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1) An alarm beeps every 30 sec & another device beeps at every 50 secs.If they beep together at 9:30 AM, what time they will beep together immediately after 9:30 AM? A) 9:35 B) 9:32:45 C) 9:32:30 D) 9:45
Answer & Explanation
Answer: 9:32:30 (L.C.M of 30 & 50 is 150.Every 2 min & 30 seconds they will beep together.
They will beep together again at 9:32:30 AM immediately after 9:30AM.)
2) The product of two numbers is 1350.If the H.C.F of the two numbers is 15, what is the sum of two numbers?A) 75 B) 70 C) 65 D) 60
Answer & Explanation
Answer: 75 (Product of two numbers = Product of their L.C.M & H.C.F.
H.C.F is 15 & L.C.M is 90. 30 & 45 are the numbers for which the L.C.M & H.C.F are 90 & 15 respectively.
The sum is 75
)
3) Five bells commence tolling together & tolls at intervals of 2,4,6,8,10 seconds respectively. In 1 hour, how many times they will toll together?A) 15 B) 16 C) 30 D) 31
Answer & Explanation
Answer: 31 (L.C.M of 2,4,6,8 & 10 is 120.The bells will toll together every 120 seconds.
They will toll together (60/2) +1= 31 times in 1 hour)
4) A chemist has 3 different liquids 68 litres, 136 litres & 170 litres respectively. Find the least number of casks of equal size required to store all the liquids without mixing them. The size of each cask needs to be an integer.A) 12 B) 15 C) 11 D) 10
Answer & Explanation
Answer: 11 (We need to fill three different types of liquids without mixing in casks of equal size.
Therefore the volume of each cask (say v litres) will be the factor of 68,136 & 170.Since we need to find the least no of casks, we need to take the highest factor to find the max no of litres the cask can hold.
H.C.F of 68,136,170 is 34.
No of casks required = (68/34 + 136/34 + 170/34) = 2+4+5 = 11)
5) The L.C.M of three different numbers is 240, which of the following cannot be their H.C.FA) 20 B) 24 C) 12 D) 25
Answer & Explanation
Answer: 25 (Since H.C.F is always a factor of L.C.M.
25 is not the H.C.F as it is not a factor of 240.)
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