2018 ___ ___ 1100 - MT - y - MATHEMATICS (71) Algebra - SET - C (E)

MT - y

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40

Q.P. SET CODE

C A.1.(A) Solve ANY Four of the following :

(i) A = { 2 } B = { x | 7x – 1 = 13} 7x – 1 = 13 ∴ 7x = 13 + 1 ∴ 7x = 14

∴ x =

147

∴ x = 2 ∴ B = { 2 } ∴ A = B 1

(ii) |15 – 2| =|13| = 13 \|15 – 2| = 13 1

(iii) Standard form : y5 + 2y4 + 3y3– y2– 7y – 1

(iv) Ratio of ` 700 to ` 308 = 700308

= 28 × 2528 × 11

= 2511

\ Ratio of ` 700 to ` 308 = 25 : 11 1 (v) (i) x + y = 3 (ii) 2x – 7y = 1 (iii) x + 3y = 10 (iv) 3x – y = 5 (v) 4x + 3y = 9 1

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(vi)

Sr. No. Individuals Age Taxable income (`)Will have to pay income tax or

not1. Miss. Mehta 44 ` 5,82,000 YeS

2. Mr. Desilva 81 ` 4,50,000 No

A.1.(B) Solve ANY TWo of the following :

(i) p(x) = x2 – 5x + 5 At x = 0 \ p(0) = (0)2 – 5 × (0)+ 5 1 \ p(0) = 5 \ The value of given polynomial when y = 0 is 5. 1

(ii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 1 \ 29 = 15 + n(B) – 7 \ 29 = 8 + n(B) \ n(B) = 29 – 8 \ n(B) = 21 1

(iii) Ratio of radius of circle to its circumference =

radius Circumference 1

= r 2pr

= 1 2p

= 1 : 2p ∴ Ratio of radius to circumference = 1 : 2p 1

A.2.(A) Choose the correct alternative answer and write.

(i) (A) 7 Dx = 49 , Dy = – 63, D = 7 1

By cramer’s rule,

DD

=xx

=

497

= 7

½

½

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(ii) (B) x2 – 8x + 15 = 0 1 Working : α=3,β=5, x2 – (α+ β)x + αβ= 0 x2 – (3+ 5)x + (3)(5)= 0 x2 – 8x + 15 = 0 (iii) (B) The share is at discount of ` 25 1 Working : Face value = ` 100 Market value = ` 75 MV = FV – Discount ∴75 = 100 – Discount ∴Discount = 100 – 75

∴ Discount = `25 (iv) (B) Median 1

A.2.(B) Solve ANY TWo of the following :

(i) Here t1 = a = 14

,

t2 =

34

, t3 = 54 , t4 =

74

For an A. P.

d = t2 – t1 = 34

14

3 14

24

12

− =−

= = 1

d = t3 – t2 = 54

34

5 34

24

12

− =−

= = 1

\

First term (a) is 14 and Common difference (d) is

12 .

(ii) x + 7y = 10 ...(i) 3x – 2y = 7 ...(ii) Multiplying equation (i) by 3 ½ 3x + 21y = 30 ...(iii) Subtracting equation (ii) from (iii),

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3x + 21y = 30 3x – 2y = 7 (–) (+) (–) 23y = 23

\ y = 2323

\ y = 1 ½ Substituting y = 1 in (iii), \ 3x – 2(1) = 7 \ 3x – 2 = 7 \ 3x = 7 + 2 ∴3x = 9 \ x = 3 ½ \ (x , y) = (3, 1) is the solution of the given simultaneous equations. ½

(iii) 3 x2 + 2 x – 2 3 = 0 Comparing with ax2 + bx + c = 0, we get, a = 3 , b = 2 , c = – 2 3 1 D = b2 – 4ac

= 22( ) – 4 × 3 × (– 2 3 ) ½

= 2 + 24 \ D = 26 As D > 0, the roots of the quadratic equation are real and unequal. ½

A.3.(A) Solve ANY TWo of the following : (i)

No. of shares

MV

of shares

Total value

Brokerage 0.2%

9% CGST on

brokerage

9% SGST on

broker-age

Total Val-ue of

shares

75 S ` 200 ` 15000 ` 30 ` 2.70 ` 2.70 ` 14964.60

2

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(ii)

½

½

1

∴ The sum of natural numbers from 1 to 140 which are divisible by 4 = (iii) g = 500, Assumed Mean (A) = 1250

fiui

No. of workers

(fi)ui =

idg

di = xi – A = xi – 1250

Class Mark

(xi)Fund

(in rupees)

–70–2801520

3528321510

–2–1012

–1000–500

05001000

250750

1250®A17502250

0 - 500500 - 10001000 - 15001500 - 20002000 - 2500

∑fiui = – 63∑fi = 120Total

u =

= −63120

= –0.525

Between 1 to 140. natural number divisible by 4

4, 8, ............, 136

How many numbers? ∴ n = 34

n = 34 , a = 4 , d = 4

tn = a + (n – 1)d

136 = 4 + (n – 1) × 4

n = 34 Sn =

2n

[2a + (n – 1)d]

S34 = 34

2 [2 × 4 + (34 – 1)4 ] = 2380

2380

1

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Mean (x ) = A + u . g

= 1250 + 500 (–0.525)

= 1250 – 262.5

= 987.50

∴ Mean of the Fund collected is ` 987.50. 1

A.3.(B) Solve ANY TWo of the following :

(i) x = 3 is one of the roots of the given quadratic equation. ½ \ It satisfies the given equation. Substituting x = 3 in given quadratic equation. ½ \ k(3)2 – 10(3) + 3 = 0. \ 9k – 30 + 3 = 0. \ 9k – 27 = 0. \ 9k = 27.

\ k = 279

.

\ k = 3. 1

(ii) The A.P. is 9, 4, –1, – 6, –11, ... Here, a = 9, d = 4 – 9 = –5, t27 = ?

tn = a + (n – 1) d 1 \ t27= 9 + (27 – 1) (–5) (

n = 27) = 9 + 26 ´ (– 5) = 9 – 130 \ t27 = –121

Thus 27th term of the A.P. is –121

1

(iii) There are 11 alphabets in the word ‘mathematics \ n (S) = 11 Let A be the event that the letter is m 1 \ n (A) = 2

P(A) = (A)

(S)nn

= 211

\ P(A) = 211 1

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A.4. Solve ANY THree of the following :

(i)

No. of studentsTime (in minutes)

1420 242216

60 - 8080 - 100100 - 120120 - 140140 - 160

Time in minutes

Y

X

ScaleOn X axis : 2 cm = 20 MinutesOn Y saxis : 1 cm = 2 Students

2

4

6

8

10

12

18

20

22

24

60 80 100 120 140 160

No.

of

Stud

ents

Y’

X’

14

16

1

2

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(ii) Here S55 = 3300, t28 = ?

Sn = 2n

[2a + (n –1) d 1

\ s55 = 552

[2a + (55 – 1)d]

\ 3300 = 552

[2a + 54d]

\ 3300 = 552

´ 2 [a + 27d]

\ 3300 = 55 (a + 27d)

\ a + 27d = 330055 1

\ a + 27d = 60 ...(i) Now, tn = a + (n – 1) d \ t28 = a + (28 – 1) d \ t28 = a + 27d \ t28 = 60 ...[from (i)] Thus, the 28th term is 60. 1

(iii)

Supplier : M/s Sahil Enterprises GSTIN: 27LMOAB1397H1Z1 B/24, Paranjape Scheme, Malad (W) Mumbai - 400 069, Maharashtra Receiver : M/s Sandeep Book Agency GSTIN: 27ABCDE7999H1Z7 2, Ground Floor,N. M. Joshi Marg, Lower Parel, Mumbai - 400 004, Maharashtra Invoice No. : GST/19 Invoice Date: 12.01.2018

Sr. No.

HSN Code

Name of

Product

Rate (`)

Quantity Taxable amount

CGST SGST Total (`)Rate Tax Rate Tax

(1) 3924 Pencil box

20 100 ` 2000 6% 120 6% 120 2240

(2) 9503 Jigsaw puzzles

100 50 ` 5000 6% 300 9% 300 5600

Grand Total : ` 7000 420 420 7840

1

1

1

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(iv) Let the numerator of the fraction be x and the denominator be y.

\ The fraction is xy

According to first condition, y = 2x + 4 \ –2x + y = 4 ...(i) ½ According to second condition, (y – 6) = 12 ( x – 6) \ y – 6 = 12x – 72 \ – 6 + 72 = 12x – y \ 66 = 12x – y \ 12x – y = 66 ...(ii) ½ Adding (i) and (ii) we get, – 2x + y = 4 12x – y = 66 10x = 70

\ x = 7010

½

\ x = 7 Substituting x = 7 in equation (i) we get – 2x + y = 4 \ – 2(7) + y = 4 \ – 14 + y = 4 ½ \ y = 4 + 14 \ y = 18 ½

\ xy =

718

\ ½

The required fraction is

718

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A.5. Solve ANY oNe of the following :

(i) Let the number of students standing in each row be x and let number of rows be y. ∴ Total number of students participating in the drill = xy Asperthefirstgivencondition, ½ (x – 3) (y + 10) = xy ∴ x (y + 10) – 3 (y + 10) = xy ∴ xy + 10x – 3y – 30 = xy ½ ∴ 10x – 3y = 30 .......(i) As per the second given condition, ½ (x + 5) (y – 10) = xy ∴ x (y – 10) + 5 (y – 10) = xy ∴ xy – 10x + 5y – 50 = xy ½ ∴ – 10x + 5y = 50 ......(ii) Adding (i) and (ii), we get 10x – 3y = 30 – 10x + 5y = 50

2y = 80 ∴ y = 80

2 ½

∴ y = 40 Substituting y = 40 in (i), we get 10x – 3 (40) = 30 ∴ 10x – 120 = 30 ∴ 10x = 30 + 120 ∴ 10x = 150

∴ x = 15010

∴ x = 15 ½ The total number of students participating in the drill = xy = 15 × 40 = 600 1

∴ 600 students were participating in the drill.

(ii) Here the class given are discontinuous they have to be made continuous. The difference between lower limit of a class and upper limit of previous class is 0.1

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∴1.2 0.052

=

Hence we subtract 0.05 from lower limit of every class and add ½ 0.05 to upper limit of every class. Class width (h) = 1

Haemoglobin % (mg / 100ml) Continuous class No. of persons13.1 - 14 13.05-14.05 8 → f1

14.1 - 15 14.05-15.05 12 → f0

15.1 - 16 15.05-16.05 10 → f2

16.1 - 17 16.05-17.05 617.1 - 18 17.05-18.05 4

Here the maximum frequency fm = 12 The corresponding class 14.05 - 15.05 is the modal class. ½ L = 14.05, f0 = 12, f1 = 8, f2 = 10, h = 1

Mode = L

2 +

0 1

0 1 2

f – fh

f – f – f 1

=

12 – 814.05 + ×1

2(12) – 8 –10

= 414.05 +24 – 18

= 414.05 +6

= 14.05 + 0.67

= 14.72

∴ Mode of haemoglobin is 14.72 mg/100 ml. 1

A.6. Solve ANY oNe of the following : (i) The Sample space for the game of chance of spinning an arrow is S = {1, 2, 3, 4, 5, 6, 7, 8} 1 \ n(S) = 8 (i) Let A be the event that the arrow will rest at 8. \ n(A) = 1

1

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P(A) = (A)(S)

nn =

18

∴ P(A) = 18 ½

(ii) Let B be the event that the arrow will rest at an odd number. B = {1, 3, 5, 7} \ n(B) = 4

P(B) = (B)(S)

nn =

48 =

12

∴ P(B) = 12 ½

(iii) Let C be the event that the arrow will rest at a number greater than 2. C = {3, 4, 5, 6, 7, 8} \ n(C) = 6

P(C) = (C)(S)

nn =

68 =

34

∴ P(C) = 34 ½

(iv) Let D be the event that the arrow will rest at a number less than 9. D = {1, 2, 3, 4, 5, 6, 7, 8} \ n(D) = 8

P(D) =

( )( )DS

nn

= 88

= 1

∴ P(D) = 1 ½

(ii) Let the marks scored by Suyash in first unit test be x. \ The marks scored by him in second unit test is (x + 10) 1 According to given condition, 5 × (x + 10) = x2 ½ \ 5x + 50 = x2

\ x2 – 5x – 50 = 0 \ x2 – 10x + 5x – 50 = 0 ½

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\ x(x – 10) + 5(x – 10) = 0 \ (x – 10) (x + 5) = 0 ½ \ x – 10 = 0 or x + 5 = 0 \ x = 10 or x = –5 x ¹ –5 as marks scored in unit test cannot be negative. \ x = 10 ½ \ The marks scored by Suyash in first test is 10.

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