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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 ?

  • 25
  • 37
  • 54
  • 65

Answer: Option A

Explanation:

The required number will be such that it leaves a reminder of 1 when divided by 2, 3 or 4 and no reminder when divided by 5. Such a number is 25 among options

Similar Questions :

1. In a group of 15 people, 7 read French, 8 read English while 3 of them read none of these two. How many of them read French and English both ?

  • 2
  • 3
  • 4
  • 5

Answer: Option B

2. First bunch of bananas has (1/4) again as many bananas as a second bunch. If the second bunch has 3 bananas less than the first bunch, then the number of bananas in the first bunch are

  • 9
  • 10
  • 12
  • 15

Answer: Option D

Explanation:

Let the number of bananas in the second bunch be x.
Then, the number of bananas in the first bunch
= x + 1/4 x
= 5/4 x
So,
5/4 x - x = 3
=> 5x - 4x = 12
x = 12

Number of bananas in first bunch
= 5/4 * 12 = 15

3. Reena is twice as old as Sunita. Three years ago, she was three times as old as Sunita. How old is Reena now ?

  • 6 years
  • 12 years
  • 14 years
  • 16 years

Answer: Option B

Explanation:

Let Sunita's present age = x years
Then Reena present age = 2x years

Three years ago
(2x-3) = 3(x-3)
2x - 3 = 3x - 9
or x = 6

Reena age = 2x = 2*6 = 12 year.

4. In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament ?

  • 12
  • 15
  • 30
  • 36

Answer: Option B

Explanation:

I. matches of first player with other 5 players
II. matches of second player with 4 players other than the first player
III. matches of third player with 3 players other than the first player and second player.
IV. matches of fourth player with 2 players other than the first player, second player and third player.
V. matches of fifth player with 1 player other than the first player, second player, third player and fourth player.

So total matches will be 5+4+3+2+1 = 15



5. A is three times as old as B. C was twice-as old as A four years ago. In four years' time, A will be 31. What are the present ages of B and C ?

  • 9, 46
  • 9, 50
  • 10, 46
  • 10, 50

Answer: Option B

Explanation:

As from the question we have,
A = 3B --- (i)
c - 4 = 2(A+4) ---(ii)

Also, A + 4 = 31
or A = 31-4 = 27

Putting A = 27 in (i) we get
B = 9
Putting A = 27 in (ii) we get
C = 50

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