ADP Placement Papers
ADP Aptitude Test
1. In a class there are 60% of girls of which 25% poor. What is the probability that a poor girl is selected is leader?
A. 20%
B. 15%
C. 25%
D. 30%
Explanation:
Assume total students in the class = 100
Then Girls = 60% of (100) = 60
Poor girls = 25% of (60) = 15
So probability that a poor girl is selected leader = Poor girls / Total students = 15/100 = 15%.
2. Of the following, which is the closest approximation of (50.2*0.49)/199.8?
A. 1
B. 0.125
C. 12.5
D. 012.5
Explanation:
For approximation (50.2×0.49)/199.8 can be taken as 50×0.5/200 = 25/200 = 1/8 = 0.125.
3. In a mixture of a, b and c, if a and b are mixed in 3:5 ratio and b and c are mixed in 8:5 ratio and if the final mixture is 35 liters, find the amount of b?
A. 14.5
B. 15.73
C. 16.25
D. 10.4
Explanation:
As b is common in both ratios, we should equate b in both ratios by multiplying suitable numbers.
a: b = 3: 5 = 24: 40
b:c = 8: 5 = 40: 25
Now a: b: c = 24: 40: 25.
Amount of b in the mixture = 4089×35 = 15.73.
4. A is twice efficient than B. A and B can both work together to complete a work in 7 days. Then find in how many days A alone can complete the work?
A. 10 days
B. 12 days
C. 11 days
D. 10.5 days
Explanation:
Let us assume A can do 2 units of work each day, then B can do only 1 unit a day. If both can complete the work in 7 days, total work done by these two together = (2 + 1) x 7 = 21 units
If these 21 units to be done by A alone, then he will take 21 / 2 = 10.5 days.
5. The water from one outlet, flowing at a constant rate, can fill the swimming pool in 9 hours. The water from second outlet, flowing at a constant rate can fill up the same pool in approximately in 5 hours. If both the outlets are used at the same time, approximately what is the number of hours required to fill the pool?
A. 2 hrs.
B. 3.2 hrs.
C. 4 hrs.
D. 5 hrs.
Explanation:
Assume tank capacity is 45 Liters.
Given that the first pipe fills the tank in 9 hours.
So its capacity is 45 / 9 = 5 Liters/ Hour.
Second pipe fills the tank in 5 hours.
So its capacity is 45 / 5 = 9 Liters/Hour.
If both pipes are opened together, then combined capacity is 14 liters/hour.
To fill a tank of capacity 45 liters, both pipes takes 45 / 14 = 3.21 Hours.
6. If 75 % of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percentage answered both correctly?
A. 60%
B. 40%
C. 50%
D. 55%
Explanation:
It is a problem belongs to sets. We use the following formula n(A∪B) = n(A) + n(B) - n(A∩B).
Here n(A∪B) is the people who answered at least one of the questions.
It was given that 20% answered neither question then the students who answered at least one question is 100% - 20% = 80%
Now substituting in the formula we get 80% = 75% + 55% - n(A∩B) => n(A∩B) = 50%.
7. A student's average (arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80?
A. 80
B. 78
C. 88
D. 60
Explanation:
We know that Average = (Sum of the observations) - (No of observations)
So, Sum of 4 test scores = 78×4=312.
Sum of 5 tests scores = 80×5=400.
Hence, 5th test score=400-312=88.
8. Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated?
A. 50%
B. 40%
C. 55%
D. 48%
Explanation:
Total seeds germinated in Plot I = 57% of 200 = 114
Total seeds germinated in Plot II = 42% of 300 = 126
Total germinated seeds = 114 + 126 = 240
The percentage of germinated seeds of the total seeds = 240500×100 = 48%.
9. The price of lunch for 15 people was 207 pounds, including a 15 percent gratuity of service. What was the average price per person, EXCLUDING the gratuity?
A. 15 pounds
B. 12 pounds
C. 18 pounds
D. 20 pounds
Explanation:
Let the net price excluding the gratuity of service = x pounds
Then, total price including 15% gratuity of service = x*(100+15)/100 = 1.15 x pounds.
So, 1.15 x = 207 pounds.
=> x = 207 / 1.15 = 180 pounds.
Net price of lunch for each person = 180 / 15 = 12 pounds.
10. Bhanu spends 30% of his income on petrol on scooter 20% of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol, then what is the expenditure on house rent?
A. Rs 100
B. Rs 110
C. Rs.150
D. Rs.140
Explanation:
Given, 30% of (Income) = 300.
=> Income = 1000.
After having spent Rs.300 on petrol, he left with Rs.700.
His spending on house rent = 20% of (700) = Rs.140.
11. A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X?
A. Rs 44.44
B. Rs 45
C. Rs 24
D. Rs 50
Explanation:
Total Cost price = 24×X
As 1/3rd of the beans spoiled, remaining beans are (2/3) * 24 = 16 kgs.
Selling price = 200 × 16 = 3200.
Profit = Selling price - Cost price = 3200 - 24×X.
Given,
Profit = 2 × Cost price.
3200 - 24×X = 2 × (24×X).
Solving X = 44.44.
12. 3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs. 144.What is the combined price of 1 apple, 1 peach, and 1 mango.
A. 37
B. 39
C. 35
D. 36
Explanation:
According to the question,
3m + 4a = 85 .......(1)
5a + 6p = 122 .......(2)
6m + 2p = 114 .......(3)
On multiplying eq(1) by 2, we get
=> 6m + 8a = 170 .......(4)
Now subtracting (3) from (4), we get :-
=> 8a - 2p = 56 .......(5)
(2) => 5a + 6p = 122
From eq(2) and eq(5) * 3 ,
On Solving we get a = 10, p = 12, m = 15
So, a + p + m = 37.
13. J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in How many days?
A. 32
B. 48
C. 96
D. 24
Explanation:
Assume the total work = 48 units.
Capacity of J = 48 / 16 = 3 units / day
Capacity of P = 48 / 24 = 2 units / day
Capacity of J, P, H = 48 / 8 = 6 units / day
From the above capacity of H = 6 - 2 - 3 = 1
So H takes 48 / 1 days = 48 days to dig the well.
14. A dog taken four leaps for every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed?
A. 16:15
B. 15:16
C. 1:4
D. 5:4
Explanation:
In terms of number of leaps, the ratio of the Dog and hare speeds are 4: 5
But Given that 3 leaps of dog = 4 leaps of hare, i.e., Leap lengths = 4: 3 (If Dog is covering in 3 leaps what hare as covered in 4 leaps then Leap lengths are inversely proportional)
So Dog speed = 4 x 4 = 16
Hare speed = 5 x 3 = 15
So speeds ratio = 16: 15.
15. Letters in the word ABUSER are permuted in all possible ways and arranged in alphabetical order then find the word at position 49 in the permuted alphabetical order?
A. ARBSEU
B. ARBESU
C. ARBSUE
D. ARBEUS
Explanation:
The best way to solve this problem is Just ask how many words starts with A. If we fix A, then the remaining letters can be arranged in 5! ways = 120. So the asked word must start with A.
Arrange all the given letters in alphabetical order. ABERSU
Let us find all the words start with AB. AB**** = 4! = 24 ways
Now we find all the words start with AE. AE****= 4! = 24 ways
So next word start with AR and remaining letters are BESU.
16. An article manufactured by a company consists of two parts X and Y. In the process of manufacturing of part X, 9 out 100 parts many be defective. Similarly, 5 out of 100 are likely to be defective in the manufacturer of Y. Calculate the probability that the assembled product will not be defective?
A. 0.6485
B. 0.6565
C. 0.8645
D. none of these
Explanation:
Probability that the part X is non defective is = 1 - 9/100=.91
Probability that the part Y is non defective is = 1 - 5/100=.95
so, Probability of non-defective product=0.91×0.95=0.8645.
17. 3 friends A, B, C went for week end party to McDonald's restaurant and there they measure there weights in some order in 7 rounds. A, B, C, AB, BC, AC, ABC. Final round measure is 155kg then find the average weight of all the 7 rounds?
A. 80.5 kgs
B. 150 kgs
C. 90.5 kgs
D. 88.5 kgs
Explanation:
Average weight = [(a + b + c + (a+b) + (b+c) + (c+a) +(a+b+c)] / 7
= 4 *(a+b+c)/ 7
= 4 x 155/7
= 88.5 kgs.
18. In a mixture of a, b and c, if a and b are mixed in 3:5 ratio and b and c are mixed in 8:5 ratio and if the final mixture is 35 liters, find the amount of b?
A. 15.73
B. 13.25
C. 16.0
D. 17.54
Explanation:
As b is common in both ratios, we should equate b in both ratios by multiplying suitable numbers.
a:b = 3 : 5 = 24 : 40
b:c = 8 : 5 = 40 : 25
Now a : b : c = 24 : 40 : 25.
Amount of b in the mixture = 4089×35 = 15.73.
19. Ray writes a two-digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the number.
A. 35
B. 42
C. 49
D. 57
Explanation:
Let the two-digit number be xy.
4(x + y) +3 = 10x + y .......(1)
10x + y + 18 = 10 y + x .......(2)
Solving 1st equation we get 2x - y = 1 .... (3)
Solving 2nd equation we get y - x = 2 .... (4)
Solving 3 and 4, we get x = 3 and y = 5.
20. A and B are running around a circular track of length 120 meters with speeds 12 m/s and 6 m/s in the same direction. When will they meet for the first time?
A. 15 secs
B. 20 secs
C. 25 secs
D. 17 secs
Explanation:
A meets B when A covers one round more than B.
A's relative speed = (12 - 6) m/s. So he takes 120 / 6 seconds to gain one extra round.
So after 20 seconds A meets B.
21. The diagonal of a square is twice the side of equilateral triangle then the ratio of Area of the Triangle to the Area of Square is?
A. √3:8
B. √3:2
C. √3:4
D. √4: √3
Explanation:
Let the side of equilateral triangle = 1 unit.
We know that area of an equilateral triangle = (√3/4) *a2
As side = 1-unit area of the equilateral triangle = √3/4.
Now Diagonal of the square = 2 (side of the equilateral triangle) = 2
We know that area of the square = (1/2) *D2 where D = diagonal.
So area of the square = (1/2) *(22)=2
Ratio of the areas of equilateral triangle and square = √3/4: 2 ⇒ √3:8.
22. A hollow cube of size 5 cm is taken, with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If 4 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted?
A. 588
B. 488
C. 500
D. 458
Explanation:
The Hallow cube volume = n3−(n−2)2, Here n is the number of small cubes lie on the big cube edge.
Now n = 5.
So Hallow cube volume = 53−(5−2)2=125−27=98.
So 98 small cubes required to make a hallow cube of size 5 cm.
Now total surfaces = 6 x 98 = 588.
Now if the bigger cube is painted 4 sides, total 4 x 25 small faces got paint.
So remaining small faces which does not have paint after cutting is 588 - 100 = 488.
23. For the FIFA world cup, Paul the octopus has been predicting the winner of each match with amazing success. It is rumored that in a match between 2 team's A and B, Paul picks A with the same probability as A's chances of winning. Let's assume such rumors to be true and that in a match between Ghana and Bolivia; Ghana the stronger team has a probability of 2/3 of winning the game.
What is the probability that Paul will correctly pick the winner of the Ghana-Bolivia game?
A. 1/9
B. 4/9
C. 5/9
D. 2/3
Explanation:
The probability that Paul correctly picks the winner = (A's Chances of winning) x (Paul's picking the winner correctly) + (A's chances of losing) x (Paul picks wrongly) = (2/3) × (2/3) + (1/3) × (1/3) =5/9.
24. Value of a scooter depreciates in such a way that its value at the end of each year is 3/4th of its value at the beginning of the same year. If the initial value of scooter is 40,000, what is the value of the scooter at the end of 3 years.
A. 23125
B. 19000
C. 13435
D. 16875
Explanation:
Value of the scooter at the end of the year = 40000× (3/4)3 = 16875.
25. The perimeter of an equilateral triangle and regular hexagon are equal. Find out the ratio of their areas?
A. 3:2
B. 2:3
C. 1:6
D. 6:1
Explanation:
Let the side of the equilateral triangle = a units and side of the regular hexagon is b units.
Given that, 3a=6b => a/b=2/1.
Now ratio of the areas of equilateral triangle and hexagon = (√3/4) *a2:(3√3/2) b2.
=> (√3/4) *(2)2:(3√3/2) (1)2
=> 2:3.
26. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?
A. 980
B. 797
C. 955
D. 618
Explanation:
Let x be the number to be added to 5678.
When you divide 5678 + x by 460 the remainder = 35.
Therefore, 5678 + x = 460k + 35 here k is some quotient.
=> 5643 + x should exactly divisible by 460.
Now from the given options x = 797.
27. In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.
A. 450
B. 420
C. 350
D. 320
Explanation:
3450=23×34×51=a×b×c.
We have to distribute three 2's to a, b, c in 3+3−1C3−1=5C2=10 ways
We have to distribute four 3's to a, b, c in 3+4−1C3−1=6C2=15 ways
We have to distribute one 5 to a, b, c in 3 ways.
Total ways = 10×15×3=450 ways.
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