Verbal Reasoning - Arithmetic Reasoning - Discussion
Discussion Forum : Arithmetic Reasoning - Section 2 (Q.No. 3)
3.
I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?
Answer: Option
Explanation:
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5. Such a number is 25.
Discussion:
5 comments Page 1 of 1.
Amandeep said:
6 years ago
4*3*2=24 and one sweet left 24 +1=25.
(2)
Vishnu said:
7 years ago
I cannot understand this please anyone explain this.
Thank you.
Thank you.
CSR said:
1 decade ago
We just have to find the second least common multiple which is 24 + 1 = 25.
Siddharth said:
1 decade ago
The question asks What is the minimum number. 65 is also an answer but we need the minimum number.
Shefali chhabra said:
1 decade ago
My answer is coming out to be 65.
lcm of (2,3,4) +1= 13
lcm of (13,5)=65
lcm of (2,3,4) +1= 13
lcm of (13,5)=65
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