Question

E 1 Exercise 3.4 1. Find the smallest number which when divided by 15, 25 and 35 leaves a remainder 7 in each case. 2. Find the least number which on adding 5 to it becomes exactly divisible by 15, 25, 30 and 45. 3. Four bells ring at intervals of 6, 8, 12 and 20 minutes. They ring simultaneously at 8 am At what time will they ring together again? 4. Find the least 5-digit number which on dividing by 4, 12, 20 and 24 leaves remainder 3 in each case. 5. Two tankers have the capacity of 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.​

Answer 1

Answer:

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HERE ARE YOUR ANSWERS

Step-by-step explanation:

1. LCM OF 15,25,35 IS 525.

add 7 in 525 so 525+7=532.

so the answer is 532

2.sobin this also LCM OF 15, 25,30,45 is 450

so 450-5= 441.

3 Required time =LCM of 6,18,12,20,

2|6, 8,12,20,

2|3,4, 6,10

3|3, 2,3,5,

1,2,1,5

required time=2×2×3×2×5

=120 minutes=2hrஃ bell will again ring together at(8+2) =10 am.

4. Least 5 digit number = 10000.

LCM(4,12,20,24) = 120

Dividing 10000 by 120 leaves a remainder 40 , Subtract 40 from 10000 = 9960

So this means 9960 is divisible by 120.

9960 + 120 = 10080.

To find the number which leaves remainder 3, we have to add 3 to 10080.

10080 + 3 = 10083.

5.So, H.C.F of 850 litres and 680 litres is 170. Therefore, the maximum capacity of the container is 170 litres