Solutions - Model Paper II

Section A

1) L.C.M. of p(x)& q(x) = (x+2) (x+3)²(x-5)
G.C.D. of p(x) & q(x) = (x+3)
P(x) = (x+2) (x+3)²
By formula:- P(x)*q(x)= (L.C.M.)(G.C.D.)
(x+2)(x+3)²*q(x) = (x+2) (x+3)² (x-5) (x+3)
\ q(x) = (x-5) (x+3) = x² + 3x - 5x - 15 = x²-2x-15.

2) (x-2)(x²+9x+20) = (x-2) (x+5) (x+4) = (x-2)
   (x+4)(x²+11x+30) (x+4) (x+6) (x+5) (x+6)

3) 15m² + 2m - 8 = 0 Þ 15m² + 12m -10m - 8 = 0
Þ 3m(5m + 4) -2(5m + 4) = 0
Þ(5m + 4) (3m - 2) = 0
\ 5m + 4 = 0 & 3m-2 = 0 \ m = -4/5 & m = 2/3 .Ans

4) x²+x = k(2-5x) x² + x = k(2 - 5x)
\ x² + x + 5kx - 2k = 0
\ x² + x(1 + 5k) - 2k = 0
\ 1 + 5k =0 5k = -1
k = -1/5 Ans

5)LHS = cosA + secA
                  cos A
= cos²A + 1 = ( secA = 1/cosA)
    cos²A
= 1 + sec²A ( 1/cos²A = sec²A)
= 1 + 1 + tan²A
= 2 + tan²A = RHS

6) A = 45⁰ LHS = cos2A
=cos2(45⁰)
=cos 90⁰
= 0
RHS = 1-2sin²45
=1-2(1/Ö2)²
=1-2/2
=1-1 =0
L.H.S.= R.H.S. Ans

.7) 1st 10 multiples of 3 are
3,6,9,12,15,18,21,24,27,30.
\Median =(5th+6th) terms
                     2
= 15+18
       2
= 33
    2
=16.5 Ans

8) No. of Girls =25
Mean weight = 40
Total weight =40x25 = 1000 kgs.
Mean weight including man (i.e.for 26 people)
=26x40.5 = 1053
\weight of man =1053-1000
=53 kgs Ans.

10) A D ABC = BC²
      A D PQR   QR²
\36 = (6.4)²
   25    (QR²)
\ QR2=(6.4)² x36
                 25
QR =(6.4x6)²
            5
= 76 cms

11) PS =10 & SQ =8
QS2 = PS x SR
(8)2 =10x SR
64 = 10 x SR
\SR =6.4

12)

Area of rhombus = d₁* d₂ = 24 * 10 = 120 cm³
                                                 2           2
The diagonals of a rhombus bisect each other
\AO = 12 cm, OD = 5cm
In D AOD, let AD = x cm
\x² = 12²+5²
= 144 +25
=169
=13 cm.
Thus the parameter of rhombus = 4x13
=52 cm Ans.

13) Volume of cube =1000 cm³
volume of cube = l³ =1000
\l = 10 cm
total surface area = 6l²
= 6x(10)²
=600 cm Ans

14) Area of rectangle = 20x15 = 300 cm²
Area of shaded region =124 cm²
\Area of circle = 300-124
= 176 cm²
Q Area of .circle = pr²
\pr² = 176
\x² = 176 x 1
                  22
x2 = 56
\x = 2Ö14 cm Ans

15) Let L.P. be Rs.100/-
\ sales tax = 6% = Rs.6.
\S.P. = Rs.106/-
LP = 100 x 583 = Rs.550 Ans
        106

Section B

Mean = 50 + åfx = 50 + 4320+70a
                     åf               96+a

\ 4800+50a = 4320 +70a \ 30a = 480 \ a = 16 Ans

17) 3x-4y=12 6x-3y=18

18)

19) (a+b+c) (ab+bc+ca) -abc
= a²b+abc+a²c+ab²+b²c+abc+abc+bc²+ac²-abc
= a²(b+c)+a(b²+2bc+c²) +bc(b+c)
= (b+c) [a²+a(b+c)+bc]
= (b+c)[a(a+b) x c(a+b)]
= (b+c)(a+b)(a+c)

20)

Given Þ 1) ABC is an equilateral D
2) BD= 1 BC
           3
Prove Þ 7 Ab² = 9AD²
Proof Þ D ABE @ D ACE
In right angled D , by pythagorus theorm
AB²= BE²+AE² ------I)
In D ADE
AD² = AE²+DE² ------II)
From  II -I
AB²-AD² = BE²-DE²
=BE²-(BE-BD)²
=(BC/2)² - (BC/2 -BC/3)²
=BC²/4 - BC²/36
=8BC²/36
\AB²-AD² = 2/9 AB²
\7AB² = 9 AD² Ans

21)LHS = (m²+n²)cos²B
= (cos²A/cos²B+cos²A/sin²B) cos²B
= cos²A (1/cos²B+1/sin²B).cos²B
=cos²A(sin²B+cos²B/cos²B.sin²B).cos²B
=cos²A/sin²B
RHS = cos²A/sin²B \LHS = RHS

22)

Cost of living index = åp₁q₀
åp₀q₀ x 100 = 1910/ 1395 x100
=136.92 Ans.

23) Given ÞIn circle of DABC touches BC,(A & AB at D,E & F respectively.
Prove Þ AF+BD+CE=AE+BF+CD=1/2(parameter of D ABC)
Proof Þ Q length of two tangents from an
External point to a circle are equal
\ AF = AD
BD=BF
CE=CD
Adding these we get
AF+BD+CE = AE+BF+CD
Now parameter = AB+BC+CA
AF+BF+BD+DC+AE+CE
=2(AF+BD+CE)
\AF+BD+CE+AE = BF+CD
= ½ (perimeter of D ABC) Ans.

24) Isoceles DABC & DEF have equal vertical angles
Also, AB = DE
AC DF
\DABC ~D DEF - (SAS Test)
AB = AC = AL = BC
DE    DF    DM    EF ------1)
Now area D ABC =1/2BC x AL = 9
       area D DEF   ½ EF x DM    16
\BC x AL = 9
   EF DM     16
\ AL x AL = 9
    DM DM   16
\AL²= 9
  DM²  16
\AL = 3
  DM    4
\ AL:DM = 3:4 Ans

25) 2prh + 2pr² = 462
2pr(r+h) =462
2prh = 1/3 2pr(r+h)
\h= 1/3(r+h)
2h = r
Put r = 2h
2p . 2h . 3h = 462
12ph² = 462
\h² = 462x7 = 12.25
          12x22
\h = 3.5 cm
volume = 22/7
= 539 cm³ Ans

Section C

26)

Left = height of tower
Tan A = h/a -----1
Tan (90 - A) = h/b -----­2
Cot A = h/b
From 1 & 2
H² = ab
h = Öab Ans

27)2(x²+1/x,²) -3 (x+1/x) -1 = 0
2[(x+1/x)²-2)] -3(x+1/x-1=0
2(x+1/x)² -4 -3(x+1/x)-5 = 0
2(x+1/x)² -3(x+1/x)-5 = 0
Let x+1/x be y
\2y²-3y-5 =0
\2y²-5y+2y-5=0
\y (2y-5/0+1(2y-5) =0
\(2y-5)(y+1) =0
\2y-5=0 & y+1 =0
\y=5/2 & y = -1
Resubstituting
x+1/x = 5/2 x+1/x =-1
x+1/x = -1 x2+1 =-x
2(x²+1)= 5x x²+x+1=0
2x²-5x+2=0
2x²-4x-x+2=0
2x(x-2)-1(x-2)=0
(2x-1)(x-2)=0
x=1/2 & x=2 Ans

28) Given ® Let L & N be mid-points of AM & MB respectively.
Let O -centre
R -radius
Radius is drawn to touch 3 semicircles at PQR
Const ® Join OL1,OM & ON, then
O,P,L are collinear
R,Q,M are collinear
O,Q,N are collinear
Prove ® radius of circle = 1/6 AB
Proof ® Let AB = x \ AL =1/4x
\PL =x/4.
Now OL =OP+PL= r +x/4
ON =OQ+QN=x+x/4
\DOLN is an isoceles Dm whose mid-point is M.
\OM ^LN
Hence D OML is a right angled D
OL²=OM²+LM²
(r+x/4)² =(RM-QR)²+(x/4)²
=(x/2-r)²+(x-4)²
\x²+2x.x+ x² =x² + r²-2x . r + x²
            4    16   4         2         16
\3rx =x²
   2      4
\r = x/6 =1/6AB Ans.

29) Gross Income = 12x8300 = Rs.99,600
(-)Std.deduction =(Rs.25000 since the income does not exceed 100000) - Rs.25,000
Taxable Income Rs.74,600
Income Tax =(As the taxable income liesBetween Rs.60,000 & Rs,1,50,000)
\Tax =Rs.1000+20% of income exceeding Rs.60,000 =1000+20% of 74,600-60,000 =1000+2920 Rs.3,920
Savings =PF 9,200 p.a. (-) 1,840
Rebate on savings =20 of 9,200 Rs.2,080
Tax already paid 11x100 (-) 1,100
Rs.980
Tax to pay Rs.980
She doesn't want to pay any more tax he must invest in NSC 5 x Rs.980 =(20% of 4900 is Rs. 980) Rs.4,900

30) Let the digit at units place = x
digit at ten's place =y
\given no. = 10x+y
Reversed no.=10y+x
\According to given condition
xy=8
10x+y = -63 =10y+x
9x-9y=63
x-y=7
put y = x-7 in equation
x(x-7)=8
x2-7x-8=0
(x-8) (x+1) =0x =8,x = -1(neglected)
\y=1
\required no =81