Quantitative Aptitude PYQs

Explanation:

Lines of regression of x on y

⇒x - 3y + 4 = 0

⇒x = 3y – 4

⇒ b_xy = 3

Line of regression of y on x

⇒ 2x – 7y+8 = 0

⇒ y =\(\frac{2}{7}x + \frac{8}{7}\)

⇒ b_yx = 2/7

Now

⇒ b _xy + 7b _yx  =  3 + 7× 2 /7 = 5

∴ Option (d) is correct

Explanation:

Mean  = \(\frac{1+ 4+ 9 ... n^2}{n}\)

⇒ 130 = \(\frac{(n(n+1)(2n+1)}{6n}\)

⇒ 780 = 2n² + 3n + 1

⇒ 2n² + 3n – 779 =0

⇒ 2n² + 41n – 38n – 779 = 0

⇒ n(2n + 41) – 19(2n + 41) = 0 

⇒ (2n + 41) (n – 19) = 0

⇒ x = 19 ..... ( n = \(-\frac{41}{2}\) is not possible)

∴ Option (b) is correct.

Explanation:

⇒n(S) = ¹⁰C₃

Favourable outcomes = (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9, 10)

⇒n(E) = 8

⇒ P(E) = \(\frac{n(E)}{n(S)} = \frac{8}{^{10}C_3}\)

= \(\frac{8\times3\times2\times1}{10\times9 \times8} =\frac{1}{15}\)

∴ Option (c) is correct

Explanation:

A, B and C are mutually exclusive and exhaustive events:

⇒  P(A) + P(B) + P(C)= 1.....(i)

Also 

⇒ 3P(B) = 4P(A)

⇒ P(A) = \(\frac{3}{4} P(B)\)

and 3P(C) = 2P(B) 

⇒ P(C) = \(\frac{2}{3}P(B)\)

From (i) 

⇒ \(\frac{3}{4}P(B) +P(B)+ \frac{2}{3}P(B) =1\)

⇒ \(\frac{9P(B)+12P(B)+8P(B)}{12} = 1 \)

⇒ \(\frac{29P(B)}{12} =1\)

⇒ P(A) = \(\frac{3}{4} \times \frac{12}{29} = \frac{9}{29}\)

∴ Option (c) is correct.

Explanation:

Number on faces on die are 4, 4, 5, 5, 5 and 6.

⇒n(S) = 6

P (getting 4 or 5) = 5/6

∴ Option (c) is correct

Explanation:

Black = 2, yellow = 4 and white = 6

⇒ Total Balls = = 2 + 4 + 6 = 12

P (all three are of the same colour).

= \(\frac{2^3}{12^3}+ \frac{4^3}{12^3} +\frac{6^3}{12^3} = \frac{1}{6}\)

∴ Option (a) is correct

Explanation:

Here, P(A) = 5/6 , P(B) = 4/5 , P(C) = 3/4

Also

P(B) = \(1 - \frac{4}{5} = \frac{1}{5}\)

Now

⇒ \(P ( A\cap \overline{B} \cap C) = P(A).P(B).P(C)\)

= \(\frac{5}{6}\times \frac{1}{5}\times \frac{3}{4} = \frac{1}{8}\)

∴ Option (a) is correct.

Explanation:

The word given is ZOLOGY. We need to find the probability that the consonants and vowels occur alternatively in all possible arrangements.

• Vowels: O, O, O
• Consonants: Z, L, G, Y

There are 3 vowels and 4 consonants.

The total number of ways the letters of the word "ZOOLOGY" can be arranged (without any restrictions) is:

Formula: Total arrangements = n! / r₁! r₂! ... r_k!

Where:

• n = Total number of letters = 7
• r₁ = Frequency of O = 3

So, the total number of arrangements is:

Total arrangements = 7! / 3! = 5040 / 6 = 840

We need to place the consonants and vowels alternately:

• Consonants will occupy the odd positions (1, 3, 5, 7).
• Vowels will occupy the even positions (2, 4, 6).

The number of ways to arrange the consonants (Z, L, G, Y) in the 4 odd positions is:

4! = 24

The number of ways to arrange the vowels (O, O, O) in the 3 even positions is:

3! / 3! = 1

Thus, the number of favorable arrangements is:

Favorable arrangements = 4! × (3! / 3!) = 24 × 1 = 24

The probability is given by the ratio of favorable arrangements to the total arrangements:

Probability = Favorable arrangements / Total arrangements = 24 / 840 = 1 / 35

The probability that the consonants and vowels occur alternatively is 1 / 35.

Explanation:

⇒n(S) = 100 

Numbers from 7 to 100 satisfy x² + x > 50

Then, n(E)= 100 – 6 = 94

⇒ \(P(E) = \frac{n(E)}{n(S)} = \frac{94}{100} = \frac{47}{50}\)

∴ Option (b) is correct

Explanation:

The mean of 10 natural numbers 

⇒ \(\overline{x} = \frac{1+2+3+4+....10}{10}\) 

= \(\frac{10\times11}{2\times10}\) = 5.5

Mean deviation = \(\frac{[1- 5.5] [2-5.5] [3- 5.5] ... [10- 5.5]}{10} = 2.5\)

∴ Option (b) is correct.

Explanation:

Given:

\(\rm \Sigma_{i=1}^9x_i^2=885\)

Variance σ = \(\rm \Sigma_{i=1}^9x_i^2 - (mean)^2\)

= \(\frac{855}{9} -(M)^2\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

⇒ σ + M² = \(\frac{855}{9} = 95\)

∴ Option (b) is correct.

Explanation:

Given:

\(\overline x = \frac{x_1+x_2+x_3+......... x_n}{n}\)

⇒ \(x_1+x_2+X_3+......x_n= n\overline x \)

⇒ Mean when x_n is replaced by k

⇒ New mean = \(\frac{x_1+x_2+x_3+......... x_{n-1}+k}{n}\)

= \(\frac{n\overline x -x_n+k}{n}\)

∴ Option (d) is correct

Explanation:

A fair coin is tossed till two heads occur in succession.

We are asked to find:

P(the number of tosses required is less than 6)

This means we need to compute the probability that the first occurrence of two consecutive heads (HH) happens at or before the 5^th toss.

Valid sequences where HH first appears at the end are:

HH (length 2)

THH (length 3)

TTHH, HTHH (length 4)

TTTHH, THTHH, HTTHH (length 5)

So the total probability is:

P = P(HH) + P(THH) + P(TTHH) + P(HTHH) + P(TTTHH) + P(THTHH) + P(HTTHH)

= \((\frac{1}{2})^2 + (\frac{1}{2})^3 + 2 \times (\frac{1}{2})^4 + 3 \times (\frac{1}{2})^5\)

= \(\frac{1}{4} + \frac{1}{8} + \frac{2}{16} + \frac{3}{32}\)

= \(\frac{8 + 4 + 4 + 3}{32} = \frac{19}{32}\)

∴ Option (4) is correct.

Explanation:

Given:

Urn A contains 2 white and 2 black balls while urn B contains 3 white and 2 black balls. 

⇒ E1 = One white ball transferred from A to B; then urn A contains white = 1 and b = 2 and urn B contains white = 4 and black = 2

⇒ E2 = One black ball transferred from A to B. Then urn A contains white = 2 and black = 1 and urn B contains white = 3 and black = 3

⇒ F = One white ball is drawn from urn B

⇒ \(P(F) =P(E_1). P(\frac{F}{E_1})+ P(E_2).P(\frac{F}{E_2})\)

= \(\frac{2}{4}\times\frac{4}{6}+\frac{2}{4}\times\frac{3}{6} = \frac{7}{12}\)

∴ Option (b) is correct.

Explanation:

Given:

\(P(A) = P(\frac{A}{B}) = 0.25\)

and \(P(\frac{B}{A}) = 0.5\)

I. \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(A)}\)

⇒ P(A∩B) = P(A) P(B|A)

⇒ P(A∩B) = 0.25 × 0.5 = 0.125

Now

⇒ \(P(\frac{B}{A}) = \frac{P(A∩ B)}{P(B)}\)

⇒ \(P(B)= \frac{P(A∩ B)}{P(\frac{A}{B})}\)

⇒ \(P(B) = \frac{0.125}{0.25} = 0.5\)

Now, P(A).P(B) = 0.25 × 0.5 = 0.125 = P(A∩B)

Thus  A and B are independent

II. \(P(\overline A\cup \overline B ) = 1 – P(A ∩ B)\)

= 1 – 0.125 = 0.875

III. \(P(\overline A∩ \overline B ) = 1 – P(A \cup B)\)

= 1 – [P(A) + P(B) – P(A ∩ B)

=  1 – [0.25 + 0.5 – 0.125]

= = 1 – 0.625 = 0.375

So all statements I, II, and III are correct.

∴ Option (d) is correct.

Explanation:

⇒n(S) = 6 × 6 = 36

⇒Sum 9 = (3, 6), (4, 5), (5, 4), (6, 3)

⇒ Sum 10 = (4, 6), (5, 5), (6, 4)

⇒P (sum 9 or 10) = 7/36

Then, P (sum neither 9 nor 10) = \(1- \frac{7}{36} =\frac{29}{36}\)

∴ Option (d) is correct

Explanation:

Given

n = 6

⇒ p = P (suffering from disease) = 20/100 = 1/5

⇒ q = P (not suffering from disease) = 1 - 1/5 = 4/5

⇒ p(x ≥ 4) = p(x = 4) + p(x = 5) + p(x = 6)

⇒ \(^6C_4(\frac{1}{5})^4 (\frac{4}{5})^2 +^6C_5 (\frac{1}{5})^5(\frac{4}{5})^1+^6C_6(\frac{1}{5})^6\)

= \(\frac{6!}{4!2!}\times\frac{16}{5^6}+ \frac{6!}{5!}\times\frac{4}{5^6}+ \frac{1}{5^6}\) 

= \(\frac{265}{15625} = \frac{53}{3125}\)

∴ Option (a) is correct

Explanation:

⇒n(S) = 6 × 5 × 4

⇒ n(E) = n (one of the faces shown is an ace.)

=  3(1 × 5 × 4) = 3 × 5 × 4 

⇒ \(P(E) = \frac{3\times 5\times 4}{6\times 5\times4} = \frac{1}{2}\)

∴ Option (d) is correct

Explanation:

Possible outcomes x < y < z

= = (1, 2, 3) (1, 2, 4), (1, 2, 5), (1, 2, 6) (1, 3, 4) (1, 3, 5) (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4),(2, 3, 5) (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6) (3, 5, 6), (4, 5, 6)

Thus, Total possible outcomes = 20

∴ Option (a) is correct.

Explanation:

Given:

⇒ Mean x = np = 6....(i)

⇒ SD = \(\sqrt{npq} = \sqrt2\)...(ii)

Dividing (ii) by (i), we get

⇒ \(\frac{npq}{np} = \frac{2}{6}\)

⇒ q= 1/3

Now

⇒ p = 1 – q = 1- 1\3 = 2/3

From (i)

⇒\(n\times \frac{2}{3} = 6\)

⇒n =9

∴ Option (d) is correct.