-
Interest for third year = Rs.363
Interest for second year = Rs.330
Difference = Rs.33
Hence the interest on Rs.330 for 1 year = Rs.33
Since the period is 1 year we can use the simple interest formula
\ r = 100 * I
P*t
= 100 * 33 = 10%
330 * 1
Let the principal be Rs.100
Interest for the first year = Rs.10
Amount for the first year = Rs.110
Principal for the 2nd year = Rs.110
Interest for the 2nd year = 110 * 1 * 10
100
= Rs.11
\ Sum for 2nd year = Rs.100
If interest for 2nd year = Rs.330
Then the sum = 100 * 330
11
= Rs.3000
-
Present value = Rs.4000
Reduction for the 1st year = 4000 * 1 * 5
100
= Rs.200
\ value after 1 year = Rs.4000 - Rs.200
= Rs.3800
Reduction for the 2nd year = 3800 * 1 * 5
100
= Rs. 190
Hence the value after 2 years = 3800 - 190
= Rs.3610
-
In DAOB and DCOD
ÐAOB @ ÐCOD opposite angles
Ð OAB @ ÐOCD alternate angles
\ DAOB @ DCOD (A - A test)
\ AO = OB = BA
CO OD DC
\ OB = BA
OD DC
\ OB = 6
15 20
\ OB = 6 * 15 = 9 = 4.5
20 2
-
i) L.H.S. = 1 = 1
1 + 1 1 + sinq
sinq sinq
= 1 * sinq
1 + sinq
= R.H.S.
ii) In DAPB : AB² = AP² + PB², LP = 90°
DDCP : DC² = DP² + PC²
\ AB² + DC² = AP² + PC² + DP² + PC²
= AD² + BC²
\ AB² + DC² = AD² + BC²
- construction .
-
i) Ö2x² - 17 = 9
squaring both the sides
2x² - 17 = 81
2x² = 98
\ x = ±7
ii) a = x1, b = 12, c = 9
3 = -b ± Öb - 4ac
2 2a
\ 3 = -(-12) ± Ö(-12)² - 4x1 * 9
2 2 * x1
\ Solving we get
a = 4.
-
i) A * B = {(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)}
B * A = {(4,1),(4,2),(4,3),(5,1),(5,2),(5,3)}
n(A * B) = n(A) * n(B)
= 3 * 2
= 6
ii) Let the two boys are at B and C. AD is the tower 100m high.
Now in triangle ABD
BD = tan57° \ BD = 100 tan57°
AD = 100 * 1.5399
= 153.99
In triangle ADC
DC = tan66°
AD
\ DC = 100 tan66° = 100 * 2.2460 = 224.60
BC = BD + DC = 153.99 + 224.60 = 378.59
\ The distance between the boys is 378.59 m.
-
a) A + B = é 1 2 ù + é5 6ù
ë 3 4 û ë 7 8û
= é 6 8 ù
ë10 12û
¼ A+B = é 6/4 8/4 ù
ë10/4 12/4 û
= é 3/2 2ù
ë 5/2 3û
b) A² = é4 -2 ù
ë6 -3 û
BC = é2 -6ù
ë-3 3û
\ A² - A + BC = é4 -2ù - é4 -2ù + é2 -6ù
ë6 -3û ë6 -3û ë-3 3û
= é2 -6ù
ë-3 3û
c) Median = (11+1) = 6th observation = 5
2
Median = (9+1) = 5th observation = 25
2
(Arrange in ascending order first)
Mode = 5 Since 5 occurs often.
Mode = 25 Since 25 occurs often.
-
a) y = 2x+4
x -1 0 1 2
y 2 4 6 8
Slope = -OA = 4 = 4
OB 2
Intercepts on the y axis = OA = 4 units
Area = 1 OA * OB = 1 * 2 * 4 = 4 sq.units
2 2
b) Slope of AB = y1-y2 = a + 1 = a + 1 = -(a + 1)
x1- x2 2-3 -1
Slope of BC = -1 + 5 = 4 = -4
3 - 4 -1
Since points are collinear
Slope of AB = slope of BC Þ -(a + 1) = -4
\ a + 1 = 4 \ a = 3
-
|
Age
|
f
|
x
|
f(x)
|
|
10-20
|
10
|
15
|
150
|
|
20-30
|
5
|
25
|
125
|
|
30-40
|
15
|
35
|
525
|
|
Total
|
30
|
|
800
|
x = 800 = 26.66
30
|
Age
|
f
|
x
|
f(x)
|
|
10-20
|
20
|
15
|
300
|
|
20-30
|
10
|
25
|
250
|
|
30-40
|
5
|
35
|
525
|
|
Total
|
45
|
|
1075
|
x = 1075 = 23.88
45
b)
As DADC ~ DBEF
\ AD = DC = AC \ 35 = DC = 42
BE EC BC 30 21 BC
As 35 = DC \ DC = 35 * 21 = 24.5
30 21 30
and 35 = 42 \ BC = 42*30 = 36
30 BC 35
\ BC = 36 and DC = 24.5
-
a) construction
b) construction
-
a) To prove: - arc ABC @ arc ADC
Proof: - seg AB @ seg CD
seg BC @ seg AD (opp. sides of rectangle)
\ arc AB @ arc CD
arc BC @ arc AD (arcs associated with congruent chords)
\ m(arc AB) + m(arc BC) = c(arc CD) + m(arc AD)
\ m(arc ABC = m(arc CDA)
\ arc ABC @ arc ADC
b). PA = PC = PB (radii)
\ DPCA and DPCB are isosceles with
ÐPAC @ ÐPCA and ÐPBC @ ÐPCB
\ mÐPAC + mÐPBC = mÐPCA + mÐPCB
= mÐACB
Since mÐPAC + mÐPBC + mÐACB = 180°
\ in DABC
mÐACB + mÐACB = 180°
\ mÐACB = 90°
\ AB² = AC² + BC² = 24² + 10² = 676
\ AB = 26
\ radius = 26 = 13
2
-
a) Here r = 1.5 m and h = 2 m
Area of the curved surface = 2prh
= 2 * 3.14 * 1.5 * 2
= 18.84 sq.m
Area of base of the barrel = pr²
= 3.14 * (1.5)²
= 7.065 sq.m
Surface area of the barrel without lid
= 18.84 + 7.0665
= 25.905 sq.m
Total surface area of the barrel to be painted
= 2 * 25.905 = 51.810 sq.m
Expenditure of painting = Rate of painting * Area to be painted
= 8 * 51.810
= Rs. 414.48
b) ÐARB is a right angle
\ in DARB
BR² = AB² - AR²
= 61² - 11²
= 3600
\ BR = 60
Since BR = BS = 60
\ perimeter of ARBS = AR + BR + BS + AS
= 11 + 60 + 60 + 11
= 142
