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1 / 291
A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?
A300 meters.
B550 meters.
C340 meters.
D240 meters.
Answer: Option D
Explanation:Here is no explanation for this answer
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2 / 291
A shopkeeper allows a discount of 10% on the marked price and still gains 17% on the whole. Find at what percent above the cost price did he mark his goods.
A35%
B25%
C30%
D45%
Answer: Option C
Explanation:Lets assume the cost price = Rs. 100.
So SP = 117 (Given gain 17% on the whole.)
Lets assume the marked price = Rs. x
So, 90% of x = 117 (10% discount on marked price)
=> 90/100 * x = 117
=> x = 130
So, shopkeeper marked his goods 30% above the CP.
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3 / 291
The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
A16 cm.
B18 cm.
C17 cm.
D19 cm.
Answer: Option B
Explanation:As per the question:
(Perimeter of rectangle)/Breadth of rectangle) = 5/1
=> 2(l + b)/b = 5/1
=> 2l + 2b = 5b => 3b = 2l =>
b = (2/3)l ----------(1)
Area of rectangle = 216 cm^2
So, l x b = 216
put the value of b from eq. (1).
=> l * (2/3)l = 216 .
l*l = 324 or l^2 = 324
So, l = 18 cm.
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4 / 291
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A276 metre
B273 metre
C270 metre
D263 metre
Answer: Option B
Explanation:Here is no explanation for this answer
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5 / 291
A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
A550 kg.
B400 kg.
C600 kg.
D650 kg.
Answer: Option C
Explanation:Profit on 1st part Profit on 2nd part:
8% 18%
\ /
Mean Profi 14%
/ \
4 6
Ratio of 1st and 2nd parts = 4 : 6 = 2 : 3
Quantity of 2nd kind = (3/5 ) * 1000 kg = 600 kg
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6 / 291
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
A33, 46
B33, 42
C32, 46
DNone of these
Answer: Option B
Explanation:Lets assume first student secured x marks.
So second student will secured (x + 9).
Sum of both students marks = x + (x + 9) = (2x + 9)
As per Question, (x + 9) was 56% of the sum of their marks.
So, (x+9) = 56/100(2x+9)
=> (x+9) = 14/25(2x+9)
=> 25x + 225 = 28x + 126
=> x = 33.
So, (x + 9) = 33 + 9 = 42
Hence, their marks are 33 and 42 respectively.
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7 / 291
If log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + 1, then x is equal to:
A9
B3
C1
D4
Answer: Option B
Explanation:Given log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + 1
=> log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + log(base 10) 10
Hint: log (base Z)Z = 1
=> log(base 10)[5 (5x + 1)] = log(base 10) [10(x + 5)] ------ (Point to Remember.)
=> 5(5x + 1) = 10(x + 5)
=> 5x + 1 = 2x + 10
=> x = 3
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8 / 291
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
A1/221
B1/212
C1
D25/57
Answer: Option A
Explanation:Lets assume the sample space = S
Then, n(S) = 52C2 = (52 x 51)/(2x1) = 1326
Let E = event of getting 2 kings out of 4
=> n(E) = 4C2 = (4 x 3)/(2 x 1) = 6
=> P(E) =n(E)/n(S) =6/1326 = 1/221
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9 / 291
How many three digit numbers abc are formed where at least two of the three digits are same.
A250
B252
C300
D280
Answer: Option B
Explanation:Total 3 digit numbers = 9 x 10 x 10 = 900
Total number of 3 digit numbers without repetition = 9 x 9 x 8 = 648
So, the number of three digit numbers with at least one digit repeats = 900 - 648 = 252.
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10 / 291
Find the missing term of the series:-
2,5,10,17,?,41
A38
B41
C28
DNone of these
Answer: Option C
Explanation:2+3=5, 5+5=10, 10+7=17,
So, the next prime number is 11
so,17+11=28.
and, 28+13=41
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