Students must start practicing the questions from RBSE 10th Maths Model Papers Set 6 with Answers in English provided here.

## RBSE Class 10 Maths Model Paper Set 6 with Answers in English

Time: 2:45 Hours

Maximum Marks: 80

General Instructions:

- All the questions are compulsory.
- Write the answer of each question in the given answer book only.
- For questions having more than one part the answers to their parts are to be written together in continuity.
- Candidate must write first his/her Roll No. on the question paper compulsorily.
- Question numbers 17 to 23 have internal choices.
- The marks weightage of the questions are as follows :

Section | Number of Questions | Total Weightage | Marks for each question |

Section A | 1 (i to xii), 2(i to vi), 3(i to xii) = 30 | 30 | 1 |

Section B | 4 to 16 =13 | 26 | 2 |

Section C | 17 to 20 = 4 | 12 | 3 |

Section D | 21 to 23 = 3 | 12 | 4 |

Section – A

Question 1.

Multiple Choice Questions :

(i) Sum of the exponents of prime factors in the prime factorization of 196 is : (1)

(a) 3

(b) 4

(c) 5

(d) 2

Answer:

(b) 4

(ii) If one zero of the quadratic polynomial x^{2} + kx – 10 = 0 is 2, then value of k is : (1)

(a) 10

(b) 3

(c) – 7

(d) – 2

Answer:

(b) 3

(iii) The value of k for which the system of equations :

2x + 3y = 5,

4x + ky = 10

has unique solution, is : (1)

(a) ≠3

(b) ≠6

(c) ≠0

(d) ≠1

Answer:

(b) ≠6

(iv) If x^{2} – 4x + k = 0 has a root x – 2, then value of k is : (1)

(a) – 1

(b) 4

(c) 2

(d) – 2

Answer:

(b) 4

(v) The 5th terms of the sequence defined by t_{1} = 2, t_{2} = 3 and t_{n} = t_{n-1} + t_{n-2} for n ≥ 3 : (1)

(a) 16

(b) 13

(c) 15

(d) 2.

Answer:

(b) 13

(vi) To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A_{1}; A_{2}, A_{3}, …. are located at equal distances on the ray AX and the point B is joined to : (1)

(a) A_{12}

(b) A_{11}

(c) A_{10}

(d) A_{0}

Answer:

(b) A_{11}

(vii) If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y ) in the ratio 3:1, then the value of y is : (1)

(a) 4

(b) 3

(c) 2

(d) 1

Answer:

(d) 1

(viii) If cot A = \(\frac{12}{5}\), then cos A : (1)

(a) \(\frac{12}{13}\)

(b) \(\frac{13}{12}\)

(c) \(\frac{13}{5}\)

(d) \(\frac{5}{12}\)

Answer:

(a) \(\frac{12}{13}\)

(ix) If sin 3A = cos (A – 10), where 3A is an acute angle, then value of A is equal to: (1)

(a) 30°

(b) 35°

(c) 25°

(d) 20°

Answer:

(c) 25°

(x) If the mode of a data is 18 and median is 22, then mean is : (1)

(a) 20

(b) 22

(c) 24

(d) 26

Answer:

(c) 24

(xi) The mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is : (1)

(a) 4

(b) 7

(c) 11

(d) 6

Answer:

(b) 7

(xii) If an event cannot occur, then its probability is : (1)

(a) 1

(b) \(\frac{3}{4}\)

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

(d) 0

Answer:

(d) 0

Question 2.

Fill in the blanks :

(i) If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) the pair of equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 is ________ (1)

Answer:

Consistent

(ii) Infinite APs do not have a ________ term. (1)

Answer:

last

(iii) Point of intersection of perpendicular, bisectos of two non-parallel ________ of a circle is centre of circle. (1)

Answer:

chord

(iv) If the area of a triangle is ________ square unit, then its vertices will be collinear. (1)

Answer:

zero

(v) If tan θ = 1, then the value of \(\frac{5 \sin \theta+4 \cos \theta}{5 \sin \theta-4 \cos \theta}\) is ________ (1)

Answer:

9

(vi) If u_{i} = \(\frac{x_{i}-5}{14}\) Σf_{i}u_{i} =35, Σf_{i} =7, then the value of x is ________ (1)

Answer:

12

Question 3.

Very Short Answer Type Questions:

(i) What is the HCF of the smallest prime and the smallest composite number?

Answer:

Smallest prime number = 2

Smallest composite number = 4

∴ 2 = 2 and 4 = 2 × 2 = 22

HCF of (2, 4) = 2.

(ii) Find the zeroes of the polynomial x^{2} – 3 = 0.

Answer:

We have, x^{2} – 3 = 0

⇒ (x + √3)(x – √3) = 0

∴x = ±√3

(iii) If α, β, γ are roots of a cubic polynomial, then find the sum of roots. (1)

Answer:

Let cubic polymial, ax^{2} + bx^{2} + cx + d = 0

Then sum of its roots

α + β + γ = \(-\frac{\text { Cofficien to } x^{2}}{\text { Cofficien to } x^{3}}=\frac{-b}{a}\)

(iv) Check the consistency the pair of equations x + y = 5 and 2x + 2y = 10. (1)

Answer:

Here, a_{1} = 1, b_{1} = 1, c_{1} = – 5 and a_{2} = 2, b_{2} = 2, c_{2} = – 10

Therefore, the given linear equations are coincident lines. So, the equations are consistent.

(v) Find the zeroes of the qufratic equation x^{2} – 4x + 3 = 0. (1)

Answer:

x^{2} – 4x + 3 = 0

⇒ x^{2} – 3x – x + 3 = 0

⇒ x^{2} – 3x – x + 3 = 0

⇒ x(x – 3) – 1 (x – 3) = 0

⇒ x = 1, 3

So, roots are 1 and 3.

(vi) Find the possible root of \(\sqrt{3 x^{2}+6}\) = 9. (1)

Answer:

The given equation is \(\sqrt{3 x^{2}+6}\) = 9

Squaring on both sides, we get

3x^{2} + 6 = 81

⇒ 3x^{2} = 81 – 6

⇒ x^{2} = \(\frac{75}{3}\) = 25

Taking square root of both sides, we get

x = \(\sqrt{25}\) = ± 5

(vii) Draw two concetric circles of raddi 2 cm and 5 cm. Take a point P on the other circle and contsruct a pair of tangents PA and PB to the smaller circle. (1)

Answer:

(viii) Find the mid point of the line segment between points P(- 1, -5, 3), and Q(6, – 2). (1)

Answer:

Mid Points

= \(\left(\frac{-1.5+6}{2}, \frac{3-2}{2}\right)=\left(\frac{4.5}{2}, \frac{1}{2}\right)\)

Thus, mid point of the line segment is \(\left(\frac{4.5}{2}, \frac{1}{2}\right)\)

(ix) In ∆ABC, right angled at B, AB = 5 cm and ∠ACB = 30°. Determine the lengths of the sides BC and AC. (1)

Answer:

In right ABC, we have AB = 5 cm and ∠ACB = 30°.

∴ tan30° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{5}{B C}\)

⇒ BC = 5√3 cm

and sin 30° = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇒ \(\frac{1}{2}=\frac{5}{\mathrm{AC}}\)

⇒ AC = 5 × 2= 10 cm

Hence, BC = cm and AC = 10 cm.

(xi) Find the mode of the following frequency distribution : (1)

Answer:

The class interval 30 – 35 has maximum frequency. So, it is the modal class.

l = 30, f_{1} = 10, f_{0} = 9, f_{2} = 3,

(xii) A die is thrown once. Find the probability of getting ‘at most’2′?

Answer:

A die is thrown once, the possible numbers are : 1, 2, 3, 4, 5, 6 The numbers ’at most’ are = 1, 2 Let E be event of getting at most 2 Number of outcomes to favourable event E = 2

Section- B

Question 4.

Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. (2)

Answer:

Let a be any positive integer and 6 = 2.

Then, by Euclid’s division algorithm there exists integers q and r such that

a = 2q + r where 0 ≤ r < 2 r = 0, 1

For r = 0

a= 2q + 0 = 2q

For r = 1

a = 2q + 1

If a = 2q

It is even integer.

If a = 2q + 1

It is odd integer.

Question 5.

Divide x^{3} – 6x^{2} + 11x – 6 by x – 2 and verify the division algorithm. (2)

Answer:

Let f(x) = x^{3} – 6x^{2} + 11x – 6 and g(x) = x – 2

Now, we divide fix) by g(x), as follows:

Hence, q(x) = x^{2} – 4x + 3 and r(x) = 0.

According to division algorithm of polynomials.

f(x)- g(x) X q(x) + r(x)

⇒ x^{3} – 6x^{2} + 11x – 6 = (x – 2)x^{2} – 4x + 3) + 0

⇒ x^{3} – 6x^{2} + 11x – 6 = x^{3} – 4x^{2} + 3x – 2x^{2} + 8x – 6

⇒ x^{3} – 6x^{2} + 11x – 6

⇒ x^{3} – 6x^{2} + 11x – 6

∴ LHS = RHS

Hence, the division algorithm is verified.

Question 6.

For what value of k, will the following pair of equations have infinitely many solutions : 2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k + 1. (2)

Answer:

The given equations are :

2a + 3y – 7= 0 …(1)

And (k + 2)x – 3 (1 – k)y – (5k + 1) = 0 …(2)

The condition for infinitely many solutions is

⇒ -6(1 – k) = 3k + 6

and -10k – 2 = -7k – 14

⇒ -6 + 6k = 3k + 6

and -10k + 7k = -14 + 2

⇒ 3k = 12 and -3k = -12

⇒ k = 4 and k = \(\frac{-12}{-3}\) = 4

Hence, k = 4

Question 7.

Solve the Equation: \(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}\); Where x ≠ 1, 2 (2)

Answer:

We have,

⇒ 3x^{2} – 5a = 6x^{2} – 18x + 12

⇒ 6x^{2} – 18x + 12 – 3x^{2} + 5a = 0

⇒ 3x^{2} – 13x + 12 = 0

⇒ 3x^{2} – (9 + 4)x + 12 = 0

⇒ 3x^{2} – 9x – 4x + 12 = 0

⇒ 3x(x – 3) – 4(x – 3) = 0

⇒ (x – 3) (3x – 4) = 0

⇒ x = 3 and x = \(\frac{4}{3}\)

Question 8.

Find the sum of the first 20 terms of the following AP : 1, 4, 7, 10, …………… (2)

Answer:

The given sequence of AP is 1, 4, 7, 10, ……………

Here, a = 1, d= 4-1 = 3, n = 20 We know that

We Know that

S_{n} = \(\frac{n}{2}\)[2a + (n – 1)d]

= \(\frac{20}{2}\)[2 × 1 + (20 – 1) × 3]

= 10[2 + 57]

= 10 × 59 = 590

Question 9.

Find the sum of all two-digit natural numbers which are divisible by 4. (2)

Answer:

9. The all two digit numbers which are divisible by 4 are :

12, 16, 20, 96

Here, a = 12, a_{n} = l = 96, d = 16 -12 = 4

a_{n} = a + (n – 1)d

⇒ 96 = a + (n – 1 )d

⇒ 96 = 12 + (n – 1) × 4

⇒ 96 = 12 + 4ra – 4

⇒ 96 – 8 = 4n

⇒ 88 = 4n

Sum of first 22 terms = S_{22}

S_{22} = \(\frac{n}{2}\) [12 + 96]

= 11 × 108 = 1188

Hence, sum of all two digit numbers divisible by 4 = 1188.

Question 10.

Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{3}{4}\) of the corresponding sides of the triangle ABC (i.e of scale factor \(\frac{3}{4}\)). (2)

Answer:

Steps of construction :

- Draw a given triangle ABC.
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

- Along BX mark four points B
_{1}, B_{2}, B_{3}and B_{4}such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}. - Join B
_{4}C. - From B
_{3}, draw B_{3}C’ B_{4}C, meeting BC at C’. - From C’ draw C’A’ ∥ AC meeting AB at A’.

Then A’BC’ is required triangle each of whose side is of the corresponding sides of ∆ABC.

Question 11.

Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{5}{3}\) of the correspondirig sides of the triangle ABC (i.e. of scale factor \(\frac{1}{2}\)). (2)

Answer:

Steps of construction :

- Draw a given ∆ABC.
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Along BX mark 5 points B
_{1}, B_{2}, B_{3}, B_{4}and B_{5}such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}. - Join B
_{3}to C.

- From B
_{5}draw B_{5}C’ ∥ B_{3}C intersecting the produced line segment BC at C’. - From C’ draw C’A’ AC, intersecting the produced line segment BA at A’.

Then A’BC’ is the required triangle each of whose side is of the corresponding sides of ∆ABC.

Question 12.

Show that : tan 48° tan 23° tan 42° tan 67° = 1 (2)

Answer:

We have,

LHS = tan 48° tan 23° tan 42° tan 67°

= (tan 48° tan 42°) (tan 23° tan 67°)

= [tan (90° – 42°) tan 42°] [tan (90° – 67°) tan 67°]

= (cot 42° tan 42°) (cot 67° tan 67°)

= (cot 42° \(\frac{1}{\cot 42^{\circ}}\)) (cot 67°\(\frac{1}{\cot 67^{\circ}}\))

= 1 × 1 = 1 = RHS

Question 13.

Prove that: \(\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{{cosec} A-1}{{cosec} A+1}\) (2)

Answer:

We know,

Question 14.

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. (2)

Find the mode of the data.

Answer:

The class 4000 – 5000 has maximum frequency. So, it is the modal class.

l = 4000, f_{1} = 18, f_{0} =4, f_{2} = 9 and h = 1000.

Hence, mode = 4608.7 runs

Question 15.

To find out the concentration of SO_{2} in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: (2)

Find the mean concentration of SO_{2} in the air.

Answer:

From the table, we have

Σf_{i} = 30, Σf_{i}x_{i} = 2.96

∴ x̄ = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}\) = 0.099

Hence, the mean concentration of SO_{2} in the air = 0.099 ppm.

Question 16.

A child has a die whose six faces show the letters as shown below: (2)

The die is thrown once. What is the probability of getting (i) A? (ii) C ?

Answer:

In throwing the die any one of the given six faces may come upward.

The number of all possible outcomes = 6

⇒ n(S) = 6

(i) Since, there are two faces with letter A.

Let E1 be event of getting faces with letter A.

Number of favourable outcomes

n(E) = 2

∴ P(A) = \(\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(ii) Number of favourable outcomes to get C = 3

⇒ n(E) = 3

P(C) = \(\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

Section – C

Question 17.

If m times the mth term of an AP is equal to n times its nth term and m ≠ n, show that the (m + n)th term of AP is zero. (3)

Or

How many terms of the AP : 55; 51; 47; must be taken so that their sum is 405. (3)

Answer:

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

We have, mam = nan

⇒ m[a + (m – 1)d] = n[a + (n – 1 )d]

⇒ ma + m^{2}d – md = na + n^{2}d – nd

⇒ m^{2}d – n^{2}d – md + nd = na – ma

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

⇒ d[m^{2} – n^{2} – (m – n)] = a(n – m)

⇒ d[(m + n) (m – n) – (m – n)] = a(m – n)

⇒ d[(m – n) (m + n – 1)] = – a(m – n)

⇒ d(m + n – 1) = \(\frac{-a(m-n)}{(m-n)}\)

⇒ d(m + n – 1) = – a

⇒ d = – \(\frac{a}{m+n-1}\)

Now, a_{(m+n)} = a + (m + n – 1)d

= a + {(m + n – 1) × \(\frac{-a}{(m+n-1)}\)}

= a – a = 0

Proved

Question 18.

Prove that the points (2, – 2), (- 2, 1) and (5, 2) are the vertices of a right angled triangle. Also, find the area of this triangle. (3)

Or

The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices. (3)

Answer:

Let the points A (2, -2), B (-2, 1) and C (5, 2) are the vertices of right angled triangle.

AB^{2} = (- 2 – 2)^{2} + (1 + 2)^{2}

= 16 + 9 = 25

BC^{2} = (5 + 2)^{2} + (2 – 1)^{2} = 49 + 1 = 50

And AC^{2} = (5 – 2)^{2} + (2 + 2)^{2} = 9 + 16 = 25

AB^{2} + AC^{2} = 25 + 25 = 50 = BC^{2}

Since, BC^{2} = AB^{2} + AC^{2},

So by converse of phythagores theorem ∠A = 90° there fore point A (2, – 2), B (- 2,1) and C (5, 2) are the vertices of a right angled triangle

Here, x_{1} = 2, y_{1} = -2, x_{2} = -2 , y_{2} = 1, x_{3} = 5, y_{3} = 2

Area of ∆ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}]

= \(\frac{1}{2}\)[2(1 – 2) + (-2)(2 + 2) + 5(-2 – 1)]

= \(\frac{1}{2}\) [2 x (- 1) + (- 2) x 4 + 5 x (- 3)]

= \(\frac{1}{2}\)[-2 – 8 – 15] = \(\frac{1}{2}\)[- 25] = \(\frac{-25}{2}\)

Hence, area of ∆ = \(\frac{25}{2}\) sq. units.

Question 19.

If 1 + sin^{2}θ = 3 sinθ cos θ, prove that tan θ = 1 or \(\frac{1}{2}\). (3)

Or

If the sum of 20 terms of an AP is 860 and its first term is 5, then find the 17th term. (3)

Answer:

We have, 1 + sin^{2}θ = 3 sin θ cos θ

⇒ 1 + tan 2θ + tan 2θ = 3 tanθ

⇒ 2 tan 2θ – 3 tan θ + 1 = 0

⇒ 2 tan 2θ – (2 + 1) tan θ + 1 = 0

⇒ 2 tan 2θ – 2 tan θ – tan θ + 1 = 0

⇒ 2 tan θ (tan θ -1) -1(tan θ – 1) = 0

⇒ (tan θ – 1) (2 tan θ – 1) = 0

⇒ tan θ – 1 = 0 or 2 tan θ – 1 = 0

⇒ tan θ = 1 or tan θ = \(\frac{1}{2}\)

Question 20.

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. (3)

Or

The frequency distribution table of agriculture landing in a village is given below : (3)

Answer:

Here, h = 3

From the table, we have a = 755, h = 3, Σf_{i} = 30, Σf_{i}u_{i} = 4

∴ x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h

⇒ x̄ = 75.5 + \(\left(\frac{4}{30}\right)\) × 3

⇒ x̄ = 755 + 0.4 = 75.9

Hence, mean heart beats per minute = 75.9.

Section – D

Question 21.

Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x axis and shade the triangular region. (4)

Or

Solve graphically the system of equations 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0 (4)

Answer:

We draw the graph of given linear equations as follows :

x – y + 1 = 0

⇒ y = x + 1

We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for equation x – y + 1 = 0

Table 1

and 3x + 2y – 12 = 0

⇒ 2y = 12 – 3x ⇒ y = \(\frac{12-3 x}{2}\)

∴ We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for equation 3x + 2y – 12 = 0

Table 2

Now, we plot the points (2, 3), (4, 5) and (5, 6) on the graph paper and we draw a graph which passes through these points.

∴ We get a graph of linear equation x – y + 1 = 0.

Again, we plot the points (0, 6), (2, 3) and (4, 0) on the graph paper and we draw a graph, which passes through these points.

∴ We get a graph of linear equation

3x + 2y – 12 = 0.

We observe that the graphs of linear equations x – y + 1 = 0 and 3x + 2y – 12 = 0 intersect at point (2, 3). The graphs of linear equations meet x-axis at points B and C. So, the coordinates of iABC formed are A(2, 3), B(-1, 0) and C(4, 0).

Question 22.

Construct a ∆ABC with AB = 6 cm, BC = 5 cm and ∠B = 60°. Now construct another triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of ∆ABC. (4)

0r

Draw a circle with centre O and radius equal to 3 cm. Take two points P and Q on one side of its extended diameter each at distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. (4)

Answer:

Steps of Construction :

- Draw a line segment BC = 5 cm.
- At B, draw ∠CBY = 60°.
- From B, draw and arc AB = 6 cm meeting by at A.

- Join AC. Thus, ∆ABC obtained
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Along BX mark 3 points B
_{1}; B_{2}, and B_{3}such that BB_{1}= B_{1}B_{2}= B_{2}B_{3} - Join B
_{3}C - From B
_{2}draw B_{2}C’ ∥ B_{3}C meeting BC at C’. - From C’ draw A’C’ ∥ AC meeting AB at A’.

Then A’ BC’ is required triangle, each of whose side is \(\frac{2}{3}\)of corresponding sides of ∆ABC

Jutification: Since A’C’ ∥ AC Therefore, A’BC’ in ∆ABC

\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{2}{3}\)

Question 23.

The following table gives production on yield per hectare of wheat of 100 farms of a village.

Change the distribution of a more than type distribution and draw its ogive. (4)

Or

The median of the following data is 50. Find the values of p and q, if sum of all the frequencies is 90 : (4)

Answer:

We prepare the cumulative frequency table by more than type method as given :

\(\frac{n}{2}=\frac{100}{2}\) = 50

We plots the points (40, 100), (45, 96), (50, 90), (55, 74), (60, 54) and (65, 24) on the graph paper. Joining these points with a free hand to obtain more than type ogive curve as shown in a graph.

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