR

(a)

(i) The milkman shifts the pH of the fresh milk from 6 to slightly alkaline because in

alkaline condition, milk does not set as curd easily.

(ii) Since this milk is slightly basic than usual milk, acids produced to set the curd are

neutralised by the base. Therefore, it takes a longer time for the curd to set.

(b) A reaction in which an acid and base react with each other to give a salt and water is

termed as neutralisation reaction. In this reaction, energy is evolved in the form of heat.

For example:

(i) NaOH + HCl NaCl + H2O

(ii) During indigestion (caused due to the production of excess of hydrochloric acid in the

stomach), we administer an antacid (generally milk of magnesia, Mg(OH)2 which is

basic in nature). The antacid neutralizes the excess of acids and thus gives relief from

indigestion. Mg(OH)2 + 2 HCl MgCl2 + 2 H2O

22](a)Amphoteric oxides are those oxides which show both acids as well as bases to form salts and water.

Ex: Aluminium metal reacts in this manner with acids and bases.

SnO2 is also an example of amphoteric oxides.



(b) Metals such as Sodium and Potassium are kept immersed is Kerosene becausethey are very reactive and have high affinity towards oxygen and will violently react with atmospheric oxygen on contact with air.

(c) Aluminium reacts readily with steam to give aluminium oxide and hydrogen gas, the reaction does not always occur. This is due to a thin but strong layer of aluminium oxide being coated onto the metal, thus preventing it from the reaction. aluminium + steam aluminium oxide + hydrogen 2Al(s) + 3H2O(g) Al2O3(s) + 3H2(g) [1]

(d) (i) Bromine (ii) Iodine

OR

(a) Diamond and graphite are the two allotropes of carbon.

Diamond –

hardest substance

electrical insulator

Graphite –

comparatively soft, it is slippery over layers

good electrical conductor

(b) Aluminium articles have a longer life and attractive finish compared to many other metals because of the formation of a thin transparent protective film cover of Aluminium oxide on the surface of Al formed due to its spontaneous reaction with oxygen.

(c) (i) Ore : An ore is a type of rock that contains minerals with important elements including metals. The ores are extracted through mining; these are then refined to extract the valuable element(s).

(ii) Gangue: In mining, gangue is the commercially worthless material that surrounds, or is closely mixed with, a wanted mineral in an ore deposit.

(d) Electronic configuration of metal atom is significant to know about the kind of bond that the metal will be forming for example in the formation of calcium chloride, chlorine only needs one electron to complete its octet so two atoms of chlorine accept one electron each lost by calcium ion.

23](a)



(i) Part in which starch digestion starts – Mouth

(ii) Part in which bile is stored – Gall Bladder

(iii) Part in which nutrients are absorbed – Small intestine

(iv) Part in which water is absorbed – Large intestine

(b) Kills bacteria in the stomach (1/2) provide acidic medium for the action of pepsin.

(c) (i) Controls the release of food from the stomach to small intestine.

(ii) Controls the release of undigested waste from the rectum through the anus.

OR

(a)

(b) Within the lungs, the passage divides into smaller and smaller tubes which finally terminate in balloon – like structures which are called alveoli.

The walls of the alveoli contain an extensive network of blood-vessels which provides 114 For Solution’s Contact : Ajay Sir : 9766607105 / Dheeraj Sir :

9766607104


surface where the exchange of gases can take place.24]

Brief explanation of activity:

Connect the circuit as shown in the figure below. Switch on the battery so that the current begins to flow. Sprinkle some fine iron filings around the current carrying wire. Tap the surface gently. The iron filings get arranged in concentric circles.

When current is decreased, field gets decrease

When the current is reversed, field also gets reversed

OR

A magnet is placed on a sheet of paper. A compass needle is placed near the North Pole. The position of its two ends is marked with the help of a sharp pencil. Now the compass is moved in such a way that its south end occupies position occupied by north end previously. Again the two ends are marked with sharp pencil. In this way, process goes on step by step till the south pole of the magnet is reached. Now all points are joined to get a smooth curve which represents a field line. In this way many field lines can be drawn

Region A has stronger magnetic field

Because the strength of the field is proportional to the relative closeness of field lines.25]

Less than 3.526]

Blue to red in A and no change in B



27]Metal Al2(SO4)3 CuSO4 FeSO4 ZnSO4

(a) Al X ? ? ?28]

Carbon dioxide29]

sodium bicarbonate > water > lemon juice30]

Hydrogen gas and iron chloride are produced31]

Student II only32]

Both (ii) and (iii)33]

Correct reading for current I incorrect reading for potential difference V34]

17.535]

A1 = A336]

1.37 ?37]

To make the plant free of starch, it is:38]

The part not covered turns blue39]

(ii), (iii), (iv), (i)40]

441]

(i), (iii), (iv), (ii)42]

A is KOH solution and B is water


sample question paper for class x – mathematics &...

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