Number System: Solved 444 Number System Questions and answers section with explanation for various online exam preparation, various interviews, Logical Reasoning 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 ?
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 Ã¢â€¡â€™ x Ã¢â‚¬â€œ y = -5 -----------(ii) From eq. (i) and (ii) , we get x = 3 and y = 8
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 Ã¢â€¡â€™ 6x + 3x + 2x = 1551 Ã¢â€¡â€™11x = 1551 Ã¢â€¡â€™ x = 141 Now, the difference between first and third number = 6x Ã¢â‚¬â€œ 2x = 4x = 4 * 141 = 564
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^{2} = 9222 ---> y^{2} + 19y - 9222 = 0 By using the formula, y = {-b ± √[b^{2} - 4ac]} / 2a We 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
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,Ã¢â‚¬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: Arithmetic Progression Formula: a_{n} = a_{1} + (n - 1)d Where, a_{n} = last term = 198 a_{1} = first term = 102 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 ---> n = 33
Formula: Sum 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
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: 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
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
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, Digit at unit's place of the given expression "43 * 69 * 551 * 9242" = 4
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 Amount he get back = 1000 - 747 = Rs. 253
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, 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.
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
15. Find the unit digit of the product of all the prime numbers between 1 and (17) ^{ 17 }
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Correct Ans:0
Explanation:
Find the unit digit of the product of all the prime numbers between 1 and (17) ^{ 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 Ã¢â‚¬Ëœ0Ã¢â‚¬â„¢ as the unit digit.
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
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