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## CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 1 with Solutions

Time : 2 Hr.

Max. Marks : 40

General Instructions :

- The question paper consists of 14 questions divided into three Sections A, B, C.
- All questions are compulsory.
- Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
- Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.

Section – A

Question 1.

Find the value of a_{25} – a_{15} for the AP : 6, 9,12,15, ………………….. (2)

OR

If 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term.

Solution:

Here a = 6, d = 3

∴ a_{25} = a + (25 – 1 )d [∵ a _{n} = a + (n – 1)d]

⇒ a_{25} = 6 + 24 × 3 = 78

and a_{15} = a + (15 – 1 )d [∵ a _{n} = a + (n – 1)d]

⇒ a_{15} = 6 + 14 × 3 = 48

Hence a_{25} – a_{15} = 78 – 48 = 30.

OR

5 × a_{5} = 7 × a_{7}

5 (a + 4 d) = 7(a + 6 d) [∵ a _{n} = a + (n – 1)d]

⇒ 5 a + 20 d = 7 a + 42 d

⇒ 0 = 2 a + 22 d

⇒ 0 =2(a + 11 d)

⇒ a + 11 d = 0

⇒ a_{12} = 0

Question 2.

Find the value of m so that the quadratic equation mx(5x – 6) = – 9 has two equal roots. (2)

Solution:

mx(5x – 6) = – 9

⇒ 5 mx^{2} – 6 mx + 9=0

∴ Standard quadratic equation is given by

ax^{2} + bx + c =0

We get, a = 5m, b = – 6m, c = 9

For equal roots, D =0

⇒ b^{2} – 4ac = 0

⇒ (-6m)^{2} -4(5m) (9) =0

⇒ 36m^{2} – 36m × 5 = 0

⇒ 36m(m – 5) = 0

⇒ m = 0, 5

Rejecting m = 0,

we get m =5.

Question 3.

From a point P, two tangents PA and PB are drawn to a circle C(0, r). If OP = 2r, then find ∠APB. What type of triangle is APB? (2)

Solution:

AP = PB (Tangents from same external point are equal)

\(\begin{aligned}

\angle \mathrm{PAB} &=\angle \mathrm{PBA} \\

∵\mathrm{OA} & \perp \mathrm{PA}

\end{aligned}\) (Radius is always perpendicular to tangent i.e., ∠PAO = 90°)

In ∆ABC \(\frac {OA}{OP}\) = sin θ (sin θ = \(\frac {P}{H}\))

⇒ \(\frac {r}{2r}\) = sin θ

⇒ sin θ = \(\frac {1}{2}\) = sin 30°

⇒ θ = 30°

∵ ∆PAO = ∆PBO

⇒∠APO =∠BPO

∴ ∠APB = 2 x 30° = 60°

∵ ∠APB + ∠PBA + ∠PAB = 180°

⇒ 60° + ∠PBA + ∠PBA = 180°

⇒ ∠PBA = 60°

∴ ∠PAB = ∠PBA = 60°

∴ ∆APB is an equilateral triangle.

Question 4.

The curved surface area of a right circular cone is 12320 cm^{2}. If the radius of its base is 56 cm, then find its height (2)

Solution:

CSA of cone = 12320 cm^{2} , r = 56 cm

∴ πrl = 12320

⇒ \(\frac {22}{7}\) × 56 × l = 12320

⇒ l = \(\frac {12320}{22 × 56}\) × 7

⇒ l = 70

⇒ l^{2} = r^{2} + h^{2}

⇒ 70^{2} = 56^{2} + h^{2}

⇒ 4900 – 3136 = h^{2}

⇒ h^{2} = 1764

⇒ h = \(\sqrt{1764}\)

∴ h = 42 cm

Question 5.

Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below: (2)

Marks Obtained | 0-20 | 20-40 | 40-60 | 60-80 | 80 -100 |

Number of Students | 5 | 10 | – | 6 | 3 |

Solution:

Modal class is 40 – 60, l = 40, h = 20, f_{1} = ?,f_{0} = 10,f_{2} = 6

Mode = l + hx(\(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\))

⇒ 45 = 40 + 20 × (\(\left[\frac{f_{1}-10}{2 f_{1}-10-6}\right]\))

⇒ \(\frac{1}{4}=\frac{f_{1}-10}{2 f_{1}-16}\)

⇒ 2 f_{1} – 16 = 4 ff_{1} – 40

⇒ f_{1} = 12

Question 6.

if Ritu were younger by 5 years than what she really is, then the square of her age would have been 11 more than five times her present age. What is her present age? (2)

OR

Solve for x : 9x^{2} – 6px + (p^{2} – q^{2}) = 0.

Solution:

Let the present age be x years.

According to question,

(x – 5)^{2} =11 + 5x

x^{2}+25 – 10x – 11 – 5x =0

x^{2} – 15x + 14 = 0

x^{2} – (14+1)x + 14 = 0

x(x – 14) – 1(x – 14) = 0

(x – 14) (x – 1) = 0

x = 14 or 1

x = 14 years (rejecting x = 1 as in that case Rit&s age 5 years ago will be negative)

OR

9x^{2} – 6px + (p^{2} – q^{2}) = 0

Here, a = 9, b = -6p, c= p^{2} – q^{2}

D = b^{2} – 4ac

= (- 6p)^{2} 4(9) (p^{2} – q^{2})

= 36p^{2} – 36p^{2} + 36q^{2}

36 q^{2}

x = \(\frac{-b \pm \sqrt{D}}{2 a}=\frac{6 p \pm 6 q}{18}=\frac{p+q}{3}\) or \(\frac {p-q}{3}\)

Section – B

Question 7.

Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median. (3)

Distance (in m) | 0-1 | 1-2 | 2-3 | 3-4 | 4-5 |

Number of Students | 40 | 80 | 62 | 38 | 30 |

Solution:

Distance (in m) | 0-1 | 1-2 | 2-3 | 3-4 | 4-5 |

Number of Students | 40 | 80 | 62 | 38 | 30 |

c.f. | 40 | 120 | 182 | 220 | 250 |

\(\frac{n}{2}=\frac{250}{2}=125\)

Median class is 2 -3, l= 2, h = 1, c.f. = 120,f = 62

Median = \(l+\frac{\frac{n}{2}-c \cdot f}{f} \times h\)

= 2 + \(\frac{5}{62}\)

= \(\frac{129}{62}\)

= \(2 \frac{5}{62}\) m or 2.08 m

50% of students jumped below \(2 \frac{5}{62}\) m and 50% above it.

Question 8.

Construct a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°. (3)

Solution:

Steps of constructions :

Step I : Draw a circle with O as center and radius 4 cm.

Step II : Draw any ∠AOB = 120° (180° – 60°)

Step III : From A and B draw ∠PAO = ∠PBO = 90° which meet at P.

∴ PA and PB are the required tangents.

Hence ∠BPA = 60°, as PBOA is a cyclic quadrilateral.

Question 9.

The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs. (3)

Runs scored | 0-40 | 40-80 | 80 -120 | 120 -160 | 160 – 200 |

Number of batsmen | 12 | 20 | 35 | 30 | 23 |

Solution:

Runs scored | 0-40 | 40-80 | 80 -120 | 120 -160 | 160 – 200 |

Number of batsmen | 12 | 20 | 35 | 30 | 23 |

xi | 20 | 60 | 100 | 140 | 180 |

f_{i}x_{i} |
240 | 1200 | 3500 | 4200 | 4140 |

Mean (x) = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{13280}{120}=110.67\) runs

Question 10.

Two vertical poles of different heights are standing 20 m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole from the foot of the first pole is 30°. Find the difference between the heights of two poles. (Take \(\sqrt{3}\) = 1.73) (3)

OR

A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The aiigle of elevation of the top of the building from his eye is 600. Calculate the height of the building. (Take \(\sqrt{3}\) = 1.73)

Solution:

In ∆ABD,

\(\frac {AB}{BD}\) = tan 60°

⇒ \(\frac{\mathrm{H}}{20}=\frac{\sqrt{3}}{1}\)

⇒ H = 20 \(\sqrt{3}\) m

In ∆BDC,

\(\frac{C D}{B D}=\tan 30^{\circ}\)

⇒ \(\frac{h}{20}=\frac{1}{\sqrt{3}}\)

⇒ \(h=\frac{20}{\sqrt{3}} \mathrm{~m}\)

H – h = \(20 \sqrt{3}-\frac{20}{\sqrt{3}}\)

= \(\frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{40 \sqrt{3}}{3}\)

= \(\frac{40 \times 1.73}{3}\)

= \(\frac {69.3}{3}\)

= 23.06

OR

In ABC

\(\frac{\mathrm{BC}}{\mathrm{AC}}=\tan 60^{\circ}\)

⇒ \(\frac{\mathrm{BC}}{50}=\frac{\sqrt{3}}{1}\)

⇒ BC = \(50\sqrt{3}\)

= 50 x 1.73

= 86.5 m

∴ Height of the building (h) = 86.5 + 1.7

= 88.2 m.

Section – C

Question 11.

The internal and external radii of a spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. Also find the total surface area of the cylinder. (4)

Solution:

Given, r_{1} = 3 cm, R =5 cm and radius of cylinder (r_{2}) = =7 cm

According to the question,

Volume of shell = Volume of cylinder

\(\frac{4}{3} \pi\left(\mathrm{R}^{3}-r_{1}^{3}\right)=\pi r_{2}^{2} h\)

\(\frac{4}{3}\left(5^{3}-3^{3}\right)=(7)^{2} \times h\)

⇒ \(\frac{4}{3}(125-27)=49 \times h\)

⇒ \(\frac{4}{3} \times \frac{98}{49}=h\)

h = \(\frac{8}{3}\) cm

T.S.A. of cylinder = \(2 \pi r_{2} h+2 \pi r_{2}^{2}\)

\(2 \pi r_{2}\left(h+r_{2}\right)\)

\(2 \times \frac{22}{7} \times 7\left(\frac{8}{3}+7\right)\)

\(44 \times \frac{29}{3}\)

\(\frac{1276}{3}\) cm^{2} or 425.33 cm^{2}

Question 12.

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 to the centre. (4)

OR

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

∠ AOB + ∠APB = 180°

Solution:

To prove: ∠AOB + ∠APB = 180°

Proof :

OA ⊥PA [∵ Radius is always perpendicular to tangent]

∠OAP = 90°

and OB ⊥PB

∠OBP = 90°

In quadrilateral OAPB by angle sum property,

∠OAP + ∠APB + ∠OBP + ∠AOB = 3600

⇒ 90°+∠APB+90°+ ∠AOB=360°

⇒ ∠APB + ∠AOB =360°- 180°

⇒ ∠APB + ∠AOB = 180°.

Construction : Join OQ.

OR

TP = TQ [Tangents from same external point are equal]

∠TPQ = ∠TQP……………….(i)

In TPQ, by angle sum property,

∠PTQ + ∠TPQ + ∠TQP = 180°

⇒ ∠FTQ + ∠TPQ + ∠TPQ = 180° [from (i)]

⇒ ∠PTQ + 2∠TPQ = 180°……………….. (ii)

OP ⊥ TP (Radius is always perpendicular to tangent)

∴ ∠OPT = 90°

⇒ ∠OPQ + ∠TPQ = 90°

⇒ ∠TPQ = 90° – ∠OPQ …………………….(iii)

From equations (ii) and (iii),

∠PTQ + 2(90° ∠OPQ) = 180°

⇒ ∠PTQ + 180° – 2∠OPQ = 180°

⇒ ∠PTQ = 2∠OPQ.

Question 13.

Case Study-1:

Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites.

A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes (tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.

(Lighthouse of Mumbai Harbour. Picture credit – Times of India Travel)

(i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower. (2)

(ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(√3 – 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower? (2)

Solution:

∠ACB = 30° (Alternate angle)

(i) In ∆ABC, = tan 30°

\(\frac{\mathrm{AB}}{\mathrm{BC}}=\tan 30^{\circ}\)

⇒ \(\frac{240}{\mathrm{BC}}=\frac{1}{\sqrt{3}}\)

⇒ BC = \(240\sqrt{3}\) m

(ii)

BD = \(240\sqrt{3}- {240\sqrt{3}-1}\) [BD = BC – DC]

= \(240 \sqrt{3}-240 \sqrt{3}+240\)

= 240 m

In ∆ABD

\(\frac{\mathrm{AB}}{\mathrm{BD}}=\tan \theta\)

⇒ \(\frac{240}{240}=\tan \theta\)

⇒ 1 = tan θ

⇒ θ = 45°

Question 14.

Case Study-2:

Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.

Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour. But he wants to achieve a target of 3900 push-ups in 1 hour for which he practices regularly. With each day of practice, he is able to make 5 more push-ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved. Keeping the above situation in mind answer the following questions:

(i) Form an A.P. representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished. (2)

(ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved. (2)

Solution:

3000, 3005, 3010 ………………… 3900

(i) a = 3000

d =5

a_{n}= 3900

a_{n}= a + (n – 1)d

⇒ 3900 = 3000 + (n – 1) x 5

⇒ \(\frac{900}{5} = n – 1\)

⇒ 180 = n – 1

⇒ n =181 days

Hence, minimum number of days of practice = n – 1 = 180 days.

(ii) S_{n} = \(\frac{n}{2}\) (a+l)

S_{181} = \(\frac{181}{2}\) (3000 + 3900)

= \(\frac{181}{2} \times 6900\)

= 181 x 3450

= 624450 push-ups.