M4maths Previous Puzzles

12 Apr 2017 Permutation

How many 3-digit numbers can be formed from the digits 1, 2, 3, 5, 6, 7 and 0 which are divisible by 5 and none of the digits is repeated?

OPtion

1) 20
2) 24
3) 30
4) 36
5) 40
6) 44
7) 48
8) 55
9) 54
10) None of these

Solution

Answers: 55
To be divisible by 5 unit digit of number should be either 5 or 0
We have total 7 digits out of which we cannot have 0 at the hundred’s place (else it will become a 2 digit number)
Unit place is filled with 5
a. Tens place can be filled with 0
Hundreds place can now be filled with (1, 2, 3, 6, 7). We have 5 ways
So we have 1*1*5 = 5
b. Tens place can be filled with any of the digits (1, 2, 3, 6, 7). We have 5 ways
Hundreds place can now be filled with remaining of 4 digits. We have 4 ways
So we have 1*5*4 = 20

Unit place is filled with 0
Tens place can be filled with any of the digits (1, 2, 3, 5, 6, 7). We have 6 ways
Hundreds place can now be filled with remaining of 5 digits. We have 5 ways
So we have 1*6*5 = 30

So Total number of ways = 5 + 20 + 30 = 55

Option 8)

Correct Option: 8 Best Solution (2)

11 Apr 2017 Divisibility

What is the largest 5 digit number which is divisible by 18, 26 and 35?

OPtion

1) 99918
2) 99450
3) 99900
4) 98280
5) 98820
6) 98918
7) 98260
8) 98200
9) 98180
10) None of these

Solution

Answer: 98280
LCM of 18, 26, 35 = 8190

99999 ÷ 8190 = 12, remainder = 1719
Hence, greatest number of 5 digits which is divisible by 18, 26 and 35
= 99999 - 1719 = 98280
Or we can also get the answer by simply = 8190*12 = 98280

Option 4)

Correct Option: 4 Best Solution (9)

10 Apr 2017 Series

Find the 8th term of below series:
2+i, 3+i, 5+2i, 7+3i, 11+5i, ...

OPtion

1) 13+7i
2) 14+7i
3) 13+8i
4) 17+8i
5) 17+13i
6) 13+13i
7) 19+21i
8) 19+13i
9) 17+21i
10) None of these

Solution

Answer: 19+21i

Each element of series can be represented as a and ib where
a(n) = nth prime number
ib(n) = nth Fibonacci number

8th prime number when first term is 2 = 19
8th Fibonacci number when first number is 1 = 21

Hence 8th term = 19+21i
Option 7)

Correct Option: 7 Best Solution (4)

07 Apr 2017 Time and Work

P is able to do a piece of work in 15 days and Q can do the same work in 10 days. If they can work together for 2 days, what is the fraction of work left?

OPtion

1) 8/15
2) 2/3
3) 1/3
4) 1/2
5) 3/4
6) 3/5
7) 3/7
8) 2/5
9) 2/7
10) None of these

Solution

Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/10
Amount of work P and Q can do in 1 day = 1/15 + 1/10 = 5/30 = 1/6
Amount of work P and Q can together do in 2 days = 2 × (1/6) = 1/3
Fraction of work left = 1 – 1/3= 2/3

Option 2)

Correct Option: 2 Best Solution (13)

06 Apr 2017 Compound and Simple Interest

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 48. The sum is:

OPtion

1) 4375
2) 5625
3) 5000
4) 7500
5) 6250
6) 6500
7) 7000
8) 7250
9) 6750
10) None of these

Solution

Given that difference between compound interest and simple interest is Rs.48
= > 48 = P(4/100)^2
= > 48 = P(1/25)^2
= > 48 = P(1/625)
= > P = 30000

Correct Option: 10 Best Solution (0)

05 Apr 2017 Simplification

If log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13 , then value of x is ?
Note:
Assume all log represented above is to the base of 10.
^ denotes power of
! denotes factorial

OPtion

1) 33/19
2) 66/19
3) 66/18
4) 33/18
5) 68/38
6) 74/38
7) 64/38
8) 32/17
9) 64/36
10) None of these

Solution

Considering base is 10, so 1 could be represented as Log10, 2 as Log 100…..
Let’s revise the formula for Logs
Log(XY) = LogX+LogY
Applying the formula we get:

= > log4 + log(17x+1) = log(3x+7) - 0^(0!) + 0^(0) + 9 Mod 11 - 243 Mod 13
= > log(68x + 4) = log(3x+7) - 0 + 1 + 9 - 9
= > log(68x + 4) = log(3x+7) + 1
= > log(68x + 4) = log(3x+7) + log 10
= > log(68x + 4) = log(30x+70)
= > log(68x + 4) = log(30x+70)
= > 68x + 4 = 30x+70
= > 38x = 66
= > 19x = 33
= > x = 33/19
Option 1)

Correct Option: 1 Best Solution (3)

04 Apr 2017 Probability

Three dices are thrown simultaneously. What is the probability of getting three numbers whose sum is 13?

OPtion

1) 14/216
2) 15/216
3) 16/216
4) 17/216
5) 18/216
6) 19/216
7) 10/116
8) 10/54
9) 5/54
10) None of these

Solution

Answer: 7/72
Total Number of possible outcomes = 6C1*6C1*6C1 = 216
Positive outcomes must satisfy D1+D2+D3 = 13
Positive outcomes which satisfy above condition:
1, 6, 6
2, 6, 5
2, 5, 6
3, 6, 4
3, 5, 5
3, 4, 6
4, 6, 3
4, 5, 4
4, 4, 5
4, 3, 6
5, 6, 2
5, 5, 3
5, 4, 4
5, 3, 5
5, 2, 6
6, 5, 2
6, 4, 3
6, 3, 4
6, 2, 5
6, 1, 6
6, 6, 1

Count of positive outcome = 21
Probability = Positive outcome/Total Possible Outcomes = 21/216 = 7/72
Option 10)

Correct Option: 10 Best Solution (6)

03 Apr 2017 Percentage

Rahane scored 38 runs in 2nd innings of test match against Australia which included 4 fours and 2 sixes. What percent of his total 2nd innings score did he make by running between the wickets?

OPtion

1) 25.12%
2) 25.32%
3) 25.52%
4) 25.82%
5) 26.02%
6) 26.22%
7) 26.42%
8) 26.32%
9) 26.62%
10)26.82%

Solution

Number of runs made by running = 38 - (4 x 4 + 2 x 6)
= 38 - (28) = 10

Required percentage = (10/38)*100 = 26.32%
Option 8)

Correct Option: 8 Best Solution (9)

31 Mar 2017 Age

15 years ago, P was half of Q's age. If the ratio of their present ages is 6:7, what will be the product of their present ages?

OPtion

1) 145
2) 210
3) 323
4) 380
5) 320
6) 378
7) 399
8) 340
9) 418
10) None of these

Solution

Answer: 378
Let the present age of P be P and Q be Q
15 Years ago
P's age P-15 and Q's age Q-15
As per question
P-15 = (Q-15)/2
=> 2P -30 = Q-15
2P- Q = 15 ---- (i)

Also its given that P/Q = 6/7
P = 6Q/7

Putting value of P in equation 1

12Q/7 - Q = 15
5Q = 15*7
Q = 21
P = 18

P*Q - 21* 18 = 378
Option 6)

Correct Option: 6 Best Solution (15)

30 Mar 2017 Series

Find the tenth term of below series:
7, 16, 36, 78, 144, .....

OPtion

1) 212
2) 798
3) 668
4) 324
5) 900
6) 438
7) 546
8) 222
9) 216
10) 232

Solution

Answer: 900

Series is formed by = (nth Prime number* (n+1)th Prime number) +1
10th term = (10th Prime No. * 11th Prime No.) +1
= (29*31) +1 = 900

Option 5)

Correct Option: 5 Best Solution (2)