Section-A
Answer according to instruction : (Question no. 1 to 16) (1 mark each)State whether the following statements are True or False.
1. The product of its zeroes : 3x2 +6x +2 is (-3).
Ans :-False
2. P(x) = 25x2 – 10x + 1, Discriminant is zero.
Ans :- True
3. 1/2,1/3,1/4,1/5..... is a arithmetic progression.
Ans :-False
4. sin (60°). cos (60°) =3/4
Ans :-False
Fill in the blanks by selecting the proper alternative from those given below each statement.
5. If two dice are throw once, then_____ event possible.
(a) 6
(b) 12
(c) 24
(d) 36
6. If a dice is throw once, then______ the probability of number more than 5 appearing
on the dice.
(a) 1/5 (b) 1/6 (c) 1/2 (d) 1/4
7. If Σfi = 100, Σfi di = 1234 then Mean (x) =
(a) 12.34 (b) 1.234 (c) 123.4 (d) 0.1234
8. If 28x + 22 y = 50 and 22x+28 y = 150 then x + y = ____
(a) 2 (b) 1 (c) 4 (d) 200
Fill in the blanks so as to make the following statements true.
9. If (3, 1) is a solution of 10x + 2ky = 20 then k = -5
10. tan(θ). cot(θ) + 1 = 2
11.If A (5/2 , 3/2 ) and B ( 3/2 ,5/2 ) are given points, then mid point of AB =(2,2)
12. If r=13 cm, and chord of circle's length 24 cm then___is the distance between chord
and centre. Ans :- 5 cm
Answer the following by a number or a word or a sentence :
13. Are of circle heaving l = 20 cm and radius of circle = 3 cm then find the area of sector
Are of circle = θ/360 x 2πr =20, θ/360 =20/2πr -----(1)
Area of sector = θ/360 x πr2 = 20/2πr / πr2-----from (1)
Area of sector = 30 cm2
14. What is the formula of volume for hemisphere?
Ans :- 2/3 πr3
15. Which of the following cannot be the probability of an event ?!
16. If P(E)= 0.909 then P(E)= ? Ans :- 0.091
Section : B
Answer the following questions in short. (Q:17 to 26) (2 Marks each)
17. prove that: 1/2-√3 is irreational.
Let us assume to the contrary that 1/2-√3 is rational.
Then, there exist positive co-primes p and q such that,
=) 1/2-√3 =p/q
=) q= p(2-√3)
=)q= 2p-p√3
=) √3= q-2p/-p
=) but this contradicts the fact that, √3 is irrational.
Hence, 1/2-√3 is irrational..
18. Find HCF (10,000, 100000) using Euclid's divison algorithm.
The quotient for by dividing 100000 by 10000 is 10 . It has been exactly divisible by providing the remainder as 0.
10000 is the Highest Common Factor of the numbers 10000 and 100000 by using Euclid’s algorithm.
19.If x-2 is the factor of x4 -16 then find the remainder?
Given :- x -2 is a factor of x4 -16 then x=2
Putting the value of x in equation we get,
= (2)4 - ( 16 )
= 16 - 16
= 0
Remainder is 0
20.Solve the pair of equations by reducing them to a pair of linear equations.
6x + 3y = 6xy , 2x + 4y = 5xy
Ans :- 6x + 3y = 6xy
divide each term with xy ,
=>6/y + 3/x = 6 ---( 1 )
2x + 4y = 5xy
divide each term with xy , we get
=> 2/y + 4/x = 5 ---( 2 )
Let 1/x = a , 1/y = b
6/y + 3/x = 6 => 6b + 3a = 6 ---( 3 )
2/y + 4/x = 5 => 2b + 4a = 5 ----( 4 )
Do [ ( 3 ) - 3 × ( 4 ) ], we get
6b + 3a = 6
6b + 12a = 15
_____________
- 9a = -9
a = ( -9 )/( -9 )
a = 1
Substitute a = 1 in equation ( 4 ),
we get
2b + 4 = 5
2b = 5 - 4
b = 1/2
Therefore ,
1/x = a = 1 =>x = 1 ,
1/y = b = 1/2 => y = 2
21.If tan(θ) = √3 then find sin(θ) + cosec(θ).
22. Evaluate :
sin2(60) - cos2 (30) + tan2 (30)
= (√3/2)2 - (√3/2)2 + (1/√3)2
=1/3
22. If A is acute angle then prove tha 1 + cot2A = cosec2A
Proof: Again a2 + c2 = b2 ( by Pythagoras Theorem) .......(i)
Dividng eq (i) by a2
a2/a2 + c2/a2 = b2 /a2
1 + (c/a)2 = (b/a)2
But cot A = c/a , cosecA = b/a
1 +(cotA)2 =( cosecA)2
1 + cot 2A = cosec2A
23.Prove that the parallelogram circumscribing a circle is a rhombus.
Answers
Given ABCD is a ||gm such that its sides touch a circle with centre O.
∴ AB = CD and AB || CD,
AD = BC and AD || BC
Now, P, Q, R and S are the touching point of both the circle and the ||gm
We know that, tangents to a circle from an exterior point are equal in length.
∴ AP = AS [Tangents from point A] ... (1)
BP = BQ [Tangents from point B] ... (2)
CR = CQ [Tangents from point C] ... (3)
DR = DS [Tangents from point D] ... (4)
On adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC]
⇒ AB = BC
Therefore, AB = BC implies
AB = BC = CD = AD
Hence, ABCD is a rhombus.
OR
23.Prove that a quadrilateral ABCD is drawn to circumscribe a circle then AB + CD = AD + BC
Ans :-DR = DS (Tangents on the circle from point D) …........... (1)
CR = CQ (Tangents on the circle from point C) …........… (2)
BP = BQ (Tangents on the circle from point B) …........… (3)
AP = AS (Tangents on the circle from point A) …............(4)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
24.Find the mean of 30 numbers if it is given that mean of then is 12 and the mean of remaining 20 is 9.
Mean of ten no:s is 12. If s1 is the sum if these ten numbers, then:
mean of ten numbers= 12
s1/10 = 12
so, s1 = 12*10
s1= 120
mean of remaining twenty no:s is 9. If s2 is the sum if these twenty numbers, then:
mean of twenty numbers= 9
s2/20 = 9
so, s2 = 9*20
s2= 180
Therefore:
mean of the 30 numbers
= sum of the 30 no:s/30
=(s1 + s2) / 30
= (120 + 180) / 30
= 300 / 30
= 10
The answer is 10
25.Is it possible to design a rectangular mango grove whose length is twice its breath and the area is 800 m2 ? If so, find its length and breadth.
Ans :- Yes it is possible.
let breadth=x
length=2x
area=2x²
according to ques.
2x²=800
x²=400
x=20
Hence, Length=40
Breadth=20
26. If the price of a book is reduced by Rs.5 ,a person can buy 5 more books for Rs 300. Find the original list price of the book
Ans :- Let the prize of one book be Rs. x
let the no. of books be y
so total amount spent = xy
300 =xy
now changed price of book = Rs. (x -5)
no. of books bought = y + 5
so total amount spent = (x - 5)(y + 5)
300 = xy +5x -5y -25
325 = 300 + 5(x - y)
25 = 5(x - y)
x - y = 5
y = x - 5
now putting this in xy = 300
(x - 5)x = 300
x2 -5x - 300 = 0
x2 -20x + 15x -300 = 0
(x - 20) ( x + 15) = 0
x = 20 or x = -15
Price of book cannot be negative so x is not equal to -15
so actual cost of book is x= Rs. 20
Section : C
Answer the following questions in short. (Q.27 to 34) (3 Marks each)
27.If p(x) ÷ q(x),where p(x) = x4 -3x2 + 4x + 5,q(x) = x2 +1-x then find remainder and quotient.
X²-x+1 ) x⁴-3x²+4x+5 (x²+x-3
x⁴+1x² -x³
__________
x³-4x²+4x
x³-1x²+ 1x
______________
-3x²+3x+5
.-3x²+3x-3
________________
..................... 8
Hence Remainder is 8
Quotient is x²-x+3
28. Find the roots of the equation:1/x + 1/ 9-x = 1/2
Ans :- 1/x + 1/ 9-x = 1/2
9-x + x/x(9-x) =1/2
9 x 2 = x(9-x)
18 = 9x - x2
9x - x2 - 18 = 0
x2 -9x + 18 = 0 ----------------multiplying by -1
x2 -6x - 3x + 18 = 0
x(x - 6)- 3(x - 6) = 0
(x -6)(x - 3) = 0
x = 6 or x = 3
29.Find the sum of first 15 multiples of 8 .
The first 15 multiples of 8
8, 16, 24, 32, 40, 48, 56,64
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S15 = ?
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960
OR
29.Find the sum of the odd numbers between 0 to 50.
Odd number between 0 and 50 are 1,3,5,7,9.....49
l=a+(n-1)d
49=1+(n-1)2
2n-2=48
2n=50
n=50/2
n=25
Sn=n/2(a+l)
=25/2(1+49)
=25/2(50)
=25(25)
=625.
Therefore the sum of all odd numbers between 0 and 50 is 625.
30. Find the point of the X -axis which is equidistant from (2,-5) and (-2,9).
let the given point be A(2,-5) and B(-2,9) and p point be p(x,0)
PA=PB=)PA2=PB2
=)(x-2)2+(0+5)2=(x+2)2+(0-9)2
=)x2+4-4x+25=x2+4x+4+81
=)-8x=56
x=-56/8
x = -7
point p will be (-7,0)
30. Find the mean
Class
5-15 6
25-35 21
35-45 23
45-55 14
55-65 5
OR
31.Find class 0 -10 ,10 -20 ,20 -30 ,30-40,40-50,50-60,60-70,70-80 and frequency are 7,14,13,12,20,11,15,8 respectively
Class f
0 -10 7
10 -20 14
20 -30 13
30-40 12
40-50 (l) 20
50-60 11
60-70 15
70-80 8
Mode = l + f1 - fo/2f1 -fo -f2 x h
Mode = 40 + 20 - 12/( 2x20 - 12 - 11) X 10
Mode = 40 + 8/(40 - 23) X 10
Mode = 40 + 8/( 17) X 10
Mode = 40 + 80/( 17)
Mode = 44.70
32.Prove that the tangents drawn at the ends of diameter of a circle are parallel.
33.In figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, Find the area of the shaded region.
Radius of larger circle,(OA) R = 7 cm
Diameter of smaller circle,(OD) = 7 cm
Radius of smaller circle = 7/2 cm
Height of ΔBCA( OC) = 7 cm
Base of ΔBCA ( AB )= 14 cm
Area of ΔBCA = 1/2 × base × height
Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm²
Area of larger semicircle with radius (OA)7 cm = 1/2 ×π ×7²= 1/2 ×22/7 ×7×7 = 77 cm²
Area of smaller circle with radius 7/2 cm = πr²= 22/7 × 7/2 × 7/2 = 77/2 cm²
Area of the shaded region =Area of smaller circle with radius 7/2 cm +Area of larger semicircle with radius 7 cm - Area of ΔBCA
Area of the shaded region = 77/2 + 77 - 49= 77/2 + 28 = (77 +56) /2= 133/2
Area of the shaded region = 66.5 cm²
34. A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is√6: √π
Let r and a be the radius of the sphere and edge of the cube respectively.
Given, Surface area of sphere = Surface area of cube
4πr2 = 6a2
(r/a)2 = 3 / 2π
r / a = √(3/2π)
Volume of sphere / Volume of cube = (4/3)πr3 / a3 = (4π/3)(r/a)3
= (4π/3)(√(3/2π))3
= (4π/3)(3/2π)(√(3/2π))
= 2√(3/2π)
= √(4x3/2π)
= √(6/π)
Thus, Volume of sphere : Volume of cube = √6 : √π.
OR
34.A height of the cone is 30 cm. A small cone is out of at the top by a plane parallel
base if its volume be of the volume of the given cone at what height about the b
the section cut.
Volume of the big cone = V = π/3 * R² H
H = 30 cm
Volume of the small cone , which is cut off = π/3 r² h
In the regular cone, the radius of the base circle and the height increase proportionately from the apex. Hence r = R/x and h = H/x, where x = a positive number
Volume of small cone = π/3 R²H/x³ = V/x³ = 1/27 * V
=> x = 3
Hence height of small cone = H/3 = 10 cm
Height of the remaining part of cone = 20 cm
Section : D
Answer the followng as required with calculations : (Question no. 35 to 39)
(4 marks each)
35. Draw a triangle ABC with side BC=7 cm, ZB = 45°, ZA = 105°, then construct triangle whose side are times the corresponding sides of triangle ABC.
∠B = 45°, ∠A = 105°
Sum of all interior angles in a triangle is 180°.
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 30°
The required triangle can be drawn as follows.
Step 1
Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.
Step 4
Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.
Step 5
Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.
Justification
The construction can be justified by proving that
In ΔABC and ΔA'BC',
∠ABC = ∠A'BC' (Common)
∠ACB = ∠A'C'B (Corresponding angles)
∴ ΔABC ∼ ΔA'BC' (AA similarity criterion)… (1)
In ΔBB3C and ΔBB4C',
∠B3BC = ∠B4BC' (Common)
∠BB3C = ∠BB4C' (Corresponding angles)
∴ ΔBB3C ∼ ΔBB4C' (AA similarity criterion)
On comparing equations (1) and (2), we obtain
⇒ This justifies the construction.
OR
35. Draw a circle of radius 6 cm, From a point 10 cm away from its centre. Construct
pair of tangents to the circle and measure their lengths.
36. A mean standing on the deck of a ship. Which is 10 m above the water level, observe
angle of elevation of the top of a hill as 60° and angle of depression of the base of the
as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Ans :-Let a man is standing on the Deck of a ship at point a such that AB = 10 m & let CE be the hill
Thus, AB = CD = 10 m
The top and bottom of a hill is E and C.
Given, the angle of depression of the base C of the hill observed from A is 30° and angle of elevation of the top of the hill observed from A is 60 °
Then ∠EAD= 60° &
∠CAE= ∠BCA= 30°. (Alternate ANGLES)
Let AD = BC = x m & DE= h m
In ∆ ADE
tan 60° = Perpendicular / base = DE/AD
√3= h/x [tan 60° = √3]
h = √3x……..(1)
In ∆ ABC
tan 30° = AB /BC
[ tan30° = 1/√3]
1/√3 = 10/x
x= 10√3 m.. …………..(2)
Substitute the value of x from equation (2) in equation (1), we have
h = √3x
h= √3× 10√3= 10 × 3= 30 m
h = 30 m
The height of the hill is CE= CD+ DE= 10 +30= 40 m
Hence, the height of the hill is 40 m & the Distance of the hill from the ship is 10√3 m.
37. The Altitude of a right triangle is 7 cm less then its base. If the hypotenuse is 13 cm.
Find the other two sides.
Ans :- Let base be x cm , then altitude will be (x-7) cm.
Now ,
By using pythagoras theorem-
132 = x2 + (x-7)2
169 = x2 + x2 + 49 - 14x
2x2 - 14x - 120 = 0
x2 - 7x - 60 = 0
x2 - 12x + 5x - 60 = 0
x ( x - 12 ) + 5 ( x - 12 ) = 0
Hence x = 12
sO , Altitude is 5 cm
AND. ,
base = 12cm
38.Water in a canal 6 m wide and 1.5 deep, is flowing with a speed of 10 km/h. How much area will be irrigate in 30 minutes, If 8 cm of standing water is needed ?
Ans :- Speed of flowing of water from canal is 10km/h , means length of water flows in 1 hour = 5 km , so length of water flows in 30 minutes = 5km
Now, volume of water flowing from canal = length of water flows in 30 minutes × breadth of canal × deep of canal
= 5000m × 6m × 1.5m = 45000 m³
Let area of field = x m²
Then, volume of water irrigates into the field = area of field × height of water during irrigation
= x m² × 8× 10⁻² m = 0.08x m³
Now, volume of water flowing from canal = volume of water in field
45000 m³ = 0.08x m³
x = 562500 m²
We know, 1 hectare = 10000 m²
So, area of field in hectare = 562500/10000 = 56.25 hectares
39.State and prove Basic propertionality theorem.
Basic Proportionality Theorem :
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Given : In a triangle ABC, a straight line l parallel to BC, intersects AB at D and AC at E.
To prove : AD/DB = AE/EC
Construction : Join BE, CD.
Draw EF ⊥ AB and DG ⊥ CA
Proof :
Step 1 :
Because EF ⊥ AB, EF is the height of the triangles ADE and DBE.
Area (ΔADE) = 1/2 ⋅ base ⋅ height = 1/2 ⋅ AD ⋅ EF
Area (ΔDBE) = 1/2 ⋅ base ⋅ height = 1/2 ⋅ DB ⋅ EF
Therefore, Area (ΔADE) / Area (ΔDBE) :
= (1/2 ⋅ AD ⋅ EF) / (1/2 ⋅ DB ⋅ EF)
Area (ΔADE) / Area (ΔDBE) = AD / DB -----(1)
Step 2 :
Similarly, we get
Area (ΔADE) / Area (ΔDCE) :
= (1/2 ⋅ AE ⋅ DG) / (1/2 ⋅ EC ⋅ DG)
Area (ΔADE) / Area (ΔDCE) = AE / EC -----(2)
Step 3 :
But ΔDBE and ΔDCE are on the same base DE and between the same parallel straight lines BC and DE.
Therefore, Area (ΔDBE) = Area (ΔDCE) -----(3)
Step 4 :
From (1), (2) and (3), we can obtain
AD / DB = AE / EC
Hence, the theorem is proved.
OR
39.State and prove Pythagorus theorem.
Pythagoras' theorem :-
→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Step-by-step explanation:
It's prove :-
➡ Given :-
→ A △ABC in which ∠ABC = 90° .
➡To prove :-
→ AC² = AB² + BC² .
➡ Construction :-
→ Draw BD ⊥ AC .
➡ Proof :-
In △ADB and △ABC , we have
∠A = ∠A ( common ) .
∠ADB = ∠ABC [ each equal to 90° ] .
∴ △ADB ∼ △ABC [ By AA-similarity ] .
⇒ AD/AB = AB/AC .
⇒ AB² = AD × AC ............(1) .
In △BDC and △ABC , we have
∠C = ∠C ( common ) .
∠BDC = ∠ABC [ each equal to 90° ] .
∴ △BDC ∼ △ABC [ By AA-similarity ] .
⇒ DC/BC = BC/AC .
⇒ BC² = DC × AC. ............(2) .
Add in equation (1) and (2) , we get
⇒ AB² + BC² = AD × AC + DC × AC .
⇒ AB² + BC² = AC( AD + DC ) .
⇒ AB² + BC² = AC × AC .
AB² + BC² = AC2
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