# 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

1. When a number is subtracted from the number 8,12 and 20, the remainders are in continued proportion, Find the number ?

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Correct Ans:4

Explanation:

Explanation :

8-x / 12 â€“x = 12-x /20 â€“x

(8-x)(20-x) = (12 â€“ x)(12 â€“x )

160 â€“ 8x â€“ 20x + x

4x = 16

8-x / 12 â€“x = 12-x /20 â€“x

(8-x)(20-x) = (12 â€“ x)(12 â€“x )

160 â€“ 8x â€“ 20x + x

^{2}= 144 â€“ 12x â€“ 12x + x^{2}4x = 16

**x = 4**
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2. The sum of the digits of a 2-digit number is 11. If we add 45 to the number, the new number obtained is a number formed by interchange of the digits. What is the number?

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Correct Ans:x = 3 and y = 8

Explanation:

Let the number is 10x + y

Given: x + y = 11 ----------(i)

According to the question,

10x + y + 45 = 10y + x

â‡’ x â€“ y = -5 -----------(ii)

From eq. (i) and (ii) , we get

Given: x + y = 11 ----------(i)

According to the question,

10x + y + 45 = 10y + x

â‡’ x â€“ y = -5 -----------(ii)

From eq. (i) and (ii) , we get

**x = 3 and y = 8**
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3. Among three numbers, the first is twice the second and thrice the third, if the average of three numbers is 517, then what is the difference between the first and the third number?

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Correct Ans:564

Explanation:

Let the first number be 6x then second and third number be 3x and 2x respectively.

Given: average of these numbers = 517

Then, their sum = 517 * 3

= 1551

â‡’ 6x + 3x + 2x = 1551

â‡’11x = 1551

â‡’ x = 141

Now, the difference between first and third number = 6x â€“ 2x

= 4x

= 4 * 141

=

Given: average of these numbers = 517

Then, their sum = 517 * 3

= 1551

â‡’ 6x + 3x + 2x = 1551

â‡’11x = 1551

â‡’ x = 141

Now, the difference between first and third number = 6x â€“ 2x

= 4x

= 4 * 141

=

**564**
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4. If the fractions 1/2, 2/3, 5/9, 6/13, and 7/9 are arranged in ascending order of their values, which one will be the fourth?

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Correct Ans:(2/3)

Explanation:

Given fractions are 1/2, 2/3, 5/9, 6/13, and 7/9

L.C.M of their denominators ie., 2, 3, 9, 13, 9 = 234

Thus, given fractions become:

1/2 = 117/234

2/3 = (2 * 78)/234 = 156/234

5/9 = (5 * 26)/234 = 130/234

6/13 = (6 * 18)/234 = 108/234

7/9 = (7 * 26)/234 = 182/234

---> On arranging the numerators in ascending order, we get

108, 117, 130, 156, 182

Therefore

---> Here, the fourth one is 2/3.

L.C.M of their denominators ie., 2, 3, 9, 13, 9 = 234

Thus, given fractions become:

1/2 = 117/234

2/3 = (2 * 78)/234 = 156/234

5/9 = (5 * 26)/234 = 130/234

6/13 = (6 * 18)/234 = 108/234

7/9 = (7 * 26)/234 = 182/234

---> On arranging the numerators in ascending order, we get

108, 117, 130, 156, 182

Therefore

**Ascending order of the given fractions is:**

6/13 < 1/2 < 5/9 < 2/3 < 7/96/13 < 1/2 < 5/9 < 2/3 < 7/9

---> Here, the fourth one is 2/3.

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5. The product of two natural numbers is 9222. If they differ by 19 then find out the sum of the number.

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Correct Ans:193

Explanation:

Let 'x' and 'y' be the two numbers where (x > y).

Given that, Product of x and y = 9222

---> x * y = 9222 ---> eqn (1)

Given that, Difference of x and y = 19

--->. x - y = 19

---> x = 19 + y ---> eqn (2)

Substitute this 'x' value in eqn (1), we get

---> (19 + y) * y = 9222

---> 19y + y

---> y

By using the formula, y = {-b ± √[b

We can find the value of y,

Where, b = +19

a = 1

c = -9222

---> y = {-19 ± √[(19)

= {-19 ± √[361 + 36888]} / 2

= {-19 ± √[37249]} / 2

= {-19 ± 193} / 2

= {-19 + 193} / 2; {-19 - 193} / 2

= {174/2}; {-212/2}

= 87; -106

'y' can't be negative number (i.e, -106) as 'y' is a natural number.

Therefore,

Hence, x = 19 + y (---> from eqn (2))

--> x = 19 + 87

-->

Therefore,

Given that, Product of x and y = 9222

---> x * y = 9222 ---> eqn (1)

Given that, Difference of x and y = 19

--->. x - y = 19

---> x = 19 + y ---> eqn (2)

Substitute this 'x' value in eqn (1), we get

---> (19 + y) * y = 9222

---> 19y + y

^{2}= 9222---> y

^{2}+ 19y - 9222 = 0By using the formula, y = {-b ± √[b

^{2}- 4ac]} / 2aWe can find the value of y,

Where, b = +19

a = 1

c = -9222

---> y = {-19 ± √[(19)

^{2}- 4 * 1 * (-9222)]} / 2 * 1= {-19 ± √[361 + 36888]} / 2

= {-19 ± √[37249]} / 2

= {-19 ± 193} / 2

= {-19 + 193} / 2; {-19 - 193} / 2

= {174/2}; {-212/2}

= 87; -106

'y' can't be negative number (i.e, -106) as 'y' is a natural number.

Therefore,

**y = 87**Hence, x = 19 + y (---> from eqn (2))

--> x = 19 + 87

-->

**x = 106**Therefore,

**sum of the numbers**= x + y = 106 + 87 =**193**
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6. What is the sum of all natural numbers between 100 and 200 which are multiples of 3?

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Correct Ans:4950

Explanation:

Multiples of 3 between 100 and 200 are 102, 105, 108,â€¦ ,198.

Here, the first term = 102

last term = 198

Let the number of Multiples of 3 between 100 and 200 = n

W.K.T:

Where, a

a

d = common difference = 105 - 102 = 3

---> 198 = 102 + (n - 1) * 3

---> 198 - 102 = (n - 1) * 3

---> 96 = (n - 1) * 3

---> (n - 1) = 96/3 = 32

---> n = 32 + 1

--->

where n = number of elements = 33

a = first term = 102

l = last term = 198

Thus, using the above formula,

= (33/2) * 300

= 33 * 150

=

Here, the first term = 102

last term = 198

Let the number of Multiples of 3 between 100 and 200 = n

W.K.T:

**Arithmetic Progression Formula:**

aa

_{n}= a_{1}+ (n - 1)dWhere, a

_{n}= last term = 198a

_{1}= first term = 102d = common difference = 105 - 102 = 3

---> 198 = 102 + (n - 1) * 3

---> 198 - 102 = (n - 1) * 3

---> 96 = (n - 1) * 3

---> (n - 1) = 96/3 = 32

---> n = 32 + 1

--->

**n = 33****Formula:**

Sum of n terms = SSum of n terms = S

_{n}= (n/2) * (a + l)where n = number of elements = 33

a = first term = 102

l = last term = 198

Thus, using the above formula,

**Sum of all natural numbers between 100 and 200 which are multiples of 3**= (33/2) * (102 + 198)= (33/2) * 300

= 33 * 150

=

**4950**
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7. If the sum of 1st 11 terms of an A.P is equal to sum of 1st 19 terms of that A.P then, Find the sum of 1st 30 terms?

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Correct Ans:0

Explanation:

W.K.T: Sum of first n terms of an Arithmetic Progression (A.P) is:

where, n = number of terms

a = first term

d = common difference

Now, sum of 1st 11 terms of an A.P = sum of 1st 19 terms of that A.P

---> (11/2)[2a + (11 - 1)d] = (19/2)[2a + (19 - 1)d]

---> 22a + 11* (10) * d = 38a + 19 *(18) *d

---> 22a + 110d = 38a + 342d

---> 38a - 22a + 342d - 110d = 0

---> 16a + 232d = 0

---> Dividing above eqn by 8, we get

--->

Then, Sum of 1st 30 terms of an A.P = (30/2)[2a + (30 - 1)d]

= (15)[2a + 29d]

= (15)[0] ----> [Since from eqn (1), 2a + 29d = 0]

= 0

Hence,

**S = (n/2)[2a + (n - 1)d]**where, n = number of terms

a = first term

d = common difference

Now, sum of 1st 11 terms of an A.P = sum of 1st 19 terms of that A.P

---> (11/2)[2a + (11 - 1)d] = (19/2)[2a + (19 - 1)d]

---> 22a + 11* (10) * d = 38a + 19 *(18) *d

---> 22a + 110d = 38a + 342d

---> 38a - 22a + 342d - 110d = 0

---> 16a + 232d = 0

---> Dividing above eqn by 8, we get

--->

**2a + 29d = 0**---> eqn (1)Then, Sum of 1st 30 terms of an A.P = (30/2)[2a + (30 - 1)d]

= (15)[2a + 29d]

= (15)[0] ----> [Since from eqn (1), 2a + 29d = 0]

= 0

Hence,

**Sum of 1st 30 terms of an A.P = 0**
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8. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is _________.

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Correct Ans:220030

Explanation:

Let the required number be 'N' which is the 'dividend'.

Given, Divisor = sum of 555 and 445 = 555 + 445 = 1000

Quotient = 2 * (555 - 445) = 2 * 110 = 220

Remainder = 30

W.K.T:

---> Dividend ie., Required Number = 1000 * 220 + 30

---> Required Number = 220000â€¬ + 30

--->

Given, Divisor = sum of 555 and 445 = 555 + 445 = 1000

Quotient = 2 * (555 - 445) = 2 * 110 = 220

Remainder = 30

W.K.T:

**Dividend = Divisor * Quotient + Remainder**---> Dividend ie., Required Number = 1000 * 220 + 30

---> Required Number = 220000â€¬ + 30

--->

**Required Number = 220030**
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9. If the numbers âˆ›9, âˆœ20, (25)

^{(1/6)}are arranged in ascending order, then the right arrangement is ________.SHOW ANSWER

Correct Ans:(25)

^{(1/6)}< âˆ›9 < âˆœ20Explanation:

Given numbers are (9)

L.C.M of 3, 4, and 6 = 12

Now,

(9)

----> (9)

----> (9)

----> (20)

----> (25)

Therefore, right arrangement in

^{(1/3)}, (20)^{(1/4)}, (25)^{(1/6)}L.C.M of 3, 4, and 6 = 12

Now,

(9)

^{(12/3)}, (20)^{(12/4)}, (25)^{(12/6)}----> (9)

^{4}, (20)^{3}, (25)^{2}----> (9)

^{4}= 6561,----> (20)

^{3}= 8000,----> (25)

^{2}= 625Therefore, right arrangement in

**ascending order**is:**(25)**^{(1/6)}< âˆ›9 < âˆœ20
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10. The unit digit in 43 * 69 * 551 * 9242 is

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Correct Ans:4

Explanation:

If we write all the given numbers in a form (10a + b), b will turn out to be the unit's digit here.

So when we multiply all the given numbers, the unit's digit in the product will be nothing but the product of all unit's digits, irrespective of what the other digits in the number are.

From the given expression,

---> unit digit of 43 * unit digit of 69 ---> 3 * 9 = 27 ---> here, unit digit = 7

7 * unit digit of 551 = 7 * 1 = 7 ---> here, unit digit =7

7 * unit digit of 9242 = 7 * 2 = 14 ---> here, unit digit = 4

Therefore,

So when we multiply all the given numbers, the unit's digit in the product will be nothing but the product of all unit's digits, irrespective of what the other digits in the number are.

From the given expression,

---> unit digit of 43 * unit digit of 69 ---> 3 * 9 = 27 ---> here, unit digit = 7

7 * unit digit of 551 = 7 * 1 = 7 ---> here, unit digit =7

7 * unit digit of 9242 = 7 * 2 = 14 ---> here, unit digit = 4

Therefore,

**Digit at unit's place**of the given expression "43 * 69 * 551 * 9242" =**4**
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11. You purchased two pieces of cloth measuring 1.2 m and 1.3 m each at Rs. 330 and Rs. 270 per meter respectively and gave Rs. 1000 at the payment counter. How much cash will you get back?

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Correct Ans:Rs. 253

Explanation:

Total cost of the two pieces of cloth = (1.2 * 330) + (1.3 * 270) = Rs. 747

Given that he paid an amount of Rs. 1000 at the counter

Given that he paid an amount of Rs. 1000 at the counter

**Amount he get back**= 1000 - 747 =**Rs. 253**
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12. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

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Correct Ans:85

Explanation:

Since the three numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So,

Given product of the first two numbers = 551

-->

Given product of the last two numbers = 1073

--->

Thus, Required sum = (19 + 29 + 37) = 85

-->

Also, the given two products have the middle number in common.

So,

**middle number**= H.C.F. of 551 and 1073 =**29**Given product of the first two numbers = 551

-->

**First number**= 551/29 =**19**Given product of the last two numbers = 1073

--->

**Third number**= 1073/29 =**37**Thus, Required sum = (19 + 29 + 37) = 85

-->

**Sum of the three numbers = 85**.
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13. The least number which must be added-to 1728 to make it a perfect square is

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Correct Ans:36

Explanation:

The least number which must be added-to 1728 to make it a perfect square is

Reference:

----> 1728 + 36 = 1764 = (42)

----> Hence 36 is to be added with 1728 to make it perfect square.

Reference:

----> 1728 + 36 = 1764 = (42)

^{2}----> Hence 36 is to be added with 1728 to make it perfect square.

**Hence the answer is : 36**
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14. Find the total number of factors of 15120

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Correct Ans:80

Explanation:

Find the total number of factors of 15120

If a composite no. N has been written in the form of

-----> N = a

Then the no. of total division or factors of

-----> N = (p+1)(q+1)(r+1)(s+1),.........

Hence,

-----> 15120 = 2

-----> = (4+1) (3+1) (1+1) (1+1)

-----> = 80

If a composite no. N has been written in the form of

-----> N = a

^{p},b^{q},c^{r},d^{s},..........Then the no. of total division or factors of

-----> N = (p+1)(q+1)(r+1)(s+1),.........

Hence,

-----> 15120 = 2

^{4}* 3^{4}* 5^{1}* 7^{1}total no of factors,-----> = (4+1) (3+1) (1+1) (1+1)

-----> = 80

**Hence the answer is : 80**
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15. Find the unit digit of the product of all the prime numbers between 1 and (17)

^{ 17 }SHOW ANSWER

Correct Ans:0

Explanation:

Find the unit digit of the product of all the prime numbers between 1 and (17)

Reference:

The set of prime numbers S = { 2,3,5,7,11,13,.....}

Since there is one 5 and one 2 present in this series. If we multiply there two numbers we get â€˜0â€™ as the unit digit.

^{ 17 }:Reference:

The set of prime numbers S = { 2,3,5,7,11,13,.....}

Since there is one 5 and one 2 present in this series. If we multiply there two numbers we get â€˜0â€™ as the unit digit.

**Hence the unit digit of given expression is 0.**
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16. If (a*b) = 6a - 4b + 3ab, then ((6 * 3) + (4 * 3)) is equals to.

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Correct Ans:126

Explanation:

If (a*b) = 6a - 4b + 3ab, then ((6 * 3) + (4 * 3)) is equals to.

Reference:

----> (6 * 6) - (4 * 3) + (3 * 6 * 3 ) + (4 * 6) - (4 *3) + (3* 4* 3)

----> = 36 -12 + 54 +24 - 12 + 36

----> = 126

Reference:

----> (6 * 6) - (4 * 3) + (3 * 6 * 3 ) + (4 * 6) - (4 *3) + (3* 4* 3)

----> = 36 -12 + 54 +24 - 12 + 36

----> = 126

**Hence the answer is : 126**
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17. Which of the following cannot be the number of zeroes at the end of any factorial?

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Correct Ans:5

Explanation:

Which of the following cannot be the number of zeroes at the end of any factorial:

Reference:

----> We know that the number of zeroes at the end depends on the pair of (2 * 5) . We know that number of zeroes in 5! Is 1.

----> (5! to 9!) = 1

----> (10! to 14!) = 2

----> (15! to 19! ) = 3

----> (20! to 24!) = 4

----> (25! to 29!) = 6

----> Here number of zeroes between 25! to 29! is 6 because 25 = 5

Reference:

----> We know that the number of zeroes at the end depends on the pair of (2 * 5) . We know that number of zeroes in 5! Is 1.

----> (5! to 9!) = 1

----> (10! to 14!) = 2

----> (15! to 19! ) = 3

----> (20! to 24!) = 4

----> (25! to 29!) = 6

----> Here number of zeroes between 25! to 29! is 6 because 25 = 5

^{2}So, 5 cannot be the number of zeroes at the end of any factorial value.**Hence the answer is : 5**
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18. If p is prime number, then which of the following may also be a prime number?

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Correct Ans:(p -2)

Explanation:

If p is prime number, then which of the following may also be a prime number:

Reference:

---> None of the prime number other than 2 is divisible by 2 and we know that 1 is not a prime.

---> p

---> 3p is also a composite number.

---> Now, for p = 5,7 and 13 we have p -2 = 3,5 and 11 respectively, which are prime numbers.

Reference:

---> None of the prime number other than 2 is divisible by 2 and we know that 1 is not a prime.

---> p

^{2}can be factorized as (p* p) which makes it a composite number.---> 3p is also a composite number.

---> Now, for p = 5,7 and 13 we have p -2 = 3,5 and 11 respectively, which are prime numbers.

**Hence the answer is : (p -2)**
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19. N is the largest 3-digit number, which when divided by 3, 4 and 6 leaves the remainder 1, 2 and 4 respectively. What is the remainder when N is divided by 7?

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Correct Ans:0

Explanation:

N is the largest 3-digit number, which when divided by 3, 4 and 6 leaves the remainder 1, 2 and 4 respectively. What is the remainder when N is divided by 7:

Reference :

Since,(3 -1) = (4 - 2) = (6 - 4) = 2

Least number which when divided by 3, 4 and 6 leaves 1, 2 and 4 as remainders is

----> LCM(3,4,6)

----> = 12-2

----> = 10

Largest three digits multiple of 12 which is under 1000 when we add 10 to it

----> (1000/12) = 4 (remainder)

----> 1000 - 1 = 996 (multiple of 12)

----> 996 - 12 = 984 (multiple of 12)

Required number = 984+10 = 994

Remainder = ((994)/7) = 0

Reference :

Since,(3 -1) = (4 - 2) = (6 - 4) = 2

Least number which when divided by 3, 4 and 6 leaves 1, 2 and 4 as remainders is

----> LCM(3,4,6)

----> = 12-2

----> = 10

Largest three digits multiple of 12 which is under 1000 when we add 10 to it

----> (1000/12) = 4 (remainder)

----> 1000 - 1 = 996 (multiple of 12)

----> 996 - 12 = 984 (multiple of 12)

Required number = 984+10 = 994

Remainder = ((994)/7) = 0

**Hence the answer is : 0**
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20. What is the greatest number that will divide 1204, 3664 and 5904 leaving the same remainder?

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Correct Ans:20

Explanation:

What is the greatest number that will divide 1204, 3664 and 5904 leaving the same remainder:

Reference :

Required number,

----> HCF [(3664 - 1024),(5904 - 3664),(5904 - 1204)]

----> HCF(2466,2240,4700)

----> 20

Reference :

Required number,

----> HCF [(3664 - 1024),(5904 - 3664),(5904 - 1204)]

----> HCF(2466,2240,4700)

----> 20

**Hence the answer is : 20**
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