Maths paper set 6

QUESTION PAPER-6
Section-A
Answer according to instruction : (Question no. I to 16) (1 mark each)
State whether the following statements are True or False.
1.For polynomial 4x3 +3x2 +4x+5, then the sum of zeroes is -3/4. True
2.Quadiatic equation:⇒ AB² + BC² = AC²
 x2 +x+1= 0 has two real roots.True
3.In an AP, nth term is a (n - 1)d.False
4.sin(90°) + cos(90°)= 1True
Fill in the blanks by selecting the proper alternative from those given below each statement:
5.A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts The number of outcomes favourable to E is
(a) 4  (b) 13   (c) 48   (d) 51
6.One coin is drawn at random from a bag containing coines numbered 1 to 40. The probability that the selected coin has a number which is a multiple of 5 is.......
(a) 1/5       (b) 3/5     (c) 4/5      (d) 1/3
7.Consider the following frequencey dictribution :
Class    0-5    5-11     12-17    18-23      24-29
  F         13        10         15          8             11
The upper limit of the median class is
(a) 17      (b) 17.5    (c) 18     (d) 18.5
8.The value of k for which the pair of equations kx-y-2 = 0 and 6x-2y-3 =0 will have infinitely many solution is
(a) 3     (b) -3    (c) -12    (d) no value
Fill in the blanks so as to make the following statements true.
9......2.....value of m, do the equations 3x-y=-8 and 6x-my +16 = 0 represent  coincident line.
10.cosecQ secQ) sin Q cosQ=.......1.....
11.The distance of the point A (-4, -5) from the Point of Y-axis is___-4_______.
12. PQ is a tangent to the circle with center O, such that PQ is twice of radius than OP = ___root 5 x___
Answer the following by a number or a word or a sentence:
13.What is the equation of total surface area of frustum of cone ?

πl (r1 + r2 ) + πr1sq + πr2sq

14.Find the area of the square that can be inscribed in a circle of radius 8 cm ?
128 cm2


15.The sum of the probabilities of all the elementary events of an expriment is ?
=>Such an event is called sure or certainevent. The sum of the probabilities of all the elementary events of an experiment is 1.

16.The probability of an event is less than or equal to ?
1

Section : B
Answer the following questions in short. (Q:17 to 26) (2 Marks each)

17.Prove that 2 +2 /2 is irrational.
=>Let us assume on the contrary that 2+2 root 2 is rational.
Then, there exist co-prime a and b(b not equal to 0), such that
2+ 2root 2= a by b
= root 2= a by b--2
= root 2 =a--2b by b
= root 2 is rational [Therefore,2, a & b are integers, a--2b by b is a rational number]
This contradicts the fact the root 2 is irrational.
The contradiction has arisen because of our incorrect assumption that 5+ root 2 is rational.
Hence, 2+2root 2 is irrational.

18.Find HCF (3002, 50) using Euclid's divison algorithın ?
3002 = 50×60 +2
60 = 2 × 30 + 2
HCF = 2
19. If (3x+ 1) is a factor of polynomial then find the quotient ?
20.If x +1 is a factor of 2x3 +ax2 +2bx +1, then find the values of a and b given that 2a - 3b = 4
=> x+1 is a factor of 2x³+ax²+2bx+1
x+1 = 0
x = -1
→ 2(-1)³+a(-1)²+2b(-1)+1 = 0
→ 2(-1) + a(1) - 2b + 1 = 0
→ -2+a-2b+1 = 0
→ a-2b-1 = 0
→ a-2b = 1
Multiply it by 2,
2(a-2b) = 2(1)
2a-4b = 2 -----(1)
Given,
2a-3b = 0 ------(2)
(1) - (2)
2a-4b = 2
-{2a-3b = 0}
––––––––
-b = 2
b = -2
2a - 3(-2) = 0
2a+6 = 0
2a = -6
a = -6/2
a = -3
Therefore, a = -3 and b = -2
OR
20.The angle of a triangle are x, y and 40°. The difference between the two angles x and y is 30°. Find x and y.
Answers    =>   x-y=30 →equation 1
x+y+40 = 180 (since sum of angles of a triangle = 180)
x+y=140 →equation 2
equation 1 → x-y=30
equation 2 → x+y=140
                  ⇒ 2x= 170  (∵ -y+y=0)
                   ⇒x = 170/2
                ⇒  x = 85
substitute x in equation 1 equation 1 → x-y=30
   85-30 = y
   55 = y
∴x=85
 y=55             
21.If cot Q = 4/3 then find the value of cosec2 Q +1
=> 1 + cot ^2 Q = cosec ^ 2Q
So
cosec ^ 2 Q + 1
1 + cot ^ 2 Q+ 1
2 + ( 4/3 ) ^2
2+ 16/9
34/9

22.Evaluate : tan(52°)- cot(48°)
= tan (90 -
22.If sin A = cos B, prove that A +B = 90°
sinA=CosB
sinA=sin(90-B)
a=90-b
a+b=90
23.Prove tlhat the centre of a circle touching two intersecting lines lines on the angle bisector
of the lines.

Assume that
O touches l1 and l2 at M and N , then we get as ,
⇒OM=ON                    ( As it is the radius of the circle )
 Therefore ,From the centre
′O′ of the circle , it  has equal distance from l1 and l2
Now , In
△OPM and OPN
⇒OM=ON           ( Radius of the circle )
⇒∠OMP=∠ONP    ( As , Radius is perpendicular to its tangent )
⇒OP=OP         ( Common sides )
Therefore ,
△OPM≅△OPN    ( S S S congruence rule )
By C.P.C.T ,
⇒∠MPO
=∠NPO
So ,l bisects ∠MPN.
Therefore ,
O lies on the bisector of the angle between l1,l2
Hence , we prove that  the centre of a circle touching two intersecting lines lies on the angle bisector of the lines
23.AB and CD are common tangents to two circles of unequal radii. prove that AB = CD.
=>   Just draw another tangent AD.
Now consider tangents AD & DC.
Since lenghts of tangents from an external point to a circle is equal,
                 AD = CD  ---------[1]
Similarly, AD = AB   ---------[2]    ( with the same reason) .
Now from [1] & [2]
                 AB = CD
        Hence proved.

24.Find the mean :
Class    100-104     104-108     108-112   112-116    116-120
  f           15                 20                32              18               15.
25.Find the roots of 6x2 - v2x -2 = 0 by factarisation of quadratic equation.
=> 6x2 - v2x -2 = 0
=> 6x2 - (3v2x) -(2v2x ) - 2= 0
=> 3x(2x -v2) -v2(2x  - v2)= 0
(2x - v2)(3x - v2)=0
X = v2/2  OR X = v2/3
26.Find two numbers whose sum is 21 and product is 108.
=> Let one integer be x
Then the second integer = 21 - x.
Their product = 108
ie. x(21-x) = 108
21x - x^2 = 108
21x - x^2 - 108 = 0
-x^2 + 21x - 108 = 0
Multiplying the whole with -1,
x^2 - 21x + 108 = 0 -------- (1)
So (1) can be written as
x^2 - 9x - 12x + 108 = 0
x(x-9) - 12(x-9) = 0
(x-9) (x-12) = 0
So either x-9 = 0 or x-12 = 0
x = 9 or x = 12
OR
26.A natural number, when incresed by 12, equals 160 times its reciprocal. Find the number.
y 12 equals 160 times its reciprocal. find the number
Answers=>
Let the natural number = c
If the number increased by 12 = x + 12
Reciprocal of the number = 1/x
According to the problem given,
x + 12 = 160 times of reciprocal of x
x + 12 = 160/ x
x( x + 12 ) = 160
x^2 + 12x - 160 = 0
x^2 + 20x - 8x - 160 = 0
x( x + 20) - 8( x + 20)= 0
( x + 20 ) ( x - 8 ) = 0
x + 20 = 0 or x - 8 = 0
x = - 20 or x = 8
But x is a natural number.
Therefore,
Required number = x = 8
Section : C
* Answer the following as required with calculations : (Questions no. 27 to 34)
on(3 marks each)
27. Divident = x3-3x2+x +2, Quotient (x-2) and Remainder (-2x +4), then find the Divisor ?
Divisor*Quotient+Remainder= Divident
g(x)(x-2)+(-2x+4)=x3-3x2+x+2
g(x)(x-2)=x3-3x2+x+2+2x-4
g(x)(x-2)=x3-3x2+3x-2
g(x)=x3-3x2+3x-2/(x-2)
g(x)=x2-x+1

28.Find the roots:1/x +1/9-x = 1/2
=>9-x + x/x × 9- x = 1/2
9/9x - x2 =1/2
18 = 9x - x2
-X2 +9x + 18 = 0
Multiplying by -1
X2 -9x - 18 = 0
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 all 3 digit numbers which leave remainder 3 when divided by 5.
ind the sum of all 3 digit numbers which leave remainder 3 when divided by 5
Answers
The smallest 3 digit no. = 100 is divisble by 5
=> 103 is the 1st 3 digit no. which will give rem. 3
The largest 3 digit no. = 999
999/5 gives rem. 4 => 998 will give rem. 3
this series is obviouly in AP , ( 103,108,111,...998 )
998 = 103 + ( n - 1 ) 5
=> n = 895/5 +1 = 180
So sum
= n/2 ( first term + last term )
= 180/2 ( 103 + 998 ) = 90 ( 1101 ) = 99090
OR
29.Find the value of n if 2n + 1, n2 +n+ 1, 3n2 - 3n + 3 are consecutive terms of an AP
=> if a,b,c are in AP
b-a=c-b
2b = a+c
2(n²+n+1) = 2n+1+3n²-3n+3
2n²+2n+2 = 3n²-n+4
n²-3n+2 = 0
n²-n-2n+2 = 0
n(n-1)-2(n-1) = 0
(n-1)(n-2)=0
n=1,n=2
if n=1 : numbers are 3,3,3
if n=2: numbers are 5,7,9

30.Show that (4, 4), (3, 5), (-1, 1) are vertices of a right angled triangle.
=> Mark the points as A, B, C
A: (4, 4)
B: (3, 5)
C: (-1, -1)
Slope of AB = (5-4)/(3-4) = -1
Slope of BC = (3+1)/(5+1) = 2/3
Slope fo AC = (4+1)/(4+1) = 1
Two lines are perpendicular if the product of their slopes is -1
-1 × 1 = -1
AB is perpendicular to AC
ABC is a right-angled triangle with a right angle at A (4, 4)
31.Medium (M) = 28.5 find x and y ?
Class    0-10    10-20    20-30    30-40    40-50     50-30
f             5           x             20        15            y              5
Answer = since median 28.5
20-30 is median class
Median = l +2n- cf /f x h
Where , l is the lower limit of the median class = 20
h = class interval = 10 - 0 = 10
n =Σfi = 60
n/2 = 60/2 = 30
Cf = x + 5
f = 20
Median = l +2n- cf /f x h
28.5 = 20 + 30- ( x + 5)/ 20 × 10
28.5 = 20 + 25 - x / 2
28.5 - 20 = 25 - x / 2
8.5 = 25 - x / 2
8.5 x 2 = 25 - x
17 = 25 - x
X = 8
31.Find the Mediun ?
Class     5-10     10-15     15-20     20-25      25-30     30-35      35-40    40-45    45-50
     F            49      133          63            15            6            7               4            2             1
=> Now N = 280.  So n /2 = 280 / 2 = 140
10----15 is the median class.
Median =  l + (n/2 - cf /f) x h
here l is lower limit of median class
 n = the total number
cf = cumulative frequency of the previous median class
f = frequency of median class
h = size of class
Now l = 10, c = 49, f = 133, h = 15 -10 = 5
Substituting in the formula we get
      Median = 10 + (140 - 49 /133) x 5
   Median = 1785 / 133
Median = 13.42
So salary is Rs 13.42
If salary is in thousands then 13.42 x 1000 =  Rs 13420.

32.In a right triangle ABC in which angle B 90°, a circle is drawn with AB as diameter  intersecting the hypotenuse AC and P. Prove that the tangent to the cricle
Answers

ΔABC is a right angled triangle.
∠ABC = 90°.
A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which intersects BC at Q.
Join BP.
PQ and BQ are tangents drawn from an external point Q.
∴ PQ = BQ   -------------- (1)  (Length of tangents drawn from an external point to the circle are equal)
⇒ ∠PBQ = ∠BPQ    (In a triangle, angles opposit to equal sides are equal)
Given that, AB is the diameter of the circle.
∴ ∠APB = 90°  (Angle in a semi-circle is a right angle)
∠APB + ∠BPC = 180°   (Linear pair)
∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°
Consider ΔBPC,
∠BPC + ∠PBC + ∠PCB = 180°  (Angle sum property of a triangle)
∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90°  ----------- (2)
∠BPC = 90°
∴ ∠BPQ + ∠CPQ = 90°    ...(3)
From equations (2) and (3), we get
∠PBC + ∠PCB = ∠BPQ + ∠CPQ
⇒ ∠PCQ = ∠CPQ  (Since, ∠BPQ = ∠PBQ)
Consider ΔPQC,
∠PCQ = ∠CPQ
∴ PQ = QC  ----------- (4)
From equations (1) and (4), we get
BQ = QC
33.A circle of radius 7.5 cm is inscribed in a fugure. Find the arch of the shaded region
=>give that r = 7.5cm.
therefore diameter= 7.5×2
=15cm.
diameter of circle = side of square = 7.5cm.
therefore area of shaded region = area of square - area of circle
= a^2 - pi * r^2
= (15)^2 - 22/7 × 7.5 ×7.5
= 225 - 176.785
= 48.23 cm^2

34.A solid metalic hemisphere of radius 8 cm is melted and recasted into a right circular cone of base radius 6 cm. Determine the height of the cone.
=>Radius of the hemisphere = 8 cm
R = 8 cm
Volume of the hemisphere
= ( V ) = ( 2 / 3) × pi × R^3 ---------(1)
Radius of the right circular cone = r
r = 6 cm
Let the height of the cone = h cm
Volume of the right circular cone
= v = ( 1/ 3 ) × pi × r ^2 × h -------(2)
According to the problem,
The solid metallic sphere is melted and recasted into a right circular cone.
Therefore,
Volume of the hemisphere and volume of the right circular cone are equal.
( 2 ) = ( 1 )
( 1/ 3) × pi × r ^2 × h = ( 2/3 ) × pi × R ^3
h = 2 × ( R ^3 / r ^2 )
Substitute R and r values
h = 2 × ( 8 × 8 × 8 )/ ( 6 × 6 )
h = ( 4 × 8 × 8 ) / ( 3 × 3 )
h = 256 / 9
h = 28. 44 cm
Height of the cone = h = 28.44 cm
OR
34.A heap of wheat in the farm of a cone of diameter 9 m and height 3.5 m. Find the volume of the wheat. How many canvas cloth is required to just cover the heap ?
Solution:-
Given : Diameter = 9 m or radius = 9/2 = 4.5 m and h = 3.5 m
Volume of cone = 1/3πr²h
= 1/3*22/7*4.5*4.5*3.5
= 1559.25/21
Volume of the cone = 74.25 cu m
Just to cover the heap, we have to find the curved surface area of the cone. For slant height 'l' is needed.
So, l = √r²+h²
l = √4.5²+3.5²
l = √20.25+12.25
l = √32.5
l = 5.7008 m
Curved surface area of the cone = πrl
= 22/7*4.5*5.7008
= 564.3792/7
CSA of the cone = 80.6256 m²
So, area of the canvas required to cover the heap of wheat is 80.6256 sq m

Section : D
Answer the followng as required with calculations : (Question no. 35 to 39)
(4 marks each)
36.A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Sol: Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
 Case (i) Speed = Distance / Time = (1500 / x) Hrs 
Case (iI) Time taken by the aeroplane = (x - 1/2) Hrs
 Distance to the destination = 1500 km Speed = Distance / Time = 1500 / (x - 1/2) Hrs   Increased speed = 250 km/hr
   ⇒ [1500 / (x - 1/2)] - [1500 / x] = 250
 ⇒ 1/(2x2 - x) = 1/6 ⇒ 2x2 - x = 6
 ⇒ (x - 2)(2x + 3) = 0 ⇒ x = 2 or -3/2
Since, the time can not be negative, The usual time taken by the aeroplane = 2 hrs and the usual speed = (1500 / 2) = 750 km/hr.

37.The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of of elevation is 45°, Find the height of the tower SQ and the distance XQ.
In triangle YRQ,
tan 45 = QR/ YR
This gives YR = x
or XP = x [As YR = XP].. (1)
Now, in triangle XPQ,
tan 60 = PQ / PX
Root 3 = x+40/x
This gives:
[Using(1)]
X = 40 / V3 - 1
x = 20 (V3 + 1) = 54.64 m
(By rationalisation)
So, height of the tower, PQ = x + 40 = 54.64 + 40 =
94.64 m
In triangle XPQ,
sin 60 = PQ / XQ
V3/2 = 94.64/ XQ
This gives XQ = 109.3 m

38.Two right circular cones A and B are made. A having 3 times the radius of B and B having half the volume of A. Find the ratio of height of A and B.
Let radius of cone Y = r
Therefore, radius of cone X = 3r
Let volume of cone Y = V
Therefore, volume of cone X = 2V
Let ,h1 be the height of the X and h2 be height of Y
Therefore volume of cone X = 1/3 π(3r2)h1 = 3πr2h1
Volume of cone Y = 1/3 πr2h2
Therefore ,2V/V = 9h1/h2
h1/h2 = 2/9
h{1}} h{2} = {2}{9}
h2: h1 = 9 : 2

39.Sides AB and BC and Median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of A PQR. Prove that  ABC ~ PQR.

Answers
Given: ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e., AB/PQ = BC/QR = AD/PM
To Prove: ΔABC ~ ΔPQR
Proof: AB/PQ = BC/QR = AD/PM
⇒ AB/PQ = BC/QR = AD/PM (D is the mid-point of BC. M is the mid point of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR ...(i)
∠ABC = ∠PQR ...(ii)
From equation (i) and (ii), we get
ΔABC ~ ΔPQR [By SAS similarity criterion]

39.State and prove the Pythagoras 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