IBPS Placement Papers

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 1. (√5-√10)^2+ (√2 + 5)^2 = (?)^3 - 22

Asked In IBPS

ANS:

expand squares using (a+b)^2=a^2+b^2+2ab, (a-b)^2=a^2+b^2-2ab

5+10-2√5√10 + 2+25+2√2(5)= x^3-22
5+10-10√2+2+25+10√2=42=x^3-22.x^3=64.therefore x=4.ans=4

2. 4, 8, 24, 60, ? , 224

Asked In IBPS

ANS

124 is the answer
8-4=4 which is 2^2
24-8=16 which is 4^2
60-24=36 which is 6^2
now
x-60=64 which is 8^2
so x=60+64=124
now
224-124=100 which is 10^2
so the missing number is 124..

3. A is sister of B. C is brother of D. D is a sister of A. How is B related to D?
Options:
1)Brother
2)Brother or Sister
3)Sister
4)Data inadequate
5)None of these

Asked In IBPS

ANS

Answer (2)Brother or Sister

As the Gender of B is not confirmed.

4. If the letters of the word ‘PROTECTION’ which are at odd numbered
position in the English alphabet are picked up and are arranged in
alphabetical order from left and if they are now substituted by Z, Y, X
and so on, beginning from left which letter will get substituted by X

Asked In IBPS

ANS:

I as PROTECTION odd numbered positions are o,e,t,o re arranging from left to right in alphabetical order c,e,i,n,o,p,r,t substituted by z,y,x from left to right so c=z;;e=y;;i=x the answer is I

5. MADE=BLVD & REAL=PULS , LOAP=?

Options
DIUB
SNLH
VEDF
AKNR
UGEM

Asked In IBPS

ANS:

the answer is SNLH as 'A' is replaced by 'L' in both the example.

6. Which of the following will be the third digit of the fourth number
after the following numbers are arranged in descending order after
reversing the positions of the digits within each number ?
645, 869, 458, 347, 981, 792

Asked In IBPS

ANS:

Ans is 6.
reversing the positions of the digits within each number
it becomes 546,968,854,743,189,297
descending order is 968,854,743,546,297,189
asked fourth number, third digit fourth number is 546 and the third digit in this number is 6

7. All arrows are bows.
All bows are swords.
Some swords are daggers.
All daggers are knives.

Conclusions:

I) all knives are bows.
II) some swords are knives.
III) all bows are arrows.
IV) all arrows are swords.

A. Only II follows.
B. Only II and IV follow.
C. Only III and IV follow.
D. Only I and III follow.
E. None of these.

Also Explain the reason.

Asked In IBPS

ANS:

Answer is B.)Only II and IV follow.
reason: for (II)
Some swords are daggers.--------------(I type)
All daggers are knives.---------------(A type)

I type statement + A type statement = I type statement.

so conclusion:some swords are knives.-----(I type statement)

similarly for(IV)...
All arrows are bows.-----------(A type)
All bows are swords.-----------(A type)

A type statement + A type statement = A type statement.
so Conclusion:all arrows are swords.-----------(A type)

8. 7 8 24 ? 460 2425 find out the missing term

Asked In IBPS

ANS;

99

7*1+1^3=7+1=8
8*2+2^3=16+8=24
24*3+3^3=72+27=99
99*4+4^3=396+64=460
460*5+5^3=2300+125=2425

9. How many such pairs of letters are there in the word GOVERNMENT each of which has as many letters between them in the word(in both forward and backward directions) as in english alphabet?

Asked In IBPS

ANS:

Answer is 4

the question generally means - to identify the no of words
in which starting letter and the ending letter must have
the same position difference as in English alphabet

here GOVERNMENT has 4 pairs

G(OVERN)M = G(HIJKL)M- G and M have 4 alphabets in between.

O(VE)R = O(PQ)R - O and R have 2 alphabets in between.
Reverse, MN & N
Number of such pairs are GM, OR, N& MN.
Answer is 4

10. In a certain code GRANT is written as UOBSH and PRIDE is written as FEJSQ.How is SOLD written in that code?

Asked In IBPS

ANS:

Answer: EMPT

reason:
Coded by next alphabet i.e
after G - H
R - S
A - B
N - O
T - U
and coded alphabet is written as reverse order:UOBSH
Similarly...
coded alphabet for PRIDE is QSJEF
coded alphabet is written as reverse order:FEJSQ

similarly..
SOLD is written as EMPT

11. a sum of money invested at certain rate of interest compounded annually becomes thrice in 2 years. in how much time will the interest become 26 times that of principal ?

Asked In IBPS

ANS

3p=p(1+(r/100))^2 ==> r=73.2

26p=p(1+(73.2/100))^t-p ==> t=3years

12. Vikas bought 100 kg of wheat at 16 per kg and 300 kg at the rate of 14 per kg. he sold the mixture to get 24% profit.At what price per kg he sold the mixture ?
a)16
b)18
c)16.42
d)16.24
e)none of these

Asked In IBPS

ANS:

100 kg * 16 =Rs.1600
300 kg * 14 =Rs.4200
--------------------
400 kg cp=5800

24% profit means (24/100)5800 =1392
Total = 5800+ 1392 =7192
400 kg=7192
1 kg = 7192/400 =17.98(18)

ans is B

13. 82 51 28 20 ?

Asked In IBPS

ANS:

Ans is 5

82 51 28 20 5
82-51=31 & 51-28=23 & 28-20=8
so 31-23=8 then the next digit be 23-8=15
20-15=5

(OR)

9 bcoz 2^2+5=9

i think 9,7, odd then 5 and 4, 2 even

14. a shopkeeper sold in air-conditioner for rs 25935 at a discount of 9% and earned a profit of 3.14%. what would have been the %age of profit earned if no discount had been offered.

Asked In IBPS

ANS:

13.34%

(100-9)% of SP =(100+3.14)% of CP
91% of SP=103.14% of CP
SP=100*(103.14/91)% of CP= 113.34% CP
So With no discount profit earned=113.34-100= 13.34%

15. in a glass of milk, the portion of pure milk and water is 3:1, how much of the mixture must be withdrawn and substituted by water so that the resulting mixture become half pure milk and half pure water

Asked In IBPS

ANS:

1/3 of mixture.

Let there be y L of milk
===> 3/4*y is pure milk
1/4*y is water

Let x be the portion of mixture withdrawn
Now we have (y-x)L of mixture
pure milk = 3/4*(y-x)
water = 1/4*(y-x)

The portion withdrawn is replaced by water (ie x Liters of water is added)
Quantity of water = 1/4*(y-x)+ x

Since the resulting solution is half pure water, quantity of water is y/2
ie,
1/4*(y-x)+ x = y/2
3/4*x = 1/4*y
3x = y
x = 1/3*y
===> 1/3 rd of the mixture must be withdrawn and replaced by water.

16. Question. 7365+(5.4)2+.J? =7437.16
1) 1894
2) 1681
3) 1764
4) 2025
5) None of these

Asked In IBPS

ANS:

ans: 5. none of these.

solving the equation, j=61.36

17. A word and number arrangement machine when given an input line of words and number rearrange them followings a particular rule in each step. The following is an illustration of input and rearrangement.
Input : world 32 73 verb 26 new desk 19
Step I : 73 world 32 verb 26 new desk 19
Step II : 73 desk world 32 verb 26 new 19
Step III : 73 desk 32 world verb 26 new 19
Step IV : 73 desk 32 new world verb 26 19
Step V : 73 desk 32 new 26 world verb 19
Step VI : 73 desk 32 new 26 verb world19
Step VII : 73 desk 32 new 26 verb 19 world
and Step VII is the last step of the above input.
As per the rules followed in the above steps, find out in each of the following questions the appropriate step for the given input.
31. Step II of an input is: 51 brown 22 36 49 cloud sky red. How many more steps will be required to complete the rearrangements?
Option
(a) Three
(b) Six
(c) Five
(d) Four
(e) None of these

Asked In IBPS

ANS:

@snehalatha
Observing the 1st Input (world 32 73 verb 26 new desk 19) and its output(Step VII : 73 desk 32 new 26 verb 19 world) we can conclude that the machine re-writes the pattern in a from decending order of numbers and ascending order of words(alphabetical) alternatively.
The input was
world 32 73 verb 26 new desk 19

In 1st step the largest number is taken from the series and placed at 1st. So the output of step 1 is
Step I : 73 world 32 verb 26 new desk 19

In the 2nd step the word which comes alphabetically 1st is taken and placed at 2nd position.
Step II : 73 desk world 32 verb 26 new 19

Again in the 3rd step the 2nd largset number is taken and placed at the 3rd position.
Step III : 73 desk 32 world verb 26 new 19

and so on.. until every number is arranged in decending order and words in ascending order one at a time in each step.

Similarly for the next input
Step II of an input is: 51 brown 22 36 49 cloud sky red.

In step 3 - 49 is placed at 3rd positon and the remaining is written as it is.
Step III : 51 brown 49 22 36 cloud sky red

In step 4 - cloud is placed at 4th position.
Step IV : 51 brown 49 cloud 22 36 sky red

In 5th step 36 is placed in 5th position.
Step V : 51 brown 49 cloud 36 22 sky red

In 6th step red is placed in 6th position.
Step VI : 51 brown 49 cloud 38 red 22 sky

Now 22 and sky are in correct positions.So no more step is required.The output is obtained in the 6th step. So 4 more steps are required.

18. Karnataka Express leaves Bangalore at 5 p.m. and reaches Dharmavaram at 9 p.m. Kurla Express leaves Dharmavaram at 7 p.m. and reaches Bangalore at 10.30 p.m. What time do the trains cross one another?
Option
a) 8.26 p.m.
b) 8 p.m
c) 7.36 p.m.
d) 7.56 p.m
e) None of these

Asked In IBPS

ANS:

@SNEHALATHA
yes the above answer is correct.

Let the total distance be x.
Time taken by 1st train = 9-5 = 4 hours
Time taken by 2nd train = 10.30 - 7 = 3.30 hours = 7/2 hours
Speed of 1st train = distance/time = x/4
Speed of 2nd train = 2x/7

To find the meeting time of 2 trains,u have to make the starting time of both trains equal.
Here the starting time of 1st train is 5pm and 2nd train is 7pm.
So calculate the position of 1st train at 7pm.

Distance travelled by 1st train in 2 hours (5pm to 7pm)
= speed*time
= x/4 * 2
=x/2

So by 7pm the 1st train has already covered a distance of x/2, which implies the distance between the 2 trains at 7pm is remaining x/2 distance(since total distance is x).

Now since both trains start at same time moving towards each other, the time travelled by both trains before meeting is same but the distance travelled is different(since they travel at different speed)
Let time taken by both trains be 't',

So distance traveled by 1st train = x/4*t
distance travelled by 2nd train = 2x/7*t

When they meet , the total distance travelled by both is x/2

x/4*t + 2x/7*t = x/2
15/28*t = 1/2
t = 14/15

't' is the time travelled by both trains before meeting after 7pm.
Time of meeting = 7pm + 14/15
= 7 hours 14/15*60minutes
=7 hours 56 minutes
7:56 pm

19. A farmer sold two of his cows for 210/-. He sold one cow at a profit of 10% and other for a loss of 10%. Totally he gained 5% on selling both the cows. What is the original cost of each cow?
Option
a) 130 and 70
b) 150 and 50
c) 120 and 80
d) 115 and 85

Asked In IBPS

ANS:

let cp of one cow is a and cp of other is b
a+b=200
sp of one cow =a*90/100....(1)
sp of another cow=a*110/100....(2)
(1)+(2)=210
solving this we get a=50 b=150

20. Q. Last digit of given expression
(36572)^145 * (30766)^132 will be..

Asked In IBPS

ANS:

(36572)^145 * (30766)^132
last digit of expression will be the last digit of (2)^145 * (6)^132

last digit of 2^145 = (2^4)^36 *2 = (16)^36 *2 = 6*2 = 12 => 2

last digit of 6^132 = 6
so, Last digit of given expression = 2*6 =12 => 2

ans 2

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