# MAT Online Test Series | MAT Mock Test | MAT Sample Papers

## MAT Online Test Series 1 | MAT Mock Test | MAT Sample Papers |

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MAT Online Test in English – Series 1, MAT Free Online Test Series 1. MAT Free Mock Test Exam 2020. MAT Exam Free Online Quiz 2020, MAT Full Online Mock Test **Series 1st** in English. MAT Free Mock Test Series in English. MAT Free Mock Test **Series 1.** MAT English Language Online Test in English **Series 1st**. 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 1**. Here we are providing** MAT Full Mock Test Paper in English. MAT **Mock Test **Series 1st** 2020. Now Test your self for MAT Exam by using below quiz…

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- Question 1 of 50
##### 1. Question

Find the H.C.F of 12, 16, 18 and 24?

CorrectIncorrect - Question 2 of 50
##### 2. Question

How much 60% of 50 is greater than 40% of 30?

Correct(60/100) * 50 – (40/100) * 30

30 – 12 = 18Incorrect(60/100) * 50 – (40/100) * 30

30 – 12 = 18 - Question 3 of 50
##### 3. Question

A jogger running at 9 km/hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?

CorrectSpeed of train relative to jogger = 45 – 9 = 36 km/hr.

= 36 * 5/18 = 10 m/sec.

Distance to be covered = 240 + 120 = 360 m.

Time taken = 360/10 = 36 sec.IncorrectSpeed of train relative to jogger = 45 – 9 = 36 km/hr.

= 36 * 5/18 = 10 m/sec.

Distance to be covered = 240 + 120 = 360 m.

Time taken = 360/10 = 36 sec. - Question 4 of 50
##### 4. Question

Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?

CorrectThey are moving in opposite directions, relative speed is equal to the sum of their speeds.

Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.

The time required = d/s = 100/35 = 20/7 sec.IncorrectThey are moving in opposite directions, relative speed is equal to the sum of their speeds.

Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.

The time required = d/s = 100/35 = 20/7 sec. - Question 5 of 50
##### 5. Question

What should be the least number to be added to the 51234 number to make it divisible by 9?

CorrectThe least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits.

Sum of digits = 15.

Nearest multiple of 9 greater than sum of digits = 18.

Hence 3 has to be added.IncorrectThe least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits.

Sum of digits = 15.

Nearest multiple of 9 greater than sum of digits = 18.

Hence 3 has to be added. - Question 6 of 50
##### 6. Question

Product of two co-prime numbers is 117. Their L.C.M should be:

CorrectH.C.F of co-prime numbers is 1.

So, L.C.M = 117/1 = 117.IncorrectH.C.F of co-prime numbers is 1.

So, L.C.M = 117/1 = 117. - Question 7 of 50
##### 7. Question

10 camels cost as much as 24 horses, 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants. If the cost of 10 elephants is Rs.170000, find the cost of a camel?

CorrectCost of the camel = P

10 camels = 24 horses

16 horses = 4 oxen

6 oxen = 4 elephants

10 elephants = Rs.170000

P = Rs.[(24 * 4 * 4 * 170000)/(10 * 16 * 6 * 10)]

P = Rs.(65280000/9600) => P = Rs.6800IncorrectCost of the camel = P

10 camels = 24 horses

16 horses = 4 oxen

6 oxen = 4 elephants

10 elephants = Rs.170000

P = Rs.[(24 * 4 * 4 * 170000)/(10 * 16 * 6 * 10)]

P = Rs.(65280000/9600) => P = Rs.6800 - Question 8 of 50
##### 8. Question

In a partnership between A, B and C. A’s capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C’s share is Rs.130, what is B’s capital?

Correct200 + 130 = 330

800 – 330 = 470

200 —- 5000

470 —- ? => 11750Incorrect200 + 130 = 330

800 – 330 = 470

200 —- 5000

470 —- ? => 11750 - Question 9 of 50
##### 9. Question

The average of 9 observations was 9, that of the 1

^{st}of 5 being 10 and that of the last 5 being 8. What was the 5^{th}observation?Correct1 to 9 = 9 * 9 = 81

1 to 5 = 5 * 10 = 50

5 to 9 = 5 * 8 = 40

5^{th}= 50 + 40 = 90 – 81 = 9Incorrect1 to 9 = 9 * 9 = 81

1 to 5 = 5 * 10 = 50

5 to 9 = 5 * 8 = 40

5^{th}= 50 + 40 = 90 – 81 = 9 - Question 10 of 50
##### 10. Question

The difference between a number and its three-fifth is 50. What is the number?

CorrectLet the number be x. Then,

x – 3/5 x = 50 => 2/5 x = 50

x = (50 * 5)/2 = 125.IncorrectLet the number be x. Then,

x – 3/5 x = 50 => 2/5 x = 50

x = (50 * 5)/2 = 125. - Question 11 of 50
##### 11. Question

Albert buys 4 horses and 9 cows for Rs. 13,400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is:

CorrectLet C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.

Then, 4x + 9y = 13400 — (i)

And, 10% of 4x + 20% of 9y = 1880

2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)

Solving (i) and (ii), we get : x = 2000 and y = 600.

Cost price of each horse = Rs. 2000.IncorrectLet C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.

Then, 4x + 9y = 13400 — (i)

And, 10% of 4x + 20% of 9y = 1880

2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)

Solving (i) and (ii), we get : x = 2000 and y = 600.

Cost price of each horse = Rs. 2000. - Question 12 of 50
##### 12. Question

The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor?

CorrectLet the length and the breadth of the floor be l m and b m respectively.

l = b + 200% of b = l + 2b = 3b

Area of the floor = 324/3 = 108 sq m

l b = 108 i.e., l * l/3 = 108

l^{2}= 324 => l = 18.IncorrectLet the length and the breadth of the floor be l m and b m respectively.

l = b + 200% of b = l + 2b = 3b

Area of the floor = 324/3 = 108 sq m

l b = 108 i.e., l * l/3 = 108

l^{2}= 324 => l = 18. - Question 13 of 50
##### 13. Question

When 5% is lost in grinding wheat, a country has to import 20 million bags; but when only 2% is lost, it has to import only 15 million bags. Find the quantity of wheat, which grows in the country?

Correct5% – 2% = 3%

3% —- 5

100% —- ? => 166 2/3Incorrect5% – 2% = 3%

3% —- 5

100% —- ? => 166 2/3 - Question 14 of 50
##### 14. Question

Eight men, ten women and six boys together can complete a piece of work in eight days. In how many days can 20 women complete the same work if 20 men can complete it in 12 days?

CorrectLet the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.

Work = 8(8m + 10w +6b)units = 12(20m)

10w +6b = 22m

b is unknown.

We cannot find the relation between m and w.

We cannot answer the question.IncorrectLet the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.

Work = 8(8m + 10w +6b)units = 12(20m)

10w +6b = 22m

b is unknown.

We cannot find the relation between m and w.

We cannot answer the question. - Question 15 of 50
##### 15. Question

A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A.

Correct(3*8 + 2*4):(4*8 + 5*4)

8:13

8/21 * 630 = 240Incorrect(3*8 + 2*4):(4*8 + 5*4)

8:13

8/21 * 630 = 240 - Question 16 of 50
##### 16. Question

(64 + 9 + 9) / (2 * 20 + 1) = ?

Correct(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2

Incorrect(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2

- Question 17 of 50
##### 17. Question

30 men can do a work in 40 days. When should 20 men leave the work so that the entire work is completed in 40 days after they leave the work?

CorrectTotal work to be done = 30 * 40 = 1200

Let 20 men leave the work after ‘P’ days, so that the remaining work is completed in 40 days after they leave the work.

40P + (20 * 40) = 1200

40P = 400 => P = 10 daysIncorrectTotal work to be done = 30 * 40 = 1200

Let 20 men leave the work after ‘P’ days, so that the remaining work is completed in 40 days after they leave the work.

40P + (20 * 40) = 1200

40P = 400 => P = 10 days - Question 18 of 50
##### 18. Question

How many times the keys of a typewriter have to be pressed in order to write first 400 counting numbers?

Correct1 to 9 = 9 * 1 = 9

10 to 99 = 90 * 2 = 180

100 to 400 = 301 * 3 = 903

———–

1092Incorrect1 to 9 = 9 * 1 = 9

10 to 99 = 90 * 2 = 180

100 to 400 = 301 * 3 = 903

———–

1092 - Question 19 of 50
##### 19. Question

The H.C.F of two numbers is 8. Which of the following can never be their L.C.M?

CorrectH.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.

IncorrectH.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.

- Question 20 of 50
##### 20. Question

The sub-duplicate ratio of 1:4 is

Correct√1:√4 = 1:2

Incorrect√1:√4 = 1:2

- Question 21 of 50
##### 21. Question

Two men Amar and Bhavan have the ratio of their monthly incomes as 6:5. The ratio of their monthly expenditures is 3:2. If Bhavan saves one-fourth of his income, find the ratio of their monthly savings?

CorrectLet the monthly incomes of Amar and Bhavan be 6x and 5x respectively.

Let the monthly expenditures of Amar and Bhavan be 3y and 2y respectively.

Savings of Bhavan every month = 1/4 (5x)

=(his income) – (his expenditure) = 5x – 2y

=> 5x = 20x – 8y => y = 15x/8

Ratio of savings of Amar and Bhavan

= 6x – 3y : 1/4 (5x) = 6x – 3(15x/8) : 5x/4

= 3x/8 : 5x/4 => 3:10IncorrectLet the monthly incomes of Amar and Bhavan be 6x and 5x respectively.

Let the monthly expenditures of Amar and Bhavan be 3y and 2y respectively.

Savings of Bhavan every month = 1/4 (5x)

=(his income) – (his expenditure) = 5x – 2y

=> 5x = 20x – 8y => y = 15x/8

Ratio of savings of Amar and Bhavan

= 6x – 3y : 1/4 (5x) = 6x – 3(15x/8) : 5x/4

= 3x/8 : 5x/4 => 3:10 - Question 22 of 50
##### 22. Question

The curved surface of a sphere is 64 π cm

^{2}. Find its radius?Correct4 πr

^{2 }= 64 => r = 4Incorrect4 πr

^{2 }= 64 => r = 4 - Question 23 of 50
##### 23. Question

A train 540 meters long is running with a speed of 54 kmph. The time taken by it to cross a tunnel 180 meters long is?

CorrectD = 540 + 180 = 720

S = 54 * 5/18 = 15 mps

T = 720/15 = 48 secIncorrectD = 540 + 180 = 720

S = 54 * 5/18 = 15 mps

T = 720/15 = 48 sec - Question 24 of 50
##### 24. Question

A, B and C started a business with capitals of Rs. 8000, Rs. 10000 and Rs. 12000 respectively. At the end of the year, the profit share of B is Rs. 1500. The difference between the profit shares of A and C is?

CorrectRatio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6

And also given that, profit share of B is Rs. 1500

=> 5 parts out of 15 parts is Rs. 1500

Now, required difference is 6 – 4 = 2 parts

Required difference = 2/5 (1500) = Rs. 600IncorrectRatio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6

And also given that, profit share of B is Rs. 1500

=> 5 parts out of 15 parts is Rs. 1500

Now, required difference is 6 – 4 = 2 parts

Required difference = 2/5 (1500) = Rs. 600 - Question 25 of 50
##### 25. Question

Ravi invested certain amount for two rates of simple interests at 6% p.a. and 7% p.a. What is the ratio of Ravi’s investments if the interests from those investments are equal?

CorrectLet x be the investment of Ravi in 6% and y be in 7%

x(6)(n)/100 = y(7)(n)/100

=> x/y = 7/6

x : y = 7 : 6IncorrectLet x be the investment of Ravi in 6% and y be in 7%

x(6)(n)/100 = y(7)(n)/100

=> x/y = 7/6

x : y = 7 : 6 - Question 26 of 50
##### 26. Question

A motorcyclist goes from Bombay to Pune, a distance of 192 kms at an average of 32 kmph speed. Another man starts from Bombay by car 2 ½ hours after the first, and reaches Pune ½ hour earlier. What is the ratio of the speed of the motorcycle and the car?

CorrectT = 192/32 = 6 h

T = 6 – 3 = 3

Time Ratio = 6:3 = 2:1

Speed Ratio = 1:2IncorrectT = 192/32 = 6 h

T = 6 – 3 = 3

Time Ratio = 6:3 = 2:1

Speed Ratio = 1:2 - Question 27 of 50
##### 27. Question

(180/37 of 14.8) / (17/60.1 of 180.30) = ?

Correct(180/37 of 14.8) / (17/60.1 of 180.30) = 72/51 = 24/17 = 1.41

Incorrect(180/37 of 14.8) / (17/60.1 of 180.30) = 72/51 = 24/17 = 1.41

- Question 28 of 50
##### 28. Question

If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is:

CorrectLet C.P. be Rs. x.

Then, 5% of x = 350 – 340 = 10

x/20 = 10 => x = 200IncorrectLet C.P. be Rs. x.

Then, 5% of x = 350 – 340 = 10

x/20 = 10 => x = 200 - Question 29 of 50
##### 29. Question

The cost price of an article is 64% of the marked price. Calculate the gain percent after allowing a discount of 12%?

CorrectLet marked price = Rs. 100.

Then, C.P. = RS. 64, S.P. = Rs. 88

Gain % = 24/64 * 100 = 37.5%.IncorrectLet marked price = Rs. 100.

Then, C.P. = RS. 64, S.P. = Rs. 88

Gain % = 24/64 * 100 = 37.5%. - Question 30 of 50
##### 30. Question

Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is:

CorrectLet the number be x. Then,

3x^{2}– 4x = x + 50

3x^{2}– 5x – 50 = 0

(3x + 10)(x – 5) = 0

x = 5IncorrectLet the number be x. Then,

3x^{2}– 4x = x + 50

3x^{2}– 5x – 50 = 0

(3x + 10)(x – 5) = 0

x = 5 - Question 31 of 50
##### 31. Question

A candidate got 35% of the votes polled and he lost to his rival by 2250 votes. How many votes were cast?

Correct35%———–L

65%———–W

——————

30%———-2250

100%———? => 7500Incorrect35%———–L

65%———–W

——————

30%———-2250

100%———? => 7500 - Question 32 of 50
##### 32. Question

Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?

Correct4/12 + x/15 = 1

x = 10Incorrect4/12 + x/15 = 1

x = 10 - Question 33 of 50
##### 33. Question

Two third of three- fifth of one fourth of a number is 24, what is 30% of that number?

Correctx * 2/3 * 3/5 * 1/4 = 24

x = 240

240 * 30/100 = 72Incorrectx * 2/3 * 3/5 * 1/4 = 24

x = 240

240 * 30/100 = 72 - Question 34 of 50
##### 34. Question

A cylinder and a cone have a same height and same radius of the base. The ratio between the volumes of the cylinder and cone is?

CorrectIncorrect - Question 35 of 50
##### 35. Question

Sum of two numbers is 15. Two times of the first exceeds by 5 from the three times of the other. Then the numbers will be?

Correctx + y = 15

2x – 3y = 5

x = 10 y = 5Incorrectx + y = 15

2x – 3y = 5

x = 10 y = 5 - Question 36 of 50
##### 36. Question

Find the roots of the quadratic equation: 2x

^{2}+ 3x – 9 = 0?Correct2x

^{2}+ 6x – 3x – 9 = 0

2x(x + 3) – 3(x + 3) = 0

(x + 3)(2x – 3) = 0

=> x = -3 or x = 3/2.Incorrect2x

^{2}+ 6x – 3x – 9 = 0

2x(x + 3) – 3(x + 3) = 0

(x + 3)(2x – 3) = 0

=> x = -3 or x = 3/2. - Question 37 of 50
##### 37. Question

Eight years ago, Ajay’s age was 4/3 times that of Vijay. Eight years hence, Ajay’s age will be 6/5 times that of Vijay. What is the present age of Ajay?

CorrectLet the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.

A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)

3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8

V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8

=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8

=> 10 – 20/3 = 10/12 A – 9/12 A

=> 10/3 = A/12 => A = 40.IncorrectLet the present ages of Ajay and Vijay be ‘A’ and ‘V’ years respectively.

A – 8 = 4/3 (V – 8) and A + 8 = 6/5 (V + 8)

3/4(A – 8) = V – 8 and 5/6(A + 8) = V + 8

V = 3/4 (A – 8) + 8 = 5/6 (A + 8) – 8

=> 3/4 A – 6 + 8 = 5/6 A + 20/3 – 8

=> 10 – 20/3 = 10/12 A – 9/12 A

=> 10/3 = A/12 => A = 40. - Question 38 of 50
##### 38. Question

Two men can complete a piece of work in four days. Two women can complete the same work in eight days. Four boys can complete the same work in five days. If four men, eight women and 20 boys work together in how many days can the work be completed?

CorrectTwo men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.

Similarly four women would take two days to complete it and 20 children would take one day to complete it.

All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day

= 2 times the unit work in a day.

They will take 1/2 a day to complete it working together.IncorrectTwo men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.

Similarly four women would take two days to complete it and 20 children would take one day to complete it.

All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day

= 2 times the unit work in a day.

They will take 1/2 a day to complete it working together. - Question 39 of 50
##### 39. Question

If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest?

Correct8 = (1*20*R)/100

R = 40%Incorrect8 = (1*20*R)/100

R = 40% - Question 40 of 50
##### 40. 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?

CorrectLet 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.IncorrectLet 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 41 of 50
##### 41. Question

A man rows his boat 85 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?

CorrectSpeed downstream = d/t = 85/(2 1/2) = 34 kmph

Speed upstream = d/t = 45/(2 1/2) = 18 kmph

The speed of the stream = (34 – 18)/2 = 8 kmphIncorrectSpeed downstream = d/t = 85/(2 1/2) = 34 kmph

Speed upstream = d/t = 45/(2 1/2) = 18 kmph

The speed of the stream = (34 – 18)/2 = 8 kmph - Question 42 of 50
##### 42. Question

A, B and C are partners in a business. Their capitals are respectively, Rs.5000, Rs.6000 and Rs.4000. A gets 30% of the total profit for managing the business. The remaining profit is divided among three in the ratio of their capitals. In the end of the year, the profit of A is Rs.200 more than the sum of the profits of B and C. Find the total profit.

CorrectA:B:C = 5:6:4

Let the total profit = 100 – 30 = 70

5/15 * 70 = 70/3

A share = 70/3 + 30 = 160/3

B + C share = 100 – 160/3 = 140/3

A-(B+C) = 160/3 – 140/3 = 20/3

20/3 —- 200

100 —- ? => 3000IncorrectA:B:C = 5:6:4

Let the total profit = 100 – 30 = 70

5/15 * 70 = 70/3

A share = 70/3 + 30 = 160/3

B + C share = 100 – 160/3 = 140/3

A-(B+C) = 160/3 – 140/3 = 20/3

20/3 —- 200

100 —- ? => 3000 - Question 43 of 50
##### 43. Question

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?

CorrectL.C.M of 252, 308 and 198 = 2772

So, A, B and C will again meet at the starting point in 2772 sec, i.e., 46 min 12 sec.IncorrectL.C.M of 252, 308 and 198 = 2772

So, A, B and C will again meet at the starting point in 2772 sec, i.e., 46 min 12 sec. - Question 44 of 50
##### 44. Question

How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?

CorrectI = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50

IncorrectI = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50

- Question 45 of 50
##### 45. Question

(112 * 5

^{4}) = ?Correct(112 * 5

^{4}) = 112 * (10/2)^{4}

= (112 * 10^{4})/24 = 1120000/16 = 70000Incorrect(112 * 5

^{4}) = 112 * (10/2)^{4}

= (112 * 10^{4})/24 = 1120000/16 = 70000 - Question 46 of 50
##### 46. Question

In a pair of fractions, fraction A is twice the fraction B and the product of two fractions is 2/25. What is the value of fraction A?

CorrectA = 2B => B = 1/2 A, so, AB = 2/25

1/2 A^{2}= 2/25

A^{2}= 4/25

A = 2/5IncorrectA = 2B => B = 1/2 A, so, AB = 2/25

1/2 A^{2}= 2/25

A^{2}= 4/25

A = 2/5 - Question 47 of 50
##### 47. Question

The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?

CorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560IncorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560 - Question 48 of 50
##### 48. Question

If 12 men do a work in 80 days, in how many days will 16 men do it?

Correct12 * 80 = 16 * x

x = 60 daysIncorrect12 * 80 = 16 * x

x = 60 days - Question 49 of 50
##### 49. Question

(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?

CorrectAs numerator is of the form a

^{2}– b^{2}= (a + b) (a – b), so the given expression simplifying becomes

(0.9^{2}– 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1IncorrectAs numerator is of the form a

^{2}– b^{2}= (a + b) (a – b), so the given expression simplifying becomes

(0.9^{2}– 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1 - Question 50 of 50
##### 50. Question

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

CorrectAverage after 11 innings = 36

Required number of runs

= (36 * 11) – (32 * 10) = 396 – 320 = 76.IncorrectAverage after 11 innings = 36

Required number of runs

= (36 * 11) – (32 * 10) = 396 – 320 = 76.