ARITHMETIC IMPORTANT QUESTION FOR IBPS PRELIMINARY EXAM.
The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
a.8,20 ,28
b.20 ,35 ,45
c.16,28,36
d.12 ,24 ,36
e.None of the above

A and B can do a job together in 7 days. A is 7/4 times as efficient as B. The same job can be done by A alone in :
a.28/3 Days
b.11 days
c.49/4 Days
d.49/3 days
e.None of the above

Moving 6/7 of its usual speed a train is 10 min late. Find its usual time to cover the journey.
a.       25 min
b.      15 min
c.       35 min
d.      60 min
e.      None of these

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
a.504
b.536
c.544
d.548
e.600

The average weight of X, Y and Z is 45 kg. If the average weight of X and Y be 40 kg and that of Y and Z be 43 kg, then the weight of Y is:
a.17 kg
b.20 kg
c.26 kg
d.31 kg
e.34 kg

Two taps A and B can fill a cistern in 15 and 30 minutes respectively. There is a third exhaust tap C at the bottom of tank. If all taps are opened at the same time, the cistern will be full in 25 minutes. In what time can exhaust tap C empty the cistern when full?
a.10 Min
b.12 Min
c.11 Min
d.8 Min
e.14 Min

Vijayawada is at a distance of 340 km from Hyderabad. A train starts from Hyderabad to Vijay Wada at 4 A.M. with a speed of 60 km/hr. Another train starts from Vijay Wada to Hyderabad at 5 A.M. with a speed of 80 km/hr. At what distance from Hyderabad will the two trains cross each othe r and also find the time when they cross each other?
a.6.30 AM
b.7.15 AM
c.7.00 AM
d7.30 AM
e.8.00 AM

A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C
a.Rs.300
b.Rs. 400
c.Rs. 500
d.Rs. 600
e.Rs. 450

A contractor undertakes to do a piece of work in 150 days, so he deployed 200 men. He found that only a quarter of work is done in 50 days. How many additional men must be deployed so that the work may be finished on time?
a.Rs.105
b.Rs. 110
c.Rs. 95
d.Rs. 100
e.Rs. 120

A rectangular field has dimensions 25 m by 15 m. two mutually perpendicular passage of 2 m width have been grown in the rest of the field.find the area under the grass.
a.       295 sq m.
b.      299 sq m.
c.       300 sq m.
d.      375 sq m.
e.      None of these

1.C
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
20x = 80
x = 4.
Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

2.B
(A's 1 day's work) : (B's 1 day's work) =7/4 :1 = 7:4
Let A's and B's 1 day's work be 7x and 4x respectively.
Then, 7x + 4x =1/7   x = 1/77
A's 1 day's work = 1 X7 /77 = 1/11
= 11 Days

3. D
New speed = 6/7 of usual speed
Now time taken = 7/6 of usual time
(7/6 of the usual time) – (usual time) = 10 min
==> 1/6 of the usual time = 10 min
==> usual time = 60 min.

4.D  Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8   = 548.

5.D
X + Y + Z = (45 x 3) = 135 .... (i)
X + Y = (40 x 2) = 80 .... (ii)
Y + Z = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: X + 2Y + Z = 166 .... (iv)
Subtracting (i) from (iv), we get : Y = 31.
Y's weight = 31 kg.

6.A
Here, X = 15, Y = 30 and Z = 25
C can empty the full tank in = (XYZ /YZ +ZX –XY) Minutes
= ( 15*30*25/ ( 30*25 +15*25 -15*30)
= 11250/1125 = 10 Minutes

7.C
s1 = 60, s2 = 80, T = time from 4 A.M. to 5 A.M. = 1 hr
Distance of meeting point from Hyderabad = S1 {( d +S2 t) / (S1 +S2)} km
= 60 {( 340 + 80X1) / (60+80)} = 180 km
Time of their meeting = {( d +S2 t) / (S1 +S2)}
{( 340 + 80X1) / (60+80)}
3 Hrs after 4 o [email protected] 7:00 AM

8.B
C's 1 day's work =
1/3−( 1/6+ 1/8) = (1 / 3 − 7 /24)
=1/ 24
=  A:B:C=1/ 6:1/8:1/ 24
=4:3:1
C′s Share=3200 * 1/8 =400

9.D
200men x 50Days = 1/4 of work ,
so Total work = 200*50*4.
Now Let extra Men required is x
Therefore, (200+x)*100(remaining Days) =3/4 of work (That is remaining work).
So , (200+x)*100 = 200*50*4*(3/4)
=> 200+x=300
=> x= 100

10. B
Here l=25 m ,b= 15 m , x= 2m
Then, area under the grass
= area of rect angular – area of passage
= lXb – x(l+b-x)
= 25X15-2(25+15-2)
= 375-76 = 299 sq m.

ARITHMETIC IMPORTANT QUESTION FOR IBPS PRELIMINARY EXAM. Reviewed by SSC IBPS on 16:40:00 Rating: 5