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A can build a wall in 30 days. B can demolish that wall in 60 days. If A and B work on alternate days, when will the wall be completed?

a) 180 days b) 90 days c) 120 days d) 60 days

Answer is : C

Solution :

Given :-
1. A build in => 30 days
2. B demolish in => 60 days

What To Find:-
How much time will it take to build the wall if A and B work in Alternative days?

Procedure :-
1. in 1 day A can build = 1/30 th part
2. in 1 day B can be fill = -1/60 th part

So,
for first 2 days A and B can build = 1/30 - 1/60 = 1/60
=> In 2 minutes A+B+C can be build = 1/60th Part.
=> Time taken to build 1/60 part = 2 days
=> Time taken to build full part = 2 * 60 days = 120 days.

(Note :- in Problem it is clear that first

A will work for 1 day and then

B will work for 1 day

and again A and B and so on......

So for first 2 days A and B will work each for one day)

Pipes A and B can fill a cistern in 20 min and 30 min and C can empty it in 15 min. If the three are kept open successively for 1 min each in the order how soon will the cistern be filled?

a) 180min b) 90min c) 120 min d) 60 min

Answer is : A

Solution :

Given :-
1. A fills in                 => 20 min
2. B fills in       => 30 min
3. C can empty in      => 15 min

What To Find:-
How much time will it take to fill the cistern if we use each for one minute in order one by one?

Procedure :-
1. in 1 minute A can be fill = 1/20th part
2. in 1 minute B can be fill = 1/30th part
3. in 1 minute C can be fill = -1/15th part

So,
for first 3 minutes A, B and C can fill = 1/20 + 1/30 - 1/15 = 1/60
=> In 3 minutes A+B+C can be fill   = 1/60th Part.
=> Time taken to fill 1/60 part       = 3 min
=> Time taken to fill full part       = 3 * 60 min = 180 minutes.

(Note :- in Problem it is clear that first

A will be opened for 1 min and then

B opened for 1 minute and then

C opened for 1 minute

and again A,B and then C and so on......

So for first 3 minutes A, B and C will be Opened each for one minute)

How many 3 digit numbers sum will have sum 18?

a) 51 b) 54 c) 61 d) 64

Answer is : B

Solution :

Given :-
1. Number is 3 digitnumber.
2. Sum should be 18.

What To Find:-
How many such numbers exist?

Procedure :-
Numbers starts with 1 ð 198, 189.................................= 2 numbers
Numbers starts with 2 ð 297, 279, 288..........................= 3 numbers
Numbers starts with 3 ð 396, 387, 378, 369...................= 4 numbers
Numbers starts with 4 ð 495, 486, 477. 468, 459............= 5 numbers
Numbers starts with 5 ð 594, 585, 576. 567, 558, 549.....= 6 numbers
similaly....
Numbers starts with 6 ð 7 numbers

Numbers starts with 7 ð 8 numbers

Numbers starts with 8 ð 9 numbers

Numbers starts with 9 ð 10 numbers

Total Numbers are ð 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 54

A, B, C are in GP and M,N,O are in AP, then (O-N)log A + (M-O)log B + (N-M)log C is equal to ?

Options :

A) (O-N)log ABC

B) 0

C) 1

D) none

Answer is : B

Solution :

Given :-
1. A, B, C are in GP. So B² = A C
2. M,N,O are in AP. So
=> O-N = N-M = x (assume),
then => (O–M) = 2(O–N) = 2x.

What To Find:-
value of (O-N)log A + (M-O)log B + (N-M)log C

Procedure :-
Remind that (O-M) = 2x,so (M-O) = -2x
Now, find value
(O-N)log A + (M-O)log B + (N-M)log C

=> X log A - 2X log B + X log C

=> X (log A - 2 log B + log C)

=> X (log A - log B² + log C)

=> X(log AC – log B²)

=> X(log B² – log B²)

=> X(0)

=> 0.

Compute the number of distinct ways in which 56 toffees can be distributed to 5 persons A,B,C,D and E so that no person receives less than 10 toffees(toffee can not be devided)?

Options : A) 210 B) 240 C) 180 D) 230

Answer is : A

Solution :

Given :-
1. 56 toffees are there
2. 5 members are there
3. no person receives less than 10 toffees

What To Find:-

No. of distinct ways to distribute.

Procedure :-
1. No person recieves less than 10 toffees.
So distribute 10 toffees to five members.
2. Remaining toffees are 6.
3. These 6 toffees we have got to share for 5 persons.

Formula :- If we share n elements to r members
then number of distinct distributions are : (n+r-1)C(r-1).

4. If we share 6 elements to 5 members.
number of distinct distributions are
(6+5-1)C(5-1) = 10C4 = 210

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How many 3 digit numbers are there in which product of their digit is 36. Ex: 236, in this.. product of digit is 36?

Options : A) 32 B) 36 C) 21 D) 15

Answer is : C

Solution :

Given :-
1. Each number is 3 digit number
2. Product Of digits is 36

What To Find:-
How many such numbers are there.

Procedure :-
1. Factors of 36 are :
1 X 36
2 X 18
3 X 12
4 X 9
6 X 6
2. Sub Factors are :

1 X 36 ---> 1 X (1 X 36) --> Not 3 digit number
---> 1 X (2 X 18) --> Not 3 digit number
---> 1 X (3 X 12) --> Not 3 digit number
---> 1 X (4 X 9) --> 3 digit number

2 X 18 ---> 2 X (1 X 18) --> Not 3 digit number
---> 2 X (2 X 9) --> 3 digit number
---> 2 X (3 X 6) --> 3 digit number

3 X 12 ---> 3 X (1 X 12) --> Not 3 digit number
---> 3 X (2 X 6) --> 3 digit number
---> 3 X (3 X 4) --> 3 digit number

4 X 9 ---> 4 X 9 X 1 --> 3 digit number
---> 4 X (3 X 3) --> 3 digit number
---> (2 X 2) X 9 --> 3 digit number
6 X 6 ---> 6 X (6 X 1) --> 3 digit number
---> 6 X (2 X 3) --> 3 digit number

4. Total 3 digit numbers we got from above list are :

149, 194, 491, 491, 914, 941 --- 6
229, 292, 922, --- 3
236, 263, 632, 623. 326, 362 --- 6
334, 343, 433 --- 3
661, 616, 166 --- 3
---------------------------------------------
Total Numbers are : --- 21
---------------------------------------------

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32400 students login everyday to elitmus website. And stay on the website for 9 minutes. If the access for website is only 18 hours in a day, how many students can be found online at any point of time?

Options : A) 324 B) 300 C) 270 D) 200

Answer is : C

Solution :

Given :-
1. 32400 students login everyday
2. every student stays for 9 minutes.
3. accessible times is 18 hours

What To Find:-
Total Students can be found at any sec ?

Procedure :-
1, accessible time is 18 hours = 18 * 60 minutes
2. how many 9 minutes exist in 18 hours = 18 * 60 / 9 = 120 (say this is period)
3. So total students have to access the site in one of these 120 periods.
4. at any point total students can be found = 32400 / 120 = 270.

So.... Simple...... :-)