Q1.
Two numbers have 18 as their HCF and 480 as their LCM, is it true ?
Solution.
False
Note : LCM is exactly divisible by HCF .
Solution
Q2.
A circular field has a circumference of 180 km . Three cyclists start together and can travel 18,12 and 15 km a day , on circular
track. when they will meet again ?
Solution.
First find time to complete one round
time =
distance/speed
So, find the
LCM of
180/18
,
180/12
,
180/15
=
LCM of 10, 12, 15 = 60 days .
All will again after 60 days .
Solution
Q3.
Which smallest number leaves remainder 10 and 14 when divided by 28 and 32.
Solution.
= LCM of 28 , 32 - 18
= 224 - 18
= 206
Solution
Q4.
A circular field has a circumference of 210 km . Three cyclists start together and can travel 30 km,42 km and 70 km an hour, on circular
track. when they will meet again ?
Solution.
Find the LCM of
210/30
,
210/42
,
210/70
find the LCM of 7, 5 , 3
= 105 hrs
they will after 105 hrs.
Or
210/HCF of 30, 70, 42
=
210/2
= 105 hrs
Solution
Q5.
Find the number, which divided by 35, 56 and 91 leaves remainder 11 in each case .
Solution.
Required number = LCM of 35, 56 & 91 + 11
= 3640 + 11
= 3651
Solution
Q6.
Three persons step off together , their steps cover 60 cm, 55 cm and 58 cm respectively . What is the minimum distance , so that
each can cover the distance in complete steps ?
Solution.
Required distance is LCM of 60, 55 & 58
= 19140 cm
= 191.4 M
Solution
Q7.
Find the smallest number which when increased by 16 is exactly divisible by 55 and 58 .
Solution.
Required number = LCM of 55 & 58 - 16
= 3190 - 16
= 3174
Solution
Q8.
The LCM and HCF of two rational numbers are equal , then numbers are :
(a) equal (b) Co-prime (c) prime (d) composite
Solution.
(a) equal
Solution
Q9.
Find the HCF and LCM of 23 and P, when P is a prime number .
Solution.
HCF = 1
LCM = 23P
Note :If P and q are two prime numbers then their HCF is 1 and LCM is Pq.
Solution
Q10.
Find the least number which is divided by 2,3,4,5,6 leaves 2 as remainder in each case(except 2) but it is exactly divisible by 7 and 2 .
Solution.
The LCM of 2,3,4,5,6 is 60
So the desired number is 60k + 2 (k is any positive integer)
= (7 x 8 + 4)k + 2 = (7 x 8)k + (4k + 2)
Hence , choose the least value of k for which 4k + 2 is divisible by 7.
(that is k = 3)
∴ The desired number = 182
Solution
