Aricent Quantitative Aptitude Questions And Answers
Aricent Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" for Aricent Exam. These Quantitative Aptitude MCQs are asked in previous Aricent placements/recruitment exams and will help you to prepare for upcoming Aricent drives.
1. A, B and C started a business by investing Rs. 12,800/-. Rs.16,800/- and Rs. 9,600/- respectively. If after 8 months B received Rs. 13,125/- as his share of profit, what amount did C get as his share of profit?
A. Rs. 7,800
B. Rs. 7,150
C. Rs. 8,250
D. Rs. 7,500
View Answer
Ans : D
Explanation: Here the profit will be distributed in the ratio of their investments
i.e. 12800:16800:9600 → 16:21:12
Let 'x' be the total profit
we know that profit received by B = 13125/-
Therefore,
21 / (16+21+12) * x = 13125
Share received by C = (12 / 49) * 30625 = 7500/-
2. The roots of the equation 4x-3*2x+2+32=0 would include:
A. 2,3
B. 5,4
C. 6,2
D. 8,5
View Answer
Ans : A
Explanation: 2^(2x) - 3 * [2^(x) * 2^2] + 32 = 0
2^(2x) - 12 * 2^(x) + 32 = 0
Let's give 2^x a value of y:
y^2 - 12y + 32 = 0
Factor it out:
(y - 8)(y - 4) = 0
y = 4, 8
Now substitute 2^x back for y and solve for each:
2^x = 4
x = 2
2^x = 8
x = 3
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Overall:
x = 2, 3
3. What least value must be assigned to * so that the number 63576*2 is divisible by 8?
A. 1
B. 2
C. 3
D. 4
View Answer
Ans : C
Explanation: The least possible value is 3
4. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
A. 4830
B. 5120
C. 6420
D. 8960
View Answer
Ans : B
Explanation: Clearly, l = (48 – 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) Cu.m = 5120 Cu.m.
5. If the speed of a train is 20 m/sec, find the speed the train in kmph.
A. 84 kmph
B. 72 kmph
C. 68 kmph
D. 55 kmph
View Answer
Ans : B
Explanation: Speed = Distance / Time
Speed = 20 m/sec
Speed = 20 x 18/5 m/sec
Speed = 72 kmph
Hence, the speed of the train is 72 kmph
6. Pure milk costs Rs. 16/- per liter. After adding water the milkman sells the mixture @ Rs. 15/- per liter and thereby makes a profit of 25% in what respective ratio does he mix milk with water?
A. 0.125694444444444
B. 0.16875
C. 0.210416666666667
D. 0.167361111111111
View Answer
Ans : A
Explanation: If SP of mixture is 15/- per liter and profit is 25%, then
CP of mixture = (100/125) * 15 = 12/- liter
As 16/- is the cost of pure 1000 ml, therefore, for 12/-, the quantity of milk =
(12/16) * 1000 = 750 ml
So for 12/- per liter, the milk is 750 ml, then water will be 250 ml.
Thus, required ratio = 750:250 = 3:1
7. A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest.
A. 0.05
B. 0.08
C. 0.06
D. 0.03
View Answer
Ans : D
Explanation: Simple interest = 27250 – 25000 = 2250
Time = 3 years.
SI = PTR / 100 → R = SI * 100 / PT
R = 2250 * 100 / 25000 * 3 → R = 3%.
8. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A. 1
B. 2
C. 3
D. 4
View Answer
Ans : B
Explanation: Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
9. A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2.
A. 43222
B. 43287
C. 43284
D. 43289
View Answer
Ans : C
Explanation: |X| can take 7 values.
To get |X|<2 ( i.e., −2 ⇒ P(|X|<2)= Favourable CasesTotal Cases = 3/7
10. A merchant bought some goods worth Rs. 6,000/- and sold half of them at 12% profit. At what profit percent should he sell the remaining goods to make an overall profit of 18%?
A. 0.2
B. 0.22
C. 0.24
D. 0.26
View Answer
Ans : C
Explanation: Let CPT = 6000/-
Therefore, cost of half of goods CP1= 3000/- and of other half goods = CP2= 3000/-
So, SP1=3000 × 1.12 = 3360/- and SPT= 6000 × 1.18 = 7080/-
Thus, SP2= 7080 – 3360 = 3720/-
Profit % = [(3720 - 3000) / (3000)] * 100 = 24%
Discussion