# ARITHMETIC IMPORTANT QUESTION FOR IBPS PRELIMINARY EXAM.

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

**Answers :**

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 clock@ 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.
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