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.The correct answer is option (1) 48 minutes past 10

In 24 hours, the clock gains 10 minutes.

In 1 hour, the clock will gain \(10\over24 \) minutes.

6 AM to 11 AM (next day) = 24 hrs + 5 hrs

Additional time = 10 min + 5 x \(\bf 10\over24 \) ≈ 12 min

so, the right time = 11 AM - Additional time = 10:48 AM or 48 minutes past 10

The correct answer is option (2)

Formula Used:

The angle between the hour hand and minutes hand

θ = (30H - 11 M/2)

Calculation:

H = 5 and angle = 0°

When minute & hour hands coincide, then θ = 0

⇒ 0 = 30 × 5 - 11 M/2

⇒ 0 = 150 - 11 M/2

⇒ 150 = 11 M/2

⇒ 300 = 11 M

⇒ M = 300/11

⇒ M = \(27\dfrac{3}{11}\) Minutes

∴ The correct answer is option (2).

The correct answer is option (3) 30 days.

Given:

Both together can finish work in 12 days and Anil alone can do it in 20 days.

Formula used:

Time taken for both together, to do work is \(\frac{xy}{x+y}\), where x is the time taken by Anil alone, and y is the time taken by Suman alone.

Calculation:

\(\dfrac{xy}{x+y}\) = 12

if we put the value of x = 20 in above equation, we get

\(\dfrac{20y}{20+y} = 12\)

or 20y = 240 + 12y

or y = \(\dfrac{240}{8} = 30 ;days\)

Hence Suman will take 30 days to complete the work alone.

The correct answer is option (2) i.e., 5 students.

Total students n(A U B) = 55

Students who like to play Cricket n(A) = 34

Students who like to play Badminton n(B) = 26

Formula used n(A U B) = n(A) + n(B) – n(A ∩ B)

students like to play both Cricket and Badminton n(A ∩ B) = 

n(A) + n(B) - n(A U B) = 34 + 26 - 55 = 5

The correct answer is option (3), 60 and 40.

Let the numbers are A and B; 

Given

AB = 2400 .....(1)

and A2 + B2 = 5200

We know that 

(A + B)2 = A2 + B2 + 2AB = 5200 + 4800 = 10000

Or, A + B = 100 .....(2)

By solving equation (1) and (2)

We get a and b = 60 and 40.

The correct answer is option (2), there will be 7 numbers between 100 and 500 divisible by 4,5 and 6.

Detailed Solution:

LCM of 4, 5 and 6 = 60

Now we have to find numbers between 100 and 500 which are exactly by 60

First number between 100 and 500 which is exactly divisible by 60 = 120

⇒ Second number between 100 and 500 which is exactly divisible by 60 = 120 + 60 = 180

∴ There are 7 numbers between 100 and 500 which are divisible by 4, 5 and 6

Numbers Between 100 and 500 which are exactly divisible by 60 = 120, 180, 240, 300, 360, 420 and 480

Alternative solution:

Quotient on dividing 100 by 60 = 1

Quotient on dividing 500 by 60 = 8

Required answer = 8 - 1 = 7

∴ 7 Numbers between 100 and 500 which are divisible by 4, 5 and 6.

Shortcut Trick

Initial Ratio (Sand : Cement) = 4 : 1 (Total = 5 units)

New Ratio (Sand : Cement) = 6 : 1 (Cement remains constant)

Difference in Sand = 6 − 4 = 2 units

Given 5 units = 70 kg ⇒ 1 unit = 14 kg

Sand to be added = 2 units = 2 × 14 kg = 28 kg

∴ The correct answer is 28 kg.

Alternate Method

Given: Total initial mixture = 70 kg, Initial Ratio = 4 : 1, Final Ratio = 6 : 1

⇒ Initial Sand = (4/5) × 70 = 56 kg

⇒ Initial Cement = (1/5) × 70 = 14 kg

⇒ Let Sand added be 's' kg.

⇒ (56 + s) / 14 = 6 / 1

⇒ 56 + s = 84

⇒ s = 84 − 56 = 28 kg

∴ The correct answer is 28 kg.

Additional Information

Constant Quantity Rule

In mixture problems where only one component is added, the quantity of the other component remains constant. Always equate the non-changing component in both ratios.

Alligation Method

This can be solved by considering added sand as 100% sand and initial mixture as 80% sand (4/5) to reach a final 85.7% (6/7).

Ratio Balancing

If the second component's ratio changes, multiply the entire ratio by a factor to make the non-changing component's parts equal.

The correct answer is option (2) Rs. 240 per dozen.

S.P. = Rs.150 per dozen

or 75% of C.P. = Rs150

⇒ \(\dfrac34 \times ;C.P.= 150\)

⇒ C.P. = 200

To get the profit of 20% 

M.P. = C.P. + 20% of C.P.

M.P. = 200 + 20% of 200 = Rs 240

The correct answer is option (1), 52 cm.

The lengths of the first diagonal (d1) = 10

The length of the second diagonal (d2) = 24 cm

⇒ a2 = (d1/2)2 + (d2/2)2  (as diagonals of rhombus bisect at 90°)

⇒ a2 = (10/2)2 + (24/2)2

⇒ a2 = 25 + 144

⇒ a = √169

⇒ a = 13

∴ Perimeter of rhombus = 4 × a  = 4 × 13 = 52 cm

Option (2) 12 seconds will be the correct answer. 

Given:

Length of the train = 140m

Speed of the Train = 60 km/h = 60 × (5/18) = \(50\over3\) m/sec.

Speed of the dog = 18 km/h = 18 × (5/18) = 5 m/sec.

Time =  T

Calculation:

Time = \(\dfrac{Lenght ;of ;the ;train } {Difference; in ;the ;speeds}\)

⇒ T = \(\dfrac{140}{{50\over3} - 5}\)

⇒ s = 420/35 m/m/sec.

⇒ s = 12 s