If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Problems on H.C.F and L.C.M
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

55/601

11/120

120/11

601/55

55/601

11/120

120/11

601/55

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Problems on H.C.F and L.C.M
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 0, 62,98 and 130 as remainders respectively is .

15210

15110

15120

11115

15210

15110

15120

11115

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Problems on H.C.F and L.C.M
The least number which when increased by 5 each divisible by each one of 24, 32, 36 and 54 is :

869

4320

427

859

869

4320

427

859

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Problems on H.C.F and L.C.M
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

15210

11115

15120

15110

15210

11115

15120

15110

ANSWER DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M
In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?

820,369

700,400

800,500

820,360

820,369

700,400

800,500

820,360

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Problems on H.C.F and L.C.M
The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change

8:27:12 hrs

9:22:08 hrs

10:13:17 hrs

7:25:14 hrs

8:27:12 hrs

9:22:08 hrs

10:13:17 hrs

7:25:14 hrs

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Problems on H.C.F and L.C.M

Problems on H.C.F and L.C.M If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Problems on H.C.F and L.C.M The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 0, 62,98 and 130 as remainders respectively is .

Problems on H.C.F and L.C.M The least number which when increased by 5 each divisible by each one of 24, 32, 36 and 54 is :

Problems on H.C.F and L.C.M The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

Problems on H.C.F and L.C.M In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?

Problems on H.C.F and L.C.M The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change

Problems on H.C.F and L.C.M
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Problems on H.C.F and L.C.M
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 0, 62,98 and 130 as remainders respectively is .

Problems on H.C.F and L.C.M
The least number which when increased by 5 each divisible by each one of 24, 32, 36 and 54 is :

Problems on H.C.F and L.C.M
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:

Problems on H.C.F and L.C.M
In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?

Problems on H.C.F and L.C.M
The traffic lights at three different road crossings change after every 48 sec., 72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00 hours,then at what time they again change