Solution:

Assuming that John has A number of apples and Jacobs has B number of apples.

As per the question,
A + 1 = B - 1
=> B - A = 2 ------ (p)

Also, A - 1 = (B + 1) / 2
=> 2A - B = 3 ------ (q)

Solving (p) and (q)
A = 5 and B = 7

Therefore, A has 5 chocolates and B has 7 chocolates right now.

Solution:

119

Explanation:
Let the minimum number of chocolates = C
When divided by 2, remainder = 1
When divided by 3, remainder = 2
When divided by 4, remainder = 3
When divided by 5, remainder = 4
When divided by 6, remainder = 5
When divided by 7, remainder = 0

Therefore, C is divisible by 7.

Whenever C is divided by any number less than 7, the remainder is 1 less than the divisor.
This implies that, C + 1 is the LCM of 2, 3, 4, 5 and 6.

Now LCM of 2, 3, 4, 5 and 6 = 60
But 60 - 1 = 59 is not divisible by 7

60 * 2 = 120
120 - 1 = 119 which is divisible by 7

Therefore, C + 1 = 120 or C = 119

Number of minimum number of chocolates in the box = 119

Solution:

18

Explanation:
Let us suppose that there were (3 * G) garlic bread sticks originally.

On dividing equally, the number of bread sticks with each of the friend = G

Each one of them ate four. The remaining sticks with each one of them = (G - 4)

Now, the total number of breadsticks left = 3 * (G - 4)
This number is equal to the breadsticks each one of them had after division.

=> 3 * (G - 4) = G
(3 * G) - 12 = G
2 * G = 12
G = 6.

Therefore, the number of breadsticks originally = 3 * G = 3 * 6 = 18.

Solution:

Suppose that we have B buses at present and S refers to the space before we expanded. This implies that
S = B - 12 ----- [1]

When we expanded, we came up with more space. It can be equated as
S + 0.4 x S = B + 12 ------ [2]

Rewriting the equation we get
1.45 = B + 12
Also, B = 1.45 - 12 ----- [3]

The first equation can also be written as
B = S + 12 ----- [4]

If we equate the third and fourth equation, we will have
1.4 S - 12 = S + 12

Let us subtract S from both the sides
0.4 S - 12 = 12

Adding 12 to both sides
0.4 S = 24

Multiplying by ten and dividing by four on both sides
S = 60

If we insert this value in the fourth equation, we will have
B = 72

Thus there are 72 buses.

Initially we had room for 60 buses after expanding the garage space by 40 percent, we will be having enough room to have 84 buses out of which twelve is for future.

Solution:

2hr 11min

Explanation:
Now this question cannot be solved unless you know a fact i.e. the hour hand is exactly on the minute mark five times every hour - on the hour, 12 minutes past hour, 24 minutes past hour, 36 minutes past hour and 48 minutes past hour.

Denoting the number of hours with X and the number of minutes as Y, let us begin with the calculation.

When we have the hour hand on the minute mark, the position of hour hand:
5X + Y/12
And the position of the minute hand = Y

When the first observation was taken,
Y = Y = 5X + Y/12 + 6
Or, 60X = 11Y - 72.

With the given facts, we know for sure that Y can only be 0, 12, 24, 36 or 48.
According to which, the possible values for X and Y can be 1 and 12 respectively. Thus the time must be 1:12 when the first observation was taken.

Similarly, in the second observation, let's make the equation:
60X = 11Y - 84
Here, the possible values for X and Y here are 3 and 24 respectively. According to this, the time is 3:24 for sure.

From 1:12 to 3:24, two hours and eleven minutes have passed.

Solution:

Impossible

Explanation:
Let us first denote everything with variables.
d= distance to your friend's place
T = time taken to get there
t = time required to get back
R = Speed while returning back

d = 20T
T = d/20

d = Rt
t = d/R

We have made equations for T and t and now we can derive an equation for the round trip.

2d = 40(T + t)
2d = 40(d/20 + d/R)
2d = 40d(1/20 + 1/R)
1 = 20(R/20R + 20/20R)
20R = 20(R+20)
R = R + 20

Do you see the paradox here? You literally have to travel back at an infinite speed if you want to make the average speed of your trip 40 mph. The faster return speed will have lesser impact of the average speed but consider that the quicker your return trip is, the faster you will make it.

We are talking about travelling double the distance in the same time that was taken for one way trip. Thus, if you travel at infinite speed, you will be able to attain an average of 40 mph.

Solution:

6

Explanation:
(2^x)(30^3) = (2^3) (3^3) (4^3) (5^3)

(2^x)(30^3) = (120^3)

(2^x) = (120^3)/ (30^3)

(2^x) = (4^3)

(2^x) = 64

ln(2^x) = ln(64)

x ln(2) = ln(64)

x = ln(64)/ln(2)

x = 6

Thus x = 6.

Solution:

7.5Km

Explanation:
Denoting Ashley with A and Cecelia with C

Speed A = 60 km/hr = 1.2 x Speed C [km/hr] = 1 km/ min

Hence Speed C = 60/1.2 [km/hr] = 50 [km/hr] = 5/6 km/ min

Time C = D [km] /Speed C [km/min] = D/Speed C [min] = D/(5/6) min = 6D/5 min = 1.2 D min

Time A = D [km] /Speed A [km/min] = D/Speed A [min] = D/(1) min = D min

Since Time C – Time A = 1.5 min

Then (1.2 – 1)D min = 0.2D min = 1.5 min



Hence D = 1.5/0.2 km = 1.5*5 = 7.5 km

The total distance as calculated above is 7.5 Km.

Solution:

4%

Explanation:
Suppose that the initial volume of the water drop was p.

p - (1/4) p = (3/4) p

(3/4) p - (1/5) [(3/4) p] = (3/4) p * [1 - (1/5)] = (3/4) p * (4/5) = (3/5) p

(3/75) p = (1/25) p

1/25 = 4%

Therefore, the required percentage is four percent.

Solution:

50%

Explanation:
When the coins are tossed there can be eight possible outcomes (2*2*2)
HHH
HTT
HTH
HHT
TTT
THH
THT
TTH

If you find out, we have only four favorable outcomes. Also, the events here are independent and thus we can calculate the probability by using these only.

Thus the probability is 50 percent.