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Q2. find LCM of the given numbers by division method
1.480,672=3360
2.6,8,45=360
3.24,40,84=840
4.20,36,63,77=13860
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Q3.Find the least number which when increased by 15 is exactly divisible by 15,35,48
=>1665 is the least number which when increased by 15 will be divisible by 15, 35 and 48.
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Q4. Find the least number which when increased by 16,15 and 18 leaves a remainder of 5 in each case
=>To find the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case, we have to find the L.C.M. of 6, 15 and 18 and then add 5 in that number. Hence, 95 is the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case.
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Q5. Find the least number which when increased by 24,46,45 and 54 leaves a remainder of 3 in each case
=>221
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Q6. find the greatest 3 digit number which is exactly divisible by 8,20 and 24
=>960
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Q7. find the smallest 4 digit number which is exactly divisible by 32,36 and 48
=>1152
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Q8. find the greatest 4 digit number which is exactly divisible by each of 8,12 and 20
=>9960
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Q9. Find the least number of a five digit which is exactly divisible by 32,36 and 45
=>10080
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Q10. three boys step off together from the same spot their steps measure 63cm,70 cm and 77 CM respectively what is the minimum distance each should cover so that all can cover the same instance in complete step
=>For finding minimum distance, we have to find L.C.M of 63, 70, 77. L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm. Therefore, the minimum distance is 6930 cm.
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Q11. traffic lights at three different road Crossings change after 48 second,72 second and 108 second respectively. advocate time they will change together again if they change simultaneously at 7 a.m.?
=>Given that traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds respectively.
So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds
48 = 2 × 2 × 2 × 2 × 3
72 = 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
Hence LCM of 48, 72 and 108 is (2 × 2 × 2 × 2 × 3 × 3 × 3) = 432
So after 432 seconds they will change simultaneously
We know that 60 seconds = 1 minute
so on dividing 432 / 60 we get 7 as quotient and 12 as reminder
Hence, 432 seconds = 7 min 12 seconds
Therefore the time is = 7 a.m. + 7 minutes 12 seconds
Hence the lights change simultaneously at = 7:07:12 a.m
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Q12. if the product of two numbers is 4032 and their HCF is 12 find LCM?
=> product of two number = HCF*LCM
LCM= product of two number
HCF
LCM=4032 =336
12
LCM=336
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Q13. the HCF and LCM of two number are 9 and 270 respectively if one of the number is 45 find the other number
=>HCF*LCM=product of two numbers
othernumber=270*9=45
2430 =54
45
other number is a 54
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Q14. find the HCF of 180 and 336 hence find their LCM
=>336=180*1+156
180=156*1+24
156=24*6+12
24=12*2+0=12
HCF=12
product of two number=HCF*LCM
LCM=product of two number
HCF
LCM=180*336=5040
12
LCM=5040
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Q15. can two number have 15 as their HCF and 110 as their LCM give reason to justify your answer
=> the LCM of two numbers is perfected divisible by the HCF
110 = 7.333
15
as the LCM is not divisible by the HCF than two numbers cannot have 15 and 110 as their HCF and LCM respectively
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Answered by
3
Q2. find the LCM of the given numbers by division method
=>1.480,672=3360
2.6,8,45=360
3.24,40,84=840
4.20,36,63,77=13860
=============================
Q3. Find the least number which when increased by 15 is exactly divisible by 15,35,48
=> 1665 is the least number which when increased by 15 will be divisible by 15, 35 and 48.
=============================
Q4. Find the least number which increased by 16,15 and 18 leaves a remainder of 5 in each case
=>To find the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case, we have to find the L.C.M. of 6, 15 and 18 and then add 5 in that number. Hence, 95 is the least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case.
=============================
Q5. Find the least number which when increased by 24,46,45 and 54 leaves a remainder of 3 in each case
=>122
=============================
Q6. find the greatest 3 digit number which is a exactly divisible by 8,20 and 24
=>960
=============================
Q7. find the smallest 4 digit number which is exactly divisible by 32,36 and 48
=>1152
=============================
Q8. find the greatest 4 digit number which is exactly divisible by each of 8,12 and 20
=>9960
=============================
Q9. Find the least number of a five digit which is exactly divisible by 32,36 and 45
=>10080
=============================
Q10.three boys step off together from the same spot their steps measure 63 cm,76cm 77 cm respectively what is the minimum distance each should cover so that all can cover the same distance in complete step
=>For finding minimum distance, we have to find L.C.M of 63, 70, 77. L.C.M. of 63, 70 and 77 = 7 x 9 x 10 x 11 = 6930 cm. Therefore, the minimum distance is 6930 cm.
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Q11. traffic lights at three different Crossings change after 48 second, 72 second and 108 seconds respectively advocate them they will change together again if they change simultaneously 7.am
=>Given that traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds respectively.
So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds
48 = 2 × 2 × 2 × 2 × 3
72 = 2 × 2 × 2 × 3 × 3
108 = 2 × 2 × 3 × 3 × 3
Hence LCM of 48, 72 and 108 is (2 × 2 × 2 × 2 × 3 × 3 × 3) = 432
So after 432 seconds they will change simultaneously
We know that 60 seconds = 1 minute
so on dividing 432 / 60 we get 7 as quotient and 12 as reminder
Hence, 432 seconds = 7 min 12 seconds
Therefore the time is = 7 a.m. + 7 minutes 12 seconds
Hence the lights change simultaneously at = 7:07:12 a.m
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Q12. the product of two numbers is 4032 and their HCF is 12 find LCM
product of two number=HCF*LCM
LCM =
.
.
LCM=336
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Q13. The Other CF and LCM of two numbers are 9 added 270 respectively if one of the number is 45 find the other
=>HCF*LCM=product of 2 numbers
other number = 270*9=45
2430 = 54
45
other number is 54
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Q14. find HCF of 180 and 336 hence find LCM
=>HCF
336=180*1+156
180=156*1+24
156=24*6+12
=12
HCF=12
LCM
product of two number = HCF*LCM
LCM=
.
LCM=180*
.
LCM=5040
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Q15. can two numbers have 15 as their HCF and 110 as a LCM give reasons to justify your answer
=>As the LCM is not divisible by the HCF then two numbers can't have 15 and 110 as their HCF and LCM respectively
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