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Practice Test: Number System- 1 - CAT MCQ


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20 Questions MCQ Test - Practice Test: Number System- 1

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Practice Test: Number System- 1 - 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?

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

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.

Practice Test: Number System- 1 - Question 2

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?

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

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 3

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.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 4

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 .

Practice Test: Number System- 1 - Question 5

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

Detailed Solution for Practice Test: Number System- 1 - Question 5

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.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 6


Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 7

The Correct Answer is C: 10

Let the total pages be ‘n’ and the number which is added twice is x;

∴ {n(n + 1)}/2 = 1000 – x

⇒ n(n + 1) = 2000 – 2x

Now, by hit and trial, we can say that the value of n = 44 i.e. initially there were 44 pages and their sum was (44 × 45)/2 = 990

Since the given sum is 1000, so we can say that the page number 10 was added twice.

Practice Test: Number System- 1 - Question 8

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

Detailed Solution for Practice Test: Number System- 1 - Question 8

Let the common remainder be x.

32506 – x is divisible by n.

34041 – x is divisible by n.

Difference of (32506 – x) and (34041 – x) = (32506 – x) – (34041 – x)

⇒ 32506 – x – 34041 + x

⇒ 32506 – 34041

⇒ 1535 

Factors of 1535 = 1 × 5 × 307 × 1535

3-digit number = 307

⇒ n = 307

∴ The value of n is 307.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 9

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

Practice Test: Number System- 1 - Question 10

How many factors of 1080 are perfect squares?

Detailed Solution for Practice Test: Number System- 1 - Question 10

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.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 11
  • 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 
Practice Test: Number System- 1 - Question 12

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?

 

Detailed Solution for Practice Test: Number System- 1 - Question 12

The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66.

Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.

66 need not to be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide

429 and also divides 693. Hence, 33 is the correct answer for the size of each row.

For how many rows would be required we need to follow the following process:

Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.

Therefore, the correct answer is A

Practice Test: Number System- 1 - Question 13

Write three rational numbers between 4 and 5?

Detailed Solution for Practice Test: Number System- 1 - Question 13
  • 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.
Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 14

However you arrange the digits, you will observe that the sum of digits is always equal to (100)(1+2+3) = 600.

Notice that 600 is divisible by 3 but not by 9.

Hence the original number is divisible by 3 but not by 3^2 which is 9.

ORIGINAL NUMBER = 3(p) where p is an integer does not have the prime factor 3. But 3 is not a perfect square!

Hence there will be no number which has 100 1’s, 100 2’s and 100 3’s which will be a perfect square.

Practice Test: Number System- 1 - Question 15

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

Detailed Solution for Practice Test: Number System- 1 - Question 15

Given:

73 × 75 × 78 × 57 × 197 × 37 is divided by 34

Calculation:

73 × 75 × 78 × 57 × 197 × 3734

We have taken individual remainder like

When 73 is divided by 34 gives remainder is 5

Similarly

Practice Test: Number System- 1 - Question 16

Four bells ring 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- 1 - Question 16

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

Calculation:

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

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours. So the correct option is B

Note: In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 17

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.

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 18

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.

Practice Test: Number System- 1 - Question 19

Find the highest power of 24 in 150!

Detailed Solution for Practice Test: Number System- 1 - Question 19

Practice Test: Number System- 1 - 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?

Detailed Solution for Practice Test: Number System- 1 - Question 20

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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