2018 ___ ___ 1100 - MT - x - MATHEMATICS (71) Algebra - SET - B (E)

MT - x

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

Q.P. SET CODE

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

(i) E = {Asia, Africa, North America, South America, Europe, Antarctica, Australia} 1 (ii) 50 = 25 2× = 25 2×

\ 50 = 5 2 1

(iii) The polynomial (1, 2, 3) contains 3 coefficients. \ The degree of the polynomial is 3 – 1 = 2

\

The index form of the given polynomial is x2+ 2x + 3 1

(iv) 15 : 25 = 15

25

= 15 × 425 × 4

= 60100

\ 15 : 25 = 60% 1

(v) x + y = 7

1

x = 1 and y = 6, x = 7 and y = 0 x = –2 and y = 9, x = –1 and y = 8 x = 0 and y = 7 are the solutions of the given equation.

(vi) Class mark = upper limit + lower limit 2

= 35 402+

set - B2 / MT - x

= 752

= 37.5

\The classmark of the class 35 - 40 is 37.5 1

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

(i) 3 12 18×

= 3 4 3 9 2× × ×

= 3 2 3 3 2× × 1 = 6 3 3 2×

= 18 6×

= 18 6

\ 3 12 18 18 6 1

(ii)

Classes Tally marks Frequency (f) (No. of students)

12 - 13

13 - 14

14 -15

15 - 16

Total N = ∑ f = 35

(iii) (2y2 + 7y + 5) + (3y + 9) + (3y2 – 4y – 3) = 2y2 + 7y + 5 + 3y + 9 + 3y2 – 4y – 3 = 2y2 + 3y2 + 7y + 3y – 4y + 5 + 9 – 3 1 = 5y2 + 6y + 11 ∴ (2y2 + 7y + 5) + (3y + 9) + (3y2 – 4y – 3) = 5y2 + 6y + 11 1

5

14

12

4

1

1

set - B3 / MT - x

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

(i) (C) 16-18 Working : 1

Distance covered per liter ( km )

Frequency( No. of cars )

Cumulative frequency(less than type)

12 - 14 11 1114 - 16 12 2316 - 18 20 4318 - 20 7 50Total N = 50

Here N = 50

∴ N

2 =

502

= 25

Cumulative frequency ( less than type ) Which is just greater than 25 is 43.

∴ Corresponding class 16 - 18 is medium class. (ii) (D) 1st July 2017 1 (iii) (B) 465 1 Working : For an A.P 1, 2, 3, 4,..... a = 1 , d = t2 – t1 = 2 – 1 = 1 n = 30

Sn = 2n

[2a + (n – 1) d]

S30= 302

[2(1) + (30 – 1) (1)]

= 15 [2 + 29]

= 15 [31]

∴ S30= 465

set - B4 / MT - x

(iv) (B) 3 1 Working : 4x + 5y =19 for x = 1 ∴ 4(1) + 5y = 19 ∴ 4 + 5y = 19 ∴ 5y = 19 – 4 ∴ 5y = 15

∴ y = 155

∴ y = 3

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

(i) Let a and b are the roots of the quadratic equation.

Let a = 12

and b = −12

½

\ a + b = 12

+ −

12

= 0 and ½

a × b = 12

× −12

= −14

½

Then the quadratic equation is ½ x2 – (a + b)x + ab = 0

x2 – (0)x + −

14

= 0

\ x2 – 14

= 0

\ 4x2 – 1 = 0 .... [Multplying both sides by 4]

(ii) x + 7y = 10 ...(i) 3x – 2y = 7 ...(ii) Multiply equation (i) by 3 we get, 3x + 21y = 30 ...(iii) ½ Subtracting equation (ii) from (iii), 3x + 21y = 30 3x – 2y = 7 \ 23y = 23 ½ \ y = 1

set - B5 / MT - x

Substituting y = 1 in equation (i), \ x + 7y = 10 \ x + 7 (1) = 10 \ x = 10 – 7 \ x = 3 ½ \ (x , y) = (3, 1) is the solution of the given simultaneous equations. ½

(iii) FV of share = ` 100 Discount = ` 30 \ MV = FV – discount = 100 – 30 ½ = ̀ 70 \ Amount received = No. of shares × MV = 300 × 70 ½ = ` 21,000 \ Total amount obtained by selling 300 shares = 300 ´ 70 = ` 21000 ½

\

The amount obtained by selling 300 shares is ` 21,000.

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

brokerage

Total Value of shares

100 B ` 45 ` 4500 ` 9 ` 0.81 ` 0.81 ` 4510.62

(ii) 2 x2 + 7x + 5 2 = 0

\ 2 x2 + 5x + 2x + 5 2 = 0 ½

\ x ( 2 x + 5 ) + 2 ( 2 x + 5) = 0 ½

\ ( 2 x + 5 ) (x + 2 ) = 0 \ 2 x + 5 = 0 or x + 2 = 0

\ x = −52

or x = – 2 ½

\ −−52and– 2 are roots of the equation. ½

2

set - B6 / MT - x

(iii) Simple interest = P R N´ ´

100 Simple interest after 1 year = 1000 10 1

100´ ´

= 100

Simple interest after 2 year = 1000 10 2100´ ´

= 200 ½

Simple interest after 3 year = 1000 10 3100´ ´

= 300

According to this the simple interest for 4, 5, 6 years will be

400, 500 , 600 respectively. ½

From this d = 100 and a = 100

Amount of simple interest after 20 years. tn = a + (n –1) d \ t20 = 100 + (20 – 1) 100 1 \ t20 = 2000

\

Amount of Simple interest after 20 years is ` 2000

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

(i) 3x2 – 5x + 7 = 0

Comparing with ax2 + bx + c = 0, we get,

a = 3, b = –5, c = 7 1 D=b2 – 4ac = (–5)2 – 4 × 3 × 7 = 25 – 84 \ D = – 59 1 \ As D < 0, the roots of the quadratic equation are not real.

set - B7 / MT - x

(ii) FV = ` 100 and Premium = ` 30 \MV = FV + Premium = 100 + 30 = ` 130 Brokerage per share = 0.3% of ` 130 ½

= 0 3100

130.´

= ` 0.39 \ Purchase price per share = MV + Brokerage ½ = 130 + 0.39 = ` 130.39 \ The purchase price of one share is ` 130.39 1

(iii) The alphabets of English are n(S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z} \ n (S) = 26 ½

Let A be the event that the card is a vowel card. A = {a, e, i, o, u} \ n (A) = 5 ½

P(A) = (A)(S)

nn

= 526

\P(A) = 526

1

A.4. Solve ANY THrEE of the following :

(i) x2 – 4x – 3 = 0 Comparing with ax2 + bx + c = 0, we get, a = 1, b = –4, c = –3. ½ \ b2 – 4ac = (–4)2 – 4 × 1 × –3 = 16 + 12 = 28 ½

\x = − ± −b b aca

2 42

(Formula) 1

\x = − − ±

×( )4 28

2 1

set - B8 / MT - x

\x = 4 4 72

± ×

\x = 4 4 72

± ×

\x = 4 2 72

± ½

\x = 2 2 72

( )±

\x = 2 7±

\x = 2 7+ or x = 2 7− \ The roots of given quadratic equation are 2 7++ and 2 7−− . ½

(ii) Age group (Years)

No. of persons Measure of Central angle

20 - 25

25 - 30

30 - 35

35 - 40

80

60

35

25

80200 × 360 = 144°

60200

× 360 = 108°

35200

× 360 = 63°

25200

× 360 = 45°

Total 200 360°

30 - 35

25 - 30

144°

108°20 - 25

35 - 40

Age group

Age group

Age group

Age group63°

45°

1

2

set - B9 / MT - x

(iii) 6x – 3y = –10 ... (i) Expressing the given equation 3x + 5y – 8 = 0 in the form ax + by = c we get, 3x + 5y = 8 ... (ii)

D = 6 33 5

= (6 ´ 5) – (–3 ´ 3) = 30 – (–9) = 30 + 9 \ D = 39 ½

Dx = 10 3

8 5

= (–10 ´ 5) – (–3 ´ 8) = – 50 – (–24) = – 50 + 24 \ Dx = – 26 ½

Dy = 6 103 8

= (6 ´ 8) – (– 10 ´ 3) = 48 – (– 30) = 48 + 30

\ Dy = 78 ½ By Cramer’s rule

x = D –26 –2D 39 3

= =x and

y = D 78 2D 39

= =y

1

\ x =

23 and y = 2 is the solution of the given simultaneous equations.

½

(iv) Given For an A.P. : t11 = 16 and t21 = 29 To find : t41= ? tn = a + (n – 1) d \ t11 = a + (11 – 1) d

\ 16 = a + 10d

\ a + 10d = 16 ...(i) ½

set - B10 / MT - x

Also t21= a + (21 – 1) d

\ 29 = a + 20d

\ a + 20d = 29 ...(ii) ½ Subtracting (i) from (ii), a + 20d = 29 a + 10d = 16 (–) (–) (–) 10d = 13

\ d = 1310

½

Substituting d = 1310

in equation (i)

a + 10d = 16

\ a + 10 ´ 1310

= 16

\a + 13 = 16 \ a = 16 – 13 ½ \ a = 3 Here, a = 3 ; d = 13

10,

To find : t41 = ?

tn = a + (n – 1) d ½

\ t41 = 3 + (41 – 1) 1310

[

n = 41]

= 3 + 40 ´ 1310

= 3 + (4 ´ 13) = 3 + 52 \ t41 = 55 \ Thus 41st term of the A.P. is 55. ½

set - B11 / MT - x

A.5. Solve ANY oNE of the following :

(i) x + y = 0 i.e. y = – x

x 0 1 2y 0 – 1 – 2

(x, y) (0, 0) (1, –1) (2, –2) when x = 0

\ y = (– 0) \ y = 0

when x = 1 \ y = –(1) \ y = –1

when x = 2

\y = (–2) \ y = –2

2x – y = 9 i.e. y = 2x – 9

x 0 1 2y –9 –7 –5

(x, y) (0, –9) (1, –7) (2, –5)

when x = 0 \ y = 2(0) – 9 \ y = 0 – 9 \ y = – 9 when x = 1 \ y = 2(1) – 9 \ y = 2 – 9 y = –7 when x = 2 \ y = 2(2) – 9 \ y = 4 – 9 \ y = – 5

½

½

set - B12 / MT - x

The lines of the two given simultaneous equations intersect each other at (3 ,–3) \The solution of given simultaneous equations is (3, –3) i.e. x = 3, y = –3 1

(ii) No. of words typed per minute Continuous

ClassesFrequency

(No. of typists)20 - 29 19.5 – 29.5 030 - 39 29.5 – 39.5 240 - 49 39.5 – 49.5 850 - 59 49.5 – 59.5 1560 - 69 59.5 – 69.5 1270 - 79 69.5 - 79.5 380 - 89 79.5 - 89.5 0

Scale1cm = 1 Uniton both axes

Y

Y’

X’ X0–1 1–2–3–4–5 2 3 4 5 6

–2

–1

–3

–4

–5

–6

2

1

3

4

5

6

(0, 0)

(1, –1)

(0, –9)

(2, –5)

(1, –7)

(x + y = 0)

(2x –

y = 9

)(2, –2)

–7

–8

–9

(3, –3)

2

1

set - B13 / MT - x

3

A.6. Solve ANY oNE of the following : (i) In a hockey team, No. of Defenders = 6 No. of Offenders = 4 No. of Goalee = 1 Total no. of players = 11. 1 \ n(S) = 11

No. of words typed per minute

Y

X

ScaleOn X axis : 2 cm = 10 wordsOn Y axis : 1 cm = 1 typist

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5

No.

of t

ypis

ts

Y’

X’

set - B14 / MT - x

(i) Let A be the event that goalee will be selected as a captain. \ n(A) = 1

P(A) =

(A) 1(S) 11

=nn

∴ P(A) = 111

1

(ii) Let B be the event that a defender will be selected. \ n(B) = 6

P(B) = (B)(S)

nn

= 611

∴ P(B) = 611 1

(ii) Rate of GST = 5 % Taxable value ( duting purchase ) = ` 24,500

Input tax = 5 24500100

×

= ` 1225 ½

Taxable value = ` 26,500

Output tax = 5 26500100

×

= ` 1325 ½

∴ GST Payable = Output tax – ITC = 1325 – 1225 ∴ GST Payable = ` 100 1

∴ CGST= UTGST = 12

× 100 = ` 50 ½

∴ GST to be paid by Malik gas agency is ` 100 & amounts of CGST & ½ UTGST are ` 50 each.

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