Aptitude Discussion

Q. |
Each of $X$ alarm tolls at regular intervals. All of them tolls together twelve times a day. No two alarm at equal intervals of time. If each alarm tolls after a whole number of minutes, what is the maximum possible value of $X$? |

✖ A. |
14 |

✔ B. |
16 |

✖ C. |
18 |

✖ D. |
20 |

**Solution:**

Option(**B**) is correct

The alarm tolls together twelves times a day. Therefore, they toll together once every 2 hours (or 120 minutes).

Since no two alarms toll at equal intervals of time, the total number of distinct factors of 120, including 1 and 120 itself $= 2^3×3×5$

The number of factors $= (3+1)×2×2 = 16$.

The maximum value of $X$ is **16**.

**Vineet**

*()
*

Taking a=6, b=8, c=10. ab combo - 12 tyms ring together, ac combo - 10 tyms, bc combo - 3 tyms, acb combo - 2 tyms..

Hence 2 out of 3 rings = (ab+bc+ac)-abc = 12+10+3-2 = 23

**Vicky**

*()
*

No. of divisors of 120= 2(3 times) * 5(1 times) * 3(1 times)

$= (3+1)\times (1+1)\times(1+1)$

$= 16$

**Maia**

*()
*

I understood everything up to the distinct factors of 120, but how do you get $(3+1)\times 2 \times 2=16$?

$120=2^3\times 3 \times 5$

no of factors = (prime no1 power +1)*(prime no power+1)*....

therefore

here No.of factors,

$=(3+1)*(1+1)*(1+1)$

$=4*2*2$

$=16$

Three bells B1, B2 and B3 started tolling together and they kept tolling at intervals of 6, 8 and 10 seconds respectively. How many times did exactly two of three bells toll together in the first 5 minutes from the start?

Options are:

1 21

2 22

3 23

4 24

5 25