Six bells commence tolling together and toll at intervals of 2,4,6,8,10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 – 399) = 9600.
Workspace
Find the least number which when divided by 12, 15, 20 and 54 leaves 8 as remainder in each case.
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
∴ Required number (9999 – 399) = 9600.
Workspace
If HCF of p and q is x and q = xy, then the LCM of p and q is = ?
There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and of the last three is 12673. Find the fourth number?
Let the four prime numbers, in the ascending order of magnitude are a, b, c and d.
Given that the product of the first three is 7429 and of the last three is 12673
So a×b×c = 7429
b×c×d = 12673
Divide them:
Thus the first number is 17
The fourth number is 29
The middle two numbers are 19 and 23.
So your answer is D. 29
Workspace
John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?
L.C.M. of 250, 300 and 150 = 1500 sec Dividing 1500 by 60 we get 25, which mean 25 minutes. John, Smith and Kate meet after
25 minutes.
Workspace
The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.
Workspace
A man was employed on the promise that he will be paid the highest wages per day. The contract money to be paid was Rs. 1189. Finally he was paid only Rs. 1073. For how many days did he actually work?
On dividing a number by 4, the remainder is 2. The quotient so obtained when divided by 5, leaves the remainder 3. Now the quotient so obtained when divided by 6, leaves the remainder 5. The last quotient is 7. The number was
The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.
Workspace
Three different containers contain 1365 litres, 1560 litres and 1755 litres of mixtures of milk and water respectively. What biggest measure that can measure all the different quantities exactly ?
containers contain mixtures of milk and water as 1365 lit, 1560 lit and 1755 lit.
Biggest measure that can measure all these different quantities exactly can be given by
HCF of (1365, 1560, 1755)
That can be found as
HCF of 1365, 1755 = 195
Now, HCF of 195 and 1560 = 195
Hence, biggest measure that can measure all the given different quantities exactly is 195 lit
Workspace
Given : Three numbers 17, 42 and 93
Find the largest number to divide all the three numbers leaving the remainders 4, 3, and 15 respectively at the end?
Here greatest number that can divide means the HCF
Remainders are different so simply subtract remainders from numbers
17 – 4 = 13; 42 – 3 = 39; 93 – 15 = 78
Now let’s find HCF of 13, 39 and 78
By direct observation we can see that all numbers are divisible by 13.
∴ HCF = 13 = required greatest number
Workspace
A milkman has 75 litres milk in one cane and 45 litres in another. The maximum capacity of container which can measure milk of either container exact number ?
Let the numbers be 13x and 13y (∵ HCF of the numbers = 13)
13x×13y=2028⇒xy=12 co-primes with product 12 are (1, 12) and (3, 4) (∵ we need to take only co-primes with product 12. If we take two numbers with product 12,
but not co-prime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers. Hence numbers are 39 and 52.
Workspace
Find the least number which when divided by 2, 3, 4 and 5 leaves a remainder 3. But when divided by 9 leaves no remainder?
Required number is of the form 840k + 3