Quantitative Aptitude Practice Question and Answer
8Q: A sum is equally invested in two different schemes on CI at the rate of 15% and 20% for two years. If interest gained from the sum invested at 20% is Rs. 528.75 more than the sum invested at 15%, find the total sum? 1872 05b5cc658e4d2b4197774bde2
5b5cc658e4d2b4197774bde2- 1Rs. 7000false
- 2Rs. 4500false
- 3Rs. 9000true
- 4Rs. 8200false
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Answer : 3. "Rs. 9000"
Explanation :
Answer: C) Rs. 9000 Explanation: Let Rs. K invested in each scheme Two years C.I on 20% = 20 + 20 + 20x20/100 = 44% Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25% Now, (P x 44/100) - (P x 32.25/100) = 528.75 => 11.75 P = 52875 => P = Rs. 4500 ┬а Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.
Q: 1/8 as a decimal? 1871 05b5cc6a6e4d2b4197774cd6f
5b5cc6a6e4d2b4197774cd6f- 10.125true
- 20.225false
- 30.325false
- 40.425false
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Answer : 1. "0.125"
Explanation :
Answer: A) 0.125 Explanation: 1/8 to convert into┬аdecimal, Find an equivalent fraction with a denominator of a power of 10. In this case, we will use 1000. 18┬аx┬а125125┬а=┬а1251000 This can now be written as a decimal with 3 decimal places (because we have thousandths. i.e, 0.125 ┬а Hence,┬а18┬аas┬аa┬аdecimal┬а=┬а0.125
Q: Two numbers are 35% and 42% are less than a third number .How much percent is the second number less than the first ? 1870 05b5cc6dce4d2b4197774e96f
5b5cc6dce4d2b4197774e96f- 110.769 %true
- 211.23 %false
- 310.4 %false
- 49.87 %false
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Answer : 1. "10.769 %"
Explanation :
Answer: A) 10.769 % Explanation: I ┬а ┬а ┬аII ┬а ┬а III 65 ┬а 58 ┬а 100 65 -------- 7100 ------ ? ┬а=> 10.769%
Q: 5 + 2 x 10/2 = ? 1870 05b5cc6a6e4d2b4197774cd33
5b5cc6a6e4d2b4197774cd33- 115true
- 235false
- 310false
- 420false
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Answer : 1. "15"
Explanation :
Answer: A) 15 Explanation: Using BODMAS Rule, First solve division, mutiplication and Addition So here 10/2 = 5 Then 2 x 5 =┬а 10 Now, 5 + 10 = 15. ┬а Hence, 5 + 2 x 10/2 = 15.
Q: Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m. ? 1869 05b5cc6dae4d2b4197774e899
5b5cc6dae4d2b4197774e899- 120 kmphfalse
- 218 kmphfalse
- 312 kmphtrue
- 416 kmphfalse
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Answer : 3. "12 kmph"
Explanation :
Answer: C) 12 kmph Explanation: Let the distance traveled be x km.Then, x/10 - x/15 = 23x - 2x = 60 => x = 60 km.Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.Required speed = 60/5 = 12 kmph.
Q: The average age of husband, wife and their child 3 years ago was 24 years and that of wife and child 5 years ago was 25 years. The present age of the husband is: 1869 05b5cc6bce4d2b4197774d8d4
5b5cc6bce4d2b4197774d8d4- 121 yrstrue
- 227 yrsfalse
- 329 yrsfalse
- 431 yrsfalse
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Answer : 1. "21 yrs"
Explanation :
Answer: A) 21 yrs Explanation: Sum of ages of husband, wife and child= (24 ├Ч 3) + 9 = 81 years Sum of ages of wife and child => 25 ├Ч 2 + 10 = 50 + 10 = 60 years Age of husband = 81 - 60 = 21 years.
Q: The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is ┬а? 1869 05b5cc6e2e4d2b4197774ec25
5b5cc6e2e4d2b4197774ec25- 13 yearsfalse
- 24 yearstrue
- 32.5 yearsfalse
- 42 yearsfalse
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Answer : 2. "4 years"
Explanation :
Answer: B) 4 years Explanation: P1+20100n┬а> 2P┬а ┬а ┬а ┬а Now, (6/5 x 6/5 x 6/5 x 6/5) > 2. ┬а ┬а ┬а So, n = 4 years.
Q: If every seconds Saturday and all Sundays are holidays in a 30 days month beginning on Saturday, then how many working days are there in that month ? (Month starts from Saturday) 1869 05b5cc6f8e4d2b4197774f054
5b5cc6f8e4d2b4197774f054- 125false
- 222false
- 324false
- 423true
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Answer : 4. "23"
Explanation :
Answer: D) 23 Explanation: As month begins on Saturday, so 2nd, 9th, 16th, 23rd, 30th days will be Sundays. While 8th and 22nd days are second Saturdays. Thus, there are 7 holidays in all. Hence, no. of working days = 30 тАУ 7 =23