ICSE Maths Paper II

ANSWER-

Principal for the 1^st year  = 3125
Interest for the 1^st year     = 3125 * 1 * 4
                                                 100
                                          = Rs. 125
Principal for the 2^nd year  = Rs. 130
Principal for the 3^rd year  =  Rs. 3380
Interest for ½ year             = 3380 * 1/2  * 4
                                                    100
                                          = Rs. 67.60
Amount after 2 ½ years     = Rs. 3447.60

Principal for the 1^st year = Rs. 5000
Interest for 1^st year         = 5000 * 1 * 5/100
                                        = Rs. 250
Sum refunded = Rs. 150
Principal for the 2^nd year = Rs. 5000 + Rs. 250
                                         = Rs. 150
Interest for the 2^nd year    = 5100 * 1 * 5/100
                                         = Rs. 255
Principal for 3^rd year       = 5100 + 255 - 150
                                          = Rs. 5205
Interest for 3^rd year = Rs. 260.25
Principal for 4th year = 5205 + 260.25 - 150
                                   = Rs. 5315.25

OA   =  2
AA₁    3
     OA          =      2            (componendo)
OA + AA₁        2 + 3 
OA    = 2 Þ Scale factor = OA₁ = 2.5
OA₁     5                           OA 
  AD     =  2
A₁D₁       5 
    3       =  2
A₁D₁        5      
\ A₁D₁ = 7.5 Units
area ABCD         =   AD²   =   22 =   4
area A₁B₁C₁D₁      A₁D₁²       5²     25

a) L.H.S= Ö1-Sinq/Ö1+Sinq
=Ö1-Sinq /Ö1+Sinq * Ö1-Sinq/Ö1-Sinq
= 1-Sinq
    Cos²q
= 1-Sinq
    Cosq
= 1-     Sinq
Cosq  Cosq
= Secq - tanq=R.H.S

     

Let the triangle be ABC
Let AB = x; BC=Ö2x and CA =Ö3x
AB²+BC²=x²+(Ö2x)²=3x²
Again CA²=(Ö3x)²=3x²
\Ab²+BC²=CA²
ÞÐABC=90⁰
Þ DABC is right angled D

```
 Construction.
```

i) x² + 1 = 18
         x²
Þ x² + 1 - 2 = 18 - 2
           x²
Þ x² + 1  - 2(x²) 1 = 16
            x²         x²
éx - 1  ù ² = 16 
ë     x  û
x - 1 = ± 4
     x

ii) Let 2x² +6x +1 = 0
Comparing with ax² + bx + c = 0
We get a = 2, b = 6, c = 1
Þ x = -6 ±  Ö36 - 4 *2 *1
                     2 * 2
        = -6 ± Ö28
                 4
x = -6 + 5.29 or -6 - 5.29
            4              4
x = -0.1775 or -2.8225
  = -0.23 or -2.8

Section - II.

i)  2 <  x  £  5, x £ N
     Þ x S{ 3,4,5}
     Again 4 £ y < 7, y S N
     Þ yS {4,5,6 }
     Þ R = { (3,4), (3,5), (4,5 ), (5,6) }
     Range = {4,5,6}
     Arrow diagram :
                                

ii) Let AP be the flagstaff on top of building PC and B is the point of observation
     ÐPCB = 60°
     In rt. D ABC,  AC = tan 63º
                          50
     AC = 50 * 1.963 = 98.15 cm.
     \ y + x = 98.15 cm.
     In rt D PBC, PC = tan 60
                        50
     \ PC = 50 * 1.732 = 86.60
     \ x = 86.6 M
     \ y = 98.15 - 86.6
     y = 11.55m

a) A will not have a multiplicative inverse
     16 * 4 - x² =0
     Þ 64- x² = 0      Þ x² = 64
     Þ x = ± 8

b) L.H.S.= éa  2 ù + é -3 b ù - é -1 2  ù      
                ë3 -1 û    ë -2  2û   ë c  d  û

                éa - 3 + 1ù é  2 + b -2  ù
                ë3 - 2 - c û ë -1+ 2 -d   û

                  é a -2 ù é b     ù
                  ë 1 -c û ë 1- d û

                  é2 -1ù = R.H.S
                  ë4  3û
     Þ a -2 = 2  Þ a = 4
     b  = -1     Þ b = -1
     1 - c = 4  Þ c = -3
     1 - d = 4  Þ d = -2
c)  
       
         
          x
        
         
          f
        
         
          d=X-A
        
         
          fd
        
      
       
         
          5
        
         
          10
        
         
          -2
        
         
          -20
        
      
       
         
          6
        
         
          12
        
         
          -1
        
         
          -12
        
      
       
         
          7
        
         
          15
        
         
          0
        
         
          0
        
      
       
         
          8
        
         
          11
        
         
          1
        
         
          11
        
      
       
         
          9
        
         
          7
        
         
          2
        
         
          14
        
      
       
         
          10
        
         
          5
        
         
          3
        
         
          15
        
      
       
         
          Total
        
         
          60
        
          
          
        
         
          8
        
      
    
S f = 60 S fd =8
\ Mean = A + Sfd = 7 + 8 = 7 + 0.13 = 7.13
            Sf          60

a) 2x - y = 3  Þ y = 2x - 3
    
       
         
          x
        
         
          0
        
         
          -1
        
         
          2
        
      
       
         
          y
        
         
          -3
        
         
          -5
        
         
          1
        
      
    

    x + 3y = 5 Þ  y = -1/3 x +5/3
    
       
         
          x
        
         
          5
        
         
          2
        
         
          -1
        
      
       
         
          y
        
         
          0
        
         
          1
        
         
          2
        
      
    

 
 b) Given, 2y = 3x + 1  Þ y = 3x + 1 Þ y = 3x + 1
                                            2     2              2 
     \ Slope = 3 Hence slope of a parallel line = 3 
                     2                                               2
     Slope of a perpendicular line =             -1                
                                                        Slope of given line 
                                                    =    -1 
                                                          3/2
                                                    = 2/3

a)
       
         
          Class(x) 
        
         
          Frequency(F)
        
         
          u =x-A
        
         
          Fu
        
      
       
         
          10 - 15
        
         
          5
        
         
          -3
        
         
          -15
        
      
       
         
          15 - 20
        
         
          6
        
         
          -2
        
         
          -12
        
      
       
         
          20 - 35
        
         
          8
        
         
          -1
        
         
          -8
        
      
       
         
          30 - 35
        
         
          6
        
         
          1
        
         
          6
        
      
       
         
          35 - 40
        
         
          3
        
         
          2
        
         
          6
        
      
       
         
          Total
        
         
          40
        
          
          
        
         
          -23
        
      
    
Clearly assumed mean A = 25+30 = 55 = 27.5 
                                           2          2 
Mean  =  A + Sfu * i   where Sf = 40, Sfu = -23 
                     Sf                   
           = 27.5 - 23 * 5  = 24.63 
                        40 
c)Seg. PQ || side BC 
    \ AP = AQ (Basic proportionality)
         PB    QC 
    \  6 =   9 
        12   QC 
    \ 6 * QC = 9 * 12  
    \ QC = 18

```
11) a) Construction.
```

 a)  
             
     Consider the above figure 
     Given:  ABCD is cyclic 
     To prove: mÐA + mÐC = 180°
                    mÐB + mÐD = 180° 
     Proof: Arc BCD is intercepted by an inscribed angle BAD
     \ mÐBAD = 1/2 m (arc BCD)
     Similarly, are BAD is intercepted by an inscribed angle BCD
     \mÐBCD = ½ m (arc BAD)
     \ m ÐBAD + m ÐBCD = ½ [ m (arc BCD) + m (arc BAD) ]
     = ½ * 360° 
     m ÐA + m ÐC = 180° 
     Similarly m ÐB + mÐD = 180° can be shown.

b)  Let seg AB be a chord of length 18.
     Let M be its midpoint - Then seg PM ^ AB 
     \ PM² = PB² - MB² 
     = 12² - 9² 
     = 63
     \ PM = 3Ö7
     \ The midpoint of chord AB is at a distance of 3Ö7 from P.
     Therefore the midpoint of every other chord of same Length is also at a distance 3Ö7 from P.
     \ All such chords will have their midpoints on a circle with centre P and radius 3Ö7.
      This new circle is concentric with the given circle.

a) Surface area of sphere = 4 pr² 
     \ S = 4 * 3.14 * (3.75)² 
     \ Log S = log [ 4* 3.14 * (3.75)² ]
     = log 4 + log 3.14 + log (3.75)² 
     = log 4 + log 3.14 + 2 log (3.75)
     = 2.2470
     S = antilog (2.2470) 
     = 176.6
     The curved surface area of the sphere is 176.6sq. cm.

     
    
Given: P is the centre of a circle. A is a point on the circle 
          Line l is perpendicular to radius PA at point A.
To Prove: Line l is a tangent to the circle.

Proof: Let, on line l, B be a point distinct from A 
          \ Seg. PB is hypotenuse of rt. D PAB
         \ PB > PA
             PB ¹ radius
         \ Point B is not on circle 
         \ No point of the line l other than A, lies on the circle.
         \ line l has one and exactly one point common with the circle. 
         \ line l is tangent to the circle.

Class(x)	Frequency(F)	u =x-A	Fu
10 - 15	5	-3	-15
15 - 20	6	-2	-12
20 - 35	8	-1	-8
30 - 35	6	1	6
35 - 40	3	2	6
Total	40		-23