Question

There are 312 ,260 , and 156 students in class VIII , IX and X respectively. Buses are to be hired to take them to the picnic. find the maximum no. of students who can sit in a bus if each bus takes equal no of students

Answer 1

In the above Question , the following information is given -

There are 312 ,260 , and 156 students in class VIII , IX and X respectively.

Buses are to be hired to take them to the picnic.

To find -

Find the maximum no. of students who can sit in a bus if each bus takes equal no of students .

Solution -

Here ,

There are 312 ,260 , and 156 students in class VIII , IX and X respectively.

Buses are to be hired to take them to the picnic.

Here , let us first find the HCF of the number of students .

To find this HCF , Euclid's Division Algorithm is used .

This HCF can also be found using other methods.

So ,

312 = 260 × 1 + 52

260 = 52 × 5 + 0

So , the HCF of 312 and 260 is 52 .

Now ,

156 = 52 × 3 + 0

So ,

the HCF of 312 , 156 and 260 is 52 .

So , each bus can take 312 students , satisfying the given conditions.

Number of buses for class VIIi

=> [ 312 / 52 ]

=> 6 buses .

Number of buses for Class 9 -

=> [ 156 / 52 ]

=> 3 buses

Number of buses for class 10

=> [ 260 / 52 ]

=> 5 buses .

Total number or busses -

=> 3 + 5 + 6

=> 14 buses .

Answer -

Maximum number of students that can sit I ln a bus is 52 and the total number of buses required is 14 .

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Answer 2

$\huge\sf\pink{Answer}$

☞ 52 Students can sit in a bus

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ There are 312, 260,156 students in class VIII , IX and X

✭ Each bus takes equal number of students

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ Number of students who can sit in a bus?

$\rule{110}1$

$\huge\sf\purple{Steps}$

Here your answer can simply be found by finding the HCF of the 3 numbers,

≫ HCF(312,260,156)

➝ $\sf 312 = 2\times2\times2\times3\times13$

➝ $\sf 260 = 2\times2\times5\times13$

➝ $\sf 156 = 2\times2\times13\times3$

➝ $\sf HCF(312,260,156) = 2^2\times13 = 52$

$\sf\orange{\therefore HCF(312,260,156) = 2^2\times13 = 52}$

$\rule{170}3$