**30/ 3/ 1 2020 Class 10 Math Standard Question Paper Solution**

**SECTION – A**

**Question numbers 1 to 10 are multiple choice questions of 1 mark each. You have to select the correct choice :**

**(1) The HCF of 135 and 225 is**

(a) 15

(b) 75

(c) 45

(d) 5

Solution: (c) 45

**(2) The exponent of 2 in the prime factorization of 144, is **

(a) 2

(b) 4

(c) 1

(d) 6

Solution: (b) 4

**(3) The common difference of an AP, whose n ^{th} term is an = (3n + 7), is (a) 3 **

(b) 7

(c) 10

(d) 6

Solution: (a) 3

**(4) The value of λ for which (x ^{2 }+ 4x + λ) is a perfect square, is**

(a) 16

(b) 9

(c) 1

(d) 4

Solution: (d) 4

**(5) The value of k, for which the pair of linear equations kx + y = k ^{2} and x + ky = 1 have infinitely many solutions is **

(a) ±1

(b) 1

(c) –1

(d) 2

Solution: (b) 1

**(6) The value of p for which (2p + 1), 10 and (5p + 5) are three consecutive terms of an AP is**

(a) –1

(b) –2

(c) 1

(d) 2

Solution: (d) 2

**OR**

**The number of terms of an AP 5, 9, 13, … 185 is **

(a) 31

(b) 51

(c) 41

(d) 40

Solution: 1 mark should be given to each candidate.

**(7) In Fig. 1, the graph of the polynomial p(x) is given. The number of zeroes of the polynomial is**

(a) 1

(b) 2

(c) 3

(d) 0

Solution: (b) 2

**(8) If (a, b) is the mid-point of the line segment joining the points A(10, –6) and B(k, 4) and a – 2b = 18, the value of k is**

(a) 30

(b) 22

(c) 4

(d) 40

Solution: (b) 22

**(9) The value of k for which the points A (0, 1), B (2, k) and C(4, –5) are collinear is**

(a) 2

(b) –2

(c) 0

(d) 4

Solution: (b) –2

**(10) If ΔABC ~ Δ DEF such that AB = 1.2 cm and DE = 1.4 cm, the ratio of the areas of ΔABC and ΔDEF is**

(a) 49 : 36

(b) 6 : 7

(c) 7 : 6

(d) 36 : 49

Solution:(d) 36 : 49

**In Q. Nos. 11 to 15, fill in the blanks. Each question is of 1 mark:**

**(11) √2 times the distance between (0, 5) and (–5, 0) is _________. **

Solution: 10

**(12) The distance between two parallel tangents of a circle of radius 4 cm is _________.**

Solution: 8 cm

**(13) In Fig. 2, PA and PB are tangents to the circle with centre O such that ****∠****APB = 50°, th****en the measure of ****∠****OAB is _________.**

Solution: 25°

**OR**

**In Fig. 3, PQ is a chord of a circle and PT is tangent at P such that ****∠****QPT = 60°, then the measure of ****∠****PRQ is _________****.**

Solution: 120°

**(14) 3 cot 40 ^{0} / tan 50^{0} – 1/2 (cos 35^{0}/ sin 55^{0}) = _________.**

Solution: 5/2

**(15) If cot θ = 7/8, then the value of (1 + sin θ) (1 – sin θ)/ (1 + cos θ) (1 – cos θ) = _________.**

Solution: 49/64

**Q.Nos. 16 to 20 are short answer type questions of 1 mark each.**

**(16) What is the value of (1/ 1 + cot2 θ + 1/ 1 + tan ^{2} θ)?**

Solution: Given expression = sin^{2} θ + cos^{2} θ

= 1

**(17) Two right circular cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1, what is the ratio of their volumes?**

Solution: V_{1} : V_{2 }= 1/3 π (3r)^{2} h: 1/3 π r^{2} (3h)

= 3 : 1

**(18) Using the empirical formula, find the mode of a distribution whose mean is 8.32 and the median is 8.05.**

Solution: Mode= = 3 x 8.05 – 2 x 8.32

= 7. 51

**(19) The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?**

Solution: Prob ( no rain tomorrow) = 1 – 0.85

= 0.15

**(20) What is the arithmetic mean of first n natural numbers?**

Solution: Sum of first n natural numbers = n(n + 1)/2

∴ Mean = n + 1/ 2

**SECTION – B**

**Q. Nos. 21 to 26 carry 2 marks each.**

**(21) Find the 11 ^{th} term from the last term (towards the first term) of the AP 12, 8, 4, …, –84.**

Solution: l = –84

d = –4

t_{11} (from the end) = –84 + 40 = – 44

**OR**

**Solve the equation : 1 + 5 + 9 + 13 + … + x = 1326**

Solution: n/2 (1 + x) = 1326 …. (i)

X = 1 + (n – 1) x 4

Solving (i) and (ii) x = 101

**(22) In Fig. 4 AB is a chord of circle with centre O, AOC is diameter and AT is tangent at A. Prove that ****∠****BAT = ****∠****ACB****.**

Solution: ∠BAC = 90° – ∠BAT … (i)

In ΔBAC, ∠B = 90°

∴ ∠BCA = 90° – ∠BAC

or ∠ACB = ∠BAT (Using (i))

**(23) If tan θ = 3/4, find the value of ( 1 – cos ^{2 } θ/ 1 + cos^{2} θ)**

Solution: sec^{2}θ = 1 + 9/16 = 25/16

∴ cos^{2}θ = 16/25

Hence 1 – cos^{2} θ / 1 + cos^{2} θ = 1- 16/25/1+ 16/25 = 9/41

**OR**

**If tan θ = √3, find the value of ( 2sec θ / 1 + tan ^{2} θ)**

Solution: sec2 θ = 1 + 3 = 4

∴ sec θ = 2

Hence 2 sec θ / 1 + tan^{2} θ = 2 x 2 /4 = 1

**(24) ****Read the following passage and answer the questions given at the end : Students of Class XII presented a gift to their school in the form of an electric lamp in the shape of a glass hemispherical base surmounted by a metallic cylindrical top of same radius 21 cm and height 3.5 cm. The top was silver coated and the glass surface was painted red.**

**(i) What is the cost of silver coating the top at the rate of ₹ 5 per 100 cm2?**

**(ii) What is the surface area of glass to be painted red?**

Solution: (i) Surface Area of the top = 2 x 22/7 x 21 x 3.5 = 462 cm^{2}

Cost of silver coating = 462 x 5/100 = Rs. 23.10

(ii) Surface Area of glass = 2 x 22/7 x 21 x 21

= 2772 cm^{2}

**(25) Find the probability that a leap year selected at random will contain 53 Sundays and 53 Mondays.**

Solution: 366 days = 52 weeks + 2 days

Total possible outcomes are 7 (SM, MT, TW, WTh, ThF, FS, SS)

Prob (having 53 Sundays & 53 Mondays) = 1/7

**(26) Find the value of p, if the mean of the following distribution is 7.5.**

Classes |
2 – 4 |
4 – 6 |
6 – 8 |
8 – 10 |
10 – 12 |
12 -14 |

Frequency (fi) |
6 |
8 |
15 |
p |
8 |
4 |

Solution:

Class |
Frequency (f) |
x |
fx |

2 – 4
4 – 6
6 – 8
8 – 10
10 – 12
12 – 14 |
6
8
15
P
8
4 |
3
5
7 9 11 13 |
18
40
105 9p 88 52 |

41 + p | 303 + 9p |

Mean= 7.5 = 303 + 9p/ 41 + p ⇒ p = 3

**SECTION – C**

**Q.Nos. 27 to 34 carry 3 marks each.**

**(27) Find a, b and c if it is given that the numbers a, 7, b, 23, c are in AP.**

Solution: a, 7, b, 23, c are in A.P

Let d be the common difference of AP.

∴ a + d = 7 … (i)

a + 3d = 23 … (ii)

Solving (i) & (ii) , d = 8

∴ a = –1, b = 15, c = 31

**OR**

**If m times the mth term of an AP is equal to n times its nth term, show that the (m + n)th term of the AP is zero.**

Solution: Given m[a + (m – 1) d] = n[a + (n – 1) d]

⇒ a(m – n) + d(m^{2} – m – n^{2 }+ n) = 0

⇒ (m – n) [a + (m + n – 1) d = 0

∵ m ≠ n ⇒ a + (m + n – 1) d = 0

⇒ a_{m }+ _{n} = 0

**(28) Find the values of k, for which the quadratic equation**

**(k + 4) x ^{2} + (k + 1) x + 1 = 0 has equal roots.**

Solution: For equal roots (k + 1)^{2} – 4(k + 4) × 1 = 0

⇒ k^{2} – 2k – 15 = 0

⇒ (k + 3) (k – 5) = 0

⇒ k = – 3, 5

**(29) On dividing x ^{3} –3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4 respectively. Find g(x).**

Solution: x^{3} – 3x^{2} + x + 2 = (x – 2) × g(x) + (–2x + 4)

⇒ (x – 2) g(x) = x3 – 3×2 + 3x – 2

⇒ g(x) = (x – 2) (x^{2} – x + 1)/ ( x – 2)

= x^{2} – x + 1

**OR**

**If the sum of the squares of zeros of the quadratic polynomial f(x) = x ^{2} – 8x + k is 40, find the value of k.**

Solution: Let the zeroes of polynomial f(x) be α and β

∴ α + β = 8 and αβ = k

∵ α^{2} + β^{2} = 40

⇒ (α + β) ^{2 }– 2αβ = 40

⇒ 64 – 2k = 40

⇒ k = 12

**(30) In what ratio does the point P(–4, y) divide the line segment joining the points A(–6, 10) and B(3, –8) if it lies on AB. Hence find the value of y.**

Solution:

Let AP : PB = k : 1

∴ -4 = 3k- 6/ k + 1

⇒ K = 2/7

Hence y = -8k + 10/ k + 1 = -8 x 2/7 + 10/ 2/7 + 1 = 6

**(31) ****Prove that, a tangent to a circle is perpendicular to the radius through the point of contact.**

Solution: Given, To prove, figure

Correct proof

**OR**

**Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.**

Solution:

∠PAO = 90° (radius ⊥ tangent)

∠PBO = 90°

Now

∠PAO + ∠AOB + ∠OBP + ∠BPA = 360°

⇒ 90° + ∠AOB + 90° + ∠BPA = 360°

⇒ ∠AOB + ∠BPA = 180°

or ∠AOB and ∠BPA are supplementary.

**(32) In a right triangle, prove that the square of the hypotenuse is equal to the sum of squares of the other two sides.**

Solution: Correct given, To prove & figure

Correct proof

**(33) If sin θ + cos θ = p and sec θ + cosec θ = q, show that q (p2 – 1) = 2p.**

Solution: LHS = q(p^{2 }– 1) = (sec θ + cosec θ) ((sin θ + cos θ) ^{2} – 1)

= sin θ + cos θ/ sin θ cos θ x 2 sin θ cos θ

= 2 (sin θ + cos θ)

= 2p = RHS

**(34) 500 persons are taking dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m ^{3}?**

Solution: Let the rise in the water level be h

∴ 500 × .04 = 80 × 50 × h

⇒ h = 500 x .04/ 80 x 50

= .005 m

**SECTION – D**

**Q.Nos. 35 to 40 carry 4 marks each.**

**(35) Show that (12) ^{n} cannot end with digit 0 or 5 for any natural number n.**

Solution: 12^{n }= (2^{2} × 3)^{n }= 2^{2n} × 3^{n}

Since there is no factor of the form 5^{m} therefore 12^{n} can not

end with digit 0 or 5 for any natural number n.

**OR**

**Prove that (√ 2+ √ 5) is irrational.**

Solution: Let us assume √2 +√5 is rational number

Let √2 + √5 = m where m is rational

⇒ (√2 + √5)^{2} = m^{2}

⇒ m^{2 }= 7 + 2 √10

⇒ √10 = m^{2 }– 7

∵ m is rational

∴ m^{2} – 7/2 is also rational

but √10 is rational

⇒ LHS ≠ RHS

It means our assumption was wrong.

Hence √2 + √5 is an irrational number.

**(36) A train covered a certain distance at a uniform speed. If the train would have been 6 km/hr. faster, it would have taken 4 hours less than the scheduled time and if the train were slower by 6 km/hr., it would have taken 6 hrs. more than the scheduled time. Find the length of the journey.**

Solution: Let usual speed of train be x km/hr and distance covered be d km.

Therefore d/x – d/x+6 = 4 ……. (i)

d/x – 6 – d/x = 6 ……. (ii)

Solving (i) and (ii) x = 30 and d = 720

∴ Length of journey = 720 km

**(37) In an equilateral triangle ABC, D is a point on the side BC such that**

**BD = 1/ 3 BC. Prove that 9 AD ^{2} = 7 AB^{2}.**

Solution:

Draw AE ⊥ BC

∵ ΔABC is an equilateral Δ

∴ BE = BC/2

Now, AD^{2} = AE^{2} + DE^{2} and AB^{2} = AE^{2} + BE^{2}

⇒ AB^{2} = AD^{2} – DE^{2} + BE^{2}

= AD^{2} + (BE + DE) (BE – DE)

= AD^{2} + BC/3 X (BC/2 + BC/2 – BC/3)

= AD^{2} + 2/9 BC^{2 }= AD2 + 2/9 AB^{2}

⇒ 7AB^{2 }= 9AD^{2}

**OR**

**Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.**

Solution:

AB^{2} + BC^{2} + CD^{2} + AD^{2}

= 4 AB^{2} (∵ABCD is a rhombus)

= 4 (OA^{2} + OB^{2 })

=4 (AC^{2}/4 + BD^{2}/4)

=AC^{2} +BD^{2}

**(38) If the angle of elevation of a cloud from a point 10 metres above a lake is 30° and the angle of depression of its reflection in the lake is 60°, find the height of the cloud from the surface of lake.**

Solution:

Let C represents the position of cloud and C` represents its reflection in the lake.

tan 30° = 1/√3 = h/x

⇒ x = h √3 ……… (i)

tan 60° =√3 = h + 20/x ….. (ii)

Solving (i) and (ii) h =10

∴ Height of cloud from surface of the lake = 20 m

**OR**

**A vertical tower of height 20 m stands on a horizontal plane and is surmounted by a vertical flag-staff of height h. At a point on the plane, the angle of elevation of the bottom and top of the flag staff are 45° and 60° respectively. Find the value of h.**

Solution:

Let AC be the tower and CD be the flag-staff.

tan 45° = 1 = AC/AB

⇒ AC = AB ….. (i)

tan 60° = √3 = AC + h/AB

⇒ √3 AB – AC + h ….. (ii)

Using (i) and (ii)

AC (√3 – 1) = h

⇒ h = (√ 3 – 1) m

**(39) A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.**

Solution: Internal radius of cylinder (r_{2 }) = 30 cm = 0.30 m

Outer radius of cylinder (r_{1} ) = 30 + 5 = 35 cm = 0.35 m

Therefore 4.4 × 2.6 × 1 = π × h × ( (0.35)^{2} – (.30)^{2})

= π × h × 1/100 x 100 x 65 x 5

⇒ h = 352/ π m or 112 m

**(40) For the following frequency distribution, draw a cumulative frequency curve of ‘more than’ type and hence obtain the median value.**

Classes | 0 – 10 | 10 – 20 | 20 – 30 | 30 -40 | 40 – 50 | 50 – 60 | 60 – 70 |

Frequency | 5 | 15 | 20 | 23 | 17 | 11 | 9 |

** **Solution: Plotting points (0, 100) (10, 95) (20, 80) (30, 60) (40, 37) (50, 20) (60, 9) and joining them.

Median = 34.3 (approx)

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