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# Test: Number System- 1

## 20 Questions MCQ Test Quantitative Aptitude | Test: Number System- 1

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Attempt Test: Number System- 1 | 20 questions in 40 minutes | Mock test for CAT preparation | Free important questions MCQ to study Quantitative Aptitude for CAT Exam | Download free PDF with solutions
QUESTION: 1

### 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?

Solution:

Let the number be X.
From the given information, 53x – 35x = 540 => 18x = 540 => x = 30
So, new product = 53*30 = 1590

QUESTION: 2

### How many factors of 25 * 36 * 52 are perfect squares?

Solution:

Any factor of this number should be of the form 2a × 3b × 5c
For the factor to be a perfect square a, b, c has to be even.
⇒ a can take values 0, 2, 4
⇒ b can take values 0, 2, 4, 6
⇒ c can take values 0, 2
So, total number of perfect squares = 3 × 4 × 2 = 24

QUESTION: 3

### 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?

Solution:

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.

QUESTION: 4

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

Which of the following options are true?

Solution:

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 .

QUESTION: 5

The sum of the first 100 natural numbers, 1 to 100 is divisible by:

Solution:

The sum of the first 100 natural numbers is:

=  (n * (n + 1)) / 2
=  (100 * 101) / 2
=  50 * 101

101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.

QUESTION: 6

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?

Solution:

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.

QUESTION: 7

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?

Solution:

If the number of pages is 44, the sum will be 44*45/2 = 22*45 = 990
So, the number 10 was added twice.

QUESTION: 8

The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?

Solution:

The difference of the numbers = 34041 – 32506 = 1535
The number that divides both these numbers must be a factor of 1535.
307 is the only 3 digit integer that divides 1535.

QUESTION: 9

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?

Solution:

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 x 2 x 3
⇒ 15 = 3 x 5
⇒ 18 = 2 x 3 x 3
⇒ 20 = 2 x 2 x 5

∴ LCM = 2 x 2 x 3 x 5 x 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 x 2 x 3 x 3 x 5 x 5
= 900

QUESTION: 10

How many factors of 1080 are perfect squares?

Solution:

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

So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.

QUESTION: 11

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 is 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?

Solution:
• 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
QUESTION: 12

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?

Solution:

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.

Therefore, the correct answer is Option A.

QUESTION: 13

Write three rational numbers between 4 and 5?

Solution:
• There are several rational numbers between 4 and 5. The numbers are between 16/4 and 20/4.
• Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.
QUESTION: 14

1 ’s are given 100 times, 2 ’s are given 100 times and 3’s are given 100 times. Now numbers are made by arranging these 300 digits in all possible ways. How many of these numbers will be perfect squares?

Solution:

Solve this question step by step:

1. Any number formed by this method is clearly divisible by 3.
2. Since it needs to be a square, it should be divisible by (3)[2*k]. k varies over the natural numbers.
3. Now consider the original number. It has hundred 1’s, hundred 2’s and hundred 3’s. Sum of these digits is 600. This is not divisible by 9. Hence number is not divisible by 9.
4. If a number is divisible by (3)[2*k], it is divisible by 3k.
5. This number is not divisible by 3k for any k > 1.

Hence it is not a perfect square for any arrangement.

QUESTION: 15

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

Solution:

Remainder = (73 x 75 x 78 x 57 x 197 x 37) / 34
= (5 x 7 x 10 x 23 x 27 x 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 x 7 x 10 x 23 x 27 x 3) / 34
= (35 x 30 x 23 x 27) / 34
= (1 x (-4) x (-11) x (-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 the negative or positive remainders at any time.)

= (1 x -4 x -11 x -7) / 34
= (28 x -11) / 34
= (-6 x -11) / 34
= 66 / 34

∴  R = 32

QUESTION: 16

There are two integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder. What is the value of n?

Solution:

Find the difference between both 34041 and 32506.

34041 - 32506 = 1535

So, N should be a factor of 1535. Factors of 1535:

1, 5, 307, 1535

Here, 307 is a three-digit factor of 1535. So, N = 307.
34041 and 32506 divided by 307 leaves remainder 271. So, 307 is the answer.

QUESTION: 17

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?

Solution:

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.

Therefore, the correct answer is Option D.

QUESTION: 18

Teacher said that there were 100 students in his class, 24 of whom were boys and 32 were girls. Which base system did the teacher use in this statement?

Solution:

We are provided with the equation (32) + (24) = (100). Let us assume our base be 'b'
Then,we can say:

⇒ 32 = 3 x b+ 2 x b0 = 3b+2
⇒ 24 = 2 x b1+ 4 x b= 2b+4
⇒ 100 = 1 x b+ 0 x b+ 0 x b0 = b2

Now, according to our question:

⇒ 32 + 24=100
⇒ (3b + 2) + (2b + 4) = (b2)
⇒ 5b + 6 = b2
⇒ b- 5b - 6 = 0
⇒ b- 6b + b - 6 = 0
⇒ b(b - 6) + 1(b - 6) = 0
⇒ (b - 6) * (b + 1) = 0
⇒ b = 6,- 1

Base can't be negative. Hence b = 6.
∴ Base assumed in the asked question must be 6.

QUESTION: 19

Find the highest power of 24 in 150!

Solution:

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:

= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3
= 48

Highest power of 3 in 150! is:

= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]
= 72

As the powers of 8 are less, powers of 24 in 150! = 48

QUESTION: 20

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?

Solution:

Factors: Beauty of the number 1001. This number is not prime, is a product of three distinct primes and does wonderful things to three-digit numbers when multiplied to them.

To start with ‘abcabc’ = ‘abc’ * 1001 or abc * 7 * 11 * 13 (This is a critical idea to remember).

‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have exactly two factors. (This is in fact the definition of a prime number)

So, ‘abcabc’ is a number like 101101 or 103103.

’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.

As we have already seen, any number of the form paqbrc will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime.

So, p * 7 * 11 * 13 will have = (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) = 16 factors

The question is "How many factors does the 6-digit number ‘abcabc’ have?"

Hence the answer is 16 factors.
Choice A is the correct answer.

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