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Number System MCQs for UPPSC (UP) Exam
It covers all Important Questions with answers on Number System for the UPPSC (UP) exam. The questions are based on important topics. Details about the questions:- Topic: Number System
- Type of Questions: MCQs with solutions
- Number of Questions: 41
- You can attempt them on EduRev to score high in UPPSC (UP) exam.
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?- a)80
- b)75
- c)41
- d)53
Correct answer is option 'D'. Can you explain this answer?
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?
a)
80
b)
75
c)
41
d)
53
 | Krishna Iyer answered |
Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K.
Let k = 1; the number becomes 53
If it is divided by 84, the remainder is 53.
Hence Option D is correct
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Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?- a)1050
- b) 540
- c)1440
- d)1590
Correct answer is option 'D'. Can you explain this answer?
Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?
a)
1050
b)
540
c)
1440
d)
1590
 | Meera Rana answered |
Let us assume that the number with which Anita has to perform the multiplication is 'x'.
Instead of finding 35x, she calculated 53x.
The difference = 53x - 35x = 18x = 540
Therefore, x = 540/18 = 30
So, the new product = 30 x 53 = 1590.
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?- a)1
- b)4
- c)5
- d)10
Correct answer is option 'A'. Can you explain this answer?
Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeroes will the product end?
a)
1
b)
4
c)
5
d)
10
| | Kavya Saxena answered |
For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.
Find the highest power of 24 in 150!- a)46
- b)47
- c)48
- d)49
Correct answer is option 'C'. Can you explain this answer?
Find the highest power of 24 in 150!
a)
46
b)
47
c)
48
d)
49
 | Anaya Patel answered |
24 = 8 × 3
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 23
Highest power of 8 in 150! is:
Therefore, we need to find the highest power of 8 and 3 in 150!
8 = 23
Highest power of 8 in 150! is:
= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3
= 48
= 48
Highest power of 3 in 150! is:
= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]
= 72
= 72
As the powers of 8 are less, powers of 24 in 150! = 48
Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?- a)She could have found the sum of pages as 1990
- b)She could have found sum of pages as 1275
- c)She could have got sum of pages as 1375
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?
a)
She could have found the sum of pages as 1990
b)
She could have found sum of pages as 1275
c)
She could have got sum of pages as 1375
d)
None of these
| | Tanishq Dey answered |
Information Given:
- The notebook has 96 leaves, which means it has 192 pages (since each leaf has two pages).
- The pages are numbered from 1 to 192.
- Tappo tore out the last 25 leaves of the notebook. Since each leaf has 2 pages, she tore out 50 pages.
Step 1: Determine the page numbers torn out
The last 25 leaves correspond to the last 50 pages in the notebook. Since the total number of pages is 192, the page numbers torn out would be from 143 to 192.
Step 2: Calculate the sum of the torn-out pages
The sum of an arithmetic series (in this case, the page numbers) is given by:
For the torn-out pages from 143 to 192:
- First term (aaa) = 143
- Last term (lll) = 192
- Number of terms (nnn) = 50
So, the sum is:
Step 3: Analyze each option
- Option 1: She could have found the sum of pages as 1990.
- To find if this is possible, subtract 1990 from the total sum of all pages (1 to 192):
- Since the remaining sum does not match with any realistic remaining pages, this option is not possible.
- Option 2: She could have found the sum of pages as 1275.
- Subtracting 1275 from the total sum: Remaining sum=18528−1275=17253
- The sum is possible and reasonable, so this option is possible.
- Option 3: She could have found the sum of pages as 1375.
- Subtracting 1375 from the total sum: Remaining sum=18528−1375=17153
- The sum is possible and reasonable, so this option is possible.
Conclusion:
Option 1: She could have found the sum of pages as 1990 is not true because the sum 1990 cannot realistically be the sum of the pages torn out in this context.
Answer: Option 1
In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:I. a + c = e,
II. b – d = d and
III. e + a = bWhich of the following options are true?- a)b = 4, d = 2
- b)a = 4, e = 6
- c)b = 6, e = 2
- d)a = 4, c = 6
Correct answer is option 'B'. Can you explain this answer?
In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:
I. a + c = e,
II. b – d = d and
III. e + a = b
II. b – d = d and
III. e + a = b
Which of the following options are true?
a)
b = 4, d = 2
b)
a = 4, e = 6
c)
b = 6, e = 2
d)
a = 4, c = 6
| | Aditya Kumar answered |
We have a + c = e so possible summation 6+4=10 or 4+2 = 6.
Also b = 2d so possible values 4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .
Also b = 2d so possible values 4 = 2 * 2 or 10 = 5 * 2.
So considering both we have b = 10 , d = 5, a= 4 ,c = 2, e = 6.
Hence the correct option is B .
How many factors of 1080 are perfect squares?- a)6
- b)4
- c)8
- d)12
Correct answer is option 'B'. Can you explain this answer?
How many factors of 1080 are perfect squares?
a)
6
b)
4
c)
8
d)
12
 | Bank Exams India answered |
The factors of 1080 which are perfect square:
1080 → 23 × 33 × 5
For, a number to be a perfect square, all the powers of numbers should be even number.
Power of 2 → 0 or 2
Power of 3 → 0 or 2
Power of 5 → 0
Power of 3 → 0 or 2
Power of 5 → 0
So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.
Hence, Option B is correct.
The sum of the first 100 natural numbers, 1 to 100 is divisible by- a)2, 4 and 8
- b)2 and 4
- c)2
- d)100
Correct answer is option 'C'. Can you explain this answer?
The sum of the first 100 natural numbers, 1 to 100 is divisible by
a)
2, 4 and 8
b)
2 and 4
c)
2
d)
100
 | Machine Experts answered |
The sum of the first 100 natural numbers is:
= (n * (n + 1)) / 2
= (100 * 101) / 2
= 50 * 101
= (100 * 101) / 2
= 50 * 101
101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.
Hence, 50 * 101 will be divisible by 2.
All the page numbers from a book are added, beginning at page 1. However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?- a)44
- b)45
- c)10
- d)12
Correct answer is option 'C'. Can you explain this answer?
All the page numbers from a book are added, beginning at page 1.
However, one page number was added twice by mistake. The sum obtained was 1000. Which page number was added twice?
a)
44
b)
45
c)
10
d)
12
| | Kavya Saxena answered |
The Correct Answer is C: 10
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?- a)16
- b)24
- c)18
- d)30
Correct answer is option 'A'. Can you explain this answer?
If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?
a)
16
b)
24
c)
18
d)
30
| | Krishna Iyer answered |
The correct option is A
16
'abc' has 2 factors.
This means 'abc' is a prime number (Only a prime number can have exactly 2 factors).
Now, 'abcabc' = 'abc'×1001
'abcabc' = 'abc' × 7 × 11 × 13
Since 'abc' is prime we can write 'abcabc' as - p1×71×111×131
No. of factors = (1+1) (1+1) (1+1) (1+1) = 16 factors.
In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?- a)5
- b)8
- c)1
- d)4
Correct answer is option 'A'. Can you explain this answer?
In a four-digit number, the sum of the first 2 digits is equal to that of the last 2 digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other 2 digits. What is the third digit of the number?
a)
5
b)
8
c)
1
d)
4
| | Arun Sharma answered |
Let the 4 digit no. be xyzw.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.
According to given conditions we have x + y = z + w, x + w = z, y + w = 2x + 2z.
With help of these equations, we deduce that y = 2w, z = 5x.
Now the minimum value x can take is 1 so z = 5 and the no. is 1854, which satisfies all the conditions. Hence option A.
Rohan purchased some pens, pencils and erasers for his young brothers and sisters for the ensuing examinations. He had to buy atleast 11 pieces of each item in a manner that the number of pens purchased be more than the number of pencils, which is more than the number of erasers. He purchased a total of 38 pieces.If the number of pencils cannot be equally divided among his 4 brothers and sisters, how many pens did he purchase?- a)11
- b)12
- c)13
- d)14
Correct answer is option 'D'. Can you explain this answer?
Rohan purchased some pens, pencils and erasers for his young brothers and sisters for the ensuing examinations. He had to buy atleast 11 pieces of each item in a manner that the number of pens purchased be more than the number of pencils, which is more than the number of erasers. He purchased a total of 38 pieces.
If the number of pencils cannot be equally divided among his 4 brothers and sisters, how many pens did he purchase?
a)
11
b)
12
c)
13
d)
14
| | Bank Exams India answered |
- Different possibilities for the number of pencils = 12 or 13.
- Since it cannot be divided into his 4 brothers and sisters, it has to be 13.
- The number of erasers should be less than the number of pencils and greater than or equal to 11. So the number of erasers can be 11 or 12.
- If the number of erasers is 12, then the number of pens = 38 - 13 - 12 = 13, which is not possible as the number of pens should be more than the number of pencils.
- So the number of erasers = 11 and therefore the number of pens = 14
Two players A and B are playing a game of putting ‘+’ and '-'signs in between any two integers written from 1 to 100. A starts the game by putting a plus sign anywhere between any two integers. Once all the signs have been put, the result is calculated. If it is even then A wins and if it is odd then B wins, provided they are putting signs by taking turns one by one and either of them can put any sign anywhere between any two integers. Who will win at the end?- a)A
- b)B
- c)Either A or B
- d)Cannot be determined
Correct answer is option 'A'. Can you explain this answer?
Two players A and B are playing a game of putting ‘+’ and '-'signs in between any two integers written from 1 to 100. A starts the game by putting a plus sign anywhere between any two integers. Once all the signs have been put, the result is calculated. If it is even then A wins and if it is odd then B wins, provided they are putting signs by taking turns one by one and either of them can put any sign anywhere between any two integers. Who will win at the end?
a)
A
b)
B
c)
Either A or B
d)
Cannot be determined
| | Meera Rana answered |
Whatever is the sign between two consecutive integers starting from 1 to 100, it will be odd. So, we are getting 50 sets of odd numbers. Now, whatever calculation we do among 50 odd numbers, result will always be even. So, A will win always.
Find the remainder when 4^96 is divided by 6.
a)0
b)2
c)3
d)4
Correct answer is option 'D'. Can you explain this answer?
 | Faizan Khan answered |
496/6, We can write it in this form
(6 - 2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6, It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4.
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.
(6 - 2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6, It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16 * 16 * 16 * 16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4 * 4 * 4 * 4............ 24 times)/6.
Hence, Required Remainder = 4.
NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.
The total number of 3 digit numbers which have two or more consecutive digits identical is:- a)171
- b)170
- c)90
- d)180
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
The total number of 3 digit numbers which have two or more consecutive digits identical is:
a)
171
b)
170
c)
90
d)
180
e)
None of these
| | Faizan Khan answered |
In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171
Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171
Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171
The integers 34041 and 32506,when divided by a three digit integer N, leave the same remainder. What can be the value of N?- a)289
- b)307
- c)317
- d)319
Correct answer is option 'B'. Can you explain this answer?
The integers 34041 and 32506,when divided by a three digit integer N, leave the same remainder. What can be the value of N?
a)
289
b)
307
c)
317
d)
319
 | Ishani Rane answered |
Let the common remainder be x. Then numbers (34041 – x) and (32506 – x) would be completely divisible by n. Hence the difference of the numbers (34041 – x) and (32506 – x) will also be divisible by n or (34041 – x – 32506 + x) = 1535 will also be divisible by n. Now, using options we find that 1535 is divisible by 307.
When a number is successively divided by 7,5 and 4, it leaves remainders of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8,5 and 6 ?- a)3,0,3
- b)2,2,4
- c)5,0,3
- d)2,4,2
Correct answer is option 'A'. Can you explain this answer?
When a number is successively divided by 7,5 and 4, it leaves remainders of 4,2 and 3 respectively. What will be the respective remainders when the smallest such number is successively divided by 8,5 and 6 ?
a)
3,0,3
b)
2,2,4
c)
5,0,3
d)
2,4,2
 | Preeti Khanna answered |
The number would be in the form of (7X+4) as when this number is divide by 7, will give remainder 4.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied.
Now, we will try hit and trial method to obtained the number.
Put, X=17, then
7X+4=7×17+4=119+4=123
Now, when 123 divided by 7, gives quotient 17 , remainder =4
17 divided by 5, quotient =3, remainder =2
3 divide by 4 gives remainder 3.
So for first condition satisfied.
Now, 123 divided by 8, quotient =15, remainder =3
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.
Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?- a)30
- b)70
- c)105
- d)315
Correct answer is option 'B'. Can you explain this answer?
Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?
a)
30
b)
70
c)
105
d)
315
 | Pritam Saha answered |
Prime numbers less than 10 = 2, 3, 5, 7.
If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.
The smallest and largest numbers are of form 2_7 and 7_2
Since it is given that the sum of the digits is >13, x will be 5.
Verifying, 752-257 = 495. Answer is option (b).
Since it is given that the sum of the digits is >13, x will be 5.
Verifying, 752-257 = 495. Answer is option (b).
as 7*5*2 = 70
A nursery has 363, 429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required? - a)33 and 45
- b)37 and 48
- c)41 and 56
- d)45 and 55
Correct answer is option 'A'. Can you explain this answer?
A nursery has 363, 429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required?
a)
33 and 45
b)
37 and 48
c)
41 and 56
d)
45 and 55
 | Debanshi Chakraborty answered |
Solution:
To find the size of each row and the number of rows required, we need to find the HCF (highest common factor) of the given numbers.
1. Find the prime factors of the given numbers:
- 363 = 3 x 11 x 11
- 429 = 3 x 11 x 13
- 693 = 3 x 3 x 7 x 11
2. Identify the common factors of the given numbers:
- The common factor is 3 x 11 = 33
3. Divide each number by the common factor:
- 363 ÷ 33 = 11
- 429 ÷ 33 = 13
- 693 ÷ 33 = 21
The size of each row is 33 plants and the number of rows required are 11, 13, and 21 for the three varieties respectively.
Therefore, the correct answer is option A: 33 and 45.
To find the size of each row and the number of rows required, we need to find the HCF (highest common factor) of the given numbers.
1. Find the prime factors of the given numbers:
- 363 = 3 x 11 x 11
- 429 = 3 x 11 x 13
- 693 = 3 x 3 x 7 x 11
2. Identify the common factors of the given numbers:
- The common factor is 3 x 11 = 33
3. Divide each number by the common factor:
- 363 ÷ 33 = 11
- 429 ÷ 33 = 13
- 693 ÷ 33 = 21
The size of each row is 33 plants and the number of rows required are 11, 13, and 21 for the three varieties respectively.
Therefore, the correct answer is option A: 33 and 45.
What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?
a)900
b)400
c)1600
d)2500
Correct answer is option 'A'. Can you explain this answer?
| | Hridoy Mehra answered |
In this type of question, We need to find out the LCM of the given numbers.
LCM of 12, 15, 18 and 20;
12 = 2*2*3;
15 = 3*5;
18 = 2*3*3;
20 = 2*2*5;
Hence, LCM = 2*2*3*5*3
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers = 2*2*3*3*5*5 = 900.
LCM of 12, 15, 18 and 20;
12 = 2*2*3;
15 = 3*5;
18 = 2*3*3;
20 = 2*2*5;
Hence, LCM = 2*2*3*5*3
Since, the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers = 2*2*3*3*5*5 = 900.
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