In this online Aptitude Test 1,we give the questions whose qualities are in high standard and practicing these questions will definitely make the students to reach their goals.
Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 |
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)+(B+C)+(A+C)= 1/12+1/18+1/24 2A+2B+2C=(6+4+3)/72 2(A+B+C)=13/72 (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 = 7(1+2+3+4+.........+20) 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% = 3X100%=300% Hence percentage of rice production increased from 1996 to 1997 is 300% |