CBSE Sample Papers For Class 12 Maths Mock Paper 1

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Q1. If A = diag(1, –2, 4) and B = diag(6, 2, -1), then the value of 3A + 5B is

Opt:

diag(7, 0, 3).

diag(7, 4, 5).

diag(27, 16, –17).

diag(33, 4, 7).

Ans:

diag(33, 4, 7).

Q2. If A is a skew-symmetric matrix of odd order, then |A| is equal to

Opt:

A’.

-A.

Ans:

Q3.

$The value of \sin \cot^{- 1} \tan (\cos^{- 1} 10) is$

Opt:

-10.

–10².

10.

10².

Ans:

10.

Q4.

$\begin{matrix} Let f : R \to R be a function given by f (x) = x^{2} + 1, then the \\ value of f^{- 1} (82) is \end{matrix}$

Opt:

{-3, 3}.

{3, 9}.

{-3, -9}.

{-9, 9}.

Ans:

{-9, 9}.

Q5.

Opt:

Ans:

Q6.

$\begin{matrix} Let f (x) = {(x + 1)}^{cotx} be continuous at x = 0, then f (0) will be \\ equal to \end{matrix}$

Opt:

1/e.

e².

e³.

Ans:

Q7.

Opt:

Ans:

Q8. A fair die is thrown twenty times. The probability that on
the tenth throw the fourth six appears is”
Opt:

Ans:

Q9. The value of objective function is minimum under linear constraints at
Opt: any point outside the feasible region. the centre of the feasible region. any vertex of the feasible region. the vertex which is minimum distance from (0, 0).
Ans: any vertex of the feasible region.

Q10.

$\begin{matrix} The value of \\ \sin^{- 1} \frac{2 x}{1 + x^{2}} - 2 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 3 \tan^{- 1} \frac{2 x}{1 - x^{2}} is equal to \end{matrix}$

Opt:

$1$ $\frac{1}{3}$ $\frac{1}{\sqrt{3}}$ $\frac{1}{2 \sqrt{3}}$

Ans:

$\frac{1}{\sqrt{3}}$

Q11.

Three persons A, B and C are to speak at a function
along with five others. If they all speak in random order,
the probability that A speaks before B and B speaks before C is
Opt:

Ans:

Q12.

$\begin{matrix} The value of x if \\ 3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3} \end{matrix}$

Opt:

$1$ $\frac{1}{3}$ $\frac{1}{\sqrt{3}}$ $\frac{1}{2 \sqrt{3}}$

Ans:

$\frac{1}{\sqrt{3}}$

Q13.

$\begin{matrix} If f (x) = \begin{matrix} \begin{matrix} 1 \\ 2 x \\ 3 x (x - 1) \end{matrix} & \begin{matrix} x \\ x (x - 1) \\ x (x - 1) (x - 2) \end{matrix} & \begin{matrix} x + 1 \\ x (x + 1) \\ x (x^{2} - 1) \end{matrix} \end{matrix}, then f (100) \\ is equal to \end{matrix}$

Opt:

100.

–100.

Ans:

Q14. If f(x) = |7x + 2| and g(x) = f(f(x)), then for x > 20, g’(x) equals

Opt:

-1.

Ans:

Q15.

$\begin{matrix} If the function f (x) = \begin{matrix} \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}}, if x \neq 2 \\ k, if x = 2 \end{matrix} \\ is continuous for all x, then k is equal to \end{matrix}$

Opt:

Ans:

Q16.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered-pairs, which, when added to R make it an equivalence relation, is

Opt:

Ans:

Q17.

Opt:

Ans:

Q18. ${Show that sin}^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x$

Opt:

$\begin{matrix} \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = 2 \sin^{- 1} x \\ let x = sinθ \\ \sin^{- 1} (2 x \sqrt{1 - x^{2}}) = \sin^{- 1} (2 sinθ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 sinθ cosθ) \\ = \sin^{- 1} (\sin 2 θ) \\ = 2 θ \\ = 2 \sin^{- 1} x \end{matrix}$

Ans:

Q19.

$\begin{matrix} Show that the functionf : \to R given by f (x) = \begin{matrix} 1 if x > 0 \\ 0 if x = 0 \\ - 1 if x < 0 \end{matrix} \\ is neither one - one nor onto . \end{matrix}$

Ans:

(a) f(1) = f(2) = 1

1 and 2 have the same images.

⇒ f(x₁) = f(x₂) = 1 for x > 0
But x₁ ≠ x₂

Similarly -1 and -2 have the same images.

Hence f is not one-one.

(b) Except -1, 0, 1 no other element of co-domain of f has any pre-image in its domain.

⇒ f is not onto.

Q20. Find two numbers whose sum is 24 and whose product is as large as possible.

Ans:

Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus,

P(x) = x(24 – x) = 24 – x²

∴ P'(x) = 24 – 2x

P”(x) = -2

Now, P'(x) = 0 ⇒ x = 12

Also, P”(12)= -2 < 0

∴ By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Q21. $Find \vec{a} and \vec{b}, if (\vec{a} - \vec{b}) . (\vec{a} + \vec{b}) = 27 and \vec{a} = 2 \vec{b} .$

Ans:

$\begin{matrix} (\vec{a} - \vec{b}) . (\vec{a} + \vec{b}) = 27 \\ \Rightarrow {\vec{a}}^{2} - {\vec{b}}^{2} = 27 \\ \Rightarrow 4 {\vec{b}}^{2} - {\vec{b}}^{2} = 27 \\ \Rightarrow 3 {\vec{b}}^{2} = 27 \\ \Rightarrow {\vec{b}}^{2} = 9 \\ \Rightarrow \vec{b} = 3 \\ Now, \vec{a} = 2 \vec{b} \\ \Rightarrow \vec{a} = 6 \\ Thus, \vec{a} = 6 and \vec{b} = 3 \end{matrix}$

Q22.

$\begin{matrix} Find a vector in the direction of vector 2 \hat{i} - \hat{j} + 3 \hat{k} \\ which has magnitude 7 units . \end{matrix}$

Ans:

$\begin{matrix} The unit vector of 2 \hat{i} - \hat{j} + 3 \hat{k} = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{2 \hat{i} - \hat{j} + 3 \hat{k}} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{2^{2} - {(- 1)}^{2} + 3^{2}}} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{4 + 1 + 9}} \\ = \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{14}} \\ Thus, the vector in the direction of (2 \hat{i} - \hat{j} + 3 \hat{k}) which has \\ magnitude 7 units is \\ = 7 \times \frac{2 \hat{i} - \hat{j} + 3 \hat{k}}{\sqrt{14}} \\ = \frac{14 \hat{i} - 7 \hat{j} + 21 \hat{k}}{\sqrt{14}} \\ = \frac{14}{\sqrt{14}} \hat{i} - \frac{7}{\sqrt{14}} \hat{j} + \frac{21}{\sqrt{14}} \hat{k} \end{matrix}$

Q23. Form the differential equation of the family of parabolas having vertex at originand axis along positive x-axis.

Ans:

The equation of parabolas having vertex at origin
and axisalong positive y-axis is as follows
y² = 4ax …(i)
Differentiating w.r.t. x, we get
2yy’ = 4a

Putting value of 4a in equation (i), we get
y² = 2yy’ x

y = 2y’ x

or2y’x – y =0

Thus, the required differential equation is
2y’x – y =0.

Q24. Given that E and F are events such that P(E) = 0.5, P(F) = 0.4 and P(E ∩ F) = 0.1, find P (E | F) and f (F | E).

Ans:

$\begin{matrix} Since, P(E | F) = \frac{P (E \cap F)}{P (F)} \\ ∴ P(E | F) = \frac{0 .1}{0 .4} \\ = \frac{1}{4} \\ And P(F | E) = \frac{P (E \cap F)}{P (E)} \\ = \frac{0 .1}{0 .5} \\ = \frac{1}{5} \end{matrix}$

Q25. Find the rate of change of the area of a circle with respect to its radius r when r = 5cm.

Ans:

The area A of a circle with radius r is given by A =

$π r^{2}$

Therefore, the rate of change of the area A with respect to its radius r is given by

dA/dr = 2

$π$

r = 10

$π$

(Since, r = 5 cm given)

Thus, the area of the circle is changing at the rate of 10

$π$

cm²/s.

Q26.

$\begin{matrix} If the line \vec{r} = (\hat{i} - 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + \hat{j} + 2 \hat{k}) and the plane \\ \vec{r} . (3 \hat{i} - 2 \hat{j} + m \hat{k}) = 14 . Find the value of m . \end{matrix}$

Ans:

$\begin{matrix} The given line is parallel to the vector \vec{b} = 2 \hat{i} + \hat{j} + 2 \hat{k} \\ and the given vector is normal to the vector \vec{n} = 3 \hat{i} - 2 \hat{j} + m \hat{k} \\ If the line is parallel to th the plane then normal is perpendicular \\ to the line . \\ ∴ \vec{b} ⊥ \vec{n} \\ \Rightarrow \vec{b} . \vec{n} =0 \\ \Rightarrow (2 \hat{i} + \hat{j} + 2 \hat{k}) . (3 \hat{i} - 2 \hat{j} + m \hat{k}) = 0 \\ \Rightarrow 6 - 2 + 2 m = 0 \\ \Rightarrow m = - 2 \end{matrix}$

Q27. Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans:

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ ∵ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q28.

$If A = \begin{matrix} 1 0 \\ - 1 7 \end{matrix} and I = \begin{matrix} 1 0 \\ 0 1 \end{matrix}, then find k so that A^{2} = 8 A + kI .$

Ans:

$\begin{matrix} A^{2} = AA = \begin{matrix} 1 0 \\ - 1 7 \end{matrix} \begin{matrix} 1 0 \\ - 1 7 \end{matrix} \\ = \begin{matrix} 1 + 0 0 + 0 \\ - 1 - 7 0 + 49 \end{matrix} = \begin{matrix} 1 0 \\ - 8 49 \end{matrix} \\ 8 A + KI = 8 \begin{matrix} 1 0 \\ - 1 7 \end{matrix} + K \begin{matrix} 1 0 \\ 0 1 \end{matrix} = \begin{matrix} 8 + k 0 \\ - 8 56 + k \end{matrix} \\ Given A^{2} = 8 A + kI \\ \Rightarrow \begin{matrix} 1 0 \\ - 8 49 \end{matrix} = \begin{matrix} 8 + k 0 \\ - 8 56 + k \end{matrix} \\ \Rightarrow 8 + k = 1 and 56+ k = 49 \\ \Rightarrow k = - 7 \end{matrix}$

Q29. A firm deals with two kinds of fruit juices – pineapple and orange juice. These are mixed and two mixtures are sold as soft drinks A and B. One tin of A requires 4 litres of pineapple and 1 litre of orange juice. One tin of B requires 2 litres of pineapple and 3 litres of orange juice. The firm has only 46 litres of pineapple juice and 24 litres of orange juice. Each tin of A and B is sold at a profit of Rs. 4 and Rs. 3 respectively. How many tins of each type should the firm produce to maximise the profit? Solve the problem graphically.

Ans:

Let the number of tins of A be x and that of B be y.
According to the given conditions, the L.P.P. is
x ≥ 0, y ≥ 0 … (1)
4x + 2y ≤ 46 ⇒ 2x + y ≤ 23 … (2)
x + 3y ≤ 24 … (3)
and to maximise Z = 4x + 3y.
Plotting the inequations (i), (ii) and (iii) on the graph,

We get the shaded portion as the optimum solution. Possible points for maximum profit are A(23/2,0), B(9,5) and C(0,8)

$\begin{array}{l} Points & Z = 4 x + 3 y & Value \\ A (\frac{23}{2}, 0) & 46 + 0 & 46 \\ B(9, 5) & 36 + 15 & 51 [maximum] \\ C(0, 8) & 0 + 24 & 24 \end{array}$

We notice profit is maximum at B (9,5), i.e., x = 9, y = 5.
Hence, 9 tins of A and 5 tins of B should be produced.

Q30.

$\begin{matrix} If a, b, care in A . P then without expanding show that \\ \begin{matrix} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ x +4 x +5 x +2 c \end{matrix} = 0 \end{matrix}$

Ans:

$\begin{matrix} Here, Δ = \begin{matrix} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ x +4 x +5 x +2 c \end{matrix} \\ = \begin{matrix} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ 1 1 2 (c - b) \end{matrix} (by applying R_{3} \to R_{3} - R_{2}) \\ = \begin{matrix} x +2 x +3 x +2 a \\ 1 1 2 (b - a) \\ 1 1 2 (c - b) \end{matrix} (by applying R_{2} \to R_{2} - R_{1}) \\ Now because a, b, c are in A . P ., we have \\ b - a = c - b \\ Hence in Δ R_{2} = R_{3} \\ If any two rows or columns of a determinant are identical \\ then the value of determinant is zero . \\ \Rightarrow Δ = 0 \end{matrix}$

Q31.

$\begin{matrix} Using properties of determinants, show that \\ \begin{matrix} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{matrix} = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = abc + bc + ca + ab . \end{matrix}$

Ans:

$\begin{matrix} Let Δ = \begin{matrix} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{matrix} \\ By applying c_{1} \to c_{2} and c_{3} \to c_{3} - c_{2}, we get \\ \begin{matrix} a 1 0 \\ - b 1 + b - b \\ 0 1 c \end{matrix} \\ Expanding along R_{1}, we obtain \\ Δ = a \begin{matrix} 1 + b - b \\ 1 c \end{matrix} - 1 \begin{matrix} - b - b \\ 0 c \end{matrix} + 0 \\ = a (c + bc + b) - (- bc) \\ = ac + abc + ab + bc \\ = abc + bc + ca + ab \\ = abc a + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \end{matrix}$

Q32. Determine whether the following relation is reflexive, symmetric and transitive: Relation R on the set N of all natural numbers is defined as
R = {(x, y): y = x + 5 and x < 4}

Ans:

R = {(x, y): y = x + 5 and x < 4}
∴ R = {(1, 6), (2, 7), (3, 8)}

Reflexivity: (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.

Symmetry: (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.

Transitivity: Since (1, 6) ∈ R and there is no ordered pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is not transitive.

Q33. Show that the Relation R on the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.

Ans:

R = {(a, b) : |a – b| is a multiple of 4}, where a, b ∈ A ={x Z : 0 ≤ x ≤ 12} = {0, 1, 2, …, 12}.

Reflexivity: For any a ∈ A, |a – a| = 0, which is a multiple of 4.

⇒ (a, a) ∈ R, for all a ∈ A.

So, R is reflexive.

Symmetry: Let (a, b) ∈ R. Then,

(a, b) ∈ R

⇒ |a – b| is a multiple of 4
⇒ |a – b| = 4k for some k ∈ N
⇒ |b – a| = 4k for some k ∈ N
⇒ (b, a) ∈ R

So, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then,
⇒ |a – b| is a multiple of 4 and |b – c| is a multiple of 4
⇒ |a – b| = 4k and |b – c| = 4m for some k, m ∈ N
⇒ a – b = ± 4k and b – c = ± 4m
⇒ a – c = ± 4k ± 4m
⇒ a – c is a multiple of 4
⇒ |a – c| is a multiple of 4
⇒ (a, c) ∈ R

So, R is transitive.

Hence, R is an equivalence relation.

Q34.

$\begin{matrix} F i n d t h e a r e a o f t h e s m a l l e r r e g i o n b o u n d e d b y t h e e l l i p s e \\ \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 a n d t h e l i n e \frac{x}{3} + \frac{y}{2} = 1 . \end{matrix}$ $\begin{matrix} \end{matrix}$

Ans:

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$\begin{matrix} ∴ A r e a (A B C A) = A r e a (O B C A O) - A r e a (O B A O) \\ = \int_{0}^{3} 2 \sqrt{1 - \frac{x^{2}}{9}} d x - \int_{0}^{3} 2 (1 - \frac{x}{3}) d x \\ = \frac{2}{3} \int_{0}^{3} \sqrt{9 - x^{2}} d x - \frac{2}{3} \int_{0}^{3} (3 - x) d x \\ = \frac{2}{3} {[\frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{x}{3})]}_{0}^{3} - \frac{2}{3} {[3 x - \frac{x^{2}}{2}]}_{0}^{3} \\ = \frac{2}{3} {\frac{3}{2} \sqrt{9 - 3^{2}} + \frac{9}{2} \sin^{- 1} (\frac{3}{3}) - \frac{0}{2} \sqrt{9 - 0^{2}} - \frac{9}{2} \sin^{- 1} (\frac{0}{3})} \\ - \frac{2}{3} {3 (3) - \frac{3^{2}}{2} - 3 (0) + \frac{0^{2}}{2}} \\ = \frac{2}{3} {\frac{9}{2} \times \frac{π}{2}} - \frac{2}{3} {9 - \frac{9}{2}} \\ = \frac{3 π}{2} - 3 = \frac{3}{2} (π - 2) s q u a r e units \end{matrix}$

Q35.

$\begin{matrix} For the differential equation xy \frac{dy}{dx} = (x +2) (y +2), find the \\ solution curve passing through the point (1, -1) . \end{matrix}$

Ans:

$\begin{matrix} The given differential equation is : \\ xy \frac{dy}{dx} = (x + 2) (y + 2) \\ \frac{ydy}{(y + 2)} = \frac{(x + 2)}{x} dx \\ \frac{(y + 2 - 2) dy}{(y + 2)} = \frac{(x + 2)}{x} dx \\ 1 dy - \frac{2}{(y + 2)} dy = (1 + \frac{2}{x}) dx \\ Integrating both sides, we get \\ \int 1 dy - \int \frac{2}{(y + 2)} dy = \int 1 dx + 2 \int \frac{1}{x} dx \\ y - 2 \log (y + 2) = x + 2 logx + C \\ y - x - C = {logx}^{2} + \log {(y + 2)}^{2} \\ y - x - C = {logx}^{2} {(y + 2)}^{2} . .. (ii) \\ Since, the curve passes through (1, - 1) . \\ - 1 - 1 - C = \log {(1)}^{2} {(- 1 + 2)}^{2} \\ - 2 - C = \log {(1)}^{2} \\ \Rightarrow C = - 2 \\ Putting the value of C in equation (ii), we get \\ y - x + 2 = {logx}^{2} {(y + 2)}^{2} \\ This is the required solution of the curve . \end{matrix}$

Q36.

$\begin{matrix} Show that height of the cylinder of greatest volume which \\ can be inscribed in a right circular cone of height h and semi \\ vertical angle θ is one - third that of the cone and the \\ greatest volume of cylinder is \frac{4}{27} {πh}^{3} \tan^{2} α . \end{matrix}$ $\begin{matrix} \end{matrix}$

Ans:

$\begin{matrix} In given right circular cone of fixed height is h and \\ semi - vertical angle be α . \\ Then a cylinder of radius R and height H is inscribed in \\ the cone . Then ∠ GAO = α, OG = r, OA = h, OE = R and CE = H . \\ We have, r = htanα \\ Now, since ΔAOG is similar to ΔCEG, we have : \\ \frac{AO}{OG} = \frac{CE}{EG} \Rightarrow \frac{h}{r} = \frac{H}{r - R} [EG = OG - OE \end{matrix}$ $\begin{matrix} \Rightarrow H = \frac{h}{r} (r - R) \\ = \frac{h}{htan α} (htan α - R) \\ = \frac{1}{tan α} (htan α - R) \\ Now, the volume (V) of the cylinder \\ V = π R^{2} h \\ = \frac{π R^{2}}{tan α} (htan α - R) \\ = π R^{2} h - \frac{π R^{3}}{tan α} \\ D i f f e r e n t i a t i n g w .r .t . R, we get \\ \frac{d V}{d R} = 2 π R h - \frac{3 π R^{2}}{tan α} \\ a n d \frac{d^{2} V}{d R^{2}} = 2 π h - \frac{6 π R}{tan α} \\ F o r maxima or minima, we have \\ \frac{d V}{d r} = 0 \Rightarrow 2 π R h - \frac{3 π R^{2}}{tan α} = 0 \\ \Rightarrow π R (2 h - \frac{3 R}{tan α}) = 0 \Rightarrow R = \frac{2 h \tan α}{3} \\ S o, \frac{d^{2} V}{d R^{2}} < 0 f o r R = \frac{2 h \tan α}{3} \\ ∴ By second derivative test, the volume of the cylinder is \\ the greatest when R = \frac{2 h \tan α}{3} . \\ T h e n, H = \frac{1}{tan α} (htan α - R) \\ = \frac{1}{tan α} (htan α - \frac{2 h \tan α}{3}) \\ = \frac{h}{3} \\ Thus, the height of the cylinder is one-third the height of the \\ cone when the volume of the cylinder is the greatest . \\ Now, the maximum volume of the cylinder \\ V = π R^{2} H \\ = π {(\frac{2 h \tan α}{3})}^{2} (\frac{h}{3}) \\ ∴ V = \frac{4}{27} π h^{3} \tan^{2} α \end{matrix}$

CBSE Sample Papers For Class 12 Maths Mock Paper 1

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