Dell Aptitude Questions and Answers with Explanation, Solutions: Searching for a good platform for placement preparation? Then you were in the right place. Here, we had given the Dell Aptitude Questions and Answers with Explanation for the sake of the candidates who are dreaming to work in MNC company like DELL. To get placed in such a company, candidates should have good knowledge in Aptitude section. Dell Aptitude Questions and Answers with Solutions are available in this article.
Moreover, Dell Aptitude Questions and Answers For Freshers of 2018, 2019, 2020 Batches along with PDF are enriched here in the below sections. You can also download the Dell Aptitude Questions and Answers in PDF available at the end of the page. Well, practice these Dell Aptitude Questions and Answers and gain a good score in the Placement examination.
Dell Aptitude Questions and Answers For Freshers (2018, 2019, 2020 Batches)
Number of Questions: 20 Questions
| Topics | Questions asked (Approximately) |
|---|---|
| Averages | 2-4 Questions |
| Profit and Loss | 3-5 Questions |
| Time and Work | 2-4 Questions |
| Percentages | 2-3 Questions |
| Time and Distance | 1-3 Questions |
| Quadratic Equations | 2-3 Questions |
| Races and Games | 1-2 Questions |
| Mixtures and Allegations | 2-3 Questions |
| Simple Interest | 3-5 Questions |
| Compound Interest | 2-4 Questions |
| Simplification | 1-2 Questions |
| Areas | 2-5 Questions |
| Simple Equations | 1-3 Questions |
| Mensuration | 2-3 Questions |
| Boats and Streams | 1-4 Questions |
Dell Aptitude Questions and Answers along with Explanation
Getting placed in such a reputed company like DELL is not an easy task. But it is possible with good practice. At first, check Dell Aptitude Questions and Answers with Explanation from this page and gain knowledge in this section. These are the sample questions collected from the previous year Dell Aptitude Questions and Answers with Solutions. So make sure that you practice these Dell Aptitude Questions and Answers well, in some case they can appear in the placement examination. Note Down all these and get your preparation started.
Dell Aptitude Questions and Answers in PDF format
1. A batsman in his 17 th innings makes a score of 85 and thereby increasing his average by 3. What is his average after the 17 thinnings?
A. 34
B. 35
C. 36
D. 37
Answer – D. 37
Explanation:
According to the given data, we have arranged
16x + 85 = 17(x + 3)
x = 34 + 3 = 37
2. The average age of P and Q 10 years ago was 20. The average age of P, Q, and R today is 30, so what will be the age of R after 5 years.
A. 25
B. 35
C. 45
D. 50
Answer – B. 35
Explanation:
P+Q = 60.
P, Q, R = 90
so Age of R = 30+5 = 35
3. The average salary of the entire staff in an office is Rs 250 per month. The average salary of officers is Rs 520 and that of non-officers is Rs. 200. If the number of officers is 15, then find the number of non–officers in the office
A. 56
B. 81
C. 87
D. 823
Answer – B. 81
Explanation:
Let us assume, the required number of non–officers = x
200x + 520 x 15 = 250 ( 15 + x )
250x – 200x = 520 * 15 – 250 x 15
50x = 4050
x = 81
4. The average age of a group of persons going for a tour to Shimla is 22 years. 25 new persons with an average age of 10 years join the group and their average age becomes 12 years. The number of persons initially going for a tour is?
A. 5
B. 7
C. 8
D. 9
Answer – A. 5
Explanation:
An initial number of persons = x
= 22x + 25 * 10 – 12(x + 25)
= 22x + 250 – 12x – 300
10x = 50
x = 5
5. A circular wire of radius 56 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6:5. The smaller side of the rectangle is
A. 70 cm
B. 75 cm
C. 80 cm
D. 85 cm
Answer – C. 80 cm
Explanation:
The perimeter of the circle, that is, the rectangle is,
P=2πr = 2 * 22/7 * 56 =16×22 cm.
Let us assume the actual length and breadth of the rectangle be, 6xand 5x
So perimeter will be,
P=2(6x+5x)=22x
16×22=22x
Therefore, x=16.
Finally, The smaller side or breadth = 5x = 80 cm
6. The length of a rectangular plot is 10 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300, what is the length of the plot in meters?
A. 50
B. 55
C. 60
D. 65
Answer – B. 55
Explanation:
Let us assume, breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = 5300/26.50 = 200 m
2[(x + 10) + x] = 200
x = 45.
Hence, length = x + 10 = 55 m.
7. A rectangular field has its length and breadth in the ratio 6:5 respectively. A man riding bicycle complete one lap of this field along its perimeter at the speed of 12kmph in 1.5m. What is the area of the field?
A. 5468
B. 5460
C. 5345
D. 5487
Answer – A. 5468
Explanation:
Speed = 12*5/18 = 3.3 m/s
Perimeter = Distance
2(6x+5x) = 90*3.3
22x= 297
X=13.5
Area = 6*13.5 *5*13.5 = 5467.5 = 5468 sq.m
8. The area of the square is three-fifths the area of a rectangle. The length of the rectangle is 25cm and breadth 10cm less than its length. What is the area of the square?
A. 12 cm
B. 15 cm
C. 17 cm
D. 20 cm
Answer – B. 15 cm
Explanation:
Area of rectangle = 25*15 = 375 sq.cm
Area of square = 3*375/5 = 225 sq.cm = a*a
Side of the square = a = 15cm
9. A solid metallic cylinder of base radius 5 cm and height 7 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones?
A. 49500
B. 51500
C. 52500
D. 53500
Answer – C. 52500
Explanation:
Number of cones = Volume of Cylinder / Volume of one cone
= π*5*5*7 / (1/3π * 1/10 * 1/10 * 1)
= 52500
10. Four horses are tethered at 4 corners of a square field of side 60 meters so that they just cannot reach one another. The area left ungrazed by the horses is:
A. 523 m
B. 612 m
C. 771 m
D. 841m
Answer – C. 771 m
Explanation:
Length of the rope = 60/2 = 30 m
The total area covered by 4 horse = 22*30*30/7
Ungrazed area = Area of Square field – the area covered by 4 horse
= 3600 – 2829 = 771 m
11. 19, 30, 23, 34, 27, ?
A. 38
D. 39
B. 53
C. 66
Answer – A. 38
Explanation:
23 – 19 = 4
34 – 30 = 4
27 – 23 = 4
38 – 34 = 4
12. 52 54 28 30 ? 18 10
A. 14
B. 15
C. 16
D. 18
Answer – C. 16
Explanation:
52 / 1 + 2 = 54
54 / 2 + 1 = 28
28 / 1 + 2 = 30
30 / 2 + 1 = 16
13. 15 151 1210 ? 29056
A. 6263
B. 7263
C. 5463
D. 4263
Answer – B. 7263
Explanation:
15 * 10 + 1 = 151
151 * 8 + 2 = 1210
1210 * 6 + 3 = 7263
7263 * 4 + 4 = 29056
14. 3 7 12 27 50 ?
A. 105
B. 100
C. 95
D. 85
Answer – D. 105
Explanation:
3 * 2 + 1 = 7
7 * 2 – 2 = 12
12 * 2 + 3 = 27
27 * 2 – 4 = 50
50 * 2 + 5 = 105
15. 10 52 253 1006 3013 ?
A. 5022
B. 6022
C. 7022
D. 8022
Answer – B. 6022
Explanation:
10*6 -8 = 52
52 *5 – 7 = 253
253 *4 – 6 = 1006
1006 *3 – 5 = 3013
3013 * 2 – 4 = 6022
16. The ratio between the M and N age is 7: 9. If the difference between the present ages of Q and P’s age after 4 years is 2 then what is the total of the present ages of P and Q?
A. 42
B. 44
C. 46
D. 48
Answer – D. 48
Explanation:
let the age of N is 9x and that of A is 7x. So
9x – (7x +4) = 2, x = 3
So, the sum will be = 27 + 21 = 48
17. Two trains are traveling in the same direction with 60 km/hr and 75 km/hr respectively. The faster train crosses a man sitting in the slower train in 30 sec. find the length of the faster train.
A. 140 meter
B. 150 meter
C. 100 meter
D. 125 meter
Answer – D. 125 meter
Explanation:
L = 15*5/18*30 = 125 meter
18. Two stations A and B are 150 km apart from each other. One train starts from A at 6 AM at a speed of 30 km/hr and travels towards B. Another train starts from station B at 7 AM at a speed of 20 km/hr. At what time they will meet.
A. 7: 34 AM
B. 8: 34 AM
C. 9: 34 AM
D. 10: 34 AM
Answer – C. 9:24 AM
Explanation:
Distance travel by the first train in one hour = 30
Now, the distance remains 120 km only.
x/30 = (120 – x)/20, so we get x = 72 km
Now, time = (30 + 72)/30 = 3hrs and 24minutes i.e. 9: 24 am
19. A boat can travel 4.2 km upstream in 14 min. If the ratio of the speed of the boat in still water to the speed of the stream is 7:1. How much time will the boat take to cover 17.6 km downstream?
A. 32 min
B. 38 min
C. 44 min
D. 48 min
Answer – C. 44 min
Explanation:
Speed = 7x: x
Downstream = 8x; upstream = 6x
Upstream speed = 4.2*60/14 = 18 kmph
6x = 18
X = 3
Downstream = 8*3 = 24
Time taken for 17.6 km = 17.6*60/24 = 44 min
20. A man swims downstream 40 km in 5 hours and upstream 24 km in 2 hours. Find his speed in still water ?
A. 10 kmph
B. 15 kmph
C. 8 kmph
D. 12 kmph
Answer – A. 10 kmph
Explanation:
Downstream = 40/5= 8 kmph
Upstream = 24/2== 12 kmph
Speed in still water = 1/2 ( 8+12) = 10 kmph
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