MAT Mock Test Series 4 | MAT Online Test Series 4 | MAT Sample Papers
MAT Mock Test Series 4 | MAT Online Test Series 4 | MAT Sample Papers
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MAT Online Test in English – Series 4, MAT Free Online Test Series 4. MAT Free Mock Test Exam 2019. MAT Mock Test for competitive examination, entrance examination and campus interview. Take various tests and find out how much you score before you appear for your next interview and written test. MAT Exam Free Online Quiz 2019, MAT Full Online Mock Test Series 4th in English. MAT Free Mock Test Series in English. MAT Free Mock Test Series 4. MAT English Language Online Test in English Series 4th. Take MAT Online Quiz. The MAT Full online mock test paper is free for all students. MAT Question and Answers in English and Hindi Series 4. Here we are providing MAT Full Mock Test Paper in English. MAT Mock Test Series 4th 2019. Now Test your self for MAT Exam by using below quiz…
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Question 1 of 50
1. Question
The average of five results is 46 and that of the first four is 45. The fifth result is?
Correct
5 * 46 – 4 * 45 = 50
Incorrect
5 * 46 – 4 * 45 = 50
-
Question 2 of 50
2. Question
Which of the following is the greatest?
Correct
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest.Incorrect
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest. -
Question 3 of 50
3. Question
A tank is filled in eight hours by three pipes A, B and C. Pipe A is twice as fast as pipe B, and B is twice as fast as C. How much time will pipe B alone take to fill the tank?
Correct
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours.Incorrect
1/A + 1/B + 1/C = 1/8 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/8
=> (1 + 2 + 4)/2B = 1/8
=> 2B/7 = 8
=> B = 28 hours. -
Question 4 of 50
4. Question
Amar, Bhavan and Chetan divide an amount of Rs.5600 among themselves in the ratio 3:6:5. If an amount of Rs.400 is deducted from each of their shares, what will be the new ratio of their shares of the amount?
Correct
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4Incorrect
Let the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.
3x + 6x + 5x = 5600
14x = 5600 => x = 400
Required ratio = 3x – 400 : 6x – 400 : 5x – 400
= 3x – x : 6x – x : 5x – x
= 2x : 5x : 4x => 2:5:4 -
Question 5 of 50
5. Question
A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?
Correct
2/5 + (2 + x)/10 = 1 => x = 4 days
Incorrect
2/5 + (2 + x)/10 = 1 => x = 4 days
-
Question 6 of 50
6. Question
The length of rectangle is thrice its breadth and its perimeter is 96 m, find the area of the rectangle?
Correct
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432Incorrect
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432 -
Question 7 of 50
7. Question
Find the one which does not belong to that group ?
Correct
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
Incorrect
In each of the groups B4, D16, E25 and F36 the number is the square of the position value of the letter. This pattern is not followed in I91.
-
Question 8 of 50
8. Question
The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is:
Correct
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
Incorrect
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
-
Question 9 of 50
9. Question
When 2 is added to half of one-third of one-fifth of a number, the result is one-fifteenth of the number. Find the number?
Correct
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60Incorrect
Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60 -
Question 10 of 50
10. Question
In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are at-least 50 years old. Find the number of males aged below 50 years?
Correct
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.Incorrect
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120. -
Question 11 of 50
11. Question
A train crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train?
Correct
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150mIncorrect
Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m -
Question 12 of 50
12. Question
A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
Correct
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100-L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CP-SP = (6250/3)- 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4%Incorrect
SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100-L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CP-SP = (6250/3)- 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100 =
0.04 * 100 = 4% -
Question 13 of 50
13. Question
Pipe A can fill a tank in 6 hours. Due to a leak at the bottom, it takes 9 hours for the pipe A to fill the tank. In what time can the leak alone empty the full tank?
Correct
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18.Incorrect
Let the leak can empty the full tank in x hours 1/6 – 1/x = 1/9
=> 1/x = 1/6 – 1/9 = (3 – 2)/18 = 1/18
=> x = 18. -
Question 14 of 50
14. Question
A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in?
Correct
(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.Incorrect
(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days. -
Question 15 of 50
15. Question
Find the greatest 4-digit number exactly divisible by 3, 4 and 5?
Correct
Greatest 4-digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
Incorrect
Greatest 4-digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 – 39 = 9960.
-
Question 16 of 50
16. Question
A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately?
Correct
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.Incorrect
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr. -
Question 17 of 50
17. Question
The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
Correct
3x + 5x + 7x = 45
x =3
3x = 9Incorrect
3x + 5x + 7x = 45
x =3
3x = 9 -
Question 18 of 50
18. Question
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
Correct
x – y = 5
4x – 6y = 6
x = 12 y = 7Incorrect
x – y = 5
4x – 6y = 6
x = 12 y = 7 -
Question 19 of 50
19. Question
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = ?
Correct
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8Incorrect
(6 – 2 * 3 + 6) / (2 + 6 * 3 – 4) = (6 – 6 + 6) / (2 + 18 – 4)
= 6/16 = 3/8 -
Question 20 of 50
20. Question
The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average of the team?
Correct
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years.Incorrect
Let the average of the whole team be x years.
11x – (26 + 29) = 9(x – 1)
= 11x – 9x = 46
= 2x = 46 => x = 23
So, average age of the team is 23 years. -
Question 21 of 50
21. Question
31.2 * 14.5 * 9.6 = ?
Correct
? = 4343.04
Incorrect
? = 4343.04
-
Question 22 of 50
22. Question
A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?
Correct
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4Incorrect
M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4 -
Question 23 of 50
23. Question
A train 150 m long running at 72 kmph crosses a platform in 25 sec. What is the length of the platform?
Correct
D = 72 * 5/18 = 25 = 500 – 150 = 350
Incorrect
D = 72 * 5/18 = 25 = 500 – 150 = 350
-
Question 24 of 50
24. Question
Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case?
Correct
Calculate the differences, taking two numbers at a time as follows:
(215-47) = 168
(365-215) = 150
(365-47) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.Incorrect
Calculate the differences, taking two numbers at a time as follows:
(215-47) = 168
(365-215) = 150
(365-47) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5. -
Question 25 of 50
25. Question
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Correct
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.Incorrect
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000. -
Question 26 of 50
26. Question
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves one-fourth of his income, find the ratio of their monthly savings?
Correct
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.Incorrect
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10. -
Question 27 of 50
27. Question
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high?
Correct
d2 = 122 + 42 + 32 = 13
Incorrect
d2 = 122 + 42 + 32 = 13
-
Question 28 of 50
28. Question
The spherical ball of lead 3 cm in diameter is melted and recast into 3 spherical balls. The diameters of two of these are 1 ½ cm and 2 cm respectively. The diameter of third ball is?
Correct
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)3 + 23 + r3]
r = 1.25
d = 2.5Incorrect
4/3 π * 3 * 3 * 3 = 4/3 π[(3/2)3 + 23 + r3]
r = 1.25
d = 2.5 -
Question 29 of 50
29. Question
In a division sum, the remainder is 0. As student mistook the divisor by 12 instead of 21 and obtained 35 as quotient. What is the correct quotient?
Correct
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20Incorrect
Number = (12 * 35) = 420
Correct quotient = 420/21 = 20 -
Question 30 of 50
30. Question
The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Correct
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.Incorrect
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9. -
Question 31 of 50
31. Question
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
Correct
x + y = 80
x – 4y = 5
x = 65 y = 15Incorrect
x + y = 80
x – 4y = 5
x = 65 y = 15 -
Question 32 of 50
32. Question
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full is?
Correct
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.Incorrect
(A + B)’s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)’s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 – 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs. -
Question 33 of 50
33. Question
The ratio between the present ages of P and Q is 6:7. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years?
Correct
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8Incorrect
Let P’s age and Q’s age be 6x and 7x years respectively.
Then, 7x – 6x = 4 => x = 4
Required ratio = (6x + 4) : (7x + 4)
28 : 32 = 7:8 -
Question 34 of 50
34. Question
A, B, C enter into a partnership investing Rs. 35,000, Rs. 45,000 and Rs. 55,000 respectively. The respective shares of A, B, C in annual profit of Rs. 40,500 are:
Correct
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500Incorrect
A:B:C = 35000 : 45000 : 55000 = 7:9:11
A’s share = 40500 * 7/27 = Rs. 10500
B’s share = 40500 * 9/27 = Rs. 13500
C’s share = 40500 * 11/27 = Rs. 16500 -
Question 35 of 50
35. Question
If (461 + 462 + 463 + 464) is divisible by ?, then ? =
Correct
461 + 462 + 463 + 464
=> 461(1 + 4 + 16 + 64)
461(85) = 17 (5 * 461) So it is divisible by 17.Incorrect
461 + 462 + 463 + 464
=> 461(1 + 4 + 16 + 64)
461(85) = 17 (5 * 461) So it is divisible by 17. -
Question 36 of 50
36. Question
A money lender finds that due to a fall in the annual rate of interest from 8% to 7 3/4 % his yearly income diminishes by Rs. 61.50, his capital is?
Correct
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600.Incorrect
Let the capital be Rs. x. Then,
(x * 8 * 1)/100 – (x * 31/4 * 1/100) = 61.50
32x – 31x = 6150 * 4
x = 24,600. -
Question 37 of 50
37. Question
Find the nearest to 25268 which is exactly divisible by 467?
Correct
Incorrect
-
Question 38 of 50
38. Question
The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?
Correct
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28Incorrect
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28 -
Question 39 of 50
39. Question
50 men can complete a work in 65 days.Five days after started the work, 20 men left the group. In how many days can the remaining work be completed?
Correct
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M1 D1 = M2 D2
=> 50 * 60 = 30 * D2
=> D2 = (50 * 60)/30 = 100 days.Incorrect
After 5 days, the following situation prevails.
50 men can complete the work in 60 days.
30 men can complete the work in ? days.
M1 D1 = M2 D2
=> 50 * 60 = 30 * D2
=> D2 = (50 * 60)/30 = 100 days. -
Question 40 of 50
40. Question
A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?
Correct
30 * 20 * x = (8 * 5.5 * 1.5)/2
Incorrect
30 * 20 * x = (8 * 5.5 * 1.5)/2
-
Question 41 of 50
41. Question
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8) = ?
Correct
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 23(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125Incorrect
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 23(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125 -
Question 42 of 50
42. Question
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream?
Correct
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours.Incorrect
Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 60/26 = 30/13 hours. -
Question 43 of 50
43. Question
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?
Correct
LCM = 1400
1400 – 6 = 1394Incorrect
LCM = 1400
1400 – 6 = 1394 -
Question 44 of 50
44. Question
Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?
Correct
4000
200 —- I
200
10 —- II
—————
4410
2210
——–
2000
110 —- III
110
5.50 —- IV
———–
2425.50
2210
———–
4635.50
4000
———-
635.50Incorrect
4000
200 —- I
200
10 —- II
—————
4410
2210
——–
2000
110 —- III
110
5.50 —- IV
———–
2425.50
2210
———–
4635.50
4000
———-
635.50 -
Question 45 of 50
45. Question
The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 13 runs. Find his average in all the 30 matches?
Correct
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31.Incorrect
Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 130.
Total score of the batsman in the 30 matches = 930.
Average score of the batsman = 930/30 = 31. -
Question 46 of 50
46. Question
Kiran travels from A to B by car and returns from B to A by cycle in 7 hours. If he travels both ways by car he saves 3 hours. What is the time taken to cover both ways by cycle?
Correct
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.Incorrect
Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours. -
Question 47 of 50
47. Question
The area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm is?
Correct
(5 * 3.5)/2 = 8.75
Incorrect
(5 * 3.5)/2 = 8.75
-
Question 48 of 50
48. Question
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
Correct
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.Incorrect
Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000. -
Question 49 of 50
49. Question
How many meters of carpet 50cm, wide will be required to cover the floor of a room 30m * 20m?
Correct
50/100 * x = 30 * 20 => x = 1200
Incorrect
50/100 * x = 30 * 20 => x = 1200
-
Question 50 of 50
50. Question
3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?
Correct
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.Incorrect
Let 1 woman’s 1 day work = x.
Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.