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)
HCF( 17,23)=1 (TRUE)
If p(x) = x2 - 7x + 10 then , no of zeros are three (FALSE)
If sinA = 1 than A = 900(TRUE)
The value of SinA and CosA can never be greater than 1(TRUE)
Select the most appropriate alternative from the given alternatives ( 5 to 10)
If p and q are positive integers where p = ab2 and q = a3b where a and b are prime numbers then LCM ( p,q) = a3b2
ab
a2b2
a3b2
a3b3
Graphically , pair of linear equations 6x -3y + 10 = 0 and 2x - y + 9 = 0 represents two lines which are Parallel
Intersection exactly at one point
Intersecting exactly at two points
Coincident
Parallel
If roots of quadratic equation ax2 + bx + c = 0 , a ≠ 0, are real and distinct than
b2 - 4ac ﹤0
b2 - 4ac = 0
b2 - 4ac ﹥0
b2 - 4ac ≠ 0
If A(0,6) and B(0,-2) , than distance between A and B is 8
6
8
4
2
Probability of getting 80 marks out of 80 marks in Mathematics is
79/80
1/80
1/81
79/81
For the given figure , if y = p(x), then no. of zeros are 2
1
2
3
0
Fill up the blanks ( 11 to 16)
The abscissa of point of intersection of less than type and of more than type cumulative frequency curves of a grouped data gives its median ( Mean , median, mode)
Probability of a sure event is __1__ (0,1,2)
A tangent to circle intersect in__1___point ( 0,1,2)
A circle which touches all sides of quadrilateral A,B,C,D , If AB=7, BC=3, CD=4 then AD =_8_ (8,7,11)
Ref : ( sum of opposites sides are equal , AB + CD = BC + AD)
Length of Minor arc is 𝛑r𝜃/180 (𝛑r𝜃/180, 𝛑r2𝜃/360,𝜃/360)
If the radius of the circle is doubled than area becomes _4__times(2,3,4)
SECTION B (20 marks)
Solve the following ( 17 to 26) , Each carries two(2) marks.
Find the zeros of quadratic equations X2 + 2X - 8 = 0
X2 + 2X - 8 = 0
Splitting the middle term ( 4-2 = 2 and 4 x 2 = 8)
X2 + 4X - 2X - 8 = 0
X(X+4) -2(X+4)=0
(X+4)(X-2)=0
X+4=0 OR X-2=0
X= -4 or the X = 2 are zeros of quadratic equation X2 + 2X - 8 = 0
OR
Find the quotient and remainder for
(3x2 - x3 - 3x + 5) / (x- 1-x2)
Answer : Quotient : (x-2) and remainder 3
Find the quadratic polynomial whose sum and product of its zero given by respectively ( ¼, -1)
Assume that the quadratic polynomial is ax2 + bx + c = 0
Therefore sum of the roots = -b/a
∴ -b/a = ¼
∴ b = -a/4 (i)
Now the product of the root = c/a
∴ c/a = -1
∴ c = -a (ii)
Putting value of b and c in the quadratic equation
∴ ax2 + bx + c = 0
∴ax2 +(-a/4)x + (-a) = 0
∴4ax2 -ax -4a = 0
∴4x2 - x - 4 = 0
4x2 - x - 4 = 0 is the quadratic polynomial whose sum and product of its zero given by respectively ( ¼, -1)
No of three digit numbers divisible by 3.
First 3 digit number divisible by 3 is 102
Last 3 digit number divisible by 3 is 999 ( nth term)
an = a1 + (n-1)d
a1 = 102
an = 999
d = 3
999 = 102 + (n-1)*3
n= 1 + ( 999 - 102)/3
n = 300
Therefore total number of 3 digits numbers divisible by 3 are 300
Evaluate 2tan245 + cos230 - sin260
we know that tan45 =1, cos30 = √3/2 , sin60 =√3/2
∴2tan245 + cos230 - sin260 = 2 (1)2 + (√3/2)2 - (√3/2)2
∴2tan245 + cos230 - sin260 = 2*1 + 0
∴2tan245 + cos230 - sin260 = 2
Answer : 2tan245 + cos230 - sin260 = 2
OR
If sin3A = cos(A-26) where 3A is an acute angel, find the value of A.
sin3A = cos(A-26)
∴ sin3A = sin[90-(A-26)] (∵ 3A is an acute angel so 3A ＜90)
Comparing the value in the bracket
∴ 3A = 90-(A-26)
∴ 3A = 90-A +26
∴ 4A = 116
∴ A = 29
Answer : A = 29 degree
Find the value of y for which distance between points P(2,-3) and Q(10,Y) is 10 units.
10 = √(2-10)2 + (-3-Y)2
100 = 64 +(3 + Y)2
36 = (3+Y)2
+/-6 = 3 +Y
Y = 3 or Y = -9
ANSWER : Y = 3 or Y = -9
An observer 1.5 meter tall is 28.5 meters away from a chimney. Angel of elevation of top of the chimney from her eyes is 450. What is the height of the chimney ?
As shown in the following figure , chimney is AC and observer is DE.
DE = BC = 1.5 metre observer height given.
AC = AB + BC
BD = 28.5 meter given.
△ABD , ∠ADB = 45 given.
Now tanD = AB / BD
tan 45 = AB / 28.5
1= AB / 28.5
AB = 28.5.
Therefore the height of chimney AB + BC = 28.5 + 1.5 meter = 30 meter.
From a point Q, the length of the tangent to circle is 24 cm. Point Q is 25 cm away from the centre of the circle. Find the radius of the circle.
As given in the question, PQ = 24,OQ =25
Radius OP = sqrt (252 - 242) = sqrt(49) =7 cm
OR
In the figure , TP and TQ are tangent to the circle with centre of O,so that angle ∠POQ= 1100 then find ∠PTQ
Since TP and TQ are tangent to circle ∠OPT and ∠OQT are right angle and hence they are 900 each
Now in quadrilateral OPTQ
∠OPT + ∠PTQ + ∠OQT + ∠POQ = 360
90 + ∠PTQ + 90 + 110 = 360
∠PTQ = 70
ANSWER : ∠PTQ = 70
The edge of cube is 5 cm then find the total surface area of the cube.
Area of one face of cube = 5 x 5 cm2
= 25 cm2
So total surface area of the cube = 6 x 25 = 150 cm2
OR
A solid shown in the figure is made of cylinder (A ) with Hemispherical depression (B) , Then write the formula to find total surface area of the the solid
Assume that the height of cylinder is h and , radius of the base is r
∴Total surface Area of the solid = Area of hemispherical depression B + Curved surface of cylinder A + Area of Base of the cylinder A
∴Total surface Area of the solid = 2𝛑r2 + 2𝛑rh + 𝛑r2
∴Total surface Area of the solid = 3𝛑r2 + 2𝛑rh
∴ Total surface Area of the solid = 𝛑r(3r + 2h)
Answer : Total surface area of the given solid can be found by using formula Area = 𝛑r(3r + 2h)
∑ fidi = -26, a = 30, ∑ fi = 13 then find X.
X ( x bar , mean ) = a + (∑ fidi / ∑ fi)
X bar = 30 + (-26/13)
X bar = 30 + (-2)
X bar = 28
A box contains 3 blue, 2 white and 4 red marbles. Alok drawn a marble at random. What is the probability that it will be
White
Not red
Total number of event 9
Probability of drawing white marble = No of white marble / total number of event (Total no of Marblesare 9)
= 2/9
Probability of drawing ball other than red colour = ( No of white white marble + No of Blue marble ) / total number of events
=( 3 +2 ) / 9
= 5/9
Answer : Probability of a) drawing white marble 2/9 and b) Probability of drawing marble other than red is 5/9
SECTION C (24 marks)
Answer the following as asked with calculation (27 to 34) each carries 3 marks.
Solve the pair of linear equations using the elimination method.
2x + 3y = 46 AND 3x + 5y = 74
2x + 3y = 46 (i)
3x + 5y = 74 (ii)
Here we will eliminate variable y, so multiply (i) with 5 and (ii) with 3 we will get
10x + 15y = 230 (iii)
And
9x + 15y = 222 (iv)
Now subtracting (iv) from (iii)
(10x + 15y) - (9x + 15y) = 230-222
x = 8
Putting value of x in the (i)
2(8) + 3y = 46
3y = 46 - 16
3y = 30
y = 10
Answer : x=8 and y = 10 is the solution of the given pair of linear equation
OR
Difference of two natural number is 5 and difference of their reciprocal is 1/10, find these two numbers.
Assume that bigger number is x and smaller number is y
The difference is 5
x - y = 5,
Difference of their reciprocal is 1/10. Reciprocal of y will be bigger
1/y - 1/x = 1/10
10(x-y) = xy
10*5 = xy
50 =xy
From (i) , y = x-5
x(x-5)=50
x2 - 5x-50 = 0
Splitting the middle term
x2 +5x-10x-50 = 0 ( 5-10 = -5 and 5 x -10 = 50)
(x+5)(x-10)=0
Therefor x=-5 or x = 10
But since x is natural number x can not be -5, so x = 10
Therefor y = x- 5 = 10-5 =5
Answer : 10 and 5 are two natural number whose difference is 5 and the difference of their reciprocal is 1/10.
Find the nature of roots of the given quadratic equation. If real root exists , find them. 2x2-6x +3 = 0
OR
A train travels 360 km at uniform speed. If the speed have been 5km / hr more it would have taken 1 hour less for the same journey. Find the speed of the train.
Assume that the speed of train is x km/hr and it takes 360/x hr to travel 360 km.
x + 5, so time taken will be 360 / (x+5) hrs.
This time is 1 hr less than the normal time
360 / (x+5) = [360 /x -1]
∴360x = (360-x)(x+5)
∴(x-360)(x+5) + 360x = 0
∴x2 - 355x - 1800 + 360x = 0
∴x2 + 5x - 1800 = 0
Splitting the middle term
∴x2 +45x - 40x - 1800 = 0
∴x(x + 45)-40(x+45) =0
∴(x + 45)(x-40)=0
∴x+45 = 0 or x-40=0
∴x= -45 or x = 40
Since speed can not be negative., x=40
Answer : the speed of the train to travel 360 km is 40 km/hr.
Find the sum of the first 40 positive integers divisible by 6.
a1 = 6, d=6, n=40
∴S = n[2a+ (n-1)d]/2
∴S = 40(2*6 + (40-1)*6)/2
∴S= 20(12 +39*6)
∴S= 4920
Answer : sum of the first 40 positive integers divisible by 6 is 4920
Find the twentieth term from the last term of AP 3,8, 13….253.
Lets put the series in reverse order 253,248…
a1 = 253, d= -5, n=20
an = a1 + (n-1)d
∴a20 = 253 + (20-1)(-5)
∴a20 = 253 - 19*5
∴a20 = 253 - 95
∴a20 = 158
Answer : 158 is the twentieth term from the last term of AP 3,8, 13….253.
Find the value of K for which point are collinear (8,1) (k,-4), (2,-5)
Assume that A(8,1), B(k,-4), C(2,-5)
The three points will be collinear if the
Δ formed by these points has an area equal to zero.
Let A(8, 1), B(k, -4) and C(2, -5) be the vertices of a triangle.
∴ The given points will be collinear, if ar (ΔABC)=0
∵ Area of triangle having coordinates (x1,y1), (x2,y2), (x3,y3) is
1/2×[(x1)(y2−y3) + (x2)(y3−y1) + (x3)(y1−y2)]
∴ 1/2[8(−4+5)+k(−5−1)+2(1+4)]=0
∴ 8−6k+10=0
∴ 6k=18
∴ k=3
Answer : Value of K = 3, for (8,1) (k,-4), (2,-5) to be collinear
OR
In what ratio does the point (-4,6) divide the line segment joining points A(-6,10) and B(3,-8)
Assume that the point (-4,6) divides the line segment joining the A(-6,10) and B(3,-8) in ratio m1 : m2
x1= -6, y1 = 10, x2 = 3, y2=-8
therefor -4 = (m1x2 + m2x1)/(m1+m2) and 6 = (m1y2 + m2y1)/(m1+m2)
-4(m1 + m2) = 3m1 - 6m2 and 6(m1+m2) = -8m1 + 10m2
-4m1 - 4m2) = 3m1 - 6m2 and 6m1+6m2 = -8m1 + 10m2
7m1-2m2 =0 14m1 - 4m2 =0
From both of above, m1 : m2 = 2:7
Answer : The point (-4,6) divide the line segment joining points A(-6,10) and B(3,-8) in the ratio 2:7
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by suitable method
Mean = ∑ fixi / ∑ fi = 5275/25 = 211
Answer : Mean expenditure of all households is 211 rs.
A die is thrown once, find the probability of getting
A prime number
A number lying between 2 and 6
An odd number.
Total number of possibilities are 1,2,3,4,5,6 ( events)
Ans a) Possibility of getting prime number
Prime numbers are 2,3 and 5 ( no of possibilities of getting prime number is 3)
Probability of getting prime number = 3/6 = ½
Reference :
"A common definition of a prime number is that it can only be divided by 1 and the number itself. For example:
13 is prime, because the only numbers that divide into 13 are 1 and 13 itself.
6 is not prime, because it can be divided by 2 and 3 as well as 1 and 6.
Using this definition, 1 can be divided by 1 and the number itself, which is also 1, so 1 is a prime number.
However, modern mathematicians define a number as prime if it is divided by exactly two numbers. For example:
13 is prime, because it can be divided by exactly two numbers, 1 and 13.
6 is not prime, because it can be divided by four numbers, 1, 2, 3 and 6.
1 can only be divided by one number, 1 itself, so with this definition 1 is not a prime number.
Ans b) Possibility of getting number between 2 and 6 are ( 3,4,5)
So probability = 3/6 = ½
Ans c) Odd numbers are 1,3,5,
So probability of getting odd number is 3/6 = ½
Answer : a) ½ b) ½ c) ½
One card is drawn from the well shuffled deck of 52 cards. Find the probability of getting
A king of red colour
Not a spade
The queen of heart
Total number of possibilities = 52
Answer a) Possibilities of getting king of red colour is 2 ( king of hearts and king of diamonds)
∴ Probability of getting King of red colour = 2/52 =1/26
Answer b) No of cards which are not spade are 52- 13 = 39
∴ Probability of getting card which is not spade = 39/52 = ¾
Answer c) No of Queen of Heart is =1
∴ Probability of getting Queen of Heart is = 1/52
Answer : a) 1/26 b) ¾ c) 1/52
SECTION D (20 marks)
Solve the following,( 35 to 39) Each carries 4 marks each.
State and prove the pythagorean theorem
Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.
The sides of the right-angled triangle are called base, perpendicular and hypotenuse .
According to Pythagoras theorem ,
(AC)2=(AB)2 + (BC)2
Proof:
Given, a triangle ABC in which ∠ABC is 90.
Construction: Draw a perpendicular BD on AC i.e. BD ⊥ AC.
In ΔABD and ΔABC
we have,
∠BAD = ∠BAC
i.e. ∠A is common in both triangles.
∠ABC = ∠ADB = 90
Therefore ΔABC∼ΔABD ( By AA similarity i.e. angle-angle similarity)
So,
⇒AD/AB=AB/AC
⇒AB2 = AD×AC ...(1)
In ΔBDC and ΔABC
we have,
∠BCD = ∠BCA
i.e. ∠C
is common in both triangles.
∠ABC = ∠ADC = 90
Therefore ΔABC∼ΔBDC
( By AA similarity i.e. angle-angle similarity)
So,⇒DC/BC=BC/AC
⇒BC2 = AC×DC ...(2)
Adding equation (1) and (2) , we get
⇒AB2 + BC2 = AD×AC + AC× DC
⇒AB2 + BC2 = AC(AD + DC)
⇒AB2 + BC2 = AC(AC)
⇒AB2 + BC2 = AC2
Hence, proved.
Note: In a right angled triangle , hypotenuse is the longest side of the triangle and is opposite to the right angle i.e. 900
. By drawing a perpendicular from point B and dividing the triangle ABC into 2 parts and using angle-angle similarity to prove the Pythagoras theorem.
OR
Prove that the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.Prove that the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
Hint: If two triangles are similar then their corresponding sides are proportional to each other. We can also say the ratio of corresponding sides in two similar triangles are equal.
Complete step-by-step answer:
Lets draw two similar triangle ABC and PQR as below:
As we know to find the area we need the height of the triangle. So we can draw a perpendicular AD from A to BC and PS from P to QR.
In triangle ABD and triangle PQS
∠B=∠Q
{∵ΔABC∼ΔPQR}
∠ADB=∠PSQ=90∘
(By construction)
As two angles are equal so the third angle of both triangles should also be equal.
∠BAD=∠QPS
So by AAA similarity
ΔABD∼ΔPQS
So we can say the ratio of corresponding sides should be equal. So we can write
AB/PQ=AD/PS …………………………………………..(i)
Now we can write area of triangle ABC as
Area(ΔABC)=1/2×AD×BC ………………………………..(ii)
Area(ΔPQR)=1/2×PS×QR…………………………………….(iii)
On dividing equation (ii) and (iii)
Area(ΔABC)Area(ΔPQR)=AD×BC/PS×QR
By using equation (i) we can write
Area(ΔABC)Area(ΔPQR)=AB×BC/PQ×QR……………………………..(iv)
As given ΔABC∼ΔPQR
So this ratio of corresponding sides should be equal. So we can write
AB/PQ=BC/QR=AC/PR
Hence we can use this value in equation (iv). So we can write
Area(ΔABC)Area(ΔPQR)=AB×AB/PQ×PQ=(AB/PQ)2
Similarly we can write
Area(ΔABC)/Area(ΔPQR)=(AB/PQ)2=(BC/QR)2=(AC/PR)2
Hence we can say the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
The median of the distribution given below is 28.5. Find the value of P and Q
Solution
Median is 28.5, N/2 = 30, this lies between interval 20-30, frequence for this interval is 20 and Cumulative frequency is 5 + P
Total number of students = 60
∴ 45 + P + Q = 60
∴ P + Q = 60 - 45
∴ P + Q = 15
Now : l= 20 , f = 20 and cf = 5 + P h = 10, n = 60
Median = l + ((n/2-cf)/f)h
28.5 = 20 +10*(30-(5+P))/20
28.5 = 20 +(30-5-P)/2
28.5 = 20 + (25-P)/2
57 = 40 + 25 - P
P = 8
Since P+ Q = 15
Q = 7
Answer : P=8 and Q = 7
Draw a triangle ABC with side BC = 6 CM, AB=5 CM and ∠ABC = 600. Then construct a triangle whose side are ¾ of the corresponding side of the triangle ABC
Steps of construction -
(i) Draw a line segment BC = 6 cm.
(ii) Construct ∠XBC= 600.
(iii) With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A.
(iv) Join AC. Thus, Δ ABC is obtained.
(v) Draw an acute angle ∠ CBY below of B.
(vi) Mark 4-equal parts on By as B1,B2,B3 and B4.
(vii) Join B4 to C.
(viii) From B3, draw a line parallel to B4
C intersecting BC at C.
(ix) Draw another line parallel to CA from C', intersecting AB at A.
(x)Δ A′BC′ is required triangle which is similar to ΔABC such that
BC′=¾(BC).
OR
Construct a tangent to circle with radius 4 cm from a point on concentric circle of radius 6 cm and measure its length.
Steps of construction
1) Draw a circle of radius 4 cm from the centre O. With same centre and radius equal to 6 cm, construct another circle.
2) Take any point P on the circumference of the outer circle and join OP.
3) Construct perpendicular bisector for the line segment PO, which intersect OP at point M
4)Now, with M as centre, construct a circle of radius equal to PM or MO.
5) This circle now intersects the inner circle at points Q and R.
Join PQ and PR.
6) Thus, tangents have been constructed from outer circle to the inner circle.
7) On measuring PR or PQ, we get PR
=PR
= 4.4cm (approx)
Verification
PO acts as a diameter for the smallest circle and hence,
∠PQO = ∠PRO= 900 [Angle in the semi circle]
Thus, OQ⊥PQ and OR⊥PR hence,PQ and PR are tangents.
Consider, right ΔPQO
⇒PO2= PQ2 + OQ2
62=PQ2 +42
∴PQ=√36−16
=√20
=4..47cm (approx)
Hence, verified.
A survey conducted on 20 household in a locality by a group of a student resulted in following frequency table for number of family members in a household.
Find the mode of this data.
The class with highest frequency is 3-5,
So 3-5 is modal class.
l = 3 ( lower limit of modal class)
f1 = 8 ( frequence of modal class)
f0 = 7 ( frequency of class preceding modal class)
f2 = 2 ( frequency of class succeeding modal class)
h = 2
Mode = l + [ (f1- f0)/(2f1- f0 - f2)]*h
Mode = 3 + [ (8-7)/(2*8- 7 - 2)]*2
Mode = 3 + [ 1/(16- 7 - 2)]*2
Mode = 3 + [ 1/(7)]*2
Mode = 3 + (2/7)
Mode = 23/7 = 3.2857 = 3.286
Answer : The mode of given data is 3.286
A metallic sphere of radius 4.2 cm is melted and recast into shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Volume of the metallic sphere = Volume of the cylinder
( 4/3)(𝛑r3) = 𝛑r2h
( 4/3)(22/7)(4.2*4.2*4.2) =(22/7)(6*6)*h
h = (4/3)(4.2*4.2*4.2/6*6)
h = 4*1.4*0.7*0.7
h = 5.6*0.49
h = (2.8 - 0.056)
h = 2.74 cm
Answer : Height of the cylinder will be 2.74 cm
OR
A Wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in the figure. If the height of cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of this article.
Surface area of the wooden article = 2 x ( surface area of hemispheres) + surface area of curved surface of the cylinder
Surface area of the wooden article = 2 x (2𝛑r2 ) + 2𝛑rh
Surface area of the wooden article = 2𝛑r(2r + h)
Surface area of the wooden article = 2(22/7)(3.5)(7 + 10)
Surface area of the wooden article = (22)(7 + 10)
Surface area of the wooden article = (22)(17)
Surface area of the wooden article = 374 cm2
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