M4maths Previous Puzzles

20 Apr 2016 Minimum number of rows

21 banana trees, 42 grapes trees and 56 papaya trees have to be planted in rows such that each row contain the same number of trees of one variety only. Minimum number of rows i which the above trees may be planted is

OPtion

1) 3
2) 10
3) 13
4) 15
5) 16
6) 17
7) 18
8) 20
9) 22
10)24

Solution

HCF of 21, 42, 56 = 7
Number of rows of banana trees, grapes trees and papaya trees are
21/7 = 3
42/7 = 6
56/7 = 8
Therefore Required number of row
3+6+8 = 17
Option 6)

Correct Option: 6 Best Solution (14)

19 Apr 2016 Number

The value of
2^n + 2^(n-1) / 2^(n+1) - 2^n is

OPtion

1) 1/2
2) 2^(n-1)/2^(n+1)
3) 2^(n+1)/2^(n-1)
4) 3/2
5) 2^(n-1)/2^(n+2)
6) 2/3
7) 2^n/2^(n+1)
8) 2^(n-1)/2^n
9) 1/3
10)None of these

Solution

Given Exp
2^(n-1) (2+1) / 2^n (2-1) = 3/2
Option 4)

Correct Option: 4 Best Solution (18)

18 Apr 2016 Sequence

Next number of the sequence
1, 3/2, 8/3, 247/48 will be

OPtion

1) 56345/5928
2) 58433/4368
3) 60433/5928
4) 60433/5467
5) 63456/5921
6) 64321/4536
7) 68733/3468
8) 70890/2341
9) 72314/2357
10) 72978/2379

Solution

1,
1*2 = 2 & 2 - 1/2 = 3/2
3/2*2 = 3 & 3 - 1/3 = 8/3
8/3*2 = 16/3 & 16/3 - 3/16 = 247/48
247/48*2 = 247/24 & 247/24 - 24/247 = 60433/5928

Correct Option: 3 Best Solution (2)

15 Apr 2016 Probability

Mohit has two coins, one of Rs. 1 denomination and the other of Rs. 2 denomination. He tosses the two coins simultaneously. What is the probability that he gets at least one head?

OPtion

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

Solution

We write H1 for head and T1 for tail of Rs. 1 coin. We write H2 for head and T2 for tail of Rs. 2 coin.
Thus, the possible outcomes are H1H2, H1T2, H2T1, T1T2
All the four outcomes are equally likely. Let the event getting at least one head be denoted by A.
Now, the three outcomes H1H2, H1T2, H2T1 favour the happening of the event A.
Therefore, P(A) = 3/4
Hence, the probability that Mohit gets at least one head is 3/4.
Option 7)

Correct Option: 7 Best Solution (22)

14 Apr 2016 Rate of flow in kilometer

A cylindrical pipe has inner diameter of 7 cm and water flows through it at 192.5 litre per minute. Find the rate of flow in kilometers per hours.

OPtion

1) 1 km/hr
2) 2.5 km/hr
3) 3 km/hr
4) 3.5 km/hr
5) 4 km/hr
6) 5 km/hr
7) 5.5 km/hr
8) 6 km/hr
9) 7 km/hr
10)7.5 km/hr

Solution

Volume of water that flows per hour = (192.5 * 60) litre
= 192.50*60*1000 cm^3 ....(i)
Inner diameter of the pipe = 7 cm
Inner radius of the pipe = 7/3 cm = 3.5 cm
Let h cm be the length of the column of water that flows in one hour
Clearly, water column is a cylinder of radius 3.5 cm and length h cm.
therefore
Volume of water that flows in 1 hours
= Volume of cylinder of radius 3.5 cm and length h cm
= 22/7 * 3.5^2 * h cm^3 ....(ii)
From (i) and (ii), we get
22/7 * 3.5*3.5*h = 192.50*60*1000
h = 300000 cm
= 3 km
Hence, the rate of flow of water is 3 km per hour.
Option 3)

Correct Option: 3 Best Solution (10)

13 Apr 2016 Which option is correct

Suppose b varies as the sum of two quantities of which one varies directly as a and the other inversely as a.
If b=6 when a=4 and b=10/3 when a=3, then which option is correct.

OPtion

1) b = 4a - 1/a
2) b = 2a - 3/a
3) b = a - 1/a
4) b = 7a - 8/a
5) b = 5a - 8/a
6) b = 2a - 8/a
7) b = 2a - 9/a
8) b = 2a - 1/a
9) b = a - 8/a
10)None of these

Solution

By using Ration and proportion
Let b = x+y where x => a and y => 1/a
Let x = k1 a and y = k2/a where k1 and k2 are constant
therefore
b = k1 a + k2/a
By using given value find k1 and k2
Putting the values we get
b = 2a - 8/a
Option 6)

Correct Option: 6 Best Solution (27)

12 Apr 2016 Quadratic equation

The root of the equation 4^x - 3.2^(x+2) + 32 = 0

OPtion

1) 1, 2 and 3
2) 2 and 3
3) 2, 3 and 4
4) 1 and 4
5) 4 and 8
6) 2 and 4
7) 1 and 2
8) 1, 3 and 4
9) 3 and 4
10)None of these

Solution

Given equation is 2^2x - 3.2^x * 2^2 + 32 = 0
2^2x - 12.2^x + 32 = 0
let y = 2^x
y - 12y + 32 = 0
solve y=8 or y=4
2^x = 8
x = 3
and 2^x = 4
x = 2
2 and 3
Option 2)

Correct Option: 2 Best Solution (10)

11 Apr 2016 Sequence

What is the next number of the following sequence.
21, 43, 65, 87, 109, ........

OPtion

1) 1112
2) 249
3) 119
4) 1113
5) 1214
6) 149
7) 228
8) 1012
9) 1211
10)168

Solution

21 = first even second odd
43 = first even second odd
65
87
109
Similarly
1211
Option 9)

Correct Option: 9 Best Solution (26)

08 Apr 2016 Probability

A bag contains 5 white, 4 red and 3 black balls. A ball is drawn at random from the bag. Find the probability that it is not black.

OPtion

1) 1/3
2) 1/5
3) 1/4
4) 3/4
5) 3/5
6) 2/3
7) 1/12
8) 1/9
9) 1/7
10)1/8

Solution

Total number of balls in the bag = 5+4+3 = 12
As we don't want a black ball, it means that the number of favorable outcomes = Total number of white and red balls = 5+4 = 9
Therefore probability of getting a ball other than black ball = 9/12
= 3/4
Option 4)

Correct Option: 4 Best Solution (21)

07 Apr 2016 Radius of the base

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. The radius of the base of the cone is

OPtion

1) Sqrt10 cm
2) sqrt15 cm
3) 2 Sqrt3 cm
4) 2 Sqrt7 cm
5) Sqrt13 cm
6) Sqrt7 cm
7) 2 Sqrt6 cm
8) Sqrt14 cm
9) Sqrt17 cm
10)2 Sqrt2 cm

Solution

Volume of Hemi-sphere = Volume of cone
Therefore
2/3*Pi*7*7*7 = 1/3 Pi R^2*49
Solve
R = Sqrt14
Option 8)

Correct Option: 8 Best Solution (12)