Question Detail

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

1. 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

2. 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.

3. 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

4. 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}

5. Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively

Answer: Option B

Explanation:

Answer will be HCF of (400-9, 435-10, 541-14)

HCF of (391, 425, 527) = 17

Question Detail

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 :

1. 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

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

3. 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

4. Find the HCF of 54, 288, 360

5. Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively

3. 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