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Day-13 | Bank Aptitude Questions and Answers with Explanation

  • Question-1
  • Explanation

If the profit is 40% and Rs. 500 is the difference between c.p and s.p then c.p is:

Options :

a)Rs. 1750
b)Rs.1250
c)Rs.1500
d)Rs.1800

Answer:      b) Rs. 1250

Let c.p be Rs. X then s.p=Rs. 140% of x=Rs. 14X/10
Now, 14X/10-X=Rs.500
4X=Rs. 5000
X=Rs. 1250
s.p=Rs. 14*1250/10=Rs.1750
Hence the c.p is Rs. 1250


  • Question-2
  • Explanation

In a room, there are some peacocks and some cats. If the total number of eyes in the room is 1246 and the total number of legs is 1406. What is the number of cats in the room?

Options :

a)65
b)80
c)54
d)70

Answer:      b) 80

Let X,Y be the number of peacocks and cats respectively, from observing the given data, the number of heads in the room will be 1246/2=623 (since each has two eyes)
i.e X+Y=623…..eqn1
peacock has two legs and cat has 4 legs,
2X+4Y=1406…eqn2.
Solving eqn 1 and eqn2,
We get Y =80 and X=543
Hence the answer is 80.


  • Question-3
  • Explanation

In a forest, there are some monkeys, tigers and parrots. If the total number of heads in the forest is 63 and total number of legs is 166 then find the number of monkeys, if the birds are twice same as the animals.

Options :

a)25
b)22
c)26
d)21

Answer:       b) 22

Let X,Y,Z be the number of monkeys, tigers and parrots respectively.
Given that total number of heads is 63, X+Y+z=63…… eq1
Also total number of legs is 166, 2X+4Y+2Z=166…… eq2
Birds are twice same as animals,
i.e, 2Z=X+Y=>x+Y-2Z=0……eq3
subtracting eq 1 from eq3 we get,
3Z=63=>z=21
Solving eq1 and eq2 we get,
Y=20
Adding eq2 and eq3 we get,
3X+5Y=166=>3X+5×20=166=>X=22
Therefore X=22,Y=20 and Z=21
Hence the answer is 22.

 

  • Question-4
  • Explanation

There are some birds, animals and insects in a forest. 4:5 1:3 are the ratios of birds to animals and animals to insects respectively. If the difference between insects and animals is 250 then find the total number of birds and insects.

Options :

a)475
b)200
c)375
d)100

Answer:       a) 475

Part: 1
As given that the ratio of animals to insects is 1:3
Then X,3X are the number of animals and insects.
Now, 3X-1X=250(given)
2X=250
X=125.
i.e…, them number of animals is 125.
And the number ofinsects is =3×125=375.
Part: 2
Also given that 4:5 is the ratio of birds and animals then 5X=125.
=>X=25
Therefore the number of birds is 4X=4*25=100
Now the total of birds and insects is =100+375=475.
Hence the answer is 475.

 

  • Question-5
  • Explanation

Find the number of ways to arrange the letters in the word “LETTER”.

Options :

a)20
b)360
c)180
d)5040

Answer:       c) 180

The word “LETTER” contains 6 letters, namely 1-L, 2-E
2-T and 1-R.
Required number of ways=6!/(1!)*(2!)*(1!)=720/4=180.

 

  • Question-6
  • Explanation

In how many different ways can the letters of the word. “SYSTEMATIC” be arranged so that the vowels always come together?

Options :

a)10080
b)60480
c)362880
d)none of these

Answer:      b) 60480

in the word “SYSTEMATIC”,consider the vowels EAI as one letter.
Thus, we have SYSTAMATIC(EAI).
The word has 8(7+1) letters of which s&T occurs 2 times and the rest are 1 time.
Number of ways of arranging these letters=8!/(2!X2!)=40320/4=10080.
Now, 3 vowels in which each are different, can be arranged in 3!=6ways.
Required number of ways=(10080*6)=60480.

 

  • Question-7
  • Explanation

How many 3-letter words can be formed with or without meaning from the letters M,Y,aS,I and N,which are ending with M and none of the letters should be repeated?.

Options :

a)20
b)18
c)25
d)27

Answer:       a)20

Since each desired word is ending with M, the least place is occupied with M. So,there is only 1 way.
The second place can now be filled by any of the remaining 5 leeters(Y,A,S,I,N). so , there are 5 ways of filling that place.
Then , the firs place can now be filled by any of the remaining 4 letters. So , there are 4 ways to fill.
Required number of words=(1*5*4)=20.


  • Question-8
  • Explanation

Find the number of different ways in which word OPTIMIZE can be rearranged?

Options :

a)20155
b)20160
c)20165
d)201660

Answer:      b) 20160

Number of rearrangements=n!(r!*r2!…..)
In the above formula, n represents the number of letters in the given word. R1. R2. R3. Represent number of repeated words.
In OPTIMIZE there are 8 words. Word is repeated twice.
Therefore, number of rearrangements=8!2!=20160


  • Question-9
  • Explanation

Find the total number of words that can be formed out of INDIA

Options :

a)360
b)175
c)183
d)188

Answer:       a) 360

Word “INDIA’ has 5 letters. ‘I’ is repeated twice and ‘T’ is repeated twice.
Therefore, number of rearrangements=6!/2!=360

 

  • Question-10
  • Explanation

Consider a word APPLE. Now, if ‘a’ is to be retained as first letter always, how many possible arrangements are possible.

Options :

a)7
b)8
c)9
d)12

Answer:       d)12

Since A has to be retained as first letter always. The number of possible arrangement out of PPLE will be our answer. In ‘PPLE’, letter ‘P’ occurs twice.
Therefore, number of words that can be formed out of ‘APPLE’ by retaining A as first letter=4!/2!=12

 

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