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Sample Paper Mathematics

**Answer 1.**

(i) (c) ₹ 1,050

Let the price of the article be ₹ x.

So, x + GST = 1,239

x + 18% of x = 1,239

x+18x/100=1,239

118x/100=1,239

⇒ x = ₹ 1,050

(ii) (b) –2 √ 6

√2/3 is a solution of the equation

So, 3 ×(√2/3)^{2} + m × √(2/3)+2=0

3 ×2/3 + m × √(2/3)+2=0

m × √(2/3)=-4

m =-4√3/√2=0

m =-2√6

(iii) (c) 1/3

p(x) = x^{3} – ax^{2} + 2x + a – 1
(x – a) is a factor of p(x)
a is the zero of p(x)
p(x = a) = 0
⇒ (a^{3}) – a(a^{2}) + 2a + a – 1 = 0

⇒ a^{3} – a^{3} + 3a – 1 = 0

⇒ 3a – 1 = 0

⇒ a = 1/3

(v) (c) 945

9, 18, 27, .................. .

Clearly, this list forms an A. P.

First term of A.P. (a) = 9

Common difference (d) = 9

We know,

S_{n} = n/2 [2a + (n – 1)d]

S_{n} = 14/2 [2×9 + (14 – 1)9]

= 7[18 + 117]

= 945

Required sum = 945

(vi) (d) (2, – 1)

(vii) (b) 6 cm

In ΔADE and DABC

∠ADE = ∠ABC [Corresponding angles]

∠A = ∠A [Common angle]

∠ADE ~ ∠ABC [AA similarity rule]

AD/AB=AE/AC

⇒ 2/6=AE/9

⇒ AE = 3 cm

CE = AC – AE

= 9 – 3 = 6 cm.

(viii) (b) 2.5

Inequality 7 – 3x **≥ **(-1/2)

⇒ – 3x **≥ **(-1/2) - 7

⇒ – 3x **≥ **(-15/2)

⇒ x ≤ 15/6

⇒ x ≤ 2.5

Greatest value of x = 2.5

(ix) (d) 8

No. of possible outcomes n(S) = n(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)

= 8.

(x) (b) 48π cm^{2}

Let r = 2a and h = 3a
Given Volume = πr^{2}h = 96π cm3
⇒ π × (2a)^{2} × 3a = 96π
⇒ 12a^{3} = 96
⇒ a^{3} = 8
⇒ a = 2
r = 2 × 2 = 4 cm and h = 2 × 3 = 6 cm
Hence, Curved surface area = 2πrh
= 2 ×π × 4 × 6
= 48π cm2.

(xi) (d) 2

and x – 2y = 6

⇒ x – 2(–2) = 6

⇒ x + 4 = 6

⇒ x = 6 – 4 = 2

(xii) (b) y = 2x

Let the equation be y = mx + c

As the slope is 2, y = 2x + c

as it is passing through (1,2), substituting this in the given equation

2 = 2 × 1 + c

⇒ c = 0 Hence, the equation is y = 2x

(xiii) (b) ∠A = ∠B

The angles at the circumference subtended by the same arc are equal.

(xiv) (c) n(n+1)/√2

Given: series is, √2+√8+√18+√32+.........

⇒ √2+2√2+3√2+4√2+......+n√2

⇒√2(1+2+3+4+.....+n)

We know that,

We know that,

(xv) (c) Mode

(iii) (cosec θ – sin θ)(sec θ – cos θ) (tan θ + cot θ) = 1

**Answer 3.**

(i) For cylinder 1 :

Diameter = 24 cm

Radius, R = 24/2 = 12 cm

Height, H = 220 cm

For cylinder 2 :

Radius, r = 8 cm

Height, h = 60 cm

Total volume of iron in two cylinders = πR^{2}H + πr^{2}h

= π(144 × 220 + 64 × 60) cm3

= 3.14 × 35520 = 111532.8 cm3

Total mass of iron = 111532.8 × 8 g

=111532.8 ×8/1000 kg

892.26 kg (approximately). Ans.

(ii) For Point P :

(x_{1}, y_{1}) = A(– 2, 6), (x_{2}, y_{2}) = B(3, – 4) and m_{1} : m_{2} = 2 : 3.

(iii) (i) A (3, 5), B (– 2, – 4) are plotted on graph.

(iv) AA′BB′ is an isosceles trapezium.

(v) Two invariant points under reflection in X-axis are C (3, 0) and D (– 2, 0).

**Answer 4.**

(i) When the product is sold from Agra to Kanpur (intra - state transaction) :

For dealer in Agra :

S.P = ₹ 20,000

CGST = 9% of ₹ 20,000 = ₹ 1,800

SGST = 9% of ₹ 20,000 = ₹1,800

When product is sold from Kanpur to Jaipur (inter- state transaction):

For the dealer in Kanpur :

Input tax credit = ₹1,800 + ₹ 1,800 = ₹ 3,600

C.P = ₹ 20,000 and Profit = ₹ 5,000

S.P = ₹ 20,000 + ₹ 5,000 = ₹ 25,000

IGST = 18% of 25,000 = ₹ 4,500

Net GST paid by the dealer at Kanpur

= Output GST – Input GST

= 4,500 – 3,600 = ₹ 900

The cost of goods/services at Jaipur

= S.P in Kanpur + IGST

= 25,000 + 18% of 25,000

= 25,000 + 4,500 = ₹ 29,500

IGST = 18% of 25,000 = ₹ 4,500

= Output GST – Input GST

= 4,500 – 3,600 = ₹ 900

The cost of goods/services at Jaipur

= S.P in Kanpur + IGST

= 25,000 + 18% of 25,000

= 25,000 + 4,500 = ₹ 29,500

(ii) Given equation is,

(k + 1)x^{2} – 4kx + 9 = 0

Here, a = (k + 1), b = – 4k, c = 9

For real and equal roots,

D = b^{2} – 4ac = 0

⇒ (– 4k)^{2} – 4(k + 1)9 = 0

⇒ 16k^{2} – 36(k + 1) = 0

16k2 – 36k – 36 = 0

Dividing both sides by 4, we get

4k^{2} – 9k – 9 = 0

⇒ 4k^{2} – 12k + 3k – 9 = 0

⇒ 4k(k – 3) + 3(k – 3) = 0

⇒ (k – 3)(4k + 3) = 0

k – 3 = 0 4k + 3 = 0

k = 3 k = − 3/4

For k = 3, the equation will be

4x^{2} – 12x + 9 = 0

⇒ 4x^{2} – 6x – 6x + 9 = 0

⇒ 2x(2x – 3) – 3(2x – 3) = 0

⇒ (2x – 3)^{2} = 0

⇒ x = 3/2, 3/2

For k = – 3/4 , the equation will be

(-3+1/4)x^{2}-4(-3/4)x+9=0

⇒ 1x^{2}/4 + 3x + 9 = 0

⇒ x^{2} + 12x + 36 = 0

⇒ x^{2} + 6x + 6x + 36 = 0

⇒ x(x + 6) + 6(x + 6) = 0

⇒ (x + 6)^{2} = 0

⇒ x = – 6, – 6.

Hence, x = 3 /2 , − 6.

(iii)

Class Interval | Frequency |

0 – 5 | 2 |

5 – 10 | 7 |

10 – 15 | 18 |

20 – 25 | 8 |

25 – 30 | 5 |

15 – 20 | 10 |

∴ Mode = 13 (using histogram).

**Answer 4.**

(ii) Given, ∠SRT = 65° and SP is tangent

∠TSR = 90° [Angle between the radius and tangent]

∴ ∠SRT + ∠STR = ∠TSR = 90°

⇒ ∠STR = ∠TSR – ∠SRT

∠STR = x° = 90° – 65° = 25°

∠y° = 2∠x°

[Angle subtended at the centre is double that of the angle subtended by the arc at same circle]

∴ ∠y° = 2 × 25° = 50°

and ∠SPO = ∠OSP – ∠SOP

∴ ∠SPO = ∠z = 90° – 50° = 40°

Hence, ∠x = 25°, ∠y = 50° and ∠z = 40°. Ans.

(iii) Given expression is 2x^{3} – x^{2} – px – 2 and x – 2 is the factor.

(a) Put x – 2 = 0 ⇒ x = 2 in expression, we get

2(2)^{3} – (2)^{2} – p(2) – 2 = 0

⇒ 16 – 4 – 2p – 2 = 0

⇒ 10 – 2p = 0

⇒ p = 5 Ans.

(b) Putting the value of p, the expression becomes 2x^{3} – x^{2} – 5x – 2.

Dividing the expression by (x – 2),

∴ 2x^{3} – x^{2} – 5x – 2 = (x – 2)(2x^{2} + 3x + 1)

= (x – 2) {2x2 + (2 + 1) x + 1}

= (x – 2) (2x^{2} + 2x + x + 1)

= (x – 2) {2x (x + 1) +1 (x + 1)}

= (x – 2) (2x + 1) (x + 1) Ans.

**Answer 6.**

(i) Given, A(7, – 3) and B(1, 9).

(a) Slope of AB = y2-y1/x2-x1

= 9-(-3) /1-7

=12/-6

=-2

(b) Slope of perpendicular bisector of AB (m) = -1/(-2)=1/2

Mid-point of AB =((x1+x2)/2,(y1+y2)/2)

=((7+1)2,(-3+9)/2)=(4,3)

Let the equation of required line be

y – y1 = m(x – x1)

⇒ y – 3 = 1/2 (x – 4)

⇒ 2y – 6 = x – 4

⇒ x – 2y – 4 + 6 = 0

⇒ x – 2y + 2 = 0, which is the required line. Ans.

(c) Since (– 2, p) lies on x – 2y + 2 = 0

Putting x = 2, y = p, we get

– 2 – 2p + 2 = 0

⇒ 2p = 0

⇒ p = 0 Ans.

(iii) Let the first term of A.P. be a and the common difference be d.

Given,

**Answer 7.**

(i) After removing king, queen and jack of clubs, 49 cards are left in the deck.

∴ Total number of possible outcomes = 49

(a) There are 3 kings left in the deck.

∴ Number of favourable outcomes = 3

∴ P (a king) = 3/49. Ans.

(b) After removing king, queen and jack of clubs, number of clubs cards left = 13 – 3 = 10.

∴ No. of favourable outcomes = 10

∴ P (a club) = 10/49. Ans.

(c) There is only one ‘10’ of hearts

∴ No. of favourable outcomes = 1

∴ P (a ‘10’ of hearts) = 1/49. Ans.

(ii) According to the question,

2/3 (Volume of hemisphere) = Volume of cone

2/3(2 πr^{3}/3) = -1πr^{2}h/3

4/9 (3.5)^{3}=1/3(3.5)^{2}.h

h =4 × 3.5 × 3.5 × 3.5 × 3/3.5 × 3.5 × 9

42.0/9

14m/3 = 4.67 m

Slant height of cone, l = √r^{2}+h^{2}

=√(3.5)^{2}+(4.67)^{2}

35m/6

Now, Surface area of buoy = Surface area of right cone + Surface area of hemisphere

= πrl + 2πr^{2}

= pr(l + 2r)

22/7 × 3.5 ((35/6)+2 × 3.5)

= 11 × (5.83 + 7)

= 11 × 12.83

= 141.13 m2 Ans.

(iii) Given, AB is a diameter.

and ∠BAC = 40°.

In ΔABC

∴ ∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles in a D is 180°]

⇒ ∠ABC + 90° + 40° = 180°

∠ABC = 180° – 130° = 50°

Now, ∠ACT = ∠ABC [Angles in alternate segment]

⇒ ∠ACT = 50°. Ans.

**Answer 8.**

(i)

Marks : Less than | Classes | Mid Values (x) | No. of students (c.f.) | Frequency (f) | fx |

10 | 0 – 10 | 5 | 2 | 2 | 10 |

20 | 10 – 20 | 15 | 7 | 5 | 75 |

30 | 20 – 30 | 25 | 15 | 8 | 200 |

40 | 30 – 40 | 35 | 18 | 3 | 105 |

50 | 40 – 50 | 45 | 20 | 2 | 90 |

Σf = 20 | Σfx = 480 |

∴ Arithmetic mean = Σfx/Σf=480/20=24

(iii) Given, ∠ABC = ∠DAC = x (say)

AB = 8 cm, AC = 4 cm, AD = 5 cm.

(a) In ΔACD andΔBCA

∠ABC = ∠DAC [Given]

∠ACD = ∠BCA [Common]

∴ ΔACD ~ ΔBCA [AA axiom]

Hence ΔACD is similar to ΔBCA. Hence Proved.

(b) ΔACD ~ ΔBCA

∴ AC/BC=CD/CA=AD/BA [Corresponding parts of similar triangles]

⇒ 4/BC=CD/4=5/8

⇒ 4/BC = 5/8

⇒ BC = 8 x 4/5=32/5=6.4 CM

and CD/4 = 5/8

⇒ CD = 5 x 4 /8

⇒ CD = 2.5 cm. Ans.

**Answer 9.**

(i) The original cost of each book = ₹ x.

Total cost = ₹ 960

No. of books purchased = 960/x

If the price was ₹ 8 less, the cost of each book = ₹ (x – 8)water.

No. of books purchased = 960/x -8

According to question,

(960/x-8)-(960/x) = 4

960(1/x-8)-(1/x) = 4

960 x (x-x+8)/x(x-8)=4

⇒ 960 × 8 = 4x (x – 8)

⇒ 4x^{2} – 32x = 7680

⇒ x^{2} – 8x – 1920 = 0

⇒ x^{2} – 48x + 40x – 1920 = 0

⇒ x (x – 48) + 40 (x – 48) = 0

⇒ (x – 48) (x + 40) = 0

⇒ x – 48 = 0 or x + 40 = 0

⇒ x = 48 or x = – 40 (not possible)

x = 48

The original cost of each book = ₹ 48.

(ii)

Marks | No. of Students | c.f. |

0 – 10 | 5 | 5 |

10 – 20 | 11 | 16 |

20 – 30 | 10 | 26 |

30 – 40 | 20 | 46 |

50 – 60 | 37 | 111 |

60 – 70 | 40 | 151 |

70 – 80 | 29 | 180 |

80 – 90 | 14 | 194 |

90 – 100 | 6 | 200 |

∴ N = 200 [even]

(a) Median = Nth/2 value

= 100th value

= 57 marks (from ogive) Ans.

(b) No. of students scoring less than 40 marks = 45.

**Answer 10.**

(i) We have,

(x^{4}+1)/2x^{2}=17/8

(x^{4}+1+2x^{2})/(x^{4}+1-2x^{2})=17+8/17+8

⇒( (x^{2})^{2}+2.x^{2}.1+1)/( (x^{2})^{2}-2.x^{2}.1+1)=25/9

⇒ (x^{2} +1)^{2}/(x^{2} -1)^{2} =25/9

⇒ (x^{2}+1)/(x^{2} - 1) =5/3

⇒ 5(x^{2} – 1) = 3(x^{2} + 1)

⇒ 5x^{2} – 5 = 3x^{2} + 3

⇒ 5x^{2} – 3x^{2} = 3 + 5

⇒ 2x^{2} = 8

⇒ x^{2} = 4

⇒ x = ± 2. Ans.

(ii) Steps of construction :

(a) Draw DABC with each side 5 cm.

(b) Bisect ∠ABC and ∠ACB.

(c) Let the bisectors meet at O.

(d) Draw a perpendicular OD from O to BC.

(e) Taking OD as radius, draw the required incircle.

In DADB, tan 60° = AD/BD

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