Maharashtra State Board Class 12th Maths Sample Paper Set 2 with Solutions Answers Pdf Download.

## Maharashtra Board Class 12 Maths Model Paper Set 2 with Solutions

**Section A**

Question 1.

Select and write the most appropriate answer from the given alternatives for each questions.

(i) If p ∧ q is F. p → q is F then the truth values of p and q are ………………

(a) T.T

(b) T.F

(c) ET

(d) EF

Answer:

(b) T.F

(ii) The principal solutions of equation cot θ = √3 are:

(a) \(\frac{\pi}{6} \cdot \frac{7 \pi}{6}\)

(b) \(\frac{\pi}{6}, \frac{5 \pi}{6}\)

(c) \(\frac{\pi}{6}, \frac{8 \pi}{6}\)

(d) \(\frac{7 \pi}{6}, \frac{\pi}{3}\)

Answer:

(a) \(\frac{\pi}{6} \cdot \frac{7 \pi}{6}\)

(iii) If acute angle between tines ax² + 2hxy + by² = 0 is, π/4, then 4h² =

(a) a² + 4ab + b²

(b) a² + 6ab + b²

(c) (a + 2b)(a + 3b)

(d) (a – 2b)(2a + b)

Answer:

(b) a² + 6ab + b²

(iv) The direction ‘ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}\) = \(\frac{y+17}{-3}\) = \(\frac{z-6}{1}\) and \(\frac{x+5}{1}\) = \(\frac{y+3}{2}\) = \(\frac{z-4}{-2}\) are:

(a) 4,5, 7

(b) 4, – 5, 7

(c) 4, – 5, – 7

(d) – 4, 5,8

Answer:

(a) 4,5, 7

v) If LPP has two optimal solutions, then it has infinite number of optimal solutions

(vi) Let f(x) and g(x) be differentiable for 0 ≤ x ≤ 1 such that f(0) = 0, g(0) = 0. f(1) = 6. Let there exists a real number c in (0,1) such that f'(c) = 2g'(c). Then the value of g(1) must be:

(a) 1

(b) 3

(c) 2.5

(d) -1

Answer:

(b) 3

(vii) \(\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^2 x . d x\) =

(a) \(\frac{7 \pi}{256}\)

(b) \(\frac{3 \pi}{256}\)

(c) \(\frac{5 \pi}{256}\)

(d) \(\frac{-5 \pi}{256}\)

Answer:

(c) \(\frac{5 \pi}{256}\)

(viii) The area bounded by the curve y = x³, the X-axis and the lines x = – 2 and x = 1 is:

(a) -9 sq. units

(b) \(\frac{-15}{4}\) sq. units

(c) \(\frac{15}{4}\) sq.units

(d) \(\frac{17}{4}\)sq.units

Answer:

(c) \(\frac{15}{4}\) sq.units

Question 2.

Answer the following questions:

(i) Check whether the following matrix is invertible or not : \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\)

Solution:

Let A = \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\)

Then, |A| = \(\left|\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right|\) = 1 – 0 = 1 ≠ 0

∴ A is a non-singular matrix.

Hence, A^{-1} exists.

(ii) State whether the equation 2 sin θ = 3 has a solution or not?

Solution:

2 sin θ = 3

∴ sin θ = \(\frac{3}{2}\)

This is not possible because – 1 ≤ sin θ ≤ 1 for any θ.

∴ 2 sin θ = 3 does not have any solution.

(iii) Elvaluate: \(\int_0^{\frac{\pi}{4}} \cot ^2 x \cdot d x\)

Solution:

\(\int_0^{\frac{\pi}{4}} \cot ^2 x \cdot d x\)

(iv) Determine the order and degree of the foll.owirig differential. equation:

\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)

Solution:

The given Differential equation is

\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)

The D.E. has highest order derivative — with power 2.

∴ \(\left(\frac{d y}{d x}\right)^2\) = 2 sin x + 3

The D.E has highest order derivative \(\frac{d y}{d x}\) with power 2.

∴ The given D.E. is of order 1 and degree 2.

**Section B**

**Attempt any EIGHT of the following questions:**

Question 3.

Examine whether the following statement pattern is a tautology or o contradiction or a .contingency.

(p V q) → (q ∧ p).

Solution:

All the entries in the last column of the above truth table are T.

∴ (p ∧ q) → (q V p) is a tautology.

Question 4.

Transform \(\left[\begin{array}{ccc}

1 & -1 & 2 \\

2 & 1 & 3 \\

3 & 2 & 4

\end{array}\right]\) into an upper trianguLar matrix by suitable row transformations.

Solution:

Given matrix is :

\(\left[\begin{array}{ccc}

1 & -1 & 2 \\

2 & 1 & 3 \\

3 & 2 & 4

\end{array}\right]\)

Question 5.

Find k, if the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product.

Answer:

Comparing the equation x² + kxy- 3y2 = 0 with ax² + 2hxy – by² = 0, we get a = 1.2h = k. b = – 3.

Let m_{1} and m_{2} be the slopes of the lines represented by x² + kxy – 3y2 = 0

∴ m_{1} + m_{2} = \(\frac{-2h}{b}\) = \(– \frac{k}{-3}\) = \(\frac{k}{3}\)

and m_{1} m_{2} = \(\frac{a}{b}\) = \(\frac{1}{-3}\) = \(– \frac{1}{3}\)

Now, m_{1} + m_{2} = 2(m_{1} m_{2})

∴ \(\frac{k}{3}\) = 2(\(\frac{1}{3}\))

∴ k = -2

Question 6.

Find the separate equation of the tine represented bg the following equation: 3x² – 2√3xy – 3y² = 0.

Answer:

3x² – 2√3xy – 3y² = 0

∴ 3x² – 3√3xy + √3xy – 3y² = 0

∴ 3x(x – √3y) + √3y(x – √3y) = 0

∴ (x – √3y)(3x +√3y) = 0

The separate equations of the lines are

x – √3y = 0 and 3x + √3y = 0

Question 7.

If \(\overrightarrow{A B}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) and initial. Point A (1, 5, 0). Find the terminal. point B.

Answer:

Let \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) be the position vectors of A and B.

Given :A(1,5.0)

a = \(\hat{i}+5 \hat{j}\)

Now, \(\overrightarrow{A B}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)

\(\overline{\mathrm{b}}\) – \(\overline{\mathrm{a}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)

\(\overline{\mathrm{b}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) + \(\overline{\mathrm{a}}\)

= \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) + \(\hat{i}+5 \hat{j}\)

Hence, the terminal points is B(3,1,7).

Question 8.

Find the vector equation of line passing through the point having position vector \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) and having direction ratios -3, 4, 2.

Answer:

Let A be the point whose position vector is a = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\)

Let \(\overline{\mathrm{b}}\) be the vector parallel to the line having direction ratios = – 3,4,2

Then, b = \(-3 \hat{i}+4 \hat{j}+2 \hat{k}\)

The vector equation of the line passing through A(\(\overline{\mathrm{a}}\)) and parallel to \(\overline{\mathrm{b}}\) is \(\overline{\mathrm{r}}\) = \(\overline{\mathrm{a}}\) + λ\(\overline{\mathrm{b}}\) , where λ is a scalar.

∴ The required vector equation of the line is \(\overline{\mathrm{r}}\) = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) + λ(\(-3 \hat{i}+4 \hat{j}+2 \hat{k}\))

Question 9.

Find the Cartesian equations of the Line passing through A(2,2,1) and B(1, 3,0).

Answer:

The cartesian equations of the line passing through the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are:

\(\frac{x-x_1}{x_2-x_1}\) = \(\frac{y-y_1}{y_2-y_1}\) = \(\frac{z-z_1}{z_2-z_1}\)

Here, (x_{1}, y_{1}, z_{1}) = (2, 2,1) and (x_{2}, y_{2}, z_{2}) = (1,3,0)

The required cartesian equation is

\(\frac{x-2}{1-2}\) = \(\frac{y-2}{3-2}\) = \(\frac{z-1}{0-1}\)

i.e., \(\frac{x-2}{-1}\) = \(\frac{y-2}{1}\) = \(\frac{z-1}{-1}\)

Question 10.

Solve graphically: x ≥ 0 and y ≥ 0.

Answer:

Consider the lines whose equations are x = 0, y = 0.

These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.

Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.

Question 11.

Differentiate the following w.r.t x. : cos (x² + a²)

Answer:

Let y = cos (x² + a²)

Differentiating w.r.t x, we get

\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [cos (x² + a²)]

= -sin(x² + a²).(2x + 0)

= -2x sin(x²+a²).

Question 12.

Find the area of the ellipse \(\frac{x^2}{25}\) + \(\frac{y^2}{16}\) = 1 using integration.

Answer:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO. Clearly for this region, the limits of integration are 0 and 5.

From the equation of the ellipse

Img 5

In the first quadrant y > 0

∴ y = \(\frac{4}{5}\) \(\sqrt{25-x^2}\)

∴ Area of the ellipse = 4 (Area of the region OABO)

Img 6

Question 13.

State if the following is not the probability mass function of a random variable. Give reasons for your answer.

Tablee 1

Answer:

P.m.f. of random variable should satisfy the following conditions:

(a) 0 < p, < 1

(b) Σp_{i}= 1

Tablee 2

P(X = 3) = – 0.1, i.e., p_{i} < 0 which does not satisfy 0 ≤ p_{i} ≤ 1

Hence, P(X) cannot be regarded as p.m.f. of the random variable X.

Question 14.

In a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Answer:

This is a problem of Bernoulli trials. As for each question, there are 2 outcomes. Let X represent the number of correct answers by guessing in the set of 5 multiple- choice questions.

Probability of getting a correct answer is, p = \(\frac{1}{3}\)

q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{1}{2}\)

Clearly, X has a binomial distribution with n = 5

and p = \(\frac{1}{3}\)

The p.m.f of X is given by

P(X = x) = ^{n}C_{x} p^{x}q^{n-x}, x = 0,1,2,4,5

i.e., p(x) = ^{n}C_{x} (\(\frac{1}{3}\))^{x} (\(\frac{2}{3}\))^{5-x} x = 0,1,2,4,5

P(four or more correct answers) = P[X ≥ 4] = P(4) + P(5)

Img 7

Hence,’the probability of getting four or more correct answer is \(\frac{11}{243}\)

**Section C**

**Attempt any EIGHT of the followings questions:**

Question 15.

Find the matrix of the co-factors for the following matrix.

\(\left[\begin{array}{ccc}

1 & 0 & 2 \\

-2 & 1 & 3 \\

0 & 3 & -5

\end{array}\right]\)

Answer:

Let A = \(\left[\begin{array}{ccc}

1 & 0 & 2 \\

-2 & 1 & 3 \\

0 & 3 & -5

\end{array}\right]\)

Here,

Img 8

Question 16.

Find the general solution of the following equation:

cos θ + sin θ = 1.

Answer:

cos θ + sin θ = 1

∴ sin θ = 1 – cos θ

Img 9

The general solution of sin θ = 0 is θ = nπ, n ∈ Z and tan θ = tan α is 0 = nπ + α, n ∈ Z

The required general solution is

\(\frac{θ}{2}\) = nπ, n ∈ Z or \(\frac{θ}{2}\) = nπ + \(\frac{π}{2}\), n ∈ Z

i.e., θ = 2nπ, n ∈ Z or θ = 2njt + \(\frac{\pi}{2}\), n ∈ Z

Question 17.

Show that the following points are collinear: P = (4,5,2), Q = (3,2,4), R = (5,8,0).

Answer:

Let a, b, c be position vectors of the points.

P = (4,5,2), Q = (3,2,4), R = (5,8,0) respectively.

Then \(\overline{\mathrm{a}}\) = \(4 \hat{i}+5 \hat{j}+2 \hat{k}\), \(\bar{b}\) = \(3 \hat{i}+2 \hat{j}+4 \hat{k}\), \(\bar{c}=5 \hat{i}+8 \hat{j}+0 \hat{k}\)

AB = \(\bar{b}\) – \(\bar{a}\)

= \(3 \hat{i}+2 \hat{j}+4 \hat{k}\) – \(4 \hat{i}+5 \hat{j}+2 \hat{k}\)

= \(– \hat{i}-3 \hat{j}+2 \hat{k}\)

= -(\(\hat{i}+3 \hat{j}-2 \hat{k}\)) …(1)

and BC = \(\bar{c}\) – \(\bar{b}\)

= \(5 \hat{i}+8 \hat{j}+0 \hat{k}\) – \(3 \hat{i}+2 \hat{j}+4 \hat{k}\)

= \(2 \hat{i}+6 \hat{j}-4 \hat{k}\)

= 2(\(\hat{i}+3 \hat{j}-2 \hat{k}\))

= 2.AB ………….[By (1)]

∴ BC is a non-zero scalar multiple of AB

∴ They are parallel to each other.

But they have point B in common.

∴ BC and AB are collinear vectors.

Hence, the points A, B and C are collinear.

Question 18.

Find the position vector of point R which divides the line joining the points P and Q whose position vectors are \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) in the ratio 3:2 is internally.

Answer:

It is given that the points P and Q have position vectors

\(\bar{p}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\bar{q}\) =-\(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) respectively.

If R(\(\bar{r}\)) divides the line segment PQ internally in the ratio 3:2, by section formula for internal division.

\(\bar{r}\) = \(\frac{3 q+2 p}{3+2}\)

Img 10

coordinates of R = \(\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)\)

Hence, the position vector of R is \(\frac{1}{5}\)(\(-11 \hat{i}+4 \hat{j}-9 \hat{k}\)) and the coordinates of R are \(\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)\)

Question 19.

Differentiate the following w.r.t. x : \(x^{\tan ^{-1} x}\)

Answer:

Let y = \(x^{\tan ^{-1} x}\)

Then log y = log(\(x^{\tan ^{-1} x}\)) = (tan^{-1} x) (log x)

Differentiating both sides w.r.t. x, we get

Img 11

Question 20.

Find the equations of tangents and normals to the following curves at the indicated points on them : x³ + y³ – 9xy = 0 at (2,4)

Solution:

x³ + y³ – 9xy = 0

Differentiating both sides w.r.t. x, we get

Img 12

= Slope of the tangent at (2,4)

∴ The equation of the tangent at (2,4) is

y – 4 = \(\frac{4}{5}\)(x – 2)

∴ 5y – 20 = 4x – 8

4x – 5y +12 = 0

The slope of normal at (2,4) = \(\frac{4}{5}\)

= \(-\frac{5}{4}\)

∴ The equation of the normal at (2,4) is

y – 4 = \(-\frac{5}{4}\)(x-2)

∴ 4y – 16 = -5x + 10

∴ 5x + 4y – 26 = 0

Hence, the equation of tangent and normal are 4x – 5y + 12 = 0 and 5x – 4y – 26 = 0 respectively.

Question 21.

Evaluate the following integrals: \(\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x\)

Answer:

Img 13

Question 22.

Integrate the following function w.r.t. :

Answer:

Img 14

= log |cosec t – cot t| + c

= log |cosec (x + log x) – cot (x + log x)| + c.

Question 23.

Evaluate the following integral as limit of a sum:

Answer:

Let f[x) = 3x² – 1, for 0 ≤ x ≤ 2.

Divide the closed interval [0,2] into n subintervals each of length h at the points.

0,0 + h, 0+ 2h,…..,0 + rh,……..,0+nh = 2

i.e. 0, h, 2h, …,rh,…….,nh = 2

∴ h = \(\frac{2}{n}\) and as n → oo, h → 0

Here, a = 0

∴ f(a + rh) = f(0 + rh) = f(rh) = 3(rh)² – 1 = 3r²h² – 1

Img 15

= 4(1+0)(2+0) – 2

= 8 – 2 = 6

Question 24.

Solve the following differential equation:

\(x \sin \left(\frac{y}{x}\right) d y\) = \(\)

Answer:

Img 16

Intergrating both sides, we get

\(\int \sin v d v\) = \(-\int \frac{1}{x} d x\)

∴ -cos v = -log x – c

cos(\(\frac{y}{x}\)) = log x + c

This general solution.

Question 25.

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.

Answer:

When a die is tossed twice, the sample space S has 6×6 = 36 sample points.

∴ n(S) = 36

Trial will be a success if the number on at least one die is 5 or 6.

Let X denote the number of dice on which 5 or 6 appears. Then X can take values 0,1,2,

When X = 0 i.e., 5 or 6 do not appear on any of the dice, then

X = {(1,1), (1, 2), (1, 3), (1, 4), (2,1), (2, 2), (2, 3). (2, 4), (3,1), (3, 2), (3, 3), (3,4). (4,1), (4, 2), (4. 3), (4,4)}.

∴ n(X) = 16.

∴ P(X = 0) = \(\frac{n(X)}{n(S)}\) = \(\frac{16}{36}\) = \(\frac{4}{9}\)

When X = 1, i.e., 5 or 6 appear on exactly one of the dice, then

X = {(1, 5), (1, 6), (2. 5), (2, 6), (3, 5). (3, 6), (4, 5), (4, 6), (5,1), (5, 2), (5, 3), (5,4), (6,1). (6.2). (6.3), (6,4)}

∴ n(X) =16

∴ P(X = 1) = \(\frac{n(X)}{n(S)}\) = \(\frac{16}{36}\) = \(\frac{4}{9}\)

When X = 2, i.e., 5 or 6 appear on both of the dice, then X = {(5,5), (5, 6), (6, 5), (6, 6)}

∴ n(X) = 4.

∴ P(X = 1) = \(\frac{n(X)}{n(S)}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Question 26.

In a large school, 80% of pupil like mathematics. A visitor. to the school asks each of 4 pupils, chosen at random, whether they like mathematics.

Find the probability that the visitor obtains answer yes from at least 2 pupils:

(a) when the number of pupils questioned remains at 4.

(b) when the number of pupils questioned is increased to 8.

Answer:

Let X = number of pupils like Mathematics

p = probability that pupils like Mathematics

∴ p = 80% = \(\frac{80}{100}\) = \(\frac{4}{5}\)

and q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)

Given, n = 4

∴ X – B(4, \(\frac{4}{5}\))

The p.m.f of X is given by

P(X = x) = ^{n}C_{x} p^{x}q^{n-x}

i.e., p(x) = ^{4}C_{x} (\(\frac{4}{5}\))^{x} (\(\frac{1}{5}\))^{4-x}

**Section D**

**Attempt any FIVE of the following questions:**

Question 27.

Using the truth table prove the following logical equivalence. – (p v q)v (~ p ∧ q) = ~ p

Solution:

Img 17

The entries in columns 3 and 7 are identical.

∴ ~ (\((p \vee q)\)) v (- p ∧ q) = ~ p

Question 28.

In ∆ABC prove that (b + c – a) tan A/2 = (c + a – b) tan B/2 = (a + b – c) tan C/2.

Solution:

(b + c – a) tan A/2

Img 18

Question 29.

Express \(-\hat{i}-3 \hat{j}+4 \hat{k}\) as the linear combination of the vectors \(2 \hat{i}+\hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(3 \hat{i}+\hat{j}-2 \hat{k}\).

Solution:

Let \(\bar{a}\) = \(2 \hat{i}+\hat{j}-4 \hat{k}\)

\(\bar{b}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\)

\(\bar{c}\) = \(3 \hat{i}+\hat{j}-2 \hat{k}\)

\(\bar{p}\) = \(-\hat{i}-3 \hat{j}+4 \hat{k}\)

Suppose \(\bar{p}\) = x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\)

Then,

\(-\hat{i}-3 \hat{j}+4 \hat{k}\) = x(\(2 \hat{i}+\hat{j}-4 \hat{k}\)) + y(\(2 \hat{i}-\hat{j}+3 \hat{k}\)) + z(\(3 \hat{i}+\hat{j}-2 \hat{k}\))

\(-\hat{i}-3 \hat{j}+4 \hat{k}\) = \((2 x+2 y+3 z) \hat{i}\) + \((x – y + z) \hat{j}\) + \((-4 x+3 y-2 z) \hat{k}\)

By equality of vectors

2 x+2 y+3 z = -1

x – y + z = -3

-4 x+3 y-2 z = 4

We have to solve these equations by using Cramer’s Rule.

Img 19

Img 20

\(\bar{p}\) = 2\(\bar{a}\) + 2\(\bar{b}\) – 3\(\bar{c}\)

Question 30.

Find the values of λ so that the lines \(\frac{1-x}{3}\) = \(\frac{7 y-14}{\lambda}\) = \(\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\) are the right angles.

Solution:

The equations of the given lines are

\(\frac{1-x}{3}\) = \(\frac{7 y-14}{\lambda}\) = \(\frac{z-3}{2}\)

and \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\)

Equation (1) can be written as:

\(\frac{-(x-1)}{3}\) = \(\frac{7(y-2)}{2 \lambda}\) = \(\frac{z-3}{2}\)

i.e., \(\frac{x-1}{-3}=\frac{y-2}{\frac{2 \lambda}{7}}=\frac{z-3}{2}\)

The direction ratios of this line are

a_{1} = -3, b_{1} = \(\frac{2 \lambda}{7}\), c_{1} = 2

Equation (2) can be written as:

\(\frac{-7(x-1)}{3 \lambda}=\frac{y-5}{1}=\frac{-(z-6)}{5}\)

The direction ratios of this line are

a_{2} = \(\frac{-3 \lambda}{7}\) , b_{2} = 1, c_{2} = – 5

since the lines (1) and (2) are at right angles,

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

∴ (-3) (\(\frac{-3 \lambda}{7}\)) + (\(\frac{2 \lambda}{7}\))(1) + 2(-5) = 0

∴ (\(\frac{9 \lambda}{7}\)) + (\(\frac{2 \lambda}{7}\)) – 10 = 0

∴ \(\frac{11 \lambda}{7}\) = 10

∴ λ = \(\frac{70}{11}\).