QUESTION PAPER -5
Section-A
Answer according to instruction : (Question no. I to 16) (1 mark each)
State whether the following statements are True or False.
1.If the zeroes of a quadratic polynomial ax2+ bx +c are both positive, then a, b and c have the same sign.
2.Every quadratic equation has at least one real root.
3.In an AP if a = 1, an = 20 and S = 399, then n is 38.
4.The value of sin Q + cosQ is always greater then 1.
Fill in the blanks by selecting the proper alternative from those given below each statement.
5.The probability that a non leap year selected at random will contain 53 sunday is =.....
(a) 1/7 (b) 2/7 (c) 3/7 (d) 5/7
6.Dhwani is asked to take a number from I to 100. The probability that it is a prime is
(a) 1/5 (b) 6/25 (c) 1/4 (d) 13/50
7.If Efi = 20, Efixi = 440, then Mean (x) =
(a) 20 (b) 21 (c) 22 (d) 40
8.The pair of equations x+2y+ 5=0 and -3x-6y +1=0 have
(a) a unique solution (c) infinitely many solution
(b) exactly two solution (d) no solution
Fill in the blanks so as to make the following statements true.
9.If the lines given by 3x+2ky =2 and 2x + 5y +1=0 are parallel, then the value k is = 0
10.Cosec2Q .sin2Q =........
11.The distance of the point A (2, 3) from the any point of X-axis is.........
12. PQ is a tangent to the circle with centre O such that OP = 4 cm and OPQ =30°, Then PQ = .....
Answer the following by a number or a word or a sentence :
13.The area of the circle that can be inscribed in a square of side 6 cm is.....
14.What is the equation volume of trustum of cone ?
15.If an event cannot occur, then its probability is ?
16.Amount is very unlikely to happen. Its probability is closest to ?
Section : B
Answer the following questions in short. (Q:17 to 26) (2 Marks each)
17.Prove that 5+2√3 is irrational.
Let us assume that 5+2√3 is rational
5+2√3 = p/q ( where p and q are co prime)
2√3 = p/q-5
2√3 = p-5q/q
√3 = p-5q/2q
now p , 5 , 2 and q are integers
∴ p-5q/2q is rational
∴ √3 is rational
but we know that √3 is irrational . This is a contradiction which has arisen due to our wrong
assumption.
∴ 5+2√3 is irrational
18.Find LCM (30, 50) using Euclid's division algorithm ?
50=30*1+20
30=20*1+10
20=10*2+0
so,LCM of 50 and 30=10
19.If one of the zeroes of the quadratic polynomical (K - 1) x2+ kx + 1 is (-3), then the
Given:-
One of the zeroes of the quadratic polynomial say = (k-1)x² + kx + 1 is ₋3
Solution:-
⇒ p (-3) = 0
⇒ (k - 1)(-3)² + k(-3) + 1 = 0
⇒ 9(k - 1) -3k + 1 = 0
⇒ 9k - 9 - 3k + 1 = 0
⇒ 6k -8 = 0
⇒ Hence, the value of k is
20. Solve the following pairs of equations.
x+y-3.3=0 and 2y-3x-0.6 = 0
Answer :-
2y-3x-0.6 = 0
so 0.6=3y-2x
so 3y-2x=0.6 or 3y=2x-0.6
again x+y=3.3
so 3x+3y=9.9
so 3x+2x-0.6=9.9
so 5x=9.9+0.6=10.5
so x= 2.1
so y= 3.3-2.1=1.2
OR
20. x/a + y/b - a - b = 0, x/a2 + y/b2 - 2 = 0
X/a + y/b = a+b
⇒y/b = a + b - x/a .................. i
x/a² + y/b² = 2 ................ii
⇒x/a² + y/b x1/b = 2
⇒x/a² + a+b/b - x/ab = 2 puttting i
⇒x(b-a)/a²b = (b-a)/b
⇒x = a²
so x = a²
using ii we get
y = b²
so x=a² and y=b²
21. sin Q = a/b then cosQ is?
Answer :- Sinθ=a/b
∴, cosθ=√(1-sin²θ) [∵, sin²θ+cos²θ=1]
or, cosθ=√(1-a²/b²)
or, cosθ=√{(b²-a²)/b²}
or, cosθ=√(b²-a²)/b
22 If sin a = 1/2 and cos B = 1/2 then find value of (a + B)
SinA =Sin 30°
A = 30° ----(1)
ii ) cosB = {1}{2}
cosB = Cos60
=> B = 60° ----(2)
Now,
Value of (A+B)
= 30°+60° [ from (1)&(2) ]
= 90°
Therefore,.
A+B = 90°
OR
22.If 4 tan Q = 3, then obtain value of ( 4sinQ - cosQ/4sinQ + cosQ) is equal to ?
4tanQ = 3 so, tanQ = 3/4
we know, tanQ = perpendicular/base
tanQ = 4/3 = perpendicular/base
So, perpendicular = 3 and base = 4
Now use Pythagoras theorem to find hypotenuse
Hypotenuse = √{perpendicular ²+base²}
∴ hypotenuse = √(3² + 4²) = √(25) = 5
Now, sinQ = perpendicular/hypotenuse = 3/5
cosQ = base/hypotenuse = 4/5
now, (4sinQ - cosQ)
= 4 × 3/5 - 4/5
= 12/5 - 4/5
= 8/5
now,(4sinQ+cosQ)=4×3/5+4/5-1
=12/5 + 4/5
=16/5
( 4sinQ - cosQ/4sinQ + cosQ)= 8/5 ÷ 16/5= 8/16
=1/2
23. If an isosceles triangle ABC; in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.
Solution:
S =ВСх АН = НВ x АН
Theorem: The angle subtended at the centre of a circle by an arc is twice any angle at the circumference standing on the same arc.
So angle AOC = 2 angle ABC
Triangle AOC is an isosceles triangle (OA and OC are radiuses)
So if OM is a perpendicular to AC, then AM=MC and angleAOM = angle MOC =1/2 angle AOC = angle ABC
Triangle AOM and triangle ABH are similar triangle (angle AOM = angle ABC, angle AMO and angle AHB are right -AA)
AM/AO = АН/AB
AM =1/2 AC = 3 cm A0 = R = 9 cm AB = 6 cm
AH = AM X AB/AO = 3х6/9 = 2 cm
Using Pythagoras' theorem
HB = VAB2– AH2 = V36 – 4 = 4/2 cm
S = 4/7 x2 = 8/2 cm²
Answer: the area of the triangle is 8V2 cm², the area of the circle is S, = TR² = 81n cm?
23. Out of the concentric circles, the raidus of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.
Answer :- Lets say centre O
OA=5cm,AC/2=4cm,
dist of O fromAC =Sqrt(5^2-4^2)=3
tangent to the2 inner circle
so radius= 3cm
24 The median of the distribution given below is 14.4. Find the value of x and y, if the total frequence is 20.
Class 0-6 6-12 12-18 18-24 24-30
F 4 x 56 y 1
Class Interval Frequency Cumulative Frequency
0 - 6 4 4
6 - 12 x 4 + x
12 - 18 5 9 + x
18 - 24 y 9 + x + y
24 - 30 1 10 + x + y
______________________________________________
20
______________________________________________
10 + x + y = 20
x + y = 20 - 10
x + y = 10
Median = 14.4
So, Median class is 12 - 18
Median = l + [(N/2 - cf)*i]/f
l = Lower limit of the Median Class = 12
Class Interval = i = 6
Cumulative Frequency (cf) of the class before the Median Class = 4 + x
N = 20
N/2 = 20/2 = 10
f = frequency of the Median Class = 5
⇒ Median = l + [(N/2 - cf)*i]/2
⇒ 14.4 = 12 + {10 - (4 + x)*6]/5
⇒ 14.4 - 12 = {(10 - 4 - x)*6}/5
⇒ 2.4 = {(6 - x)*6}/5
⇒ 2.4 = (36 - 6x)/5
⇒ 2.4*5 = 36 - 6x
⇒ 12 = 36 - 6x
⇒ - 6x = 12 - 36
- 6x = - 24
⇒ 6x = 24
x = 24/6
x = 4
Since, x + y = 10
4 + y = 10
y = 10 - 4
y = 6
So, the value of x is 4 and the value of y is 6.
25. Find the real roots of 6x2 -7x+2=0, by the method of completing the squares.
6x2 -7x+2=0
x2 -7/6x+2/6=0 x2 -7/6x+2/6=0 x2 -2.7/2.6x+1/3=0
Adding and subtracting (7/12)2
x2 -2.7/2.6x+ (7/12)2 - (7/12)2 +1/3=0
(X - 7/12)2 = (7/12)2 -1/3
(X - 7/12)2 = 49/144 - 1/3
(X - 7/12)2 = 1/144
X - 7/12 = 1/12 .................. taking squareroot on both sides
X = 1/12 +7/12 OR X = 7/12 -1/12
X= 8/12 OR X = 6/12
X = 2/3 OR X = 1/2
26.If Diwani were younger by 5 years then what she really is, then the square of her age (in
years) would have been 11 more than five times her actual age. What is her age now ?
Answer :- (x - 5)2 = 11 +5x
(x )2 - 10x + 25 = 11 +5x
(x )2 - 10x - 5x + 25 - 11 = 0
(x )2 - 15x + 14 = 0
(x )2 - 14x - x + 14 = 0
x(x - 14) - 1(x - 14) = 0
(x - 14)(x - 1) = 0
x - 14 = 0 OR x -1 = 0
x = 14 OR x = 1
OR
26. Find the real roots of:63/x +72/x+6 -3=0;
63/x +72/x+6 -3=0;
63(x+6)+72x /x(x+6) =3
63x +378 +72x = 3(x2 +6x)
135x +378 = 3x2 +18x
3x2 +18x- 135x - 378 =0
3x2 - 117x - 378 =0
x2 - 39x - 126 =0.........dividing by 3 on both sides
x2 - 42x +3x - 126 =0.
x(x - 42) + 3(x - 42) - 126 =0
(x - 42)(x -3) =0
(x - 42)=0 OR (x -3) = 0
x = 42 OR x = 3
Section : C
Answer the following as required with calculations : (Questions no. 27 to 34)
(3 marks each)
27. Find the sum of zeroes and product of zeroes: x3+2x2+3x+4
Solution :
If α,β and γ are the zeroes of a cubic polynomial then
x 3 - (α + β+γ)x 2 + (αβ + βγ+αγ)x - αβγ ____________ (1)
x3+2x2+3x+4 = 0 ____________ (comparing (1) & (2))
α + β+γ = 2
αβ + βγ+αγ = -7
αβγ = -14
x 3 -(2 )x 2 + (-7)x – (-14)
x 3 -2 x 2 -7x + 14
28. Find the roots: 3√2x2-5x-√2= 0
3√2x²-5x-√2=0
3√2x²-(6-1)x -√2 = 0
3√2x²- 6x + x-√2=0
3√2x(x-√2)+1(x-√2) = 0
(x-√2)(3√2x+1) = 0
.•. x=√2 or x = -1/3√2
so the roots are √2 & -1/3√2.
Later rationalizing - √2/6
29. Find the sum of last ten terms of AP: 8, 10, 12,.. ., 126
Answer :- a/c to question ,
we invert this series ,
126 , 124 , 122 ,........8
now,
first term of new AP is = 126
common difference = -2
use formula,
Sn =n/2 {2a +(n-1)d
S10=10/2{2 × 126 +(10-1)(-2)}
=5{252 -18}
=5 × 234
=1170
OR
29. In an AP, If Sn = 3n2 +5n and ak = 164. Find the value of k.
Solution:-
Given: Sn = 3n² + 5n
S₁ = 3(1)² + 5(1)
= 3 + 5
T₁ = 8
S₂ = 3(2)² + 5(2)
= 12 + 10
= 22
So, T₁ + T₂ = 22
8 + T₂ = 22
T₂ = 22 - 8
T₂ = 14
Now,
a = 8 and d = 14 - 8 = 6
As ak = 164
or a + (k-1)d = 164
8 + (k-1)6 = 164
6k - 6 = 164 - 8
6k = 164 + 6 - 8
6k = 162
k = 162/6
k = 27
30. Name the type of triangle formed by the verticies A (-5, 6), B (-4, -2) and C(7, 5).
31.Find the mode:
Class 0-5 5-10 10-15 15-20 20-25 25-30 30-35
f 10 15 30 80 40 20 5
Mode class = 15-20
Lower limit (l) = 15
f1 = 80
f0 = 30
f2 = 40
Mode = l + ( f1 + f0 )/( 2f1 - f0 - f2 )h
= 15 +(80 - 30)/(2(80) - 30 - 40 )5
= 15 +(50)/((160) - 70)5
= 15 + 50/90×5
15 + 25 / 9
= 135 +25 /9
= 160 /9
= 17.77
OR
31.Find the mode.
Class 0-10 10-20 20-30 30-40 40-50 50-60
f 22 10 8 15 5 6
Class 0-10 10-20 20-30 30-40 40-50 50-60
f 22 10 8 15 5 6
Cf 22 32 40 55 60 66
n =66
f = 8
l = 20
Cf = 32
h = 10
Median = l + ( n/2 - cf/f)h
Median = 20 + ( 66/2 - 32/8)10
Median = 20 + ( 33 - 32/8)10
Median = 20 + ( 1/8)10
Median = 20 + ( 10/8)
Median = 20 + 1.25
Median = 21.25
32.Let S denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c.If a circle touches the sides BC, CA, AB at D, E, F respectively, Prove that BD = s -b.
Answer :-
Given :ABC is a triangle
BC = a, CA = b, AB = c
To prove : BD = S - b
construction : O is a center of circle and OD and OE are perpendicular on BA & BC respectively
Proof : In triangle BDO & BEO
OB = OB ..........Common side
Angle BDO = Angle BEO.......each of 90
OD = EO.........radius of circle
Triangle BDO =~ BEO
BD = BE ........by CPCT
Similarly, BD = BE =CE=CF =AF=AD
BC = a, CA = b, AB = c.......given
Hence , BD = BE =CE=CF =AF=AD = a/2=b/2=c/2.......(1)
S = a + b + c /2
S = a/2 + b/2 + c/2
S =b/2 + c/2 + BD.......from (1)
S = b/2 + b/2 + BD
S = b + BD
BD = S - b
Hence, proved
33.Find the diameter of the circle whose area is equal to the sum of the areas of the two circles of diameters 20 cm and 48 cm.
To find diameter of circle [ d1 ]
given diameter of 2nd circle= 20cm
radius [ r₂] = 10 cm
diameter of 2rd circle = 48 cm
radius [r₃] = 24 cm
πr₁² =πr₂² + πr₃²
лr₁² = л(r₂² + r₃²)
лr₁²/л = (10)² +(24)²
r₁² = 100 + 576
r₁² = 676
r₁= √676
r₁ = 26 cm
diameter= 2r₁ = 2*26 = 52 cm
34.Find the area of the folower bed (with Semi-circular ends) shown in figure.
Solution:
Length and breadth of a circular bed are 38
cm and 10 cm.
... Area of rectangle ACDF = Length x Breadth = 38 x 10 = 380 cm2
Both ends of flower bed are semi-circles.
Radius of semi-circle =
DF/2 = 10/2 = 5 cm
Area of one semi-circles = π2 r/2= π2 (5)2/2 =25 π/2 cm
Area of two semi-circles = 25 π/2 x 2 =25
Total area of flower bed
= Area of rectangle ACDF + Area of two semi-circles
= (380 + 25 π) cm
Section : D
Answer the followng as required with calculations : (Question no. 35 to 39)
(4 marks each)
35.Draw line segment 6 cm and divide it in 3:4.
Steps of Construction:
1.Draw a line segment AB of length 6cm.
2.Draw a ray AZ making an acute angle with the line AB.
3.As we have to divide the line in the ratio of 3:4. So we will make 3 + 4 = 7 points along AZ.
4. Along AZ we mark 7 arcs taking A as starting point for the first, A1 as a starting point for second and so on till A7.
5.join the A7B
6.With the help of these arcs, this line can be divided into 7 equal points.
7.In our case of 3:4, we join the 3rd point i.e. A3 with the line AB such that the line A3P is parallel to line A7B.
8. P is the required point, the point that divides the line in ratio of 3:4.
OR
35. Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ABC = 60° Construct a triangle similar to triangle ABC with scale factor 5/7 . Justify the construction.
Steps of construction:
(i) Draw a A ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°.
(ii) Draw a ray BX making an acute angle with BC on
the opposite side of vertex A.
(iii) Locate 4 points (as 4 is greater in 3 and 4), B1,
B2, B3, B4, on line segment BX such that BB1 = B,B2
= B2B3 = B3B4.
(iv) Join B4C and draw a line through B3, parallel to
B4C intersecting BC at C'.
(iv) Draw a line through C' parallel to AC intersecting AB at A'. A A'BC' is the required triangle.
36.At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was
found to be 3 minutes less then t2 /4 minutes; Find t.
Answer :- The minute hand is at t
Time needed by minute hand to show 3PM = 60 -t
According to the question ---
60 - t = t^2/4 - 3
60 - t =( t^2 - 12)/4
4(60 - t) = t^2 - 12
240 - 4t = t^2 - 12
- 4t = t^2 - 12 - 240
t^2 +4t - 252 = 0
D = b^2 - 4ac = 4^2 - 4*1*(-252) = 16 +1008 = 1024
Now , -b +/- √D/2a
= -4 +/- 1024/ 2
there will be two roots
= -4 +1024/2 and -4 - 1024/2
= 14 and - 18
but time cannot be negative.
So t = 14
37.From a balloon vertically above a straight roud, the angles of depression of two cars at an instant are found to be 45° and 60°. If the cares are 100 m apart, Find the height of the ballow.
Solution:-
Let the two angle be x = 45° and y = 60° respectively.
And the distance between the two cars is d = 100 m
Height of balloon is AB = h meter (See the figure)
Let BC = x meter
In triangle ABO,
tan 45° = AB/BO ⇒ 1 = h/(x+100) ⇒ h = x + 100 ........(1)
In triangle ABC,
tan 60° = AB/BC
⇒ √3 = h/x
⇒ h = x√3 ...... (2)
As h = x + 100
⇒ x + 100 = x√3
⇒ 100 = √3 x - x
⇒ 100 = x(√3 - 1)
⇒ x = 100/(√3 - 1)
Now rationalising
⇒ x = 100 /(√3 - 1) (√3 + 1)/(√3 +1)
X=100(√3 +1)/3-1
X=100(√3 +1)/2
X=50(√3 +1)
38.A solid toy is in the form of a hemisphere surounded by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy If a cube circumscribes the toy. Then find difference of the volumes of cube and the toy Also find the total surface area of the toy.
Radius of the hemisphere and the cone = 4 cm
Height of the cone = 4 cm
Volume of the toy = volume of the hemisphere +volume of the cone
=2/3 π r3 + 1/3 πr2h
= 2x22/(3x7)x(4)3 + 1x 22(4)3/( 3x7)x 4
=1408/7 cm
A cube circumscribes the given solid. Therefore,
edge of the cube should be 8 cm.
Volume of the cube = 83 = 512 cm3
Difference in the volumes of the cube and the toy =
512 -1408/7 = 310.86 cm3
Total surface area of the toy=curved surface area of cone
= πrl + 2 πr2, where l= √h2 +r2 =√42 +√42= 4 √2
= π r (I+ 2r)
=22/7 x4 x (4 √2 + 2 x 4)
=88 x 4(√2 +8)
= 171.68 cm2
39.Prove that, ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Step 1:
Given Data: Δ ABC ~ Δ PQR
To Prove: (ΔABC) / (ΔPQR) = ({AB} / {PQ})2=({BC} / {QR})2=({CA} / {RP})2
Step 2:
Draw AM ⊥ BC, PN ⊥ QR
(ΔABC) / (ΔPQR) = (½ × BC × AM) / (½ × QR × PN)
= BC/QR × AM/PN........................................... [I]
In Δ ABM and Δ PQN,
Step 3:
∠B = ∠Q (Δ ABC ~ Δ PQR)
∠M = ∠N (each 90°)
Step 4:
So, Δ ABM ~ Δ PQN
AM/PN = AB/PQ ... ………………. [ii]
AB/PQ = BC/QR = CA/RP (Δ ABC ~ Δ PQR)..................... [iii]
Step 5:
Therefore Equation (i)
(ΔABC) / (ΔPQR) = BC/QR × AM/PN
= AB/PQ × AB/PQ [From Equation (ii) and Equation (iii)]
=({AB} / {PQ})2
Step 6:
Using Equation (iii)
(ABC) / (PQR) => ({AB} / {PQ})2=({BC} / {QR})2=({CA} / {RP})2
39.State and prove the pythagoras theorem.
Given: A right-angled triangle ABC.
To Prove- AC2 = AB2 + BC2
Proof: First, we have to drop a perpendicular BD onto the side AC
We know, △ADB ~ △ABC
Therefore, ADAB=ABAC (Condition for similarity)
Or, AB2 = AD × AC ……………………………..……..(1)
Also, △BDC ~△ABC
Therefore, CDBC=BCAC (Condition for similarity)
Or, BC2= CD × AC ……………………………………..(2)
Adding the equations (1) and (2) we get,
AB2 + BC2 = AD × AC + CD × AC
AB2 + BC2 = AC (AD + CD)
Since, AD + CD = AC
Therefore, AC2 = AB2 + BC2
Hence, the Pythagorean thoerem is proved.
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