What approximate value should come in place of question mark (?) in the following questions? (Note: You are not expected to calculate the exact value.)

1. 0.003 * 0.9 * 0.005 * 0.2 + 0.008 * 0.5 + 23.85 21.05 = ?a. 17b. 11c. 3d. 5e. 6

Solution :- (3) 0.003 * 0.9 * 0.005 * 0.2 + 0.008 * 0.5 + 23.85 21.05= 0.0027 * 0.0001 + 0.0004 + 23.85 21.05= 0.0000027 + 0.004 + 2.8= 2.8040027 3

2. (2356.237 * 4.5) 1357.895 + 1124.237 425.231 + (35 * 0.23) = ?

a. 9052b. 9952c. 11735d. 10952e. 9852

Solution :- (B) (2356.237 * 4.5) 1356.895 + 1124.237 425.231 + (35 * 0.23)10603 1357 + 1124 425 +8.0511735 1782 9952

3. 2222.1 * 11 + 3333.1 * 11.01 + 4444 * 11 + 5555 * 11 6666.1 * 11 + 333 * 121 = ?

a. 130861b. 136161c. 138061d. 149061e. 159061

Answer :- C

4. 472.05 *101.32+337+472137* 0.5/2 = ?

a. 48447b. 55342c. 58947d. 40132e. 35000

Answer :- ASolution:-?= 472.05*101.32+337+472137*0.5/2472.05*101.32+337+472137* 0.2547672 + 809 34 48447

In each question a number series is given. In each series only one number is wrong. Find out that number

5. 25/12, 41/20, 61/30, 85/40, 113/56, 145/72, 181/90a. 25/12b. 61/30c. 113/56d. 181/90e. 85/40

Solution :- E

The series is (3^2 + 4^2)/3*4, (4^2+5^2)/4*5, (5^2+6^2)/5*6, . So, 25/12, 41/20, 61/30, 85/42, 113/56, 145/72, 181/90

Hence, it shud be 85/42 in place of 85/40.

6. 14, 39, 84 ,156 ,258 ,399 ,584a. 14b. 156c. 84d. 258e. 584

Solution :- B The series is 2^3+2^2+2, 3^3+3^2+3, 4^3+4^2+4, 5^3+5^2+5, 6^3+6^2+6, so on Thus, 14, 39, 84, 155, 258, 399, 584.Hence, there should be 155 in place of 156.

7. 421875, 438976, 456533 ,474551 ,493039 ,512000 ,531441

a. 421875b. 493039c. 474551d. 531441e. 512000

Answer :- C The Series Is (75)^3, 76^3, 77^3, 78^3, 79^3, 80^3, 81^3

There Shud Be 474552 in place of 474551

8. 9/5, 35/13, 94/25, 189/41, 341/61, 559/85, 855/113a. 9/5b. 35/13c. 94/25d. 341/61e. 559/85

Answer :- CThe Series Is (1^3+2^3)/(1^2+2^2), (2^3+3^3)/(2^2+3^2), (3^3+4^3)/(3^2+4^2), and so onThe series is 9/5, 35/13, 91/25, 189/41, 341/61, 559/85, 855/113Hence, there should be 91/25 in place of 94/25 9. A bag contains 8 red and 6 blue balls. If 5 balls are drawn at random, what is the probability that 3 arered and 2 are blue?a. 60/143b. 50/143c. 30/143d. 20/143e. None Of These

Answer :- AUsing formula,Probability =( No. of selections with restriction)/(No. of selection without restriction)= (n1Cr1 * n2Cr2)/ {(n1+n2)C(r1+r2)}[Selections of red and blue balls are taking place togethr so, these are multiplied]

Where , n1=8, r1=3 , n2=6, r2=2Solving Probability = 60/143

10. Two cards are drawn from a pack of 52 cards. Find the probability that both are red cards.a. 10/102b. 20/102c. 30/102d. 25/102e. None Of These

Answer :- D

Required Probability = (26C2)/(52C2)= (26*25)/(52*51)= 25/102

11. I. X^2 1 = 0II. y^2 + 4y + 3 = 0

1. If x > y 2. If x>y 3. If x < y4. If x=Y

12. I. x^2 7x + 12 = 0II. y^2 12y + 32 = 0 1. If x > y 2. If x>y 3. If x < y4. If xy 3. If x < y4. If x y 2. If x>y 3. If x < y4. If x y 2. If x>y 3. If x < y4. If x