Explanation : 

A simple mathematical operation

If we subtract 1 from each number in the original sequence, we get the following sequence:

13, 17, 19, 23, 29, 31

Notice that this is simply the sequence of prime numbers starting from 13.

The next prime number after 31 is 37. Therefore, the next number in the original sequence, which corresponds to the prime number 37 after subtracting 1, is:

37 + 1 = 38

Therefore, the correct answer is (c) 38

Explanation :

x+1x+5x+xx+x1=1xx

x + 1x + 5x + xx + x1 = 1xx

=> x + 10+ x + 50 + x +10x + x + 10x + 1 = 100 + 10x + x

=> 24x + 61 = 100 + 11x

=> 13x = 39

=> x = 3

Model Answer:

Let XYZ be a 3 digit number

Let’s write XYZ in expanded form, where X, Y, and Z are the digits in the hundreds, tens, and units places, respectively.

XYZ = 100X + 10Y + Z

Similarly, we can write YZX and ZXY in expanded form:

YZX = 100Y + 10Z + X
ZXY = 100Z + 10X + Y

Adding these three expressions, we get:

XYZ + YZX + ZXY = 100X + 10Y + Z + 100Y + 10Z + X + 100Z + 10X + Y
= 111X + 111Y + 111Z=111(X + Y + Z)

Factors of 111 are 3 and 37. That eliminate option 3 and 37.

Explanation : 

Let x y be the volumes …

Let’s solve this step by step.

1) Let’s say the side length of cube P is ‘a’
Then volume of P (x) = a³
Mass of P is m

2) For cube Q:
Side length is 2a (given: each side of Q is two times that of P)
Volume of Q (y) = (2a)³ = 8a³
Mass of Q (n) = 2m (given: mass of Q is two times that of P)

3) Now, let’s calculate u = m/x
u = m/a³

4) Let’s calculate v = n/y
v = 2m/(8a³)
v = (2m)/(8a³)
v = m/(4a³)
v = (m/a³)/4
v = u/4

5) Therefore:
u = 4v

Looking at the options, (a) u = 4v is correct.

The answer is (a).

Explanation : 

Integers between 1 and 100 which have 4 as a digit are:

To determine how many integers between 1 and 100 contain the digit 4 but are NOT divisible by 4, follow these steps:

1. Identify Numbers Containing the Digit 4:
Single-digit: 4
Two-digit: 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94

Total numbers with the digit 4: 19

2. Identify Numbers Divisible by 4 Among Them:
4 ÷ 4 = 1 → 4
24 ÷ 4 = 6 → 24
40 ÷ 4 = 10 → 40
44 ÷ 4 = 11 → 44
48 ÷ 4 = 12 → 48
64 ÷ 4 = 16 → 64
84 ÷ 4 = 21 → 84

Total numbers divisible by 4: 7

3. Calculate Numbers with Digit 4 but Not Divisible by 4:
19 (total with digit 4) – 7 (divisible by 4) = 12

Answer: 12

Explanation : 

Overall Average marks in Hindi

Let’s solve this step by step:

1. Given:
x = number of girls
y = number of boys

2. English marks equation:
Total marks = (9x + 8y)
Average = 8.8
Therefore: (9x + 8y)/(x + y) = 8.8 … (1)

3. Hindi marks:
Total marks = (8x + 7y)
From equation (1), we found: x = 4y

4. Substituting x = 4y in Hindi marks:
Total marks = 8(4y) + 7y
= 32y + 7y
= 39y

5. Final calculation:
Average Hindi marks = 39y/(5y)
= 7.8

Therefore, the overall average marks in Hindi is 7.8.

Explanation : 

Let a3bc and de2f be four digits

As per the given condition in the question, each letter represents a different digit greater than 3.
So we can replace the letters with 4, 5, 6, 7, 8, or 9.
A3BC + DE2F = 15902
Step 1: Unit digit
If we add C & F, then we should get 12. Only then can we get 2 at the unit place in the sum (15902).
So, C, F can be (4, 8) or (5, 7)
Step 2: Tens digit
We got a carry of 1 from 12. Now, we know that the tens digit of the sum, 15902 is 0.
So, B + 2 = 9
Or B = 7
Hence, C, F cannot be (5, 7). They must be (4, 8).
Step 3: Hundreds digit
We got a carry of 1 from 10. Now, we know that the hundreds digit of the sum, 15902 is 9.
So, E + 3 = 8
Or E = 8 – 3 = 5
Hence, we found that B =7, C = 4/8, E = 5 and F = 4/8
So, A/D = 6/9
So, difference between A and D = 9 – 6 = 3

Explanation : 

Torn pages numbers from a booklet ??

Let’s solve this step by step.

1) In a booklet, pages are numbered consecutively starting from 1.
2) When one page (which has numbers on both sides) is torn out, the sum of remaining numbers is 195.
3) Let’s say the last page number in the booklet is n.

4) Sum of all numbers from 1 to n = n(n+1)/2

5) When one page is torn out, it removes two consecutive numbers (front and back). Let’s call these numbers x and x+1.

6) So: [n(n+1)/2] – (x + (x+1)) = 195
Or: [n(n+1)/2] – (2x + 1) = 195

7) Let’s try some values:
If n = 20:
20×21/2 = 210
210 – (2x + 1) = 195
210 – 195 = 2x + 1
15 = 2x + 1
14 = 2x
x = 7

8) When x = 7, the torn page would have numbers 7 and 8.

9) Let’s verify:
– Total sum for n=20 is 210
– Removing 7 and 8 gives 210 – 15 = 195 ✓

Therefore, the torn page contains the numbers 7 and 8.

The answer is (b) 7, 8

Explanation : 

5 digit prime numbers

To determine how many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4, and 5 without repetition, we can use the fact that a number is prime if and only if it is divisible only by 1 and itself.

Next, consider the sum of the digits of any five-digit number formed using these digits. The sum is 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 3. Therefore, Zero is the answer.

Explanation : 

Sum of Digits of Intergers

1) For numbers 10-99 (excluding 100):
Tens digits: Each digit (1 to 9) appears 10 times
Sum of tens digits = (1+2+3+4+5+6+7+8+9) × 10
Sum of tens digits = 45 × 10 = 450

2) Units digits: Each digit (0 to 9) appears 9 times
Sum of units digits = (0+1+2+3+4+5+6+7+8+9) × 9
Sum of units digits = 45 × 9 = 405

3) For number 100:
Sum of its digits = 1 + 0 + 0 = 1

4) Total sum = Sum of tens digits + Sum of units digits + Sum of digits in 100
Total = 450 + 405 + 1 = 856

Therefore, the sum of all digits which appear in all the integers from 10 to 100 is 856. The answer is (b) 856.