Class 10

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 8.

CBSE Sample Papers for Class 10 Maths Paper 8

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 8 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections A, B, C andD.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each,
• There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.

Section-A

Question 1.
Find the smallest positive rational number by which [latex s=2]\frac { 1 }{ 7 } [/latex] should be multiplied so that its decimal expansion terminates after 2 places of decimal.

Question 2.
If the vertices ofa triangle are (1,2), (4, -6) and(3, 5) then, findthearea oftriangle (NTSE2014,2015)

Question 3.
If [latex s=2]\left( \tan { \theta } +\frac { 1 }{ \tan { \theta } } \right) [/latex] = 2, then find the value of [latex s=2]\tan ^{ 2 }{ \theta } +\frac { 1 }{ \tan ^{ 2 }{ \theta } } [/latex](NTSE 2015)

Question 4.
Evaluate: [latex]\sqrt { 6+\sqrt { 6 } +\sqrt { 6+ } } [/latex]…..

Question 5.
What is the common difference of an A. P. in which a₂₁ -a₇ = 84 ?

Question 6.
The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is
12.1 cm, find the corresponding median of the other.

Section-B

Question 7.
Two bills of ₹ 6075 and ₹ 8505 respectively are to be paid separately by cheques of same amount. Find
the largest possible amount of each cheque.

Question 8.
Find the value of70 + 68 + 66 +……..+ 40.

Question 9.
Find the chance that a non-leap year contains 53 Saturdays.

Question 10.
Ajar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue
marble and a green marble at random from the jar is [latex s=2]\frac { 1 }{ 3 } [/latex] and [latex s=2]\frac { 4 }{ 9 } [/latex] respectively. How many white marbles does the jar contain?

Question 11.
If the lines given by 3x + 2ky = 2 and 2x+5y= 1 are parallel, then find the value of k

Question 12.
Find ‘k’ so that the points (7, -2), (5,1) and (3, k) are collinear.

Section-C

Question 13.
If a² – b² is a prime number, show that a² – b² = a + b, where a, b are natural number.

Question 14.
If the polynomial 6x⁴ + 8x³ + 17x² + 21x + 7 is divided by another polynomial 3x² + 4x+ 1, the remainder comes out to be (ax + b ), find the values of a and b.

Question 15.
If tan θ + sin θ = m and tan θ – sin θ = n, then prove that m² – n² = 4[latex]\sqrt { mn } [/latex]
OR

Question 16.
The following table gives the number of pages written by Sarika for completing her own book for 30 days:

Find the mean number of pages written per day.

Question 17.

Question 18.
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

Question 19.
In figure, two concentric circles with centre O, have radii 21 cm and 42 cm. If ZAOB = 60°, find the area of the shaded region [ Use π = [latex s=2]\frac { 22 }{ 7 } [/latex] ].

Question 20.
If volumes of two spheres are in the ratio of64 : 27, then what is the ratio of their surface areas?
OR
A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. What is the capacity of the capsule?

Question 21.
If A is the area of a right angled triangle and ‘b’ is one of the sides containing the right angle, prove that
the length of the altitude on the hypotenuse is [latex s=2]\frac { 2Ab }{ \sqrt { { b }^{ 4 }+4{ A }^{ 2 } } } [/latex]
OR
ABC is a right. A, right angled at B. AD and CE are the two medians drawn from A and C respectively. If
AC = 5 cm and AD = [latex s=2]\frac { 3\sqrt { 5 } }{ 2 } [/latex] cm, find CE

Question 22.
If the points A (k + 1,2k), B (3k, 2k+ 3) and C (5k- 1,5k) are collinear, then find the value of k.
OR
Show that the points A (1, 0), B (5,3), C (2,7) and D (-2,4) are the vertices of a parallelogram.

Section-D

Question 23.
An aeroplane left 30 minutes later than its scheduled time. The pilot decided to increase its speed. In order to reach its destination 1500 km away in time, pilot has to increase its speed by 250 km/hr from its usual speed. Determine the usual speed.

Question 24.

Question 25.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Question 26.
If h, c, v are respectively the height, the curved surface area and the volume of a cone, prove that 3πvh³– c²h²+ 9v²=0

Question 27.

Question 28.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
OR
If a line divides any two sides of a triangle in the same ratio, then prove that the line is parallel to the third side.

Question 29.
From the top of a tower, the angles of depression of two objects on the same side of the tower are found to be α and β ( α > β). If the distance between the objects is P; show that the height ‘h’ of the tower is given by h = [latex s=2]\frac { P\tan { \alpha } \tan { \beta } }{ \tan { \alpha } -\tan { \beta } } [/latex] .
OR
A vertical tower stands on a horizontal plane and is surmounted by vertical flag staff of height ‘h’ At a point on the plane, the angles of elevation of the bottom and the top of the flag- staff are α and β respectively. Prove that the height of the tower is [latex s=2]\frac { h\tan { \alpha } }{ \tan { \beta -\tan { \alpha } } } [/latex]

Question 30.
The weekly expenditure of5000 families is tabulated below :

Find the median expenditure.
OR
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.

Solutions
Section-A

Solution 1.
Since: [latex s=2]\frac { 1 }{ 7 } [/latex] × [latex s=2]\frac { 7 }{ 100 } [/latex] = [latex s=2]\frac { 1 }{ 100 } [/latex] = 0.01
The smallest rational number is [latex s=2]\frac { 7 }{ 100 } [/latex] (1)

Solution 2.

Solution 3.

Solution 4.
Let x = [latex]\sqrt { 6+\sqrt { 6 } +\sqrt { 6+ } } [/latex] ….Then, x = [latex]\sqrt { 6+x } [/latex]
On squaring both the sides, we get
x² = 6+ x ⇒x²-x-6 = 0
⇒ (x-3)(x + 2) = 0 ⇒ x = 3 or x = -2.
∵ x > 0
So, x = 3 (1)

Solution 5.
Since, a₂₁ – a₇ = 84 …(i)
Let the first term and common difference of
given A.P. be a and d respectively.
So, a₂₁ = a+ (21 – 1)d = a + 20d
Now, a₇= a + (7 – 1)d = a + 6d
From equation (i),
(a + 20d) – (a + 6d) = 84
⇒ (20-6)d= 84 ⇒ 14d= 84 ⇒d = 6
Hence, the common difference of given A.P. is 6. (1)

Solution 6.
Let ABC and DEFbe two triangles such that ∆ABC ~ ∆DEF

Let AL, DM be their medians respectively

Hence required corresponding median = 8.8 cm.

Section-B

Solution 7.
Largest possible amount of cheque will betheHCF (6075,8505).
Applying Euclid’s division lemma to 8505 and 6075, we have,
8505 = 6075 × 1 +2430 (1/2)
Since, remainder 2430 ≠ 0 again applying division lemma to 6075 and 2430
6075 = 2430 × 2+1215 (1/2)
Again remainder 1215 ≠ 0
So, again applying the division lemma to 2430 and 1215
2430=1215 × 2 +0 (1/2)
Here the remainder is zero
So,H.C.F= 1215 (1/2)
Therefore, the largest possible amount of each cheque will be 1215.

Solution 8.
Series: 70 + 68 + 66 + +40
Here a = 70, d = 68 – 70 = -2, tn = 1 = 40.
Now, tn = a + (n-1)d =70 + (n-1)(-2) ⇒ 40 = 70 – 2n + 2 ⇒n= 16 (1)
∴ S_n =[latex s=2]\frac { n }{ 2 } [/latex][a + l] ⇒S_n = [latex s=2]\frac { 16 }{ 2 } [/latex][70 + 40] = 880 (1)

Solution 9.
S = {S, M, T, W, Th, F, Sa} (1/2)
n(S)=7
A non-leap year contains 365 days,
i.e., 52 weeks + 1 day.
E={Sa}
n(E)=1 (1/2)
∴ P(E) = [latex s=2]\frac { n(E) }{ n(S) } [/latex] = [latex s=2]\frac { 1 }{ 7 } [/latex] (1)

Solution 10.
Let there be x blue, y green and z white marbles in the jar.
Then, x+y+ z=54 …(i) (1/2)
∴ P(selecting a blue marble) = [latex s=2]\frac { x }{ 54 } [/latex] ⇒[latex s=2]\frac { x }{ 54 } [/latex] = [latex s=2]\frac { 1 }{ 3 } [/latex] ⇒ x = 18 (1/2)
Similarly, /’(selecting a green marble) = [latex s=2]\frac { y }{ 54 } [/latex] ⇒ [latex s=2]\frac { y }{ 54 } [/latex]= [latex s=2]\frac { 4 }{ 9 } [/latex] ⇒y = 24 (1/2)
Substituting thevaluesofxandyin(i),z= 12
Hence, the jar contains 12 white marbles. (1/2)

Solution 11.
Given lines,
3x + 2ky- 2 = 0
and 2x+ 5y-1 =0
Here, a₁ =3,b₁ =2k, c₁ =-2
a₂ = 2, b₂ = 5, c₂ =-1 (1)
Condition for parallel lines is [latex s=2]\frac { 3 }{ 2 } [/latex] = [latex s=2]\frac { 2k }{ 5 } [/latex] ⇒k = [latex s=2]\frac { 15 }{ 4 } [/latex] (1)

Solution 12.
Since points are collinear,
so, x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0 (1/2)
Taking (7, -2) as (x₁, y₁), (5,1) as (x₂, y₂) and (3, k) as (x₃, y₃), we have
7(1 – k) + 5(k + 2) + 3 (-2 -1) = 0 (1/2)
or, 7 – 7k + 5k + 10 – 9 = 0 ⇒ 2k = 8 ⇒k = 4. (1)

Section-C

Solution 13.
a² – b² = (a-b)(a +b) ….(1)
∵ a² -b² is a prime number (1)
∴One of the two factors = 1
So, a — b= 1 [∵a – b< a+b]
∵ The only divisors of a prime number are 1 and it self. (1)
(1) become a² -b²= 1 (a + b)
or a² – b² = a + b
e.g., 3² – 2² = 5 (which is prime)
⇒ 3² – 2² = 3 +2 (1)

Solution 14.

Solution 15.

OR

Solution 16.

Solution 17.

Solution 18.
Let there be a circle with centre O whereas AB, BC, CD and DA are tangents at P, Q, R and S respectively.

Here,
AP = AS …(i)
BP = BQ …(ii)
CR=CQ …(iii)
DR=DS …(iv)
[Tangents drawn from a point (outside the circle) on a given circle are equal in lengths]
From equations (i), (ii), (iii) and (iv), (1)
(AP +BP) + (CR + DR) = (BQ + CQ) + (DS + AS)
AB + CD = BC + DA (Hence proved.) (1)

Solution 19.

Solution 20.


OR

Solution 21.

OR

Solution 22.

OR

Since, the coordinates of the mid-points of diagonals AC and BD are same.
∴ They bisect each other.
Hence, ABCD is a parallelogram. (1)

Section-D

Solution 23.

Solution 24.

Solution 25.

Steps of construction:
(i) Drawa line AB of length 8 cm.
(ii) With centre A and radius 4 cm, draw a circle, which intersect AB at point P.
(iii) With centre B and radius 3 cm, draw the second circle.
(iv) With centre P and radius PA or PB draw third circle, which intersects the circle drawn in step (ii) at C & D and the circle drawn in step (iii) at E & F.
(v) Draw rays AE,AF,BC and BD.
(vi) Thus AE,AF,BC and BD are required tangents. (1)

Solution 26.

Solution 27.

Solution 28.

OR

Solution 29.

OR
Let AB be the tower and BC be the flag.
LetAB = H
BC = h

Solution 30.


OR

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These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 7.

CBSE Sample Papers for Class 10 Maths Paper 7

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 7 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each,
•  There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.

Section-A

Question 1.
What type of decimal expansion does a rational number has? How can you distinguish it from decimal expansion of irrational numbers?

Question 2.
Coordinates of P and Q are (4,-3) and (-1, 7). What is the abscissa of a point R on the line segment PQ such that [latex]\frac { PR }{ PQ } [/latex] = [latex]\frac { 3 }{ 5 } [/latex] ? (NTSE 2012)

Question 3.
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P. ?

Question 4.
For the equation 3×2 + px + 3 = 0, if one of the roots is the square of the other then find p. (NTSE 2014-2015)

Question 5.
Find the value of sin (45° + θ) – cos (45° – θ)

Question 6.
In Fig., if PQ || RS, prove that ∆ POQ ~ ∆ SOR

Section-B

Question 7.
When 2²⁵⁶ is divided by 17 then find the remainder.

Question 8.
In what ratio, the line segment joining the points (3,5) & (-4,2) is divided byy-axis?

Question 9.
Find the solution of the pair equations [latex s=2]\frac { x }{ 10 } [/latex] + [latex s=2]\frac { y }{ 5 } [/latex] – 1= 0 and [latex s=2]\frac { x }{ 8 } [/latex] + [latex s=2]\frac { y }{ 6 } [/latex] = 15. Hence, find λ, if y = λx + 5.

Question 10.
Find the sum of first 24 terms of the sequence whose n^th term is a_n = 3+[latex s=2]\frac { 2n }{ 3 } [/latex]

Question 11.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If6000 tickets are sold,

Question 12.
A book containing 100 pages is opened at random. Find the probability that a doublet page is found.

Section-C

Question 13.
Show that any positive odd integer is of the form 8q ± 1 and 8q ± 3, where q is some integer.

Question 14.
If a and 8 are the zeroes of the quadratic polynomial f(x) = ax² + bx + c then evaluate [latex]\frac { { \alpha }^{ 2 } }{ { \beta }^{ 2 } } +\frac { { \beta }^{ 2 } }{ { \alpha }^{ 2 } } [/latex]

Question 15.
In the given fig, BD ⊥ AC and CE ⊥ AB. Prove that
(i) ∆AEC ~ ∆ADB
(ii) [latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex]

OR
In the given fig, [latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex] prove that ∠A = ∠C and ∠B = ∠D

Question 16.
Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t.
OR
The three vertices of a parallelogram, taken in order, are (1, -2), (3,6) and (5,10). Find the coordinates of its fourth vertex .

Question 17.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Question 18.
Calculate the mean of the following frequency distribution :

Question 19.
In figure, ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. Find area of the shaded region.

Question 20.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Roohi paid ₹ 27 for a book kept for seven days, while Shushama paid₹ 21 for the book she kept for five days. What is the fixed charge?

Question 21.
Without using trigonometric tables, prove that:- tan7° tan 23° tan 60° tan 67° tan 83° = [latex]\sqrt { 3 } [/latex]
OR
If 5 tan θ = 4, then find the value of [latex s=2]\frac { 5\sin { \theta } -3\cos { \theta } }{ 5\sin { \theta } +2\cos { \theta } } [/latex]

Question 22.
A copper rod of diameter 1cm and length 8 cm, is drawn into a wire of length 18 m of uniform thickness. Find the thickness of wire.
OR
If a solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. Find the radius of the sphere

Section-D

Question 23.
In a right triangle, prove that the square of the hypoten’use is equal to the sum of the squares of the other two sides.
OR
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.

Question 24.
What is the value of x if
cot x = [latex s=2]\frac { 5 }{ 3 } [/latex] tan 130 tan 37° tan 45 ° tan 53° tan 77° -[latex s=2]\frac { 2 }{ 3 } [/latex] cosec² 58° + [latex s=2]\frac { 2 }{ 3 } [/latex] cot 58° tan 32°.

Question 25.
Solve the equation: [latex s=2]\left( \frac { 2x-3 }{ x-1 } \right) -4\left( \frac { x-1 }{ 2x-3 } \right) [/latex] = 3 x ≠ 1, 3/2

Question 26.
Find the median of the following data :

OR
The median ofthe following data is 525. Find the values of* and y if the total frequency is 100.

Question 27.
Find the volume and total surface area of a tumbler in the form of a frustum of a cone, if the diameter of the ends are 6.50 cm and 3.50 cm and the perpendicular height ofthe tumbler is 7.80 cm.

Question 28.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.

Question 29.
If the ratio ofthe sum ofthe first n terms oftwoA.Ps is (7n+ 1): (4n + 27), then find the ratio of their 9^th terms.

Question 30.
An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and
60° respectively. Find the width ofthe river. [Use [latex]\sqrt { 3 } [/latex] = 1.732]
OR
The angle of elevation of a cloud from a point 60 m above the surface ofthe water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height ofthe cloud from the surface of water.

Solutions
Section-A

Solution 1.
A rational number is either terminating or non-terminating repeating.
An irrational number is non-terminating and non-repeating. (1)

Solution 2.
P (4, -3) and Q (-1,7)
PR: RQ = 3:2
4 × 2 + (-1) × 3
Abscissa = [latex s=2]\frac { 4\times 2+(-1)\times 3 }{ 3+2 } [/latex]= 1 (1)

Solution 3.
Given k + 9,2k – 1 and 2k + 7 are in AP.
∴ 2(2k – 1) = (k + 9)+(2k + 7) [By Arithmetic mean] (1/2)
⇒ 4k-2 = 3k+16 ⇒ k=18 (1/2)

Solution 4.
α + α2 = [latex s=2]\frac { -p }{ 3 } [/latex] ………..(i)
α3 = [latex s=2]\frac { 3 }{ 3 } [/latex] = 1 =>(α- l)(α3+ α+ 1) = 0 ⇒ α = 1
∴ p= – 6 (∵From (i)) (1)

Solution 5.
Given that
sin (45° + θ) – cos (45° – θ) = cos [90° – (45° + θ)] – cos (45° – θ) [∵ cos (90° – θ) = sin θ]
= cos (45° – θ) – cos (45° – θ) = 0 (1)

Solution 6.
PQ || RS (Given)
So, ∠P = ∠S (Alternate angles)
and ∠Q=∠R (1/2)
Also, ∠POQ = ∠ SOR (Vertically opposite angles)
Therefore, ∆POQ ~ ∆SOR (AAA similarity criterion) (1/2)

Section-B

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.
P (winning) = 0.08
Total tickets sold = 6000
Let the number of tickets she bought be x, then probability of winning =[latex s=2]\frac { x }{ 6000 } [/latex] (1)
⇒[latex s=2]\frac { x }{ 6000 } [/latex] = 0.08 ⇒ x = 6000 × 0.08 ⇒ x = 480 (1)
Thus the girl bought 480 tickets.

Solution 12.
S= {1,2,3, ,100} (1/2)
n(S)= 100
E ={ 11,22,33,44, 55,66,77, 88,99} (1)
n(E) = 9
∴P(E) = [latex s=2]\frac { 9 }{ 100 } [/latex] (1/2)

Section-C

Solution 13.
Let a and b be two positive integers where a is odd.
Applying division lemma a = 8q + r where 0 < r < 8
So, r can take any of the values 0, 1,2,3,4,5,6,7
Therefore, a= 8q, 8q + 1, 8q + 2, 8q + 3, 8q + 4,
8q + 5, 8q + 6, 8q + 7 (1)
Since, a is odd.
Therefore, a cannot take values 8q, 8q + 2, 8q + 4, 8q + 6 since they can expressed as multiples of 2.
So, a will take values 8q+ 1,8q + 3, 8q + 5, 8q + 7. (1)
Also, 8q + 5 = 8q + 8 – 3 = 8 (q + 1) – 3 = 8q’ — 3
where q’ = q + 1, 8q + 7 = 8q + 8 – 1 =8q’- 1
So, every positive odd integer is of the form 8q ± 1, 8q ± 3. (1)

Solution 14.

Solution 15.
Given: In the fig., BD ⊥ AC and CE ⊥AB
To prove: (i) ∆AEC ~ ∆ADB
(ii) [latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex] (1)

Proof (i) In ∆AEC and ∆ADB
∠1 = ∠2 (each 90°)
∠A = ∠A (common)
∴ ∆AEC ~ ∆ADB (by AArule)
(ii) ∆AEC ~ ∆ADB
[latex s=2]\frac { CA }{ AB } [/latex] = [latex s=2]\frac { CE }{ DB } [/latex] (∵ Angles are similar ∴ corresponding sides are proportional) (1)
Hence proved
OR
[latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex]

To prove: ∠A = ∠C and ∠B = ∠D (1)
Proof: In ∆AOD and ∆BOC
[latex s=2]\frac { OA }{ OC } [/latex] = [latex s=2]\frac { OD }{ OB } [/latex] (Given)
and ∠AOD = ∠BOC (Vertically opposite angles)
∴ ∆AOD ~ ∆BOC (by SAS) (1)
∴ ∠A = ∠C and ∠B = ∠D (C.P.C.T.) (1)

Solution 16.
Let A (t, t – 2), B (t + 2, t + 2) and C (t + 3, t) be the vertices of the given triangle. (1/2)
Since the area of the triangle having vertices (x₁, y₁), (x₂, y₂) and(x₃, y₃)

Therefore, the area of the triangle with specified vertices is independent of t. (1/2)
OR
Let fourth vertex be D (a, b)
Let the diagonals AC and BD intersect at E,

We know that the diagonals of a parallelogram bisect each other.
∴ E is the mid-point of AC as well as that ofBD.
mid-point of AC is

Hence, The fourth vertex of the given parallelogram is D (3,2). (1)

Solution 17.
Given : PA and PB are tangents to the circle from an external point P. CD is another tangent at Q. PA= 12 cm, QC = QD = 3 cm

To find: PC + PD (1/2)
Proof: PA = PC + AC
12 = PC+ 3
[∵ QC = AC = 3 cm, tangents from external point to a circle are equal in length]
PC = 9 cm …(i) (1)
Similarly, BD = QD = 3 cm
and PB = PA =12 cm (1/2)
PB = PD + BD
12 = PD +3
PD = 9 cm
Now, PC + PD = 9 + 9 = 18 cm (1)

Solution 18.

Solution 19.
We mark the area of unshaded regions as A₁,A₂A₃ and A₄ as shown in the figure.
Area of the shaded region = Area of the square ABCD – Area of the unshaded portion (A₁+A₂+A₃+ A₄).

Area of unshaded region A₁+A₃= Area of square ABQD – Area of semi circle on BC as diameter- Area of semi-circle on AD as diameter.

Similarly, area of unshaded region A₂ – A₄ = Area of square ABCD – Area of the semi-circles on diameter AB and DC
So, Area A₂+A₄ = 10² -2× [latex s=2]\frac { \pi \times { 5 }^{ 2 } }{ 2 } [/latex] =100-25π.
So,Area A₁ + A₂ + A₃ + A₄ = 200-50π.
Area of the shaded region = Area of square ABCD – Area (A₁ + A₂ + A₃ + A₄) (1)
= 10²-(200-50π)
= (100-200 + 50 π)sq. cm.
= 50 π -100 = 50 (π – 2) sq. cm.
= 50 (3.14-2)= 50 × 1.14 = 57 cm². (1)

Solution 20.
x = F ixed charge
y = Additional charge for each day
x + 4y = 27 …(i)
x + 2y = 21 …(ii) (1)
⇒ (x + 4y)-(x+ 2y) = 27-21 => 2y = 6 ⇒y = 3 (1)
∵ x + 2y=21 =>orx + 2(3) = 21 ⇒ x= ₹ 15 (1)

Solution 21.
L.H.S.
= tan 7° tan 23° tan 60° tan 67° tan 83°= tan 7° tan 83° tan 23° tan 67° tan 60° (1)
= tan 7° cot 7° tan 23° cot 23° tan 60°(1)

OR

Solution 22.
Volume of rod = πr²h = π × (1/2)² × (8) = 2πcm³ ….(i) (1)
Length of wire of same volume = 18m = 1800 cm.
Let ‘r’ be the radius of cross-section of wire (1/2)
∴ Volume = πr²h
Volume of wire = πr² × 1800 ….(ii)
From (i) & (ii), πr² × 1800 = 2π
r² = 2/1800 =>r²= 1/900 =>r=1/30cm. (1)
Thickness of wire = diameter of cross-section = 2r = 2 × [latex s=2]\frac { 1 }{ 30 } [/latex] = [latex s=2]\frac { 1 }{ 15 } [/latex] = 0.067 cm (approximately) (1/2)
OR
Given, dimensions of the cuboid
= 49 cm x 33 cm x 24 cm
Now, volume of the cuboid
= 49x33x24 = 38808 cm³
Let the radius of the sphere is r, then (1/2)
Volume of the sphere = [latex s=2]\frac { 4 }{ 3 } [/latex]πr³ (1/2)
According to the given condition
Volume of the sphere = Volume of the cuboid (1/2)

Section-D

Solution 23.
Given: A triangle ABC in which ZB = 90°.

To prove : ACsup>2 = ABsup>2 + BCsup>2 or (Hypotenuse)sup>2 = (Base)sup>2 + (Perpendicular)sup>2
Construction : From B, drawBD⊥ AC. (1)
Proof: Since BD ⊥AC.
∴∆ADB ~ ∆ABC
∴[latex s=2]\frac { AD }{ AB } [/latex] = [latex s=2]\frac { AB }{ AC } [/latex] ⇒AB² =AC ×AD …(i) (1)

OR

Solution 24.

Solution 25.

Solution 26.

OR

Solution 27.

Solution 28.
Steps of construction:
1. Draw a circle of radius 4 cm, with O as centre.
2. Take a pointAon the circumference ofthe circle andjoin OA. Draw a perpendicular to OA at A.
3. Draw a radius OB, making an angle of 120° with OA.
4. Draw a perpendicular to OB at B. Suppose these perpendiculars intersect at P.

Here, PA and PB are two tangents drawn to the circle inclined at an angle of 60° to each other.
Justification
The construction can be justified by proving that
∠APB = 60°.
∠OAP = 90° (By Construction)
∠OBP = 90° (By Construction)
∠AOB=120° (By Construction)
The sum of all interior angles of a quadrilateral is 360°,.
∠OAP + ∠AOB + ∠OBP + ∠APB = 360° .
⇒ 90°+120°+ 90° +∠APB = 360° (1)
⇒ ∠APB = 60°

Solution 29.

Solution 30.

OR

We hope the CBSE Sample Papers for Class 10 Maths paper 7 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 7, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 5.

CBSE Sample Papers for Class 10 Maths Paper 5

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80

General Instructions

• All questions are compulsory.
• The question paper consists of 30 questions divided into four sections A, B, C andD.
• Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each,
• There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculators is not permitted.

Section-A

Question 1.
Can the number 6ⁿ, n being number, end with the digit 5? Give reasons.

Question 2.
ABC is an isosceles triangle in which AB = AC = 10 cm. BC = 12 cm. PQRS is a rectangle inside the
isosceles triangle. Given PQ = SR = 2x cm and PS = QR = y cm. Prove that xb = 6 – [latex s=2]\frac { 3y }{ 4 } [/latex]

Question 3.
Find the middle term oftheA.P. 6,13,20,……..,216.

Question 4.
Represent in the form of quadratic equation : (x – 3) (2x + 1) = x (x + 5).

Question 5.
Find the third vertex of a triangle, iftwoofits vertices are at (-3, 1) and (0,-2) and the centroid is at the origin.

Question 6.
Prove: [latex s=2]\frac { \sin ^{ 4 }{ \theta } -\cos ^{ 4 }{ \theta } }{ \sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } } [/latex] = 1

Section-B

Question 7.
Use Euclid’s algorithm to find the HCF of 6812 and 289 f6.

Question 8.
Find the relation between x and y if the points A(x, y), B (-5,7) and C (-4,5) are collinear.

Question 9.
From a normal pack of cards, a card is drawn at random, find the probability of getting a jack or a king.

Question 10.
Find the value of K for which the system of equations has no solution.
3x +y = 1; (2k – 1)x + (k – 1)y = (2k + 1)

Question 11.
Which term of the progression 20, 19 [latex s=2]\frac { 1 }{ 4 } [/latex],18[latex s=2]\frac { 1 }{ 2 } [/latex],17 [latex s=2]\frac { 3 }{ 4 } [/latex],…… is the first negative term?

Question 12.
In a single throw of two dice, find the probability of getting a sum of 10.

Section-C

Question 13.
Prove that if x and y are odd positive integers, then x² + y² is even but not divisible by 4.

Question 14.
If a, B are the roots of the polynomials^) = 2x² + 5x + k satisfying the relation α² + β² + αβ = [latex s=2]\frac { 21 }{ 4 } [/latex], then find the value of k for this to be possible.

Question 15.
In the figure, ABC is a right triangle, right angled at B. AD and CE are two medians drawn from A and C
respectively. If AC= 5 cm and AD = [latex s=2]\frac { 3\sqrt { 5 } }{ 2 } [/latex] cm, find the length of CE.

OR
In ∆ABC, AD⊥BC and point D lies on BC such that 2DB = 3CD. Prove that 5AB² = 5AC² + BC²

Question 16.
Solve the following equations: [latex s=2]\frac { 1 }{ 7x } +\frac { 1 }{ 6y } [/latex] = 3, [latex s=2]\frac { 1 }{ 2x } -\frac { 1 }{ 3y } [/latex] = 5

Question 17.
If cos θ +[latex]\sqrt { 3 } [/latex] sinθ = 2sinθ . Show that sin θ – [latex]\sqrt { 3 } [/latex]cos θ = 2cos θ.
OR

Question 18.
If A (-1,-1), B (-1,4), C (5,4) andD (5,-1) are the vertices of a quadrilateral ABCD, find its area.
OR
ABCD is a rectangle formed by the points A(-l, -1), B(- 1,4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Question 19.
A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.

Question 20.
The numbers 5, 7, 10, 12, 2x-8, 2x+ 10, 35, 41, 42, 50 are arranged in ascending order. Iftheir median is 25 then find the value of x.

Question 21.
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Question 22.
A well of diameter 4 m is dug 14 m deep. The earth taken out is spread evenly all around the well to form a 40 cm high embankment. Find the width of the embankment.
OR
A sphere of diameter 12.6 cm is melted and recast into a right circular cone of height 25.2 cm. Find the diameter of the base of the cone.

Section-D

Question 23.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

Question 24.
In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.

Question 25.
Solve the equation by using qudratic formula : (x + 4) (x + 5) = 3(x + 1) (x + 2) + 2x
OR
If the roots of the equation (c² – ab) x² – 2(a² – bc) x + (b² – ac) = 0 are equal, prove that either a = 0 or a³ + b³ + c³ = 3abc

Question 26.
If the areas of two similar triangles are equal, prove that they are congruent.
OR
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of triangle PQR. Prove that ∆ABC ~ ∆PQR

Question 27.
The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height
ofthe flagstaff. [Use [latex]\sqrt { 3 } [/latex] = 1.73]
OR
From a point P on the ground the angle of elevation ofthe top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length ofthe flag staff is 5m, find the height of the tower.

Question 28.
The mean of three positive numbers is 10 more than the smallest ofthe numbers and 15 less than the largest of the three. If the median of the three numbers is 5, then find the means of squares of the numbers.(NTSE 2016)

Question 29.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.

Question 30.
The diameter ofthe internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm, respectively.
Ifit is melted and recast into a solid cylinder of length 2[latex s=2]\frac { 3 }{ 2 } [/latex] cm. Find the diameter ofthe cylinder.

Solutions
Section-A

Solution 1.
If 6ⁿ ends with 5, then it must have 5 as a factor. Prime factor of 6ⁿ are 2 and 3.
∴ 6ⁿ = (2 × 3)n = 2ⁿ × 3ⁿ
The fundamental theorem of arithmetic, the prime factorization of every composite numbers is unique.
∴ 6ⁿ can never end with 5. (1)

Solution 2.

Solution 3.
The given A.P. is 6, 13, 20, ………..,216
Here, d = 7, a = 6
a_n = a + (n-1)d
∴216 = 6 + (n- 1) × 7 ⇒n = 31

∴a₁₆= 6+ 15 × 7 = 111 (1)

Solution 4.
(x-3)(2x+l)-x(x + 5)
2x² – 5x – 3 = x² + 5x
x² – 10x – 3 = 0 (1)

Solution 5.
Let th e third vertex of the triangle PQR be given by P (x, y).
By definition of centroid of a triangle, we have [latex s=2]\frac { -3+0+x }{ 3 } [/latex] = 0 [∵The centroid is at the origin]
and [latex s=2]\frac { 1-2+y }{ 3 } [/latex] = 0
From above : x = 3 and y = 1

∴ The required third vertex is (3,1)

Solution 6.
Expressing sin⁴θ -cos⁴θ as (sin²θ)² – (cos²θ)²
⇒ sin⁴ θ -cos⁴ θ
= (sin²θ -cos²θ). (sin²θ + cos²θ)
Now, we know that sin²θ + cos²θ = 1
∴ [latex s=2]\frac { \sin ^{ 4 }{ \theta } -\cos ^{ 4 }{ \theta } }{ \sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } } [/latex] = 1 = R.H.S. (Hence Proved) (1)

Section-B

Solution 7.
Since, 28916 > 6812, we apply division lemma to 28916 and 6812, to get
28916 = 6812 × 4+1668
6812=1668 × 4+140 (1/2)
1668= 140 × 11 + 128
140= 128 × 1 + 12 (1/2)
128= 12 × 10 + 8
12= 8 × 1 +4 (1/2)
8 =4 × 2 + 0
The remainder has became zero. Now our procedure stops. At this stage the divisor is 4. So the HCF(6812,28916) is 4. (1/2)

Solution 8.
By ar (∆ABC) = 0 for collinear points A, B and C.
∴x(7 – 5)-5(5 – y)-4(y – 7) = 0 (1)
So, 2x – 25 +5y – 4y + 28 = 0 (1/2)
Hence, 2x + y + 3 = 0 (1/2)

Solution 9.
n(S) = 52, n(E) = 4 + 4 = 8 (1)
P(E) = [latex s=2]\frac { n(E) }{ n(S) } [/latex] = [latex s=2]\frac { 8 }{ 52 } [/latex] = [latex s=2]\frac { 2 }{ 13 } [/latex] (1)

Solution 10.

Solution 11.

Solution 12.
n(S) = 6x 6= 36, E = {(4,6), (5,5), (6,4)} n(E) = 3
n(E) = 3 (1)
P(E) = [latex s=2]\frac { n(E) }{ n(S) } [/latex] = [latex s=2]\frac { 3}{ 36 } [/latex] = [latex s=2]\frac { 1 }{ 12 } [/latex] (1)

Section-C

Solution 13.
We know that any odd positive integer is of the form 2q + 1 for some whole number q.
So, let x = 2m + 1 and y = 2n + 1 for some whole number m and n.

Solution 14.

Solution 15.



OR

Solution 16.

Solution 17.


OR

Solution 18.

OR

Solution 19.

Solution 20.
N= 10 (even).
Median = [latex s=2]\frac { 1 }{ 2 } [/latex] (5^th obs. + ^th obs.) (1)
= [latex s=2]\frac { 1 }{ 2 } [/latex] (2x – 8 + 2x+ 10) =2x+1 (1)
2x+1=25 ⇒ x=12. (1)

Solution 21.
Given : Tangents AB and DC are parallel
To Prove : PQ passing through centre O
Const: DrawEO || AB

Proof :AB || DC
Sum of the angles on the same side of transversal is 180°
∠APO + ∠EOP=180° ⇒ ∠EOP= 180°-90° =90° (1)
Similarly, ∠EOQ = 90°
∴ ∠EOP + ∠EOQ = 90° + 90° = 180°
∴ PQ is a straight line
Elence PQ passing through the centre O. (1)

Solution 22.

Section-D

Solution 23.
L.H.S. = (sinA + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 + cos² A + sec² A + 2 (2)
= 5 + 1 + cot² A + 1 + tan² A (1)
= 7 + cot² A + tan² A = R.H.S. (1)

Solution 24.

Solution 25.

OR

Solution 26.

OR

Solution 27.

OR

Solution 28.

Solution 29.

Steps:
(i) Draw AABC and perpendicular BD on AC
(ii) Draw a circle with BC as a diameter.
(iii) Circle passing through B, C and D.
(iv) Let Q be the mid point of BC. Join AQ.
(v) Draw a circle with AQ as diameter. This circle cuts the previous circle at point P and B.
(vi) Now, join AP and AB are desired tangents drawn from A to the circle passing through B, C
and D. (1)

Solution 30.

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