Question Detail

Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest terms.

\begin{aligned} \frac{30}{25} \end{aligned}
\begin{aligned} \frac{28}{29} \end{aligned}
\begin{aligned} \frac{28}{29} \end{aligned}
\begin{aligned} \frac{16}{25} \end{aligned}

Answer: Option D

Explanation:

We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25

1. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is

\begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
\begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
\begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
\begin{aligned} 2 \times 3 \times 5 \end{aligned}

Answer: Option C

Explanation:

As in HCF we will choose the minimum common factors among the given.. So answer will be third option

2. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

Answer: Option D

Explanation:

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.

Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31

3. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

Answer: Option B

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127

4. There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

Answer: Option C

Explanation:

As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;

So first number is : 551/29 = 19
Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85

5. What will be the LCM of 8, 24, 36 and 54

Answer: Option C

Explanation:

LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216

Question Detail

Reduce \begin{aligned} \frac{368}{575} \end{aligned} to the lowest terms.

1. HCF of \begin{aligned} 2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned} is

2. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

3. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

4. There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

5. What will be the LCM of 8, 24, 36 and 54

Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest terms.

1. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is