IIT JEE Practice Question and Answer
8Q: A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red. 1381 05b5cc7d0e4d2b4197775121d
5b5cc7d0e4d2b4197775121d- 123/42false
- 219/42true
- 37/32false
- 416/39false
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Answer : 2. "19/42"
Explanation :
Answer: B) 19/42 Explanation: A red ball can be drawn in two mutually exclusive ways (i) Selecting bag I and then drawing a red ball from it. (ii) Selecting bag II and then drawing a red ball from it. Let E1, E2 and A denote the events defined as follows: E1 = selecting bag I, E2 = selecting bag II A = drawing a red ball Since one of the two bags is selected randomly, therefore P(E1) = 1/2 and P(E2) = 1/2 Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7 PAE2 = Probability of drawing a red ball when the second bag has been selected = 2/6 Using the law of total probability, we have P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 = 12×47+12×26=1942
Q: Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected? 1451 05b5cc7d0e4d2b41977751222
5b5cc7d0e4d2b41977751222- 15/7false
- 21/5false
- 32/7true
- 42/35false
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Answer : 3. "2/7"
Explanation :
Answer: C) 2/7 Explanation: P( only one of them will be selected) = p[(E and not F) or (F and not E)] = PE∩F∪F∩E = PEPF+PFPE =17×45+15×67=27
Q: A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is : 1565 05b5cc7d0e4d2b41977751213
5b5cc7d0e4d2b41977751213- 135/96false
- 219/90true
- 319/96false
- 4None of thesefalse
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Answer : 2. "19/90"
Explanation :
Answer: B) 19/90 Explanation: ASSISTANT→AAINSSSTT STATISTICS→ACIISSSTTT Here N and C are not common and same letters can be A, I, S, T. Therefore Probability of choosing A = 2C19C1×1C110C1 = 1/45 Probability of choosing I = 19C1×2C110C1 = 1/45 Probability of choosing S = 3C19C1×3C110C1 = 1/10 Probability of choosing T = 2C19C1×3C110C1 = 1/15 Hence, Required probability = 145+145+110+115= 1990
Q: 8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is : 2905 05b5cc7d0e4d2b41977751218
5b5cc7d0e4d2b41977751218- 18/39false
- 215/39true
- 312/13false
- 4None of thesefalse
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Answer : 2. "15/39"
Explanation :
Answer: B) 15/39 Explanation: P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) = 1-16C1× 14C1×12C1×10C116C4=1539
Q: What is the number of digits in 333? Given that log3 = 0.47712? 5527 05b5cc7d0e4d2b41977751209
5b5cc7d0e4d2b41977751209- 112false
- 213true
- 314false
- 415false
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Answer : 2. "13"
Explanation :
Answer: B) 13 Explanation: Let Let x=333 = 333 Then, logx = 33 log3 = 27 x 0.47712 = 12.88224 Since the characteristic in the resultant value of log x is 12 ∴The number of digits in x is (12 + 1) = 13 Hence the required number of digits in 333is 13.
Q: Find value of log27 +log 8 +log1000log 120 1698 15b5cc7d0e4d2b4197775120e
5b5cc7d0e4d2b4197775120e- 11/2false
- 23/2true
- 32false
- 42/3false
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Answer : 2. "3/2"
Explanation :
Answer: B) 3/2 Explanation: = log 33 + log 23+ log 103log10×3×22 =log33 12+log 23+log 10312log(10×3×22) =12log 33+3 log 2+12 log103log10+log3+log22 =32log 3 + 2 log 2 + log 10log 3 + 2 log 2 + log 10 = 32
Q: For x∈N, x>1, and p=logxx+1, q=logx+1x+2 then which one of the following is correct? 6917 25b5cc7d0e4d2b419777511ff
5b5cc7d0e4d2b419777511ff- 1p < qfalse
- 2p = qfalse
- 3p > qtrue
- 4can't be determinedfalse
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Answer : 3. "p > q"
Explanation :
Answer: C) p > q Explanation: kl>k+1l+1 for (k,l) > 0 and k > l Let k = x+1 and l = x Therefore, x+1x>(x+1)+1(x)+1 (x + 1) > x Therefore, log(x+1)log(x)>log(x+2)log(x+1) ⇒logxx+1 >logx+1x+2
Q: The Value of logtan10+logtan20+⋯⋯+logtan890 is 1447 15b5cc7d0e4d2b41977751204
5b5cc7d0e4d2b41977751204- 1-1false
- 20true
- 31/2false
- 41false
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Answer : 2. "0"
Explanation :
Answer: B) 0 Explanation: = log tan10+log tan890 + log tan20+ log tan880+⋯⋯+log tan450 = log [tan10 × tan890] + log [tan20 × tan880 ] +⋯⋯+log1 ∵ tan(90-θ)=cotθ and tan 450=1 = log 1 + log 1 +.....+log 1 = 0.