Question Detail

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

1. Find the HCF of 2.1, 1.05 and 0.63

0.44
0.64
0.21
None of above

Answer: Option C

Explanation:

To solve this question quickly, first remove decimal by multiplying each term with 100,
Then terms become 210, 105, 63
Then HCF of above terms is 21,
So Answer is 0.21

2. 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

3. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

Answer: Option B

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number = \begin{aligned} \frac{693 \times 11}{99} \end{aligned} = 77

4. 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

5. 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