9th MATHS- MODEL TEST PAPER

SECTION A

Question 1

Represent |x-3|<2 on the number line.
Ans.
| x-3|<2 Consider | x-3|=2
if x-3 >0, x-3=2 => x = 5
If x-3<0, -(x-3) = 2 => -x = 2-3= -1
X = 1
____________________
l l l l l l l l l l l l
-4 -3 -2 -1 0 1 2 3 4 5 6 7

Question 2

Evaluate |3|+|-2-3| -3-|-7|.
Ans. 2.
|3| + |-2-3| - 3 - |-7 |= 3 + 5 - 3 - 7 = -2

Question 3

Express 3/2 Ö(8) as pure surel of order 4.
Ans.
3/2 Ö8 = Ö32X8 = Ö9 X 2 = Ö18 = 18½ = (182) ¼
Ö22

= 4Ö182 = 4Ö324

Question 4

______ _______
Öa2-b2 + a + Öa2+b2 - b
Öa2+b2 + b a-Öa2-b2

Simplify
Ans.
______ ______
Öa2 - b2 + a + Öa2 + B2 - b =
Öa2 + Öb2 + b a - Ö a2 - Öb2
_____ _____ _____ _______
= (a+Ö a2-b2) (a- Ö a2-b2) + [Öa2+b2 - b] [Ö a2 + b2 + b]
(b+ Öa2+Öb2) (a - Öa2 - Öb2)

= a2-(a2-b2) + (a2 + b2-b2) = a2 - b2________________
[ b+ Öa2 + Öb2 ] [ a- Öa2 - Öb2] = [b + Öa2 + Öb2] [ a -Ö a2-Öb2]

Question 5

Find the value of a if x-2 is a factor of x5-3x4-ax3+3ax2+2ax+4.
Ans.
Given x - 2 is a factor of x5-3x4-ax3+3ax3+2ax+4
25 - 3 x 24 - a x 23 + 3a x 23 + 2a x 2 + 4 = 0
32 - 48 - 8a + 12a + 4a + 4 = 0
- 12 + 8a = 0 = 8a = 12 = a = 12/8 = 3/2

Question 6

Resolve into factors (x+2)(x2+25)+10 x2 + 20 x.
Ans.
(x+2 ) (x2 + 25) + 10x2 + 20x
= (x+2) (x2+25) + 10x (x+2)
= (x+2) [x2 + 25 + 10x]
= (x + 2) (x + 5) 2

Question 7

Using factor theorem, show that a-b, b-c and c-a are the factors of a (b2-c2) + b (c2-a2) + c (a2-b2).

Ans.
If a-b is a facts, when we put a = b, expression should become zero. Similarly when we put b=c and c=a, experience should become zero.

1. Let a=b \ a(b2-c2) + b (c2 -a2 ) + c (a2 - b2)
= b(b2 - c2) - B (c2 - b2) + c (b2 - b2)
= b3 - bc2 + bc2 - b3 + 0 = o

2. Let b = c expression becomes
a (c2 - c2) + c (c2 - a2) + c (a2 - c2)
= 0 + c3 - ca2 + ca2 - c3 = 0

Let a = c expression becomes
C (b2 - c2) + b (c2 - c2) + c (c2 - b2)
= c b2 - c3 + 0 + c3 - cb2 = 0
a - b, b - c, c - a are factors of the expression
a (b2 - c2) + b (c2 - a2) + c (a2 - b2)

Question 8

Solve x2+5x+4 = 3
x2+3x+2 2

Ans.
x2 + 5x + 4 = 3
x2 + 3x + 2 = 2

=> 3x2 + 9x + 6 = 2x2 + 10x + 8
=> x2 - x - 2 = 0
=> x2 - 2x + x - 2 = 0
x (x-2) + 1 (x-2) = 0
=> (x - 2) (x+1) = 0 = x = 2
x = -1

Question 9

Solve 2x-3 + x+3 = 4x+1
5 4 7

Ans.
2x - 3 + x + 3 = 4x + 1
5 4 7

8x - 12 + 5x + 15 = 4x + 1
20 7

56 - 84 + 35x + 105 = 80x + 20
57= 45x
\ x = 57 = 19
45 15

Question 10

Simplify [81/16]-3/4 . [25/9]-3/2 ÷ [5/2]-3.

Ans.
(81/16)-3/4 . (25/9)-3/2 ÷ (5/2)-3

= [(3/2)4]-3/4 . [(5/3)2]-3/2 = (3/2)-3 . (5/3)-3
(5/2)-3 (5/2)-3

= 3/2 x 5/3
5/2

= [3/2 x 5/3 x 2/5]-3 = 1-3 =1

Question 11

If log 2 = 0.3010, log 3 = 0.4771 and log 7 = 0.8457, find the value of log 42.
Ans.
Log 2 = 0.3010, Log3 = 0.4771 Log7 = 0.8457
42 = 2 x 3 x 7
log 42 = log (2 x 3 x 7 )
= log 2 + log 3 + log 7
= 0.3010 + 0.4771 + 0.8457 = 1.6238

Question 12

Show that 2 (Cos4 60° + Sin4 30° ) - (tan2 60° + Cot2 45° ) + 3 Sec2 30 = 1/4.
Ans.
2 (Cos4 60° + Sin4 30° ) - (tan2 60° + Cot2 45° ) + 3 Sec2 30 = 1/4
= 2[(1/2)4 + (1/2)4] - [(Ö3)2 + (1)] + 3 x (2/Ö3)2
= 2[1/16 + 1/16] - [3 + 1] + 4
= 2 x 2/16 - 4 + 4 = 1/4 = RHS

Question 13

If x = 30°, Verify that Sin x = Ö(1-Cos 2x)/2.
Ans.
x = 30° LHS = Sin 30 = ½
__________ ________ ______
RHS = Ö1-Cos2x/2 = Ö1-Cos2x (30)/2 = Ö1 - Cos 60/2 = Ö1 - ½/2
= Ö½ = Ö¼ = ½
Ö2
\LHS = RHS => Sinx = Ö1-Cos2x
Ö 2

Question 14

The class marks of a distribution are 47, 52, 57, 62, 72. Determine the class zise, the class limits and the true class limits.
Ans.
Class marks = 47, 52, 57, 62, 67, 72
Class interval = 5
: Class limits = 44.5 - 49.5
49.5 - 54.5
54.5 - 59.5
59.5 - 64.5
64.5 - 69.5
69.5 - 74.5

since the classes are exclusive, so the class limit and true class limit are the some.

Question 15

ABC is a right triangle, right angled at C. If A=30° and AB=40 units, find the remaining two sides and ÐB pf Triangle ABC.

Ans.

Let D ABC lie right led at C
ÐA = 30° and AB = 40 unit
\ ÐB = 180 - (90 + 30 ) = 180 - 120 = 60°

In right D ABC,
Sin 30 = BC ie 1 = BC = BC = 20 Unit
AB 2 40

Cos 30 = AC ie Ö3 = AB \ AB = 20 Ö3 Units
AB 2 40

SECTION B

Question 16

Simplify If 7Ö(3) - 5Ö(2)/Ö(48) + Ö(18) = a+b Ö(6), Find the values of a and b.

Ans.
7Ö3 - 5Ö2 = 7Ö3 - 5Ö2 = (7Ö3 - 5Ö2 ) (4Ö3 - 3Ö2)
Ö48 + Ö18 4Ö3 + 3Ö2 (4Ö3 + 3Ö2) (4Ö3 - 3Ö2)

= 84 - 2Ö16 - 20Ö6 + 30
(4Ö3)2 - (3Ö2)2
= 114 - 41 Ö6 = 114 - 41Ö6 = a + bÖ6
48 - 18 30
\ a = 114, b =- 41
30 30

Question 17

Factorize x3-9x2+27x-27.
Ans.
x3 - 9x2 + 27x - 27
= (x3 - 27) - 9x (x - 3)
= (x - 3) (x2 + 9 + 3x) - 9x (x-3)
= (x - 3) [ x2 + 9 + 3x = 9x]
= (x - 3) [x2 - 6x +9] = (x - 3) (x - 3)2 = (x - 3 )3

Question 18

The length of a rectangle is 4 cm. more than the breadth, and the perimeter is 11 cm more than the breadth. Find the length and breadth of the rectangle.

Ans.

let the bread to = x
length = x + 4
perimeters = 2(l + b)
= 2 [x+4+x] = 2[2x+4]
\By question 2 [2x + 4) = x + 11
4x + 8 = x + 11
3x = 3 => x = 1
\Length = 4 + 1 = 5 cm
bread lb = 1cm

Question 19

A boy is now one - third as old as his father. Twelve years hence he will be half as old as his father. Determine the present age of the boy and that of his father.

Ans.

Let father's present age = x
\Son's present age = x
3
12 years hence Father's age = x + 12

Son's age = x + 12
3
by question , x + 12 = 1 [x + 12]
3 2
x + 36 = 1 (x + 12)
3 2
2x + 72 = 3x + 36
x = 36
\Father's present age = 36
\Son's present age = 36 =12 years
3

Question 20

Evaluate (73.56)3 x (0.0371)2 / 68.21 Using Logarithms.
Ans.
(73. 56)3 x (0.0372)2
68.21
considers long [ (73.56)3 x (0.0371)2]
68.21
= log (78.56)3 + log (0.0371)2 - Log 68.21
= 3 log 78.56 + 2 Log 0.0371 - log 68.21
_
= 3 x 1.8952 + 2 x 2.5694 - 1.8339
_
= 5.6856 + 3.1388 - 1.8339
= 2.8244 - 1.8339
= 0.9905
\(73.56)3 x ( 0.0371)2 = Anti log 0.9905
68.21 = 9.783 x 10o
= 9.783

Question 21

In a rectangle ABCD, AB = 20 cm, Ang BAC = 60 °. Calculate side BC and diagonals.

Ans.

In right D ABC
= BC = tan 60 + Ö3
AB
\BC = AB = Ö3
=20 Ö3
= 20 x 1.73 = 34.6 cm
AB = Cos 60 = 1
AC 2
\Ac = 2AB = 2 x 20 = 40 cm
\AC = BD = 40 cm [ Diagonals are equal in rectangle]
BC = 34.6cm

Question 22

An observer 1m. tall is 2m. away from a pole of height 3m. Find the angle of elevation of the top of the pole as observed by him.
Ans.

Let AB tea the man of height 1m., CD be the pole 3 m
Height Draw AE ^ CD
AE = BC= 2m.
From big weight DE = 3 - 1 = 2 mts. [ EC = 1 Mts).

Let Q be the angle of elevation
\tan Q= DE = 2 = 1
AE 2
= Q = 45°
\ His angle of elevation is 45°

Question 23

A ladder 10 m long reaches a point on a wall which is 10 m. from the top of the wall. The angle of depression of the foot of the ladder as observed from the top of the wall is 60 °. Find the height of the wall.
Ans.

Let ABC be the wall where AB = 10 mts
PB in the ladder 10 m long
DAPC = 60° = ÐXAP (Angle of depression)
\In D ABP, Ð PAB = 90 Ð XAP [Ð XAB = 90°]
= 90 - 60° = 30°
But Ð ABP in Isolation
\ÐAPB = 30°
\ ÐBPC = 60 - 30 = 30°
\ In right Ð BPC, BC = Sin 30 = 1
BP 2
i.e. BC = 1 \ BC = 5mts.
10 2
\ Total height of wall = (10 + 5) = 15 mts.

Question 24

Draw a Histogram and a frequency polygon for the following data.Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No. of Students 5 10 4 6 7 3 2 2 3 9

Ans.

For Histoyam and frequency polygon
X areas - Class limit
Y areas - Frequency
Done at last

Question 25

A pharamaist needs to strengthen a 15 % alcohol solution to one of 32% alcohol. How much pure alcohol should be added to 400 ml of the 15 % of solution.
Ans.
Let there are x, Rs 5 notes in the purse
Then there will be (90 - x) no. of Rs.10 notes are there
\Total value of there notes
= 5x + (90 - x) 10.
By question we here
5x + (90 - x) 10 = 500
5x + 900 - 10x = 500
5x = 400
x = 80
= There are 80 Rs. 5 notes and
(90 - 80) = 10, Rs. 10 notes are there

Question 26

Rationalize the denominator 1
Ö(7)+ Ö(6) - Ö(13).
Ans.

1 = (Ö7 + Ö6) + Ö13
Ö7 + Ö6 - Ö13 [(Ö7 + Ö6) - Ö13][(Ö7 + Ö6)+ Ö13]

= Ö7 + Ö6 + Ö13 = Ö7 + Ö6 + Ö13 = Ö7 + Ö6 + Ö13
(Ö7 + Ö6)2 - Ö13 7 + 6 + 2Ö42 - 13 2Ö42
______ _____ _____
= (Ö7 + Ö6 + Ö13 ) Ö42 = Ö7 x 42 + Ö6 x 42 + Ö13 x 42
2Ö42 x Ö42 2 x 42

= 7Ö6 + 6Ö7 + Ö546
84

Question 27

A man invested Rs 35000/- a part of it at an annual rate of 12 % and the rest at 14% if he received a total annual interest of Rs 4460, how much did he invest at each rate?
Ans.

Let the amount he invested in 12% interest be x.
Rest of the amount (35,000 - x) in 14% rate
Interest earned = 12 x x + 14 (35,000 - x) = 4460
100 100
12x + 490000 - 14x = 446000
2x = 44000
x = 22000 Rs.

\ He invested Rs.22,000/- in 12% interest and
Rs. ( 35,000 - 22,000) = 13,000 Rs. In 14% interest

Question 28

The value of a macine depreciates 10% annually. Its present value is Rs 45450/-. Find is value of 5 years ago.

Ans.

Depreciation rate r = -10%
Present value = 45,450 Rs.
Let its value 5 yrs. ago = P
\ P(1 - r/100)5 = 45,450

P(1 - 10/100)5 = 45,450

P(9/10)5 = 45,450

P = 45,450
.95

Consider logP = log 45,450 - log .95
= 4.6576 - 5 x T.9542 = 4.8866

\ P = Antilog 4.8866
= 7.702 x 104 = Rs. 77,020

Question 29

The angle of depression of the foot of a 9 m. high pole from the top of a building is 60°. and the anlge of depression of the foot of the building at the top of the pole is 30°. Find the height of the building.
Ans.

Let AB be the building, let its height
h mts. PC in the Pole 9m height
Ð ABC = 60o°; ÐPAC = 30°
In right D APC,
AC = Cot 30
PC
AC = Ö3
9
\AC = 9 Ö3 mts.

in right D ABC, h = ten 60 = 3
AC
\ h = AC x Ö3 = 9Ö3 x Ö3 = 27 mts.
\ Height of building = 27 mts.

Question 30

How many kilograms of tea at Rs. 50/- per Kg should be mixed with 35 Kg of tea costing Rs 60/- per Kg So as to Sell the mixture at Rs 57/- per kg without gaining or loosing any thing in the transaction.

Ans.

Let x Kg of Rs. 50 per kg tea should be added with 35 kg of tea costing Rs.60/ kg.
Total weight of tea after mixing = (x + 35) Kg
Selling price of (x + 35) Kg.
= (x = 35) 57 Rs.
But here SP = CP
C.P. of the tea = (50x + 35 x 60) Rs.
\(x+35) 57 = 50x + 2100
57x + 1995 = 50x + 2100
\ 7x = 105 \ x= 15 kg
\ 15 Kg of Rs. 50 per kg should be added.

9th MATHS- MODEL TEST PAPER

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