# CBSE Sample Papers for Class 10 Maths Paper 7

## CBSE Sample Papers for Class 10 Maths Paper 7

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

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 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• Use of calculator is not permitted.

SECTION-A

Question 1.
Find the smallest number which is divisible by 85 and 119.

Question 2.
Write a quadratic polynomial, the sum and product of whose zeroes are 3 and – 2 respectively

Question 3.
Is x = – 4 a solution of the equation 2x² + 5x – 12 = 0 ?

Question 4.
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 5.
Find the distance between (a, b) and (- a, – b).

Question 6.
If tan θ = cot (30° + θ), find the value of θ.

SECTION-B

Question 7.
Show that any positive odd integer is of the form 4m + 1 or 4m + 3, where m is some integer.

Question 8.
What is the quotient and the remainder, when x² + 3x +1 divides 3x4 + 5x3 – 7x² + 2x + 2 ?

Question 9.
Find the 10th term from the end of the A.P. 8, 10, 12, …, 126.

Question 10.
In what ratio does the point P(2, – 5) divide the line segment joining A(- 3, 5) and B(4, – 9) ?

Question 11.
A bag contains 8 green balls, 5 red balls and 7 white balls. A ball is drawn at random from the bag. Find the probability of getting :
(i) neither a green ball nor a red ball.
(ii) a white ball or a green ball.

Question 12.
For the following distribution : Find the sum of class marks of the median class and modal class.

SECTION-C

Question 13.
On a morning walk, three boys step off together and their steps measure 45 cm, 40 cm and 42 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?

OR

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Question 14.
Find the value of k for which 2x² – (k – 2)x + 1 = 0 has equal roots.

Question 15.
For which values of a and b, will the following pair of linear equations have infinitely many solutions ?
x + 2y = 1; (a – b)x + (a + b)y = a + b – 2

Question 16.
A circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle. Question 17.
The diagonals of a trapezium ABCD, with AB || DC, intersect each other at the point O. If AB = 2 CD, find the ratio of the area of ∆AOB to the area of ∆COD.

Question 18.
If A(4, – 8), B(3, 6) and C(5, – 4) are the vertices of a ∆ABC, D is the mid-point of BC and P is a point on AD joined such that $$\\ \frac { AP }{ PD }$$ = 2, find the coordinates of P.

OR

Find the area of a rhombus if its vertices, taken in order, are (3, 0), (4, 5), (- 1, 4) and (- 2, – 1).

Question 19.
Without using the trigonometric tables, evaluate the following :
$$\frac { 11 }{ 7 } \frac { \sin { { 70 }^{ O } } }{ cos{ 20 }^{ O } } -\frac { 4 }{ 7 } \frac { cos{ 53 }^{ O }cosec{ 37 }^{ O } }{ { tan15 }^{ O }tan{ 35 }^{ O }tan{ 55 }^{ O }tan{ 75 }^{ O } }$$

OR

If A, B, C are interior angles of a ∆ABC, then show that :
$$cos\left( \frac { B+C }{ 2 } \right) =sin\frac { A }{ 2 }$$

Question 20.
ABCD is a rectangle such that AB = 10 cm and AD = 7 cm. A semi-circle is drawn with AD as diameter such that the rectangle and semi-circle don’t overlap. Find the perimeter of this figure. (Use π = 22/7)

Question 21.
A cone, a cylinder and a hemisphere are of equal base and have the same height. What is the ratio of their volumes ?

OR

The radii of the ends of a frustum of a cone 40 cm high are 20 cm and 11 cm. Determine its total surface area and volume. (Use π = 3.14)

Question 22.
Two different dice are rolled together. Find the probability of getting :
(i) the sum of numbers on two dice to be 5.
(ii) even numbers on both dice.

SECTION-D

Question 23.
Determine, graphically, the vertices of the triangle formed by the lines y = x, 3y = x and x + y = 8.

Question 24.
If the pth term of an A.R is 1/q and qth term is 1/p, prove that the sum of first pq terms of the A.P is $$\left( \frac { pq+1 }{ 2 } \right)$$

Question 25.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

OR

Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A= 105°. Then, construct a triangle whose sides are 3/4 times the corresponding sides of ∆ABC.

Question 26.
State and prove the Pythagoras theorem.

Question 27.
Draw “more than ogive” for the following distribution : Question 28.
If tan θ + sec θ = l, then prove that:
$$sec\theta =\frac { { l }^{ 2 }+1 }{ 2l }$$

OR

If tan θ + sin θ = m and tan θ – sin θ = n, show that (m² – n²) = √mn.

Question 29.
From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°. If the cars are 100 m apart, find the height of the balloon.

OR

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 10 metres towards the tower, the angle of elevation doubles. Find the height of the tower.

Question 30.
The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm³, then find the value of r.

SOLUTIONS

SECTION-A

Solution 1:
Smallest number which is divisible by 85 and 119 = L.C.M. of numbers 85 and 119
∴85 = 17 x 5
119 = 17 x 7
∴ LCM (85, 119) = 17 x 5 x 7 = 595. Ans.

Solution 2:
Given, Sum of zeroes = 3
Product of zeroes = – 2
We know, quadratic polynomial is given as,
= k[x² – (Sum of zeroes)x + Product of zeroes]
= k[x² – 3x – 2] where k ≠ 0.

Solution 3:
Let p(x) = 2x² + 5x – 12 = 0
For x = – 4
p(x) = 2( – 4)² + 5( – 4) – 12
= 32 – 20 – 12 = 0
p(x = – 4) =0
x = – 4 is the solution of equation 2x² + 5x – 12 = 0.

Solution 4:
Given : PA = 12 cm
and QC = QD = 3 cm
We know, tangents drawn from an external point to a circle are equal.
∴PA = PB
QC =CA
and DQ = DB
Now,
PA + PB = 2 x 12
PC + AC + BD + PD = 24 [∵AC=QC,BD=QD]
PC + QC + QD + PD = 24
PC + 3 + 3 + PD = 24
PC + PD =24 – 6
PC + PD = 18 cm.

Solution 5:
Given points are A(a, b) and B(- a, – b)
By distance formula,
AB = $$\left| \sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \right|$$
= $$\left| \sqrt { { (-a-a) }^{ 2 }+{ (-b-b) }^{ 2 } } \right|$$
= $$\left| \sqrt { { 4a }^{ 2 }+{ 4b }^{ 2 } } \right|$$
AB = $$2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }$$ units

Solution 6:
Given
tan θ = cot (30° + θ)
cot (90° – θ) = cot (30° + θ)
90° – θ = 30° + θ
2θ = 90° – 30°
θ = $$\\ \frac { 60 }{ 2 }$$ = 30°
θ = 30°.

SECTION-B

Solution 7:
Let the number be a = 4q + r where r = 0,1, 2, 3.
Case I: When r = 0
a = 4q
=> a = 2(2(q)
It is an even number.

Case II: When r = 1
a =4q + 1
It is an odd number.

Case III: When r = 2
a =4q + 2
=> a = 2(2q + 1)
It is an even number.

Case IV:
a = 4q + 3
=> a = 2(2q + 1) + 1
It is an odd number.
Hence, any positive odd number is the form of (4m + 1) or (4m + 3).

Solution 8: Solution 9:
Given A.P. is 8, 10, 12, …, 126.
Here l = 126, d = 10 – 8 = 12 – 10 = 2
10th term from the end = l + (n – 1) (- d)
= 126 + (10 – 1) (- 2)
= 126 – 18 = 108
Hence 10th term from the end = 108.

Solution 10:
We have A( – 3, 5), B(4, – 9) and P(2, – 5). By section formula,
$${ p }_{ x }=\frac { { m }_{ 1 }{ x }_{ 2 }+{ m }_{ 2 }{ x }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } }$$
$$2=\frac { { m }_{ 1 }(4)+{ m }_{ 2 }(-3) }{ { m }_{ 1 }+{ m }_{ 2 } }$$
2m1 + 2m2 = 4m1 – 3m2
2m2 + 3m2 = 4m1 – 2m1
5m2 = 2m1
$$\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 5 }{ 2 }$$
∴ m1 : m2 = 5 : 2
So, point P divides line segment AB in the ratio 5 : 2.

Solution 11:
We have, Green balls = 8, Red balls = 5, White balls = 7
Total balls in the bag = 8 + 5 + 7 = 20
(i) Number of balls which are neither a green ball nor a red ball
= No. of white balls = 7
Probability of neither a green ball nor a red ball = $$\frac { No.\quad of\quad white\quad balls }{ total\quad balls } =\frac { 7 }{ 20 }$$
(ii) A white ball or a green ball = 7 + 8 = 15
Probability of a white ball or a green ball nor a red ball = $$\frac { white+green\quad balls }{ total\quad balls } =\frac { 15 }{ 20 } =\frac { 3 }{ 4 }$$

Solution 12: Median = $$\\ \frac { N }{ 2 }$$ th term = $$\\ \frac { 76 }{ 2 }$$ = 38th term
Frequency just above 38 is 67 corresponding to class 15 – 20.
Median class = 15 – 20
Class mark = $$\\ \frac { 15+20 }{ 2 }$$ = 17.5
30 is the highest frequency corresponding to class 15 – 20.
∴ Modal class = 15 – 20
∴ Class mark = $$\\ \frac { 15+20 }{ 2 }$$ = 17.5
Sum of class marks of median class and modal class = 17.5 + 17.5 = 35.

SECTION-C

Solution 13:
Given, three boys step off together and their steps measure 45 cm, 40 cm and 42 cm.
Minimum distance each should walk in complete steps = L.C.M. of the numbers 45, 40 and 42
∵ 45 = 3 x 3 x 5
40 = 2 x 2 x 2 x 5
42 = 2 x 3 x 7
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520
Hence, minimum distance each should walk = 2520 cm = 25.20 m. Ans.

OR

Let the number be a = 3q + r where r = 0,1, 2.
Case I: When r = 0
a = 3q
a² – 9q² = 3(3q²)
a² = 3m, where 3q² = m

Case II: When r = 1
a =3q + 1
a² = (3q + 1)² = 9q² + 6q + 1
= 3(3q² + 2q) + 1
a² = 3m + 1, where m = 3q² + 2q

Case III: When r = 2
a = 3q + 2
a² = (3q + 2)²
a² = 9q² + 12q + 4
a² = 9q² + 12q + 3 + 1
a² = 3(3q² + 4q + 1) + 1
a² = 3m + 1, where m – 3q² + 4q + 1
Hence, square of any integer is of the form 3m or 3m + 1. Hence Proved.

Solution 14:
Given equation is,
2x² – (k – 2)x + 1 = 0
On comparing the equation with ax² + bx + c = 0, we get
a = 2, b = – (k – 2) and c = 1
For equal roots
b² – 4ac = 0
{- (k – 2)}² – 4 x 2 x 1 = 0
(k- 2)² – 8 = 0
(k – 2)² = 8
Taking square root both sides
k – 2 = ± √8
k = 2 ± 2√2

Solution 15:
Given equations are
x + 2y – 1 =0
and (a – b)x + (a + b)y – (a + b – 2) = 0
On comparing the given equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, respectively, we get
a1 = 1, b1 = 2 and c1 = – 1
a2 = a – b, b2 = a + b and c2 = – (a + b – 2)
For infinitely many solutions Solution 16:
In the figure, OP = OQ = r cm
BP = BQ
[Tangents from point B]
∠B = ∠P = ∠Q = 90°
∴ OQBP is a square
OP = BQ = BP = OP = r cm
now,DR = DS = 5 cm [tangent from D]
∴ AR = AD – DR = 23 – 5 = 18 cm
=> AR = AQ = 18 cm [tangent from A]
now,AB = AQ + BQ
=> 29 = 18 + r
∴ r = 29 – 18 = 11 cm
Hence radius of circle = 11 cm

Solution 17:
Given : ABCD is a trapezium in which AB || CD and AB = 2CD.
In trapezium ABCD, AB || CD
∠DCO = ∠OAB
(Alternative angles)
∠CDO = ∠OBA
(Alternative angles)
∆AOB ~ ∆COD
$$\frac { AO }{ CO } =\frac { BO }{ OD } =\frac { AB }{ CD }$$
$$\frac { ar(\triangle AOB) }{ ar(\triangle COD) } =\frac { { AB }^{ 2 } }{ { CD }^{ 2 } } =\frac { { (2CD) }^{ 2 } }{ { CD }^{ 2 } } =\frac { { 4CD }^{ 2 } }{ { CD }^{ 2 } } =\frac { 4 }{ 1 }$$
∴ ar (∆AOB) : ar (∆COD) = 4 : 1.

Solution 18:
Given : In ∆ABC,
$$\frac { AP }{ PD } =2$$
AP : PD = 2 : 1
Let the coordinates of P be (x, y).
Since, D is the mid-point of side BC of ∆ABC,
Therefore, AD is the median and point P is the centroid of ∆ABC.  Solution 19: Solution 20:
Given : ABCD is a rectangle and AED is semi-circle of radius r , drawn on side AD
Clearly, $$r=\frac { AD }{ 2 } =\frac { 7 }{ 2 } cm$$ Perimeter of the figure = AB + BC + CD + arc DEA
= 10 + 7+ 10 + πr
[∵ AB = CD, AD = BC]
= $$27+\frac { 22 }{ 7 } \times \frac { 7 }{ 2 }$$
= 27 + 11
= 38 cm.

Solution 21:
Let the radius of hemisphere be r unit
Radius and height of the cylinder = r unit respectively,
and Radius and height of the cone = r unit respectively.
Volume of the cone : Volume of the cylinder : Volume of hemisphere Solution 22:
When two dice are rolled together, possible outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
n(S) = 36
(i) Sample space for sum of numbers on two dice is 5 = (1, 4), (2, 3), (3, 2), (4, 1).
∴n(E) = 4
P(E) = $$\frac { n(E) }{ n(S) } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 }$$
(ii) Sample space for even number on both dice = (2, 2), (4, 4), (6, 6), (2, 4), (2, 6), (4, 2), (4, 6), (6, 2), (6, 4)
n(E) = 9
P(E) = $$\frac { n(E) }{ n(S) } =\frac { 9 }{ 36 } =\frac { 1 }{ 4 }$$

SECTION-D

Solution 23:
Given equations are,
y = x…(i)
3y = x…(ii)
and x + y = 8…(iii)
On solving, we get
y = x  Solution 24:
Let the first term of A.P. be a and common difference be d.
Now, ap = $$\\ \frac { 1 }{ q }$$ (given)
=> a+(p-1)d = $$\\ \frac { 1 }{ q }$$…(i)
Also aq = $$\\ \frac { 1 }{ p }$$
=> a + (q – 1)d = $$\\ \frac { 1 }{ p }$$…(ii)
On subtracting equation (i) from equation (ii), we get
(p – 1)d – (q – 1)d = $$\frac { 1 }{ q } -\frac { 1 }{ p }$$
(p – 1 – q + 1)d = $$\\ \frac { p-q }{ pq }$$ Solution 25:
Steps of construction : (1) Draw BC = 8 cm and make a right angle at point B such that ∠XBC = 90°.
(2) Cut BA = 6 cm on BX and join AC.
(3) With B as centre and any radius, draw arcs intersecting AC at S and Q.
(4) With S and Q as centre and same radius, draw two arcs which are intersecting at the point R.
(5) Join BR which intersect AC at the point D. Hence BD ⊥ AC.
(6) Draw a perpendicular bisector of side BC which intersect side BC at point O.
(7) Taking O as centre and radius equal to side BO = OC, draw a circle which passes through the point B, D and C.
(8) Taking A as centre and radius 6 cm (v AB = 6 cm), draw an arc intersecting the circle at the point T.
(9) Join AT and extend it to the point P.
∴ AP is also tangent from the point A.
So, AP and AB are the required tangents.

OR

Given : BC = 7 cm
∠B = 45°, ∠A = 105°
In ∆ABC,
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° – 150° = 30° Steps of construction :
(1) Draw a line segment BC = 7 cm.
(2) Make ∠B = 45° and ∠C = 30° to obtain ∆ABC.
(3) Draw a ray BX such that ∠CBX is an acute angle.
(4) Mark 4 points B1, B2, B3 B4 on line BX such that BB1 = B1B2 – B2B3 – B3B4.
(5) Join B4C.
(6) Through B3 draw a parallel line to B4C which meets BC at the point C
(7) Through C’, draw a line parallel to AC which meets the line BA at the point A’.
A’BC’ is the required triangle.

Solution 26:
Pythagoras theorem : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides
Given : In ∆ABC, ∠B = 90°
To prove : AC² = AB² + BC²
Construction : Draw BD ⊥ AC
Proof : In ∆ABC and ∆BDC  Solution 27: Solution 28:
Given
sec θ + tan θ = l
We know that,
sec² θ – tan² θ = 1
=> (sec θ + tan θ) (sec θ – tan θ) = 1
=> (l) (sec θ – tan θ) = 1
=> sec θ – tan θ = $$\\ \frac { 1 }{ l }$$
On adding equations (i) and (ii), we get Solution 29:
Let P be the balloon at a height h m from the ground,
and position of the cars be A and B respectively.
∴ OP = h m
AB = 100 m and OA = x m
In ∆OPA,    Solution 30:
Given, the radii of bucket are 14 cm and r cm.
Its height, h = 15 cm
Volume = 5390 cm³
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