Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 10 Maths BasicTerm 2 Set 4 with Solutions

Time allowed: 2 hours

Maximum Marks: 40

General Instructions:

- The question paper consists of 14 questions divided into 3 sections A, B, C.
- 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 questions.
- Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one questions. It contains two case study based questions.

Section – A (12 Marks)

Question 1.

If for the quadratic equation 2x^{2} + kx – 6 = 0, 2 is one of its root, then find the value of k. Also, find the value of the other root.

OR

The sum of squares of three consecutive positive integers are 50. Then what are the three integers? (2)

Answer:

Since, x = 2 is a root of the equation. 2x^{2} + kx – 6 = 0

∴ 2 × 2^{2} + k × 2 – 6 = 0

⇒ 8 + 2k – 6 = 0

⇒ k = – 1

Put k = – 1 in the equation.

2x^{2} + kx – 6 = 0

We get

2x^{2} – x – 6 = 0

⇒ 2x^{2} – 4x + 3x – 6 = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ x = 2, \(\frac{-3}{2}\)

Caution:

First, find the value of k, to formulate the equation. Then, find the roots of the quadratic equation

OR

Consider three consecutive integer as x, x + 1 and x + 2.

A.T.Q x^{2} + (x + 1)^{2} + (x + 2)^{2} = 50

⇒ x^{2} + x^{2} + 1 + 2x + x^{2} + 4 + 4x = 50

⇒ 3x^{2} + 6x – 45 =0

⇒ 3(x^{2} + 2x – 15) = 0

⇒ x^{2} + 2x – 15 = 0

⇒ x^{2} + 5x – 3x – 15 = 0

⇒ x(x + 5) – 3(x + 5) =0

⇒ x = 3, – 5

x ≠ – 5 (rejected)

Hence, integers are 3, 4, 5.

Question 2.

The surface area of a sphere and a cube are equal. Then, find the ratio of their volumes. (2)

Answer:

Let, surface area of sphere be x cm^{2}

⇒ 4πr^{2} = x

⇒ r = \(\sqrt{\frac{x}{4 \neq}}\) cm,

where, r is the radius of the sphere Also, surface area of cube be x cm^{2}

⇒ 6 × side^{2} = x

⇒ side = \(\sqrt{\frac{x}{6}}\) cm

Caution:

Apply the exponents property in the end to solve the question and get the answer.

Question 3.

The daily minimum steps climbed by a man during a week were as under.

Days of week | No. of steps |

Monday | 35 |

Tuesday | 30 |

Wednesday | 27 |

Thursday | 32 |

Friday | 23 |

Saturday | 28 |

What are the mean of the steps? (2)

Answer:

Number of steps climbed in a week:

35, 30, 27, 32, 23, 28

Mean = \(\frac{\text { Sum of observations(Steps) }}{\text { Total no. of observations }}\)

= \(\frac{35+30+27+32+23+28}{6}\)

= \(\frac{175}{6}\)

= 29.17

Question 4.

The sum of first 6 terms of on AP. is 42. The ratio of its 11th term is 1: 3. CaLcuLate the first term of an AP. (2)

Answer:

Let the first term of the AP be a and common difference be ‘d’.

Now, S_{6} = 42

⇒ \(\frac{6}{2}\)(2a + 5d) = 42

2a + 5d = 14

Also, \(\frac{a_{11}}{a_{33}}\) = \(\frac{1}{3}\)

⇒ \(\frac{a+10 d}{a+32 d}\) = \(\frac{1}{3}\)

⇒ 3a + 30 d = a + 32 d

⇒ 2a = 2d

⇒ a = d

Put a = d in equation (i), we get 2a + 5 = 14

a = 2

∴ First term of an AP = 2

Question 5.

Consider the following distribution:

Marks obtained | No. of Students |

0 or more | 63 |

10 or more | 58 |

20 or more | 55 |

30 or more | 57 |

40 or more | 48 |

50 or more | 42 |

(A) Calculate the frequency of the class 30-40.

(B) Calculate the class mark of the class (2)

Answer:

Class | Frequency | f |

0-10 | 63 | 5 |

10-20 | 58 | 3 |

20-30 | 55 | 4 |

30-40 | 51 | 3 |

40-50 | 48 | 6 |

50-60 | 42 | 4 |

(A) So, frequency of the class 30-40 is 3.

(B) Class mark of the class 10-25

= \(\frac{10+25}{2}\) = \(\frac{35}{2}\) = 17.5

Related Theory:

Class mark can be calculated by dividing the sum of upper and lower limits by 2.

Question 6.

In the given figure, ∠APB = 90°. Then what is the length of OP?

In the figure, PQ and RS are the common tangents to two circle intersecting at O. prove that PQ = RS. (2)

Answer:

In the given figure, OA = r. AP and PB are tangents to the circle.

Then tangent is perpendicular to the radius at the point of contact.

∴ ∠OAP = 90°

And ∠APB = 90°

∴ ∠APO = 45°

(Since OP bisects ∠APB)

∴ In ∆APO

sin 45° = \(\frac{\mathrm{OA}}{\mathrm{OP}}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{r}{\mathrm{OP}}\)

⇒ OP = √2r units

Related Theory:

For the trigonometric ratio sine of an angle is equal to the \(\frac{Perpendicular}{\text { Hypotenuse }}\)

Or

Given: RQ and RS are common tangents to 2 circles intersecting at 0.

To Proof: PQ = RS

Since, tangents from an external point to a circle are equal.

∴ OP = OR …… (i)

And OS = OQ …(ii)

⇒ PQ = OP + OQ

= OR + OS

[From (i) and (ii)] = RS

Hence, proved.

Section – B (12 marks)

Question 7.

Poonam saved a portion of the pocket money. In first week, she saves ₹ 5 and then increased her weekly savings by ₹ 1.75 each week. In which week, will her weekly saving be ₹ 20.75? (3)

Answer:

Suppose amount of weekly savings will be ₹ 20.75 in the nth week. Clearly, amount of weekly savings form an AP with first term, a = 5, common difference, d = 1.75

∵ nth term = 20.75

a + (n – 1) × 1.75 = 20.75

5 + (n – 1) × 1.75 = 20.75

⇒ (n – 1) × 1.75 = 20.75 – 5

⇒ (n – 1) × 1.75 = 15.75

(n – 1) = \(\frac{15.75}{1.75}\)

(n – 1) = 9

⇒ n = 10

Hence, Poonam’s weekly saving will be ₹ 20.75 in 10 week.

Question 8.

A vertical tower stands on a horizontal plane is surmounted by a vertical flag staff of height ‘h’. At a point on the ground, the angles of elevations at the bottom and the tops of the flag staff are a and p respectively. Prove

that the height of the tower is \(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\)

OR

The figure shows two different positions P and Q of a helicopter flying horizontally at a uniform speed of 72 km/hr.

It is found that it takes the helicopter 10 min. to fly from P to Q. The angles of elevation of P and Q from point A on the ground are 45° and 60° respectively.

Find :

(A) The distance PQ.

(B) The distance AQ

(C) The height of the helicopter above the ground. (3)

Answer:

In the figure, AB and BC represent the tower and flagstaff respectively.

In ∆OAB,

OR

(A) As the helicopter is flying at the speed of 72 km/sec, So in 10 min it will fly

\(\left(\frac{72}{60} \times 10\right)\)km = 12 km

(B) Let, the height of the helicopter from the ground be ‘h’ km.

Then,

\(\frac{h}{\mathrm{AM}}\) = tan 60° and \(\frac{h}{\mathrm{AM}}\) = tan 45°

\(\frac{h}{\mathrm{AM}}\) = √3 and \(\frac{h}{\mathrm{AN}}\) = 1

h = √3 AM and h = AN

∴ √3 AM = AN

Since, AN – AM = MN = PQ = 12 cm

√3 AM = AM + 12

⇒ AM (√3 – 1) = 12

AM = \(\frac{12}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

= \(\frac{12(\sqrt{3}+1)}{2}\)

= 16.4 km

Now in ∆AMQ

\(\frac{A M}{A Q}\) = cos 60°

= \(\frac{1}{2}\)

⇒ AQ = 2 AM

⇒ AQ = 2 × 16.4

= 32.8 km

(C) From above, we have h = AN

= AM + 12

= 16.4 + 12

= 28.4 km

Related Theory:

Do the necessary constructions in the given figure to get the answer.

Question 9.

In the given figure, AB and CB are common tangents to the two circles of radii a and b(a > b). Prove that AB = CD (3)

Answer:

Given: Two circles of different radii such that radii of circle with centre O is a and with centre O’ is b

Construction: Join AD and BC

To Prove: AB = CD

Proof: The tangent drawn form an external point to a circle are equal in length.

If A is external point for circle having centre O’

then, AB = AD …(i)

If C is external point for circle having centre O’

then BC = CD …(ii)

Now, B is external point for circle having centre O

then, AB = BC ,..(iii)

So, from (i) (ii) and (iii), we get

AB = BC = CD

So, AB = CD

Hence proved.

Caution:

Here, circles are at a distance from each other and radii are also not equal So, use the property of tangents to get the answer.

Question 10.

What are the roots of quadratic equation 5x^{2} – 6x – 2 = 0, by completing the square method? (3 )

Answer:

Given quadratic equation:

5x^{2} – 6x – 2 =0 …(i)

On multiplying eq. (i) by 5, we get

25x^{2} – 30x – 10 = 0

⇒ (5x)^{2} – 2 × 5x × 3 – 10 = 0

On adding and subtracting 32 we get,

⇒ (5x)^{2} – 2 × 5x × 3 + 3^{2} – 10 – 3^{2} = 0

⇒ (5x – 3)^{2} – 10 – 9 = 0

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

⇒ (5x – 3)^{2} = 19

On taking square root on both sides, we get

⇒ (5x – 3) = ± √19

5x = 3 ± √19

On taking positive sign, we get .

x = \(\frac{3+\sqrt{19}}{5}\)

On taking negative sign we get

x = \(\frac{3-\sqrt{19}}{5}\)

Hence, \(\frac{3+\sqrt{19}}{5}\) and \(\frac{3-\sqrt{19}}{5}\) are the roots of the quadratic equations.

Related Theory:

Completing the square method is used for converting a quadratic expression of the form ax^{2} + bx + c to the form a(x – h)^{2} + k.

Section – C (16 marks)

Question 11.

Draw a circle of radius 5 cm. Draw two tangents to a circle, which are perpendicular to each other.

OR

Draw a circle of radius 2 cm. From a point 6 cm away from its centre, construct the pair of tangents to the circle Measure the lengths of tangents. (4)

Answer:

Steps of construction:

- Draw a circle of radius 5 cm with centre O
- Take, point A on the circle and join OA.
- Draw perpendicular to OA at A
- Draw OB, making an angle of 90° with OA.
- Draw perpendicular to OB at point B. Let these perpendicular intersect at C.

Hence, CA and CB are the required tangents inclined at an angle of 90°

Related Theory:

Here, quadrilateral OBCA formed is a cyclic quadrilateral. Therefore, for forming tangents inclined at 90°. The angle between the radii in the circle should also be 90°

OR

Steps of construction:

- Draw a circle of radius 2 cm, with centre O.
- Mark a point P at a distance of 6 cm from the centre of the circle.
- Draw a perpendicular bisector of Line OP at P
- From P as centre and OP’ as radius draw a circle intersecting the previous circle at M and N.
- Join MP and NP

MP and NP are the required tangents.

Related Theory:

Here, we apply 2 properties of circles for construction. As, tangent is perpendicular the radius and angle in a semicircle is a right angle.

Question 12.

If the median of the following distribution is 58 and sum of all the frequencies is 140. What is the value of x and y? (4)

Variable | Frequency |

15-25 | 8 |

25-35 | 10 |

35-45 | x |

45-55 | 25 |

55-65 | 40 |

65-75 | Y |

75-85 | 15 |

85-95 | 7 |

Answer:

Given, distribution is:

Variable | Frequency | c.f. |

15-25 | 8 | 8 |

25-35 | 10 | 18 |

35-45 | X | 18 +x |

45-55 | 25 | 43 +x |

55-65 | 40 | 83 +x |

65-75 | y | 83 + x + y |

75-75 | 15 | 98 + x + y |

85-95 | 7 | 105 + x + y |

Total | 105 + x + y |

And, 105 + x + y = 140

⇒ x + y = 35 (i)

Here, Median = 58

Then, median class is 55-65,

l = 55,

\(\frac{N}{2}\) = \(\frac{140}{2}\) = 70

Then, c.f. = 43 + x

Median = l + \(\left(\frac{\frac{N}{2}-c . f}{f}\right)\) × h

⇒ 58 = 55 + \(\left(\frac{70-43-x}{40}\right)\) × 10

⇒ BD = 100 m.

⇒ 3 = \(\frac{27-x}{4}\)

⇒ 12 = 27 – x

⇒ x = 27 – 12 = 15

∴ y = 35 – 15 = 20

Hence, value of x and y are 15 and 20.

Question 13.

Case Study—1

Light house is a pilLar type structure in the sea, used to give directions to ships. These are often built on the island, arts cell on cliffs. Lighthouse on water surface act as a navigational aid to the mariners and send warning to boats and ships for dangers. Traditionally, light house have been bacons of navigation for mariners for centuries, but in 21st century lighthouse are going through a major transformation as tourist destination also. India has 189 lighthouses dotting its vast coast line. A man is standing on the top of the light house.

He observes that two boats P and Q are approaching to light house from opposite direction.

He finds that angle of depression of boat P is 45° and angle of depression of boat Q is 30°.

(A) What is the distance of the boat which is nearer to the light house, at a moment? (2)

Answer:

Here, the boat at C is nearer to light house as it has lesser angle of depression i.e., angle of elevation.

∴ In ∆ACD

tan 45° = \(\frac{A D}{B D}\)

1 = \(\frac{100}{B D}\)

⇒ BD = 100 m.

(B) What is the distance between the two boats? (2)

Answer:

Distance between the 2 boats = BC

i.e., BC = BD + CD

∴ In ∆ADB, tan 30° = \(\frac{A D}{B D}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{100}{B D}\)

⇒ BD = 10o√3 m

∴ Distance between boats

= CD + BD

= 100 + 100 √3

= 100 (1 + √3) m.

Question 14.

Case Study—2

Devashish’s mother bought him a ball. He being very intellectual and always interested in finding new things, Devashish try to evaluate the radius of the ball foil calculating its volume and surface area. He tried to do it by compress seace, etc. But he was not successful. So, he saw a video on you tube, regarding the practical use of volume and surface area. Devashish decided to conduct an experiment to find the radius r of a sphere. For this he took a cylindrical container with radius, R = 7 cm and height 10 cm. He filled the container almost half by water as shown in the figure. Now he dropped the ball whose radius is to be measured in the container. After dropping the ball, the water level in the container raised by 3.4 cm.

(A) What is the approximate radius of the spherical ball dropped in the container? (2)

Answer:

5 cm

Explanation: Here, height of water raised by 3.40 cm by dropping the sphere.

The radius of container, R = 7 cm

Then, volume of the height raised = Volume of sphere

πR^{2}h = \(\frac{4}{3}\)πr^{3}

⇒ 7 × 7 × 3.4 = \(\frac{4}{3}\) × r^{3}

⇒ r^{3} = \(\frac{7 \times 7 \times 3 \times 3.4}{4}\)

≈ 125

⇒ r = 5 cm

Hence, the radius of sphere is 5 cm.

(B) If this spherical ball is cut into two hemispherical shapes, then what is surface area of two hemispheres combined ? (2)

Answer:

Radius of sphere = 5 cm

∴ Radius of each hemisphere = 5 cm

Surface area of one hemisphere = 2πr^{2} + πr^{2} = 3πr^{2}

∴ Surface area of 2 hemisphere = 2 × 3πr^{2}

= 6πr^{2}

= 6 × \(\frac{22}{7}\) × 5 × 5

= 471.43 cm^{3}