Elitmus Page

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
ð O-N = N-M = x (assume),
then ð (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

ð X log A - 2X log B + X log C

ð X (log A - 2 log B + log C)

ð X (log A - log B² + log C)

ð X(log AC – log B²)

ð X(log B² – log B²)

ð X(0)

ð 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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A and B started from the same point on a circular track. The ratio of their speed is n:1 (n is a whole number). If 5th and 17th time they meet at the same point, what can not be the value of n?

Options : A) 2 b) 3 c) 4 d) 6

Answer is : D

Solution :

If speed of two bodies moving in same direction in circular path are in the simplest ratio of a:b, then no. of pts they will meet = |a-b|
Let circumference of circle be 'L'.
Time for 1st meeting at any of the |a-b| pts = L/(a-b).
Time for 1st meeting at Starting Pt = LCM of (L/a, L/b).
Time for nth meeting at Starting Pt = n*LCM of (L/a, L/b).

Checking with the options :-
1 . Check Option A :
When n = 2 :
a=2,b=1.
No:of meeting pts = (2-1)=1pt.
Only meeting pt is L.
Time for first meeting = L/1.
Time for first meeting at starting pt = LCM of (L/2, L/1) = L.
Time for 5th meeting = 5L.
Time for 17th meeting = 17L.

Both tyms, they meet at pt L.
So, n=2 is valid.

Check Option B :

When n = 3 :
a=3,b=1.
No:of meeting pts = (3-1)=2pts.
Meeting pts are : L/2, L .
Time for first meeting = L/2.
Time for first meeting at starting pt = LCM of (L/3, L/1) = L.
Time for 5th meeting = 5*L/2.
Time for 17th meeting = 17*L/2.

Both tyms, they meet at pt L/2.
So, n=3 is valid.

Check Option C :

When n = 4 :
a=4,b=1.
No.of meeting pts = (4-1)=3pts.
Meeting pts are : L/3, L/2, L .
Time for first meeting = L/3.
Time for first meeting at starting pt = LCM of (L/4, L/1) = L.
Time for 5th meeting = 5*L/3.
Time for 17th meeting = 17*L/3.

Both times, they meet at pt L/3.
So, n=4 is valid.

Check Option D :

When n = 6 :
a=6, b=1.
No:of meeting pts = (6-1)=5pts.
Meeting pts are : L/5, L/4, L/3, L/2, L .
Time for first meeting = L/5.
Time for first meeting at starting pt = LCM of (L/5, L/1) = L.
Time for 5th meeting = 5*L/5 = L.
Time for 17th meeting = 17*L/5.

Both tyms, they meet at different pts namely L and L/5.

So, n=6 is not valid.

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...... :-)

Saturday, 17 October 2015

How many 3 digit numbers sum will have sum 18?

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

Answer is : B

Solution :

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 :

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 :

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 :

A and B started from the same point on a circular track. The ratio of their speed is n:1 (n is a whole number). If 5th and 17th time they meet at the same point, what can not be the value of n?

Options : A) 2 b) 3 c) 4 d) 6

Answer is : D

Solution :

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 :