Practice Test: Number System- 2 - Mechanical Engineering MCQ

Calculation:

Let, 2nd number be m,

⇒ m × 77 = 11 × 616

⇒ m = 616/7

⇒ m = 88

∴ The 2nd number is 88.

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

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.

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.

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.

 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.

Information Given:

• The notebook has 96 leaves, which means it has 192 pages (since each leaf has two pages).
• The pages are numbered from 1 to 192.
• Tappo tore out the last 25 leaves of the notebook. Since each leaf has 2 pages, she tore out 50 pages.

Step 1: Determine the page numbers torn out

The last 25 leaves correspond to the last 50 pages in the notebook. Since the total number of pages is 192, the page numbers torn out would be from 143 to 192.

Step 2: Calculate the sum of the torn-out pages

The sum of an arithmetic series (in this case, the page numbers) is given by:

For the torn-out pages from 143 to 192:

• First term (aaa) = 143
• Last term (lll) = 192
• Number of terms (nnn) = 50

So, the sum is:

Step 3: Analyze each option

1. Option 1: She could have found the sum of pages as 1990.

   • To find if this is possible, subtract 1990 from the total sum of all pages (1 to 192):

1. • Since the remaining sum does not match with any realistic remaining pages, this option is not possible.

2. Option 2: She could have found the sum of pages as 1275.

   • Subtracting 1275 from the total sum: Remaining sum=18528−1275=17253
   • The sum is possible and reasonable, so this option is possible.

3. Option 3: She could have found the sum of pages as 1375.

   • Subtracting 1375 from the total sum: Remaining sum=18528−1375=17153
   • The sum is possible and reasonable, so this option is possible.

Conclusion:

Option 1: She could have found the sum of pages as 1990 is not true because the sum 1990 cannot realistically be the sum of the pages torn out in this context.

Answer: Option 1

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.

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

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. 

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.

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

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.

Number = ababab

=ab×10000+ab×100+ab

=ab(10000+100+1)

=ab(10101)

Correct Answer :- a

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

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.

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.

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.

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.