Quiz 1B

Its given that B needs 25% more time than A

So, if Time taken by A = 4x days, Time taken by B = 5x days

A does half the work, So A works alone for 2x days --- (1)

A and B work for 4 days --- (2)

B works alone for 5x / 20 = x/4 days.

Total working days = 13

(1) + (2) + (3) = 13

2x + 4 + x /4 = 13 we get,x = 4

So, B alone can do the work in 5 × 4 = 20 days 

12 men + 16 boys can do the work in 5 days.

5 x (12 men + 16 boys) can do the work in 1 day.

Similarly 4 x (13 men + 24 boys) can do the same work in 1 day.

=> 52 men + 96 boys = 60 men + 80 boys.

=> 8 men = 16 boys or 1 man = 2 boys, i.e. 2 : 1

Ratio of boys to girls in the class can be obtained using the method of alligation. Ratio of boys to girls is

(60-56) : (72-60) 

= 4 : 12 

= 1 : 3

In the second test, the average marks obtained by boys and girls are 64 and 80 respectively

Average marks obtained by the class = [(64*1)+(80*3)]/4 = 304/4 = 76 

Hence, Cost price of rice = 100/1.3.

By alligation formula, the ratio for mixing the two rices

= (90 -100/1.3)/(100/1.3-40)= (17/1.3)/(48/1.3) = 17:48

Exactly one of ab, bc and ca is odd => Two are odd and one is even.

 

abc is a multiple of 4 => the even number is a multiple of 4.

The arithmetic mean of a and b is an integer => a and b are odd.

and so is the arithmetic mean of a, b and c. => a + b + c is a multiple of 3.

 

c can be 4 or 8.

c = 4; a, b can be 3, 5 or 5, 9

c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible.

Number of rearrangements of word DEMAND = 6! / 2! = 360

Number of rearrangements of word DEMAND where 2 D’s appear together = 5! = 120

Number of rearrangements of word DEMAND where 2D’s do not appear together =

360 – 120 = 240

 

Words starting with ‘A’; without two D’s adjacent to each other

Words starting with A: 5!/2! = 60

Words starting with A where 2 D’s are together = 4! = 24

Words starting with ‘A’, without two D’s adjacent to each other = 36

Next we have words starting with D.

Within this, we have words starting with DA: 4! words = 24 words

Then words starting with DE

Within this, words starting with DEA => 3! = 6 words

Then starting with DED – 3! = 6 words

Then starting with DEM

      => First word is DEMAND

      => Second is DEMAND

 

Rank of DEMAND = 36 + 24 + 6 + 6 + 2 = 74

A can win in the following scenarios:

A strike in the first shot.

A misses in the first shot, B misses in the second, A strikes in the third.

A misses in the 1st shot, B misses in the 2nd, A misses the 3rd, B misses the 4th and A strikes the 5th.

And so on…

P(A in 1st shot) = 0.6

P(A in 3rd shot) = 0.4 * 0.2 * 0.6 {A misses, then B misses and then A strikes}

P(A in 5st shot) = 0.4 * 0.2 * 0.4 * 0.2 * 0.6 {A misses, then B misses and then A misses, then B misses, then A strikes}

Overall probability = Sum of all these

 

= 0.6 + 0.4 * 0.2 * 0.6 + 0.4 * 0.2 * 0.4 * 0.2 * 0.6 + …

Which is nothing but

0.6 + (0.4 * 0.2) * 0.6 + (0.4 * 0.2)2 * 0.6 + (0.4 * 0.2)3 * 0.6…

This is an infinite geometric progression with first term 0.6 and common ratio 0.4 * 0.2

Required Probability =  a/1-r = 15 / 23

For this type of question, we need to consider only the internal arrangement within the M and 2As.

M and 2As can be rearranged as AMA, AAM, or MAA.

So, the probability that M will feature between the 2As is 1/3

Now, let us think why we need to consider only the M and 2As.

Let us start by considering a set of words where the M and 2 As are placed at positions 2, 3 and 5.

The other three letters have to be in slots 1, 4 and 6

Three letters can be placed in three different slots in 3! = 6 ways.

 

Now with __ M A __ A __ there are 6 different words.

With __ A M __ A __ there are 6 different words.

With __ A A __ M __ there are 6 different words.

For each selection of the positions for A,A and M, exactly one-third of words will have M between the two A’s.

This is why only the internal arrangement between A, A and M matters.

So, probability of M being between 2 As is 1/3

Age of new member= Previous average +  (Increase in average X Number of members including new member)

Therefore,

Age of new member = 36 + (1 × 25)

Age of new member = 36 + 25 = 61

It is clear from the question that the ratio of distances between (Delhi-Chandigarh): (Chandigarh-Shimla) = 3:4

Also, the ratio of the speeds between (Delhi-Chandigarh): (Chandigarh-Shimla) = 3:2.

Let the distances be 3x and 4x respectively and speeds be 3s and 2s.

So, the time taken from Delhi to Chandigarh will be = (x/s)

Also, time taken from Chandigarh to Shimla will be = (2x / s)

Average speed = Total distance / Total time taken. = (3x + 4x) / [(x/s) + (2x/s)] = 7s/3

Now, 7s / 3 = 49

Thus, s = 21

Average speed from Chandigarh to Shimla = 2s = 2*21 = 42 kmph