Cbse Sample Papers For Class 11 Maths Mock Paper 1

Q1. The equation of the line which makes equal intercepts on the axes and passes through the point (4, 5) is

Opt: x + y = 3. x + y = 5. x + y = 7. x + y = 9.

Ans: x + y = 9.
Q2. Different words have been formed with all the letters of the word EQUATION (use each letter exactly once) so that the words begin and end with a consonant. The number of such words, is

Opt:

40320.

4320.

1440.

720.

Ans: 4320.

Q3. If ‘P(n) = n² – n + 41’ is prime, then it is not true for

Opt:

P(41).

P(40).

P(39).

P(38).

Ans: P(41).

Q4.

Opt: a purely imaginary number. a purely real number.

Ans: a purely real number.

Q5.

Opt:

Ans:

Q6. The foot of the perpendicular from the origin to a straight line lies at the point (7, -5). Then the equation of the line is

Opt:

7x – 5y  + 74 = 0. 7x + 5y + 74 = 0. 7x + 5y – 74 = 0. 7x – 5y – 74 = 0.

Ans: 7x – 5y – 74 = 0.
Q7. The ratio in which xy-plane divides the join of (1, 2, 3) and (4, 5, 6) is

Opt:

3:1 internally. 3:1 externally. 2:1 internally. 1:2 externally.

Ans: 1:2 externally.
Q8.

Opt:

1. 3. 6. 9.

Ans: 6.
Q9. The number of real solutions of the equation 2sin² x + 7 sin x + 3 = 0 in the interval [0, 4π] is

Opt:

0.

1.

2.

3.

Ans: 2.

Q10. The set (A ∪ B ∪ C) ∩ (A ∩ B^C ∩ C^C)^C ∩ C^C is equal to

Opt:

A ∩ C^C.

A ∩ C.

B ∩ C^C.

B ∩ C.

Ans: B ∩ C^C.

Q11. The expression

1 – sinα1 + sinα+1 + sinα1 – sinα

is equal to

Opt:

1/cos α.

cos α.

2/cos α.

–1/cos α.

Ans: 2/cos α.

Q12. The probability that a student will receive A, B, C or D grade is 0.40, 0.35, 0.15 and 0.10 respectively. The probability that a student will receive C or D grade is

Opt:

0.10.

0.25.

0.40.

0.50.

Ans: 0.25.

Q13.

Opt:

-2.

0.

1/2.

∞.

Ans: -2.

Q14.

The domain of the real function f defined byf(x)=x2+2x+1x3+6x2+12x+8 is

Opt:

{2}.

(- ∞. ∞).

R-{-2}.

R-{2}.

Ans: R-{-2}.

Q15. A set of lines ax + by + c = 0, where 5a + 7b + 4c = 0 is
concurrent at the point

Opt:

(5/4, 3/4). (5/4, 1/2). (5/4, 7/4). (3/4, 1/2).

Ans: (5/4, 7/4).

Q16.

Opt:

¹⁸C₆.2⁶.

¹⁸C₆.2⁵.

¹⁸C₆.2⁴.

¹⁸C₆.2².

Ans: ¹⁸C₆.2⁶.

Q17.

Opt:

isosceles right-angled triangle.

equilateral triangle.

acute-angled scalene triangle.

obtuse-angled triangle.

Ans: equilateral triangle.

Q18. If the arithmetic mean of a distribution is 20 and the standard deviation is 7, then the coefficient of variation is

Opt:

35.

30.

27.

14.

Ans: 35.

Q19. ASSERTION – REASON BASED QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(c) A is true but R is false.

(d) A is false but R is true.

​Assertion (A): If (12, x), (13, y), (14, z) are in A × B and n(A) = 3 and n(B) = 3, then A = {12, 13, 14} and B = {x, y, z}.

Reason (R): If n(A) = p and n(B) = q then n(A × B) = pq.

Ans: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:

We know that if n(A) = p and n(B) = q then n(A × B) = pq.

If (12, x), (13, y), (14, z) are in A × B, this means the first element of each ordered pair belongs to A and the second element belongs to B.

Thus, A = {12, 13, 14} and B = {x, y, z}.

Hence, n(A) = 3 and n(B) = 3.

Q20. If 2x + i (x – y) = 5, where x and y are real numbers, find the values of x and y.

Ans:

We have  2x+ i(x – y) = 5
or            2x+ i(x – y) = 5 + 0.i
Comparing the real and imaginary parts, we get
2x = 5 and x – y = 0
⇒   x = 5/2 and x = y
Thus x = y = 5/2.

Q21. Write the negation of each of the following statements:

(i) For every natural number x, x + 0 = x = 0 + x.
(ii) There exists a capital for every state in India.

Ans:

(i) There exists a natural number x such that x + 0 ≠ x = 0 + x.
(ii) There exists a state in India which does not have its capital.

Q22. Find the equation of the parabola with focus (2, 0) and directrix x = –2.

Ans: Since the focus lies on the x-axis, thus x-axis it self is the axis of the parabola.
Since the directrix is x = –2 and the focus is (2, 0), the parabola is to be of the form
y² = 4ax with a = 2.
Hence the required equation is : y² = 4(2)x = 8x.

Q23. Find the integral value of x:7x−5≤3x+7

Ans:

We have,7x−5≤3x+7⇒7x−3x≤7+5⇒4x≤12⇒x≤3⇒x=−∞…−3,−2,−1,0,1,2,3

Q24. Find the equation of the straight line which makes an angle of 60° with the positive direction of x-axis and cuts an intercept of 6 units on the y – axis.

Ans: We have, m = tan 60° = (1/√3) and c = + 6.
The equation of the line is given by : y = mx + c Putting values , y = (1/√3)x + 6
(√3)y = x + 6√3
x – (√3)y + 6√3 = 0

Q25. Find the term independent of x in the expansion of

x+1x6.

Ans:

Suppose (r+1) th term is independent of x in the expansion of x+1x6.Tr+1=6Crx6−r1xrTr+1=6Crx6−r−rTo be independent of x, we have6−2r=0⇒r=3∴T3+1=6C3x6−31x3=20Hence, 4th term is independent of x1 i.e., 20.

Q26. Show that r does not have integral value for 65Pr=76Pr−1.

Ans:

We have 65Pr=76Pr−1or 6×5!(5−r)!=7×6!(6−r+1)!or6!(5−r)!=7×6!(6−r+1)(6−r)(6−r−1)!or (7−r)(6−r)=7or r2−13r+35=0, which does not give integral value of r.

Q27. A coin is tossed and a die is thrown. Find the probability that the outcome will be a head and a number greater than 4.

Ans:

Probability (head) = 1/2
Probability (number greater than 4) = 2/6 = 1/3  ( since there are two numbers greater than 4 in a die i.e., 5 and 6)
Required probability = 1/2 × 1/3 = 1/6.

Q28. A, B and C are mutually exclusive and exhaustive events associated with a random experiment.
Find P(A) if P(B) = (5/2)P(A) and P(C) = (3/2)P(A).

Ans:

Let P(A) = p
P(B) = (5/2) P(A)
= (5/2)P
and
P(C) = (3/2)P(A)
= (3/2)P
Since A, B and C are mutually exclusive and exhaustive events associated with a random experiment, thus A ∪ B ∪ C = S
P( A ∪ B ∪ C)  =  P(S) = 1
P(A) + P(B) + P(C) = 1
p + (5/2)p + (3/2)p = 1
(10/2)p =1
p = 1/5
P(A) =  p
= 1/5.

Q29. If three angles A, B, C, are in A.P. Prove that:

cotB=sinA–sinCcosC–cosA

Ans:

R.H.S=sinA–sinCcosC–cosA =2sinA–C2cosA+C22sinA+C2sinA–C2 =cotA+C2=cotB=L.H.S ∵A,B,C are in A.P ∴ 2B=A+C

Q30. Prove that :

tan 3A·tan 2A·tan A=tan 3A–tan 2A–tan A

Ans:

We have, 3A=2A+A⇒tan 3A=tan2A+A⇒tan 3A=tan 2A+tan A1–tan 2A·tan A⇒tan 3A1–tan 2A·tan A=tan 2A+tan A⇒tan 3A–tan 3A·tan 2A·tan A=tan 2A+tan A⇒tan 3A–tan 2A–tan A=tan 3A·tan 2A·tan A

Q31. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Ans: The intersection point of two lines 2x – y – 3 = 0 and 3x + y – 2 = 0
can be obtained by solving these equations :  
3x + y – 2 = 0
2x – y – 3 = 0
———————–
5x  = 5
x = 1 and y = 2 – 3x = 2 – 3 = –1.
Since the point (1, –1) lies on the line px + 2y – 3 = 0,
it will satisfy the equation of the line.
p(1) + 2(–1) – 3 = 0
p – 2 – 3 = 0
p = 5

Q32. Find the equation of a circle which is concentric to the given circle x² + y² – 4x – 6y – 3 = 0 and which touches the x axis.

Ans: The centre of the given circle is given by
(–g, –f) = (–1/2 coff. of x, –1/2 coff. of y) = (2,3) 
The centre of the given circle is (2,3). The centre of concentric circles are same , therefore the centre of the required circle is (2,3).
Since the circle touches the x-axis therefore the radius of the required circle is 3 units.
The equation of the required circle is
(x – 2)² + (y – 3)² = 3²
⇒ x² + y² – 4x – 6y + 4 = 0

Q33.

Solve (2x−1)3≥(3x−2)4−(2−x)5.

Ans:

We have, (2x−1)3≥(3x−2)4−(2−x)5 or(2x−1)3≥{5(2x−2)−4(2−x)}20⇒20(2x−1)≥3(19x−18)⇒34≥17xTherefore, 2≥x or x≤2Hence, the solution set is (2,∞).

Q34. Prove that: 1 + 2 + 3 +…+ n = n(n + 1)/2 for all n ∈ N.

Ans:

Let Pn=1+2+3+……+n=nn+12P1=1=11+12⇒1=1⇒P1 is True.Assuming Pk is True.Pk=1+2+3+……+k=kk+12Now Pk+1=1+2+3+……+k+k+1=k+1k+22 =kk+12+k+1 =k2+k+2k+22 =kk+1+2k+12 =k+1k+22⇒Pk+1 is TrueHence by principle of mathematical induction Pn is true for all n∈N.

Q35. Find the domain of the function

fx=x2+2x+1x2–7x+12.

Ans:

Since, x² – 7x +12 = (x – 4)(x – 3), the function is defined for all real numbers
except x = 4 and x = 3. Hence, the domain of f is R – {3, 4}.

Q36. The following is the record of goals scored by team A in a football session:

No. of goals scored	0	1	2	3	4
No. of matched	1	9	7	5	3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Ans:

Mean=∑i=1n(f)i(x)in=5025=2Variation((σ)2)=∑i=1n(f)i(((x)i−xˉ))2n=3025=1.2Standarddeviation=1.2=1.09StandarddeviationofteamB=1.25goalsMeanofteamB=2Theteamwithlowerstandarddeviationismoreconsistent.Therefore,teamAismoreconsistent.{Mean = } \frac{\sum_{{i=1}}^{{n}} \left({f}\right)_{{i}} \left({x}\right)_{{i}}}{{n}} \\ { = } \frac{{50}}{{25}} \\ { = 2} \\ {Variation} \left(\left({σ}\right)^{{2}}\right) \\ {= } \frac{\sum_{{i=1}}^{{n}} \left({f}\right)_{{i}} \left(\left(\left({x}\right)_{{i}} {-} \bar{{x}}\right)\right)^{{2}}}{{n}} \\ {= } \frac{{30}}{{25}} \\ {= 1.2} \\ {Standard\ deviation} \\ {= } \sqrt{{1.2}} \\ {= 1.09} \\ {Standard\ deviation\ of\ team\ B} \\ {= 1.25\ goals} \\ {Mean\ of\ team\ B} \\ {= 2} \\ {The\ team\ with\ lower\ standard deviation\ is\ more\ consistent.} \\ {Therefore,\ team\ A\ is\ more\ consistent.}

Mean=n∑i=1n​(f)i​(x)i​​=2550​=2Variation((σ)2)=n∑i=1n​(f)i​(((x)i​−xˉ))2​=2530​=1.2Standard deviation=1.2​=1.09Standard deviation of team B=1.25 goalsMean of team B=2The team with lower standarddeviation is more consistent.Therefore, team A is more consistent.

Q37. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Ans:

Number of given digits = 5
Number of digits in each number = 4
Number of 4-digit numbers = ⁵P₄
= 5!
= 120
Now, we find 4 digit even numbers.
Only 2 numbers are possible at units place (2, 4) as we need even number.
If no digit is repeated, then number of ways to fill remaining three places of 4-digit numbers = ⁴P₃
= 4! /(4 – 3)!
= 4! / 1!
= 24
Number of ways to arrange 2 even numbers at one’s place = 2
Total formed 4-digit even numbers = 24 x 2
= 48
Thus, there are 48, 4 digit even numbers.

Q38. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Ans:

$\begin{array}{l}\mathrm{Number}{ of vowels in the English alphabets}=5\\ \mathrm{Number}{ of consonants in the English alphabets}=21\\ \mathrm{Number}{ of words formed by 2 vowels and 2 consonants}\\ ={{\hspace{0.17em}}}^{5}{\mathrm{C}}_{2}\times {{\hspace{0.17em}}}^{21}{\mathrm{C}}_{2}\\ ={\hspace{0.17em}}\frac{5!}{2!(5-2)!}\\ \times \frac{21!}{2!(21-2)!}\\ =\frac{5!}{2!3!}\times \frac{21!}{2!{\hspace{0.17em}}19!}\\ =10\times 210\\ =2100\\ \mathrm{These}{ 2100 words has 4 letters only. }\\ {Number of ways to arrange 4 letters}={{\hspace{0.17em}}}^{4}{\mathrm{P}}_{4}\\ =\frac{4!}{(4-4)!}\\ =4!=24\\ \mathrm{So},{ the number of total words formed by 2 vowels and }\\ {2 consonants}=2100\times 24\\ {\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}                  \hspace{0.17em}}=50400\end{array}$

Number of vowels in the English alphabets=5Number of consonants in the English alphabets=21Number of words formed by 2 vowels and 2 consonants=5C2​×21C2​=2!(5−2)!5!​×2!(21−2)!21!​=2!3!5!​×2!19!21!​=10×210=2100These 2100 words has 4 letters only. Number of ways to arrange 4 letters=4P4​=(4−4)!4!​=4!=24So, the number of total words formed by 2 vowels and 2 consonants=2100×24                  =50400​

Q39. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Ans:

Let Volume of added water to acid= x litres Volume of 45% solution of acid= 1125 litres Total solution of acid=1125+x litresTherefore,45% of1125>25%1125+xand45% of1125<30%1125+xor45100×1125>25100×1125+xand45100×1125<30100×1125+xor45×1125>25×1125+xand445×1125<30×1125+xor50625>28125+25xand50625<33750+30xor50625−28125>25xand50625−33750<30xor22500>25xand16875<30xor900>xand562.5<x⇒ x∈562.5,900Thus, added water should be more than 562.5 litres but less than 900 litres.

Q40.

Ans:

Q41. In a school, selection of 11 players cricket team is to be done for the upcoming interschool cricket championship from the existing 17 players in the school cricket team. Rohit is the captain and fast bowler of the team so his place in team is reserved.

Based on this information answer the following questions:

(i) In how many ways team selection can be done?

(ii) If Soham is an all-rounder player, so he must be in the team, then find the number of ways team selection can be done?

(iii) If in team 6 players can bowl and the team must include exactly 4 bowlers, then find the number of ways team selection can be done?

Ans:

(i) Given Rohit’s place in the team is reserved.
⸫ The remaining 10 players are selected from 17 – 1 = 16 players.
⸫ The required number of ways = ¹⁶C₁₀ = 8008

(ii) Given Rohit is the captain and his place in the team is reserved and Soham must be in the team,
⸫ The remaining 9 players are selected from 17 – 2 = 15 players.

⸫ The required number of ways = ¹⁵C₉ = 5005

(iii)

​Given Rohit is captain and fast bowler and his place in the team is reserved.
Also, it is given that out of 17 players, 6 players can bowl.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.

∴ 3 bowlers can be selected out of remaining 5 bowlers in ⁵C₃ ways.
And, the remaining 7 players are selected out of 11 players in ¹¹C₇ ways.
Thus, by multiplication principle, the required number of ways of
selecting cricket team is

=5C3×11C7=5!3!(5−3)×11!7!(11−7)!=5!3!2!×11!7!4!=3300\begin{array}{l}{\rm{ = }}\;{}^{\rm{5}}{{\rm{C}}_{\rm{3}}}\;{\rm{ \times }}\;{}^{{\rm{11}}}{{\rm{C}}_{\rm{7}}}\;{\rm{ = }}\frac{{5!}}{{3!(5 – 3)}} \times \frac{{11!}}{{7!(11 – 7)!}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{5!}}{{3!2!}} \times \frac{{11!}}{{7!4!}} = 3300\end{array}

=5C3​×11C7​=3!(5−3)5!​×7!(11−7)!11!​=3!2!5!​×7!4!11!​=3300​​​

Q42. In a school, during health check-up of class XI students, age, weight, and height of 5 students were represented by sets A, B and C respectively as follows:

A = {16 yr, 16.25 yr, 16.5 yr, 16.75 yr, 17 yr}, B = {171 cm, 171.5 cm. 172 cm, 172.5 cm, 173 cm} and C = {53 kg, 53.25 kg, 53.5 kg, 53.75 kg, 54 kg},

Based on the above information answer the following questions:

(i) Elements from set B are related to the elements of set C as shown in arrow diagram. Identify that, is the given relation formed between weights and heights is a function? If yes, identify the function.

​(ii) What are the numbers of non-empty relations from set A to set B?

Ans:

(i) Since, in the given arrow diagram for each element of set A there is a unique image in set C,

Hence, the given relation is a function.

Let the function be, y = f(x).

The ordered pairs of the function are, (53, 173), (53.25, 172.5), (53.5, 172), (53.75, 171.5), (54, 171).

The above ordered pairs have same difference between the elements of elements. So, corresponding to linear change in the weight, the height varies linearly. Hence it is a linear function. Let the function, y = mx + c. So, satisfying any two ordered pairs to find m and c.

Using, (53, 173) : 173 = 53m + c              …(1)

Using, (54, 171) : 171 = 54m + c              …(2)

Subtracting (1) from (2) we get m = – 2 and by substituting m = 2 in (1) we get c = 279.

So, the function f(x) = – 2x + 279.

(ii)

Since,A={16,16.25,16.5,16.75,17},B={171cm,171.5cm,172cm,172.5cm,173cm},⇒n(A)=5,n(B)=5,∴NumberofsubsetsofA×B=25×5=225⇒NumberofrelationsfromsetAtosetB=225,As,nullsetϕisanemptysubsetofA×BaswellemptyrelationfromsetAtosetB,∴Numberofnon−emptyrelationsfromsetAtosetB=225−1.\begin{array}{l}{\rm{Since, A = }}\left\{ {16,\;16.25,\;16.5,\;16.75,\;17} \right\}{\rm{,}}\;{\rm{B = }}\left\{ {171{\rm{ cm,}}\;171.5{\rm{ cm}},\;172{\rm{ cm,}}\;172.5{\rm{ cm}},\;173{\rm{ cm}}} \right\}{\rm{, }}\\ \Rightarrow {\rm{ n}}\left( {\rm{A}} \right){\rm{ = 5,}}\;{\rm{n}}\left( {\rm{B}} \right){\rm{ = 5,}}\\\therefore {\rm{ Number}}\;{\rm{of}}\;{\rm{subsets}}\;{\rm{of}}\;{\rm{A}}\; \times \;{\rm{B}}\;{\rm{ = }}\;{{\rm{2}}^{5\; \times \;5}} = {2^{25}}\\ \Rightarrow {\rm{ Number}}\;{\rm{of}}\;{\rm{relations}}\;{\rm{from}}\;{\rm{set}}\;{\rm{A}}\;{\rm{to}}\;{\rm{set}}\;{\rm{B}}\;{\rm{ = }}\;{{\rm{2}}^{25}},\\{\rm{ As,}}\;{\rm{null}}\;{\rm{set}}\;\phi \;{\rm{is}}\;{\rm{an}}\;{\rm{empty}}\;{\rm{subset}}\;{\rm{of}}\;{\rm{A}}\;{\rm{ \times }}\;{\rm{B}}\;{\rm{as}}\;{\rm{well}}\;{\rm{empty}}\;{\rm{relation}}\;{\rm{from}}\\{\rm{ set}}\;{\rm{A}}\;{\rm{to}}\;{\rm{set}}\;{\rm{B,}}\\\therefore {\rm{ Number}}\;{\rm{of}}\;{\rm{non – empty}}\;{\rm{relations}}\;{\rm{from}}\;{\rm{set}}\;{\rm{A}}\;{\rm{to}}\;{\rm{set}}\;{\rm{B}}\;{\rm{ = }}\;{{\rm{2}}^{25}}{\rm{ – }}1.\end{array}

Since,A={16,16.25,16.5,16.75,17},B={171cm,171.5cm,172cm,172.5cm,173cm},⇒n(A)=5,n(B)=5,∴NumberofsubsetsofA×B=25×5=225⇒NumberofrelationsfromsetAtosetB=225,As,nullsetϕisanemptysubsetofA×BaswellemptyrelationfromsetAtosetB,∴Numberofnon−emptyrelationsfromsetAtosetB=225−1.​​​​