M4maths Previous Puzzles

26 Nov 2009 encode the number.

If 2.0178 = 107
3.1980 = 681
4.6212 = 1568
5.2120 = 434
Then 6.1074 =

OPtion

1) 689
2) 207
3) 197
4) 1534
5) 1475
6) 1240
7) 42510
8) 12034
9) 457
10) 99

Solution

Here 2.0178 = 0178/2 + (2+0+1+7+8) = 89 + 18 = 107
Similarly for others

Correct Option: 3 Best Solution (2)

25 Nov 2009 five digits number

How many numbers of five digits can be formed by using 1, 2, 3, 4, 5, if 4 is necessarily taken at hundred's
place and 2 is not allowed at unit's place. Repetition of numbers is not allowed.

OPtion

1) 72
2) 16
3) 12
4) 120
5) 96
6) 24
7) 60
8) 30
9) 15
10) 18

Solution

4 is necessarily taken at hundred's place and 2 is not allowed at unit's place,
hence number of methods = 1*3*3*2*1 = 18
You can also check the answer by making different numbers under given conditions.

Correct Option: 10 Best Solution (7)

24 Nov 2009 Find the exact value

Exact value of 3.789789789789789789789........ is

OPtion

1) 1261/332
2) 1262001/333000
3) 126201/33301
4) 12619701/3330000
5) 12619967/3330000
6) 1262/333
7) 12610/3329
8) 12630/3335
9) 125874/33300
10) 1261404/333000

Solution

Let x = 3.789789789789........ (1)
Then 1000x = 3789.789789789789........ (2)
Subtracting (2) from (1), we get
999x = 3786
x = 3786/999 = 1262/333

Correct Option: 6 Best Solution (5)

23 Nov 2009 xvYBUz ?

If AxyBzC = ZcbYaX
And BbyCDw = YybXWd
Then xvYBUz =

OPtion

1) acBYFe
2) ecBYFa
3) cfBYEa
4) ebCYFa
5) yeBCFa
6) ecBYDf
7) ceBYFa
8) caYBFe
9) eaFBYc
10) aeDYBc

Solution

Rules for making these codes are
1. Capital letter should be capital.
2. Small letter should be small.
3. Letters are changed according to the sequence given below
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
z y x w v u t s r q p o n m l k j i h g f e d c b a
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
Z Y X W V U T S R Q P O N M L K J I H G F E D C B A

Correct Option: 7 Best Solution (4)

21 Nov 2009 examination result.

In an examination of 500 marks (5 subjects & 100 marks for each subject), Amit got first class
certificate with 67.8 % . If he got perfect square marks in all the subjects, then what are
the number of possible methods for taking these marks in all the subjects.

OPtion

1) 28
2) 20
3) 4
4) 120
5) 15
6) 12
7) 60
8) 30
9) 32
10) 243

Solution

The possible combination of marks is 49, 64, 64, 81, 81 Total number of methods for providing
81 marks for any 2 subjects out of 5 subjects is 10, then number of methods for providing
64 marks for any 2 subjects out of remaining 3 subjects is 3, then 49 marks is provided
for the last subject. Hence total number of methods are 10*3*1 = 30

Correct Option: 8 Best Solution (0)

20 Nov 2009 code for 4864

Some numerical numbers are coded in integers as given below
Number Code
4956 7
3878 8
9765 6
4135 5
Then code for 6846 is

OPtion

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

Solution

On taking factors of all the numbers of 4956, we get 2*2, 3*3, 5, 2*3
Total numbers = 7
Similarly in 3878, we get 3, 2*2*2, 7, 2*2*2
Total numbers = 8
Similarly for others

Correct Option: 1 Best Solution (0)

19 Nov 2009 Fruit Price?

If in a coded manner prices of following fruits are as given in the list below
Fruit Price (per Kg)
ORANGE 10
BANANA 5.50
MANGO 10
GRAPE 9.40
Then the price (per Kg) of GUAVA is

OPtion

1) 12
2) 12.20
3) 9.60
4) 8
5) 10.40
6) 13.60
7) 7.80
8) 5.50
9) 9.40
10) 6

Solution

Sum of positional values of all the letters of the word ORANGE is 15+18+1+14+7+5 = 60
Average per letter is 60/6 = 10, hence the price
Similarly for others,

Correct Option: 5 Best Solution (6)

18 Nov 2009 AREA

A sector of circle of radius equal to one fourth of the diagonal of square is cut from the square
taking one corner of the square as centre. If side of the square is 4 cm. then area of the remaining part is

OPtion

1) 15.68
2) 14.43
3) 12.78
4) 10.34
5) 14.73
6) 13.14
7) 14.14
8) 11.88
9) 11.83
10) 9.83

Solution

Diagonal of the square = 4√2 cm.
Hence radius of the circle = 4√2/4 cm. = √2 cm.
Area of the sector of circle = 1/4 (π* radius2)
= 1/4 (π* (√2)2 )
= π/2
Area of the square = 4*4 = =16 cm.2
Hence area of remaining part = 16- (π/2) cm.2

Correct Option: 2 Best Solution (5)

16 Nov 2009 Code for ARE

Five words IS, AM, ARE, WAS, WERE are coded in the following manner.
Word Code
IS 2
AM 6
ARE ?
WAS 49
WERE 84
Code for ARE is

OPtion

1) 12
2) 20
3) 25
4) 48
5) 8
6) 19
7) 14
8) 18
9) 34
10) 29

Solution

On arranging all the letters of the words IS, AM, ARE, WAS, WERE in increasing order we get
AA, AE, EEI, MRR, SSWW as the code for corresponding words. On taking their positional equivalent, we get
IS = AA = 1+1 = 2
AM = AE = 1+5 = 6
Similarly for others

Correct Option: 6 Best Solution (0)

13 Nov 2009 Successive Number

The sum of squares of three successive integers is 27650. Sum of these integers is

OPtion

1) 297
2) 256
3) 198
4) 234
5) 304
6) 288
7) 312
8) 252
9) 228
10) 346

Solution

(a-1)^2 + a^2 + (a+1)^2 = 27650 , solve the equation

Correct Option: 6 Best Solution (4)