M4maths Previous Puzzles

29 Mar 2017 Race

In a 100 m race, A can give B start of 30 m and C start of 37 m. In the same 100 m race B can give C start of how many meters:

OPtion

1) 7 m
2) 10 m
3) 21 m
4) 90 m
5) 20 m
6) 11 m
7) 9 m
8) 18 m
9) 12 m
10)None of these

Solution

Answer : 10 m
When A runs 100 m, B runs 70 m and
When A runs 100 m, C runs 63 m

This means that when B runs 70 m, C runs 63 m
So when B runs 100 m, C will run = (63/70)*100 = 90 m

So B can give C a start of 100 - 90 = 10 m

Option 2)

Correct Option: 2 Best Solution (11)

28 Mar 2017 Divisibility

What is the smallest number of five digits which is divisible by 15, 27 and 35?

OPtion

1) 10230
2) 10240
3) 10800
4) 10085
5) 10090
6) 10220
7) 10385
8) 10235
9) 10250
10) None of these

Solution

LCM of 15, 27, 35 = 945

9999 ÷ 945 = 10, remainder = 549
Hence, greatest number of four digits which is divisible by 15, 27 and 35
= 9999 - 549 = 9450
So Smallest 5 digit number divisible by 15, 27 and 35 = 9450+945 = 10395

Option 10)

Correct Option: 10 Best Solution (16)

27 Mar 2017 Mixture

A merchant has 175 kg of sugar, part of which he sells at 5% profit and the rest at 15% profit. He gains 108% on the whole. The quantity sold at 5% profit is:

OPtion

1) 52.5 kg
2) 50.5 kg
3) 54.5 kg
4) 57.5 kg
5) 120 kg
6) 122.5 kg
7) 121 kg
8) 121.5 kg
9) 122 kg
10)None of these

Solution

(Quantity of Cheaper)/(Quantity of Dearer) = (Price of Dearer - Mean Price)/(Mean Price - Price of Cheaper)

(Quantity of Cheaper)/(Quantity of Dearer) = (115 - 108)/(108-105) = 7/3

Total Quantity = 175 KG

The quantity sold at 5% profit is = (7/10) * 175 = 122.5 Kg

Option 6)

Correct Option: 6 Best Solution (10)

24 Mar 2017 Probability

With Pujara and Ishant at the crease Cricket Team of India needs to score 4 runs in last 2 balls to win the match. What is the probability that India doesn't wins the match? Assume that:
- There were no wide balls or no balls bowled in these 2 balls.
- No Wicket has fallen in these 2 balls
- No Overthrows or extra runs were scored in these 2 balls
- In case if run(s) are scored on any of these 2 balls it can only be one of these values: 1, 2, 3, 4 & 6
This also implies that it’s quite possible that zero runs are scored on a particular ball

OPtion

1) 1
2) 9/36
3) 11/36
4) 10/35
5) 5/19
6) 4/1
7) 5/36
8) 5/18
9) 7/36
10)10/26

Solution

On any of these 2 balls the possible runs which can be scored are: 0,1,2,3,4,6
So total possible number of runs which can be scored in 2 balls are = 6C1*6C1 = 36
If a 4 or 6 are scored on 1st ball 2nd balls outcome doesn't matter, so actual number of possible outcomes = 36-10 = 26
Let’s assume that run scored on first ball is x and 2nd ball is y
India will not win the match if x+y <= 3.
Possible pair of values of x and y so that India doesn’t wins the match
(x, y)
(0, 0), (1, 0), (2, 0), (3, 0) , (0, 1) , (1, 1) , (2, 1) , (0, 2) , (1, 2) and (0, 3)
Count of set of values for India doesn’t wins the match = 10
Probability = Positive outcome/Total Possible Outcomes = 10/26 = 5/13
Option 10)

Correct Option: 10 Best Solution (0)

23 Mar 2017 Time, Distance & Speed

Ramanujam could either walk or drive to office. The time taken to walk to the office is 8 times the driving time. One day, his wife took the car making him walk to office. After walking 2.5 km, he reached a temple when his wife called to say that he can now take the car. Ramanujam figured that continuing to walk to the office will take as long as walking back home and then driving to the office. Calculate the distance between the temple and the office.

OPtion

1) 13/7
2) 7/3
3) 45/14
4) 16/7
5) 40/7
6) 24/7
7) 17/7
8) 43/14
9) 9/7
10)None of these

Solution

Suppose the distance between his home and office is x km. Then, distance between his office and temple is (x – 2.5) km.
If he walks at a speed of y km/hr. Then his driving speed = 8y km/hour, as he is 8 times faster in car.

Then,
2.5/y + x/8y = (x – 2.5)/y
Or, 2.5 + x/8 = x – 2.5
or 7x/8 = 5
Or, x = 40/7
Finally, x – 2.5 = 40/7 - 5/2 = (80-35)/14 = 45/14
Option 3)

Correct Option: 3 Best Solution (1)

22 Mar 2017 Radius of Sphere

2 solid spheres of radius 2 m, 1 solid cylinder of radius 2 m, height 4 meter and 3 solid cones of radius 2 m, height 6 m are melted to form a bigger solid sphere. What is the radius of this bigger solid sphere in meters?
Please note that symbol ^ denotes power of

OPtion

1) 4.17
2) 4.12
3) 4.06
4) 4.01
5) 3.87
6) 3.81
7) 3.73
8) 3.66
9) 3.58
10)3.51

Solution

Answer: 3.58m
Volume of 2 smaller sphere's = 2*(4/3)*Pi*r^3
Volume of one Cylinder = Pi*r^2*h1
Volume of 3 Cones = 3*(1/3)*Pi*r^2*h2
Volume of Larger Sphere = (4/3)*Pi*R^3
Here r=h=2m, h1=4m, h2=6m and R=?
Volume of Larger Sphere = Volume of 2 smaller sphere's + Volume of one Cylinder +Volume of 3 Cones
(4/3)*Pi*R^3 = 2*(4/3)*Pi*r^3 + P1*r^2*h1 + 3*(1/3)*Pi*r^2*h2
=> 4/3*R^3 = (8/3)*8 + 16 + 24
=> R^3/3 = 16/3 + 4 + 6 = (16+12+18)/3 = 46/3
=> R^3 = 46
R = 46^(1/3) =3.58 m

Option 9)

Correct Option: 9 Best Solution (2)

21 Mar 2017 Age

A is three years older than B who is thrice as old as C. The total of the ages of A, B and C is 59. How old is C?

OPtion

1) 24 years
2) 27 Years
3) 10 Years
4) 7 Years
5) 9 Years
6) 6 Years
7) 8 Years
8) 11 Years
9) 5 Years
10)None of these

Solution

Let’s assume Age of B be B years
Age of A = B+3
Age of C = B/3

Total age = B + 3 + B + B/3 = 59
Solving 7B/3 = 56
B = 24
C = B/3 = 24/3 = 8 years

Option 7)

Correct Option: 7 Best Solution (13)

20 Mar 2017 Compound Interest

At what rate of compound interest per annum will a sum of Rs. 6000 become Rs. 6615 in 2 years?

OPtion

1) 7.5
2) 7.25
3) 7.0
4) 6.5
5) 6.25
6) 6.0
7) 5.50
8) 5.25
9) 5.0
10)None of these

Solution

Suppose that rate of interest is R
Using Formula:
A = P (1+r/100)^T

A = 6615, P = 6000 and t= 2
So 6615 = 6000 (1+r/100) ^2
6615/6000 = (1+r/100)^2

1.05 = (1+r/100)
r = 5%

Option 9)

Correct Option: 9 Best Solution (6)

17 Mar 2017 Coding Decoding

If
MANGO = 51
GUAVA = 32

Then PEACH = ?

OPtion

1) 29
2) 33
3) 39
4) 54
5) 69
6) 84
7) 99
8) 104
9) 109
10)114

Solution

Each word is coded as = (Rank of the given word in the dictionary of words formed by the letters of the word)

So Rank of the word MANGO in the dictionary of words formed by the letters of MANGO = 51
Rank of the word GUAVA in the dictionary of words formed by the letters of GUAVA = 32

Similarly Rank of the word PEACH in the dictionary of words formed by the letters of PEACH = 109


Option 9)

Correct Option: 9 Best Solution (0)

16 Mar 2017 Distance

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 60° and 45° respectively. If the lighthouse is 123 m high, the distance between the two ships is:

OPtion

1) 146.67 m
2) 197.67 m
3) 153.9 m
4) 199.3 m
5) 171.3 m
6) 193.1 m
7) 182.5 m
8) 185.2 m
9) 190.4 m
10)194.0 m

Solution

Let AB be the lighthouse where A is the bottom point of light house and C and D be the positions of the ships.
Then, AB = 123 m, Angle ACB = 60° and angle ADB = 45°.
AB/AC = tan 60° = sq root(3) AC = AB / sq root(3) = 123/3^(1/2) = 41*3^(1/2)
AB/AD = tan 45° = 1 AD = AB = 123 m.
CD = (AC + AD) = 41*3^(1/2) + 123
= 41(3^1/2 + 3)
= 194 m
Option 10)

Correct Option: 10 Best Solution (12)