General Aptitude Test 3

The series is obtained by multiplying by 4 and dividing by 2 alternatively

⇒ 24 × 4 = 96

⇒ 96 ÷ 2 = 48

⇒ ? = 48 × 4 = 192

⇒ 192 ÷ 2 = 96

⇒ 96 × 4 = 384

⇒ 384 ÷ 2 = 192

⇒ 192 × 4 = 768

⇒ 768 ÷ 2 = 384​

A person wanted to travel from Charbag to Alambag with an average speed of 60 km/h by car.

The time takes to travel a distance of 2 km with an average speed of 60 km/h = \(\frac{2}{{60}}h = 2\;min\)

Due to heavy traffic, he could travel at 30 km/h for the first kilometer of his journey

Time is taken to cover a distance of 1 kilometer of his journey at 30 km/h \( = \frac{1}{{30}}h = 2\;min\)

He needs to cover the 2 km in 2 mins, but he able to cover only 1 km within 2 mins.

136 + 5B7 = 7A3

⇒ 100 + 30 + 6 + 500 + 10 B + 7 = 700 + 10 A + 3

⇒ B – A = 6

7A3 is divisible by 3.

So, A can be 2, 5, 8, …

For A = 2, B = 8

For A = 5, B = 11 which is not valid.

For any other value of A, B is not valid.

41 → 4, 42 → 16 (unit 6), 43 → 64 (unit 4), 44 → 256 (unit 6)

4 → 6  (cycles)

71 → 7, 72 → 49 (unit 9), 73 → 343 (unit 3), 74 → 2401 (unit 1)

7 → 9 → 3 → 1 (cycles)

(184)396 ✕ (37)44 ✕ (121)56

To get remainder

≡ (4)2×198 ✕ (7)4×11 ✕ (1)56

≡ (4)2 × (7)4 × 1

≡ 6 × 1 ≡ 6  

Let the number of questions attempted correct = r

Number of questions attempted wrong = w

Number of questions not attempted = u

We can write:

r + w + u = 65

w = 65 – r – u        ---(1)

Given that \( - \frac{1}{4}\) is given for each wrong answer and \(\frac{{ - 1}}{8}\) for unattempted questions, we can write:

\(r - \frac{w}{4} - \frac{u}{8} = 37\)    ---(2)

Using equation (1), the above equation becomes:

\(r - \frac{{\left( {65 - r - u} \right)}}{4} - \frac{u}{8} = 37\)

Solving the above, we get:

We require ‘u’ to be least possible according to the question.

∴ The value of r for which this is possible is 42, i.e.

10 × 42 + u = 426

u = 426 – 420

u = 6,

Also, w = 65 – r – u = 17

∴ The least possible number of questions that the student has not answered is 6.

Cross-Check:

Using equation (2), we get:

\(42 - \frac{{17}}{4} - \frac{6}{8} = 37\)

A subgroup can have a maximum of 5 members and a minimum of 2.

The following cases can be considered:

Case 1 (1 Doctor):

Subgroups can be:

Doctor and 1 Engineer

Doctor and 2 Engineers

Doctor and 3 Engineer

The total possible subgroups for case 1 will be:

\(3{C_1} \times 3{C_1} + 3{C_1} \times 3{C_2} + 3{C_2} \times 3{C_3}\)

= 3 × 3 + 3 × 3 + 3 × 1

= 9 + 9 + 3

= 21

Case 2 (2 Doctors):

The possible subgroups can be:

2 Doctors and 0 Engineers

2 Doctors and 1 Engineer

2 Doctors and 2 Engineers

2 Doctors and 3 Engineers.

The total possible subgroups for the above Case 2 will be:

\(= 3{C_2} \times 3{C_0} + 3{C_2} \times 3{C_1} + 3{C_2} \times 3{C_2} + 3{C_3} \times 3{C_3}\)

= 3 × 1 + 3 × 3 + 3 × 3 + 3 × 1

= 3 + 9 + 9 + 3

= 24

Case 3 (3 Doctors):

The possible subgroups can be:

3 Doctors and 0 Engineers

3 Doctors and 1 Engineers

3 Doctors and 2 Engineers

The total possible subgrops for case 3 can be:

\(= 3{C_3} \times 3{C_0} + 3{C_3} \times 3{C_1} + 3{C_3} \times 3{C_2}\)

= 1 × 1 + 3 × 1 + 1 × 3

= 1 + 3 + 3

= 7

∴ The total number of subgroups with at least 1 doctors will be:

= 7 + 21 + 24

= 52

The total time, including the rest time according to the watch at home, is:

From 9:00 am to 10:25 am = 1 hr 25 min

Total travel time excluding the rest time (i.e. the time stayed in office) is:

1 hr 25 min - 25 min = 1 hr = 60 min

Given that the speed during the return to home is twice that while going to home. ∴ The time taken by the person to go to office from home will be twice of that while going home from office.

So, the ratio of time for going to office and coming from the office is 2:1.

Since the total travel time is 60 min, we can write:

2x + x = 60

x = 20 min

The time taken to reach the office is, therefore:

2x = 2 × 20 = 40 min

Thus, the time when he reached office must have been:

9:00 AM + 40 min = 9:40 AM

But since the office clock was showing 9:35 AM, we conclude that it is slow by 5 minutes.

Let m² = n² + 96

Where m is a positive integer and m² is a perfect square.

⇒ m² – n² = 96

⇒ (m + n) (m – n) = 96

Let x = m + n and y = m – n

The possible pairs of x and y are: (x, y) = (96, 1), (48, 2), (32, 3), (24, 4), (16, 6), (12, 8)

x – y = m + n – (m – n) = 2n

For the pair (x, y) = (96, 1): 2n = 96 – 1 = 95 ⇒ n = 47.5, it is not valid as it is not an integer.

For the pair (x, y) = (48, 2): 2n = 48 – 2 = 46 ⇒ n = 23, it is valid as it is an integer.

For the pair (x, y) = (32, 3): 2n = 32 – 3 = 29 ⇒ n = 14.5, it is not valid as it is not an integer.

For the pair (x, y) = (24, 4): 2n = 24 – 4 = 20 ⇒ n = 10, it is valid as it is an integer.

For the pair (x, y) = (16, 6): 2n = 16 – 6 = 10 ⇒ n = 5, it is valid as it is an integer.