IIT JEE Practice Question and Answer
8Q: India is a Tropical country but has a Monsoonal climate because 2476 15b5cc7d3e4d2b4197775134e
5b5cc7d3e4d2b4197775134e- 1Its real extent is vastfalse
- 2It is surrounded by the seafalse
- 3The Himalayas provide the barrier effecttrue
- 4Land and sea breeze create monsoonsfalse
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Answer : 3. "The Himalayas provide the barrier effect"
Explanation :
Answer: C) The Himalayas provide the barrier effect Explanation: The whole of India, south of the Himalayas can be climatically treated as a tropical country as the Himalayas provide the barrier effect. Indian subcontinent is separated from the rest of Asia by the lofty Himalayan ranges which block the cold air masses moving southwards from Central Asia. As a result, during winters, the northern half of India is warmer by 3°C to 8°C than other areas located on same latitudes. During summer, due to over the head position of the sun, the climate in the southern parts resemble equatorial dry climate. The seasonal reversal of winds in Arabian Sea and Bay of Bengal give India a typical tropical monsoon climate. So Indian climate, to be precise, is Tropical monsoon type (a distinct wet and dry climate) rather than just a tropical or half temperate climate.
Q: C U B A E D E D A B E B A U C D B C A D B D U B C A C B E D A If all the A's are dropped from the above arrangement, which of the following will be eleventh from the left end of the above arrangement? 2558 05b5cc7d2e4d2b419777512e0
5b5cc7d2e4d2b419777512e0- 1Efalse
- 2Cfalse
- 3Dfalse
- 4Utrue
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Answer : 4. "U"
Explanation :
Answer: D) U Explanation: If all the A's are dropped from the above arrangement, we get C U B E D E D B E B U C D B C D B D U B C C B E D. Here the eleventh from the left end of the above arrangement is U
Q: 5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible? 1703 15b5cc7d1e4d2b4197775129f
5b5cc7d1e4d2b4197775129f- 12880true
- 21440false
- 3720false
- 42020false
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Answer : 1. "2880"
Explanation :
Answer: A) 2880 Explanation: There are total 9 places out of which 4 are even and rest 5 places are odd. 4 women can be arranged at 4 even places in 4! ways. and 5 men can be placed in remaining 5 places in 5! ways. Hence, the required number of permutations = 4! x 5! = 24 x 120 = 2880
Q: How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4? 2743 05b5cc7d1e4d2b41977751295
5b5cc7d1e4d2b41977751295- 1120false
- 2360true
- 3240false
- 4424false
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Answer : 2. "360"
Explanation :
Answer: B) 360 Explanation: There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times. Number of 7 digit numbers = 7!3!×2! = 420 But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'. =6!3!×2! = 60 Hence the required number of 7 digits numbers = 420 - 60 = 360
Q: A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady? 1460 05b5cc7d1e4d2b41977751286
5b5cc7d1e4d2b41977751286- 1123false
- 2113false
- 3246true
- 4945false
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Answer : 3. "246"
Explanation :
Answer: C) 246 Explanation: A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. (i) 1 lady out of 4 and 4 gentlemen out of 6 (ii) 2 ladies out of 4 and 3 gentlemen out of 6 (iii) 3 ladies out of 4 and 2 gentlemen out of 6 (iv) 4 ladies out of 4 and 1 gentlemen out of 6 In case I the number of ways = C14×C46 = 4 x 15 = 60 In case II the number of ways = C24×C36 = 6 x 20 = 120 In case III the number of ways = C34×C26 = 4 x 15 = 60 In case IV the number of ways = C44×C16 = 1 x 6 = 6 Hence, the required number of ways = 60 + 120 + 60 + 6 = 246
Q: In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf. 2959 05b5cc7d1e4d2b4197775128b
5b5cc7d1e4d2b4197775128b- 12 x (17!)false
- 22 x (18!)true
- 3(3!) x (18!)false
- 4(17!)false
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Answer : 2. "2 x (18!)"
Explanation :
Answer: B) 2 x (18!) Explanation: A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. Required number of permutations = 18 x (17!) x 2 = 2 x 18!
Q: In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines. 1743 05b5cc7d1e4d2b41977751281
5b5cc7d1e4d2b41977751281- 1525false
- 2535true
- 3545false
- 4555false
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Answer : 2. "535"
Explanation :
Answer: B) 535 Explanation: The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B. Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535
Q: A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved? 1323 05b5cc7d0e4d2b4197775122c
5b5cc7d0e4d2b4197775122c- 11/4false
- 21/2false
- 33/4true
- 47/12false
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Answer : 3. "3/4"
Explanation :
Answer: C) 3/4 Explanation: Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event ∴A, B, C are independent events Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4 PA=12, PB=23, PC= 34 ∴ P( none solves the problem) = P(not A) and (not B) and (not C) = PA∩B∩C = PAPBPC ∵ A, B, C are Independent = 12×23×34 = 14 Hence, P(the problem will be solved) = 1 - P(none solves the problem) = 1-14= 3/4