M4maths Previous Puzzles

26 Jun 2009 Orange Puzzle (Friday Again)

4 oranges are distributed among 4 friends named as Ankur, Parag, Milan and Deepak.
There are certain conditions for the distribution.
* If Ankur receives 1 orange then it is necessary to give 1-1 orange to all others.
* Parag has disagreed to receive 3 oranges.
* If Parag obtains 2 oranges then Ankur does not receive any orange.
* If Milan gets 3 oranges then Deepak gets 1 orange necessarily.
* Only Deepak can take all the 4 oranges.
How many methods are possible for the distribution of oranges?

OPtion

1) 35,
2) 16,
3) 12,
4) 22,
5) 25,
6) 19,
7) 27,
8) 20,
9) 21,
10) 30,

Solution

Category and no. of methods in each category are given in the table below

Category No. of methods in the category
4 orange to one person 4
3 to one and 1 to other 12
2-2 oranges 6
2 to one and 1-1 to other two persons 12
1-1 to four persons 1
Now according to condition 1, If Ankur receives 1 orange then it is necessary to give
1-1 orange to all others,
Therefore out of 10 methods in which Ankur receives 1 orange only that 1 is correct
in which all
boys get 1-1 oranges.
According to condition 2, Parag has disagreed to receive 3 oranges,
Therefore out of 3 methods in
which Parag gets 3 oranges, 2 are rejected. (One method has been rejected in the
previous condition)
According to condition 3, If Parag obtains 2 oranges then Ankur does not receive any orange,
Therefore out of 6 methods in which Parag receives 2 oranges only 3 are correct.
In remaining 3 methods
2 are rejected in first condition hence 1 method is rejected in this step.
According to condition 4, If Milan gets 3 oranges then Deepak gets 1 orange necessarily
Therefore out of 3 methods in which Milan gets 3 oranges only 1 method is correct.
Out of other 2 methods 1 is rejected. (Other 1 has been rejected in first condition)
According to condition 5, Only Deepak can take all the 4 oranges.
Therefore out of 4 methods
as given in the table above, 3 are rejected
Total no. of rejected methods is 9+2+1+1+3=16

Correct Option: 6 Best Solution (0)

25 Jun 2009 Compare the word.

Some words and their coded words are given in the table below
Word Coded word
Star Hair
World Book
Moon Taj Mahal
Then triad used for the set of three words (Sky, Country, Sun) is

OPtion

1. Mobile, Song, Paper
2. Man, America, China
3. Face, Book, India
4. Sun, Earth, Bombay
5. Paper, Cover, Taj Mahal
6. Mouse, Computer, CPU
7. Head, Page, Furnace
8. Sun, Moon, Earth
9. Teacher, Court, Pen
10. Foot, Finger, Dust

Solution

Sky contains stars similarly head contains hair
World has many countries similarly book has many pages
Moon and Taj Mahal are beauties similarly Sun and Furnace are symbols of intensities.

Correct Option: 7 Best Solution (0)

24 Jun 2009 Bill and tax puzzle.

A man pays for 4 purchased items according to the conditions given below:
No. Items price including vat vat %
1 T.V. 16000 4
2 Mobile 18000 8
3 Drugs 4000 12
4 Stationary 12000 0

Calculate the total marketed cost of all the articles purchased by man.

OPtion

1) 50000,
2) 52560,
3) 40323,
4) 47623,
5) 62000,
6) 53000,
7) 47170,
8) 47694,
9) 52480,
10) 48077

Solution

Marketed cost of first item is 16000×100/104 =15385
Marketed cost of second item is 18000×100/108 =16667
Marketed cost of third item is 4000×100/112 = 3571
Marketed cost of fourth item is 12000×100/100 =12000
On adding these numbers, we get solution.

Correct Option: 4 Best Solution (0)

23 Jun 2009 Day was.

If 10th April 2008 was Saturday then 12th June 2004 was

OPtion

1. Monday
2. Tuesday
3. Wednesday
4. Thursday
5. Friday
6. Saturday
7. Sunday

Solution

10 April 2008 - Saturday (Suppose)
Then 10 April 2007 - Thursday (Leap Year gap of 2 days)
10 April 2006 - Wednesday (general gap of 1 day)
10 April 2005 - Tuesday
10 April 2004 - Monday
Total days from 10 April 2004 to 12 June 2004 is 21 + 31 + 12 = 64
On dividing 64 by 7 reminder is 1,
hence the required day is also Monday.

Correct Option: 1 Best Solution (0)

22 Jun 2009 1 rupee

How many methods are possible to make 1 rupee by using 10 paisa, 20 paisa and 50 paisa coin.
It is not necessary to use all type of coins and any coin can be used more than 1 times.

OPtion

1) 10,
2) 9,
3) 8,
4) 7,
5) 6,
6) 11,
7) 12,
8) 13,
9) 14,
10) 15

Solution

0.10*10
0.10*8+0.20*1
0.10*6+0.20*2
0.10*4+0.20*3
0.10*2+0.20*4
0.20*5
0.50*1+0.10*5
0.50*1+0.20*1+0.10*3
0.50*1+0.20*2+0.10*1
0.50*2

Correct Option: 1 Best Solution (1)

Deepak   15 years AGO

10x+20y+50Z=100,10x+20y=100,20y+50z=100,10x+50z=100,x+y+z>=0,x<=0,y>=0,z>=0 which gives 10 possible conditions.

Like? Yes (1) | No   

19 Jun 2009 coded language

In a coded language
ABHISHEK is written as 34567889 and
SURENDRA is written as 122456αβ
In this language
Which one of this code represents RAJESHWARA?

OPtion

1) 2444568ααδ
2) 33444568γδ
3) 22444568γδ
4) 245556ααβγ
5) 2226678βγδ
6) 1112456788
7) 122234559δ
8) 1224567βββ
9) 122444αβγδ
10) 224445678δ

Solution

From first statement H and 8 are revised word hence H=8 similarly from second statement R=2
due to the same reason. Besides it A, S, E are common in words and 4, 5, 6 are common in coded
language therefore are equivalent yet exact values can not be obtained for A, S, E.
Other comparisons provide
(B, I, K) = (3, 7, 9) and (U, N, D) = (1, α, β)
In word RAJESHWARA
R is used twice which is equivalent to 2 hence 2 will come twice
A is used thrice => either 4, 5 or 6
are used thrice
J, E, S, H and W are used once in these 5 letters H is 8. S and E will be any 2 out of
4, 5, and 6 after selecting the value of A. J and W are new letters hence 2 new notations
should be used.

Correct Option: 3 Best Solution (0)

18 Jun 2009 different 1 is?

Out of these 10 codes, different 1 is

OPtion

1) TSRQPONMLK
2) SYEKQWCIOU
3) JMPSVYBEHK
4) GECAYWUSQO
5) LUDMVENWFO
6) AFKPUZFKPU
7) XCHMRWBGLQ
8) BWRMHCXSNI
9) KOSWAEIMQU
10) BYVSPMJGDA

Solution

code has incorrect sequence.

Correct Option: 6 Best Solution (1)

Deepak   15 years AGO

difference between the sucsessive alphabet is not constant

Like? Yes (2) | No (1)  

17 Jun 2009 Milk and water

A jar of 10 litres contains 4 litres milk and another of 6 litres is filled completely with water.
6 litre mixtures are removed to second jar after pouring all the water of second jar in first. After
then the process is revised with 4 other jars of 6 litres filled with water completely. What will be
the amount of milk remained in 10 litre jar finally.

OPtion

1) 0.52675
2) 0.01024
3) 0.12500
4) 0.04096
5) 4.00000
6) 0.40000
7) 0.66667
8) 1.50000
9) 0.06400
10) 0.80000

Solution

In every mixing process 4 part of milk is present in 10 part of mixture. Hence for this
process of 5 steps remaining milk in 10 litres jar is
4×(4/10)×(4/10)×(4/10)×(4/10)×(4/10)
=0.04096

Correct Option: 4 Best Solution (1)

Anuradha   15 years AGO

Strength of the solution is decreasing from 40%,16%,6.4%,2.56% to 1.024% hence milk left after the last transaction=4*1.024/100=0.04096L

Like? Yes (1) | No   

15 Jun 2009 letters and envelopes

Number of methods to post 4 letters in 4 envelopes such that only 1 letter goes to the
correct envelope is:-

OPtion

1) 3,
2) 4,
3) 5,
4) 6,
5) 7,
6) 8,
7) 9,
8) 10,
9) 11,
10) 12,

Solution

a b c d
A C D B
A D B C
Similarly with B, C, D

Correct Option: 6 Best Solution (1)

bhuwan adhikari   15 years AGO

no. of ways selecting a letter with correct envelope=4
no.of ways of posting rest of 3 letters=2*1=2
total no of ways=4*2=8

Like? Yes (4) | No (1)  

12 Jun 2009 Decode the word

The list given below contains some words and their ordered codes

Word Code
MAN AHB
YOU GHGE
SAID FHD
(All the codes are formed by using letters A, B, C, D, E, F, G, H, I)
Write a code for the word "SUM"?

OPtion

1) FGH
2) GHAJ
3) HAG
4) EAHG
5) BEDG
6) FKA
7) EFAB
8) EBH
9) BCKF
10)FAH

Solution

SUM =19×21×13 =5187 (According to positions of letters)
Now 5187= EAHG (According to positions of letters)

Correct Option: 4 Best Solution (0)