Question 1

1/-0

What should come at the place of ? in the following number series?

24, 96, 48, ?, 96, 384, 192, 768, 384

SOLUTION

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​

Question 2

1/-0

A person wanted to travel from Charbag to Alambag with an average speed of 60 km/h by car. The distance between Charbag and Alambag is 2 km. Due to heavy traffic, he could travel at 30 km/h for the first kilometre of his journey. What should his speed be for the remaining journey to achieve his average speed target of 60 km/h?

SOLUTION

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.So, he cannot achieve his target with any finite speed.

Question 3

1/-0

Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is

SOLUTION

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.So, the only possible value of B is 8.

Question 4

1/-0

What is the unit place digit in the below given expression?

(184)396 ✕ (37)44 ✕ (121)56

SOLUTION

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  

Question 5

1/-0

A student is answering a multiple-choice examination with 65 questions with a marking scheme as follows:

i) 1 mark for each correct answer.

ii) \( - \frac{1}{4}\) for a wrong answer.

iii) \( - \frac{1}{8}\) for a question that has not been attempted.

If the student gets 37 marks in the test then the least possible number of questions the student has not answered is:

SOLUTION

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:

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

Question 6

1/-0

In a group of 6 people, there are 3 engineers and 3 doctors. The number of subgroups from this group that can be chosen such that every subgroup has at least one doctor?

SOLUTION

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

Question 7

1/-0

One morning, Mr. Malik left his home at 9:00 am according to the watch at home. He reached the office at 9:35 am according to the watch at the office. He stayed there for 25 minutes and moved home at a speed that is twice the previous and reached home at 10:25 am according to the watch at home.

By how much time, the office watch is either fast or slow in comparison to the watch at home?

SOLUTION

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.

Question 8

1/-0

The number of all positive integer values of n for which n2 + 96 is a perfect square is:

SOLUTION

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.For the pair (x, y) = (12, 8): 2n = 12 – 8 = 4 ⇒ n = 2, it is valid as it is an integer.

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