M4maths Previous Puzzles

09 Nov 2012 distribution

Total no. of methods to distribute 6 oranges among 4 persons?
It is compulsory to give minimum 1 orange to each person.

OPtion

1) 8
2) 10
3) 12
4) 15
5) 18
6) 20
7) 21
8) 24
9) 25
10) none of these

Solution

after giving 1 orange to each remaining oranges = 2
so methods to distribute these oranges among 4 persons = C(4+2-1,4-1)
=C(5,3) = 5*4*3*2*1/3*2*1*2*1 = 10

Correct Option: 2 Best Solution (0)

07 Nov 2012 cost

Cost of 20 apples of same size is more than 100 rupee but less than 150 rupee. If cost of one apple is an integer then no. of possible costs of one apple will be

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) none of these

Solution

only 6 and 7 rupee are possible costs

Correct Option: 2 Best Solution (0)

06 Nov 2012 Find the similer number

9, 16, 24, 25,

OPtion

1) 14
2) 21
3) 32
4) 36
5) 50
6) 12
7) 28
8) 20
9) 22
10) none of these

Solution

following numbers can be splitted in the following manner:
9*1 sum of numbers = 10
8*2 sum of numbers = 10
7*3 sum of numbers = 10
6*4 sum of numbers = 10
choose the number accordingly

Correct Option: 2 Best Solution (0)

05 Nov 2012 sequence

next number of the given sequence
1, 0, -2, -9, -40, -205, ............. will be

OPtion

1) -438
2) -564
3) -675
4) -956
5) -987
6) -1123
7) -1236
8) -1546
9) -1856
10) none of these

Solution

1*1 - 1 = 0
0*2 - 2 = -2
-2*3 - 3 = -9
-9*4 - 4 = -40
-40*5 - 5 = -205
-205*6 - 6 = -1236

Correct Option: 7 Best Solution (12)

02 Nov 2012 integer

A two digit non zero number x is equal to the sum of a number y and its square. If sum of the digits of x is equal to y and tens digit of x is half of its unit digit, then product of x and y will be

OPtion

1) 30
2) 36
3) 48
4) 24
5) 16
6) 81
7) 75
8) 90
9) 18
10) none of these

Solution

let x = 10a+b (1)
then according to the given condition
x = y + y^2 (2)
a+b = y (3)
a = b/2 (4)
on solving these equations we get x = 12 and y = 3

Correct Option: 2 Best Solution (0)

01 Nov 2012 post box

There are 4 post box and 4 letters (same). Multiplication of the no. of ways to post the letters in the post box
1) if all post box are same.
2) if all post box are different.

OPtion

1) 24
2) 36
3) 48
4) 64
5) 92
6) 108
7) 136
8) 160
9) 180
10) none of these

Solution

no. of ways if all post box are same
= 5 [4,0,0,0],[3,1,0,0],[2,2,0,0],[2,1,1,0],[1,1,1,1]
no. of ways if all post box are different
= 35
details
cases of [4,0,0,0] 4*3*2*1/3*2*1 = 4
cases of [3,1,0,0] 4*3*2*1/2*1 = 12
cases of [2,1,1,0] 4*3*2*1/2*1 = 12
cases of [2,2,0,0] 4*3*2*1/2*1*2*1 = 6
cases of [1,1,1,1] 1 = 1

Correct Option: 10 Best Solution (0)

31 Oct 2012 even integers

Sum of three positive even integers is 30 and their product is 960. Sum of the two smaller integers will be?

OPtion

1) 6
2) 8
3) 10
4) 12
5) 14
6) 16
7) 18
8) 20
9) 22
10) none of these

Solution

let the integers are 2a, 2b and 2c
according to the given condition
2a+2b+2c = 30 so a+b+c = 15 (1)
and 2a*2b*2c = 960
so abc = 120 (2)
on putting the value of a from (1) in (2)
we get (15-b-c)bc = 120
now check b = 1,2,3,4,5,6,7,8,9,10,11,12,13,14 after that 15-b-c will be negative
out of these 14 numbers 14,13,11,9 and 7 are not multiple of 120 so rejected. besides it 1,2 and 3 will give imaginary roots so rejected.
For b = 8,10 and 12 roots are not integers. Finally for b = 4,5 and 6 we get a and c as integer
values of a,b and c are 4,5 and 6
so integers are 8,10 and 12

Correct Option: 7 Best Solution (6)

30 Oct 2012 solve

x + y = 8 and 1/x + 1/y = 2/3 then product of x and y will be

OPtion

1) 8
2) 9
3) 10
4) 12
5) 14
6) 15
7) 18
8) 20
9) 28
10) none of these

Solution

1/x + 1/y = 2/3
implies (x+y)/xy = 2/3
so 8/xy = 2/3
so xy = 12

Correct Option: 4 Best Solution (0)

29 Oct 2012 sequence

1, 1.6, 2.8, 5.2, 10, ..................
next number of the sequence will be

OPtion

1) 12.6
2) 14.8
3) 16.2
4) 16.4
5) 16.6
6) 17.6
7) 18.8
8) 19.6
9) 20.4
10) none of these

Solution

1*2-0.4 = 1.6
1.6*2-0.4 = 2.8
2.8*2-0.4 = 5.2
5.2*2-0.4 = 10
10*2-0.4 = 19.6

Correct Option: 8 Best Solution (0)

26 Oct 2012 probability

There are 52 cards in a pack. We have to select 4 cards of different suits without replacement. What will be its probability.

OPtion

1) 1254/23475
2) 1/24
3) 1/120
4) 2197/20825
5) 39/51
6) 169/2652
7) 3475/11111
8) 12/169
9) 1/2
10) none of these

Solution

required probability = [52/52]*[39/51]*[26/50]*[13/49]
= 13*13*13/17*25*49
= 2197/20825

Correct Option: 4 Best Solution (6)