- Question
What is the remainder if 1920-2019 is divided by 7:
a)1
b)5
c)7
d)0
e)3
Explanation
SOLUTION[B]
Using Fermat’s theorem:
If p is a prime number and a, p are co primes (ap-1) mod p = 1
Remainder when 1920 is divided by 7 = 192 mod 7=4. (Here 19 ^ 20 = ((19) ^ 6) ^ 3 * (19) ^ 2 Since the remainder for 196 is 1 the remainder for 1920 is equivalent to the 192 = 4.
Remainder when 2019 is divided by 7 = 20¹ mod 7 =6.(Here 20 the remainder is 1 and since
20 ^ 19 = (20 ^ 6) ^ 3 * (20) ^ 1 = (1 * 20)/7 The remainder is 6.
Remainder when 1920-2019 is divided by 7-4-6=-2=> 5.
Question-2
Determine the mean of all 4-digit even natural numbers of the form ‘aabb’, where a > 0
a)4840
b)5544
c)5050
d)4466
e)4864
Explanation
SOLUTION[B]
The four digit even numbers will be of form:
1100, 1122, 1144 … 1188, 2200, 2222, 2244 … 9900, 9922, 9944, 9966, 9988
Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+
(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)
=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)
=> S=5*1100(1+2+3+…9)+9(22+44+66+88)
=>S=5*1100*9*10/2 + 9*11*20
Total number of numbers are 9*5=45
.. Mean will be S/45 = 5*1100+44-5544
Question-3
From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?
a)48
b)84
c)72
d)60
e)96
Explanation
SOLUTION[D]-
Any multiple of 12 must be a multiple of both 4 and 3.
First, let us look at the constraint for a number being a multiple of 3. Sum of the digits should be a multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit and still retain a multiple of 3, we should drop either 3 or 6. So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
when the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 36 different numbers.
When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 24 different numbers. Thus a total of 60 such numbers can be formed.
Question-4
Determine as to how many numbers are there which are less than 100 and that cannot be written as a multiple of a perfect square greater than 1:
a)62
b)59
c)60
d)64
e)61
Explanation
SOLUTION[E]61
List all multiples of perfect squares (without repeating any number) and subtract this from 99.
4- there are 24 multiples of 4 (4,8,12,…..96)
9- there are 11 multiples, 2 common with 4 (36 and 72) so, add 9 multiples
160 new multiples
25-3 new multiples (25,50,75)
360 new ones
49249,98}
64-0
81-0
Total multiples of perfect squares are 38. There are 99 numbers in total. So, there are 61 numbers that are not multiples of perfect squares.
Question-5
There are three numbers such that four times the first number is equal to three times the second number and six times the second number is equal to four times the third number. If the first number is nine less than the third number, find the second number?
a)21
b)12
c)18
d)15
e)9
Explanation
SOLUTION[B]-12 We have three numbers: x, y and z.
Given:
1. 4x = 3y
2. 6y = 4z
3. x = z – 9
We need to find y.
From equation 1, we can express y in terms of x: y = 4/3 * x
Now, let’s substitute this expression for y into equation 2: 6(4/3 * x) = 4z
8x = 4z
2x = z
Now, using equation 3, we find x:
x = z – 9
x = 2x – 9
x = 9
Now, we can find y using 2:
y = 4/3 * (9)
y = 12
So, y = 12
Question-6
If Sudhir wants to write on a blank paper, all numbers from 100 to 10,000, then how many times would the digit 3 be written by him on the paper ?
a)3880
b)3840
c)3980
d)4020
e)3780
Explanation
SOLUTION[C]
Step 1: Find the total number of digits printed in the range 100 to 10,000.
The range contains the following number of digits:
Numbers from 100 to 999 have 3 digits each. There are 900 such numbers. – Numbers from 1000 to 9999 have 4 digits each. There are 9000 such numbers. Number 10,000 has 5 digits.
Therefore, the total number of digits printed is: 900 x 3 + 9000 x 4 + 5 = 36,305
Step 2: Find the number of times the digit 3 appears in the units place.
Every 10th number (i.e., 100, 110, 120, etc.) has a 3 in the units place. There are 900/10 = 90 such numbers.
Also, the number 3 appears in the units place in every number between 3000 and 3999. There are 1000 such numbers.
Therefore, the total number of times the digit 3 appears in the units place is:
90+1000 = 1090
Step 3: Find the number of times the digit 3 appears in the tens, hundreds, and thousands places.
We can use the same logic as in Step 2 to find the number of times the digit 3 appears in the tens, hundreds, and thousands places. We get:
Tens place: 90 * 10 = 900
– Hundreds place: 90 * 100 = 9000
– Thousands place: 9000
Therefore, the total number of times the digit 3 appears in the tens, hundreds thousands places is:
900 + 9000 + 9000 = 18, 900
Step 4: Find the total number of times the digit 3 appears.
The digit 3 appears in both the units place and either the tens, hundreds, or thousands places.
Therefore, the total number of times the digit 3 appears is:
1090 * 3 + 18.9 = 21.96
The correct option is (d) 3980.
