Quiz- 2

4 men will do the work in 8 days

So for 1 day, men required = 4*7 = 28 days

So, in 2 days, men required is 28/4 = 7 men.

So we have to bring in 7-4 = 3 more men to do that work in 4 days

Let the two conditioners be A and B

'A' cools at 40min

'B' at 45min

Together = (a x b)/(a + b)

= (45 x 40)/(45 + 40)

= 45 x 40/85

= 21.1764

= 22 min (approx).

Mary M and John J can finish a piece of work = 24 days

(M+J)'s 1 day job = 1/24

John J and Vino V can finish the same amount of work = 30 days

(J+V)'s 1 day job = 1/30

Vino V and Mary M can finish the same piece of work = 40 days

(V+M)'s 1 day job = 1/40

Now

(M+J+V)'s 1 day job = 1/24 + 1/30 + 1/40

2(M+J+V) = (5+4+3)/120

2(M+J+V) = 12/120

2(M+J+V) = 1/10

(M+J+V) = 1/20

They altogether can finish the work in 20 days.

It is given that A can do a piece of work in 100 days. The part of work done by A in 1 day is 1/100

The part of work done by B and C in one day is  respectively.

It is given that B and C together can do the same work in 20 days.

1/b + 1/c = 1/20..... (1)

B can do the work in same time as that of C and A together.

1/b = 1/a + 1/c

1/b - 1/c = 1/100     ..... (2)

Subtract equation (2) from (1).

=>  1/20 - 1/100

=>  2(1/c) = 4/100

=> 2(1/c) = 1/25

=> 1/c = 1/50

=>c = 50

c=50

Solution :-
MAN : 1/12

WIFE : 1/15

MAN + WIFE : 1/12 + 1/15 =3/20,,,. ie 20/3

20/3

8 hours 15 minutes
Ronald’s 1 hour’s work 
Pages typed in 6 hrs. = 32, Pages typed in 1 hr 
Elan’s 1 hour’s work = 8 pages
1 hour’s work of the both 

∴ Required Time 

explanation:

Two content Managers can complete developing questions for a module in 18

days.

=> one manager will take 18 * 2 = 36 Days to complete module

=> one manger one day work = 1/36

Work done in 7 Days = 7/18

Remaining work = 1 - 7/18 = 11/18

10 Interns join so total Man power available = 2 + 10 = 12

12 People one day work = 12 * 1/36 = 1/3

Number of days required = (11/18)/(1/3) = 11/6 ≈ 2 Days

 

The pump takes 6 hours to fill the tank

⇒ 1 hour = 1/6 of the tank

Find the amount of water filled after 2 hours:

1 hour = 1/6 of the tank

2 hours = 1/6 x 2 = 1/3 of the tank

Find the amount of water to be filled:

Amount of water to be filled = 1 - 1/3 = 2/3 of the tank

Find the number of hours needed for the standby pump to fill the tank:

The standby pump takes 10 hours to fill the tank

1 tank = 10 hour

2/3 of the tank = 2/3 x 10 = 20/3 hours = 6 hours 40 mins

 

Explaination:
The tank has the capacity of 50 L.

The required proportion is 3 : 2.

Thus for optimum result, in the tank, chemical a should occupy 30 L and b should occupy 20 L volume.

Now,chemical a occupies 50 L volume in 15 minutes

Thus, time required for it to fill upto 30 L = (15 X 30/50) minutes = 9 minutes.

Now,chemical a occupies 50 L volume in 30 minutes

Thus, time required for it to fill upto 20 L = (30 X 20/50) minutes = 12 minutes.

A's 1 hour work = 1/12,

B's 1 hour work = 1/14,

A and B work together = 1/12 + 1/14,

=> (7 + 6)/84 = 13/84,

A & B works alternatively.

So, (A + B)'s 2 hours work = 13/84,

They work 6 pair of hours.

Work = 6 * 13/84 = 13/14,

Remaining work = 1 - 13/14 = 1/14,

13th hour is A's turn.

A's 1 hour work = 1/12,

Total hour work by A = 6 + 1 (to do 1/12) = 7,

A's earning amount = 50 * 7 = Rs. 350.