IIT JEE Practice Question and Answer
8Q: Why are yeast cells frequently used as hosts for cloning? 2961 05b5cc690e4d2b4197774c438
5b5cc690e4d2b4197774c438- 1they are eukaryotic cellstrue
- 2they do not have plasmidsfalse
- 3only yeast cells allow the gene to be clonedfalse
- 4they easily form coloniesfalse
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Answer : 1. "they are eukaryotic cells"
Explanation :
Answer: A) they are eukaryotic cells Explanation: If you were to produce a protein in bacterial cloning systems, you would not get the same post transcriptional and post translational modifications because the prokaryotic and eukaryotic systems work differently on the modifications. Thus human recombinant protein genes are usually cloned in yeast cells. Also since they have a short generation time, they can be easily culture.
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: Three friends, P, Q and R started a partnership business investing money in the ratio of 5 : 4 : 2 respectively for a period of 3 years. What is the amount received by P as his share in the total profit ? a. Total amount invested in the business in Rs. 22,000. b. Profit earned at the end of 3 years is 3/8 of the total investment. c. The average amount of profit earned per year is Rs. 2750. 2958 05b5cc70be4d2b4197774f2e1
5b5cc70be4d2b4197774f2e1- 1Only c is sufficientfalse
- 2Both a & b are sufficientfalse
- 3Both A & B gives resulttrue
- 4Nonefalse
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Answer : 3. "Both A & B gives result"
Explanation :
Answer: C) Both A & B gives result Explanation: a and b give, profit after 3 years = Rs.(3/8 x 22000) = Rs.8250. From c also, profit after 3 years = Rs. (2750 x 3) = Rs. 8250. ∴ P's share = Rs.(8250 x 5/11) = Rs. 3750. Thus, (either C is redundant) or (a and b are redundant).
Q: A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss ? 2945 05b5cc71be4d2b4197774f4fe
5b5cc71be4d2b4197774f4fe- 1Rs. 404.80true
- 2Rs. 536.80false
- 3Rs.440false
- 4Rs. 160false
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Answer : 1. "Rs. 404.80"
Explanation :
Answer: A) Rs. 404.80 Explanation: Let C1 be the cost price of the first article and C2 be the cost price of the second article.Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%. We know, C1 + C2 = 600.The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2. The total selling price of the first and second article = 1.22C1 + 0.92C2. As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2. Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get1.22C1 + 0.92(600 - C1) = 600or 1.22C1 - 0.92C1 = 600 - 0.92*600or 0.3C1 = 0.08*600 = 48or C1 = 48/(0.3) = 160.If C1 = 160, then C2 = 600 - 160 = 440.The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.
Q: The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8,748, its purchase price was ? 2942 05b5cc75ce4d2b4197774fc2f
5b5cc75ce4d2b4197774fc2f- 1Rs.14,000false
- 2Rs.12,000true
- 3Rs. 12,800false
- 4Rs. 15,000false
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Answer : 2. "Rs.12,000"
Explanation :
Answer: B) Rs.12,000 Explanation: We have P = 8748, R = -10 and n = 3Therefore, the purchase price of the machine= P1+R100n= 87481-101003= 8748 x (100/90) x (100/90) x (100/90)= 12,000. Therefore, the purchase price of the machine was Rs. 12,000.
Q: In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ? 2931 05b5cc74fe4d2b4197774fb35
5b5cc74fe4d2b4197774fb35- 1145false
- 2165false
- 3175true
- 4135false
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Answer : 3. "175"
Explanation :
Answer: C) 175 Explanation: Given that, working days = 5 working hours = 8 A man get rupess per hour is Rs.2.40 So in one day the man get total rupees is 2.40 x 8 = 19.2 So in 5 days week the man get total rupees is 19.2 x 5 = 96 So in 4 week the man get total rupees is 96 x 4 = 384 So the man worked for = 160hours in 4 weeks But given that the man earned Rs.432 Hence remaning money is (432-384 = 48)which is earn by doing overtime work Overtime hours = 48/3.20 = 15 So total worked hours is = 15 + 160 = 175.
Q: A does half as much work as Band C does half as much work as A and B together. If C alone can finish the work in 40 days, then together ,all will finish the work in ? 2925 05b5cc728e4d2b4197774f714
5b5cc728e4d2b4197774f714- 117 + 4/7 daysfalse
- 213 + 1/3 daystrue
- 315 + 3/2 daysfalse
- 416 daysfalse
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Answer : 2. "13 + 1/3 days"
Explanation :
Answer: B) 13 + 1/3 days Explanation: C alone can finish the work in 40 days.As given C does half as much work as A and B together => (A + B) can do it in 20 days(A + B)s 1 days wok = 1/20.A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]B's 1 days work = (1/20) x (2/3) = 1/30(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40 All the three together will finish it in 40/3 = 13 and 1/3 days.
Q: If 6 engines consume 24 metric tonnes of coal, when each is working 8 hours day, how much coal will be required for 9 engines, each running 13hours a day, it being given that 2 engines of former type consume as much as 3 engines of latter type ? 2925 05b5cc723e4d2b4197774f672
5b5cc723e4d2b4197774f672- 145 metric tonnestrue
- 247 metric tonnesfalse
- 355 metric tonnesfalse
- 434 metric tonnesfalse
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Answer : 1. "45 metric tonnes"
Explanation :
Answer: A) 45 metric tonnes Explanation: 2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton i.e. 3 engines of latter type consumes 1 ton for one hour Hence 9 engines consumes 3 tons for one hour For 15 hours it is 15 x 3 = 45 metric tonnes.