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Question 1 of 5
1. Question
While writing all the numbers from 400 to 699, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?
Correct
Answer: (c)
Solution:
Given that,
From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.
Now,
Considering 400 to 499
Hundreds digit = 4
So we need: 4 > tens > unitsPossible pairs (tens, units):
tens can be 1, 2, 3If tens = 1 → units = 0 → 1 number
If tens = 2 → units = 0, 1 → 2 numbers
If tens = 3 → units = 0, 1, 2 → 3 numbersTotal = 1 + 2 + 3 = 6
Considering 500 to 599
Hundreds digit = 5
tens can be 1, 2, 3, 4Total = 1 + 2 + 3 + 4 = 10
Considering 600 to 699
Hundreds digit = 6
tens can be 1, 2, 3, 4, 5Total = 1 + 2 + 3 + 4 + 5 = 15
Total possible numbers = 6 + 10 + 15 = 31
Incorrect
Answer: (c)
Solution:
Given that,
From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.
Now,
Considering 400 to 499
Hundreds digit = 4
So we need: 4 > tens > unitsPossible pairs (tens, units):
tens can be 1, 2, 3If tens = 1 → units = 0 → 1 number
If tens = 2 → units = 0, 1 → 2 numbers
If tens = 3 → units = 0, 1, 2 → 3 numbersTotal = 1 + 2 + 3 = 6
Considering 500 to 599
Hundreds digit = 5
tens can be 1, 2, 3, 4Total = 1 + 2 + 3 + 4 = 10
Considering 600 to 699
Hundreds digit = 6
tens can be 1, 2, 3, 4, 5Total = 1 + 2 + 3 + 4 + 5 = 15
Total possible numbers = 6 + 10 + 15 = 31
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Question 2 of 5
2. Question
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 144. The torn page contains which of the following numbers?
Correct
Answer: (b)
Solution:
Given that,
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.
The sum of the numbers on the remaining pages is 144.
Now,
Let the torn page contain consecutive numbers P and (P + 1)
So, sum of torn page numbers = P + (P + 1) = 2P + 1
Total sum of pages = 144 + (2P + 1)
Sum of first n natural numbers = (n/2)(n + 1)
So,
144 + (2P + 1) = (n/2)(n + 1)Hit and trial method
The value of n must be such that the sum should be near 144
n = 17, sum = 153
So,
144 + (2P + 1) = 153
2P + 1 = 9
2P = 8
P = 4Torn page contains 4 and 5
Incorrect
Answer: (b)
Solution:
Given that,
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.
The sum of the numbers on the remaining pages is 144.
Now,
Let the torn page contain consecutive numbers P and (P + 1)
So, sum of torn page numbers = P + (P + 1) = 2P + 1
Total sum of pages = 144 + (2P + 1)
Sum of first n natural numbers = (n/2)(n + 1)
So,
144 + (2P + 1) = (n/2)(n + 1)Hit and trial method
The value of n must be such that the sum should be near 144
n = 17, sum = 153
So,
144 + (2P + 1) = 153
2P + 1 = 9
2P = 8
P = 4Torn page contains 4 and 5
-
Question 3 of 5
3. Question
. What is the sum of all digits which appear in all the integers from 10 to 99?
Correct
Answer: (c)
Solution:
Given that,
Sum of all digits which appear in all the integers from 10 to 99
Now,
For tens digit
10 to 19 tens digit = 1
20 to 29 tens digit = 2
30 to 39 tens digit = 3
…
90 to 99 tens digit = 9Each tens digit appears 10 times.
So, (1 + 2 + 3 + … + 9) × 10
= 45 × 10
= 450For unit digits
0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99)
So, (0 + 1 + 2 + … + 9) × 9
= 45 × 9
= 405Total sum of digits = 450 + 405 = 855
Hence option (c) is correct.
Incorrect
Answer: (c)
Solution:
Given that,
Sum of all digits which appear in all the integers from 10 to 99
Now,
For tens digit
10 to 19 tens digit = 1
20 to 29 tens digit = 2
30 to 39 tens digit = 3
…
90 to 99 tens digit = 9Each tens digit appears 10 times.
So, (1 + 2 + 3 + … + 9) × 10
= 45 × 10
= 450For unit digits
0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99)
So, (0 + 1 + 2 + … + 9) × 9
= 45 × 9
= 405Total sum of digits = 450 + 405 = 855
Hence option (c) is correct.
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Question 4 of 5
4. Question
Two Statements are given followed by two Conclusions:
Statements:
All prime numbers are natural numbers.
All natural numbers are integers.Conclusions:
Conclusion-I: All prime numbers are integers.
Conclusion-II: All integers are prime numbers.Which of the above Conclusions logically follows/follow from the two given Statements?
Correct
Answer: (a)
Solution:
Given that,
Statements:
All prime numbers are natural numbers.
All natural numbers are integers.Now,
Conclusion-I: All prime numbers are integers.
This logically follows as prime numbers ⟶ natural numbers ⟶ integers.
Hence Conclusion-I is correct.
Conclusion-II: All integers are prime numbers.
This does not follow. Integers include composite numbers and zero.
Hence Conclusion-II is incorrect.
Therefore, option (a) is correct.
Incorrect
Answer: (a)
Solution:
Given that,
Statements:
All prime numbers are natural numbers.
All natural numbers are integers.Now,
Conclusion-I: All prime numbers are integers.
This logically follows as prime numbers ⟶ natural numbers ⟶ integers.
Hence Conclusion-I is correct.
Conclusion-II: All integers are prime numbers.
This does not follow. Integers include composite numbers and zero.
Hence Conclusion-II is incorrect.
Therefore, option (a) is correct.
-
Question 5 of 5
5. Question
Each digit of an 8-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number?
Correct
Answer: (a)
Solution:
Given that,
Each digit of an 8-digit number is 1.
It is multiplied by itself.
Now,
11111111 × 11111111 = 123456787654321
Thus sum of digits
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= 64Hence option (a) is correct.
Incorrect
Answer: (a)
Solution:
Given that,
Each digit of an 8-digit number is 1.
It is multiplied by itself.
Now,
11111111 × 11111111 = 123456787654321
Thus sum of digits
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= 64Hence option (a) is correct.
