MAT Online Test Series 2  MAT Mock Test Series 2  MAT Sample Papers 
MAT Online Test Series 2  MAT Mock Test Series 2  MAT Sample Papers 
Finish Quiz
0 of 50 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
Information
MAT Online Test Series 2, MAT Free Online Test Series 2. Management Aptitude Test is one of the most reputed MBA examination for admission in top management colleges.MAT Free Mock Test Exam 2019, Prepare on our portal for MAT through online test papers including Mock, series and Practice tests. MAT Exam Free Online Quiz 2019, MAT Full Online Mock Test Series 2nd in English. MAT Free Mock Test Series in English. MAT Free Mock Test Series 2. MAT English Language Online Test in English Series 2nd. 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 2. Here we are providing MAT Full Mock Test Paper in English. MAT Mock Test Series 2nd 2019. Now Test your self for MAT Exam by using below quiz…
This paper has 50 questions.
Time allowed is 50 minutes.
The MAT Online Test Series 2nd, MAT Free Online Test Exam is Very helpful for all students. Now Scroll down below n click on “Start Quiz” or “Start Test” and Test yourself.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 50 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 Answered
 Review

Question 1 of 50
1. Question
A sum of money becomes double itself in 8 years at simple interest. How many times will it become 10 years at the same rate?
Correct
P — 2P — 8 years
2 1/4 P — 10 yearsIncorrect
P — 2P — 8 years
2 1/4 P — 10 years 
Question 2 of 50
2. Question
The bus fare for two persons for travelling between Agra and Aligarh id fourthirds the train fare between the same places for one person. The total fare paid by 6 persons travelling by bus and 8 persons travelling by train between the two places is Rs.1512. Find the train fare between the two places for one person?
Correct
Let the train fare between the two places for one person be Rs.t
Bus fare between the two places for two persons Rs.4/3 t
=> 6/2 (4/3 t) + 8(t) = 1512
=> 12t = 1512 => t = 126.Incorrect
Let the train fare between the two places for one person be Rs.t
Bus fare between the two places for two persons Rs.4/3 t
=> 6/2 (4/3 t) + 8(t) = 1512
=> 12t = 1512 => t = 126. 
Question 3 of 50
3. Question
A bag contains five white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?
Correct
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9Incorrect
Two balls can be picked from nine balls in ⁹C₂ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done ⁵C₁ . ⁴C₁ ways.
The required probability = (5 * 4)/⁹C₂ = 20/36 = 5/9 
Question 4 of 50
4. Question
On dividing 2272 as well as 875 by 3digit number N, we get the same remainder. The sum of the digits of N is:
Correct
Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3digit number is 127, the sum of whose digit is 10.Incorrect
Clearly, (2272 – 875) = 1397, is exactly divisible by N.
Now. 1397 = 11 * 127
The required 3digit number is 127, the sum of whose digit is 10. 
Question 5 of 50
5. Question
If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?
Correct
8 = (1*20*R)/100
R = 40%Incorrect
8 = (1*20*R)/100
R = 40% 
Question 6 of 50
6. Question
The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
Correct
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.Incorrect
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years. 
Question 7 of 50
7. Question
Two pipes A and B can fill a cistern in 37 1/2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after?
Correct
Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 – x) min = 1.
x(2/75 + 1/45) + (30 x) 2/75 = 1
11x + 180 – 6x = 225 => x = 9Incorrect
Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 – x) min = 1.
x(2/75 + 1/45) + (30 x) 2/75 = 1
11x + 180 – 6x = 225 => x = 9 
Question 8 of 50
8. Question
The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the principal?
Correct
P = 144(100/5)^{2} => P = 3600
Incorrect
P = 144(100/5)^{2} => P = 3600

Question 9 of 50
9. Question
Find the length of the wire required to go 15 times round a square field containing 69696 m^{2}.
Correct
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840Incorrect
a^{2} = 69696 => a = 264
4a = 1056
1056 * 15 = 15840 
Question 10 of 50
10. Question
The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?
Correct
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9Incorrect
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9 
Question 11 of 50
11. Question
A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 16 cm. What is the length of the wire?
Correct
Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)^{2} * h = (4/3)π (12)^{3} => h = 9 cmIncorrect
Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)^{2} * h = (4/3)π (12)^{3} => h = 9 cm 
Question 12 of 50
12. Question
A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
Correct
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000Incorrect
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000 
Question 13 of 50
13. Question
If the numerator of a fraction is increased by 20% and its denominator is diminished by 25% value of the fraction is 2/15. Find the original fraction.
Correct
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12Incorrect
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12 
Question 14 of 50
14. Question
A bag contains a total of 93 coins in the form of one rupee and 50 paise coins. If the total value of coins in the bag is Rs.56, find the number of 50 paise coins in the bag?
Correct
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74Incorrect
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74 
Question 15 of 50
15. Question
A and B enter into partnership with capital as 7:9. At the end of 8 months, A withdraws. If they receive the profits in the ratio of 8:9 find how long B’s capital was used?
Correct
7 * 8 : 9 * x = 8:9 => x= 7
Incorrect
7 * 8 : 9 * x = 8:9 => x= 7

Question 16 of 50
16. Question
A can do a piece of work in 21 days and B in 28 days. Together they started the work and B left after 4 days. In how many days can A alone do the remaining work?
Correct
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days.Incorrect
Let A worked for x days.
x/21 + 4/28 = 1 => x/21 = 6/7 => x = 18
A worked for 18 days. So, A can complete the remaining work in 18 – 4 = 14 days. 
Question 17 of 50
17. Question
Five men and nine women can do a piece of work in 10 days. Six men and twelve women can do the same work in 8 days. In how many days can three men and three women do the work?
Correct
(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days.Incorrect
(5m + 9w)10 = (6m + 12w)8
=> 50m + 90w = 48w + 96 w => 2m = 6w => 1m = 3w 5m + 9w = 5m + 3m = 8m
8 men can do the work in 10 days.
3m +3w = 3m + 1w = 4m
So, 4 men can do the work in (10 * 8)/4 = 20 days. 
Question 18 of 50
18. Question
The speed of a train is 90 kmph. What is the distance covered by it in 10 minutes?
Correct
90 * 10/60 = 15 kmph
Incorrect
90 * 10/60 = 15 kmph

Question 19 of 50
19. Question
Find the one which does not belong to that group ?
Correct
G^{1}F^{+3}I, Q^{2}O^{+3}R, L^{1}K^{+3}N, Y^{1}X^{+3}A and T^{1}S^{+3}V.
Except QOR, all other groups follows similar pattern.Incorrect
G^{1}F^{+3}I, Q^{2}O^{+3}R, L^{1}K^{+3}N, Y^{1}X^{+3}A and T^{1}S^{+3}V.
Except QOR, all other groups follows similar pattern. 
Question 20 of 50
20. Question
252 can be expressed as a product of primes as:
Correct
Clearly, 252 = 2 * 2 * 3 * 3 * 7
Incorrect
Clearly, 252 = 2 * 2 * 3 * 3 * 7

Question 21 of 50
21. Question
Find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5.
Correct
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders) = (208) – (11) = 197.Incorrect
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders) = (208) – (11) = 197. 
Question 22 of 50
22. Question
A pipe can fill a cistern in 20 minutes whereas the cistern when fill can be emptied by a leak in 28 minutes. When both pipes are opened, find when the cistern will be full?
Correct
1/20 – 1/28 = 1/70
70 minutesIncorrect
1/20 – 1/28 = 1/70
70 minutes 
Question 23 of 50
23. Question
(17^{14} * 17^{16}) / 17^{8} = ?
Correct
? = 17^{14+168} = 17^{22}
Incorrect
? = 17^{14+168} = 17^{22}

Question 24 of 50
24. Question
An officer was appointed on maximum daily wages on contract money of Rs. 4956. But on being absent for some days, he was paid only Rs. 3894. For how many days was he absent?
Correct
HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3Incorrect
HCF of 4956, 3894 = 354
(4956 – 3894)/354 = 3 
Question 25 of 50
25. Question
Dacid obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks?
Correct
Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.
Incorrect
Average = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75.

Question 26 of 50
26. Question
A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is .
Correct
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.Incorrect
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42. 
Question 27 of 50
27. Question
If a man lost 4% by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44%?
Correct
96% — 12
144% — ?
96/144 * 12 = 8Incorrect
96% — 12
144% — ?
96/144 * 12 = 8 
Question 28 of 50
28. Question
If 3 workers collect 48 kg of cotton in 4 days, how many kg of cotton will 9 workers collect in 2 days?
Correct
(3 * 4)/48 = (9 * 2)/ x
x = 72 kgIncorrect
(3 * 4)/48 = (9 * 2)/ x
x = 72 kg 
Question 29 of 50
29. Question
HCF of 3/16, 5/12, 7/8 is:
Correct
HCF of numerators = 1
LCM of denominators = 48
=> 1/48Incorrect
HCF of numerators = 1
LCM of denominators = 48
=> 1/48 
Question 30 of 50
30. Question
The sum of three numbers is 98. If the ratio of the first to the second is 2:3. And that of the second to the third is 5:8, then the second number is:
Correct
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30.Incorrect
Let the three parts A, B, C.
Then, A:B = 2:3 and B:C = 5:8
= (5 * 3/5) : (8 * 3/5) = 3:24/5
A:B:C = 2:3:24/5
= 10:15:24 => B = 98 * 15/49 = 30. 
Question 31 of 50
31. Question
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs. ) is?
Correct
Let the sum be Rs. x. Then,
[x (1 + 4/100)^{2} – x] = (676/625 x – x) = 51/625 x
S.I. = (x * 4 * 2)/100 = 2x/25
51x/625 – 2x/25 = 1 or x = 625.Incorrect
Let the sum be Rs. x. Then,
[x (1 + 4/100)^{2} – x] = (676/625 x – x) = 51/625 x
S.I. = (x * 4 * 2)/100 = 2x/25
51x/625 – 2x/25 = 1 or x = 625. 
Question 32 of 50
32. Question
Income and expenditure of a person are in the ratio 5:4. If the income of the person is Rs.18000, then find his savings?
Correct
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.Incorrect
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600. 
Question 33 of 50
33. Question
Tanya’s grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. Eight years ago, what was the ratio of Tanya’s age to that of her grandfather?
Correct
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53Incorrect
16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53 
Question 34 of 50
34. Question
Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in?
Correct
1/10 + 1/15 – 1/x = 1/18
x = 9Incorrect
1/10 + 1/15 – 1/x = 1/18
x = 9 
Question 35 of 50
35. Question
A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?
Correct
Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.Incorrect
Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m. 
Question 36 of 50
36. Question
The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second?
Correct
r = 3 πr^{2} = 9
r = 5 πr^{2} = 25
25 π – 16 π
100 — ? => 64%Incorrect
r = 3 πr^{2} = 9
r = 5 πr^{2} = 25
25 π – 16 π
100 — ? => 64% 
Question 37 of 50
37. 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 38 of 50
38. Question
Which one of the following is the least number of four digits divisible by 71?
Correct
1000/71 = 14 6/71
1000 + 71 – 6 = 1065Incorrect
1000/71 = 14 6/71
1000 + 71 – 6 = 1065 
Question 39 of 50
39. Question
Find the 37.5% of 976 =
Correct
37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366.Incorrect
37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366. 
Question 40 of 50
40. Question
Rajan got married 8 years ago. His present age is 6/5 times his age at the time of his marriage. Rajan’s sister was 10 years younger to him at the time of his marriage. The age of Rajan’s sister is:
Correct
Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.Incorrect
Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years. 
Question 41 of 50
41. Question
Find the one which does not belong to that group ?
Correct
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58.Incorrect
20 = 4^{2} + 4, 42 = 6^{2} + 6, 58 = 7^{2} + 9, 72 = 8^{2} + 8 and 90 = 9^{2} + 9.
20, 42, 72 and 90 can be expressed in n^{2} + n form but not 58. 
Question 42 of 50
42. Question
A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6AM and they work alternately for one hour each. When will the work be completed?
Correct
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.Incorrect
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days. 
Question 43 of 50
43. Question
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
Correct
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10.Incorrect
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10. 
Question 44 of 50
44. Question
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?
Correct
Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x^{2} – 18x + 45 = 0
(x 15)(x – 3) = 0 => x = 15Incorrect
Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x – 5) and (x – 9) hours respectively to fill the tank.
1/x + 1/(x – 5) = 1/(x – 9)
(2x – 5)(x – 9) = x(x – 5)
x^{2} – 18x + 45 = 0
(x 15)(x – 3) = 0 => x = 15 
Question 45 of 50
45. Question
Find the one which does not belong to that group ?
Correct
41, 43, 47 and 53 are prime numbers, but not 57.
Incorrect
41, 43, 47 and 53 are prime numbers, but not 57.

Question 46 of 50
46. Question
Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit’s present salary?
Correct
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57
6x = 68000 => 3x = 34000
Sumit’s present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000.Incorrect
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, (2x + 4000)/(3x + 4000) = 40/57
6x = 68000 => 3x = 34000
Sumit’s present salary = (3x + 4000) = 34000 + 4000 = Rs. 38,000. 
Question 47 of 50
47. Question
A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
Correct
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023Incorrect
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 4^{5} – 1.
= 2^{10} – 1 = 1024 – 1 = 1023 
Question 48 of 50
48. Question
The proportion of copper and zinc in the brass is 13:7. How much zinc will there be in 100 kg of brass?
Correct
7/20 * 100 = 35
Incorrect
7/20 * 100 = 35

Question 49 of 50
49. Question
A man can swim in still water at 4.5 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?
Correct
M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 – x)2
4.5 + x = 9 2x
3x = 4.5
x = 1.5Incorrect
M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 – x)2
4.5 + x = 9 2x
3x = 4.5
x = 1.5 
Question 50 of 50
50. Question
A man gains 20% by selling an article for a certain price. If the sells it at double the price, the percentage of profit will be:
Correct
Let C.P. = Rs. x.
Then, S.P. = Rs. (12% of x) = Rs. 6x/5
New S.P. = 2 * 6x/5 = Rs. 12x/5
Profit = 12x/5 – x = Rs. 7x/5
Profit = 7x/5 * 1/x * 100 = 140%.Incorrect
Let C.P. = Rs. x.
Then, S.P. = Rs. (12% of x) = Rs. 6x/5
New S.P. = 2 * 6x/5 = Rs. 12x/5
Profit = 12x/5 – x = Rs. 7x/5
Profit = 7x/5 * 1/x * 100 = 140%.