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| Number System (Part 5 of 5) |
Mostly Asked Questions
It contains all the similar types of questions that are frequently asked in the competitive examinations from the Number System !!
1. The sum of first 45 natural numbers is:
Ans. Formula to be used: n (n+1)/2Here n=45 so 45(45+1)/2 = 1035
2. The sum of two numbers is 25 and their difference is 13. Find their product?
Ans. Formula to be used: 4xy = (x + y)2 – (x – y)2
Let the numbers be x and y. Then, x+y = 25 and x-y = 13.
We know that 4xy = (x + y)2 – (x – y)2
252 – 132 = 625 – 169 = 456
Now we have 4xy=456 therefore xy=456/4=114
3. 325325 is a six-digit number. It is divisible by ____
Ans . 7 , 11 ,13 (As per Test of Divisibility)
4. (856 + 167)2 + (856 - 167)2 / 856 x 856 + 167 x 167 =____
Ans. We know that
(a + b)2 + (a - b)2 / (a2 + b2) =2(a2 + b2) / (a2 + b2) = 2
5. If the number 481 * 673 is completely divisible by 9, then the smallest whole number in place of * will be ____
Ans. Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9
Therefore x = 7 (because 29 + 7 = 36 Which is div. by 9)
6. Find the unit's digit in the product (2467)153 x (341)72
Ans. Unit's digit in the given product = unit's digit in 7153 x 172
Now, 74 gives unit digit 1. So7152 gives unit digit 1, 7153 gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence, unit's digit in the product = (7 x 1) = 7.
7. Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?
Ans. Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.
Hence, digits in place of * and $ are 4 and 0 respectively.
8. What least number must be subtracted from 2000 to get a number exactly divisible by 17?
Ans. On dividing 2000 by 17, we get 11 as remainder. Required number to be subtracted = 11.
9. Find the smallest number of 6 digits which is exactly divisible by 111.
Ans. Smallest number of 6 digits is 100000. On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 - 100) - 11. Hence, required number = 100011.
10. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.
Dividend - Remainder 15968-37
Ans. Divisor = -------------------------- = ------------- = 179.
Quotient 89
11. How many numbers between 11 and 90 are divisible by 7?
Ans. The required numbers are 14, 21, 28, 35, .... 77, 84.
This is an A.P. with a = 14 and d = (21 - 14) = 7.
Let it contain n terms. Then, Tn = 84 => a + (n - 1) d = 84 => 14 + (n - 1) x 7 = 84
On Solving the above equation we get n = 11.
Hence Required number of terms = 11.
12. 2 + 22+ 23+ ... + 28=?
Ans. Given series is a G.P. with a = 2, r = 2 and n = 8.
So , Sum = a(rn -1)/(r-1) => 2 (28 -1)/ (2-1) => 2(255) = 510
13. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number?
Ans. Let the smaller number be x.
Then larger number = (x + 1365).
x + 1365 = 6x + 15 5x = 1350 x = 270
Smaller number = 270.
14. 4500 x ? = 3375
Ans. Let the ? be A . Therefore , 4500 x A = 3375
Therefore A = 3375/4500 = 3/4. Hence A = 3/4
15. What will be remainder when 17200 is divided by 18?
Ans. When n is even, (xn - an) is completely divisible by (x + a)
=> (17200 - 1200) is completely divisible by (17 + 1), i.e., 18.=> (17200 - 1) is completely divisible by 18.
On dividing 17200 by 18, we get 1 as remainder.
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