Quantitative Aptitude PYQs

Concept:

For the observations \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)

Mean (\(\rm\overline x\)) = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)

Calculation:

Given mean of the n observation is M

M = \(\rm x_1 - k +x_2 - k+ x_3 - k +..........+x_n-k\over n\)

M = \(\rm (x_1 +x_2 + x_3 +..........+x_n)-nk\over n\)

M = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}-{nk\over n}\)

M = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\) - k

M + k = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)

∴ The required mean of x1, x2, x3, _ _ _, xn = M + k

Concept:

For the observations \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)

Mean (\(\rm\overline x\)) = \(\rm {x_1 +x_2 + x_3 +..........+x_n\over n}\)

Deviation of the observation from the mean = x_i - \(\rm\overline x\)

where i ∈ [1, n]

Calculation:

Mean = \(\rm {73+85+92+105+120\over 5}\)

\(\rm\overline x\) = \(\rm {475\over 5}\) = 95

 

The sum of the deviation = -22 +(-10) + (-3) + 10 + 25 = 0

Concept:

For any n numbers \(\rm x_1 ,x_2 ,x_3 ,..........,x_n\)

GM = \(\rm \sqrt[n]{x_1\times x_2\times x_3\times ..........\times x_n}\)

HM = \(\rm {\left[{1\over x_1}+{1\over x_2}+{1\over x_3}+..........+{1\over x_n}\over n\right]}^{-1}\)

Calculation:

GM of m and n = y

\(\rm \sqrt{mn}\) = y

HM of m and n = x

\(\rm {\left[{1\over m}+{1\over n}\over 2\right]}^{-1}\) = x

\(\rm {2mn\over m+ n}\) = x

Now given 5x = 4y

\(\rm {10mn\over m+ n}\) = \(\rm 4\sqrt{mn}\)

\(\rm 10\sqrt{mn}\) = 4(m + n)

100mn = 16(m² + n² + 2mn)

16m2 + 16n2 - 68mn = 0

\(\rm16{m\over n}+16{n\over m} -68= 0\)

Let \(\rm m\over n\) = t

16t + \(\rm 16\over t\) - 68 = 0

16t² - 68t + 16 = 0

t = \(\rm {68 \pm \sqrt{(-68)^2-4(16)(16)} \over 2\times16}\)

t = \(\rm {68 \pm \sqrt{4624-1024} \over 32}\)

t = \(\rm {68 \pm60\over 32}\) = 4 or \(\rm 1\over 4\)

\(\rm m\over n\) = 4 or \(\rm 1\over 4\)

∴ m = 4n or n = 4m

Concept:

Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

Variance = (Standard Deviation)²

Calculation:

Given coefficient of variation = 45% = 0.45

And mean = 100

As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)

0.45 = \(\rm\text{Standard Deviation} \over100\)

Standard Deviation = 100 × 0.45

SD = 45

∴ Variance = 45² = 2025

Concept:

\(\rm P(A|B)=\dfrac{P(A∩ B)}{P(B)}\)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation:

Given P(A) = L and P(B) = M

Here P(A ∪ B) ≤ 1 (∵ Max value of probability is 1)

P(A) + P(B) - P(A ∩ B) ≤ 1

L + M - P(A ∩ B) ≤ 1

P(A ∩ B) ≥ L + M - 1

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

\(\rm P(A|B)\geq\dfrac{L+M-1}{M}\)

Calculation:

Option 2 has 

Mean = \(6+18+18+18+30\over5 \) = 18

Median = 18 {As it is the middle value of the arrangement}

Mode = 18 {As it appears the most}

∴ Mean = Median = Mode

Option 2 is correct

Calculation:

Let the discarded values be x and y 

Sum of all the 12 values = 12 × mean

S₁ = 12 × 75 = 900

Sum of the 10 values after discarding = 10 × new mean

S₂ = 10 × 65 = 650

Now 650 + x + y = 900

x + y = 900 - 650 = 250

Average of x and y =\(\rm x+y\over2\)

Average = \(\rm 250\over2\) = 125

Concept:

For a quadratic equation ax2 + bx + c = 0

The sum of the roots = \(\rm-{b\over a}\)

The product of the roots = \(\rm{c\over a}\)

Calculation:

Let the other root be β  

Given equation is x(x + 1) + 1 = 0

⇒ x² + x + 1 = 0

a = 1, b = 1 and c = 1

As k is the root of the equation

⇒ k2 + k + 1 = 0

⇒ k2 = -1 - k     .....(i)

The sum of the roots = \(-{1\over1}\) = -1

⇒ β + k = -1

⇒ β = -1 - k      .....(ii)

From equation (i) and (ii), we get

⇒ β = k²

∴ The other root = k²

 

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Given equation is x(x + 1) + 1 = 0

Factor of (x2 + x + 1) = 0

\({\rm{x}} = {\rm{;}}\frac{{ - 1{\rm{;}} \pm {\rm{;}}\sqrt {{1^2} - 4{\rm{;}} \times 1{\rm{;}} \times 1} }}{{2{\rm{;}} \times 1}} = {\rm{;}}\frac{{ - 1{\rm{;}} \pm {\rm{i}}\sqrt 3 }}{2}\)

\(⇒ {\rm{x}} = {\rm{;}}\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}{\rm{;or;;}}\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\)

⇒ x = ω or ω2

Consider k = ω 

∴ The other root = ω² = k2

Concept:

If GM of series x_i = a, then 

• GM of λx_i = λa
• GM of \(\rm {x_i}^n\) = aⁿ

Calculation:

Given GM of x_i = 10

Then GM of \(\rm {x_i}^4\) = 10⁴ = 10000

Now GM of \(\rm 3{x_i}^4\) = 3 × 10000 = 30000

Concept:

For any events A and B

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• Events are independent if P(A ∩ B) = P(A) × P(B)
• Events are mutually exclusive if P(A ∩ B) = 0
• P(\(\rm\bar A\)) = 1 - P(A)

 

Calculation:

Given \(\rm P(A\cup B)=\dfrac{5}{6}, P(A∩ B)=\dfrac{1}{3}:and:P(\bar A)=\dfrac{1}{2}\)

P(A) = 1 - \(1\over2\) = \(1\over2\)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

\(\rm{5\over6} = {1\over2}+P(B)-{1\over3}\)

P(B) = \(2\over3\)

Given P(A ∩ B) ≠ 0

P(A ∩ B) = P(A) × P(B) = \({1\over2}\times{2\over3} = {1\over3}\)

∴ A and B are independent events and are not mutually exclusive events.

Only statement 1 is correct.

Formula Used:

Average = \(\frac{X_i}{n}\)

Here, X_i is the sum of observation and n is the number of reading

Calculation:

Let the average initially be x

Then the sum of the observations = 15x

As the tens digit is recorded 5 more, therefore the sum is 50 more than the correct sum. 

Correct sum of the observations = 15x - 50

The correct average = \(\rm 15x-50\over15\) = x - \(10\over3\)

∴ The average is reduced by \(10\over3\)

Concept:

For the independent events A and B, 

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A ∩ B) = P(A) × P(B)

 

Calculation:

Probability of getting head on first toss P(E) = \(1\over2\)

Probability of getting head on first toss P(F) = \(1\over2\)

Since the events are independent P(E ∩ F) = P(E) × P(F)

P(E ∩ F) = \(1\over2\) × \(1\over2\) = \(1\over4\)

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

P(E ∪ F) = \(1\over2\) + \(1\over2\) - \(1\over4\)

P(E ∪ F) = \(3\over4\)

Concept:

For the binomial distribution of probability

• Mean = np
• Variance = npq

Where n is the total number of cases, p is probability of favorable cases and q is probability of unfavorable cases(1 - p)

Calculation:

Given mean = np = \(2\over3\) and variance = npq = \(5\over9\)

\(\rm variance \over mean\) = \({5\over9}\over{2\over3}\)

q = \(5\over6\)

p = 1 - q = \(1\over6\)

np = \(2\over3\)

n = 4

∴ The probability that random variable D = 2

P = \(\rm {^nC_2}p^2q^{n-2}\)

P = \(\rm {^4C_2}\times\left({1\over6}\right)^2\times\left({5\over6}\right)^2\)

P = \(\rm 6\times{1\over36}\times{25\over36}\)

P = \(25\over216\)

Concept:

The mode of the n observation is the number that has the highest frequency.

Calculation:

Given scores 10, 12, 13, 15, 15, 13, 12, 10, x and mode = 15

Frequency of score '12' = 2

Frequency of score '15' = 2

As Mode = 15, its frequency has to be more than 2, if the value of x = 15

Then the frequency of score 15 = 3 & is highest.

So, x should be '15'

Concept:

For any two events A and B

• The minimum value of P(A ∪ B) = max [P(A), P(B)]
• The maximum value of P(A ∩ B) = min [P(A), P(B)]

 

Calculation:

Given \(\rm P(A)=\dfrac{3}{4} :and: P(B)=\dfrac{5}{8}\)

P(A) > P(B)

∴ The minimum value of P(A ∪ B) = \(\dfrac{3}{4}\)

The maximum value of P(A ∩ B) = \(\dfrac{5}{8}\)

Hence both 1 and 2 are correct

Concept:

The differentiation of f(x) wrt g(x) = \(\rm f'(x)\over g'(x)\)

f(x) and g(x) be the function of x

Calculation:

f(x) = e^x 

f'(x) = \(\rm {d\over dx}e^x = e^x\)

g(x) = x^e 

g'(x) = \(\rm {d\over dx}x^e = ex^{e-1}\)

Required derivative D = \(\rm f'(x)\over g'(x)\)

D = \(\rm e^x\over ex^{e-1}\) = \(\rm xe^x\over ex^e\)

Calculation:

Given: \(\rm\mathop {\lim }\limits_{x \to - 1} \dfrac{f(x)+1}{x^2-1}=-\dfrac{3}{2}\)

Denominator becomes zero on putting the limits

∴ The numerator must also be zero to have a certain value of the given limit, i.e.,

\(\rm \mathop {\lim }\limits_{x \to - 1} [f(x)+1]=0\)

\(\rm \mathop {\lim }\limits_{x \to - 1} f(x)\) + 1 = 0

\(\rm \mathop {\lim }\limits_{x \to - 1} f(x)\) = -1

Concept:

For the function to be continuous:

LHL = RHL = f(x)

Where LHL = \(\rm\mathop {\lim }\limits_{α \to 0}\)f(x - α) and RHL = \(\rm\mathop {\lim }\limits_{α \to 0}\)f(x + α)

Calculation:

Given that f(x) is continuous function

LHL = f(x) = RHL

\(\rm\mathop {\lim }\limits_{α \to 0}\) f(1 - α) = f(1)

\(\rm\mathop {\lim }\limits_{α \to 0}\) [a + b(1 - α)] = 5

\(\rm\mathop {\lim }\limits_{α \to 0}\) [a + b - bα] = 5

a + b = 5

Concept:

The function f(x):

• Increases for the values x of having f'(x) > 0
• Decreases for the values of x having f'(x) < 0

 

Calculation:

Given function is f(x) = sin x

∴ f'(x) = cos x

For x ∈ (0, \(π\over2\)), cos x > 0 and for x ∈ (\(π\over2\), π), cos x < 0

∴ For x ∈ (0, π), f(x) increases to max at \(π\over2\) and then decreases to min at π.

For x ∈ \(\left(\dfrac{5\pi}{2},3\pi\right)\), cos x < 0

∴ f(x) decreases for x ∈ \(\left(\dfrac{5\pi}{2},3\pi\right)\)

statement 2 is correct Only.

Concept:

The domain is the set of all possible value of x which have a finite value of f(x).

Calculation:

Given function f(x) = 3^x

The function will have a finite value for all x ∈ (-∞, ∞)

Mistake PointsThe range of the given function will be from (0,∞). 0, when x = -∞, and ∞ when x = ∞.