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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 :

  • 80
  • 82
  • 85
  • 87

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

Similar Questions :

1. 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

  • 400
  • 256
  • 120
  • 420

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

2. Find the largest number of four digits which is exactly divisible by 27,18,12,15

  • 9700
  • 9710
  • 9720
  • 9730

Answer: Option C

Explanation:

LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720

3. 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

4. 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.

  • 15
  • 16
  • 30
  • 31

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

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

  • 128
  • 126
  • 146
  • 434

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

Read more from - HCF and LCM Questions Answers
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