CAT Practice Question and Answer
8Q: The basic operations performed by a computer are 3261 05b5cc651e4d2b4197774bcea
5b5cc651e4d2b4197774bcea- 1Logical operationfalse
- 2Storage and relativefalse
- 3Arithmetic operationfalse
- 4All the abovetrue
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Answer : 4. "All the above"
Explanation :
Answer: D) All the above Explanation: The basic operations performed by a computer are Arithmetic operation, Logical operation and Storage and relative.
Q: The intersection of a column and a row 3257 05b5cc651e4d2b4197774bcdb
5b5cc651e4d2b4197774bcdb- 1Celltrue
- 2Menufalse
- 3Keyfalse
- 4Fieldfalse
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Answer : 1. "Cell"
Explanation :
Answer: A) Cell Explanation: The intersection of a column and a row in a spread sheet is called as a cell or a bon.
Q: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained ? 3257 05b5cc6dce4d2b4197774e960
5b5cc6dce4d2b4197774e960- 15:3false
- 21:4false
- 34:1false
- 49:1true
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Answer : 4. "9:1"
Explanation :
Answer: D) 9:1 Explanation: Milk = 3/5 x 20 = 12 liters, water = 8 liters If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters. Remaining milk = 12 - 6 = 6 liters Remaining water = 8 - 4 = 4 liters 10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters. The ratio of milk and water in the new mixture = 16:4 = 4:1 If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 x 10 = 8 liters. Amount of water removed = 2 liters. Remaining milk = (16 - 8) = 8 liters. Remaining water = (4 -2) = 2 liters. Now 10 lts milk is added => total milk = 18 lts The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
Q: Programs stored in ROM are called 3256 15b5cc6a9e4d2b4197774ceef
5b5cc6a9e4d2b4197774ceef- 1Hardwarefalse
- 2Firmwaretrue
- 3Freewarefalse
- 4Softwarefalse
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Answer : 2. "Firmware"
Explanation :
Answer: B) Firmware Explanation: The programs which are as permanent as hardware and stored in ROM are known as Firmware. Data stored in ROM can only be modified slowly, with difficulty, or not at all, so it is mainly used to store firmware (software that is closely tied to specific hardware, and unlikely to need frequent updates) or application software in plug-in cartridges.
Q: Complete the next number in the given number series? 21 56 102 160 231 316 3255 05b5cc5ffe4d2b4197774b436
5b5cc5ffe4d2b4197774b436- 1416true
- 2415false
- 3414false
- 4413false
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Answer : 1. "416"
Explanation :
Answer: A) 416 Explanation: The given number series is 21 56 102 160 231 316 21 56 102 160 231 316 416 +35 +46 +58 +71 +85 +100 +11 +12 +13 +14 +15 Hence, the next number in the given number series is 316 + 100 = 416.
Q: The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles ? 3253 05b5cc6dee4d2b4197774ea79
5b5cc6dee4d2b4197774ea79- 12413 sq.mfalse
- 21234 sq.mfalse
- 34312 sq.mtrue
- 42143 sq.mfalse
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Answer : 3. "4312 sq.m"
Explanation :
Answer: C) 4312 sq.m Explanation: Let the radii of the smaller and the larger circles be 's' m and 'l' m respectively. 2πs = 264 and 2πl = 352 s = 2642π and l = 3522π Difference between the areas =πl2-πs2 = π1762π2-1322π2 = 1762π-1322π = (176 - 132)(176 + 132) / π = (44 x 308) / (22/7) = 4312 sq.m
Q: Where does the series number on a map appear? 3252 05b5cc6a0e4d2b4197774ca8b
5b5cc6a0e4d2b4197774ca8b- 1Upper right marginfalse
- 2Lower left marginfalse
- 3Both A & Btrue
- 4Not Existfalse
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Answer : 3. "Both A & B"
Explanation :
Answer: C) Both A & B Explanation: The series number is found in both the upper right margin and the lower left margin. It is a sequence reference expressed either as a four-digit numeral (1125) or as a letter, followed by a three- or four-digit numeral (M661, T7110).
Q: Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest ? 3249 05b5cc6ebe4d2b4197774eeb9
5b5cc6ebe4d2b4197774eeb9- 120% p.atrue
- 215% p.afalse
- 318% p.afalse
- 424% p.afalse
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Answer : 1. "20% p.a"
Explanation :
Answer: A) 20% p.a Explanation: Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year. Rate of interest = (100 x 240) / (1200) = 20% p.a

