Model Paper -4 (2016-17)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
BLUE PRINT
S.No / Unit / Chapter / VSA / SA-1 / SA-2 / LA / Total1 Mark Each / 2 Marks Each / 3 Marks Each / 4 Marks Each
1 / ALGEBRA / QUADRATIC EQUATIONS / 1(1) / 1(3) / 2(8) / 8(23)
ARITHMETIC PROGRESSIONS / 1(1) / 2(6) / 1(4)
2 / GEOMETRY / CIRCLES / 1(2) / 3(12) / 5(17)
CONSTRUCTIONS / 1(3)
3 / TRIGONOMETRY / HEIGHTS AND DISTANCES / 1(1) / 1(3) / 1(4) / 3(8)
4 / STATISTICS AND PROBABILITY / PROBABILITY / 1(1) / 2(4) / 1(3) / 4(8)
5 / COORDINATE GEOMETRY / COORDINATE GEOMETRY / 2(4) / 1(3) / 1(4) / 4(11)
6 / MENSURATION / AREAS RELATED TO CIRCLES / 1(3) / 7(23)
SURFACE AREAS AND VOLUMES / 1(2) / 2(6) / 2(8)
1*(4)
Total / 4(4) / 6(12) / 10(30) / 11(44) / 31(90)
Model Paper -4 (2016-17)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
Time: 3 Hrs Max. Marks: 90
General Instruction:-
1) All questions are Compulsory
2) The question paper consists of 31 questions divided into 4 sections, A,B,C and D. Section – A comprises of 4 questions of 1 mark each. Section-B comprises of 6 questions of 2 marks each. Section – C comprise of 10 questions of 3 marks each and Section- D comprises of 11 questions of 4 marks each.
3) There is no overall choice.
4) Use of calculator is not permitted.
Section-A
Question number 1 to 4 carry 1 mark each.
1) If 45, a, 2 are three consecutive terms of an A.P, then the value of a.
2) Find the discriminant of the quadraticequation33x2+10x+3=0 .
3) A pole 6m high casts a shadow 2√3 m long on the ground, then find the sun’s elevation .
4) If two coins are tossed together, then find the probability of getting at most one head ?
Section-B
Question number 5 to 10 carry 2 marks each
5) In figure below,ΔABC is circumscribing a circle. Find the length of BC.
6) Find a point on x-axis which is equidistant from the points (-2,5) and (2,-3).
7) Determine the ratio in which the point P(m, 6) divides the join of A( -4, 3) and B (2,8).
8) Spherical ball of diameter 21 cm is melted and recasted into cubes, each of side 1 cm. Find the number of cubes thus formed.
9) A die is thrown once, find the probability of getting
(i) A prime number (ii) a number divisible by 2
10) All cards of ace, jack and queen are removed from the deck of playing cards. One card is drawn at random from the remaining cards. Find the probability that the card drawn is
(a) a face card (b) not a face card
Section-C
Question number 11 to 20 carry 3 marks each
11)A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random,find the probability that it is
a) black b)red c) not green
12) Determine the AP whose 3rd term is 16 and when 5th term is subtracted from 7th term, we get 12.
13) Construct a triangle similar to a given ∆ABC in which AB =4cm, BC =6cm and ∠ABC = 600,such that each side of the new triangle is 34 of the corresponding sides of given ∆ABC.( Steps of constructions not required)
14. In the figure, find the perimeter of the shaded region where ADC, AEB and BFC are semi- circles on the diameter AC, AB, and BC respectively.
15) Largest sphere is carved out of the cube of side 7cm. Find volume of the sphere.
16) The curved surface area of the cone is 12320 cm2. If the radius of the base is 56cm, find its height.
17)Find the sum of the first 31 terms of an AP whose nth term is given by 3+3n2.
18) An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from his eye is 450. What is height of the chimney?
19) Find the value of ‘p’ for which the points (-5,1), (1,p) and (4,-2) are collinear.
20) Find the roots of quadratic equation a2b2x2+b2x-a2x-1=0
Section-D
Question number 21 to 31 carry 4 marks each
21) A motor boat whose is speed is 18 km/hr in still water takes 1 hour more to go 24 Km upstream than to return downstream to same spot . Find the speed of the stream.
22) OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. If the radius of the circle is 10cm, find the area of the rhombus.
23) Prove that the lengths of the tangents drawn from an external point to a circle are equal.
24) From top of a hill the angles of depression of two consecutive kilometer stones due east are found to be 300 and 600. Find the height of the hill.
25) Find the ratio in which the y axis divides the line segment joining the points(5,-6) and (1,-4). Also find the coordinates of the point of their intersection.
26) A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5cm but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π=3.14).What value of juice seller is depicted by selling the juice in said glasses?
27) A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all around to awidth of 5m to form an embankment. Find the height of the embankment.
28) Sum of the areas of the two squares is 468 m2. If the difference of their perimeters is 24m. Find the sides of the two squares.
29) Find the 11th from the last term (towards the first term) of the AP : 10, 7, 4,….,-62.
30) The incircle of ΔABC touches the sides BC, CA and AB at D, E and F respectively. Show that
AF+BD+CE = AE+BF+CD= 12 (Perimeter of ΔABC)
31) A rectangular sheet of paper of dimensions 44 cm x 18 cm is rolled along its length and a
cylinder is formed. Find the volume of the cylinder so formed. Use π=227
Model Paper -4 (2016-17)
SUMMATIVE ASSESSMENT - 2
CLASS X
MATHEMATICS
Marking Scheme
1) Ans75 1Mark
2) Ans641Mark
3) Ans6001Mark
4) Ans34 1Mark
5) AQ = 4cm (∵ Tangents from same external point to same circle)
AC = AQ + QC (1)
11 = 4 + QC
QC = 7cm
PC = QC = 7cm (Tangents from same external point to same circle)
BP = 3cm (Tangents from same external point to same circle)
BC = PC + BP
= 7 + 3
= 10cm (1)
A(-2,5)
6) Let the point on x-axis be P(a,0) P(a,0) X-axis
According to question
AP = PB B (2, -.3)
a+22+52=a-22+-32 (1)
=>a2+4a+4+25=a2-4a+4+9
=>8a=-16
=>a=-2
Therefore the required point is (-2,0) (1)
7)
A(-4,3) k P(m,6) 1 B(2,8)
Let required ratio = k:1
Using section formula mx1+nx2m+n,my1+ny2m+n (1)
For y-coordinate 6 = 8k+3k+1
Or 6(k+1) = 8k + 3
Or 6k + 6 = 8k +3
Or 6k – 8k = 3 – 6
Or -2k = -3
k = 3/2 (1)
Therefore required ratio = 3:2
8) Let the number of cubes formed = N
N × Vol of cube = Vol of sphere
N × 13 = 43πr3 (1)
=>N = 43×227×(212)3
=>N = 388088
=>N = 4851 (1)
9) Total outcomes = {1,2,3,4,5,6}=6
i) Total prime numbers ={2,3,5}=3 (1)
P(E) = 3/6
= ½
ii) Total numbers divisible by 2 ={ 2,4,6}=3
P(E) = 3/6
= ½ (1)
10) Number of cards removed = 12 i.e [4A +4J +4Q]
Remaining cards = 52 – 12 = 40
a) Number of face cards = 4 [kings] (1)
P( E) = 4/40 = 1/10
b) Probability[not a face card] = 1 – P( E)
= 1 – 1/10
= 9 /10 (1)
11) Total number of balls =5+8+4+7=24
Total elementary events=24
a)No. of black bags=7 (1)
Favourable No. of elementary events =7
P(Getting a black balls) =7/24
b)No. of red balls =5 (1)
No. of Favourable events=5
P(red ball) =5/24
c)No. of balls not green=5+8+7=20 (1)
No. of elementary events=20
P(Getting a green ball)=20/24=5/6
12) Given a3 = 16 i.e a + 2d = 16 (1)
Also a7-a5 = 12
=> a + 6d – (a + 4d) =12
=>a +6d – a – 4d = 12
=> 2d = 12
=> d = 6 (1)
Put d = 6 in (1)
a + 2(6) = 16
=>a + 12 = 16
=> a = 4
Therefore required AP 4, 10 ,16 ,22,… (1)
13)
3Marks
14) Required perimeter = Perimeter(Semi-circleADC)+Perimeter(Semi-circleAEB)+
Perimeter(Semi-circleBFC) (1)
= π× (4.2)/2 + π×( 2.8)/2 + π×(1.4)/2 (1)
=6.6 + 4.4 + 2.2
= 13.2 cm (1)
15) Diameter of the largest sphere = side of the cube (1)
Therefore diameter = 7 cm
Volume =43πr3 (1)
= 43×227×(3.5)3
=539 / 3 cm3 (1)
16) Given curved surface area = 12320 cm3
π rl = 12320 (1)
π(56) l = 12320
l = 12320 / (π X 56)
l = 12320 X7 /(22X56)
l = 70 cm (1)
Now h2 =r2 + l2
Or h = √[ l2 – r2]
= √[ 702 – 562]
= √[ 4900 – 3136]
= √1764
= 42 cm (1)
17) Given an= 3 + 2n/3
Put n = 1, 2 , 3 ……
a =11/3, a2 =13/3, a3 = 15/3 and so on (1)
Therefore a =11/3 and d = 13/3 – 11/3 = 2/3
Sn= n/2[ 2a + (n-1)d] (1)
= 31/2 [ 2 X11/3 + 30X 2/3]
= 31/2[ 22/3 + 60/3]
= 31/2 X 82/3
=1271/3 (Ans) (1)
18)
x
D 450
B
1.5m
E
28.5m C
Let the height of the chimney = x + 1.5 (1)
In DADB
Tan 450 = x / DB
1 = x / 28.5 { EC = DB}
X = 28.5cm (1)
Therefore height of the chimney = 28.5 + 1.5
= 30 m(Ans) (1)
19) Given that points A( 5,1) , B ( 1,p) and C (4,-2) are collinear
Therefore area DABC = 0 (1)
i.e 1/2[ x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ] = 0
5(p +2) +1 (-2-1) +4(1-p) =0
5p +10 -3+ 4 -4p =0 (1)
P+11 = 0
P = -11(Ans) (1)
20) a2b2x2 + b2x – a2x – 1. = 0 (1)
b2x(a2x +1) -1 (a2x +1) =0
a2x +1=0 or b2x -1 =0 (1)
x = -1/a2 or x = 1/b2 (1)
21) let Speed of the stream = x km/hr
Speed of the boat upstream: (18-x) Km /hr
Speed of the boat Downstream: (18+x) Km /hr (1)
According to question:2418-x - 2418+x =1 (1)
x2+48x-324=0 (1)
By using quadratic formula
X=6 , -54
Speed of the stream 6 km/hr. (1)
22) A
O
B
C (1)
As its given that OABC is a rhombus
Therefore OA = AB = BC = CO = Radius =10cm[ As the sides of rhombus are equal] (1)
More over OB =10cm
Area OABC = Area DOAB + area DOBC (1)
= √3/4 X 102 +√3/4 X 102{ area of an equilateral triangle = √3/4 X Side2 }
=50√3 cm2 (1)
23) Given:- A circle with center O. Two tangents PT and PQ are drawn from external point P.
To Prove :- PT = PQ
Construction :- Join TO, OP and OQ
T
P
Q (2)
Proof:- In ∆POT and ∆POQ
∟OTP = ∟OQP [ Each 900 as radius and tangents are perpendicular to each other]
OP = OP [Common hypotenuse]
OT = OQ [ Radii of the same circle]
Therefore ∆POT =᷈ ∆POQ [ by RHS]
=>PT = PQ ( c.p.c.t) (2)
24) A
300
600
h
600 300
D C B
h/√3 1km (1)
Let AD is hill of height h km and C and B are two kilometre stones
Therefore CB = 1 km
In DADC
Tan 600 = h/DC
√3 = h / DC
DC = h /√3------(1) (1)
In DADB
Tan 300 = h / (DC + 1)
1/√3 = h / (DC +1)
DC+1 = h√3
1 = h√3 - h /√3 [ from 1] (1)
1 = 2h/√3
h = √3/2 km (1)
25)
K 1
A(5,-6) p(0,y) B(-1,-4) (1)
Let the required ratio be k:1
Using section formula , the coordinates of P which divides AB in the ratio k:1 are
(-k+5k+1 , -4k-6k+1 )
The point lies on the y -axis and abscissa is 0.
-k+5k+1=0
K=5
Point of intersection are ( 0, -133)
(1)
26) . Since the inner diameter of the glass = 5cm and height = 10cm
The apparent capacity of the glass = πr2h (1)
= 3.14X2.5X2.5X10 cm3
= 196.25cm3 (1)
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e. it is less by 2/3πr3 = 2/3X3.14X2.5X2.5X2.5cm3
= 32.71cm3 (1)
Actual capacity of the glass = apparent volume – volume of hemisphere
= 196.25 – 32.71 cm3
=163.54cm3 (1)
The bad value depicted by the juice seller is greediness of Money by befooling the customers.
27) Diameter of the well = 10m , height/depth = 14m
Inner radius of the embankment = 5m
Outter radius of the embankment = 10m
As the mud dug out of well is used to make the embankment