Latest TCS Aptitude Question SOLUTION: Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by rememberin

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(#M40021411) TCS Numerical Ability question Keep an EYE

Professor absentminded has a very peculiar problem, in that he cannot remember numbers larger than 15. However, he tells his wife, I can remember any number up to 100 by remembering the three numbers obtained as remainders when the number is divided by 3, 5 and 7 respectively. For example (2,2,3) is 17. Professor remembers that he had (1,1,6) rupees in the purse, and he paid (2,0,6) rupees to the servant. How much money is left in the purse?
option
A. 59
B. 61
C. 49
D. 56

Asked In TCS asxa asxaxax (10 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

Vaibhav Mishra  (10 years ago)

@Metasploit
He had 76 rupees,according to questions, he remember any no. larger than 15 as remainder when divided by 3,5 and 7 respectively.
Divide 76 by 3,5 and 7, we will get (1,1,6)as remainder.
He paid 20 rupees bcoz when 20 divided by 3,5 & 7 , we will get (2,0,6) as remainder.
Now money left=76-20
=56 (Ans)

Is this solution Helpfull? Yes (34) | No (20)  

sandy  (10 years ago)

@ SARANYA & RAGUNATHKC
i dont know the correct procedure but i got the answer by a process
it might be helpful to u
LET THE AMOUNT HE HAD BE X
IN THE QUESTION IT IS GIVEN THAT WHEN X IS DIVIDED BY 3,5,7 LEAVES REMAINDERS 1,1,6
LETS TAKE THE LAST NUMBER I.E 7
X%7=6...I.E THE NUM IS 7Y+6
POSSIBLE VALUES OF Y ARE FROM 1 TO 13
THEN X MAY BE FROM 13,20,27,....76,83,90,97
THEN DIVIDE THE ABOVE NUMS BY 3 AND 5 THE NUM WHICH LEAVES REMAINDERS 1,1 IS THE REQUIRED NUM
this might look like a big process but you can calculate the num in your mind.
HOPE THIS IS HELPFUL
THANK U

Is this solution Helpfull? Yes (27) | No (6)  

suryakant singh  (8 years ago)

Answer: D)56
According to question, he had (1,1,6) and paid (2,0,6)
if we devide both by 7 then we get remainder 6
Rs he had :7k+6
he paid : 7d +6
then money left is: 7(k-d)
number is divisible by 7
option A,B is invalid
then we have to decide C or D
again money left is :(5x+1) - (5y+0) = 5(x-y) + 1
then suitable option is D ,so money left is 56

Is this solution Helpfull? Yes (27) | No (3)  

mahesh  (7 years ago)

(1,1,6)-(2,0,6)=(-1,1,0)money left in his purse
so verify options which gives -1,1,0 as remainders when divided by 3,5,7
i.e,56

Is this solution Helpfull? Yes (24) | No (3)  

Aman   (10 years ago)

ans : 56

he had : 76 rupees
he gave : 20 rupees
now he has : 56 rupees

Is this solution Helpfull? Yes (3) | No (21)  

Kunal Verma  (10 years ago)

By hit and trial......starting from 27 (21(multiple of 7)+ 6 as remainder)and traversing 34,41,28,...69,76 gives the valid answer.. :)

Is this solution Helpfull? Yes (3) | No (7)  

chandan kumar mishra  (10 years ago)

the total money thet professor has is 76
as 3*25+1=76,5*15+1=76 and 7*10+6=76.......the money given to the servant was 20 as 20/3 gives 2 as remainder.......220/5 gives 0 as remainder.....and 20/7 gives 6 as remainder........ so money left is 76-20=56
ANS =56

Is this solution Helpfull? Yes (2) | No (4)  

RagunathKC  (10 years ago)

PLEASE TELL HOW TO PROCEED BY CRT METHOD [cryptography method]
IT CAN BE SOLVED BY CRT(CHINESE REMAINDER THEOREM)
R=1(mod 3)
R=1(mod 5)
R=6(mod 7)
R=?

Is this solution Helpfull? Yes (1) | No (2)  

harinee  (8 years ago)

Answer: 56
Explanation:
Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
(Explanation: See LCM formula 1 and 2: Click here)
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
⇒ 15x + 1 = 7c + 6 ⇒ c = 15x−57 ⇒ c = 2x+x−57
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N - M = 56

Is this solution Helpfull? Yes (1) | No (4)  

sa satish  (8 years ago)

Answer: 56
Explanation:
Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.

When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
⇒ 15x + 1 = 7c + 6 ⇒ c = 15x−57 ⇒ c = 2x+x−57
Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
M is a multiple of 5 from equation 2 so x and z must satisfy x = (5y - 2) / 3 and z = (5y - 6) / 7
so we get if y = 4 , the x and z also gets satisfied as x = 6 and z= 2
Therefore we get M = 20.
(125 also satisfies but this is next number)
Now N - M = 76-20 = 56

Is this solution Helpfull? Yes (1) | No (1)  

asxa asxaxax  (10 years ago)

@aman-explain.....

Is this solution Helpfull? Yes | No (2)  

kalpana murugesan  (10 years ago)

already he had 76rs wit him,he paid 20rs to servant,nw the amt in his purse is 56rs

Is this solution Helpfull? Yes | No (3)  

MD HASAN SIDDIQUI  (10 years ago)

professor had 76 rs and hence the remainder(1,1,6)...and he paid rs 20 to hyis servant which gives us the remainder of (2,0,6)...so rs(76-20) will be 56 which is D.

Is this solution Helpfull? Yes | No (2)  

saranya  (10 years ago)

how do find 76..i'm unable to understand..can anyone please explain it clearly

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sundareswari  (9 years ago)

@ sandy.. your method is simply superb..

Is this solution Helpfull? Yes | No   

PRABIR   (8 years ago)

possibilities
he had 20 27 34 41 48 55 62 69 76 83 90 97(%7=6)
paid 20 27 34 41 48 55 62 69 76 83 90 97 (%7=6 )

so he had 76
paid 20
left 56

Is this solution Helpfull? Yes | No (1)  

Rajesh kumar  (6 years ago)

Let the money with the professor = N
Then N = 3a +1 = 5b + 1 = 7c + 6.
Solving the above we get N = 181
(Explanation: See LCM formula 1 and 2: Click here)
When a number is divided by several numbers and we got same remainder in each case, then the general format of the number is LCM (divisors).x + remainder.
In this case 3, 5 are divisors. So N = 15x + 1. Now we will find the number which satisfies 15x + 1 and 7c + 6.
⇒ 15x + 1 = 7c + 6 ⇒ c =
15
x
−
5
7
⇒ c =
2
x
+
x
−
5
7

Here x = 5 satisfies. So least number satisfies the condition is 5(15)+1 = 76.
(x = 12 also satisfies condition. So substituting in 15x + 1 we get, 181 which satisfies all the three equations but this is greater than 100)
Similarly Money given to servant = M = 3x + 2 = 5y = 7z + 6
Solving we get M = 25.
(125 also satisfies but this is next number)
Now N - M = 56

Is this solution Helpfull? Yes | No   

phani  (5 years ago)

Answer: D)56
According to question, he had (1,1,6) and paid (2,0,6)
if we devide both by 7 then we get remainder 6
Rs he had :7k+6
he paid : 7d +6
then money left is: 7(k-d)
number is divisible by 7
option A,B is invalid
then we have to decide C or D
again money left is :(5x+1) - (5y+0) = 5(x-y) + 1
then suitable option is D ,so money left is 56

Is this solution Helpfull? Yes | No   

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