Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 110)
110.
A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. The number is:
Answer: Option
Explanation:
4 | x z = 6 x 1 + 4 = 10 ----------- 5 | y -2 y = 5 x z + 3 = 5 x 10 + 3 = 53 ----------- 6 | z - 3 x = 4 x y + 2 = 4 x 53 + 2 = 214 ----------- | 1 - 4 Hence, required number = 214.
Discussion:
30 comments Page 2 of 3.
D Priya said:
8 years ago
@Sarthaksaxen.
This is successive division not normal division.
This is successive division not normal division.
Sarthaksaxena said:
8 years ago
I guess this is wrong question as how it is possible to get 3 as a remainder when 214 is divisible by 5? Please tell me.
(3)
Shadab said:
8 years ago
The Number is 4 5 6 And their Remainder in 2 3 4 On just multiplying we got the answer like this [(6+4) * (5)+(3) * (4)+(2)] = 214.
(1)
Bikas khan said:
8 years ago
I can't understand it please explain simple.
Vijay mittal said:
9 years ago
In this question 214 is correct if only 214 is divided by 4, 5, 6. Orderly using the quotient of the previous solution. But in question, there is no explanation about this.
Manikandan said:
9 years ago
How is the s value equal to 1? Please explain.
Nikhil kumar said:
9 years ago
Let, p be the number.
Suppose,
p ÷ 4 = q
p ÷ 4 = q, remainder = 2.
q ÷ 5 = r
q÷5=r, remainder = 3.
r ÷ 6 = s
r ÷ 6=s, remainder = 4.
Then,
r = 6s + 4
q = 5r + 3
q = 5r + 3= 5(6s + 4) + 3 = 30s + 23.
p = 4q + 2 = 4 (30s + 23) + 2 = 120s + 94
Suppose the final quotient,
s = 1.
Then,
p = 120 * 1 + 94 = 214
214 is one of the choices given and hence it is the answer.
Suppose,
p ÷ 4 = q
p ÷ 4 = q, remainder = 2.
q ÷ 5 = r
q÷5=r, remainder = 3.
r ÷ 6 = s
r ÷ 6=s, remainder = 4.
Then,
r = 6s + 4
q = 5r + 3
q = 5r + 3= 5(6s + 4) + 3 = 30s + 23.
p = 4q + 2 = 4 (30s + 23) + 2 = 120s + 94
Suppose the final quotient,
s = 1.
Then,
p = 120 * 1 + 94 = 214
214 is one of the choices given and hence it is the answer.
(3)
Vivek gupta said:
9 years ago
Wrong answer.
How can remainder will be 3 when 214 is divided by 5?
The Correct answer should be 60k-2.
Where k is any integer.
How can remainder will be 3 when 214 is divided by 5?
The Correct answer should be 60k-2.
Where k is any integer.
Manju said:
9 years ago
Dividend = divisor * quotient + remainder.
So let the number be X.
=> X = 4a + 2--->1st eq (here a is quotient of 1st division , this will be divided by 5).
=> a = 5b + 3 (b is 2nd quotient, similarly).
=> b = 6c + 4
Considering 1st equation.
X = 4[5(6c + 4) + 3] + 2.
X = 120c + 94 ( here I did'nt get about 94).
Can anyone please explain it?
And rest calculations are like as Divya explained
So let the number be X.
=> X = 4a + 2--->1st eq (here a is quotient of 1st division , this will be divided by 5).
=> a = 5b + 3 (b is 2nd quotient, similarly).
=> b = 6c + 4
Considering 1st equation.
X = 4[5(6c + 4) + 3] + 2.
X = 120c + 94 ( here I did'nt get about 94).
Can anyone please explain it?
And rest calculations are like as Divya explained
Manju said:
9 years ago
@Divya could you please elaborate your explanation?
How a = 5b + 3?
How a = 5b + 3?
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