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 x2 + 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 x2 – 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 t1 = 2, t2 = 3 and tn = tn-1 + tn-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 A1; A2, A3, …. are located at equal distances on the ray AX and the point B is joined to : (1)
(a) A12
(b) A11
(c) A10
(d) A0
Answer:
(b) A11
(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 a1x + b1y + c1 = 0 and a2x + b2y + c2 = 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 ui = \(\frac{x_{i}-5}{14}\) Σfiui =35, Σfi =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 x2 – 3 = 0.
Answer:
We have, x2 – 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, ax2 + bx2 + 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, a1 = 1, b1 = 1, c1 = – 5 and a2 = 2, b2 = 2, c2 = – 10
Therefore, the given linear equations are coincident lines. So, the equations are consistent.
(v) Find the zeroes of the qufratic equation x2 – 4x + 3 = 0. (1)
Answer:
x2 – 4x + 3 = 0
⇒ x2 – 3x – x + 3 = 0
⇒ x2 – 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
3x2 + 6 = 81
⇒ 3x2 = 81 – 6
⇒ x2 = \(\frac{75}{3}\) = 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, f1 = 10, f0 = 9, f2 = 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 x3 – 6x2 + 11x – 6 by x – 2 and verify the division algorithm. (2)
Answer:
Let f(x) = x3 – 6x2 + 11x – 6 and g(x) = x – 2
Now, we divide fix) by g(x), as follows:
Hence, q(x) = x2 – 4x + 3 and r(x) = 0.
According to division algorithm of polynomials.
f(x)- g(x) X q(x) + r(x)
⇒ x3 – 6x2 + 11x – 6 = (x – 2)x2 – 4x + 3) + 0
⇒ x3 – 6x2 + 11x – 6 = x3 – 4x2 + 3x – 2x2 + 8x – 6
⇒ x3 – 6x2 + 11x – 6
⇒ x3 – 6x2 + 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,
⇒ 3x2 – 5a = 6x2 – 18x + 12
⇒ 6x2 – 18x + 12 – 3x2 + 5a = 0
⇒ 3x2 – 13x + 12 = 0
⇒ 3x2 – (9 + 4)x + 12 = 0
⇒ 3x2 – 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
Sn = \(\frac{n}{2}\)[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, an = l = 96, d = 16 -12 = 4
an = 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 = S22
S22 = \(\frac{n}{2}\) [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 B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.
- Join B4C.
- From B3, draw B3C’ B4C, 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 B1, B2, B3, B4 and B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
- Join B3 to C.
- From B5 draw B5C’ ∥ B3C 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, f1 = 18, f0 =4, f2 = 9 and h = 1000.
Hence, mode = 4608.7 runs
Question 15.
To find out the concentration of SO2 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 SO2 in the air.
Answer:
From the table, we have
Σfi = 30, Σfixi = 2.96
∴ x̄ = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}\) = 0.099
Hence, the mean concentration of SO2 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 + m2d – md = na + n2d – nd
⇒ m2d – n2d – md + nd = na – ma
⇒ d(m2 – n2) – d(m – n) = a(n – m)
⇒ d[m2 – n2 – (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.
AB2 = (- 2 – 2)2 + (1 + 2)2
= 16 + 9 = 25
BC2 = (5 + 2)2 + (2 – 1)2 = 49 + 1 = 50
And AC2 = (5 – 2)2 + (2 + 2)2 = 9 + 16 = 25
AB2 + AC2 = 25 + 25 = 50 = BC2
Since, BC2 = AB2 + AC2,
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, x1 = 2, y1 = -2, x2 = -2 , y2 = 1, x3 = 5, y3 = 2
Area of ∆ABC
= \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2]
= \(\frac{1}{2}\)[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 + sin2θ = 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 + sin2θ = 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, Σfi = 30, Σfiui = 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 B1; B2, and B3 such that BB1 = B1B2 = B2B3
- Join B3C
- From B2 draw B2C’ ∥ B3C 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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