•   The  card number is ABCDEFG.

Case 1. It is given that number is divisible by 9 i.e. sum of the digits will be divisible by 9.

•   Thus numbers can be 1+2+ 4+5+7+8+9= 36 which is divisible by 9

Case 2. Deleting two numbers from right number is divisible by 5 means last number is 5 or 0.

•   Since there is no 0 among the digits as per 1^st step, E=5.

Case 3. Deleting one number from right resulting number is divisible by 6 i.e. numbers which are divisible by both 2 and 3 are divisible by 6.

•   This means, last digit of the given number is even and the sum of its digits is a multiple of 3.

•   We have three even numbers (2, 4 and 8) and these should be at places of F, B and D in any order. No other digit would be even.

•   Thus Possible value of CD are 12, 72 and 92.

•   Now We have to find C + D + E.

•   If CD = 12, C + D + E = 1 + 2 + 5 = 8,

•   if CD = 72, C + D + E = 7 + 2 + 5 = 14,

•   if CD = 92, C + D + E = 9 + 2 + 5 = 16.

•   Only matched option is a.

•   As per question

1. Z = X + 8,

2. P + Q = 21,

3. Q = P – 5, 

4. X + Y = 28,

5. P = X + 5.

•   Solving 2 and 3 we get P = 13, Q = 8 (numbers of books purchased)

•   Putting value of P in 5 we get X = 8

•   This way we also find value of Y and Z from other equations as 20 and 16 respectively.

•   So total number of books (P + Q + X + Y + Z) = 21 + 28 + 16 = 65.

•   Each book cost 40, total cost = 65 × 40 = 2600.

•   As per the question  A = 2B, A = 3C, A = 4D, A = 5E and A = 6F.

•   Thus B = A/2, C = A/3, D = A/4, E = A/5, F = A/6.

•   Total number of pens:  A + B + C + D + E + F = A + A/2 + A/3 + A/4 + A/5 + A/6 = 147A/60.

•   Out of the given options, if we take 147A/60 = 294 i.e., A = 120,

•   we get B = 60, C = 40, D = 30, E = 24 and F = 20 all even.

•   A, B and C are three distinct non-zero digits.

•   Least such value would be possible for A, B and C as 1, 2 and 3.

•   So, number formed would be 123, 132, 213, 231, 312, 321.

•   Sum of these = 1332 .

•   Thus, statement 1 is true.

•   As the least number is in four-digit, a 3-digit number will not be possible.

•   Hence, statement 2 is not true.

•   We can calculate the population of animals after 12 years. P = 50 × 2 = 50 × (2)1 = 50 × (2)^(12/ 12)

•   After 24 years P = 50 × 2 × 2 = 50 × (2)^2 = 50 × (2)^(24/ 12).

•   Thus, we can write the equation as P = 50 × (2)^(n/ 12).

•   Thus, option (d) is correct answer.

The correct answer is Option B: 5.
To solve this problem, we need to determine how many times the ball hits the ground before it stops bouncing. The ball is initially dropped from a height of 1.5 meters and each time it bounces, it reaches 4/5th of the previous height. The ball stops bouncing when the height is less than 50 cm (0.5 meters).
     Let's calculate the height after each bounce:
     ● First Bounce:  
           ○ Initial height = 1.5 meters
           ○ Height after first bounce = \( \frac{4}{5} \times 1.5 = 1.2 \) meters
     ● Second Bounce:  
           ○ Height after second bounce = \( \frac{4}{5} \times 1.2 = 0.96 \) meters
     ● Third Bounce:  
           ○ Height after third bounce = \( \frac{4}{5} \times 0.96 = 0.768 \) meters
     ● Fourth Bounce:  
           ○ Height after fourth bounce = \( \frac{4}{5} \times 0.768 = 0.6144 \) meters
     ● Fifth Bounce:  
           ○ Height after fifth bounce = \( \frac{4}{5} \times 0.6144 = 0.49152 \) meters
     After the fifth bounce, the height is approximately 0.49152 meters, which is less than 0.5 meters. Therefore, the ball will not bounce again after the fifth hit.
     Thus, the ball hits the ground 5 times before it stops bouncing.

•   Let number of correct answered question = x

•   Let number of wronged answered question = y

•   As per the question. We get

•   x + y = 90    

•   5x - 2y = 387

•   On solving above questions, we get x= 81 and y= 9

•   Hence 9 questions are wrongly attempted.

As per question:

Expenditure = Rs. 1,200/-

Changes: Price increase of 25%, = 1200 × (125/100) = 1500/-

Person buys 6 kg less.

Suppose person buys x kg of rice at Rs. 1,200/-

Hike lead to reduce the quantity (x -6) kg.

Difference in both price is of 300/-

That means 6kg rice is of 300/-

⇒1 kg = 50/- 

Rs. 50 is price after hike and orginal price is asked.

⇒ 125% = 50

⇒ 100% = 50 × (100/125) = 40

∴ Thus orignial price of rice per kg was Rs. 40/-.

•  Let the cost of each ball is Rs. X. Then cost of each racket will be 3X.

•  Cost of 10 balls = 10X, and cost of 10 rackets = 30X.

•  So total cost = 10X + 30X = 40X.

•  By the condition given in question, we have

•  40X = 1300 + 1500 or 40X = 2800 or X = 70. Price of each racket = Rs. 210.

•  Given that x is greater than or equal to 25. Also, y is less than or equal to 40.

•  Let’s try to find the various values of y – x.

•  If y = 40, x can take various values like 25, 26, 27,...

•  In that case y – x will take values like 15, 14, 13, 12,......0, - 1 etc.

•  If y = 39, x can take various values 25, 26, 27,...

•  In that case y – x will take values like 14, 13, 12, 11 ..... 0, - 1 etc.

•  Similarly, if y = 38, values of y – x will be 13, 12, 11, ..... 0, -1 etc. and so on.

•  So, we can say that value of y – x is less than or equal to 15 in all cases.

•   Then If X is between -3, -1

•   X² is between 1 to 9

•   If Y is between -1, 1

•   Then Y² is between 0, 1

•   ⇒ X² – Y² is between 0, 9.

•   One third of the population migrates every year. If initially the population was x, after the first year it will be x – x/3 = x (1 – 1/3) = 2x/3

•   After second year, population = (2x/3) × (1 – 1/3) = (2/3)² x

•   Similarly, after sixth year, population = (2/3)⁶ x = (64/729)x, i.e. 64/729th part of the original population.

•   Let number of girls in the class be X

•   Number of boys in the class will be 2X (2/3 being boys, total being X + 2X = 3X)

•   Number of very tall boys = 3/4 * 2X = 18

•   So X = 12.

Let the Rs. 1 and Rs. 2 coins be x and y in number respectively. Given x + y = 50 and x + 2y = 75 Solving the above two equations, you get x = 25 and y = 25.