Question

a) Find the greatest number that will divide 135, 401 and 268 leaving the remainder 2 in
each case.

b) Dolly purchases two bags of fertilizer weighing 75 kg and 105 kg. Find the maximum
value of weight which can measure the two bags exact number of times ​

Answer 1

Answer:

a) 133 is the greatest number that will divide 135, 401 and 268 to leave 2 as the remainder each time.

b) 15 is maximum value of weight which can measure the two bags exact number of times.

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Answer 2

$\huge \blue {QUESTION}$

a) Find the greatest number that will divide 135, 401 and 268 leaving the remainder 2 in

a) Find the greatest number that will divide 135, 401 and 268 leaving the remainder 2 ineach case.

$\huge \red {ANSWER}$

In order to make 137, 182 and 422 completely divisible by the greatest number , we need to subtract 2 from each term, so

$137-2 =>135, 182-2 =>180$$and$$422-2 => 420$

Now HCF of  135, 180 and 420 is

$135 =>3×3×3×5$

$180 =>2×2×3×3×5$

$420 =>2×2×3×5×7$

$Hence$$HCF$$of$ $135, 180$$and$ $420 is 3×5 => 15$

Hence the greatest number that will divide 137, 182 and 422 leaving the remainder 2 in each case is 15.

$\huge \blue {QUESTION}$

b) Dolly purchases two bags of fertilizer weighing 75 kg and 105 kg. Find the maximum

b) Dolly purchases two bags of fertilizer weighing 75 kg and 105 kg. Find the maximumvalue of weight which can measure the two bags exact number of times

$\huge \red {ANSWER}$

For finding maximum weight, we have to find H.C.F. of 75 and 105

Factors of $75 => 3 × 5 ×5$

Factors of $105 => 3×5×7$

$H.C.F. => 15$

Therefore the required weight is 15 kg.

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