ANSWER

1. The amount at the end of 2 years will be the principal for third year.
Amount = Rs.1389.15
Principal = Rs.1323.00
Interest = Rs.66.15
Pnr/100 * I = 66.15
\ r = 5%
Again to find P, A = 1323, t = 2, r = 5%
Let the principal be Rs.100
For the 1st year = Rs.100
Interest for 1st year = 100 * I * 5/100 = Rs.5
Amount for the 1st year = 100+5 = Rs.105
Principal for the 2nd year = 105*1*5/100 = 21/4 = Rs.5.25
\ Amount for 2nd year = 105+5.25 = Rs.110.25
If the amount is Rs.110.25 then the sum = Rs.100
If the amount is Rs.1323
Then the sum = 100 * 1/110.25 * 1323
= Rs.1200

2. Principal for the first year = Rs.400
Interest for the first year = 400 * 1 * 5/100 = Rs.20
Amount for the 1st year
= Principal for the first year + interest for the first year
= Rs.400 + Rs.20 = Rs.420
principal for the 2^nd year = Rs.420
interest for the 2^ndyear = Rs.441
compound interest for 2 years = Rs.441 - Rs.400
= Rs.41

3. DDFE ~ DDBC (AAA)
Þ DF²/DB² = area DDFE/area DDBC
Now, DF/FB = 3/1 \FB/DF = 1/3
Þ FB/DF + 1 = 1/3 + 1
Þ FB+DF / DF = 4/3 Þ DF/DB = ¾
Hence A DDFE/A DDBC = 9/16
Now EF/AD = EF/BC (opp. sides of 11gm.)
\ EF/AD = EF/BD = DF/DB Þ
A DOFE/a DODA = 9/16

4. i) (sin49°/cos41°)² + (cos41°/ sin49°)²
= ((sin49°/cos(90°-49°))² + (cos(90°-49°)/ sin49°)²
= (sin49°/ sin49°)² + (sin49°/ sin49°)²
= 1² + 1² = 2

(ii) From figure :
AD²+BD² = AB² (ÐADB = 90)
AD² + 9² = 15²
\ AD = 12
Let DC =x and AC = y
AC² = AD² + DC²
y² = 144 + x²
From DABC,
AC² + AB² = AB² (ÐBAC = 90°)
\ y² = (9+x)²-15²
\ y² = x²+18x-144
x²+18x-144 = 144+x²
\ x = 16cm
\CD = 16cm

5. construction

6. (i) 2/x+1 - 4/2x-7 = 3/2-x
Þ 2(2x-7)-4(x+1)/(x+1)(2x-7)
= 3/2-x-18/2x2-5x-7 = 3/2-x-36+18x
= 6x2-15x-21
2x²-11x+5 = 0(x-5)(2x-1) = 0
\ x = 5 or ½
(ii) Let x² = m
\ 4m²-25m+36 = 0
Here a = 4, b = -25, c = 36
\ m = -(-25) ± Ö(-25)2-4(4)(36)
= 25±Ö49/8
= 32/8 or 18/8
= 4 or 9/4
Resubstituting for m = x²
x² = 4 or 9/4
\ x = 2 or 3/2

7. (i) put x = 0 then 0+y = 4 Þ y = 4
put x = 1 then 2+y = 4 Þ y = 2
put x = 2 then 4+y = 4 Þ y = 0
for all other values of x e w we do not get y e w
Hence the range of R = {4,2,0}
The domain of r = {0,1,2}
R = {(0,4),(1,2),(2,0)}
(ii)AP represents the upper part of tree APB
Height of tree = AP+PB
In rt DABP, PB/20 = tan38°
\ PB = 20 * tan30° = 20 * 0.7813 = 15.626
\ 20/PB = cos38° \ AP = 20 * 1.2690 = 25.38
\ AP + PB = 15.626 + 25.38 = 41.006
\ Height of tree was 41m.

   

8. (a) Þ é-a -bù = é1 0ù equating corresponding elements a = -1, b = 0, c = 0, d = -1
             ëc d û         ë0 1û

(b) For matrix multiplication the number of columns of 1st matrix is equal
to the number of 2nd matrix.
\ M is having 2 columns.
Since the product of the above matrix is 1 * 2
\ The number of rows in M is 1
\ M has the order 1 * 2
Now, M é1 1ù = [1,2]
ë1 2û
Þ [a,b] é1 1ù = [1,2]
ë1 2û
\ [a+b a+2b] = [1,2]
Thus a+b = 1 and a+2b = 2
\ a = 0 and b = 1
Hence M = [0,1]
(c) Mean = Sx/n = 5+3+8+7+6+1/6 = 30/6 = 5
If 3 is added to each number
New mean = original mean + 3
= 5+3
= 8

9. (a) x = 3 is a line parallel to y axis at distance of 3 units from the
y axis (line BC)
y = 4 is a line parallel to the x axis at a distance of 4 units from the x
axis (line DE)
y = x is a line through the origin and having slope = 1 (line OA)
From figure, the triangle enclosed between lines PQR. Hence co-ordinates of
DPQR are P (3,3) Q (3,4) R (4,4)
(b) y = 2x+3 and y = px/2 + 4
Slope of line, m₁ = 2 and m₂ = p/2
Since lines are parallel m₁ = m₂
\ p = 4

10. (a) Mean = Sfx/Sf = 10*100 + 20*200 + 30*300 + 40*400/100+200+300+400
= 100042000+9000+16000/1000
= 30,000/1000
Mean = 30
(b) AB/DE = BC/EF = AC/DF
12/6 = 16/EF = 14/DF
Now 12/6 = 16/EF \ EF = 8
and 12/6 = 14/DF \ DF = 7
DABC ~ DDEF
\ the corresponding angles are congruent
\ ÐB @ ÐE
but ÐB = 60° \ ÐE =60°

11. (a) construction

12. (a) Tangent segment PA @ tangent segment PC
\ m ÐPCA = MÐPAC
Since ÐAPC = 70°
\ MÐPCS+MÐPAC = 180° - 70° = 110°
2MÐPCA = 110° \ MÐPCA = 55°
By the tangent secant property
mÐPCA = mÐABC
\ mÐABC = 55°
(b)
Given : - P is the centre of the circle r is the radius of the circle
seg AB is a chord of the circle and it does not pass through P
Aim : - AB<2r
Proof : - Let M be the mid point of chord AB
\ seg PM ^ seg AM
Join point A to point P
Because the hypotenuse of a right angled triangle is its greatest side,
\ In DAPM , AM < AP
\ ½ AB < r
\ AB < 2r

    

13. (a) Suppose the dimensions of the cuboid are l = 4x , b = 3x and h = 2x
Surface area of vertical faces
= 2[b*h + h*l]
\ 448 = 2[3x*2x + 2x*4x]
\ 448 = 28x2
\ x2 = 16
\ x = 4
Hence l = 16 , b = 12 and h = 8
(b) Let CR = x , AS = y \ DR = x and SD = y
Since PB = 5 and QC = 3
\ PQ+SR = 5+3+y+x
and PS+QR = 5+y+3+x
\ PQ+SR = PS+Q
As opposite sides of a parallelogram are @
PQ = SR AND PS = QR
\ 2PQ = 2PS
\ PQ = PS
But PQ = 5+3 = PS = 8