Practice Test: Number System- 2 - UPSC MCQ

# Practice Test: Number System- 2 - UPSC MCQ

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## 20 Questions MCQ Test CSAT Preparation - Practice Test: Number System- 2

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Practice Test: Number System- 2 - Question 1

### The HCF of two numbers is 11 and their LCM is 616. If one of the numbers is 77, find the other number.

Detailed Solution for Practice Test: Number System- 2 - Question 1

Calculation:

Let, 2nd number be m,

⇒ m × 77 = 11 × 616

⇒ m = 616/7

⇒ m = 88

∴ The 2nd number is 88.

Practice Test: Number System- 2 - Question 2

### The total number of 3 digit numbers which have two or more consecutive digits identical is:

Detailed Solution for Practice Test: Number System- 2 - Question 2

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.

Practice Test: Number System- 2 - Question 3

### Find the last non zero digit of 96!

Detailed Solution for Practice Test: Number System- 2 - Question 3

Finding the Last Non-Zero Digit of 96!

• We need to find the last non-zero digit of 96!.
• To find the last non-zero digit, we need to determine the prime factors of 10 in 96!.
• 10 = 2 * 5, so we need to find the number of pairs of 2 and 5 in the prime factorization of 96!.
• Since there are more 2's than 5's in the prime factorization of 96!, we only need to focus on the number of 5's.
• There are 19 multiples of 5 in 96! (5, 10, 15, ..., 95).
• Therefore, there are 19 trailing zeros in 96!.
• Removing the trailing zeros, we are left with 96!/10^19.
• Now, we need to find the last non-zero digit of 96!/10^19.
• Since 10 = 2 * 5, we can remove one pair of 2 and 5 from each trailing zero to get the last non-zero digit.
• Therefore, we need to find the last non-zero digit of 96!/2^19.
• After dividing 96! by 2^19, we are left with a number that ends in 6.
• Hence, the last non-zero digit of 96! is 6.
Practice Test: Number System- 2 - Question 4

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?

Detailed Solution for Practice Test: Number System- 2 - Question 4

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.

Practice Test: Number System- 2 - Question 5

How many divisors of 105 will have at least one zero at its end?

Detailed Solution for Practice Test: Number System- 2 - Question 5

In order for a divisor (or any number) to have a zero at its end, it must have a 10 as a factor, i.e., a 2-and-5 pair. Notice that
105 = 25 x 55 = (24 x 54) x (2 x 5)
Therefore, any divisors of 24 x 54 will have a zero at its end when multiplied by 2 x 5. Since 24 x 54 has (4 + 1)(4 + 1) = 25 divisors, 105 has 25 divisors that have at least one zero at its end.

Practice Test: Number System- 2 - Question 6

Find the remainder when 496 is divided by 6.

Detailed Solution for Practice Test: Number System- 2 - Question 6

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.

Practice Test: Number System- 2 - Question 7

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?

Detailed Solution for Practice Test: Number System- 2 - Question 7

When we are adding the sum of page numbers of 25 pages, it will always be an odd number. So, she could not have found the sum of pages as 1990.

Practice Test: Number System- 2 - Question 8

There are 50 integers a1, a2,........, a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer is referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.
Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than:

Detailed Solution for Practice Test: Number System- 2 - Question 8

The smallest integer of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.

As this is a variable based question: the word "ANY" can be used.

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.

S1=1,2,3,4,.........24, S2=25,26,27,..........50.

Practice Test: Number System- 2 - Question 9

Srini wrote his class 10th board examination this year. When the result came out he searched for his hall ticket to see his roll number but could not trace it. He could remember only the first three digits of the 6 digit number as 267. His father, however, remembered that the number was divisible by 11. His mother gave the information that the number was also divisible by 13. They tried to recollect the number when all of a sudden Srini told that the number was a multiple of 7. What was the unit digits of the number?

Detailed Solution for Practice Test: Number System- 2 - Question 9

His roll number is divisible by 1001 and the first three digits are 267. Hence the last three digits will also be 267.

Practice Test: Number System- 2 - Question 10

When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?

Detailed Solution for Practice Test: Number System- 2 - Question 10

When 7179 and 9699 are divided by another natural number N, remainder obtained is same.
Let remainder is R, then (7179 — R) and (9699 — R) are multiples of N and {(9699 — R) — (7179 — R)} is multiple of N. Then 2520 is multiple of N or the largest value of N is 2520. Total factors of N which are multiples of 10 is 18.

Practice Test: Number System- 2 - Question 11

Twenty-five boxes of sweets are delivered to Mr Roy’s home. Mr Roy had ordered sweets of three different types. What is the minimum number of boxes of sweets which are having sweets of same type?

Detailed Solution for Practice Test: Number System- 2 - Question 11

According to the question,

Given information is, Twenty-five boxes of sweets are delivered to Mr Roy's home. Mr Roy had ordered three different types of sweets. What is the minimum number of boxes of sweets that are having sweets of same type?

So, This is one classic example of the pigeonhole principle.

Since Mr Roy has ordered for 25 boxes and there different types of sweets, minimum 8 boxes of sweets will have the same type of sweets.

Practice Test: Number System- 2 - Question 12

Find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34.

Detailed Solution for Practice Test: Number System- 2 - Question 12

Remainder,
(73 * 75 * 78 * 57 * 197 * 37)/34 ===> (5 * 7 * 10 * 23 * 27 * 3)/34
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
(5 * 7 * 10 * 23 * 27 * 3)/34 ===> (35 * 30 * 23 * 27)/34 [Number Multiplied]
(35 * 30 * 23 * 27)/34 ===> (1 * -4 * -11 * -7)/34
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]

(1 * -4 * -11 * -7)/34 ===> (28 * -11)/34 ===> (-6 * -11)/34 ===> 66/34 ===R===> 32.
Required remainder = 32.

Practice Test: Number System- 2 - Question 13

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?

Detailed Solution for Practice Test: Number System- 2 - Question 13

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).

as 7*5*2 = 70

Practice Test: Number System- 2 - Question 14

What will be remainder when 1212121212... 300 times, is being divided by 99?

Detailed Solution for Practice Test: Number System- 2 - Question 14

This number 1212121212... 300 times is divisible by 9. So, we can write 1212121212...300 times = 9 N, where N is the quotient obtained when divided by 9. Now this question is like -
1212121212... 300 times / 99
= Remainder 9 / 9 x 134680...written 50 times / 11
[121212 = 13468 x 9]

Now we will have to find the reminder obtained when 134680134680.. . 50 times is divided by 11.
For this, we are supposed to use the divisibility rule of 11 from right hand side. [Using the divisibility rule from left hand side might give us the wrong remainder, like if we find out the remainder obtained when 12 is divided by 11, remainder = 1 = (2-1)≠(1 - 2)]
Remainder 134680...written 50 times / 11 = 2
So, the total remainder = 18

Alternatively, divisibility rule of 10" - 1, n = 2 can be used to find the remainder in this case.

Practice Test: Number System- 2 - Question 15

Which of the following would always divide a six-digit number of the form ababab?

Detailed Solution for Practice Test: Number System- 2 - Question 15

Number = ababab

=ab×10000+ab×100+ab

=ab(10000+100+1)

=ab(10101)

Practice Test: Number System- 2 - Question 16

Find the unit digit:
(76476756749)8754874878

Detailed Solution for Practice Test: Number System- 2 - Question 16

Explanation : The unit digit of the number will depend on the last digit.

As we know that 91 = 9

92  = 81

93 = 729

94  = 6561

The unit digit of the number is 1 and 9, from the options we can pick the answer

Hence option a) is correct

Practice Test: Number System- 2 - Question 17

Find the unit digit:
1719∧13

Detailed Solution for Practice Test: Number System- 2 - Question 17

17 is raised to the power of 19 and 19 is raised to the power of 13.
To find the last digit of the number of this kind we will start with the base, and the base here is 17.
To get the unit digit of a number our only concern is the digit at the unit place i.e.7.
The cyclicity of 7 is 4.
Dividing 1913  by 4.
Remainder will be 3.
7 raised to power 3 (73), the unit digit of this number will be 3.

Practice Test: Number System- 2 - Question 18

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 ?

Detailed Solution for Practice Test: Number System- 2 - Question 18

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, 123 divided by 8, quotient =15, remainder =3
15 divided by 5, quotient =3, remainder =0
3 divided by 6, remainder =3.

Practice Test: Number System- 2 - Question 19

Four bells ringing together and ring at an interval of 12 sec, 15 sec, 20 sec, and 30 sec respectively. How many times will they ring together in 8 hours?

Detailed Solution for Practice Test: Number System- 2 - Question 19

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

No. of Times bell rings in 1 min (60 seconds) = 1 .

No. of times bell rings in 1 hr (60 minutes) = 1 x 60 = 60

No. of Times bell rings in 8 hr = 8 x 60 = 480

No. of Times bell rings in 8 hr + the first time all bell rings also needs to be counted

⇒ 480 + 1

∴ The bell ringing 481 times in 8 hours.

Practice Test: Number System- 2 - Question 20

What would be the greatest number that divides 14, 20, and 32 leaving the same remainder?

Detailed Solution for Practice Test: Number System- 2 - Question 20

Here, the number which divides 14, 20, and 32 leaves the same remainder.
∴ We will be using HCF model 2
The required number will be the HCF of (20 - 14), (32 - 20), and (32 - 14).
i.e. HCF (6, 12, 18)
which will be 6.
Therefore, the required number is 6.

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