Quantitative Aptitude PYQs

Concept:

Total number of days in a leap year = 366

Total number of days in February = 29

Total no. of days in a week = 7

Total no. of weeks= 29/7= 4 weeks and 1 day

 

Calculation:

To find: Probability that a leap year selected at random will contain five Sundays

As we know, in a leap year in the february months there will be 4 weeks and 1 day. 

These 1 days can be:

{Sun}, {Mon}, {Tue}, {Wed}, {Thu}, {Fri}, {Sat} 

Total outcomes = 7

Favourable outcomes = {Sun} = 1

Hence the probability that a leap year selected at random will contain five Sundays = 1/7

Concept:

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.

Calculations:

Given, the arithmetic mean of 100 observation is 40.

The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.

⇒ Sum of 100 observation = 100 \(\times ;40\) = 4000

Later, it was found that an observation '53' was wrongly read as '83

⇒ Sum of 100 observation = 4000 - 83 + 53 = 3970

New Mean = \(\dfrac {3970}{100}\)= 39.70

Calculations:

Given , a husband and wife appear in an interview for two vacancies for the same post.

Let A= be the event that the husband is selected.

B = be the event that the wife is selected. 

P(A) = \(\dfrac{1}{7}\) and P(B) = \(\dfrac{1}{5}\)

Now, 

A̅ = be the event that the husband is not selected.

B̅  = be the event that the husband is not selected.

P(A̅) = 1 - P(A) = \(\dfrac{6}{7}\) and P(B̅) = 1 - P(B) = \(\dfrac{4}{5}\)

P(At least one of them will be selected) = P(A) P(B̅) + P(A̅) P(B) + P(A) P(B)

= \(\frac17 \times \frac 45 + \frac67 \times \frac 15 + \frac17 \times \frac 15 \)

= \(\dfrac{11}{35} \ \)

Concept:

Product rule:

If a probability of event A happening is x and probability of event B happening is y then the probability of happening event s A and B is xy.

Probability:

If S is a sample space and A is a favourable event then the probability of A is given by:

\(\rm P=\dfrac{n(A)}{n(S)}\)

Combination formula:

The number of possible selection (combinations) of r objects out of n objects is given by:

\(\begin{align*} \rm ^nC_r &=\rm \dfrac{n!}{(n-r)!r!} \end{align*}\)

 

Calculation:

First we will decide the sample space.

The sample space will contain all possible combinations where we select 4 coins out of 15 coins.

Therefore,

\(\begin{align*} ^{15}C_4 &= \dfrac{15!}{11!\times 4!}\ &= \dfrac{12\times13\times 14\times 15}{4\times 3\times 2\times 1}\ &= 91\times 15 \end{align*}\)

Now the favourable event is selecting 4 coins out of the counterfeit 6 coins which is given as follows:

\(\begin{align*} ^6C_4 &= \dfrac{6!}{2!\times 4!}\ &= \dfrac{5\times 6}{2\times 1}\ &= 15 \end{align*}\)

Therefore, the required probability is:

\(\begin{align*} P &= \dfrac{15}{91\times 15}\ &= \dfrac{1}{91} \end{align*}\)

Concept:

Let A be any event in sample space S, then P(A) = \(\rm\dfrac{n(A)}{n(S)}\)

Calculations:

A committee of 3 is to be formed from a group of 2 boys and 2 girls.

Here are two boys B₁ and B₂ and two girls G₁ and G₂.

Sample space = S = {B₁B₂G₁ , B₁B₂G₂, G₁G₂B₁, G₁G₂B₂}

⇒n (S) = 4

A: The event that the committee consists of 2 boys and 1 girl

A = {B1B2G1 , B1B2G2}

⇒n(A) = 2

probability that the committee consists of 2 boys and 1 girl = P(A) = \(\rm\dfrac{n(A)}{n(S)}\)= \(\rm \dfrac 2 4 \)

Hence, A committee of 3 is to be formed from a group of 2 boys and 2 girls, the probability that the committee consists of 2 boys and 1 girl \(\dfrac 12 \)

Concept:

Let A and B are independent events, then 

P(A \(\rm\cap\) B)= P(A). P(B)

Calculations:

Give, total number of tickets = 10

Number of tickets drawn = 2

Prime number between 1 to 10 = {2, 3, 5, 7} = 4

A : The event of getting a prime number in the first draw

B : The event of getting a prime number in the second draw

The event A and B are independent events.

P(A) = Probability of getting a prime number in the first draw = \(\rm \dfrac{4}{10} = \dfrac {2}{5}\)

Number of remaining cards in pack = 9

P(B) = Probability of getting a prime number in the second draw = \(\rm \dfrac{3}{9} = \dfrac {1}{3}\)

The probability that both the tickets drawn have prime numbers P(A \(\rm\cap\) B)= P(A). P(B)

= \(\rm \dfrac{2}{5} \times \dfrac {1}{3}\)

= \(\rm\dfrac {2}{15}\)

Hence, In a lottery of 10 tickets numbered 1 to 10, two tickets are drawn simultaneously. then the probability that both the tickets drawn have prime numbers is \(\rm\dfrac {2}{15}\)

Concept:

Equation of regression of Y on X:

The equation of regression of Y on X with given values of \(\rm \bar X, \bar Y,\sigma_{x}\mbox{ and }\sigma_{y}\) is given as follows:

\(\rm \rm \dfrac{y-\bar Y}{\sigma_y} =r_{xy} \dfrac{x-\bar X}{\sigma_x} \)

 

Calculation:

It is given that X̅ = 65, Y̅ = 67, σX = 2.5, σY = 3.5 and r(X, Y) = 0.8 .

Using the equation of regression of Y on X we write:

\(\rm \rm \rm \frac{y-\bar Y}{\sigma_y} =r_{xy} \frac{x-\bar X}{\sigma_x} \ \left(\frac{y-67}{3.5}\right) =0.8 \times \left(\frac{x-65}{2.5}\right)\ y - 67 = 1.12(x-65)\y-67=1.12x-72.8\y=1.12x-5.8\)

Therefore, the required equation is y = 1.12x - 5.8

Concept:

Binomial distribution:

The binomial distribution for P(X = r) with parameters n and p is given as follows:

\(\rm P(X=r) = ^nC_rp^rq^{n-r}\)

Where q is the complementary event.

Also, the property of any random variable X,

E(aX + b)  = aE(X) + b

V(aX + b)  = (a²)V(X) 

Calculation:

Given: n = 10 and \(\rm p = \dfrac{1}{5}\)

Therefore, q = 1 - p 

\(\rm q = \dfrac{4}{5}\)

For random variable X, 

Mean = E(X) = np = 10 × (1/5) = 2

Variance = V(X) = npq = 10 × (1/5) ×(4/5)  = 8/5

Since, Y = 10 - X

⇒ E(Y) = E(10 - X)

⇒ E(Y) = 10 - E(X) = 10 - 2

⇒ 8 = n'p'      ....(1)

Also, V(Y) = V(10 - X) = 0 + (-1)² V(X)

⇒ 8/5 = n'p'q'      ....(2)

Solving 1 and 2, we will get n', p', and q' which will be,

⇒ 8/5 = 8'q'   

⇒ q' = 1/5   

⇒ p' = 4/5   

Now,

⇒ 8 = n'  (4/5)

⇒ n' = 10

Concept:

Probability of complementary event:

Let A be an event such that \(\rm \bar A\) is the complementary event then \(\rm P(A)=1-P(\bar A)\)  .

Probability of union of two events:

Let A and B be two events then the following relation holds:

\(\rm P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Conditional Probability :

For two events the conditional probability is given by:

\(\rm P(A|B)=\dfrac{P(A\cap B)}{P(B)}\)

 

Calculation:

It is given that \(\rm P(A)=0.6,P(B)=0.5\mbox{ and }P(A\cap B)=0.4\) .

We can calculate the probability of complementary events as follows \(\rm P(\bar A)=0.4,P(\bar B)=0.5\mbox{ and }P(\overline{A\cap B})=0.6\).

We know that 

\(\rm P(\bar A\cup B)=P(\bar A)+P(B)-P(\bar A\cap B)\)

Note that \(\rm P(\bar A\cap B)=P(B)-P(A\cap B)\) .

Therefore in the given case:

\(\rm P(\bar A\cap B) =0.5-0.4\ = 0.1 \)

Using this we write:

\(\rm P(\bar A\cup B) = 0.4+0.5-0.1\ = 0.8 \)

Therefore, the first statement is wrong.

Using the formula for conditional probability we write:

\(\rm P(\bar B|\bar A)=\dfrac{P(\bar A\cap \bar B)}{P(\bar A)}\)

We know that \(\rm P(\bar A\cap \bar B)=P(\overline{A\cup B})\) .

Using the given data we will first determine the value of \(\rm P(A\cup B)\).

\(\rm P(A\cup B) = P(A)+P(B)-P(A\cap B)\ = 0.6 + 0.5 - 0.4\ = 0.7 \)

Now using the complementary event formula we get:

\(\rm P(\overline{A\cup B}) = 1 - P(A\cup B)\ = 1 - 0.7\ = 0.3 \)

Substituting in the formula for the conditional formula:

\(\rm P(\bar B|\bar A) =\dfrac{P(\bar A\cap \bar B)}{P(\bar A)}\ = \dfrac{0.3}{0.4}\ = 0.75\)

Therefore, the second statement is also wrong.

Concept:

Consider, 

F : The event that the cake fails to raise.

A : The event that Cook X  bake a special kind of cake

B: The event that Cook Y  bake a special kind of cake

C : The event that Cook Z  bake a special kind of cake

The proportion of failures caused by X is given by P(A|F).

P(A|F) = \(\rm \dfrac {P(F|A)P(A)}{P(F|A)P(A)+P(F|B)P(B)+P(F|C)P(C)}\)

Calculations:

Consider, 

F : The event that the cake fails to raise.

A : The event that Cook X  bake a special kind of cake

B: The event that Cook Y  bake a special kind of cake

C : The event that Cook Z  bake a special kind of cake

Three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise.

P (F|A) = 0.02

P (F|B) = 0.03

P (F|C) = 0.05

In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. 

P(A) = 0.5

P(B) = 0.3

P(C) = 0.2

To find : The proportion of failures caused by X is given by P(A|F).

P(A|F) = \(\rm \dfrac {P(F|A)P(A)}{P(F|A)P(A)+P(F|B)P(B)+P(F|C)P(C)}\)

Substitute the values, we get

P(A|F) = \(\rm \dfrac {(0.02)(0.5)}{(0.02)(0.5)+(0.03)(0.3)+(0.05)(0.2)}\)

P(A|F) = 34.48 = \(\dfrac{10}{29}\)

Hence, three cooks X, Y and Z bake a special kind of cake, and with respective probabilities 0.02, 0.03 and 0.05, it fails to rise. In the restaurant where they work, X bakes 50%, Y bakes 30% and Z bakes 20% of cakes. The proportion of failures caused by X is \(\dfrac{10}{29}\)

Concept:

\(\begin{array}{l} \text { Distance of point } \rm \left (x_{1}, y_{1}, z_{1}\right) \text { from } a x+b y+c z+d=0 \text { is, D} =\rm \left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \end{array}\)

 

Calculation:

Here, centre of sphere is at (1, -2, 3) and equation of plane 6x - 3y + 2z - 4 = 0

So the distance from th point (1, -2, 3) to 6x - 3y + 2z - 4 = 0

\(\begin{array}{l} \rm r=\left|\frac{(1 \times 6)+(-2\times (-3))+(3 \times 2)-4}{\sqrt{6^{2}+(-3)^{2}+2^{2}}}\right| \ \rm r=\left|\frac{6+6+6-4}{\sqrt{49}}\right|=\frac{14}{7}\=2 \end{array}\)

Length of diameter = 2(2) = 4 units  

Hence, option (4) is correct.

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x₁, y_1, z₁) and (x₂, y₂, z₂) is given by, \(\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

 

Calculation:

Let M be the foot of perpendicular drawn from the the point P(2, 3, 4)

Let , \(\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k\)

x = k, y  = 0, z = 0

So M = (k, 0, 0)

Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0

PM is perepedicular to the given line so,

(2 - k) (1) + 3(0) + 4 (0) = 0

∴ k = 2

M = (2, 0, 0)

Perpendicular distance PM =

 \(\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\ =\sqrt{9+16}\ =5\)

Hence, option (2) is correct.

Concept:

If l, m, n are direction cosines and a, b, c are direction ratios, then \(\rm l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \text { and } n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Calculation:

A line has direction ratios a + b, b + c, c + a

Let, direction cosines be l, m, and n 

So,

\(\rm l=\frac{a+b}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}\)

\(m=\frac{b+c}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}} \)

\(n=\frac{c+a}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}\)

\(\rm l^2+m^2+n^2=\rm [\frac{a+b}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}]^2+[\frac{b+c}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}]^2 +[\frac{c+a}{\sqrt{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}}]^2\)

\(\ =[\frac{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}{(a+b)^{2}+(b+c)^{2}+(c+a)^{2}}]\)

= 1

Hence, option (4) is correct.

Explanation:

Two-dimensional coordinate axes divide the plane into four quadrants; 

Three-dimensional coordinate axes will divide the space into eight "compartments" known as octants.

Let (x, y, z) be an ordered triad. Take the point L of coordinate x on \(\overset \leftrightarrow{X'X}\) , the point M of coordinate y on \(\overset \leftrightarrow{Y'Y}\) and the point N of coordinate z on \(\overset \leftrightarrow{Z'Z}\). Through the point L, M, N draw planes π₁ , π₂ , π₃  perpendicular to the coordinate axes \(\overset \leftrightarrow{X'X}\),\(\overset \leftrightarrow{Y'Y}\), \(\overset \leftrightarrow{Z'Z}\) respectively.

The three mutually perpendicular planes π₁ , π₂ , π₃ intersect at a unique point of P.

Since P ϵ π₁ , the projection of P on \(\overset \leftrightarrow{X'X}\) is L and hence the x - coordinate of P is equal to L. Similarly the y - coordinate of P is equal to the y - coordinate of M and the z -coordinate of p is equal to the z -coordinate of N. The coordinates of P are x, y, z in that order. Thus for the ordered triad (x, y, z) we have a unique point P in space.

Hence a one-to-one correspondence is established between the set of points in space and the set of ordered triads of real numbers. This space is called 3D space or R³ space.

The three coordinate planes divide the space into eight compartments, each of which is called an octant.

Concept:

Plane parallel to xy- plane so, x-intercept = y-intercept = 0

 

Calculation:

Here, z-intercept = 5, and parallel to xy-plane

∴Equation of plane: z = 5

Hence, option (2) is correct.

Concept:

If vector making an angle θ with x-axis in xy plane, is given by, \(\rm cos \theta \hat i+sin\theta \hat j\)

 

Calculation:

Here, â is a unit vector in the xy-plane making an angle 30° with the positive x-axis

 \(\rm \vec a=cos 30 \hat i + sin 30 \hat j\ =\frac{\sqrt3}{2}\hat i+\frac{1}{2}\hat j\ =\frac{\sqrt 3\hat i+\hat j}{2}\)

Hence, option (1) is correct.

Concept: 

\(\rm cos^2α +cos^2β+cos^2γ =1\), where α, β, γ are the angles which inclined with X, Y, and Z axis respectively.

 

Calculation: 

Here, OA is inclined at angles 45° and 60° with x-axis and y-axis respectively.

 \(\rm cos^245+cos^260+cos^2\gamma=1\ \Rightarrow \frac{1}{2}+\frac{1}{4}+cos^2\gamma=1\ \Rightarrow cos^2\gamma =1-\frac{3}{4} =\frac{1}{4}\ \ \Rightarrow cos \gamma=\pm\frac{1}{2}\ So, \ \gamma =\pm60 \)

\(\rm\overrightarrow{O A}=|OA|(\cos 45^{\circ} \hat i+\cos 60^{\circ} \hat j\pm\cos 60^{\circ} \hat{k})\ =12(\frac{1}{\sqrt2}\hat i+\frac{1}{2}\hat j\pm\frac{1}{2}\hat k)\ \)

= \(\rm 6\sqrt{2} \hat{i}+6\hat{j}\pm 6\hat{k}\)

Hence, option (3) is correct.

Concept:

Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

 

⇒\(\rm\vec R=\vec a+\vec b\)

Parallelogram law of vector addition: 

If two vectors are acting simultaneously at a point, then it can be represented both in magnitude and direction by the adjacent sides drawn from a point. Therefore, the resultant vector is completely represented both in direction and magnitude by the diagonal of the parallelogram passing through the point

\(\rm\vec c=\vec a+\vec b\)

 

Calculation:

Here, vector a and b are adjacent sides of parallelogram

a = 2î - 4ĵ + 5k̂ and b = î - 2ĵ - 3k̂.

Let vector c and d are digonals of parallelogram 

Using parallelogram law, 

\(\rm\vec c=\vec a+\vec b\)

= (2î - 4ĵ + 5k̂) + (î - 2ĵ - 3k̂)

=.\(\rm 3\hat i-6\hat j+2\hat k\)

Now using triangle law,

\(\rm \vec b+\vec d = \vec a\ \Rightarrow \vec d=\vec a-\vec b\)

= (2î - 4ĵ + 5k̂) - (î - 2ĵ - 3k̂)

= \(\rm \hat i-2\hat j+8\hat k\)

\(\rm \vec c. \vec d=(\rm 3\hat i-6\hat j+2\hat k).(\rm \rm \hat i-2\hat j+8\hat k)\ =3+12+16\ =31\)

Hence, option (3) is correct.

Concept:

Let \(\rm\vec {a}\) and \(\rm \vec {b}\) be the two vectors,

Dot product of two vectors  is given by:  \(\rm \vec{a}.\vec{b} = |\vec{a}|.|\vec{b}| cos θ\)

Cross product of two vectors  is given by: \(\rm \vec{a}× \vec{b} = |\vec{a}|.|\vec{b}| \sin θ; \hat{n}\)

 

Calculation:

Given:\(\rm |\vec{a}× \vec{b}|^2 + |\vec{a}\cdot \vec{b}|^2=144\) and \(\rm |\vec{a}|=4\)

To Find: \(\rm |\vec{b}|\)

\(\rm |\vec{a}× \vec{b}|^2 + |\vec{a}\cdot \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 \sin^2 \theta +|\vec{a}|^2|\vec{b}|^2 \cos^2 \theta \=|\vec{a}|^2|\vec{b}|^2 [\sin^2 \theta+\cos^2 \theta]\=|\vec{a}|^2|\vec{b}|^2\)

Therefore, \(\rm |\vec{a}× \vec{b}|^2 + |\vec{a}\cdot \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2\)

⇒ 144 = 4² × \(\rm |\vec{b}|^2\)

⇒ \(\rm |\vec{b}|^2\) = 9

∴ \(\rm |\vec{b}|\) = 3

Concept:

 \(\begin{aligned} &\text { Let } \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \overrightarrow{\mathrm{i}}+\mathrm{b}_{1} \overrightarrow{\mathrm{j}}+\mathrm{c}_{1} \overrightarrow{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\mathrm{a}_{2} \overrightarrow{\mathrm{i}}+\mathrm{b}_{2} \overrightarrow{\mathrm{j}}+\mathrm{c}_{2} \overrightarrow{\mathrm{k}} \text { and } \overrightarrow{\mathrm{c}}=\mathrm{a}_{3} \overrightarrow{\mathrm{i}}+\mathrm{b}_{3} \overrightarrow{\mathrm{j}}+\mathrm{c}_{3} \overrightarrow{\mathrm{k}} \text { be the three vectors }\ &\text { Condition for coplanarity: } \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\left|\begin{array}{lll} \rm a_{1} & \mathrm{b}_{1} & \mathrm{c}_{1} \ \mathrm{a}_{2} & \mathrm{b}_{2} & \mathrm{c}_{2} \ \mathrm{a}_{3} & \mathrm{b}{3} & \mathrm{c}_{3} \end{array}\right|=0 \end{aligned}\)

Calculation:

Here, \(\rm \vec{a}=2\hat{i}-3\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-3\hat{k}\) and \(\rm \vec{c}=\hat{j}+p \hat{k}\) are coplanar 

\(\begin{array}{l} \left|\begin{array}{ccc} 2 & -3 & 1 \ 1 & 2 & -3 \ 0 & 1 & p \end{array}\right|=0 \ \rm \Rightarrow 2(2 p+3)+3(p)+1(1)=0 \ \rm \Rightarrow 4p+6+3p+1=0 \ \rm \Rightarrow 7p+7=0 \ \rm \Rightarrow p=-1 \end{array}\)

Hence, option (2) is correct.