Quantitative Aptitude Practice Question and Answer
8Q: fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ? 2303 05b5cc731e4d2b4197774f7dd
5b5cc731e4d2b4197774f7dd- 13/91true
- 22/73false
- 31/91false
- 43/73false
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Answer : 1. "3/91"
Explanation :
Answer: A) 3/91 Explanation: In a circle of n different persons, the total number of arrangements possible = (n - 1)!Total number of arrangements = n(S) = (15 – 1)! = 14 !Taking three persons as a unit, total persons = 13 (in 4 units)Therefore no. of ways for these 13 persons to around the circular table = (13 - 1)! = 12!In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =n(E) = 12! X 3!Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!= 12!x3x2 / 14x13x12! = 6/14x13 = 3/91
Q: A box contains 90 screws each of 100 gms and 100 bolts each of 150 gms. If the entire box weighs 35.5 kg, then the weight of the empty box is ? 1849 05b5cc731e4d2b4197774f7e2
5b5cc731e4d2b4197774f7e2- 110.5 kgsfalse
- 212 kgsfalse
- 39.6 kgsfalse
- 411.5 kgstrue
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Answer : 4. "11.5 kgs"
Explanation :
Answer: D) 11.5 kgs Explanation: Screws weight = 90x100 = 9000gmsbolts weight = 100 x 150 = 15000gmsweight = screws + bolts => 24000 gms =>24 kgGiven entire box weight = 35.5kgempty box = entire box weight - weight => 35.5kg - 24kg => 11.5kgso the empty box is 11.5kg.
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 ? 2224 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: A man spend Rs. 810 in buying trouser at Rs. 70 each and shirt at Rs. 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased ? 3078 05b5cc731e4d2b4197774f7b5
5b5cc731e4d2b4197774f7b5- 11:3false
- 22:1false
- 33:2true
- 42:3false
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Answer : 3. "3:2"
Explanation :
Answer: C) 3:2 Explanation: Let us assume S as number of shirts and T as number of trousersGiven that each trouser cost = Rs.70 and that of shirt = Rs.30Therefore, 70 T + 30 S = 810=> 7T + 3S = 81......(1)T = ( 81 - 3S )/7 We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of TSimplifying by taking 3 as common factor i.e, 3(27-S) / 7In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7 Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9 Hence for S, put T in eq(1), we getS = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.The ratio of T:S = 9:6 = 3:2.
Q: A wholesale dealer allows a discount of 20 % on the marked price to the retailer. The retailer sells at 5% below the marked price. If the customer pays Rs.19 for an article, what profit is made by the retailer on it ? 1284 05b5cc72ce4d2b4197774f7ab
5b5cc72ce4d2b4197774f7ab- 1Rs. 4false
- 2Rs. 3true
- 3Rs. 5false
- 4Rs. 1false
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Answer : 2. "Rs. 3"
Explanation :
Answer: B) Rs. 3 Explanation: Let wholesaler dealers marked price = 100%, Retailer's C.P = 80% And the retailer sells at 5% less than the marked price => S.P = 95%If S.P of 95% of the retailer costs Rs.19 to customer,so its C.P of 80% will cost 80 x 19/95 = 16 Profit made by the retailer = 19-16 = Rs.3
Q: Total number of students in 3 classes of a school is 333 . The number of students in class 1 and 2 are in 3:5 ratio and 2 and 3 class are 7:11 ratio . What is the strength of class that has highest number of students ? 2137 05b5cc72ce4d2b4197774f7a6
5b5cc72ce4d2b4197774f7a6- 1125false
- 2155false
- 3135false
- 4165true
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Answer : 4. "165"
Explanation :
Answer: D) 165 Explanation: Ratio C1:C2 = 3:5 and C2:C3 = 7:11So C1:C2:C3 = 21 : 35 : 55Let the strength of three classes are 21x, 35x and 55x respectively, then Given that 21x + 35x + 55x = 333=> 111x = 333 or x=3So strength of the class with highest number of students =55x = 55x3 = 165.
Q: The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ? 1717 15b5cc72ce4d2b4197774f79c
5b5cc72ce4d2b4197774f79c- 142.2 mtsfalse
- 233.45 mtsfalse
- 366.6 mtstrue
- 458.78 mtsfalse
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Answer : 3. "66.6 mts"
Explanation :
Answer: C) 66.6 mts Explanation: From above diagramAC represents the hill and DE represents the tower Given that AC = 100 m angleXAD = angleADB = 30° (∵ AX || BD ) angleXAE = angleAEC = 60° (∵ AX || CE) Let DE = h Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1) tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)∵ BD = CE and Substitute the value of CE from equation 1 100/√3 = 100−h(√3) => h = 66.66 mts The height of the tower = 66.66 mts.
Q: 5/9 of the part of the population in a village are females. If 30 % of the females are married. The percentage of unmarried males in the total males is ? 1486 05b5cc72ce4d2b4197774f7a1
5b5cc72ce4d2b4197774f7a1- 162.5 %true
- 2125 %false
- 384.32 %false
- 446.87 %false
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Answer : 1. "62.5 %"
Explanation :
Answer: A) 62.5 % Explanation: Let total population = p number of females = 5p/9number of males =(p-5p/9) = 4p/9married females = 30% of 5p/9 = 30x5p/100x9 = p/6married males = p/6unmarried males =(4p/9-p/6) = 5p/18Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%