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PRACTICE PAPER- 4 2017-18
CLASS X
MATHEMATICS
BLUE PRINT
Unit / Chapters / VSA (1mark) / SA (2mark) / SA (3mark) / LA (4Marks) / TotalNUMBER SYSTEM / NUMBER SYSTEM / 1 / 1 / 1 / 6
ALGEBRA / POLYNOMIALS / 1 / 1 / 1 / 8
LINEAR EQUATIONS / 1 / 1 / 6
QUADRATIC EQUATIONS / 1 / 1 / 6
CO-ORDINATE GEOMETRY / CO-ORDINATE GEOMETRY / 1 / 1 / 1 / 6
GEOMETRY / TRIANGLES / 1 / 1 / 1 / 8
CIRCLES / 1 / 4
CONSTRUCTIONS / 1 / 3
TRIGONOMETRY / INTRODUCTION / 1 / 1 / 1 / 6
APPLICATIONS / 1 / 1 / 6
MENSURATION / AREA RELATED TO CIRCLES / 1 / 3
SURFACE AREA AND VOLUMES / 1 / 1 / 7
STAT. AND PROBABILITY / STATSTICS AND PROBABILITY / 1 / 2 / 1 / 11
6(1) / 6(2) / 10(3) / 8(4) / 30(80)
PRACTICE PAPER- 4 2017-18
CLASS X
MATHEMATICS
Time -3hrsM.M.-80
General Instructions:
1.All questions are compulsory.
2.The question paper consists of 30 questions divided into four sections A,B,C and D. Section A comprises of 6 Questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions of 4 marks each.
4.There is no overall choice
5.Use of calculators is not permitted.
SECTION –A
Q1. From the given figure find the angle of elevation ‘a’
Q2. A card is dropped from a pack of 52 playing card find the probability of an ace .
Q3 The end poins of a diameter of a circle are P (-3, 8) and Q (7, 4). Find the centre of circle.
Q4 If and β are the zeroes of the quadratic polynomial then find+ β
Q5. Write a quadratic polynomial the sum and product of whose zeros are 3 and -2.
Q6. Sides of a triangle are 7cm, 24 cm., and 25 cm. Find whether it is a right triangle, if so
Find the length of its hypotenuse.`
SECTION –B
Q7. Using Euclid’s division Lemma find the H.C.F. of 135 and 225.
Q8. Provethatcos4+sin4+2 sin² cos² = 1.
Q9. Find the co-ordinates of the point which divides the line segment joining the points (6, 3) and
(-4, 5) in the ratio 3:2 .
Q10 . Find the value of k for which the quadratic equation (K+4) x2+(K+1)x+1=0 has equal roots.
Q11. Find the 12th term of A.P 9, 13, 17, 21, 25, ------
Q12. If the mean of 6, 4, 7,p and 10 is 8, find the value of p.
SECTION –C
Q13. A train travels a distance of 300 km at constant speed. If the speed of the train is increased by 5 km an hour, journey would have taken 2 hours less. Find the original speed of the train.
Q14. Draw a circle of radius 4cm. and construct a pair of tangents to the circle which are inclined to each other at 300.
Q15. Find the value of K for which the points (7, -2), (5, 1) and (3, K) are collinear.
Q16. Find the mode of the following data:
Marks / 0 – 10 / 10 – 20 / 20 - 30 / 30 - 40 / 40 - 50No. of students / 3 / 12 / 32 / 20 / 6
Q17. Prove that √is irrational.
Q18.
Q19. The area of an equilateral triangle is 17320.5 cm2. About each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. (Useπ=3.14 and √3= 1.73205)
Q20. . Prove thatp; - = -
Q21. A bag contains 5 red balls, 8 white balls, 4 green balls & 7 black balls. If one ball is drawn at random, find the probability that is
i) black ballii) red balliii) not a green ball.
22. For what values of a and b does the following pairs of linear equations have an infinite number of solutions:
2x + 3y = 7;
a (x + y) -b (x - y) = 3a + b – 2
SECTION –D
Q23. Find all the zeros of 2x4-3x3-3x2+6x-2.If two of its zeros are √2 and-√2.
Q24. Solve graphically the pair of equations 2x + 3y = 11 and 2x - 4y = -24. Hence find the value of coordinate of the vertices of the triangle so formed.
Q25. The angle of elevation of a jet plane from print A on the ground is 60˚. After a flight of 15 seconds the angle of elevation changes to 30˚. If the Jet plane is flying at constant height of 3600√3 m. Find the speed of Jet plane.
Q26. One Card is drawn from a well shuffled deck of 52 cards. Find the probability of getting.
(i) a king of red colour. (ii) a face card (iii) A spade (iv) a black face card.
Q27. A bucket is in the form of a frustum of a cone and hold 28. 490 litres of water. The radii of the top and bottoms are 28cm and 21 cm respectively. Find the height of the bucket.7
Q28. State and prove basic proportionality theorem
Q29. Sum of the areas of two squares is 468m2. If the difference of their perimeter is 24m. Find the sides of the two Squares.
Q30. Prove that the length of tangents drawn from an external point to a circle, are equal.
Time -3hrs M.M.-80
CLASS X
MATHEMATICS
PRACTICE PAPER- 4 2017-18
Marking Scheme
SECTION A
- (b) sol : tanA=perpendicular/base
=BC/AB
= =
tanA = tan300
A = 30 ̊ 1 Mark
- p(ace card)=4/52=1/13 1 Mark
- P 1 R(x,y) 1 Q 1 Mark
x= x1+x2 = (-3 +7)= 4/2 =2
2 2
Y= y1+y2 /2= 8+4/2 = 12/2=6
4. α+β=-2 1 mark
5. x2-3x+2 1mark
6. Yes , hypotenuse=25 1 mark
SECTION –B
7. 225=135x1+90½
135=90x1+45½
90=45x2+0½
H.C.F.=45½
8. We know that sin2 + cos2 =1 Taking Square of both sides (sin2 + cos2 )2 =12(1Mark)
(sin2 )2 + (cos2)2 + 2 sin2 .cos2)= 1
Therefore, sin4 + cos4+ 2sin2cos2=1(1Mark)
9. Let P(x, y) divides the line segment joining A(6, 3) and B(-4, 5) in the ratio 3 :2
∴ x= 3 x -4 + 2x6 =-12+12 = 01mark
3+25 3 2
Y= 3x5 + 2x3 =15+6 = 21 . .
5 5 5 A (6,3) P(x,y) B (-4,5)
∴ Required point is P(0, 21/5)1mark
Q10. a = k+4 b= k+1, c = 1
For equal roots, D=0 1 Mark
⇒ b2-4ac=0
⇒ (k+1)2-4x(k+4)x1=0
⇒k2+12+2k-4k-16=0
⇒ k2-2k-15=0
⇒ k2-5k+3k-15=0
⇒k(k-5)+3(k-5)=0
⇒ (k-5) (k+3)=0
⇒ k = 5, k = -3 Marks
Q11). a = 9, d = 13-9 = 4
∴ a12 = a+(n-1)x d 1 Mark
= 9 + 11x4 = 9 + 44 = 53 1 Mark
Q12) (1Mark)
27+p = 40
p = 13 (1Mark)
SECTION-C
Q13)Let speed of the train is x km/h 1.
∴ Increased speed = (x+5) km/h
Distance =S x T
Distance = 300 km
According to question
300/x – 300/x+5 = 2 2 Mark
⇒ 300[1/x– - 1/(x+5)] = 2
⇒=
⇒ ==
+5x= 750
+5x-750 =0
+30x-25x-750=0
x(x+30)-25(x+30)=0
(x+30)( x-25)=0
x= -30 is rejected 2 Mark
∴speed of train is 25 km/h
Q14. For correct construction 3 marks
.
Q15. Since A(7, -2), B(5,1) & C(3,K) are collinear
∴ar (∆ABC)=0 (1 Mark)
⇒ 1/2 [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=0
⇒1/2 [7(1-K)+5(K-(-2))+3(-2-1)]=0
⇒1/2 [7-7K+5K+10-6-3]=0
⇒ -2K+17-9=0
⇒ -2K=-8
⇒ K=8/2=4 (2 Mark)
Q16).
Modal Class = 20-30, l = 20, f0 = 12, f1 = 32, f2 = 20, h = 10 1 markMode = ½ mark
1 mark
= = 20+6.25 = 26.25 ½ mark
Q17 )Let be a rational number
Let = ½ mark
= p
11q2 = p2
11 divides p2 hence 11 divides P.
Let p = 11c
11q2 = 121 c2 1.5 mark
Or q2 = 11c2
11 divides q2 Hence 11 divides q
From (1) and (2) p and q have a common factor 11 which contradicts our assumption.is irrational 1 mark
.
Q18 ) In ABC and DAC
BAC = ADC
C = C
2 mark
1 mark
Q19).
. 1 Mark
Area of a Δ = 17320.5 cm2
⇒ √3/4 side2 = 17320.5
⇒ side2 = 173205/10 X 4/√3
= 173205/10 X 4/1.73205
Side = 2X100=200cm. 1 Mark
∴radius = 200/2 = 100 cm.
∴ Area of shaded portion = area of Δ – 3x area of sectors
= 17320.5 – 3x (60/360 x 3.14 x 100 x 100)
= 17320.5-15700
= 1620.5 cm2 1 Mark
Q20.)
LHS = 1 mark
1 mark
= = 2 cosec A = RHS 1 mark
Q21). Total balls = 5+8+4+7 = 24
(i) ∴ P (getting black balls) = 7/24
(ii) P (getting red balls) = 5/24
(iii) P (not getting green balls i.e getting red, black, white)
= 5+7+8 = 20/24= 5/6 ( 1Mark for each part)
Q22). The system has infinitely many solutions:
1/2 mark
1.5 mark
Equating (1) and (2), we get a = 5b
Equating (2) and (3), we get 2a - 4b = 6 1 mark
On solving, we get b = 1 and a = 5.
Q23. X2-2 is a factor of given polynomial 1mark
Division 2marks
Other two zeros 1mark
Q24. We have to solve the pair of equations graphically
2x + 3y = 11 … (1)
2x - 4y = -24 … (2) 1 mark for each table
X / 1 / 4 / -2y / 3 / 1 / 5
X / -12 / 0 / -10
y / 0 / 6 / 1
point of intersection x = -2, = 5
The triangle formed is shaded as ABC coordinates are
A (-2,5) B (-12,0) C(5.5,0). 2 marks
Q25.
1 mark
Let AB be the building and CD be the tower
Let CD = h m, AB = 60m
In rt. Δ BAC
Tan 600 = AB/AC
⇒ √3 = 60/AC
⇒AC = 60/√3 ……….(1) 1 Mark
In rt. Δ BED
Tan 300 = BE/DE
⇒1/√3 = 60-h/AC
⇒ AC = (60-h) √3 ………….(2)1 Mark
From (1) & (2)
60/√3 = (60-h) √3
⇒ 60 = (60-h) √3 x √3
⇒ 60 = (60-h) x 31 mark
Q26. 1 mark for each correct ans.
Q27. Let height of the bucket be h cm
r1 = 28 cm r2 = 21 cm
and Volume = 28.490 l = 28.490 × 1000 cm3
= 28490 cm3 ( 1 Mark)
V = 28490 cm3
=> 1/3 π h (r12 + r1.r2 +r22) = 28490 ( 1 Mark)
=> 1/3 × 22/7 × h × (282 +28 × 21 + 212) = 28490
=> 22/21 × h × (784+588+441) = 28490
=> 22/21 × h × 1813 = 28490
=> h = 28490 × 21
22 × 1813
= 15 cm
∴height of the bucket is 15 cm. 2Marks
Q28.
Basic proportionality theorem
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 ratio. 1 mark
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. 1 mark
Proof:
Now, area of ∆ADE = x base x height = x AD x EN 2 marks
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..
29. Let side of smaller square be x m.
Perimeter = 4x m.
Perimeter of larger square =24+4x
Side = Perimeter /4 =(6+x) m. ( 1 Mark)
ATQ
x2+ (6+x)2= 468 ( 1 Mark)
x2+ 36+x2 +12x -468=0
x2+ 6x – 216 =0
x2+ 18x-12x – 216 =0
x(x+18)-12(x+18)= 0
(x+18)(x-12) =0
x=-18 or 12
Reject -18 ( 2 Mark)
Sides 12m and 18m
Q30
1 Mark
Given:- A circle c(o, r) AP and BP are two tangents drawn from an external point P.
To Prove:- PA=PB
Construction:- Join OP, PA & OB 1 Mark
Proof:- Since OA | PA
And OB | PB radius through the point of the contact is | to the tangent
In rt. ΔsOAP & OBP
Hypotenuse OP=OP [common]
OA=OB [radii of same circle]
∴ ΔOAP ≅ ΔOBP [By RHS congruence rule]
Hence PA=PB [by cpct] 2 Mark