M4maths Previous Puzzles

15 Oct 2009 Calculate the number of sheets

A sheet has dimensions 80 cm.* 60 cm.* 1 mm. How many such sheets take a total volume of 0.4632 m3.

OPtion

1) 895
2) 985
3) 995
4) 975
5) 965
6) 885
7) 875
8) 865
9) 855
10) 845

Solution

Volume of one sheet = length* width* thickness
= 80 cm.* 60 cm.* 0.1 cm.
= 480 cm.3
Total volume = 0.4632 m3
No. of sheets = 0.4632 m3/ 0.000480 m3
= 965

Correct Option: 5 Best Solution (3)

14 Oct 2009 ascending descending

If DOGFISH = 3
CROCODILE = 4
MATHEMATICS = 4
COCONUT = 3
Then NOVEMBER =

OPtion

1) 5
2) 6
3) 7
4) 8
5) 9
6) 0
7) 1
8) 2
9) 3
10) 4

Solution

In DOGFISH, total number of increasing alphabetical sets are 3 - DO, FI, IS hence code is 3 similarly for others

Correct Option: 5 Best Solution (0)

13 Oct 2009 code for thursday

If in a coded language SUNDAY = 54
MONDAY = 42
TUESDAY = 65
WEDNESDAY = 70
Then THURSDAY =

OPtion

1) 86
2) 83
3) 75
4) 78
5) 49
6) 62
7) 69
8) 58
9) 53
10) 95

Solution

Here SUNDAY = SUN =19+21+14 = 54 (sum of positional values of S, U, N in English alphabet)
similarly THURSDAY = THURS = 20+8+21+18+19 = 86

Correct Option: 1 Best Solution (5)

12 Oct 2009 find out the code

If in a coded language
7351 = 5
439 = 8
12456 = 0
Then 2462 =

OPtion

1) 5
2) 6
3) 7
4) 8
5) 9
6) 0
7) 1
8) 2
9) 3
10) 4

Solution

7351 = 7*3*5*1 = 105 = 5 (only unit place is taken), similarly for others

Correct Option: 2 Best Solution (6)

09 Oct 2009 coins combination

How many combinations are possible if 3 one rupee coin, 4 fifty paise coin and 2 twenty five paise coin
are available. It is compulsory to select any one of the coin

OPtion

1) 23
2) 24
3) 9
4) 55
5) 60
6) 26
7) 12
8) 59
9) 6
10) 48

Solution

Out of 3 one rupee coin, total possible combinations are
0 one rupee coin, 1 one rupee coin,
2 one rupee coin, 3 one rupee coin,
Out of 4 fifty paise coin, total possible combinations are
0 fifty paise coin, 1 fifty paise coin, 2 fifty paise coin, 3 fifty paise coin, 4 fifty paise coin,
Similarly out of 2 twenty five paise coin, total possible combinations are
0 twenty five paise coin, 1 twenty five paise coin, 2 twenty five paise coin,
Therefore total number of methods = 4*5*3 – 1= 60-1 = 59
(1 case is deducted because in a case no coin is selected)

Correct Option: 8 Best Solution (0)

08 Oct 2009 evaluate

1.67 * 2.31= 129
2.39 * 4.31= 202
6.78 * 4.87= 816
Then 7.16 * 9.13 is equal to

OPtion

1) 506
2) 16.29
3) 228
4) 464
5) 229
6) 426
7) 217
8) 2136
9) 45
10) 276

Solution

1.67 * 2.31 = 1× 67 + 2× 31= 129
2.39 * 4.31 = 2× 39 + 4× 31= 202
Similarly for others.

Correct Option: 5 Best Solution (5)

07 Oct 2009 coded language

If in a coded language
6^3 =15 4^2 =14 8^4 =16
Then 9^5 = ?

OPtion

1) 16
2) 10
3) 19
4) 12
5) 14
6) 13
7) 15
8) 17
9) 18
10) 11

Solution

6^3 = three power of six = 15 (total letters used in written expression)
Similarly 4^2 = two power of four = 14
8^4 = four power of eight = 16
And 9^5 = five power of nine = 15

Correct Option: 7 Best Solution (0)

06 Oct 2009 what will be the next

Next number in the sequence
15, 161, 1771, 19481, 214291 is

OPtion

1) 1942651
2) 135611101
3) 2424561
4) 1357101
5) 345372101
6) 1357201
7) 23651121
8) 1764924
9) 125461
10) 234255

Solution

First number is 15
Second number is 1 |1 + 5| 1 = 1|6|1=161
Third number is 1 |1 + 6| 6 + 1 | 1 = 1|7|7|1=1771
Fourth number is 1 |1 + 7|7 + 7|7 + 1| 1
= 1| 8 | 14 | 8 | 1
= 1 |8 + 1|4 | 8 | 1 (here 1 is carry)
= 1|9|4|8|1 = 19481
Fifth number is 1 |1 + 9 | 9 + 4| 4 + 8 | 8 + 1 | 1
= 1 | 10 | 13 | 12 | 9 | 1
= 1 + 1 | 0 + 1| 3 + 1 | 2 | 9 | 1
= 2 | 1 | 4 | 2 | 9 | 1 = 214291
Next number is 1 | 2 + 1 | 1 + 4 | 4 + 2 | 2 + 9 | 9 + 1 | 1
= 1 | 3 | 5 | 6 | 11 |10 |1
= 1 | 3 | 5 | 6 + 1 | 1 + 1 | 0 | 1
= 1 | 3 | 5 | 7 | 2 | 0 | 1
= 1357201 (Answer)

Correct Option: 6 Best Solution (1)

05 Oct 2009 31 zero..

Upto how many integers we should take the product 1*2*3*4*5.......... to make 31 zeros at last of the number.

OPtion

1) 140
2) 130
3) 115
4) 110
5) 100
6) 105
7) 135
8) 125
9) 145
10) 120

Solution

For obtaining 31 zeros at last of the number, 31 powers of 5 is necessary. Product 1*2*3*4*5……………upto 125
contain 5, 10 ,15………. 125. These 25 numbers are multiple of 5, besides it 25, 50, 75, 100, 125 contain
a second 5 and 125 contains a third 5 hence total number of 5 = 25+5+1 = 31

Correct Option: 8 Best Solution (0)

02 Oct 2009 Carom Board Puzzle

A carom board is of size 2.5 m × 2.5 m. Striker is placed on a side at a position 0.5 m from the left corner.
For overcoming the condition of foul it is necessary to touch the queen placed on right side at a position 0.5 m
from the side where striker is placed. If player wants to make a two collision rebound then he should make
an impact on the first side at a distance of

OPtion

1) 0.8 m from corner
2) 1 m from corner
3) 0.5 m from corner
4) 0.75 m from corner
5) 0.6 m from corner
6) 1.25 m from corner
7) 1.4 m from corner
8) 1.3 m from corner
9) 1.5 m from corner
10) 0.25 m from corner

Solution

Let striker makes first impact at a distance x from the corner and second at a distance y from other corner.
First draw a plot for this condition. Then according to plot
x/0.5 = (2.5-x)/y => xy = 0.5(2.5-x) ..... Equation 1
and y/(2.5-x) = (2.5-y)/2 => 2y =(2.5-x) (2.5-y) .....Equation 2
On solving these equations we get x = 0.75 m

Correct Option: 4 Best Solution (0)