Aptitude Test 1 :
The test given in this section can be taken without any login credentials. At the end of the test, you can check you score as well as detailed answer for each question.
Let the number be ‘x’
Then x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’
In the above sentence we have 296 is multiplied by the constant "k", 75 is added to that. In this form , we consider the number 75 as remainder when the number x is divided by 296.
We want to find the remainder when we divide the number "x" by 37. To do this, we need to have 37 at the place where we have 296 in the above equation.
So we can write 296 as 37 times 8 and 75 as 37 times 2 plus 1. It has shown below.
x = 37 × 8k + 37 × 2 + 1
x = 37(8k + 2) + 1
When the number ‘x’ is divided by 37,the remainder is ‘1’
Let "x" and "y" be the lengths of the train and platform respectively
Relative speed of the train to man = 54-6 = 48 kmph
= 48X5/18 m/sec = 40/3 m/sec
When train passes the man, it covers its own length in the above relative speed, length of the train = Relative Speed X Time
= (40/3)X12 = 160 m
And,speed of the train = 54 kmph = 54X5/18 m/sec = 15 m/sec
The train takes 20 seconds to cross the platform.
That is, the train takes 20 seconds to cover (x+y) m distance
Distance /Speed = Time
(x+y)/15=20--->160+y = 300---> y = 140 m
Hence the lengths of the train and platform are 160 m and
140 m respectively
From the given information, we can have
(A+B) can complete(1/12) part of the work in 1 day
(B+C) can complete (1/18) part of the work in 1 day
(A+C) can complete (1/24) part of the work in 1 day
By adding the three equations, we get,
(A+B+C) can complete 13/144 part of the work in 1 day
Therefore (A+B+C) can together complete the work in 144/13 days
That is 11(1/13) days
If the original speed is 100%,
speed after increment is 133 1/3%.
Ratio of the speeds is 100%: 133 1/3%-->100%:(400/3)%
So, ratio of the speeds is 1:4/3
If the ratio of the speed is 1:4/3, ratio of time taken would be 1:3/4
When the speed is increased by 33 1/3%, 3/4 of the original time is enough to cover the same distance.
That is,when the speed is increased by 33 1/3%, 1/4 of the original time will be decreased.
The question says that when speed is increased by 33 1/3%, time is decreased by 15 minutes.
Therefore, 1/4 of the original time = 15 minutes
Original time = 4X15 = 60 minutes
Hence, time taken by him initially = 60 mins or 1 hour
Let "x" be the present age of the man and "y" be the sum of the present ages of two sons.
The present age of the man is three times the sum of the ages of two sons
x = 3y ------(1)
5 years hence, age of the man will be double the sum of the ages of his two sons
x+5 = 2(y+5+5)
x+5 = 2(y+10)
3y+5 = 2y+20........using equation(1)
Solving the above equation, we get y = 15
Plugging y=15 in equation(1),
x = 3(15)
x = 45 yrs
Hence the present age of the man is 45 years.
Clearly, the first natural number which is divisible by 7 is 7. The next numbers which are divisible by 7 are 14, 21.....
Let us write the first twenty natural numbers which are divisible by 7. They are 7,14,21,28........ up to 20 terms.
Sum of all the above numbers
= 7+14+21+28.........up t0 20 term
Since all of the above numbers are divisible by 7, we can factor 7
Therefore, sum = 7(210)
Average = (sum of all 20 numbers)/20
Average = (7X210)/20 = 73.5
Hence, the average of first 20 natural numbers which are divisible by 7 is 73.5
Let "x" be the speed upstream.
Then the speed downstream = 3x
Rate in still water = 1/2(3x+x)= 2xkm/hr
Therefore 2x = 18 ===> x = 9
Speed upstream = 9 km/hr
Speed downstream = 3X9 = 27 km/hr
rate of the stream = 1/2(27-9) = 9 km/hr
From the given ratio, ages of the three boys are 3x, 5x and 7x.
Average of the ages = 5x
Average of the ages = 25 (given)
Then we have 5x=25 ===> x=5
Hence, the age of the youngest boy = 3x5 = 15 years.
From the given information, we have
cost price of the cheaper (c) = $9.30
cost price of the dearer (d) = $10.80
cost price of the mixture (m) = $10
Rule to find the ratio for producing mixture is (d-m):(m-c)
(d-m):(m-c) = (10.8-10):(10-9.3) = 0.8:0.7 =8:7
Hence the ratio in which the first kind and second kind to be mixed is 8:7
Let 100 tons be the production of rice in 1995
1995 ===> 100 tons
1995-1996 ===> 150 toms (because production has been increased by 50%)
1996-1997 ===> 600 tons (six times production in 1995)
When we look in to the above calculations, it is very clear that the production of rice has been increased 450 tons in 1996 - 97 from 150 tons in 1996.
Percentage of increase in 1996-1997 = (450/150)X100%
Hence percentage of rice production increased from 1996 to 1997 is 300%
After having practiced answering the above questions, we hope that the students would have understood, how to solve quantitative problems easily.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
APTITUDE TESTS ONLINE
Aptitude test online
ACT MATH ONLINE TEST
TRANSFORMATIONS OF FUNCTIONS
ORDER OF OPERATIONS
Nature of the roots of a quadratic equation worksheets
MATH FOR KIDS
Time and work word problems
Pythagorean theorem word problems
Converting repeating decimals in to fractions
L.C.M method to solve time and work problems