Cbse Maths PaperIII Answers

Model Paper - III Solutions

17x + 31y = 113 and 31x + 17y = 127
Adding both we get
17x + 31y = 113
31x + 17y = 127
50x + 50y = 240
\ 5x + 5y = 24 .............(I)

Subtracting
17x + 31y = 113
-31x - 17y = -127
-14x + 14y = -14
\ x - y = 1 ..................(II)

Multipling equation II with 5 and adding with I
5x + 5y = 24
5x - 5y = 1
10x = 25 \ x = 2.5

Substituting x = 2.5 in equation I
5(2.5) + 5y = 24
5y = 11.5 \ y = 2.3
The given equation is 6x² - 11x + 3 = 0
Sum of roots = coefficient of x
                      coefficient of x²
= -11 = 11
     6     6
and product of roots = constant term
                               coefficient of x²
= 3 = 1
   6    2
Let p(x)/q(x) be the required rational expression.
Then p(x) = a quadratic polynomial with zeros 2 and -3
= (x-2) (x+3)
and q(x) = a cubic polynomial with zeros -2, 1 and 4
= (x+2)(x-1)(x-4)
hence the required rational expression
= p(x) = (x-2)(x+3)
q(x) (x+2)(x-1)(x-4)
x²- 3x - 4 * x2 - x - 6
x² - 9           x2 + 3x + 2
\ x²-3x-4 = (x-4)(x+1)
x²-9 = (x-3)(x+3)
and,
x²-x-6 = (x-3)(x+2)
x²+3x+2 = (x+1)(x+2)

\ the given expression
= x²-3x- 4 * x²-x-6
    x²-9        x²+3x+2
= (x-4)(x+1 ) * (x-3)(x+2)
    (x-3)(x+3)     (x+1)(x+2)
= (x-4)
   (x+3)
(2x³+3x²+3x+1) 3x³+x²+x-2 (3
2
6x³+2x²+2x-4
6x³+9x²+9x+3 ....... Subtracting we get
- 7x²-7x-7 (-7)
x²+x+1
x²+ x + 1) 2x³+3x²+3x+1(2x +1
2x³+2x²+2x
- - -
x²+ x+1
x²+ x+1
´
\ The required H.c.f = x²+ x + 1
a = 25, b =17, c = 12
s = a + b + c = 25 + 17 +12 = 27
2 2
D = Ö s (s-a) (s-b) (s-c)
= Ö 25 (2 * 10 * 15)
= 90 sq. cm
Area = ½ * Side * corresponding altitude
90 = ½ * 25 * h
\ h = 180/25 = 7.2 cm
Area = 176 cm²
l = 16 cm ; b = 11 cm
Circumference of base of cylinder = 11
Let r be radii of base then 2pr = 11
\ r = 11 = 7
2 * p 4
Volume = pr²h
= p * 7/4 * 7/4 * 16
= 154 cm²
V = abc
S = 2(ab + bc + ca)
RHS = 2/S [1/a + 1/b +1/c]
= 2 [ bc +ca +ab / abc]
2(ab + bc + ca)
= 1/abc = 1/V = LHS
5 tan A= 12
tan A = 12 SinA = 12 \ Sin A = 12
5 CosA 5 CosA = 5
2Cos A + Sin A = 2(5) + 12 = 10 + 12 = 22
Sin A - Cos A 12 - 5 7 7
Find AC
AC² = AB² + BC²
AC² = 15² + 8²
= 225 + 64
= 289
\ AC = 17
To Prove ÐACB = ÐADB
Construction -> Join OA and OB
Proof -> ÐAOB = 2ÐACB ........(I)
ÐAOB = 2ÐADB .................(II)
From (I) & (II)
ÐACB = ÐADB
Const Join OC & OD Bisect BO at P
Proof : BC is tangent & OC is the radius through the point of contact c
\ ÐOCB = 90⁰
The circle with center P & diameter OB passes through C
\ BP = PO = PC
In DOBC & DOBD
OC = OD - radius of same circle
OB = OB - common
BC = BD - tangent seg. are equal
DOBD=DOBD
ÐOBC = ÐCBB
But ÐDBC = 120⁰
\ ÐOBC = ÐOBD =60⁰
Now in DBPC & BP = PC
\ ÐPCB = ÐPBC
ÐPBC = ÐOBC = 60⁰
\ D BPC is an equilateral D
Now BO = 2BC
\ BP = BC
\ BO = 2BC
n₁ = population of town A = 768942
n₂ = population of town B = 609272
x₁ = Death rate of town A = 1.56 per 100 i.e 15.6 per 1000
x₂ = Death rate of town B = 1.28 per 100 i.e 12.8 per 1000
\ x = n₁x₁ + n₂x₂ = 768942 *15.6 + 609272 * 12.8
           n₁ + n₂                 768942 + 609272
= 11995495 + 7798681
         1378214
= 19794176 = 14.362 per thousand
    1378214
= 14 per thousand
Sum of 40 observations = 160 * 40 = 6400
Actual sum of 40 observations = 6400 - 125 + 165
= 6440
Correct Mean = 6440 = 161
40
A (DABC) = (AB)² .........(DABC ~ DDEF)
A (DDEF)     (DE)²
36 = AB²
64   (6.2)²
AB2 = 36 * (6.2)²
              64
AB² = (6 * 6.2 / 8)²
AB = 6 * 6.2 / 8 = 37.2 / 8 = 4.65

Let the number of camels = x
Number of camels seen in the forest = x/4
Number of camels gone to mountain slops = 2Öx
Number of camels on the bank of the river = 15
As per condition
x/4 + 2Öx + 15 = x
3x - 8Öx - 60 = 0
Let Öx = y
3y² - 8y - 60 = 0
3y² - 18y + 10y - 60 = 0
3y (y-6) + 10 (y-6) = 0
(y-6) (3y +10) =0
\ y = 6 or y = -10/3 (neglected)
y = Öx
\ Öx = 6 \ x = 36
\ Number of camels = 36

On Graph paper
2x - 6y = 12 3x - 6y = 24

x	0	6	x	0	8
y	2	0	y	4	0
(x,y)	(0,2)	(6,0)	(x,y)	(0,4)	(8,0)

Minimum balance for January 1999 = Rs 1250
Minimum balance for February 1999 = Rs 700
Minimum balance for March 1999 = Rs 3275
Minimum balance for April 1999 = Rs 3275
Minimum balance for May 1999 = Rs 3275
Minimum balance for June 1999 = Rs 3275
Minimum balance for July 1999 = Rs 3275
Minimum balance for August1999 = Rs 3275
Minimum balance for September 1999 = Rs 3275
Minimum balance for October 1999 = Rs 3275
Minimum balance for November 1999 = Rs 1775
Minimum balance for December 1999 = Rs 1775
Total = Rs 34,700
P = Rs 34,700, R = 5%, T = 1/12 years
Interest = P * R * T = 34700 * 5 * 1
100 100 * 12
= Rs 144.58
x Sin³A + y Cos³A = SinA . Cos A
x(SinA) Sin²A + (y CosA) Cos²A = SinA . CosA
x(SinA) Sin²A + (x SinA) Cos²A = SinA . Cos A
x(SinA) [Sin²A + Cos²A] = SinA . CosA
x SinA = SinA . CosA
x = Cos A
Given : x Sin A = y Cos A
CosA . SinA = y Cos A
\ y = Cos A
Prove that : x² + y² = 1
i.e Sin² A + Cos² A =1
\ 1 = 1 hence proved
ÐAOB = 360/18 = 20⁰
AB = 2MB
MB = ½AB = ½ * 60 = 30 cm
In right angle triangle MOB
Cot a/2 = OM / MB
Cot a/2 = r/30
r = 30 * Cot 20/2
= 30 * Cot 10
= 30 * 5.671
= 170.13 cm
If a line is drawn parallel to one side of a triangle intersecting the other two sides,
it then divides these sides in the same ratio.

Given : D ABC in which PQ | | BC,
Intersecting AB in P and AC in Q
To Prove : AP/ PB = AQ/ QC
Construction : Join BQ, CP and draw QN ^ AB
Proof : area (D APQ) = ½ AP * QN [Area of D = ½ * b * h]
area (D BPQ) ½ PB * QN
= AP/ PB ........(I)
Similarly area (D APQ) = AQ ........(II)
area (D BPQ) QC
But DBPQ and DCPQ stand on the same base BC,
and lie between the same parallels BC and PQ
\ area (D BPQ) = area (D CPQ) .........(III)
Hence from (I), (II), & (III) we get
AP / PB = AQ / QC

Let R = External radius of the base
r = Internal radius of the base
h = height
Thickness = (R-r)
Total surface area of a hollow cylinder
= 2 p (R + r) (h + R - r) = 4620
Area of base ring = p (R² - r²) = 115.5
R² - r² = 73.5 .........(I)
2 * 22/7 * (R + r) (7 + R - r) = 4620
(R + r) (7 + R - r) = 735 .........(II)
Dividing (II) by (I)
7 + R - r / 2(R - r) = 10
R - r = 7/19 cm
ABCD is a parallelogram
Now in D BCE
S= 25+17+26 = 34
2
Area of D BCE = Ös(s-a) (s-b) (s-c)
= Ö 34 (9 * 17 * 8)
= 204 ........................................(I)
Also area of D BEC = ½ * 17 * h ......................(II)
From (I) and (II) 17/2 h = 204 \ h = 24 cm
Area of trapezium = ½ (77 + 60) * 24 = 1644 cm²
Ramlal's total income = Rs. 1,85,000
less : Standard Deduction = Rs 20,000
\ Taxable income = Rs 1,65,000

Payable tax = 19000 + 30/100 * 15000
= 19,000 + 4,500
= 23,500

Savings PF = 3000 * 12 = Rs 36,000
add : LIC = Rs 32,000
Total savings = Rs 68,000

Rebate = 20/100 * 68,000 = Rs 13,600
but is subjected to Rs 12,000
Net tax = 23,500 - 12,000 = 11,500
Amount of income tax paid in 11months = Rs 11,000
Tax to be paid in the last month = Rs 500

Section C

a(b-c)³ + b(c-a)³ + c(a-b)³
= a (b³ - 3b²c + 3bc² - c³) + b(c³ - 3c²a + 3ca² - a³) + c(a³ - 3a²b + 3ab² - b³)
= ab³ - 3ab²c + 3abc² - ac³ + bc³ + 3bc²a + 3a²bc - ba³ + a³c - 3a²bc + 3ab²c - ab³
= (ab³ - ac³) + (bc³ - a³b) + (bc³ - b³c)
= a³(c - b) + a(b³- c³) + (bc³ - b³c) [ descending area of a ]
= -a³ (b - c) + a(b - c)(b²+ bc + c²) - bc(b² - c²)
= -a³ (b - c) + a(b - c)(b² + bc + c²) - bc(b - c)(b + c)
= (b - c) [ -a³ + a (b² + bc + c²) - bc (b + c) ]
= (b - c) [ -a³ + ab² + abc + ac² - b²c - bc²]
= (b - c) [ b²(a - c) + bc(a - c) + a(c²- a²) ]
= (b - c) [ -b²(c-a) -bc(c - a) + a(c - a)(c + a) ] ( arranging in descending power of b)
= (b - c)(c - a) [ -b² - bc + ac +a²]
= (b - c)(c - a) [ c(a - b) + (a² - b²) ]
= (b - c)(c - a) [ c(a - b) + (a - b)(a + b) ]
= (b - c)(c - a)(a - b)(c + a + b)
= (a - b)(b - c)(c - a)(a + b + c)
Rate of walking = x km/hr.
Distance = 2 km.
Time taken = Distance / Speed = 2/x hrs
Speed = (x + 1) km/hr.
Time taken = 2/x + 1
Given condition : 2 - 2 = 1
x x + 1 6
= 2x + 2 - 2x = 1
x²+x 6
= 12 = x² + x
\ x² + x - 12 = 0
\ (x + 4) (x - 3) = 0
\ x = 3 & x = -4 (neglected)
\ Rate of walking = 3 km/hr.
Let AB be the first tower and CD be the second tower
Given : CA = 70m, AB = 120m and ÐBDF = 30⁰
\ DF = CA = 70m
Let CD = h, then BF = BA - FA = BA - CD = 120 - h
Now from D BDF,
Tan 30⁰ = BF/DF = 120 - h/70 \ 1/Ö3 = 12 - h/70
Ö3(120-h) = 70 i.e. 120Ö3 - Ö3h = 70
\ Ö3 h = 120Ö3 - 70
3h = 120 * 3 - 70Ö3
3h = 360 - 70 * 1.732 = 360 - 121.24
3h = 238.76 \ h = 79.58
Hence height of the first tower is 79.58m.
To Prove : RS² = (PP¹)² +AB²
Proof : RP is a tangent and RAB is a recant
\ (PP¹)² = RA * RB .............(1)
PP¹ is a tangent and RAB is a recant
\ (PP¹)² = RA * RB ..............(2)

From (1) & (2)
RP = PP¹
\ R is mid point of PP¹
Similarly S is the mid point of QQ¹
But PP¹ = QQ¹(direct common tangent)
\ ½ PP¹ = ½ QQ¹
\ RP = SQ
\ RP²= SQ²
\ RA * RB = SA * SB
\ RA * (RS - SB) = (RS - RA) * SB
\ RA = SB
Now, RS² - AB² = (RS + AB)(RS - AB)
= [ (RA + AB + SB) + AB ][(RA +AB + SB) - AB ]
= (RA + AB + RA + AB)(RA + RA)
= (2RA + 2AB)(2RA)
= 4(RA + AB)(RA)
= 4RB * RA
= 4 RP²
=(2RP)² = (PP¹)²
\ RS² - AB² = (PP¹)²
\ RS² = AB² + (PP¹)² hence Proved.

Age Group	Population(S)	No. of Deaths(N)	Standardised Population(S1)	D = N/S * 1000	S1*D
under 10	30000	450	24000	15.0	360000
10 - 25	40000	100	32000	2.5	80000
25 - 50	30000	180	30000	6.0	180000
50 - 80	40000	200	20000	5.0	100000
above 80	25000	250	5000	10.0	50000
Total			111000		770000

Standardized Death = SSD = 770000 = 6.936
SS 111000
= 6.94