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Under this post " Number System: Types of Questions Asked in Competitive Exams " we will discuss various types of questions asked in competitive exams on number system.
Please go through all types of questions carefully, so that you can easily crack it in your exams.
Type 1: Find units digit of a number in form ab
To find the unit digit of number which is in the form ab
1) If b is not divisible by 4
Step 1: Divide b by 4, if it is not divisible then find the remainder of b when divided by 4.
Step 2: Units digit = ar, r is the remainder.
2) If b is multiple of 4
Units digit is 6: When even numbers 2, 4, 6, 8 are raised to multiple of 4.
Units digit is 1: When odd numbers 3, 7 and 9 are raised to multiple of 4.
Q 1. Find the unit digit of (4137)754
Solution: Divide b by 4, if it is not divisible fully, write the remainder
The number is in the form ab i.e (4137) 754
4 × 188 = 752, therefore we get remainder as 2
Units digit = (4137)2 = 17114769
Type 2: Divisibility of Numbers
Q 2. Find the largest 4 digit number which is divisible by 88.
Solution: We know that the largest 4 digit number is 9999.
Simply divide 9999 by 88.
After dividing 9999 by 88 we get 55 as remainder.
The number is said to be completely divisible, only if the remainder is zero.
Hence, we can find the required answer by subtracting the remainder obtained from the 4 digit number.
Therefore, required number = 9999 – 55 = 9944
Type 3: Place Value and Face Value
Q 3. Find the difference between the place value and face value of digit 6 in numeral 2960543.
Solution : Place value of digit 6 in 2960543 = 60000
Solution: In this type of questions, try to simplify the numbers, to find the answer easily.
16 16
Face value of digit 6 in 2960543 = 6
Their difference = 60000 - 6 = 59994
Type 4: Find the result of different operations (additions, subtractions, multiplications, divisions, etc) on given integers.
Q 4. Find the solution of (935421 × 625) =?Type 4: Find the result of different operations (additions, subtractions, multiplications, divisions, etc) on given integers.
Solution: In this type of questions, try to simplify the numbers, to find the answer easily.
We are asked to multiply (935421 × 625)
625 is divisible by 5 and 54= 625
This 54 can be written as (10 /2) 4
625 is divisible by 5 and 54= 625
This 54 can be written as (10 /2) 4
Hence, 935421 × | 104 | = | 9354210000 | = 584638125 |
Type 5: Number a when divided by b gives remainder r, what will be the remainder when a is divided by c.
Q 5. The remainder is 29 when a number is divided 56. If the same number is divided by 8, then what is the remainder?
Solution : Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 29 when a number (dividend) is divided 56(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
Therefore,
X = 56 × Y + 29
56 is completely divisible by 8, but 29 is not completely divisible and we get remainder as 5, which is the required answer.
OR
X = 56 × Y + 29
= (8 × 7Y) + (8 × 3) + 5
5 is the required remainder.
Type 6: Arithmetic and Geometric Progression
Q 6. If 6 + 12 + 18 + 24 + --- = 1800, then find the number of terms in the series.
Solution : This is an Arithmetic Progression, in which x = 6, y = 6, sum of terms = 1800
Sum of n terms = | n | [2x + (n – 1)y] |
2 |
1800 = | n | [2 × 6 + (n – 1)6] |
2 |
1800 = 3n (n + 1)
n(n +1) = 600
n2 + n = 600
25 × 24 = 600
Therefore,
n2 + 25n – 24n – 600 = 0
(n + 25) (n – 24) = 0
24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24
The number of terms in the series = 24
n2 + 25n – 24n – 600 = 0
(n + 25) (n – 24) = 0
24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24
The number of terms in the series = 24
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