Quantitative Aptitude Practice Question and Answer
8Q: An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%? 1504 05b5cc6aee4d2b4197774d179
5b5cc6aee4d2b4197774d179- 145 gmfalse
- 250 gmtrue
- 355 gmfalse
- 460 gmfalse
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Answer : 2. "50 gm"
Explanation :
Answer: B) 50 gm Explanation: Initial quantity of copper =80100 x 50 = 40 g And that of Bronze = 50 - 40 = 10 g Let 'p' gm of copper is added to the mixture => 50 + p x 90100 = 40 + p => 45 + 0.9p = 40 + p => p = 50 g Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
Q: A man walks at a speed of 2 km/hr and runs at a speed of 6 km/hr. How much time will the man require to cover a distance of 20 1/2 km, if he completes half of the distance, i.e., (10 1/4) km on foot and the other half by running ? 1501 05b5cc6dfe4d2b4197774eaf8
5b5cc6dfe4d2b4197774eaf8- 112.4 hrsfalse
- 211.9 hrstrue
- 310.7 hrsfalse
- 49.9 hrsfalse
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Answer : 2. "11.9 hrs"
Explanation :
Answer: B) 11.9 hrs Explanation: We know that Time = Distance/speedRequired time = (10 1/4)/2 + (10 1/4)/6 = 41/8 + 41/6= 287/24 = 11.9 hours.
Q: The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is 1501 05b5cc7b4e4d2b419777508a1
5b5cc7b4e4d2b419777508a1- 17! x 7!false
- 27! x 6!true
- 36! x 6!false
- 47! x 5!false
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Answer : 2. "7! x 6!"
Explanation :
Answer: B) 7! x 6! Explanation: The students should sit in between two teachers. There are 7 gaps in between teachers when they sit in a roundtable. This can be done in 7P6ways. 7 teachers can sit in (7-1)! ways. Required no.of ways is = 7P6.6! = 7!.6!
Q: The length of two superfast trains are 140 mts and 160 mts respectively. If they run at the speed of 60 km/hr and 80 km/hr respectively in opposite direction, find the time in which they will cross each other ? 1501 05b5cc6cde4d2b4197774e1ae
5b5cc6cde4d2b4197774e1ae- 17.71 sectrue
- 210.48 secfalse
- 39.36 secfalse
- 48.45 secfalse
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Answer : 1. "7.71 sec"
Explanation :
Answer: A) 7.71 sec Explanation: Given L1 = 140 m L2 = 160 m S1 = 60 km/hr S2 = 80 km/hr From the question we get, S1 + S2 = (L1 + L2) / T => (60 + 80) 5/18 m/s = 140 + 160/T => T = 54/7 = 7.71 sec
Q: Question : What is the sum which earned interest ? Statements : a. The total simple interest was Rs. 9000 after 9 years.b. The total of sum and simple interest was double of the sum after 6 years. 1497 05b5cc767e4d2b4197774fe10
5b5cc767e4d2b4197774fe10- 1Only a is sufficientfalse
- 2Neither a nor b is sufficientfalse
- 3Only b is sufficientfalse
- 4Both a and b sufficienttrue
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Answer : 4. "Both a and b sufficient"
Explanation :
Answer: D) Both a and b sufficient Explanation: Let the sum be Rs. x a. gives, S.I = Rs. 9000 and time = 9 years. b. gives, Sum + S.I for 6 years = 2 x Sum --> Sum = S.I for 6 years. Now, S.I for 9 years = Rs. 9000 S.I for 1 year = Rs. 9000/9 = Rs. 1000. S.I for 6 years = Rs. (1000 x 6)= Rs. 6000. --> x = Rs. 6000 Thus, both a and b are necessary to answer the question.
Q: P is 30% more efficient than Q. How much time will they, working together, take to complete a job which P alone could have done in 23 days? 1496 05b5cc6bde4d2b4197774d910
5b5cc6bde4d2b4197774d910- 116 daysfalse
- 213 daystrue
- 315 daysfalse
- 412 daysfalse
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Answer : 2. "13 days"
Explanation :
Answer: B) 13 days Explanation: Ratio of times taken by P & Q = 100 : 130 = 10:13 Let Q takes x days to do the work Then, 10:13 :: 23:x => x = 23x13/10 => x = 299/10 P's 1 day's work = 1/23 Q's 1 day's work = 10/299 (P+Q)'s 1 day's work = (1/23 + 10/299) = 23/299 = 1/13 Hence, P & Q together can complete the work in 13 days.
Q: Lasya invested certain amount for two rates of simple interests at 5% p.a. and 4% p.a. What is the ratio of Lasya's investments if the interests from those investments are equal ? 1495 05b5cc6e2e4d2b4197774ec1b
5b5cc6e2e4d2b4197774ec1b- 14 : 5true
- 25 : 4false
- 37 : 6false
- 46 : 7false
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Answer : 1. "4 : 5"
Explanation :
Answer: A) 4 : 5 Explanation: Let x be the investment of Lasya in 5% and y be in 4% x(5)(n)/100 = y(4)(n)/100 => x/y = 4/5 x : y = 4 : 5
Q: Arun’s weight is 140% of Akhil’s weight. Nani’s weight is 90% of Shreyon’s weight. Shreyon’s weight is twice as much as Akhil’s. What percentage of Arun’s weight is Nani’s weight? 1494 05b5cc6b3e4d2b4197774d3f3
5b5cc6b3e4d2b4197774d3f3- 176%false
- 277%false
- 378%true
- 479%false
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Answer : 3. "78%"
Explanation :
Answer: C) 78% Explanation: Let Akhil's weight = p kg => Shreyon's weight = 2p kg Nani's weight = 0.9 x 2p = 1.8p kg => Arun's weight = 1.4p kg Required percentage = 1.4p1.8p x 100 = 77.8% ~= 78%