GATE Practice Question and Answer
8Q: Kamal consistently runs 240 meters a day and on Saturday he runs for 400 meters. How many kilometers will he have to run in four weeks ? 2217 05b5cc6c5e4d2b4197774dd23
5b5cc6c5e4d2b4197774dd23- 15.75 kmsfalse
- 27.36 kmstrue
- 38.2 kmsfalse
- 46.98 kmsfalse
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Answer : 2. "7.36 kms"
Explanation :
Answer: B) 7.36 kms Explanation: Total running distance in four weeks = (24 x 240) + (4 x 400) = 5760 + 1600 = 7360 meters = 7360/1000 => 7.36 kms
Q: A can build a wall in 16 days while B can destroy it in 8 days. A worked for 5 days. Then B joined with A for the next 2 days. Find in how many days could A build the remaining wall ? 2217 05b5cc6cee4d2b4197774e208
5b5cc6cee4d2b4197774e208- 112 daysfalse
- 214 daysfalse
- 313 daystrue
- 416 daysfalse
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Answer : 3. "13 days"
Explanation :
Answer: C) 13 days Explanation: Now, Total work = LCM(16, 8) = 48 A's one day work = + 48/16 = + 3 B's one day work = - 48/8 = -6 Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15 2days B joined with A in working = 2(3 - 6) = - 6 Remaining Work of building wall = 48 - (15 - 6) = 39 Now this remaining work will be done by A in = 39/3 = 13 days.
Q: Rajeev's age after 15 years will be 5 times his age 5 years back. What is the present age of rajeev? 2205 05b5cc655e4d2b4197774bd70
5b5cc655e4d2b4197774bd70- 112false
- 214false
- 322false
- 410true
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Answer : 4. "10"
Explanation :
Answer: D) 10 Explanation: Let Rajeev's present age be x years. Rajeev's age after 15 years = (x + 15) years. Rajeev's age 5 years back = (x - 5) years Then ATQ, x + 15 = 5 (x - 5) x + 15 = 5x - 25 => x = 10 Hence, Rajeev's present age = x = 10 years.
Q: Here are some words translated from an artificial language. briftamint means 'militant'uftonel means 'occupied'uftonalene means 'occupation' Which word could mean "Occupant"? 2202 05b5cc76ae4d2b4197774fe7f
5b5cc76ae4d2b4197774fe7f- 1elamintfalse
- 2briftalenefalse
- 3uftonaminttrue
- 4elbriftafalse
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Answer : 3. "uftonamint"
Explanation :
Answer: C) uftonamint Explanation: Brift means the root word mili; the suffix amint means the same as the English suffix –tant; the root word ufton– means occupy; el means the suffix –ied of occupied; and alene means the suffix –tion. (Because ufton means occupy, choices a, b, and d can be easily ruled out.)
Q: Which of these is unique to flowering plants? 2198 05b5cc67be4d2b4197774c1d6
5b5cc67be4d2b4197774c1d6- 1double fertilizationtrue
- 2an embryo surrounded by nutritive tissuefalse
- 3haploid gametophytesfalse
- 4pollen productionfalse
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Answer : 1. "double fertilization"
Explanation :
Answer: A) double fertilization Explanation: In flowering plants one sperm nucleus fertilizes the egg and the other sperm nucleus fuses with two other nuclei found within the ovule, thus forming triploid endosperm. Hence, double fertilization is unique to flowering plants.
Q: Which of the following is generally true about the stratosphere? 2198 05b5cc6a8e4d2b4197774ce80
5b5cc6a8e4d2b4197774ce80- 1It is very drytrue
- 2It is very moistfalse
- 3The dryness level varies with the seasonfalse
- 4The dryness level varies with the moon cyclefalse
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Answer : 1. "It is very dry"
Explanation :
Answer: A) It is very dry Explanation: The stratosphere layer of the atmosphere is very dry.
Q: Missing Letter Challenge 2194 05b5cc6d3e4d2b4197774e554
5b5cc6d3e4d2b4197774e554- 1Ifalse
- 2Mfalse
- 3Pfalse
- 4Ltrue
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Answer : 4. "L"
Explanation :
Answer: D) L Explanation: Here the letters follow the rule (AxZ)+(A+Z) The number corresponding to the letters are used in the rule A=1, Z=26 => 26x1 + 1 + 26 = 26+27 = 53 = 053 (LxP)+(L+P) = (12 x16) + 12 + 16 = 192+28 = 220 Similarly (11x12) + 11 + 12 => 132 + 23 = 155
Q: Select a suitable figure from the four alternatives that would complete the figure matrix. 2193 05b5cc76de4d2b4197774ff03
5b5cc76de4d2b4197774ff03- 12true
- 24false
- 33false
- 41false
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Answer : 1. "2"
Explanation :
Answer: A) 2 Explanation: In each row, the second figure is obtained from the first figure by adding two mutually perpendicular line segments at the centre and the third figure is obtained from the first figure by adding four circles outside the main figure.

