SolutionsModel Paper - V
Section - A
Section - B
Section - C
SECTION A
18(4x²-25) = 2 * 2 * 3(2x + 5)(2x -5)
12(2x² + 11 x + 15) = 2 * 3 * 3(2x + 5)(x + 3)
GCD = 2 * 3(2x + 5)
= 6(2x + 5)
= 12x + 30
Let present age of 2 boys be x years and y years.
\
x/y = 4/5 \ x = 4y/5 ........... (I)
5 years ago ages of 2 boys will be (x - 5) and (y - 5)
\
(x - 5)/(y - 5) = 7/9
\
9x - 45 = 7y -35
\
9x = 10 + 7y
\
x = (7y + 10)/9 .............. (II)
from I & II
(7y + 10)/9 = 4y/5
35y + 50 = 36y
\
y = 50
x = 4y/5 = 4 * 50/5 = 40
LHS = cot²A - tan²A
= (cosec²A - 1) - (sec²A - 1)
= cosec²A - 1 - sec²A + 1
= cosec²A - sec²A = RHS
(sinA + cosA) / (sinA - cosA)
Dividing by cosA
= (sinA + cosA / cosA)/(sinA - cosA / cosA)
= (sinA/cosA + cosA/cosA)/(sinA/cosA - cosA/cosA)
= (tanA + 1 / tanA - 1)
= (4/3 + 1) / (4/3 - 1)
= 4+3/4-3 = 7
cot20 * cot25 * cot65 * cot70 = 1
LHS = cot20 * cot25 * cot65 * cot70
= cot(90 - 70) * cot(90 - 65) * cot65 * cot70
= tan70 * tan65 * cot65 * cot70
= (1/cot65) * (1/cot70) * cot65 * cot70
= 1
Let the list price be Rs.x
Sales tax = 8x/100
As per condition
x + 8x/100 = 22680
108x = 2268000
\
x = 21000
Rs. 21000
m
Ð
AOB = 120°
2m
Ð
ASB = m(arc AB)
2(m
Ð
ASB) = 120° [ m(arc AB) = m
Ð
AOB ]
m
Ð
ASB = 60°
ASBD is a quadirlateral
Ð
ADB +
Ð
ASB = 180°
Ð
ADB = 120°
AD² = AB² - BD²
AD² = 25 - 16 = 9
\
AD = 3
AD * DE = BD * DC
3 * DE = 4 * 9
DE = 36/3 = 12cm
3,5,7,8,12,14,15,17
Median = (3 + 5 + 7 + 8 + 12 + 14 + 15 + 17)/ 8
= 80/8 = 10
D
AOQ ~
D
BOP
\
A(
D
AOQ)/A(
D
BOP) = (OQ)²/(OP)²
\
A(
D
AOQ)/150 = 49/25
\
A(
D
AOQ) = 49 * 150/25 = 294 sq.cm.
Lengt
h of largest
rod = length of diagonal
=
Ö
l² + b² + h²
=
Ö
122+82+92
=
Ö
144+64+81
= 17 cm²
Area of rhombus = ½ * d
1
* d
2
72 = ½*18* d
2
d
2
= 144/18 = 8 cm
(a) 62 + 82 = 10²
36 + 64 = 100
100 = 100
Square of hypotenuse is sum of square of other 2 sides.
\
by converse of pythagoras theorem It is right angled D
(b) 122 + 132 = 15²
144 + 169 = 225
313
¹
225
\
It is not a right angled D
x² - x = k(3x - 5)
x² - x = 3kx - 5k
x² - x(1 + 3k) + 5k = 0
1 + 3k = 0
3k = -1
k = -1/3
a/b = c/d
2a/3b = 2c/3d
Applying componenda and divedendo
(2a + 3b)/(2a - 3b) = (2c + 3d)/(2c - 3d)
(2a + 3b)(2c - 3d) = (2a - 3b)(2c + 3d)
Section B
Since A and B are the roots of the equation ax² + bx + c = 0,
\
A + B = -b/a and AB = c/a
(i) (A+B)³ = A³ + B³ + 3AB(A + B) ................. (I)
A³ + B³ = (A + B)³ - 3AB(A + B)
= (-b/a)³ - 3c/a(-b/a)
= -b³/a³ + 3bc/a²
= (-b³ + 3abc)/a³ = (3abc - b³)/a³
(ii) A³/B +B²/A = (A³ + B³)/AB
= (3abc - b³/a³)/c/a (From I)
= (3abc - b³)/a³ * a/c = 3abc - b³/a²c
2b²c² + 2c²a² + 2a²b² - a4 - b4 - c4
= -(a
4
+ b
4
+ c
4
- 2b²c² - 2c²a² - 2a²b²)
= -[(a4 + b4 + c4 - 2b²c² - 2c²a² + 2a²b²) - 4a²b²]
= -[(a² + b²- c²)² - (2ab)²]
= -[(a² + b² - c²) + 2ab][(a² + b²- c²) - 2ab]
®
[x² - y² = (x + y)(x - y)]
= -[(a² + b² + 2ab) - c²][(a² + b² - 2ab) - c²]
= -[(a + b)² - c²][(a - b)² - c²]
= -[(a + b + c)(a + b - c)][(a - b + c)(a - b - c)]
= (a + b + c)(b + c - a)(a - b + c)(a + b - c)
3x + 4y = 12 8x + 5y = 40
x
0
3
x
0
8
y
4
0
y
5
0
(x,y)
(0,4)
(3,0)
(x,y)
(0,5)
(8,0)
x = (a cosA + b sinA)
x² = (a cosA + b sinA)²
= a² cos²A + 2ab cosAsinA + b² sin²A .............. (I)
y = (a sinA - b cosA)
y² = (a sinA - b cosA)²
= a² sin²A - 2ab sinAcosA + b² cos²A .............. (II)
Adding (I) & (II)
x² + y² = a²(cos²A + sin²A) + b²(sin²A+cos²A)
= a² * 1 + b² * 1
= a² + b²
Let A,B,C,D be the centres of the equal circles.
Given Each side of the square = 28 cm
Now area of the shaded region :
= [ area of the square ] - 4 * [area of quadrant] ............... (I)
But area of square ABCD = 28² = 784
and 4(area of the quadrant) = area of the circle whose radius is 14 cm.
= pr² = 22/7 * 14 * 14 = 616 sq.cm.
hence from (I) area of shaded region
= 784 - 616
= 168 sq.cm
Let ABC be an isosceles triangle in which AB = AC = 13cm
Let BC = 2x, Draw AD + BC Then BD = x
From
D
ADB, AD² = AB² - BD² = 132 - x²
\
AD =
Ö
13²-x²
Now the area of the
D
ABC = ½ BC * AD
60 = ½ * 2x *
Ö
13² - x² = x
Ö
13² - x²
60² = x²(169 - x²)
3600 = 169 x² - x
4
x
4
- 169x² + 3600 = 0
let x² = u
u² - 169u + 3600 = 0
(u - 25)(u -144) = 0
u = 25 or 144
x² = 25 or x² = 144
\
x = 5 or x = 12
2x = 10 or 24
\
base = 10 cm or 24 cm
Reena wants to pruchase the radio at Rs.2568
Let cost after reduction be Rs.x
\
x + 7x/100 = 2568
107x/100 = 2568
x = 2400
\
Reduction = 2568-2400 = 168
\
Reduction = Rs.168.
Volume of wall = (24 * 6 * 0.4) m³ = 54.6 m³
Volume occupied by mortar = 10/100 * 57.6 = 5.76 m³
Volume occupied by bricks = 57.6 - 5.76 = 51.84 m³
Volume of each brick = (25/100) * (16/100) * (10/100) = 0.004 m³
\
Number of bricks required to brick the wall
= 51.84/0.004 = 12960
cost of 6480 bricks = 350/1000 * 12960
= Rs.4536
To prove : PC = PD
Proof : PC is tangent and PAB is a secant
\
PC² = PA * PB ................. (I)
similarly PD² = PA * PB .......(II)
From (I) & (II)
PC² = PD²
\
PC = PD proved.
(x + 2)(x - 5)(x - 6)(x + 1) = 144
[(x + 2)(x - 6)][(x - 5)(x + 1)] = 144
(x² - 4x - 12)(x² - 4x - 5) = 144
let x² - 4x = y
(y - 12)(y - 5) = 144
y² - 17y - 84 = 0
(y - 21)(y + 4) = 0
y = 21 y = -4
x² - 4x = 21 x² - 4x = -4
x²- 4x - 21 = 0 x² - 4x + 4 = 0
(x -7)(x + 3) = 0 (x - 2)² = 0
x = 7 or x = -3 x = 2
{7,-3,2}
Section C
q is mean proportional between p and r
q² = pr
LHS = pqr(p + q + r)³
RHS = (pq + qr + rp)³
= (pq + qr + q²)³
= [q(p + q + r)]³
= q3 (p + q + r)³
= qpr(p + q + r)³
= pqr(p + q + r)³
LHS = RHS
(i) Area of right DABC
= ½ CB*AC
= ½ ab ....................... (I)
again taking AB as base
Area of DABC = ½ AB * CD
= ½ cp ...................... (II)
From (I) & (II)
½ ab = ½ pc
pc = ab proved ...........(III)
(ii) By Pythagoras theorem.
c² = a² + b² ...................(IV)
c = ab/p ................from (III)
(ab/p)² = a² + b²
a²b²/p² = a² + b²
Divide both side by a²b²
(a²b²/p²) * (a²b²) = (a²/a²b²) + (b²/a² b²)
1/p² = 1/b² + 1/a²
\
1/p² = 1/a² + 1/b² ............. proved
Let AB = 50 m be the tower and PQ be the pole.
From the question,
Ð
BQR = 45° and
Ð
BPA = 60°
Let PQ = h and PA = x = QR
\
BR = BA-RA = 50-h
from
D
BQR, tan45° = BR/QR = (50 - h)/x
\
1 = (50 - h)/x
\
x = 50 - h ...........(1)
Again, from
D
BPA, tan 60° = BA/PA = 50/x
\
Ö
3 = 50/x \ x = 50/
Ö
3 ...........(2)
Hence from (1) & (2) we get
50/
Ö
3 = 50 - h
\
h = 50 - 50/
Ö
3 = 50[1-1/
Ö
3]
\
h = 50 *
Ö
3-1/
Ö
3 = 50
Ö
3(
Ö
3-1)/3
= 50(3 -
Ö
3)/3 = (50 * 1.268)/3
= 1/3 * 63.4 = 21.13
Hence the height of the pole = 21.13 m
Here, quantities consumed are not given but percentage expenditure on various items are given.
Thus these percentages are taken as weightage instead of quantities. Thus :
Items
Weightage(%)q0
Price in 1975 (Base Year)p0
Price in 1980 (Current Year) p1
p0q0
p1q0
Food
40
140
165
5600
6600
Fuel
10
20
23
200
230
Clothing
20
60
70
1200
1400
Rent
20
50
80
1000
1600
Misc.
10
30
35
300
350
Total
8300
10180
cost of living index number -
Sp1q
0
/Sp
0
q
0
* 100
= 10180/8300 * 100
= 122.65
Let L and N be the mid-point of AM and MB respectively.
Let O be centre and r be the radius of the circle which is drawn to touch the three semicircles,
and let P, Q, R be their points of contact. Join OL, OM and ON.
Then the points O, P, L are collinear,
Points R, O, M are collinear,
Points O,Q, N are collinear.
Let AB = x
\
AL =1/4AB = 1/4x
\
PL = x/4
Now OL = OP + PL = x + x/4
and, ON = OQ + QN = r + x/4
\
D
OLN is an isosceles triangle whose mid point is M
\
OM
^
LN
Hence,
D
OML is a right angled
D
.
\
OL² = OM² + LM²
\
[r + x/4]² = (RM - OR)² + [x/4]² = (x/2 - r)² + (x/4)²
\
r² + 2r.x/4 + x²/16 = x²/4 + r² - 2x/2.r² + x²/16
\
3rx/2 = x²/4
\
r = x/6 = 1/6AB