Time : 3 hrs. Total Marks : 80
Write in clear and legible handwriting.
This question paper has four sections A,B,C and D and question number from 1 to 39
All questions are compulsory. Internal options are given
Numbers to the right represents the mark to for the question
Draw the neat diagram , wherever necessary
New section should be written on new page and write the answer in the numerical order
SECTION A(16 marks)
Answer the following as per the instruction given.This section has 16 questions and each carries 1 mark each.
Write true or false for the following questions ( 1 to 4)
8sec2π - 8tan2π = 8 (TRUE)
8(sec2π - tan2π)
8(1) = 8
7 x 11 x 13 + 13 is a prime number(FALSE)
7 x 11 x 13 + 13=13(7 X 11 + 1) = 13 ( 77 +1) = 13 X 78 which is not a prime number
½, ⅓, ¼ are arithmetic progression(FALSE)
(⅓ - ½) ≠ (¼ -⅓ ), therefore it is not AP
If one of the root of the quadratic equation X2 - 4X + m = 0 is 3 then m=3 (TRUE)
Fill in the blanks to make eac h statement true. ( 5 to 8)
If H.C.F ( 12, k) = 6 and LCM(12,k) = 36 then k= 18
12 X k = HCF(12,k) X LCM(12,k)
12 X k = 6 x 36
k = 18
If πΆ and π· are zeros of polynomial 3x - X2 + 8 then πΆπ· = -8
In above quadratic equation a = -1, b =3, and c =8
Now πΆπ· = c/a = 8/(-1) = -8 ( negative 8)
If 27x + 63y = 45 and 63x + 27y = 135 then x + y = 2
Adding both of above linear equations.
90x + 90y = 180
x + y = 2
Coordinates of midpoint of line segment AB joining A(2a - b, b) and B(b, 2a-b) is (a,a)
Co ordinate of midpoint ( (2a-b +b) / 2, (b+2a-b)/2)
(a,a)
Answer in one sentence, one word or one number
Which is the median class for the following frequency distribution.
Answer : Median class : 20-30
N/2 = 31, therefore cumulative frequency just greater than 31 is 35 , which belongs to class 20-30
If P(A) - P(A) = 0.8 than find the value of P(A)
We know that P(A) + P(A bar ) is 1, therefore adding this two equation.
2P(A) = 1.8, therefore , P(A) = 0.9
ANSWER: Value of P(A) = 0.9
Find the diameter of the circle whose circumference and area are equal in number.
AREA = CIRCUMFERENCE
πr2 = 2πr
r = 2
Therefore the diameter will be 4
For what value of acute angle π, cot2π.cot7π =1
cot2π.cot7π =1
cot2π= 1/ cot7π
cot2π= tan7π
cot2π= cot (90-7π)
2π= 90-7π
9π= 90
π= 10
Select the proper alternative to make each statement true.
If a sphere of radius r divided into 4 equal parts, than total surface area of each part is [answer : (b)]
πr2
2πr2
3πr2
(½)πr2
If the pair of linear equation 2x + 2y +2 =0 and 4x + ky + 8= 0 has unique solution, then k ≠ [answer : (a)]
4
2
-4
8
In the given figure PA and PB are tangent to the circle with centre O. ∠AOP = 550 than ∠APB= [answer : (b)]
350
700
1250
1100
An unbiased coin is tossed thrice,what is the probability of getting at least two heads. [answer : (c)]
⅜
⅛
½
⅔
Total possibilities : HHH, HHT, HTH,THH, TTT, TTH, THT, HTT ( 8 ), out of these, condition when we have at least two head are HHH, HHT, HTH,THH (4 nos),
Therefore the probability = 4/8 = ½
SECTION B (20 marks)
Solve the following questions showing calculations. ( 17 to 26) , Each carries two(2) marks.
Prove that 5 + 2√7 is irrational
Let, 5 + 2√7 be rational.
So 5 + 2√7 =π/π,
π€βπππ′π′πππ′π′πππ πππ‘πππππ πππ π ≠ 0
2√7 = [a/b] – 5
√7 = [a – (5b)] / 2b
Since ‘a’ and ‘b’ are integers a – 5b is also an integer.[a – (5b)] / 2b is rational. So RHS is rational.
LHS should be rational. but it is given that √7 is irrational .Our assumption is wrong.
So 5 + 2√7 is an irrational number.
Find the zeros of quadratic equations 6X2 - 13X + 6 = 0
6X2 - 13X + 6 = 0
Splitting the middle term
6X2 - 9X-4x + 6 = 0 ( 6 x 6 = 36 and -9 x -4 = 36)
(2X-3)(3x -2)= 0
x= 3/2 or x = 2/3
Verify whether the linear pair of equations (4/3)x + 2y = 8 and 2x + 3y = 12 are consistent or not.
Comparing above equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 =0
a1= 4/3 b1= 2 c1 = -8 , a2=2 , b2=3 c2 =-12
a1 / a2=⅔, b1/b2=2/3 c1/c2 = -8 /-12 = ⅔
Since a1/a2= b1/b2= c1/c2 = ⅔, the pair of equation is dependent consistent and has infinite solutions as they represents coincident lines.
OR
Solve the pair of equations by substitution method.
x + y = 4 and 2x = 8 - 3y
x + y = 4 ………………(i)
2x = 8 - 3y……………..(ii)
From (i), y = x -4 , and putting this value y into (ii)
2x = 8 - 3y
2x = 8 - 3(x - 4)
2x = 8 -3x +12
5x = 20
x = 4
Now x + y = 4
Therefore 4 + y = 4
Therefore y=0
x=4 and y=0 is the solution of the linear pair of equations x + y = 4 and 2x = 8 - 3y
If P, Q, R are interior angle of a triangle , prove that sec[(P+Q)/2] = cosec(R/2)
∠P + ∠Q + ∠R = 1800 ( ∵ P, Q, R are interior angle of a triangle)
∠P + ∠Q = 1800 - ∠R
LHS = sec[(P+Q)/2]
= 1/ cos[(P+Q)/2]
= 1/ cos[(1800 - R)/2]
= 1/ cos[900 - R/2]
= 1/ sin(R/2)
= cosec(R/2)
= RHS
Since LHS = RHS, sec[(P+Q)/2] = cosec(R/2) proved for the , interior angle P,Q, R or the triangle.
If 2 sinπ + cosπ = 2, find the value of tanπ
2 sinπ + cosπ = 2
Dividing by cosπ
∴ 2 (sinπ / cosπ) + 1 = 2 /cosπ
∴ 2 tanπ + 1 = 2secπ
∴ 2 (secπ - tanπ) = 1
∴ secπ - tanπ = ½ ………..(i)
Now , we know that sec2π - tan2π = 1
∴ (secπ - tanπ)(secπ + tanπ) =1
∴ (½)(secπ + tanπ) =1
∴ (secπ + tanπ) =2………..(ii)
Subtracting (i) from (ii)
∴ 2 tanπ) = 3/2
∴ tanπ) =¾
Answer tanπ = ¾
OR
2tan245 + x - sin260 = 2, then find the value of x
2tan245 + x - sin260 = 2
2( 1)2 + x - (√3/2)2 = 2
x = ¾
Answer : x = 3/4
As shown in the figure , quadrilateral PQRS is drawn to circumscribe a circle, prove that
Prove that PQ + RS = QR + SP
As we know that , length of tangents to a circle from an external point is equal.
PA = PB….(i)
QA = QD…(ii)
RC = RD …(iii)
SB = SC….(IV)
Now LHS = PQ + RS
= (PA + AQ) + (RC + CS)
= (PA + QA) + (RC + SC)
= PB + QD + RD + SB
= ( QD + RD) + (PB + SB)
= (QR) + ( SP)
= QR + SP
= RHS
LHS = RHS , Hence Proved
OR
Two concentric circles are of radii 29 cm and 21 cm. Find the length of chord of the larger circle which touches the smaller circle.
In the above figure, we have to find the length of CD
Now,
AC2 = AB2 + BC2
292 = 212 + BC2
BC2 = 292 - 212
BC2 = (29 - 21)(29 + 21)
BC2 = (8)(50)
BC2 = 400
BC = 20
CD = 2*BC
CD = 2 *20
CD = 40
ANSWER :Length of chord of the larger circle which touches the smaller circle is 40 cm.
For the following grouped frequency distribution find the mode.
The class with highest frequency is 40-55,
So 40-55 is modal class.
l = 40 ( lower limit of modal class)
f1 = 7 ( frequence of modal class)
f0 = 3 ( frequency of class preceding modal class)
f2 = 6 ( frequency of class succeeding modal class)
h = 15
Mode = l + [ (f1- f0)/(2f1- f0 - f2)]*h
Mode = 40 + [ (7-3)/(2*7- 3 - 6)]*15
Mode = 40 + [ 4/(14- 3 - 6)]*15
Mode = 40 + [ 4/(5)]*15
Mode = 40 + (12)
Mode = 52
Answer : The mode of given data is 52
Salma and Mona are friends. What it the probability that both have
Different birthday
Total no of possibility for birthdates are 365,
Now , Possibility of different birth date are 364
Therefor probability of having different birthday = 364/365
Same birthday in the year 2019.
Possibility of same birthdate = 1
Therefore probability of having same birthday = 1/365
Find the root of the quadratic equation 5x = 6 + 2/x by completing the square method.
Here the quadratic equation is 5x = 6 + 2/x ⇒5x2 = 6x + 2
⇒25x2 = 30x + 10
⇒25x2 -30x + 9 = 10 + 9
⇒(5x - 3)2 = 19
⇒5x - 3 = +/- √19
x = (3 +/- √19)/5
x = (3 +√19)/5 and x = (3 - √19)/5
Answer : (3 +√19)/5 and (3 - √19)/5 are roots of quadratic equation 5x = 6 + 2/x
OR
Find the root of quadratic equation √2x2 + 7x + 5√2 = 0
√2x2 + 7x + 5√2
Splitting the middle term
√2x2 + 2x + 5x + 5√2 ( 5√2 x √2 = 10, 5 x 2 =10 and 5 +2 =7)
√2x(x + √2) + 5 (x +√2) = 0
(x + √2)(√2x+5)=0
x + √2 = 0 or √2x+5 = 0
x = - √2 or x = -5/√2
- √2 and -5/√2 are roots of the given quadratic equation √2x2 + 7x + 5√2 = 0
In the given figure if PQ ∥ BC , find the value of AB
As per the proportionality theorem
AP/PB = AQ/QC
AP = PB( AQ/QC)
AP = 7.2 ( 1.8/5.4)
AP =2.4
.
AB = AP + PB
AB = 2.4 + 7.2 = 9.6
ANSWER : AB =9.6 CM
SECTION C (24 marks)
Answer the following questions showing calculation (27 to 34) each carries 3 marks.
On dividing 3x3 + x2+ 2x + 5 by a polynomial g(x), the quotient and remainder were 3x - 5 and 9x + 10 respectively, find the g(x)
Here the dividend = 3x3 + x2+ 2x + 5, divisor = g(x), quotient = (3x-5) and remainder = (9x + 10)
Dividend = Divisor x Quotient + remainder
3x3 + x2+ 2x + 5 = g(x) (3x-5) + ( 9x +10)
g(x) (3x-5) = 3x3 + x2+ 2x + 5 - ( 9x +10)
g(x) (3x-5) = 3x3 + x2 - 7x - 5
g(x) = (3x3 + x2 - 7x - 5)/(3x-5)
g(x) = x2 + 2x + 1
Answer : the divisor g(x) = x2 + 2x + 1
Sum of areas of two squares is 468 sqm. If the difference of their perimeter is 24 m. find the sides of two squares
Assume the sides of squares are x and y
Sum of areas of two squares is 468 sqm
x2 + y2 = 468
Perimeters of squares with sides x and y will be 4x and 4y respectively
The difference of perimeters is 24m
4x - 4y = 24
x - y = 6
x = y + 6 ….(i)
(y + 6)2 + y2 = 468
y2 + 12x + 36 + y2 =468
2y2 + 12x - 432 = 0
y2 + 6x - 216 = 0
Splitting the middle term
y2 + 18x-12x - 216= 0 ( 18 x -12 = 216, 18-12=6)
(y+18)(y-12)=0
y+18 = 0 or y-12=0
y= -18 or y =12
As, the side can not be negative, y=12
Putting value of y in (i)
x = y +6
x = 12 + 6 =18
Therefore the sides of squares will be 18 and 12 meters respectively.
For what value of n , nth term of two AP 65, 67,69 ,.... AND 10,17, 24 equal.
Nth term is given by the formula an = a +(n-1)d
For AP 65, 67,69 ,..,
a1 = 65, d1=2
For 10,17, 24
a2=10, d2=7
For nth term to be equal
a1 +(n-1)d1 =a2 + (n-1)d2
65 + (n-1)*2 = 10 + (n-1)*7
7n-7 -2n +2 = 55
5n = 60
n = 12
Answer : 12th term of APs will be same.
OR
Find the sum of all terms of AP -2, -5,-8…-227..
Value of an = -227, a = -2, d= -3
an = a +(n-1)d
-227 = -2 +(n-1)(-3)
3(n-1) = 227 -2
3(n-1) = 225
n-1 = 75,
n = 76
Now sum S of first 76 term of the AP
S = n(a+l)/2, where a = -2, l= -227, n=76
= 76(-2-227)/2
= 38(-229)
= -8702
Answer : Sum of all terms of AP -2, -5,-8…-227 is -8702
If P(2,3), Q(3,-2), R(-3,-5) and S(-4,-2) are vertices of a quadrilateral. Find the area of quadrilateral PQRS.
The length of the minute hand of a clock is 14 cm. Find the area swept by minute hand in 15 minutes. Find the distance to be swept to complete one revolution.
OR
In the given figure OACB is quadrant of a circle with centre O and having diameter of 7 cm. If OD is 2 cm, Find the area of the
Quadrant OACB
Shaded region.
Area of quadrant section
Radius r = 7/2 cm
Area = πr2/4 = (¼)(22/7)(7/2)2
= 22 * 7 / 16
= 77/8
=9.63 cm2…………..(i)
Area of shaded region = Area of quadrant - area of triangle ODB
Area of triangle ODB = ½(OB)(OD)
= ½(7/2)(2)
= 3.5 cm2 ……….(ii)
From (i) and (ii)
Area of shaded region = Area of quadrant - area of triangle ODB
Area of shaded region = 9.63 - 3.5
= 6.13
Answer : a) Area of quadrant section = 9.63 cm2
b) Area of shaded region = 6.13 cm2
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 28 cm and total height of the vessel is 26 cm.Find the inner surface area of the vessel.
Cylinder : r = 14 cm ( Given the dia meter of the hemisphere)
h = 26 cm
Hemisphere = r = 14cm
Inner surface area of the vessel = Surface area of the Base of cylinder + curved surface area of cylinder + Surface area of the hemisphere.
= πr2 + 2πrh + 2πr2
= πr( r + 2h + 2r)
= πr( 3r + 2h)
= 22*14(3*14 + 2*26)/7
= 44(42 + 52)
= 4136 cm2
Answer : Inner surface area of the vessel is 4136 cm2
Following frequency distribution shows the age of 100 persons. Find the median of the data.
Solution
N/2 = 50, this lies between interval 40-50, frequence for this interval is 38 and Cumulative frequency is 31
Now : l= 40 , f = 38 and cf = 31 h = 10, n = 100
Median = l + ((n/2-cf)/f)h
Median = 40 + ((100/2-31)/38)10
Median = 40 + ((50-31)/38)10
Median = 40 + (19/38)10
Median = 40 + 5 = 45
ANSWER : Median of the given data.is 45
OR
The mean of following frequency distribution is 18. Find the missing frequency.
Mean = ∑ fixi / ∑ fi
18 = (1008 + 16f) / (55 + f)
990 + 18f = 1008 + 16f
2f = 18
f = 9
Answer : The missing frequency is 9
Prove that the length of the tangents drawn from an external point to a circle are equal.
SECTION D (20 marks)
Solve the following,( 35 to 39) Each carries 4 marks each.
A boat goes 40 km upstream and 49 km downstream in 15 hours. In the same river it can go 25 km upstream and 35 km downstream in 10 hours. Determine the speed of the stream and that of the boat in still water.
Assume that speed of boat in still water is x km/hr and that of river stream is y km/hr
So speed of boat while going upstream will be (x - y) km/hr
While speed of boat while going downstream will be x + y km/hr
boat goes 40 km upstream and 49 km downstream in 15 hours
[40/(x-y)]+[49/(x+y)] = 15….(i)
boat goes 25 km upstream and 35 km downstream in 10 hours
[25/(x-y)]+[35/(x+y)] = 10
[5/(x-y)]+[7/(x+y)] = 2……(ii)
Putting 1/(x-y) = a and 1/(x+y) = b in (i) and (ii)
40a + 49b = 15….(iii)
5a + 7b = 2…..(iv)
Now multiply (iv) by 7 for elimination of b term
35a + 49b = 14….(v)
Now subtract (v) from (iii) will get
5a =1
A = ⅕, putting this value in (iv)
1 + 7b = 2
7b = 1 , b=1/7
1/(x-y) = a = ⅕
x-y =5 (vi), similarly x + y = 7..(vii)
Therefore addition of (vi)
and (vii) will become
2x = 12
x = 6 km/hr
y = x -5
y = 6-5 =1 km/hr
Answer : The speed of the stream is 1 km / hr and that of the boat in still water is 6 km/hr
As observed from the top of 100 m high hill, the angle of depression to the top of a tower is 300 and angle of depressions to the bottom of the tower is 450.Find the height of the tower and distance between base of tower and base of the hill. (take 1/√3 = 0.58)
In the above figure AB is hill and AB = 100 meter. CD is the tower.
Assume that the height of tower is h
Therefor AE = 100- h
In triangle ABD, ∠ADB = 45
tan(∠ADB) = AB/BD
tan(45)= 100/BD
1=100 /BD
BD = 100 = EC
Now in triangle AEC, ∠ACE = 30
tan(∠ACE) = AE/EC
tan(30) = (100-h) / 100
½ / (√3/2) = (100-h) / 100
1/√3 = (100-h) / 100
0.58 = (100-h) / 100
58 = 100 - h
h = 100-58 = 42
Answer : Height of the tower is 42 meter and distance between base of tower and base of the hill is 100 meter.
A container shaped like a right circular cylinder having radius 6 cm and height 15 cm is full of ice cream. The ice cream is to be filled in cone of height 12 cm and radius 3 cm, with a hemispherical shape on the top. Find the number of such cones which can be filed with the ice cream.
Cylinder : Height h = 15cm, radius 6 cm
Volume of ice cream = πr2h
= π x 6 x 6 x 15…………(i)
Cone Height = 12 cm, radius = 3 cm
Volume of Ice cream cone = ⅓ x πr2h
= ⅓ x π x 3 x 3 x 12……(ii)
No. of cones which will can be filled will be = (i) / (ii)
= (π x 6 x 6 x 15)/(⅓ x π x 3 x 3 x 12)
= 15
Answer 15 cones can be filled with the icecream from the cylinder.
In ΞXYZ if XY2 + XZ2 = YZ2 , then prove that ∠X = 900
OR
BL and CM are medians of ΞABC right angled at A.
Prove that 4(BL2 + CM2) = 5BC2
Draw a right triangle with sides (other than hypotenuse) are of length 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times corresponding sides of the given triangle.
OR
Draw a circle radius of 4.5 cm. From a point 7.5 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Write steps of construction.
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