Question Detail

Find the HCF of 54, 288, 360

Answer: Option A

Explanation:

Lets solve this question by factorization method.

\begin{aligned}
18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
\end{aligned}

So HCF will be minimum term present in all three, i.e.
\begin{aligned}
2 \times 3^2 = 18
\end{aligned}

1. Find the LCM of \begin{aligned} \frac{2}{3}, \frac{4}{6}, \frac{8}{27} \end{aligned}

\begin{aligned} \frac{2}{27} \end{aligned}
\begin{aligned} \frac{8}{3} \end{aligned}
\begin{aligned} \frac{2}{3} \end{aligned}
\begin{aligned} \frac{8}{27} \end{aligned}

Answer: Option B

Explanation:

Whenever we have to solve this sort of question, remember the formula.
LCM = \\begin{aligned} \\frac{HCF of Denominators}{LCM of Numerators} \\end{aligned}
So answers will be option 2,
Please also give attention to the difference in formula of HCF and LCM

2. 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

3. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is

10:28 am
10:30 am
10:31 am
None of above

Answer: Option C

Explanation:

L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes

They will beep together at 10:31 a.m.

Sometimes questions on red lights blinking comes in exam, which can be solved in the same way

4. 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

5. 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