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Aptitude
What will be the cost of gardening 1 meter broad boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs.10 per square meter?
Rs.3400
Rs.1700
Cannot be determind
None of these
Perimeter of the boundary = 2(L + b) = 340
Area of the boundary = [(L + 2)(b + 2) - Lb]
= [Lb + 2L + 2b + 4 - Lb]
= 2(L + b) + 4
= 340 + 4
=344.
Cost of gardening = Rs. (344 X 10)
= Rs.3440
Given that 27th February 2003 is a Thursday. What day of the week was 27th February 1603?
Monday
Thursday
Sunday
Tuesday
After every 400 years, the same day occurs.Thus, if 27th February 2003 is Thursday, before 400 years i.e., on 27th February 1603 has to be Thursday.
2 meters broad pathway is to be constructed around a rectangular plot on the inside. The area of the plot is 96 sq.m. The rate of construction is Rs.50 per square meter. Find the total cost of the construction?
Rs.4000
Rs.2400
Rs.4800
Data inadequate
Area of the plot = Lb = 96 sq. m.
Area of pathway = [(L - 4)(b - 4) - Lb]
= [Lb - 4L - 4b +16 - Lb]
= 16 - 4(L + b)
And here we don not have value of (L + b)
So,Data is inadequate.
My father was 30 years of age when my sister was born. My mother was 38 years of age when I was born. My sister was 6 years of age when my brother was born who is 3 years elder to me. What was the age of my father and my mother during the birth of my brother?
41,36
24,28
28,24
36,35
My brother was born 6 years after my sister was born and 3 years before I was born. Age of father during my brother’s birth = (30 + 6) = 36 years. Age of mother during my brother’s birth = (38 – 3) = 35 years. Hence, during the birth of my brother, my father was 36 years old and my mother was 35 years old.
The ratio of the current ages of Ajay and Vijay is 7:4. The ratio between Ajay's age 6 years ago and Vijay's age 6 years from now is 1:1. Find the ratio between Ajay's age after 12 years and Vijay's age 12 years ago.
12:1
10:1
1:1
Let the current age of Ajay and Vijay be 7x and 4x respectively.`(7x – 6) / (4x + 6) = 1/1` 7x – 6 = 4x + 6 3x = 12 x = 4 Therefore, `(7x + 12) / (4x – 12) = (7**4 + 12) / (4 ** 4 -12) = 40/4 = 10/1`
Functions f and g are defined by:
f(x) = 1/x + 3x and g(x) = -1/x + 6x - 4
The domain of (f+g)(x) is:
(-infinity , 4) U (0 , + infinity)
(-infinity , 0) U (0 , + infinity)
(-infinity , 0) U (0 , 9)
(f + g)(x) = f(x) + g(x) = (1/x + 3x) + (-1/x + 6x - 4)
(f + g)(x) = 9 x - 4
Domain of f + g is given by the interval (-infinity , 0) U (0 , + infinity)
The range of the function f(x)= x2 - 4x + 9 is:
[5 , 9]
[-5 , +infinity)
[5 , + infinity)
h(x) = x 2 - 4 x + 9 = x 2 - 4 x + 4 - 4 + 9 = (x - 2) 2 + 5
(x - 2) 2 >= 0
(x - 2) 2 + 5 >= 5
Hence minimum value h(x) can have is 5 and maximum can go upto + infinity. So the range is [5 , + infinity)
The avergae of weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg.What might be the weight of new person?
76 kg
76.5 kg
85 kg
Let Average of the 7 people who are same be 'X'
The person who is replaced is of 65kg.
Then, given problem is:
`('X' + 65)/8 + 2.5 = ('X' + 'New Person')/8`
`('X' + 85)/8 = ('X' + 'New Person')/8`
Therefore, 'New Person' = 85;
The captain of a cricket team of 11 players is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded the average age of remaining players is one year less than the average age whole team. What is the average age of the team?
23 Years
24 Years
25 Years
Let the average age of the team is x years.
11x - (26+29) = 9(x-1)
x = 23
How many days are there in a leap year?
365
366
364
363
Leap year Contains 366 days.
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