Quantitative Aptitude Practice Question and Answer
8Q: How many prime numbers exist in 65×359×113 ? 3028 05b5cc77fe4d2b419777501ab
5b5cc77fe4d2b419777501ab- 129false
- 231true
- 333false
- 427false
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Answer : 2. "31"
Explanation :
Answer: B) 31 Explanation: Given 65×359×113 ⇒2×35×5×79×113 ⇒25×35×59×79×113 Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.
Q: If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___? 3019 05b5cc6bee4d2b4197774d997
5b5cc6bee4d2b4197774d997- 114 × 13 × 13!false
- 215 × 14 × 14!false
- 314 × 12 × 12!false
- 415 × 13 × 13!true
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Answer : 4. "15 × 13 × 13!"
Explanation :
Answer: D) 15 × 13 × 13! Explanation: 15! - 14! - 13! = (15 × 14 × 13!) - (14 × 13!) - (13!) = 13! (15 × 14 - 14 - 1) = 13! (15 × 14 - 15) = 13! x 15 (14 - 1) = 15 × 13 × 13!
Q: There are 44 students in a hostel, due to the administration, 15 new students has joined. The expense of the mess increase by Rs. 33 per day. While the average expenditure per head diminished by Rs. 3, what was the original expenditure of the mess ? 3018 05b5cc74ae4d2b4197774fa9f
5b5cc74ae4d2b4197774fa9f- 1Rs. 404false
- 2Rs. 514false
- 3Rs. 340false
- 4Rs. 616true
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Answer : 4. "Rs. 616"
Explanation :
Answer: D) Rs. 616 Explanation: Let the average expenditure per head be Rs. pNow, the expenditure of the mess for old students is Rs. 44pAfter joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33Therefore, 59(p-3) = 44p + 3359p - 177 = 44p + 3315p = 210 => p = 14Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.
Q: A supplier supplies cartridges to a news paper publishing house. He earns a profit of 25% by selling cartridges for Rs. 1540. Find the cost price of the Cartridges ? 3017 05b5cc741e4d2b4197774f9a9
5b5cc741e4d2b4197774f9a9- 1Rs. 1100false
- 2Rs. 1793.4false
- 3Rs. 1440false
- 4Rs. 1232true
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Answer : 4. "Rs. 1232"
Explanation :
Answer: D) Rs. 1232 Explanation: Let Cost Price(C.P) = Pgain% = {(S.P-C.P)/C.P} x 10025 = {(1540-P)/P} x 10025/100 = (1540-P)/P=> P = 4(1540)-4P=> 5P = 4(1540)=> P = 1232So, Cost Price = Rs. 1232
Q: A bus is running at 75 kmph without any stoppage. But, with stoppages its average speed is 60 kmph. Find the time of stoppage by bus per hour. 3016 05d15ed1d5fdb2c68495d9dee
5d15ed1d5fdb2c68495d9dee- 110 minutesfalse
- 215 minutesfalse
- 312 minutestrue
- 420 minutesfalse
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Answer : 3. "12 minutes"
Q: A person takes a loan of Rs. 200 at 5% simple interest. He returns Rs.100 at the end of one year. In order clear his dues at the end of 2 years, he would pay : 3015 05b5cc77fe4d2b419777501d8
5b5cc77fe4d2b419777501d8- 1Rs. 100false
- 2Rs. 105false
- 3Rs. 115true
- 4Rs. 110false
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Answer : 3. "Rs. 115"
Explanation :
Answer: C) Rs. 115 Explanation: Amount to be paid = Rs.100+200×5×1100+100×5×1100= Rs. 115.
Q: Simplify the following : 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28 ? 3012 05b5cc6cae4d2b4197774dffc
5b5cc6cae4d2b4197774dffc- 1100true
- 21false
- 34false
- 40false
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Answer : 1. "100"
Explanation :
Answer: A) 100 Explanation: Given 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28 It is in the form of a3+b3+c3-3abca2+b2+c2-ab-bc-ca= a + b + c Here a = 37, b = 35 & c = 28 => a + b + c = 37 + 35 + 28 = 100 Therefore, = 100
Q: Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y ? 3006 05b5cc6d8e4d2b4197774e7c6
5b5cc6d8e4d2b4197774e7c6- 1360 mfalse
- 2420 mtrue
- 3460 mfalse
- 4320 mfalse
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Answer : 2. "420 m"
Explanation :
Answer: B) 420 m Explanation: Let the length of the stationary train Y be LY Given that length of train X, LX = 300 m Let the speed of Train X be V. Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively. => 300/V = 25 ---> ( 1 ) (300 + LY) / V = 60 ---> ( 2 ) From (1) V = 300/25 = 12 m/sec. From (2) (300 + LY)/12 = 60 => 300 + LY = 60 (12) = 720 => LY = 720 - 300 = 420 m Length of the stationary train = 420 m