Home » Bank Aptitude Practice » Day-15 | Bank Aptitude Questions and Answers with Explanation

Day-15 | Bank Aptitude Questions and Answers with Explanation

  • Question-1
  • Explanation

-2,4,-8,16 ?

Options :

a)-32
b)32
c)-64
d)64

Answer:      a) -32

-2*2=4
4*-2=-8
-8*2=16
16*-2=-32
Every number is got by multiplying the previous number by -2. Hence -32 is the answer.


  • Question-2
  • Explanation

2,3,5,8,?,21

Options :

a)10
b)11
c)13
d)12

Answer:      c) 13

2+3=5
3+5=8
5+8+13
13+8=21
This is a Fibonacci series, where every number is obtained by adding the previous two numbers. Hence answer is 13.


  • Question-3
  • Explanation

Sum of seven consecutive odd numbers is 133. Find the sum of the first and third number in the series.

Options :

a)18
b)38
c)48
d)28

Answer:      b)38

Since difference between any two successive odd numbers is 2, we can assume 7 consecutive odd numbers to be X-6,X-4,X-2,X,X+2,X+6. ( In our assumed series, difeerence between any two numbers is 2.)
Sum of the numbers in the series =133=X-6+X-4+x-2+x+X+2+x+4+X+6=7X
Therefore, x=133/7=19
Sum of first term and last term=X-6+X+6=2X
Substituting X=19 ( which we found earlier), we can get the answer to be 2X 19=38.

 

  • Question-4
  • Explanation

Sum of six consecutive numbers that are divisible by 3 is 99. Find the sum of the first two numbers.

Options :

a)30
b)21
c)14
d)15

Answer:       b) 21

Difference between successive numbers divisible by 3 will be 3. Therefore, the series takes the form as below.
X-6,X-3,X,X+3,X+6,X+9
Sum of all terms in the series=99=x-6+X-3+X+X+3+x+6+X+9=6X+9
6X+9=99
Or 6X=90
Or X=15
Sum of firs two numbers=X-6+X-3=2X-9=2(15)-9=21

 

  • Question-5
  • Explanation

Sum of five consecutive numbers that are divisible by 5 is 275. Find the last number in the series.

Options :

a)65
b)60
c)75
d)70

Answer:       a) 65

Difference between two adjacent numbers divisible by 5 is 5.
Series takes the form X-10,X-5,X,X+5,X+10
Sum =275+X-10+X-5+X+X+5+X+10=5X
5X=275
Or X=55
Last number in the series =X+10=55+10=65

 

  • Question-6
  • Explanation

Sum of 7 consecutive natural numbers is 91. Find the last but one number.

Options :

a)15
b)19
c)21
d)25

Answer:      a) 15

Difference between two successive natural numbers is 1.
Therefore series takes the form, x-3,x-2,x-1,x,x+1,x+2,x+3
Sum=91=x-3+x-2+x-1+x+x+1+x2+x+3=7x
7x=91
Or x=13
Last but one number =x+2=13+2=15

 

  • Question-7
  • Explanation

Section a of class VIII had boys and girls in the ratio of 2:1 . The current number of boys in the class is 40. Few girl students where shifted from section B to Section A. This changed the ratio of boys and girls to 2:3 Find the number of girls who got shifted from section B.

Options :

a)30
b)40
c)10
d)5

Answer:       b) 40

Let the original number of boys be B and that of girls be g
Then B/G=2/1.
It is given that B =40.
Therefore, 40G=2, Or G=20
Let the number of girls who were shifted from Section B be X. It is given that the new ratio of boys. B to the increased number of girls i.e G+Xis 2/3.
Then B/(G+x)=2/3
Or 40/20+x=2/3
20+X=60
Or X=40


  • Question-8
  • Explanation

Number of players in sports namely badminton, tennis and table tennis where in the ratio of 2:3:4. Originally there were 16 badminton players. If 25% of the badminton players withdrew due to some reason, What is the new ratio of the players.

Options :

a)3:6:8
b)3:7:8
c)3:6:4
d)6:6:8

Answer:       a) 3:6:8

Let the number of players in Badminton, tennis and table tennis be P,Q and R respectively.
It is given that P:Q:R =2:3:4
From the above ratio, we can easily infer that p/Q=2/3…(1) and P/R=2/4=1/2….(2)
It is given that p=16
From equation 1, P/Q=2/3 or 16/Q=2/3 . Therefore Q=16*3/2=24.
Similarly from equation. 2, P/R=1/2 or 16/R=1/2.
Therefore, R=32.
Let the number of badminton players present after 25% withdrew be P1.
Then p1=75% of p or 75/100*16
Therefore p1=12
New ratio is p1:Q:R or 12:24:32 or 3:6:8


  • Question-9
  • Explanation

Let the ratio of water and milk be 1:4 in a filled can of capacity 100 litres. The mixture was added with more water so as to change the ratio of 3:8. Find the quantity of water present originally and the amount of water (in litres) added thereafter.

Options :

a)15
b)8
c)12
d)10

Answer:       d) 10

Let the original quantities of water and milk be w and M respectively.
It is given that W/M=1/4….(1)
Also it is given that , total capacity of the can=W+M+100….(2)
If a fraction exists of the from a/b=c/d. Then we can very well write a/a+b=c/c+d.
Applying similar logic to equation 1 we get, W/(W+M)=1/(1+4)=1/5
But we know W+M=100
Therefore, W/100=1/5 or W=20 litres.
Substituting W=20 litres in equation 2, we get
M=100-20=80 litres.
Let X quantity of water be added so that the ratio becomes. 3/8.
Therefore W+X/M=38…(3)
But we know w=20 and M=80
Therefore, equation 3 becomes, 20+X/80=3/8
Or 20+x=30
X=10 litres.
We have found W=20 litres and x=10 litres

 

  • Question-10
  • Explanation

Assume the ratio of monthly wages to team leaders and team members in a factory be 2:1. If salary of team leaders is hiked by 25 % and that of team members is hiked by 30%. What would be the new ratio.

Options :

a)25/13
b)13/25
c)25/26
d)26/25

Answer:       a) 25/13

Let monthly wages of team leaders and team members be L and M respectively.
It is given that L/M=2/1
Let the hiked salary ( by 25%) of team leaders be L1.
Therefore L1=125/100*L=1.25L
Let the hiked salary (by 30%) of team members be M1.
Therefore M1=130/100*M=1.3M.
New ratio L1/M1=1.25L/1.3M
But we already know L/M=2/1
Therefore. L1/M1=1/25/1.3*2/1=2.5/1.3=25/13

 

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