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. 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

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. Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest terms.

\begin{aligned} \frac{11}{12} \end{aligned}
\begin{aligned} \frac{23}{24} \end{aligned}
\begin{aligned} \frac{26}{27} \end{aligned}
\begin{aligned} \frac{4}{7} \end{aligned}

Answer: Option A

Explanation:

HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12

4. Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}

Answer: Option B

Explanation:

HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
2 * 3*3 * 7

5. Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm

45cm
255cm
244cm
55cm

Answer: Option A

Explanation:

Convert first all terms into cm.
i.e. 1575 cm, 1125cm, 765cm.

Now whenever we need to calculate this type of question, we need to find the HCF. HCF of above terms is 255.

Question Detail

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

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

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. Reduce \begin{aligned} \frac{803}{876} \end{aligned} to the lowest terms.

4. Find the HCF of \begin{aligned} 2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7 \end{aligned}

5. Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm

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

3. Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest terms.

4. Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}