M4maths Previous Puzzles

30 Sep 2011 seconds

Total seconds between 2:10 AM to 3:25 PM will be

OPtion

1) 47700
2) 49000
3) 51200
4) 56900
5) 65000
6) 70000
7) 76500
8) 80000
9) 82600
10) 86400

Solution

Total seconds between 2:10 AM to 2:10 PM = 86400/2 = 43200
total seconds between 2:10 PM to 3:25 PM = 3600 + 900 = 4500
so total seconds = 43200 + 4500 = 47700

Correct Option: 1 Best Solution (23)

29 Sep 2011 Umbrella

How many words can be formed by using 5 letters of the word UMBRELLA
at a time

OPtion

1) 4260
2) 4220
3) 2640
4) 2460
5) 6240
6) 6420
7) 4460
8) 4640
9) 2660
10) none of these

Solution

Words having all the letters different = 7*6*5*4*3*2*1/2*1 = 2520
Words having repetition of l = [7*6*5*4*3*2*1/(3*2*1)(4*3*2*1)]*[5*4*3*2*1/2*1] = 2100
so total methods = 2520 + 2100 = 4620

Correct Option: 10 Best Solution (4)

28 Sep 2011 subject

The % obtained by a student is 74. If we take % on the basis of 4 subjects (leaving the marks of fifth subject) then % is 76. How many marks, he got in the fifth subject.

OPtion

1) 69
2) 66
3) 70
4) 71
5) 75
6) 79
7) 59
8) 56
9) 65
10) 58

Solution

74*5 = 370
76*4 = 304
so his marks in fifth subject = 370-304 = 66

Correct Option: 2 Best Solution (14)

27 Sep 2011 cost

The total cost of 10 copies and 5 pens is 125 rupee. It is given that the cost (per unit) of two articles are 2 successive integers. Total cost of 1 copy and 1 pen is

OPtion

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

Solution

According to the given condition
either 10x+5(x+1)=125
or 10(x+1)+5x=125
in one of the two cases x is not integer, so solve the other x=8
so answer = 8+9 = 17

Correct Option: 8 Best Solution (20)

26 Sep 2011 sum

1,4,9,16,......
sum of this sequence upto 20 terms will be

OPtion

1) 1980
2) 2050
3) 2870
4) 2980
5) 3160
6) 3290
7) 3340
8) 3560
9) 3920
10) none of these

Solution

1+4+9+16+.....................
= 1^2+2^2+3^2+4^2+..............
= sum of squares of first 20 integers
according to sigma summation sum of squares of first n integers
= (1/6)*n*(n+1)*(2n+1)
on putting n = 20, answer will be obtained

Correct Option: 3 Best Solution (16)

23 Sep 2011 x y z ?

x+y+z>10
y+z>10
x+z>10
possible set for x,y,z will be

OPtion

1) 3,8,4
2) 5,6,4
3) 6,4,6
4) 4,7,5
5) 7,4,5
6) 5,5,6
7) 5,6,5
8) 5,5,5
9) 8,4,4
10) more then one are possible

Solution

5,5,6 satisfies all the three conditions

Correct Option: 6 Best Solution (10)

22 Sep 2011 Find the sum

Find the sum of all possible product taken 2 digit at a time by using 2,3,4,5 will be?

OPtion

1) 65
2) 71
3) 75
4) 60
5) 61
6) 50
7) 28
8) 81
9) 82
10) none of these

Solution

The possible products are:
2*3 = 6
2*4 = 8
2*5 = 10
3*4 = 12
3*5 = 15
4*5 = 20
sum is 71

Correct Option: 2 Best Solution (10)

21 Sep 2011 zero

total number of zeros in the right in the product 1*2*3............24*25 will be

OPtion

1) 0
2) 1
3) 2
4) 3
5) 4
6) 5
7) 6
8) 7
9) 8
10) none of these

Solution

1*2*3*4.........24*25 has 5=5*1,10=5*2,15=5*3,20=5*4,25=5*5 so total number of 5 is equal to 6,
so total number of zeros in the right will be 6

Correct Option: 7 Best Solution (8)

20 Sep 2011 area

Area ( in cm.^2) of the largest rectangle formed by the string of length 24 cm. is

OPtion

1) 36
2) 40
3) 144
4) 576
5) 32
6) 100
7) 80
8) 200
9) 24
10) none of these

Solution

for getting max. area we should make an square by this string,
side of that square = 24/4 = 6 cm.
so area = 6*6 = 36 cm.^2

Correct Option: 1 Best Solution (6)

19 Sep 2011 sphere

Water from 1000 spherical vessels of radius 10 cm. is poured to a vessel of bigger radius. If bigger vessel is completely filled by the water of smaller spheres, then what will be the radius of bigger.

OPtion

1) 10000 cm.
2) 20 cm.
3) 50 cm.
4) 1000 cm.
5) 100 cm.
6) is not possible to solve
7) 60 cm.
8) 80 cm.
9) 120 cm.
10) none of these

Solution

volume depends on 3 powers of radius so new radius = 10 cm. * (1000)^1/3
= 10*10 cm. = 100 cm.

Correct Option: 5 Best Solution (8)