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9th MATHS MODEL TEST PAPER

SECTION A
Question 1
Represent x3<2 on the number line. Ans.  x3<2 Consider  x3=2 if x3 >0, x3=2 => x = 5 If x3<0, (x3) = 2 => x = 23= 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+23 37. Ans. 2. 3 + 23  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
______ _______ Öa2b2 + a + Öa2+b2  b Öa2+b2 + b aÖa2b2
Simplify Ans. ______ ______ Öa2  b2 + a + Öa2 + B2  b = Öa2 + Öb2 + b a  Ö a2  Öb2 _____ _____ _____ _______ = (a+Ö a2b2) (a Ö a2b2) + [Öa2+b2  b] [Ö a2 + b2 + b] (b+ Öa2+Öb2) (a  Öa2  Öb2)
= a2(a2b2) + (a2 + b2b2) = a2  b2________________ [ b+ Öa2 + Öb2 ] [ a Öa2  Öb2] = [b + Öa2 + Öb2] [ a Ö a2Öb2]
Question 5
Find the value of a if x2 is a factor of x53x4ax3+3ax2+2ax+4. Ans. Given x  2 is a factor of x53x4ax3+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 ab, bc and ca are the factors of a (b2c2) + b (c2a2) + c (a2b2).
Ans. If ab 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(b2c2) + 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 (x2) + 1 (x2) = 0 => (x  2) (x+1) = 0 = x = 2 x = 1
Question 9
Solve 2x3 + 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 = 13 =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(1Cos 2x)/2. Ans. x = 30° LHS = Sin 30 = ½ __________ ________ ______ RHS = Ö1Cos2x/2 = Ö1Cos2x (30)/2 = Ö1  Cos 60/2 = Ö1  ½/2 = Ö½ = Ö¼ = ½ Ö2 \LHS = RHS => Sinx = Ö1Cos2x 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 x39x2+27x27. Ans. x3  9x2 + 27x  27 = (x3  27)  9x (x  3) = (x  3) (x2 + 9 + 3x)  9x (x3) = (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 010 1020 2030 3040 4050 5060 6070 7080 8090 90100 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.

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