Question Detail

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

1. 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

2. Find the HCF 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 A

Explanation:

Whenever we have to solve this sort of question, remember the formula.
HCF = \begin{aligned} \frac{HCF of Numerators}{LCM of Denominators} \end{aligned}
So answers will be option 1

3. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is

16cm
25cm
15cm
35cm

Answer: Option D

Explanation:

So by now, you must be knowing this is a question of HCF, right.
H.C.F. of (700 cm, 385 cm, 1295 cm) = 35 cm.

4. 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

5. The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number

Answer: Option D

Explanation:

Solve this question by using below formula.
Product of 2 numbers = product of their HCF and LCM

144 * x = 12 * 5040

x = (12*5040)/144 = 420

Question Detail

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

1. 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 :

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

3. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is

4. 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

5. The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number

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

4. 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