Aptitude Test 1 :
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Step-1: Take the last two digits in the power and unit digit in the base. They are 43 and 2
Step-2: Divide the last two digits of the power by 4 (in all the problems) and get the remainder. That is, when 43 is divided by4, the remainder is 3.
Step-3: Now this remainder 3 has to be taken as exponent of the unit digit 2
That is, 23 = 8
Hence the unit digit of the given number is 8
Let the speeds of the two trains be S1 and S2
Total distance covered to cross each other = 250 + 200 = 450 m
(when they run in opposite direction or same direction)
When they run in opposite direction, relative speed --> S1 + S2 = 450/10
S1 + S2 = 45 -----(1)
When they run in the same direction, relative speed --> S1 - S2 = 450/30
S1 + S2 = 15 -----(2)
Solving the above two equations, we get
S1 = 30 m/sec = 30X18/5 kmph = 108 kmph
S2 = 15 m/sec = 15X18/5 kmph = 54 kmph
Hence, speeds of the two trains are 108 kmph and 54 kmph.
Clearly, pipe B is faster than pipe A.
When two pipes are opened together, the tank will emptied.
So the right choice would be (A) or (B)
Total capacity of the tank = 60 units. (L.C.M of 10,6)
The tank is already two-fifth full.
That is,quantity of water is in the tank =(2/5)X60 = 24 units
If both the pipes are opened together, this 24 units will be emptied.
work done by pipe A = 60/10 = 6 units/min
work done by pipe B = 60/6 = -12 units/min(emptying the tank)
Adding the above two equations,we get,(A+B)= -4 units/min
That is 4 units will be emptied per minute when both the pipes are opened together
Time taken to empty 24 units = 24/4 = 6 minutes.
Hence, it will take 6 minutes to empty the tank.
Let "x" be the distance which has to be found.
Difference between the times in walking at different speed 12 minutes = 12/60 hr = 1/5 hr
When the speed is 5kmph, time = x/5
When the speed is 6kmph, time = x/6
Difference in time taken = 1/5 hr.
(x/5) - (x/6) = 1/5
By simplification, we get x = 6 km
Hence,the distance covered by him to reach the station is 6 km
From the given ratio,age of the man is 4x and his wife is 3x
4 years hence, ratio of their ages is 9:7
(4x+4):(3x+4) = 9:7
Solving the above equation, we get x = 8
Present age of the man = 4X8 = 32 yrs
Present age of the man = 3X8 = 24 yrs
Let them get married before "t" years from the present
for the above information, we have the ratio
(32-t):(24-t) = 5:3 (Because, at the time of marriage, their ages are in the ratio 5:3)
96-3t = 120-5t
Solve the above equation, we get t=12 years
Hence they got married 12 years before.
For section A, average weight = 40 kg
That is, (sum of the weights of 36 students) /36 = 40
Sum of the weights of 36 students = 1440
For section B, average weight = 35 kg
That is, (sum of the weights of 44 students) /44 = 35
Sum of the weights of 44 students = 1540
Total weight of (for whole class = 44+36)80 students = 1440+1540 = 2980
Average weight of the whole class = 2980/80 = 37.25 kg
Let "x" and "y" be speed upstream and downstream respectively
From the given information, we have
24/x + 36/y = 6 ------(1)
36/x + 24/y = 13/2 ------(2)
Adding (1) and (2), we get
60/x + 60/y = 25/2 ===> 60(1/x+1/y) = 25/2
1/x + 1/y = 5/24 ------(3)
Subtracting (1) from (2), we get
12/x - 12/y = 1/2 ===>12(1/x-1/y) = 1/2
1/x - 1/y = 1/24 ------(4)
Adding (3) and (4), we get 2/x = 6/24 ===> x = 8
(3)===> 1/8 + 1/y = 5/24 ===> 1/y = (5/24)-(1/8)
1/y = 1/12 ===> y = 12
velocity of the current = 1/2(y-x) ===> 1/2(12-8)
=1/2(4) = 2 kmph
Hence the velocity of the current is 2 km/hr.
Sum of the terms in the given ratio = 3+5 = 8
for the given ratio,
No. of boys in the school = 720x(3/8)= 270
No. of girls in the school = 720x(5/8)= 450
Let "x" be the no. of new boys admitted in the school.
18 new girls are admitted (given)
After the above new admissions,
no. of boys in the school = 270+x
no. of girls in the school = 450+18 = 468
From the given information,
(270+x):468 = 2:3
810+3x = 936
Hence the no. of new boys admitted in the school is 42
Let the cost price of 1 liter pure milk be $1
Milk in 1 liter of mixture in A = 4/7 liter
Milk in 1 liter of mixture in B = 2/5 liter
Milk in 1 liter of mixture in C = 1/2 liter
C.P of 1 liter mixture in A (c)= $4/7
C.P of 1 liter mixture in B (d)= $2/5
C.P of 1 liter mixture in C (m)= $1/2
(cost of water is $0)
Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = (2/5-1/2):(1/2-4/7) = 1/10:1/14 = 7:5
Hence the required ratio is 7:5
Let the length and width of the rectangle be 100 cm each.
Area of the rectangle = lXW = 100X100 = 10000 cm2
The length of the rectangle is increased by 50%
Let the width of the rectangle be decreased by P% to maintain the same area
After changes, length = 150, width = (100-P)% of 100 = 100-P
Even after the above two changes, area will be same
Therefore 150X(100-P) = 10000
15000 - 150P = 10000 ===> 150P = 5000 ===> P = 33.33%
Hence, the width has to be decreased by 33.33% to maintain the same area
After having practiced answering the above questions, we hope that the students would have understood, how to solve quantitative problems easily.
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