Quantitative Aptitude PYQs

Concept:

If A, B and C are 3 independent events, then P (A ∩ B ∩ C) = P(A) × P(B) × P(C)

Calculation:

Given coin is biased heads comes up thrice as likely as tails.

\( \Rightarrow P\left( H \right) = \frac{3}{4};and;P\left( T \right) = \frac{1}{4}\)

Let, T be a random variable for the no. of tails for 3 independent tosses.

\( \Rightarrow P\left( {T = 0} \right) = P\left( {HHH} \right) = {\left( {\frac{3}{4}} \right)^3} = \frac{{27}}{{64}}\)

\( \Rightarrow P\left( {T = 1} \right) = P\left( {HHT,;THH,;HTH} \right) = 3 \times \left( {\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}} \right); = \frac{{27}}{{64}}\)

\( \Rightarrow P\left( {T = 2} \right) = P\left( {TTH,;THT,;HTT} \right) = 3 \times \left( {\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4}} \right) = \frac{9}{{64}}\)

\( \Rightarrow P\left( {T \le 2} \right) = ;P\left( {T = 0} \right) + ;P\left( {T = 1} \right) + ;P\left( {T = 2} \right) = \frac{{27}}{{64}} + \frac{{27}}{{64}} + \frac{9}{{64}} = \frac{{63}}{{64}} = 0.98\)

Concept:

Let A be an event, then \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( S \right)}}\) where S is the sample space of an experiment.

Calculation:

Given: Out of 20 books 5 books are defective.

Let A → Defective book is selected

Probability of selecting a defective book in 1^st attempt = 5 / 20

Probability of selecting a defective book in 2^nd attempt without replacement = 4 / 19

Probability of selecting a defective book in 3^rd attempt without replacement = 3 / 18

Probability of selecting a defective book in 3 attempts without replacement \(= \frac{1}{4} \times \frac{4}{{19}} \times \frac{1}{6} = \frac{1}{{114}} = 0.009\)

Concept:

If a₁, a₂ ,….., a_n are n observations arranged in ascending order. Then median of the given observation is:

1. If n is odd then there will be only one middle term which will be the median for the given observation.
2. If n is even then there will be two middle terms, so the average of two middle terms will be the median for the given observation.

Calculation:

Given: The median of the observations 22, 24, 33, 37, x + 1, x + 3, 46, 47, 57, 58 in ascending order is 42.

∵ there 10 observations ⇒ n = 10.

Hence, there are two middle terms which are (x + 1) and (x + 3).

So, the median for the given observation \( = \frac{{\left( {x + 1} \right) + \left( {x + 3} \right)}}{2} = x + 2\)

∵ It is given that the median is 42.

⇒ x + 2 = 42 ⇒ x = 40.

⇒ The 5^th observation is (x + 1) = 41 and the 6^th observation is (x + 3) = 43.

Concept:

The standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times ;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean.

Calculation:

Given: μ = 60, N =10 and \(\mathop \sum \limits_{i = 1}^{10} {\left( {{x_i} - 50} \right)^2} = 5000\)

As we know, \(\mu = \frac{{\mathop \sum \nolimits_{i = 1}^{10} {x_i}}}{{10;}} = 60 \Rightarrow ;\mathop \sum \limits_{i = 1}^{10} {x_i} = 600\) 

\( \Rightarrow \mathop \sum \limits_{i = 1}^{10} {\left( {{x_i} - 50} \right)^2} = ;\mathop \sum \limits_{i = 1}^{10} x_i^2 - 100;\mathop \sum \limits_{i = 1}^{10} {x_i} + 25000 = 5000$\) -----(1)

By substituting the value of \(\mathop \sum \limits_{i = 1}^{10} {x_i} = 600\) in equation (1), we get

\(\Rightarrow ;\mathop \sum \limits_{i = 1}^{10} x_i^2 = 40000\)       -----(2)

As we know that, the standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times ;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean.

\(\Rightarrow {\sigma ^2} = \frac{1}{N}; \times ;\mathop \sum \limits_{i = 1}^N {\left( {{x_i} - \mu } \right)^2};\)

\(\Rightarrow {\sigma ^2} = \frac{1}{N} \times \left( {\mathop \sum \limits_{i = 1}^{N;} x_i^2 + {\mu ^2} \times ;\mathop \sum \limits_{i = 1}^N 1 - 2\mu \times ;\mathop \sum \limits_{i = 1}^N {x_i}} \right);\)    ----(3)

By substituting the values of μ = 60, N =10, \(\mathop \sum \limits_{i = 1}^{10} x_i^2;and;\mathop \sum \limits_{i = 1}^{10} {x_i}\) in equation (3), we get

\( \Rightarrow {\sigma ^2} = 400\)

⇒ σ = 20$

Concept:

If α and β are the roots of the quadratic equation, ax² + bx + c = 0. Then

\(\alpha + \beta = ; - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}\)

Calculation:

Given: p and q are the roots of the equation x² – 30x + 221

By comparing the given equation with the standard quadratic equation ax² + bx + c = 0, we get a = 1, b = - 30 and c = 221.

As we know that, if α and β are the roots of the quadratic equation, ax² + bx + c = 0. Then

\(\alpha + \beta = ; - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}\)

\(\Rightarrow p + q = ;;30;and;pq = 221\)

\(\Rightarrow {p^3} + {q^3} = \left( {p + q} \right) \times \left( {{p^2} - pq + {q^2}} \right) = ;\left( {p + q} \right) \times \left[ {{{\left( {p + q} \right)}^2} - 3pq} \right]\)(\Rightarrow {p^3} + {q^3} = ;\left( {p + q} \right) \times \left[ {{{\left( {p + q} \right)}^2} - 3pq} \right] = 30 \times \left[ {900 - 663} \right] = 7110\)

Concept:

The line of regression of y on x is given by: \(y - ;\bar y = {b_{yx}};\left( {x - ;\bar x} \right)\) where b_yx is called the regression coefficient of y on x.

Similarly, the line of regression of x on y is given by: \(x - ;\bar x = {b_{xy}};\left( {y - \bar y} \right)\) whereb_xy is called the regression coefficient of x on y.

The correlation coefficient r² = b_yx × b_xy

The two lines of regression intersect each other at \(\left( {\bar x;,;\bar y} \right)\)

Calculation:

Given: Two regression lines are 6x + y = 30 and 3x + 2y = 25.

As we know that, the two lines of regression intersect each other at \(\left( {\bar x;,;\bar y} \right)\)

By solving these two equations: 6x + y = 30 and 3x + 2y = 25

We get \(\left( {\bar x;,;\bar y} \right) = \left( {\frac{{35}}{9},\frac{{20}}{3}} \right)\)

We can write 6x + y = 30 as line of regression of x on y: \(x - \frac{{35}}{9} = ; - \frac{1}{6} \times \left( {y - \frac{{20}}{3}} \right)\)   ------(1)

By comparing equation (1), with line of regression of x on y which is given by: \(x - ;\bar x = {b_{xy}};\left( {y - \bar y} \right)\) we get \({b_{xy}} = ; - \frac{1}{6}\) 

Similarly, we can write 3x + 2y = 25 as line of regression of y on x: \(y - \frac{{20}}{3} = ; - \frac{3}{2};\left( {y - \frac{{35}}{9}} \right)\)    ------(2)

By comparing equation (2), with line of regression of x on y which is given by \(;y - ;\bar y = {b_{yx}};\left( {x - ;\bar x} \right)\): we get \({b_{yx}} = ; - \frac{3}{2}\)

As we know that, r² = b_yx × b_xy

\( \Rightarrow {r^2} = ; - \frac{1}{6} \times ; - \frac{3}{2} = \frac{1}{4}\)

\( \Rightarrow r = ; \pm \frac{1}{2} = ; \pm 0.5\)

As we know that, sign of \(r,;{b_{xy}};and;{b_{yx}};is;always;same\)

⇒ r = - 0.5

Concept:

Class mark is the mid value of a class interval. i.e for a class interval c – d the class mark is given by:

c + d / 2.

Calculation:

Given: class marks in a frequency table are given to be 5, 10, 15, 20, 25, 30, 35, 40, 45, 50

So, option 1, 3 and 4 are discared ∵ the class width of each class interval is not uniform although the class mark of each of the class interval is same as required.

Hence, the correct option is 2.

Concept:

Variance is given by: \({\sigma ^2} = \frac{1}{N}; \times ;\mathop \sum \limits_{i = 1}^N {\left( {{x_i} - \mu } \right)^2}\), where μ is the mean and N is the no. of observations.

Calculation:

Given: mean of 5 observations is 4.4 and variance is 8.24. 3 observations are given to us 1, 2 and 6.

Let the other two observations are a and b.

\( ⇒ \mu = 4.4 = \frac{{1 + 2 + 6 + a + b}}{5} ⇒ a + b = 13\)

From the given options only one option satisfies the above condition.

Hence, option 2 is the correct option.

Alternate Method Consider X and Y is remaining two observation.

⇒ Mean = 4.4 = \(\frac{1+2+6+X+Y}{5}\)

⇒ 9 + X + Y = 22

⇒ X + Y = 13 ....

(1) We have,

variance = 8.24

⇒ \(\frac{1}{5}(1^2+2^2+6^2+X^2+Y^2)-4.4^2=8.24\)

⇒ \(\frac{41+X^2+y^2}{5}-19.36=8.24\)

⇒ X² + Y² + 41 = 138

⇒ X² + Y² = 97 .... (2) (x − y)² + (x + y)² = (x − y)² + 169

⇒ (x − y)² +169 = 2 × 97

⇒ (x − y)² = 25

⇒ x − y = 5 .... (3) From equation (1) & (3),

X = 9, Y = 4

Concept:

• We know, that a coin has two sides: Head and Tail.
• When a coin is tossed once, the sample space is given by: {H, T}

⇒ 2 = 2¹ possibilities in which the favorable event will be only one i.e. {H}

• When a coin is tossed twice, the sample space is given by: {HH, TT, HT, TH}

⇒ 4 = 2² possibilities in which the favorable event will be only one i.e. {TH} as we will go for the second toss if and only if we get Tail in the first toss.

• When a coin is tossed thrice, the sample space is given by: {HHH, TTT, TTH, THT, THH, HTT, HTH, HHT}

⇒ 8 = 2³ possibilities in which the favorable event will be only one i.e. {TTH} as we will go for the third toss if and only if we get Tail in the first and the second toss.

• When a coin is tossed N times, the sample space is given by: {HHHH..(N times), TTTT...(N times),.........}
• ⇒ 2ⁿ possibilities.

Explanation:

• In the given question, When a coin is tossed we can get head in the 1^st attempt or 2^nd attempt OR for the first toss only, if we would have got the head, we would stop there.
• So, the required sample space

S = {H, TH, TTH, TTTH, ……….} giving us infinite elements.

Concept:

In order to find the median of a discreet frequency distribution follow these steps:

1. First arrange the data in ascending or descending order and find the cumulative frequency (cf)
2. Find N / 2, where \(N = ;\mathop \sum \limits_{i = 1}^n {f_i}\)
3. See the cumulative frequency just greater than N / 2. The corresponding value of x is median.

Calculation:

Step 1:

Step 2:

\(N = ;\mathop \sum \limits_{i = 1}^n {f_i} = 240\)

⇒ N / 2 = 120

Step 3:

The cumulative frequency just greater than 120 is 170 and the corresponding x is 5.

Hence the median for the given discreet frequency distribution is 5.

Concept:

Probability of an event A is given by

\(P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}}\)

Calculation:

The sample space for the two dice thrown simultaneously is :

S = {(a, b) | a = 1, 2, ….., 6 and b = 1, 2, ……, 6}

⇒ n (S) = 36

Let A be a subset of S defined as: A = {(a, b) | a + b is a prime no. ∀ a and b = 1, 2, 3, …., 6}

A = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}

⇒ n (A) = 15

As we know that, \(P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}}\)

\(P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}} = \frac{{15}}{{36}} = \frac{5}{{12}}\)

Concept:

Probability of an event A is given by: \(P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}}\)

Calculation:

Given: There are 10 delivery trucks out of which 5 trucks do not meet the emission standards.

Let A = The chosen truck will not meet the emission standards.

No. of favourable cases for event A \( = {;^5}{C_3}\)

Total no. of cases \(= {;^{10}}{C_3}\)

 As we know that, probability of an event A is given by: \(P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}}\)

\( \Rightarrow P\left( A \right) = \frac{{n;\left( A \right)}}{{n;\left( S \right)}} = \frac{{{;^5}{C_3}}}{{{;^{10}}{C_3}}} = \frac{1}{{12}}\)

Calculation:

Let E₁, E₂ and E₃ be the respective events of choosing a two headed coin, a biased coin and an unbiased coin.

⇒ P (E₁) = P (E₂) = P (E₃) = 1 / 3.

Let A be the event that coin shows head.

⇒ P (A | E₁) = 1, P(A | E₂) = 75 / 100 = 3 / 4 and P (A | E₃) = 1 / 2.

We have to find P (E₁ | A). So, by using baye’s theorem , we obtain:

\(\Rightarrow P;\left( {{E_1};|;A} \right) = \frac{{P;\left( {{E_1}} \right) \times P\left( {A;|;{E_1}} \right)}}{{P;\left( {{E_1}} \right) \times P\left( {A;|;{E_1}} \right) + ;P;\left( {{E_2}} \right) \times P\left( {A;|;{E_2}} \right) + ;P;\left( {{E_1}} \right) \times P\left( {A;|;{E_3}} \right)}}\)

\( \Rightarrow P;\left( {{E_1};|;A} \right) = \frac{{\frac{1}{3} \times 1}}{{\frac{1}{3} \times \left( {1 + \frac{3}{4} + \frac{1}{2}} \right)}} = \frac{4}{9}\)

Concept:

Statement 1:

If A and B are mutually exclusive events, then P (A ∩ B) = 0.

We know that P (A ∪ B) = P(A) + P(B) – P (A ∩ B) = 0.6 + 0.6 = 1.2 > 1

This contradicts ∵ probability of any event say A: 0 ≤ P (A) ≤ 1.

Hence statement 1 is wrong.

Statement 2:

Given:  P(A|B) = 1

As we know, \(P;\left( {A;|;B} \right) = \frac{{P\left( {A \cap B{\rm{;}}} \right)}}{{P;\left( B \right)}} = 1\)

⇒ B ⊆ A

\( \Rightarrow P;\left( {\bar B;|;\bar A} \right) = \frac{{P;\left( {\bar A \cap ;\bar B} \right)}}{{P;\left( {\bar A} \right)}} = \frac{{P\left( {\overline {A \cup B} } \right)}}{{P\left( {\bar A} \right)}} = \frac{{P\left( {\bar A} \right)}}{{P\left( {\bar A} \right)}} = 1\)

Hence, statement 2 is true.

Concept:

Binomial distribution:

\(b;\left( {x;n;,;p} \right) = ;{;^n}{C_x} \times {p^x} \times {q^{n - x}}\) where p is the probability of success, q is the probability of failure, n is the total no. of attempts and x is the no. of successful attempts.

Calculation:

Given: A fair die is rolled 4 times.

Let p represent the probability of getting 6 when a dice is rolled = 1 / 6

Let q represent the probability of not getting 6 when a dice is rolled

= 1 – (1 / 6) = 5 / 6

As we know that, according to binomial distribution:

\(b;\left( {x;n;,;p} \right) = ;{;^n}{C_x} \times {p^x} \times {q^{n - x}}\)

Here, n = 4, x = 2, p = 1 / 6 and q = 5 / 6.

So, the probability of getting exactly 2 sixes when a fair dice is rolled 4 times

⇒ \(b;\left( {x;n;,;p} \right) = ;{;^4}{C_2} \times {\left( {\frac{1}{6}} \right)^2} \times {\left( {\frac{5}{6}} \right)^2} = \frac{{25}}{{216}}\)

Concept:

The standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times ;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean

Calculation:

Given: Mean of 100 observations is 50 and standard deviation is 10.

If 5 is added to each observation and ∵ there are total 100 observations ⇒ 500 is added to sum of the observations and now dividing it by the no. of observations , we get that the mean is increased by 5.

As we know that, the standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times ;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean

∵ Every observation increased by 5 and mean also increased by 5 and standard deviation is the square root of the difference of mean and each observation divided by the number of observations.

So standard deviation will remain same.

Hence, new mean is 55 and new standard deviation is still 10.

Concept:

If the range of a set of observations on a variable X is d and if Y = a + bX , then the range of the set of corresponding observations on Y is given by: b × d.

Calculation:

Given: Range of a set of observations on a variable X is 25 and Y = 40 + 3 X.

As we know that, if the range of a set of observations on a variable X is d and if Y = a + bX , then the range of the set of corresponding observations on Y is given by: b × d

Here, d = 25, a = 40 and b = 3.

So, the range of the set of corresponding observations on Y is b × d = 3 × 25 = 75.

Concept:

The standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} × ;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}} \) where μ is the arithmetic mean and it is given by: \(\frac{{\mathop \sum \nolimits_{i = 1}^n {x_i}}}{N}\)

Variance is given by: \({\sigma ^2} = \frac{1}{N}; × ;\mathop \sum \limits_{i = 1}^N {\left( {{x_i} - \mu } \right)^2}\), where μ is the mean and N is the no. of observations.

Calculation:

Given: V is the variance and M is the mean of first 15 natural numbers

Here, N = 15 and the observations are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15.

As we know that mean is given by: \(\frac{{\mathop \sum \nolimits_{i = 1}^n {x_i}}}{N}\)

\( ⇒ M = \frac{{1 + 2 + 3 + \ldots + 15}}{{15}} =\frac{ \frac{15× 16}{2}}{15}\)

⇒ M = \( \frac{{15× 16}}{{2× 15}} = 8\)

As we know that, variance is given by: \({\sigma ^2} = \frac{1}{N}; × ;\mathop \sum \limits_{i = 1}^N {\left( {{x_i} - \mu } \right)^2}\), where μ is the mean and N is the no. of observations.

\( ⇒ V= \frac{1}{{15}}; × ;\mathop \sum \limits_{i = 1}^{15} {\left( {{x_i} - 8} \right)^2} \)

\( ⇒ V= \frac{1}{{15}} × \Big[7^2+6^2+5^2+4^2+3^2+2^2+1^2+0^2+1^2 +.......+7^2\Big]\)

\( ⇒ V= \frac{1}{{15}} × \Big[2\times 7^2+2\times 6^2+2\times 5^2+2\times4^2+2\times3^2+2\times2^2+2\times1^2\Big]\)

\( ⇒ V= \frac{2}{{15}} × \Big[7^2+ 6^2+ 5^2+4^2+3^2+2^2+1^2\Big]\)

\( ⇒ V= \frac{2}{{15}} × \Big[49+ 36+ 25+16+9+4+1\Big]\)

\(\Rightarrow V =\frac{2\times 140}{15}= \frac{56}{3}\)

\( \therefore V+ {M^2} = \frac{{56}}{3} + 64 = \frac{{248}}{3};\)

Concept:

If an object covers D km at a speed of S₁ km/hr and next D km at a speed of S₂ km / hr. Then the average speed of the object is given by: \( = \frac{{2 \times {S_1} \times {S_2}}}{{{S_1} + {S_2}}}\)

Calculation:

As we know that, if an object covers D km at a speed of S₁ km / hr and next D km at a speed of S₂ km / hr. Then the average speed of the object is given by: \( = \frac{{2 \times {S_1} \times {S_2}}}{{{S_1} + {S_2}}}\)

Here, D = 60 km, S₁ = 3v km / hr, S₂ = 2v km/hr

The average speed of the car \(= \frac{{2 \times {S_1} \times {S_2}}}{{{S_1} + {S_2}}} = \frac{{2 \times 3v \times 2v}}{{5v}} = 2.4v~km/hr\)

Concept:

 Mean of N observations is given by: \(\frac{{\mathop \sum \nolimits_{i = 1}^n {x_i}}}{N}\)

Calculation:

It is given that mean weight of 150 students in a certain class is 60 kg.

Let M be the mean weight of 150 students

\( \Rightarrow M = \frac{{\mathop \sum \nolimits_{i = 1}^{150} {x_i}}}{{150}} = 60 \Rightarrow ;\mathop \sum \limits_{i = 1}^{150} {x_i} = 9000\)        -----(1)

Let there are x no. of boys ⇒ The no .of girls = 150 – x

It is given that mean weight of the boys is 70 kg and that of the girls is 55 kg

Let M₂ = 70 and M₃ = 55.

The total weight of all the boys = 70x

The total weight of all the girls = 55 × (150 - x) = 8250 – 55x

The total weight of all the students = 8250 – 55x + 70x = 8250 + 15x ----------(2)

Using equation (1) and (2), we get

⇒ 8250 + 15x = 9000 ⇒ x = 50.

∴ No. of boys = 50 and No. of girls = 150 – 50 = 100.