Directions (1-5): What will come in place in ? in the following questions?

- 38% of 4500 – 25% of ? = 1640

A) 240

B) 265

C) 280

D) 320

E) None of these

- (0.64)
^{4}÷ (0.512)^{3}× (0.8)^{4}= (0.8)^{? + 3}

A) 2

B) 1

C) 0

D) 4

E) None of these

- 3648.24 + 364.824 ÷ ? – 36.4824 = 3794.1696

A) 2

B) 6

C) 4

D) 5

E) 8

- 42% of ? = 3147 + 4258 – 7212 ÷ 12

A) 16200

B) 15400

C) 18600

D) 12800

E) 14400

- 16121.50 – 48.25 + 3423.75 – 12016 + 512 ÷ 8 = ?

A) 7545

B) 7555

C) 7575

D) 7495

E) 7515

- A box contains 6 black and 14 white balls, out of which 3 black and 5 white balls are defective. If we choose two balls at random, what is the probability that either both are white or both are non-defective?

A) 141/190

B) 123/190

C) 121/190

D) 123/199

E) None of these

- In a class, the average age of some boys is 16 years, and average age of 16 teachers is 56 years. If the average age of the combined group of all the teachers and boys is 20, then the number of students is

A) 156

B) 144

C) 136

D) 88

E) 168

- If the CI on a certain sum for 2 yrs at 10% per annum is Rs. 3150, what would be the SI on same rate for same time?

A) 2500

B) 3000

C) 2800

D) 2200

E) None of these

- A started a business with initial investment of rs.12000, after 3 month B invest rs.15000 in this business. After 8 month from starting A withdrew one-fourth of his investment and B further invest 1/15 part of his investment. If at the end of one year the difference between the shares of profit of both is 700, what is the B’s profit share?

A) 14550

B) 13550

C) 13900

D) 42000

E) 41250

- There are three taps A, B, and C. A takes thrice as much time as B and C together to fill the tank. B takes twice as much time as A and C to fill the tank. In how much time can the Tap C fill the tank individually, if they would require 10 hours to fill the tank, when opened simultaneously?

A) 16 hours

B) 12 hours

C) 22 hours

D) 20 hours

E) 24 hours

Hello sir pls explain Q-6 ….pls reply KRIYEGA…. 3 ko exam h

there are total 14 white balls so – 14C2/20C2

there are total 12 non-defective balls so – 12C2/20C22

Now we have to subtract that probability in which both white and non-defective balls are there.

So there are total 9 non-defective white balls – 9C2/20C2

14C2/20C2 + 12C2/20C22 – 9C2/20C2 = 121/190

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