M4maths Previous Puzzles

24 Jul 2009 break this representation (friday again)

If 168=1/24
207=1/9
And 24=0
Then 278 is equal to

OPtion

1) 0
2) 1/16
3) 2/19
4) 1/13
5) 3/64
6) 1/38
7) 1
8) 1/42
9) 1/14
10) 5/58

Solution

Here 168 is equal to 168 minute = 2 hour /48 min = 1/24
And 207 is equal to 207 minute = 3 hour/27 min = 1/9
Similarly 24 is equal to 24 minute = 0 hour/24 min = 0
Therefore 278 is equal to 278 minute = 4 hour/38 min = 2/19

Correct Option: 3 Best Solution (0)

23 Jul 2009 arrange the boys

Names of 10 friends are
RAHUL, MOHAN, AKBER, SUNIL, LALIT, AASHU, GOKUL, IMRAN, KANIK & SUMIT
These names are arranged in increasing order according to a coded method.
If out of these 10 names
RAHUL is coded as 7
GOKUL is coded as 8 and
IMRAN is coded as 6
then in this coded language MOHAN is

OPtion

1) 5
2) 6
3) 7
4) 8
5) 9
6) 10
7) 1
8) 2
9) 3
10) 4

Solution

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
For RAHUL sum of positions of alphabets = 18+1+8+21+12=60 7
For MOHAN sum of positions of alphabets = 13+15+8+1+14=51 4
For AKBER sum of positions of alphabets = 1+11+2+5+18=37 1
For SUNIL sum of positions of alphabets = 19+21+14+9+12=75 9
For LALIT sum of positions of alphabets = 12+1+12+9+20=54 5
For AASHU sum of positions of alphabets = 1+1+19+8+21 = 50 3
For GOKUL sum of positions of alphabets = 7+15+11+21+12= 66 8
For IMRAN sum of positions of alphabets = 9+13+18+1+14= 55 6
For KANIK sum of positions of alphabets = 11+1+14+9+11= 46 2
For SUMIT sum of positions of alphabets = 19+21+13+9+20 = 82 10

On arranging them in increasing order, MOHAN should be coded as 4.

Correct Option: 10 Best Solution (1)

22 Jul 2009 How many rectangle

8 sticks are available to make a rectangle. One side of the rectangle is taken along east
direction whereas second side is taken along north direction. How many rectangles can be
made by using these sticks. It is not compulsory to use all the sticks in a trial.

OPtion

1) 6
2) 5
3) 4
4) 3
5) 2
6) 1
7) 7
8) 8
9) 9
10) 10

Solution

By using 4 sticks, there is 1 possibility.

By using 6 sticks, there are 2 possibilities.

First contains two sticks along north & one stick along east.
Second contains one stick along north & two sticks along east.

By using 8 sticks, there are 3 possibilities.
First contains two sticks along north & two sticks along east.
Second contains three sticks along north & one stick along east.
Third contains one stick along north & three sticks along east.

Correct Option: 1 Best Solution (0)

21 Jul 2009 next number is

In the sequence given below
986, 7248, 14832, 432246, 1264824
next number is

OPtion

1) 3983434
2) 21243281
3) 490758
4) 2243216852
5) 626828924
6) 71645342
7) 24162343
8) 2122432168
9) 297685421
10) 7542677

Solution

First number is 986
Then 9*8/8*6 = 72/48 = 7248 (it is second number)
Then 7*2/2*4/4*8 = 14/8/32 = 14832 (it is third number)
Then 1*4/4*8/8*3/3*2 = 4/32/24/6 = 432246 (it is fourth number)
Then 4*3/3*2/2*2/2*4/4*6 = 12/6/4/8/24 =1264824 (it is fifth number)
Similarly 1*2/2*6/6*4/4*8/8*2/2*4 = 2/12/24/32/16/8 = 2122432168

Correct Option: 8 Best Solution (1)

20 Jul 2009 Calculate this

If 123+698 = 7121
And 745+769 = 4151
Then 346+594 is

OPtion

1. 1841
2. 8140
3. 8130
4. 8940
5. 8311
6. 1138
7. 5792
8. 6967
9. 2717
10. 8752

Solution

123+698 = 7121
This sum can be obtained by the following way
On taking digits in reverse order we get 321+896= 1217
On reversing the result we get 7121
Similarly for second sum 547+967 = 1514
On reversing the result we get 4151

In last sum 643+495 = 1138
On reversing the result we get 8311

Correct Option: 5 Best Solution (1)

17 Jul 2009 How can we get this value...

Consider the following cases
4659/223=1153
6983/332=1141
According to the rules applied in these exercises, the value of 8692/323 is

OPtion

1) 4323
2) 3423
3) 1114
4) 1411
5) 3432
6) 4111
7) 4321
8) 2341
9) 1111
10) 2431

Solution

4 6 5 9
First taking l.c.m. with 2 we get
2 3 5 9
Then l.c.m. with 2 gives
1 3 5 9
l.c.m. with 3 gives finally
1 1 5 3

Similarly for
6 9 8 3
First taking l.c.m. with 3 we get
2 3 8 1
Then l.c.m. with 3 gives
2 1 8 1
l.c.m. with 2 gives finally
1 1 4 1

For 8 6 9 2
Taking l.c.m. with 3 we get
8 2 3 2
With 2 we get
4 1 3 1
And then with 3 finally we get
4 1 1 1

Correct Option: 6 Best Solution (1)

16 Jul 2009 play with numbers

If
49-33=123
78-32=214
94-72=632
Then
96-53=?

OPtion

1) 448
2) 452
3) 453
4) 567
5) 678
6) 679
7) 764
8) 763
9) 824
10) 825

Solution

Here
49-33=(4*3)*10+(9/3)*1=120+3=123
78-32=(7*3)*10+(8/2)*1=210+4=214
And
94-72=(9*7)*10+(4/2)*1=630+2=632
Similarly
96-53=(9*5)*10+(6/3)*1=450+2=452

Correct Option: 2 Best Solution (0)

15 Jul 2009 Earth Density

Radius of earth is 6400 km. If mass of earth remains conserved and radius of earth would be 1cm
and then its density would be (approximately) :-

OPtion

1. 10^26 times the present value of density of earth
2. 10^24 times the present value of density of earth
3. 10^22 times the present value of density of earth
4. 10^20 times the present value of density of earth
5. 10^28 times the present value of density of earth
6. 10^30 times the present value of density of earth
7. 10^8 times the present value of density of earth
8. 10^16 times the present value of density of earth
9. 10^9 times the present value of density of earth
10. 10^18 times the present value of density of earth

Solution

Since mass is conserved hence
d2/d1 = V1/V2
= R13/R23
= 6400km*6400km*6400km/ 1cm*1cm*1cm
= 6400000m*6400000m*6400000m/0.01m*0.01m*0.01m
= 6.4*6.4*6.4*10^18/10^-6
= 262.144*10^24
approx 2.6*10^26
appro 10^26

Correct Option: 1 Best Solution (0)

14 Jul 2009 room of Bricks

In a room of dimensions l*b*h = 16*14*12, bricks of dimensions 6*4*2 are placed
in such a way that the room is occupied completely. Total number of bricks in a layer on the floor is :-

OPtion

1) 12
2) 56
3) 24
4) 8
5) 28
6) 82
7) 14
8) 48
9) 128
10) 192

Solution

On the basis of dimensions of brick it is necessary to place the brick in the room in such a way that
l = 4
b = 2
h = 6
( If we place the brick in any other manner then it is not possible to fill the room completely)
Now area of floor = 16*14 = 224
And area of brick = 4*2 = 8
Hence Number of bricks = 224/8 = 28

Correct Option: 5 Best Solution (0)

13 Jul 2009 solve the equation

If 16^2=31
And 17^2=61
Then 13^2 is

OPtion

1. 11
2. 25
3. 49
4. 21
5. 01
6. 52
7. 53
8. 51
9. 64
10. 92

Solution

16^2 = 256 = (5-2)*10+(6-5) = 31
17^2 = 289 = (8-2)*10+(9-8) = 61
Similarly
13^2 = 169 = (6-1)*10+(9-6) = 53

Correct Option: 7 Best Solution (0)