# Number System Questions and Answers updated daily – Aptitude

Number System Questions: Solved 444 Number System Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Number System Questions

41. A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and the divided it by 6, getting 112 as the answer. The correct answer should have been

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Correct Ans:122

Explanation:

Let the number be X.

When first added 12 to the number and then divide by 6, the answer is 112.

(X + 12)/6 = 112

(X + 12) = 672

X = 672 - 12

X = 660

Therefore, when number is first divided by 6 and added by 12 we get,

= (660/6) + 12

= 110 + 12

= 122.

When first added 12 to the number and then divide by 6, the answer is 112.

(X + 12)/6 = 112

(X + 12) = 672

X = 672 - 12

X = 660

Therefore, when number is first divided by 6 and added by 12 we get,

= (660/6) + 12

= 110 + 12

= 122.

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42. A number when divided by 136, leaves remainder 36. If the same number is divided by 17, the remainder will be

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Correct Ans:2

Explanation:

If the first divisor be a multiple of the second divisor, then required number = Remainder obtained by dividing the first remainder by the second divisor.

Since, 17 is a factor of 136.

Therefore, 36 divided by 17 is 2.

Since, 17 is a factor of 136.

Therefore, 36 divided by 17 is 2.

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43. The smallest number in the set of five consecutive even numbers is 1 more than the largest number in the set of six consecutive odd numbers. What is the difference between the square of the second largest even number and the square of the second smallest odd number, if the sum of the smallest odd number and second smallest even number is 75?

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Correct Ans:1215

Explanation:

Let us take five consecutive even numbers x, x+2, x+4, x+6, x+8 and six consecutive odd numbers y, y +4, y +6, y +8, y+10

Given,

X-y=llâ€”(l)

Simplify the above equation (I) and (2), we get

x=42

Required difference = (482-332)=1215

Given,

X-y=llâ€”(l)

Simplify the above equation (I) and (2), we get

x=42

Required difference = (482-332)=1215

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44. If 3 is added to the numerator while 1 is subtracted to the denominator ratio becomes 5 : 4 while if one is added to the denominator and 5 is subtracted from the numerator the ratio becomes 2 : 3. What is the original fraction?

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Correct Ans:17/17

Explanation:

Assume the fraction = a/b

Condition 1: 3 is added to the numerator while 1 is subtracted to the denominator ratio becomes 5 : 4

(a+3)/(b-1) = 5/4

4a - 5b = -17

Condition 2: 1 is added to the denominator and 5 is subtracted from the numerator the ratio becomes 2 : 3

(a+1)/(b-5) = 2/3

3a - 2b = 17

solving both the equation

a = b = 17

Fraction = 17/17

Condition 1: 3 is added to the numerator while 1 is subtracted to the denominator ratio becomes 5 : 4

(a+3)/(b-1) = 5/4

4a - 5b = -17

Condition 2: 1 is added to the denominator and 5 is subtracted from the numerator the ratio becomes 2 : 3

(a+1)/(b-5) = 2/3

3a - 2b = 17

solving both the equation

a = b = 17

Fraction = 17/17

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45. 1/10 of a rod is coloured red, 1/20 orange, 1/30 yellow, 1/40 green, 1/50 blue, 1/60 black and the rest is violet. If the length of the violet portion of the rod is 12.08 m, then the length of the rod is __________.

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Correct Ans:16 m

Explanation:

Assume the total parts as 1

Other colours + Violet = 1

1 - (1/10 + 1/20 + 1/30 + 1/40 + 1/50 + 1/60) = Violet

1 - (49/200) = Violet

151/200 parts is Violet

It is given that the length of the violet portion of the rod is 12.08 m

Let the length of the rod be L,

Therefore,

151 * L/200 = 12.08

L = 200 * 12.08/151

= 16 m

Length of the rod = 16 m

Other colours + Violet = 1

1 - (1/10 + 1/20 + 1/30 + 1/40 + 1/50 + 1/60) = Violet

1 - (49/200) = Violet

151/200 parts is Violet

It is given that the length of the violet portion of the rod is 12.08 m

Let the length of the rod be L,

Therefore,

151 * L/200 = 12.08

L = 200 * 12.08/151

= 16 m

Length of the rod = 16 m

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46. A fraction having denominator 30 and lying between 5/8 and 7/11 is ________.

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Correct Ans:19/30

Explanation:

Now, when the 5/8 is simplified we get,

5/8= 0.625 (approximately 0.63)

Similarly, 7/11 = 0.63

Now considering the options,

21/30 = 0.7

20/30 = 0.67

19/30 = 0.63

18/30 = 0.6

So, a fraction that lies between 5/8 and 7/11 is 19/30

5/8= 0.625 (approximately 0.63)

Similarly, 7/11 = 0.63

Now considering the options,

21/30 = 0.7

20/30 = 0.67

19/30 = 0.63

18/30 = 0.6

So, a fraction that lies between 5/8 and 7/11 is 19/30

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47. A fraction becomes 1/3, when 1 is subtracted from both the numerator and the denominator. The same fraction becomes 1/2, when 1 is added to both the numerator and the denominator. The sum of numerator and denominator of the fraction is __________.

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Correct Ans:10

Explanation:

Let the original fraction be x/y

It is given that,

(x-1)/(y-1)=1/3

= 3x â€“ 3 = y â€“ 1

= 3x â€“ y = 2 ----------(1)

It is also given that,

(x+1)/(y+1)=1/2

= 2x + 2 = y + 1

= 2x â€“ y = -1 ----------(2)

On subtracting equation (2) from equation (1), we get

(3x â€“ y) â€“ (2x â€“ y) = 2 â€“ (-1)

= 3x â€“ 2x = 3

= x = 3

Replacing the value of x in equation (1), we get

3x â€“ y = z

= 9 â€“ y = 2

y = 9 â€“ 2 = 7

âˆ´ Sum of numerator and denominator = x + y

= 7 + 3 = 10

It is given that,

(x-1)/(y-1)=1/3

= 3x â€“ 3 = y â€“ 1

= 3x â€“ y = 2 ----------(1)

It is also given that,

(x+1)/(y+1)=1/2

= 2x + 2 = y + 1

= 2x â€“ y = -1 ----------(2)

On subtracting equation (2) from equation (1), we get

(3x â€“ y) â€“ (2x â€“ y) = 2 â€“ (-1)

= 3x â€“ 2x = 3

= x = 3

Replacing the value of x in equation (1), we get

3x â€“ y = z

= 9 â€“ y = 2

y = 9 â€“ 2 = 7

âˆ´ Sum of numerator and denominator = x + y

= 7 + 3 = 10

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48. When two numbers are respectively divided by 33, the remainders are 21 and 28 respectively. If the sum of the two numbers is divided by 33, the remainder will be?

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Correct Ans:16

Explanation:

The numbers will be in the form of

1st number = 33a + 21

2nd number = 33b + 28

Sum of two numbers = 33a + 21 + 33b + 28

= 33(a + b) + 49

So, irrespective of value a and b, 33(a + b) will always be divisible by 33. So the remaining 49 is left.

When 49 is divided by 33, we get 16 as remainder.

Therefore,

1st number = 33a + 21

2nd number = 33b + 28

Sum of two numbers = 33a + 21 + 33b + 28

= 33(a + b) + 49

So, irrespective of value a and b, 33(a + b) will always be divisible by 33. So the remaining 49 is left.

When 49 is divided by 33, we get 16 as remainder.

Therefore,

**sum of two numbers divided by 33, we get 16 as a remainder.**
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49. (1

^{2}+ 2^{2}+ 3^{2}+ ………… + 10^{2}) is equal to how much?SHOW ANSWER

Correct Ans:385

Explanation:

**1**

^{2}+ 2^{2}+ 3^{2}+...+ n^{2}= [n*(n+1)*(2n+1)]/6Here, n =10

= [10*(10+1)*(2(10)+1)]/6

= [10*11*21]/6

= 2310/6

=

**385.**

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50. If in a three-digit number the last two digits places are interchanged, a new number is formed which is greater than the original number by 45. What is the difference between the last two digits of that number?

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Correct Ans:5

Explanation:

Let the three digit number,

When position of second and third digit interchanged,

According to the question,

xyz - xzy = 45

(100x + 10y + z) - (100x + 10z + y) = 45

100x + 10y + z - 100x - 10z - y = 45

9y - 9z = 45

Therefore,

**xyz = 100x + 10y + z**When position of second and third digit interchanged,

**xzy = 100x + 10z + y**According to the question,

xyz - xzy = 45

(100x + 10y + z) - (100x + 10z + y) = 45

100x + 10y + z - 100x - 10z - y = 45

9y - 9z = 45

**y - z = 5**Therefore,

**difference between last two digit of that number is 5.**
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51. If 1/6 th of wall is painted blue, 1/3rd is painted yellow and remaining 9m is painted white, what is the length of the wall?

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Correct Ans:18m

Explanation:

Given:

1/6 th part painted in blue and 1/3 rd part painted in yellow.

Remaining part painted in white = 9m

Remaining part of the wall = 1 - (1/6 + 1/3)

= 1- [(1+2)/6]

= 1 - (3/6)

= 1 - 1/2

= 1/2

Therefore, remaining 1/2 part painted in white which is 9m.

So, Length of the wall = 9 + 9 = 18m.

1/6 th part painted in blue and 1/3 rd part painted in yellow.

Remaining part painted in white = 9m

Remaining part of the wall = 1 - (1/6 + 1/3)

= 1- [(1+2)/6]

= 1 - (3/6)

= 1 - 1/2

= 1/2

Therefore, remaining 1/2 part painted in white which is 9m.

So, Length of the wall = 9 + 9 = 18m.

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52. What is the sum of all natural numbers between 100 and 400 which are divisible by 13?

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Correct Ans:5681

Explanation:

The first number divisible by 13 between 100 and 400 = 104

The last number divisible by 13 between 100 and 400 = 390

WKT, nth term = a + (n -1 )d

Where, a = first term, d - common difference

Here, nth term = 390; a = 104; d =13

390 = 104 + (n - 1)13

(390 - 104)/13 = n - 1

286/13 = n - 1

22 = n - 1

n = 23

The last number divisible by 13 between 100 and 400 = 390

WKT, nth term = a + (n -1 )d

Where, a = first term, d - common difference

Here, nth term = 390; a = 104; d =13

390 = 104 + (n - 1)13

(390 - 104)/13 = n - 1

286/13 = n - 1

22 = n - 1

n = 23

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53. What is the unit digit of the sum of first 131 whole numbers?

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Correct Ans:5

Explanation:

Given:

Sum of first 130 numbers= 1 + 2 + 3 + ....... + 130

WKT,

Here, n = 130

The sum of first 130 numbers = 130(130 + 1)/2

= 130(131)/2

= 8515

Therefore,

Sum of first 130 numbers= 1 + 2 + 3 + ....... + 130

WKT,

**1 + 2 + 3 + ......+ n = n(n + 1)/2**Here, n = 130

The sum of first 130 numbers = 130(130 + 1)/2

= 130(131)/2

= 8515

Therefore,

**unit digit of the sum of first 131 whole numbers is 5.**
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54. Find three consecutive numbers such that twice the first, three times the second and four times the third together make 191.

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Correct Ans:20, 21, 22

Explanation:

Let x, x + 1, x + 2 are three consecutive numbers.

Given that

2x + 3(x + 1) + 4(x + 2) = 191

2x + 3x + 3 + 4x + 8 = 191

9x = 191 - 11

9x = 180

x = 20

Therefore, three consecutive numbers = 20, (20 + 1), (20 + 2)

= 20,21, 22.

Given that

2x + 3(x + 1) + 4(x + 2) = 191

2x + 3x + 3 + 4x + 8 = 191

9x = 191 - 11

9x = 180

x = 20

Therefore, three consecutive numbers = 20, (20 + 1), (20 + 2)

= 20,21, 22.

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55. The sum of two numbers is 150. If one fourth of first number is 5 more than the one sixth part of the other number, find the smaller number.

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Correct Ans:72

Explanation:

Let the first number be x.

Therefore, second number be (150 - x).

According to the question,

1/4 th of first number = 5 more than 1/6 th of second number

x/4 = [(150 - x)/6] + 5

x/4 = (150 - x + 30)/6

x/4 = (180 - x)/6

6x = 720 - 4x

6x + 4x = 720

10x = 720

x = 72

Second number = 150 - 72 = 78

Therefore, smallest number is 72.

Therefore, second number be (150 - x).

According to the question,

1/4 th of first number = 5 more than 1/6 th of second number

x/4 = [(150 - x)/6] + 5

x/4 = (150 - x + 30)/6

x/4 = (180 - x)/6

6x = 720 - 4x

6x + 4x = 720

10x = 720

x = 72

Second number = 150 - 72 = 78

Therefore, smallest number is 72.

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56. A number when divided by 91 gives remainder 17. When the same number is divided by 13, the remainder will be:

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Correct Ans:4

Explanation:

X is a number

Given that,

When X is divided by 91 gives remainder 17

So, X = 91 + 17

= 108

108 divided by 91 leaves 17 as remainder

108 / 13 = 4

âˆ´ The remainder is 4

Given that,

When X is divided by 91 gives remainder 17

So, X = 91 + 17

= 108

108 divided by 91 leaves 17 as remainder

108 / 13 = 4

âˆ´ The remainder is 4

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57. Which of the numbers given below is not a square number?

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Correct Ans:2525

Explanation:

1225 = 35*35

2025 = 45*45

4225 = 65*65

Here, 2525 is not the square number.

2025 = 45*45

4225 = 65*65

Here, 2525 is not the square number.

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58. What is the sum of the first 12 terms of an arithmetic progession, if the first term is 3 and the last term is 47?

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Correct Ans:300

Explanation:

Given: a = 3, a

Therefore, required sum = (n/2)[a + a

= (12/2)[3 + 47]

= 6 * 50

= 300

Therefore, sum of first 12 terms is 300.

_{n}= 47Therefore, required sum = (n/2)[a + a

_{n}]= (12/2)[3 + 47]

= 6 * 50

= 300

Therefore, sum of first 12 terms is 300.

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59. The second lowest number of five consecutive odd number series is four more than the 5/12th of the third highest number of a five consecutive even number series. If the average of five consecutive even number series is 60, then find the difference between the highest number of both the series?

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Correct Ans:29

Explanation:

Let five consecutive even number series = a, (a+2), (a+4), (a+6), (a+8)

Let five consecutive odd number series = b, (b+2), (b+4), (b+6), (b+8)

Given, average of five consecutive even number series = 60

[a + (a+2) + (a+4) + (a+6) + (a+8)]/5 = 60

5a + 20 = 300

a = 56

Third highest number of even number series = (a+4) = 56 + 4 = 60

Second lowest number of odd number series = Four more than the 5/12th of the third highest number of even number series

= [(5/12)*60] + 4

= 29

Highest number of even number series = (a+8) = 56 + 8 = 64

Highest number of odd number series = (b+8) = (second lowest -2) + 8

= (29 - 2) + 8 = 35

Required difference = 64 - 35 = 29

Let five consecutive odd number series = b, (b+2), (b+4), (b+6), (b+8)

Given, average of five consecutive even number series = 60

[a + (a+2) + (a+4) + (a+6) + (a+8)]/5 = 60

5a + 20 = 300

a = 56

Third highest number of even number series = (a+4) = 56 + 4 = 60

Second lowest number of odd number series = Four more than the 5/12th of the third highest number of even number series

= [(5/12)*60] + 4

= 29

Highest number of even number series = (a+8) = 56 + 8 = 64

Highest number of odd number series = (b+8) = (second lowest -2) + 8

= (29 - 2) + 8 = 35

Required difference = 64 - 35 = 29

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60. If the arithmetic mean of 3a and 4b is greater than 50, and a is twice of b, then the smallest possible integer value of a is:

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Correct Ans:21

Explanation:

Given that, a is twice of b.

Letâ€™s take b = x and a = 2x

According to the question,

(3a + 4b)/2 > 50

3a + 4b > 100

3(2x) + 4(x) > 100

6x + 4x > 100

10x > 100

x > 10

Multiplying 2 on both sides,

2x > 20

a > 20 (Since, 2x = a )

Therefore, the smallest possible integer value of a is 21.

Letâ€™s take b = x and a = 2x

According to the question,

(3a + 4b)/2 > 50

3a + 4b > 100

3(2x) + 4(x) > 100

6x + 4x > 100

10x > 100

x > 10

Multiplying 2 on both sides,

2x > 20

a > 20 (Since, 2x = a )

Therefore, the smallest possible integer value of a is 21.

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