GATE Practice Question and Answer
8Q: That which cannot be corrected 2598 05b5cc6a6e4d2b4197774cd7a
5b5cc6a6e4d2b4197774cd7a- 1Incorrigibletrue
- 2Illegiblefalse
- 3Illegalfalse
- 4Indeliblefalse
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Answer : 1. "Incorrigible"
Explanation :
Answer: A) Incorrigible Explanation: The One-word substitute for That which cannot be corrected is Incorrigible.
Q: 7, 11, 19, 35, ? Find the next number in the given number series? 2595 05b5cc6ace4d2b4197774d0bb
5b5cc6ace4d2b4197774d0bb- 1131false
- 294false
- 383false
- 467true
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Answer : 4. "67"
Explanation :
Answer: D) 67 Explanation: Here the given series 7, 11, 19, 35, ? follows a pattern that (x 2 - 3) i.e, 7 7 x 2 - 3 = 11 11 x 2 - 3 = 19 19 x 2 - 3 = 35 35 x 2 - 3 = 67 67 x 2 - 3 = 131 Hence the next number in the given number series is 67.
Q: If log72 = m, then log4928 is equal to ? 2590 05b5cc761e4d2b4197774fd5d
5b5cc761e4d2b4197774fd5d- 11/(1+2m)false
- 2(1+2m)/2true
- 32m/(2m+1)false
- 4(2m+1)/2mfalse
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Answer : 2. "(1+2m)/2"
Explanation :
Answer: B) (1+2m)/2 Explanation: log4928 = 12log77×4 = 12+122log72= 12+log72= 12 + m= 1+2m2.
Q: Which unit holds data temporarily? 2590 05b5cc64de4d2b4197774bcb2
5b5cc64de4d2b4197774bcb2- 1Secondary storage unitfalse
- 2Output Unitfalse
- 3Primary Memory Unittrue
- 4Input unitfalse
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Answer : 3. "Primary Memory Unit"
Explanation :
Answer: C) Primary Memory Unit Explanation: Memory is the part of the computer that temporarily holds data and instructions before and after they are processed by the ALU. Memory is also known as primary storage, primary memory, main storage, internal storage, and main memory. Manufacturers often use the term RAM, which stands for random-access memory.
Q: Fill in the blank with suitable preposition in the following sentence. ____ the two I prefer tea. 2583 05b5cc642e4d2b4197774bb86
5b5cc642e4d2b4197774bb86- 1Fromfalse
- 2Infalse
- 3Betweentrue
- 4Amongfalse
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Answer : 3. "Between"
Explanation :
Answer: C) Between Explanation: Between the two I prefer tea is the correct sentence with suitable preposition.
Q: A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball ? 2581 05b5cc6f4e4d2b4197774f004
5b5cc6f4e4d2b4197774f004- 124 - 1false
- 22425-1false
- 3(24-1)(23-1)25true
- 4Nonefalse
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Answer : 3. "(24-1)(23-1)25"
Explanation :
Answer: C) C Explanation: It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects. Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection. Hence, we can select 1 black ball from 4 black ballsor 2 black balls from 4 black balls.or 3 black balls from 4 black balls.or 4 black balls from 4 black balls. Hence, number of ways in which we can select the black balls = 4C1 + 4C2 + 4C3 + 4C4= 24-1 ........(A) Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection. Hence, we can select 1 red ball from 3 red ballsor 2 red balls from 3 red ballsor 3 red balls from 3 red balls Hence, number of ways in which we can select the red balls= 3C1 + 3C2 + 3C3=23-1........(B) Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)or 1 blue ball from 5 blue ballsor 2 blue balls from 5 blue ballsor 3 blue balls from 5 blue ballsor 4 blue balls from 5 blue ballsor 5 blue balls from 5 blue balls. Hence, number of ways in which we can select the blue balls= 5C0 + 5C1 + 5C2 + … + 5C5= 25..............(C) From (A), (B) and (C), required number of ways= 2524-123-1
Q: The circumference of a circular base of Cylinder is 44 cm and its hight is 15 cm. Then the volume of the cylinder is? 2579 05b5cc6ade4d2b4197774d109
5b5cc6ade4d2b4197774d109- 12759 cub. cmfalse
- 22247 cub. cmfalse
- 32614 cub. cmfalse
- 42311 cub. cmtrue
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Answer : 4. "2311 cub. cm"
Explanation :
Answer: D) 2311 cub. cm Explanation: Given Circumference of a circular base of a cylinder = 44 cm We know that Circumference of a circle = 2πr = 44⇒r = 22π Now given the height of a cylinder h = 15 cm Volume of a cylinder V = πr2h = π22π215 = 22 x 22 x 153.14 = 2311 cm3.
Q: Size of the primary memory of a pc ranges between 2579 05b5cc694e4d2b4197774c4a2
5b5cc694e4d2b4197774c4a2- 1256 kb to 640 kbtrue
- 264 kb to 256 kbfalse
- 3256 kb to 512 kbfalse
- 42 kb to 8 kbfalse
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Answer : 1. "256 kb to 640 kb"
Explanation :
Answer: A) 256 kb to 640 kb Explanation: Size of the primary memory of a PC ranges between 256 KB to 640 KB.

