Practice Test: Number System- 1 - Mechanical Engineering MCQ

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.

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.

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 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 .

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.

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.

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.

• To solve the problem, we need to find the least common multiple (LCM) of the numbers 12, 15, 18, and 20.


• First, we find the prime factorization:
  
  
  
  • 12 = 2² x 3
  
  
  • 15 = 3 x 5
  
  
  • 18 = 2 x 3²
  
  
  • 20 = 2² x 5
  
  
  
  


• The LCM is obtained by taking the highest powers of all prime factors:
  
  
  
  • 2², 3², and 5.
  
  
  
  


• Calculating LCM = 2² x 3² x 5 = 180.


• To form a perfect square, multiply by 5: 180 x 5 = 900.


• Thus, the least number of soldiers is 900.

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.

• 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 

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

• 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.

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.

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

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.

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.

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

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

Now, according to our question:

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

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.