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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 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.

2. LCM of two numbers is 7700 and HCF is 11. If one number is 308 then other number is

  • 269
  • 285
  • 300
  • 275

Answer: Option D

3. 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

4. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

  • 48
  • 22
  • 56
  • 27

Answer: Option A

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.

5. 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

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