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. 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

2. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

Answer: Option B

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127

3. 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

4. Reduce $\begin{aligned} \frac{368}{575} \end{aligned}$ to the lowest terms.

$\begin{aligned} \frac{30}{25} \end{aligned}$
$\begin{aligned} \frac{28}{29} \end{aligned}$
$\begin{aligned} \frac{28}{29} \end{aligned}$
$\begin{aligned} \frac{16}{25} \end{aligned}$

Answer: Option D

Explanation:

We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25

5. 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

Question Detail

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

1. 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

2. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

3. 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 :

4. Reduce $\begin{aligned} \frac{368}{575} \end{aligned}$ to the lowest terms.

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

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Question Detail

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

1. HCF of 22×32×52,24×34×53×11\begin{aligned} 2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned} is

2. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

3. 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 :

4. Reduce 368575\begin{aligned} \frac{368}{575} \end{aligned} to the lowest terms.

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

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1. 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. Reduce $\begin{aligned} \frac{368}{575} \end{aligned}$ to the lowest terms.