Aptitude Hack#103(6-6-19)

 Question:

If the ratio of the areas of two squares is 1:4, the ratio of their perimeters is

A. 1:2
B. 1:4
C. 1:6
D. 1:8

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Aptitude Hack#85(19-5-19)

Question:

A crate of mangoes contains one bruised mango for every thirty mangoes in the crate. If three out of every four bruised mango are considerably unsaleable and there are 12 unsaleable mangoes in the crate then how many mangoes is there in the crate?

   A )   450
   B )   480
   C )   500
   D )   490

Answer: B) 480

Explanation: 

Let the total no of mangoes in the crate be x

Then the no of bruised mango = 1/30 * x

Let the no of unsalable mangoes =3/4 *(1/30 x)

1/40 *x =12

x=480

Hence 480 mangoes are there in the crate.

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Aptitude Hack#84(18-5-19)

Question:

 In a caravan, in addition to 50 hens, there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers?

  A )   17

  B )   16

  C )   14

  D )   15

Answer : D) 15

Explaination: 

Let the number of keepers be x then,

Total number of heads 

=(50 + 45 + 8 + x)= (103 + x).

Total number of feet 

= (45 + 8) *4 + (50 + x) * 2

=(312 +2x).

(312 + 2x)-(103 + x) =224

 x =15.

Hence, the number of keepers =15.

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Aptitude Hack#82(16-5-19)

Question:

Find the unit’s digit in (264)^102 + (264)^103

   A )   0
   B )   1
   C )   2
   D )   3

Answer: A) 0

Explaination :

 Required unit’s digit = unit’s digit in
 (4)^102 + (4)^103.

now, 4^2  gives unit digit 6.

(4)^102 gives unit digit 6.

(4)^103 gives unit digit of the product 
(6 x 4) i.e., 4.

Hence, unit’s digit in (264)^102 + (264)^103
 = unit’s digit in (6 + 4) = 0

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Aptitude Hack#81(15-5-19)

Question:

From a group of boys and girls,15 girls leave. There are then left 2 boys for each girl. after this,45 boys leave.there are then 5 girls for each boy. Find the number of girls in the beginning?

  A )   50
  B )   51
  C )   41
  D )   40

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Aptitude Hack#81(15-5-19)

Question:

From a group of boys and girls,15 girls leave. There are then left 2 boys for each girl. after this,45 boys leave.there are then 5 girls for each boy. Find the number of girls in the beginning?

  A )   50
  B )   51
  C )   41
  D )   40

Answer : D) 40

Explanation:

 At present, there be x boys.

Then, no of girls at present=5x

Before the boys had left: no of boys=x+45

And no of girls=5x

x+45=2*5x

9x=45

x=5

No of girls in the beginning=25+15=40

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Aptitude Hack#68(2-5-19)

Question:

If 137 + 276 = 435 how much is 731 + 672?
A) 534
B) 1403
C) 1623
D) 1513

Answer: C)1623.

Explanation: 

If we try adding the numbers directly (i.e. considering decimal)

    1 
   137
 +276 
  ------  
       3
Since 7+6=13 and the given answer 435 has 5 in the one's place, therefore, the numbers are in base 8;
 Because 13 will correspond to 15 (7+8) in base 8 system.

 Now converting 731 and 672 in decimal
731=> 7x8^2 + 3x8^1 + 1x8^0 = 473 
672=>  6x8^2 + 7x8^1 + 2x8^0 = 442 

Adding 473 + 442 = 915 
Converting 915 into base 8 = 1623.

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Aptitude Hack#66(30-4-19)

Question:
A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo? 

A) 40

B) 30

C) 50

D) 60

E) None of These


 Answer: C) 50

Explanation:

Given , Zoo had either pigeons or horses
Heads of the animals in Zoo = 80
=> pigeons + horses = 80
Let p = number of pigeons
h = number of horses
=> p = 80 - h
Given , Legs of the animals = 260
Each pigeon has 2 legs and each horse has 4 legs
=> 2p + 4h = 260
Substitute p = 80 - h
=> 2 ( 80 - h ) + 4h = 260
=> 160 - 2h + 4h = 260
=> 2h = 260 - 160
=> 2h = 100
h = 50
So , number of horses in the Zoo = 50

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Aptitude Hack#60(24-4-19)

Question:

What is the sum of two consecutive odd numbers, the difference of whose squares is 56?

A) 30

B) 28

C) 34

D) 32


Answer: B) 28 

Explanation:

Let the no. be x and (x +2).
Then (x +2)^2 – x^2 = 56
4x + 4 = 56
x + 1 = 14
x = 13
Sum of numbers = x + (x +2) = 28


Alternate solution:-

Let two numbers be x and y.
x^2 + y^2 =56
(x-y)(x+y)=56
//x-y would be 2 as the numbers odd and they are consecutive.
//we have asked sum of numbers i.e x+y
(x+y)=56/2=28.

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Aptitude Hack#59(23-4-19)

Question:

 A) 1.45

  B) 1.88

  C) 2.9

  D) 3.7

Answer: C) 2.9

Explanation:

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Aptitude Hack#50(14-3-19)

Question:
Which number does not belong in the series below? 
2, 5, 10, 17, 26, 37, 50, 64
A) 17
B) 37
C) 64
D) 26

Answer D) 64

 Explanation: 
Given series is 2,5,10,17,26,37,50,64.

Here the total number of terms are 8..i.e. A(1) to A(8).

For the 1st 7 terms, you can observe the relationship  between the terms which is 
A(n) = [ A(n-1) - {A(n-1) + A(n-2)} + 2 ] 
where n >= 3

Now for the 8th term i.e. A(8), this relationship violates.
According to the relationship,
A(8) = [A(7) + {A(7) - A(6)} + 2 ]
= [50 + {50 - 37} + 2 ]
= 50 + 13 + 2
= 65.
But in the series, it is 64. 
Hence 64 doesn't belong to the series. 

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Aptitude Hack#46(10-4-19)

Question:

Find out the wrong number in the given sequence of numbers.
36, 54, 18, 27, 9, 18.5, 4.5

A) 4.5

B) 18.5
  
C) 54

D) 18

 Answer: B) 18.5
Explanation:
The terms are alternatively multiplied by 1.5 and divided by 3. However, 18.5 does not satisfy it.

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Aptitude Hack#45(9-4-19)

Question:
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: 
    A) 544
    B) 548
    C) 504
    D) 536

Answer B) 548

Explanation:

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

= 540 + 8

= 548.

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Aptitude Hack#44(8-4-19)

Question:
What is the average of the first 21 multiples of 7?
A) 49
B) 77
C) 66
D) 57

Answer: B) 77

Explanation:

Required average  = 7(1+2+....+21) / 21
=> (7/21)X ( (21x22) / 2 )  (because sum of first 21 natural numbers => (n*n+1)/2 )
= 77

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Aptitude Hack#42(6-4-19)

Question:

The least perfect square, which is divisible by each of 21, 36 and 66 is:

A) 213444  

B) 214344

C) 214434

D) 231444

Answer: A) 213444

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 22 x 32 x 72 x 112 = 213444

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Divisibility Rules

Divisibility by 2

A number which ends with even number or 0 is always divisible by 2.
For Example 44, 120, 56, 70 etc are divisible by 2.
23, 57, 79 etc are not divisible by 2.

Divisibility by 3
For a given number, if sum of its digits is divisible by 3, then the number will be divisible by 3.
For Example: 5676 = 5 + 6 + 7 + 6 = 24, which is divisible by 3
8912 = 8 + 9 + 1 + 2 = 20, not divisible by 3




Divisibility by 4
Given number will be divisible by 4, if the last two digits are divisible by 4 and if a number with
two or more zeros at the end are also divisible by 4
For Example 4500, 134000 are divisible by 4
5228 = last two digits i.e. 28 is divisible by 4, therefore the number is divisible by 4  


Divisibility by 5  
Numbers ends with 5 or 0, will be divisible by 5.
For Example 465, 670, 565 will be divisible by 5.
556, 877 are not divisible by 5.


Divisibility by 6
A number divisible by 3 and 2 both will be divisible by 6. For divisibility by 3, you will use
the same rule of 3 as discussed above.
For Example 154: Divisible by 2.
1 + 5 + 4 = 10 which is not divisible by 3? Therefore, the number is not divisible by 6.
3924: Divisible by 2. Also, 3 + 9 + 2 + 4 = 18, which is divisible by 3? Therefore, the number divisible by 6




Divisibility by 7
To check divisibility by numbers like 7, 13, 17, 19, we always use one key digit known as
Osculator. Osculators for 7 is -2 i.e. negative osculator 2. Let’s see how we will use
osculator to check divisibility.
Suppose, we need to check divisibility of number 119,

The last digit of a number is to be multiplied with -2 (osculator) and then added to the remaining
part of the number. Check whether the resultant number is divisible by 7 or not.
Therefore, the number will be divisible by 7, if resultant is divisible by 7
  

Divisibility by 8
Given number will be divisible by 8 if the last three digits are divisible by 8 and number with
three zeros at the end are also divisible by 8
For Example 4589000, 1256 are divisible by 8



Divisibility by 9
For a given number, if sum of its digits is divisible by 9, then the number will be divisible by 9.
For Example: 5679 = 5 + 6 + 7 + 9 = 27, which is divisible by 9
8912 = 8 + 9 + 1 + 2 = 20, not divisible by 9


Divisibility by 10
A number ends with 0 is always divisible by 10
For Example 100, 2890, 4560 are divisible by 10



Divisibility by 11
If sums of digits at odd places and even places are equal or differ by a number divisible by
11, then the number will be divisible by 11
For Example 3245693
Sum of odd digits = 3 + 4 + 6 + 3 = 16
Sum of even digits = 2 + 5 + 9 = 16
Sums are equal; therefore give number is divisible by 11



Divisibility by 12
Number divisible by 4 and 3 both will be divisible by 12. For divisibility by 4 and 3, you
will use the same rule of 4 and 3 as discussed above. 
For Example: 156: 1 + 5 + 6 = 12 which is divisible by 3?
Last two digits 56 are divisible by 4. Therefore, the number is divisible by 12
3924: Divisible by 4. Also, 3 + 9 + 2 + 4 = 18, which is divisible by 3? Therefore, the number
divisible by 12.

Divisibility by 13

Osculator for 13 = 4 The method is the same as 7 as discussed above.

See how we check the divisibility by 13, the last digit is multiplied with osculator (4), then
added to the remaining number.


Divisibility by 14
A number divisible by 7 and 2 both will be divisible by 14. To check divisibility by 7 and 2, you will use the same rules as discussed above.



Divisibility by 15
A number divisible by 5 and 3 both will be divisible by 15. Divisibility by 5 and 3 will be the same as discussed above.


Divisibility by 16
A number divisible by 8 and 2 both will be divisible by 16. Divisibility by 8 and 2 will be the same as discussed above.


Divisibility by 17
Osculator for 17 is -5 (negative osculator). Method is the same as discussed in divisibility by 7 and 13.


Divisibility by 18
Number divisible by 9 and 2, will be divisible by 18.


Divisibility by 19
Osculator for 19 is 2 (negative osculator). Method is similar as discussed in divisibility  by 7 and 13.




So, Remember that
Osculator of 7 =- 2
Osculator of 13 = + 4
Osculator of 17 =- 5
Osculator of 19 = + 2  

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Aptitude Hack#35(30-3-19)

Question:
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many
distinct 4 digit numbers greater than 3000 can
be formed?

A) 50
B) 51
C) 52
D) 54

Answer B) 51

Explanation:

As the number is greater than 3000.

So Thousand's place can have 3 or 4.

Let's consider the following two cases

Case (I) when a thousand's place is 3.

      3 a b c

 If there is no restriction on the number of two's,

Three’s and four's. Then each of a, b, c can be

Filled with 2 or 3 or 4 each in 3 ways.

So 3 x 3 x 3 = 27 numbers are there. Out of

Which 3222, 3333 are invalid as 2 can be used

Twice & 3 can be used thrice only so the number of such valid

 numbers beginning with 3 are 27 - 2 = 25.

Case (Il) When the thousand's place is 4

    4 a b c

Without restriction on the number of 2's, 3's and 4's

a, b, c (as explained in case I) can be filled in 27 ways.

Out of these 27 numbers, 4 2 2 2 is only invalid

as two have to be used twice only.

So valid numbers are 27 - 1 = 26.

Total numbers from Case (I) & Case (Il)  

=>25 + 26 = 51.

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Aptitude Hack#16(11-3-19)

Question:

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

A. 12     

B. 16

C. 24

D. 48

 Answer:  D) 48

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Hence The Correct Option is D

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Aptitude Hack#15(10-3-19)

Question:
Find the sum of the expression

A)7
B)8
C)9
D)10

Answer: A) 8


Explanation:
The Series can be re-written as (√2-√1)/(√2+√1)(√2-√1) + (√3-√2)/(√2+√2)(√3-√2) + .......... which simplifies to (√2-√1) + (√3-√2) + ..... (√81-√80) which again simplifies to √81 - √1 which is 8

Hence the Correct Option is A

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Aptitude Hack#14(9-3-19)

Question:
103 x 103 + 97 x 97 = ?
A)  21348
B) 20018
C) 19648
D)21428

Answer: B) 20018

Explanation:

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