CUET PG 2023 Question Paper COQP12 Shift-2

Firstly, we can solve this question by assuming that the length of the man is negligible and that only the length of the train matters. 
If the man and the train move in opposite directions, the relative velocity will increase.

200m is 0.2km. The speed of the train is 50 kmph, and that of the man is 5 kmph. 
Effective relative velocity will be 55 kmph.  

Time taken to cover 0.2km will be 0.2/55=1/275 hours. 

An hour has 3600 seconds. 3600/275=144/11 seconds. 

Hence, the answer is Option C. 

P and B can do the work in 10 and 15 days, respectively. 

Let the total units of work be LCM(10,15) = 30 units.

So, P can do 30/10= 3 units per day

B can do 30/15 = 2 units per day

If they work together, they would do 3+2 =5 units per day.

Therefore, in 5 days, they will do $$5\times\ 5=25\ $$units

Then Work left = 30-25 = 5 units.

R can do this work in 2 days, therefore he does 5/2 = 2.5 units per day.

If they are getting Rs.1500 for 30 units of work, then they get $$\ \frac{\ 1500}{30}=Rs.\ 50$$ per unit.

B and R will do 2 + 2.5 = 4.5 units per day. 

Therefore, they will earn $$4.5\times\ 50$$= Rs. 225

Let us say the work is 36 units big. 
This means that P, who can finish the work in 9 days, does 4 units a day, and Q, who can finish the work in 12 days, does 3 units a day. 

Working alternatively means that every two days (4+3) units of work will be done. 
So, 5 cycles of this will ensure 35 units of work is completed, 5 cycles means 10 units of work. 
The last unit is completed by the person who starts work first, which is P. P does 4 units a day so that P can finish 1 unit in 0.25 days. 
Hence, the answer will be 10.25 days. 

Assume that A, B and C can do $$a$$, $$b$$ and $$c$$ units of work per day.

So, the total work is $$10\times\left(a+b\right)=10a+10b$$ . . . (1)

Similarly, the total work is $$15\times\left(b+c\right)=15b+15c$$ . . . (2)

Equate eqn(1) and eqn (2) we get

$$10a+10b=15b+15c$$

Given that A is twice as good a workman as C or $$a=2c$$

Putting in above equation we get

$$20c+10b=15b+15c$$

or, $$5c=5b$$

or, $$b=c$$

So, the efficiency of B and C is the same

If they together complete the work in 15 days, B alone will complete it in 30 days.

The average of ten numbers is 8. If each number is multiplied by 15, then the average of the set of numbers is:

Let us assume the numbers are $$x_1,x_2,......x_{10}$$

Thus, $$\dfrac{x_1+x_2+x_3+....x_{10}}{10}=8$$

Each number is multiplied by 15, So the average will also be 15 times.

$$\dfrac{15\left(x_1+x_2+x_3+....x_{10}\right)}{10}=15\times\ 8\ =\ 120$$

The average of the men is solely changing due to the change in the weight of the person being replaced. No other person in the group is coming in or going out. 
This means the change in the average of 1.5kg, which is calculated over 10 men, is solely due to the heavier person coming in. 
The change in average is 1.5kg, which means the change in the total amount of weight will be 15 kg.

This means the new man coming in is 15kgs heavier than the old man. The old man weighs 70kgs, this means the new man will weight 85kgs.   

Let the four consecutive even numbers be A, B, C and D.

Since they are consecutive even numbers, we can express them in terms of A as:

B = A + 2

C = A + 4 

D = A  + 6 

We are given that their average is 61.

Hence, ( A + B + C + D)/4 = 61

Or, A + B + C + D = 244

Replacing B, C and D

A + A + 2 + A + 4 + A + 6 = 244

4A = 232

A = 58

Hence, B, C and D are respectively 60, 62 and 64.

We need to find A * D

Hence, 58 * 64 = 3712

Let the cost price of 1 litre of milk be C

Now, assume that we mix x litres of milk with y litres of water.

The total cost price of the mixture = C × x (water is free).

The selling price must be 16.67% more than the cost price, i.e. 7/6 of the cost price

For the mixture to sell at the cost price per liter, the selling price should be equal to the cost price per liter.

Solving the equation, we get, x : y = 6 : 1

To find the probability of getting at most three heads, we will use the complement rule (P(not A)

At most three heads means we can get 0, 1, 2, or 3 heads.

The complement of this event is getting exactly 4 heads.

Since four unbiased coins are tossed, the total number of possible outcomes is:

2^4 = 16

The only way to get 4 heads is HHHH, which is 1 outcome.

P(exactly 4 heads)=1/16

P(at most 3 heads)=1−P(exactly 4 heads)

=1−1/16 = 15/16

DETAIL is a six-letter word with three vowels and three consonants. 
This means there will be three odd places and three even places. 
This means we must fit the three vowels in the three odd places. And the three consonants can fit into the three even places. 
The number of ways to arrange three items in three places will be 3!. 

We can arrange three vowels in three odd places in 3! ways, and three consonants in three even places in 3! ways. 

Total ways to arrange it will be (3!)(3!)=36.