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Wednesday, 1 July 2015

HCF and LCM Practice for IBPS

1. In a fire range, 4 shooters are firing at their respective targets. The first, the second, the third and
the fourth shooter hit the target once in every 5 s, 6 s, 7 s and 8 s, respectively. If all of them hit their
targets at 9:am, when will they hit their targets together again?
1) 9:04 am   2) 9:08 am  3) 9:14 am  4) None of these
Solution: (3).
Time after which they will hit the target again together = LCM ( 5,6,7 and 8) = 840 seconds
LCM of (5,6,7,8) = 840 seconds
They will hit target together = 840/60 = 14 min
They will hit together again at 9:14 am.

 

2. Five bells begin to toll together at intervals of 9s, 6s, 4s,10s and 8s, respectively. How many times
will they toll together in the span of 1 h (excluding the toll at the start)?
1) 5  2) 8  3) 10  4) Couldn’t be determined  5) None of these
Solution: (3)
The bells will toll together after time in seconds equal to the LCM of 9,6,4, 10 and 8.
LCM of 9, 6, 4, 10 and 8 is = 360 seconds
In one hour, the rings will toll together 3600/360 times = 10 times .

 

3. A General can draw up his soldiers in the rows of 10, 15 or 18 soldiers and he can also draw them
up in the form of a solid square. Find the least number of soldiers with the General.
1) 100  2) 3600  3) 900  4) 90  5) None of these
Solution: (3).
LCM of 10, 15 and 18 is
LCM = 2 x 3 x 5 x 3 = 90
To make it perfect square, we multiply it with 2 x 5 = 10
Therefore required number of soldiers
90 x 10 = 900

4. Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm
and 12 m 95 cm.
1) 15 cm  2) 25 cm  3) 35cm  4) 42cm  5) 45cm
Solution: (3)
Required length = HCF of 7 m, 3 m 85 cm, 12 m 95 cm
= HCF of 700 cm, 385 cm, 1295 cm is 35 cm
Hence, required length is 35 cm

 

5. Find the side of the largest possible square slabs which can be paved on the floor of a room 2 m 50
cm long and 1 m 50 cm broad. Also find the number of such slabs to pave the floor.
1) 25,20  2) 30,15  3) 50,15  4) 55,10  5) None of these
Solution: (3).
HCF of 250 cm and 150cm is 50 cm
HCF = 50
Therefore Number of slabs = Area of floor / Area of slab
= (250 x 150) / (50 x 50) = 15
(Word Problems Based on Numbers)

Also read

IBPS PO CWE V Notification out

IBPS RRB CWE IV Notification released

IBPS PO Preliminary Practice Set Download Here

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