GMAT Mock Test 4

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GMAT Mock Test 4




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. 

Example Quiz 1

1. Find the unit digit of 24382643.
                (A) 5                   (B) 6
                 (C) 7                   (D) 8


2. Two trains of length 250m and 200m run on parallel lines. When they run in the same direction, it will take 30 second to cross each other. When they run in opposite direction, it will take 10 seconds to cross each other. Find the speeds of the two trains (in kmph).
       (A) 112, 62              (B) 110, 58
        (C) 108, 54              (D) 106, 52


3. A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?
    (A) 6 min to empty     (B) 7 min to empty
    (C) 6 min to full          (D) 7 min to full


4. If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station..
           (A) 4 km             (B) 5 km
            (C) 6 km             (D) 7 km


5. The ratio of the age of a man and his wife is 4:3. At the time of marriage the ratio was 5:3 and After 4 years this ratio will become 9:7. How many years ago were they married?
            (A) 12 yrs                (B) 14 yrs
             (C) 16 yrs                (D) 18 yrs


6. There are two sections A and B of a class, consisting 36 and 44 students respectively. If the average weight of the section A is 40 kg and that of section B is 35 kg, find the average weight of the whole class (in kg).
            (A) 37.25                     (B) 38.25
             (C) 39.25                     (D) 40.25


7. A boat covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km down stream in 6.5 hours. The velocity of the current is
         (A) 2 kmph                 (B) 3 kmph
          (C) 4 kmph                 (D) 5 kmph


8. The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.
                 (A) 48                     (B) 45
                 (C) 50                     (D) 42


9. The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
           (A) 7:8                          (B) 8:7
            (C) 5:7                          (D) 7:5


10. The length of a rectangle is increased by 50%. By what percent would the width have to be decreased to maintain the same area?
         (A) 32.33%              (B) 33.33%
          (C) 34.33%              (D) 35.33%




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


jQuery UI Accordion - Default functionality
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


jQuery UI Dialog functionality
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.


jQuery UI Dialog functionality
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.


jQuery UI Dialog functionality
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


jQuery UI Dialog functionality
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.


jQuery UI Dialog functionality
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


jQuery UI Dialog functionality
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.


jQuery UI Dialog functionality
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


jQuery UI Dialog functionality
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


jQuery UI Accordion - Default functionality
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



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