Quantitative Aptitude Practice Question and Answer
8Q: Sushma joins a job at 20 yrs. First 3 years salary is 10,000 p.a. Afterwards every year increment of 2,000 per year for 10 years. Then salary becomes constant till retirement. At retirement average salary is 25,000. (throughout career). At what age Sushma retires ? 2240 05b5cc711e4d2b4197774f3be
5b5cc711e4d2b4197774f3be- 150 yrstrue
- 247 yrsfalse
- 345 yrsfalse
- 442 yrsfalse
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Answer : 1. "50 yrs"
Explanation :
Answer: A) 50 yrs Explanation: Let age of retirement = Sso according to the given condition 25000(x-20)=10000 x 3 + 12000 + 14000 + 16000 + 18000 + 20000 + 22000 + 24000 + 26000 + 28000 + 30000 + 30000(S-33)= 30000 + 210000 + 30000S-990000= 30000S - 750000or 25000S - 500000 = 30000x - 750000or 5000x = 250000or S = 50 yrs.
Q: Two trains of 440 metre and 260 metre lengths are running in the same direction and crosses each other in 35 sec. If the speed of slower train is 63 km/hr., then what is the speed of the faster train? 2239 05d15f90f5fdb2c68495d9df4
5d15f90f5fdb2c68495d9df4- 1135 km/hourtrue
- 2125 km/hourfalse
- 3138 km/hourfalse
- 4250 km/hourfalse
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Answer : 1. "135 km/hour"
Q: 5% people of a village in India died by dengue, 15% of the remainder left the village on account of fear. If now the population is reduced to 3553, how much was it in the beginning ? 2225 05b5cc6e0e4d2b4197774eb5c
5b5cc6e0e4d2b4197774eb5c- 14400true
- 23100false
- 32800false
- 45600false
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Answer : 1. "4400"
Explanation :
Answer: A) 4400 Explanation: Let 'P' be the original population at the beginning P x (95/100) x (85/100) = 3553 P = 4400.
Q: A contractor undertook to complete a piece of work in 120 Days and employed 140 men upon it. At the end of 66 days only half of the work was done,so he put on 25 extra men. By how much time did he exceed the specific time ? 2225 05b5cc709e4d2b4197774f2ad
5b5cc709e4d2b4197774f2ad- 12 daystrue
- 23 daysfalse
- 34 daysfalse
- 45 daysfalse
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Answer : 1. "2 days"
Explanation :
Answer: A) 2 days Explanation: work done=total number of person x number of days;half of work done = 140 x 66;For half of remaining work 25 extra men are added.Therefore, total men for half work remaining = 140 + 25 = 165;Let 2nd half work will be completed in K days;140 x 66 = 165 x KK = 122;Hence, extra days => 122-120 = 2days.
Q: Find the next number in the given number series? 96, 95, 87, 78, ? 2223 05b5cc60ae4d2b4197774b5c9
5b5cc60ae4d2b4197774b5c9- 164false
- 257false
- 329false
- 414true
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Answer : 4. "14"
Explanation :
Answer: D) 14 Explanation: The given number series is 96, 95, 87, 78, ? It follows a pattern that, 9696 - 12 = 9595 - 23 = 95 - 8 = 8787 - 32 = 87 - 9 = 7878 - 43 = 78 - 64 = 14 Hence, the next number in the given number series is 14.
Q: How many 6-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the digits should not repeat and the second last digit is even ? 2223 05b5cc731e4d2b4197774f7bf
5b5cc731e4d2b4197774f7bf- 1521false
- 2720true
- 3420false
- 4225false
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Answer : 2. "720"
Explanation :
Answer: B) 720 Explanation: Let last digit is 2when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.so for last digit = 2, total numbers=240 Similarly for 4 and 6When last digit = 4, total no. of ways =240and last digit = 6, total no. of ways =240so total of 720 even numbers are possible.
Q: The minimum number of half years in which a sum of money kept at 60% compound interest will be quadrupled is ? 2214 05b5cc77fe4d2b4197775016f
5b5cc77fe4d2b4197775016f- 12false
- 24false
- 33false
- 46true
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Answer : 4. "6"
Explanation :
Answer: D) 6 Explanation: Let us assume Amount be 100 Rs and we calculate in CI First year 60% of 100 = 60 amount (100+60) is 160 Second year 60% of 160 = 96 amount (160+96) is 256 Third year 60% of 256 =153.6 amount (256+153.6) is 409.6 Here the Amount of 100 Rs is quadrapled in 3 years. One year contains 2 half years Three year has 6 half years. Therefore, It takes 6 half years.
Q: How many zeros are there from 1 to 10000 ? 2200 05b5cc741e4d2b4197774f9e0
5b5cc741e4d2b4197774f9e0- 12893true
- 24528false
- 36587false
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Answer : 1. "2893"
Explanation :
Answer: A) 2893 Explanation: For solving this problem first we would break the whole range in 5 sections 1) From 1 to 9Total number of zero in this range = 0 2) From 10 to 99Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer) 3) From 100 to 999 - three type of numbers are there in this range a) x00 b) x0x c) xx0 (here x represents a non zero number)Total possibilities for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18for x0x = 9*1*9 = 81, hence total zeros = 81similarly for xx0 = 81 total zeros in three digit numbers = 18 + 81 +81 = 180 4) From 1000 to 9999 - seven type of numbers are there in this rangea)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xxTotal possibilities for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27 for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162 for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162 for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)= 2893