Arun Sharma Test: Number System- 3 - CAT MCQ

# Arun Sharma Test: Number System- 3 - CAT MCQ

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## 15 Questions MCQ Test Quantitative Aptitude (Quant) - Arun Sharma Test: Number System- 3

Arun Sharma Test: Number System- 3 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Arun Sharma Test: Number System- 3 questions and answers have been prepared according to the CAT exam syllabus.The Arun Sharma Test: Number System- 3 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Arun Sharma Test: Number System- 3 below.
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Arun Sharma Test: Number System- 3 - Question 1

### 0.xyxyxyxy.... + 0.yxyxyxyx = 4/3, if x and y are natural numbers what would be the value of x + y?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 1

= x + y = 12

Arun Sharma Test: Number System- 3 - Question 2

### The sum of first five, 3 digit prime numbers is:

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 2

► First 5, 3 digit prime numbers are listed below:

► Their Sum is 533

Arun Sharma Test: Number System- 3 - Question 3

### If a, (a+2), (a+4)  are all prime numbers, how many such values can 'a' take?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 3

► The given terms are: a, a + 2, a + 4
► For Less than 5, We know of such a set: 3, 5 and 7
► For Prime numbers Greater than 5 we have two cases
► Case 1 : The Prime number on division by 6 gives a remainder of 1
• Let us assume a is a prime number
• So a = 6p + 1
• a + 2 = 6p +3. (This will be divisible by 3. Thus not a prime)
• a + 4 = 6p + 7

► Case 2 : The Prime number on division by 6 gives a remainder of 5
• Let us assume a is a prime number
• So a = 6p + 5
• a + 2 = 6p + 7
• a + 4 = 6p + 9 (This will be divisible by 3. Thus not a prime.)

► Hence there is only one such set.

Arun Sharma Test: Number System- 3 - Question 4

What is the sum of all the negative two digit prime numbers?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 4

► Prime numbers is a property of natural numbers, There are no negative prime numbers.

Arun Sharma Test: Number System- 3 - Question 5

What is the negative of sum of all the two digit prime numbers?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 5

► All the two digit Prime numbers are listed below:
►The sum is 1043
►The negative of this sum is - 1043

Arun Sharma Test: Number System- 3 - Question 6

For the equation (2a + 3b + 9c) to be even, we must have

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 6

► Option A : b is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ This does not give a sure shot result.

► Option B : c is Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.

► Option D : Either of the two (b or c) is Even.
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x any number = Even + Even + Odd/Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x any number + 9 x Even = Even + Odd/Even + Even
⇒ This does not give a sure shot result.

► Option C = Both b and c are Even
⇒ 2a + 3b + 9c = 2 x any number + 3 x Even + 9 x Even = Even + Even + Even = Even
⇒ So only option C gives an even response.

Arun Sharma Test: Number System- 3 - Question 7

What is the difference between the highest and smallest 3 digit prime number?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 7

► Highest 3 digit prime number = 997
► Smallest 3 digit prime number = 101
► Difference = 997 - 101 = 896

Arun Sharma Test: Number System- 3 - Question 8

Suppose the sum of n consecutive integers is x+(x+1) +(x+2) +(x+3)+...+(x+(n-1)) =1000, then which of the following cannot be true about the number of terms n?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 8

Solve this question through the options. For n terms being 16 (option 1), we would need an AP with 16 terms and common difference 1, that would add up to 1000. Since, the average value of a term of this AP turns out to be 1000÷16=62.5, we can create a 16 term AP as 55,56,57....62,63,64...70 that adds up to 1000. Hence, a series of16 consecutive terms is possible. Likewise, a series of 5 terms gives the average as 200 & the 5 terms can be taken as 198,199,200,201,202. It is similarly possible for 25 terms with an average of 40 but is not possible for 20 terms with an average of 50.

Hence, option (d) is correct.

Arun Sharma Test: Number System- 3 - Question 9

What would the remainder when the square of the largest 4 digit prime number is divided by 6?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 9

► It is +1 with exceptions of prime numbers below 6 obviously.

► Any prime number above 6 is of the form:

6k + 1 or 6k - 1 ( A lot of people don't know this!))

► So therefore squares is of the form (6k ± 1)2. Expand this and clearly the only term which is not a multiple of 6 will be 1. Which again makes it of the form 6K + 1( now only + because we squared the whole term, Remember?)!.

► So therefore the remainder is of the form +1.

► So for every prime no. above 6, the remainder 1.

So, for 2, it is 4

For 3, it is 3.

For 5, it is 1.

► For every other prime no. Square, It’s 1.

Arun Sharma Test: Number System- 3 - Question 10

Let us consider a fraction whose denominator is smaller than the square of the numerator by unity. If we add 2 to the numerator and the denominator, the fraction will exceed1/3​; now if we subtract 3 from the numerator and the denominator, the fraction remains positive but smaller than 1/10​. Find the fraction.

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 10

According to the question the given fraction is N2−1N​

⟹ Adding 2 to Numerator and denominator gives N2+1N+2​>31​

⟹N2−3N−5<0

⟹Now subtracting 3 gives N2−2N−1​<101​

⟹N2−10N+8>0

There is no such Integer value of N that can satisfy the conditions

So, the correct answer is D

Arun Sharma Test: Number System- 3 - Question 11

If we divide a two-digit number by the sum of its digits, we get 4 as a quotient and 3 as a remainder. Now if we divide that two-digit number by the product of its digits, we get 3 as a quotient and 5 as a remainder and the two-digit number.

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 11

Let n(a,b) is two digit number.

n(a,b)=10a+b

Given  sum of its digitsnumber​ will give 4 as quotient and 3 as remqinder.

n(a,b)=4(a+b)+3

10a+b=4a+4b+3

6a−3b=3⇒2a=1+b

n(a,b)=3(ab)+5

10a+b=3ab+5

10a+2a−1=3a(2a−1)+5

12a−1=6a2−3a+5

6a2−15a+6=0

2a2−5a+2=0

(2a−1)(a−2)=0

a=2(∵a=21​)

∴b=2a−1=3

23 is required number.

So, the correct option is B.

Arun Sharma Test: Number System- 3 - Question 12

What two-digit number is less than the sum of the squares of its digits by 11 and exceed their doubled product by 5?

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 12

Arun Sharma Test: Number System- 3 - Question 13

The last three - digits of the multiplication 12345 × 54321 will be

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 13

If we multiply the last three digits of each terms we get the last three digits.
345 × 321 = 110745
So, the last three digits are 745

So, the correct answer is A

Arun Sharma Test: Number System- 3 - Question 14

Find the gcd (111....11 hundred ones; 11....11 sixty ones).

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 14

The correct option is C 111..... twenty ones
The required GCD would be 1111...111 twenty ones.

Arun Sharma Test: Number System- 3 - Question 15

The remainder when 1010+10100+101000+....+1010000000000 is divided by 7 is

Detailed Solution for Arun Sharma Test: Number System- 3 - Question 15

The remainder of (1010+10100+101000+....+1010000000000)÷7→(310+3100+31000+.....+310000000000)÷7→(34+34+34+34+34+34+34+34+34+34)÷7=Remainder of 40÷7→5.

So, the correct option is C

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## Quantitative Aptitude (Quant)

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