**Aptitude Test 1 : **

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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 |
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, 2 ^{3} = 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 7(4x+4):9(3x+4) 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 3(270+x)=468x2 810+3x = 936 3x=126 x=42 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 cm ^{2}
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 |

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