In this online GMAT mock test 4,we give the questions whose qualities are in high standard and practicing these questions will definitely make the students to reach their goals.
Explanation
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 |