# RBI Assistant Online Test, RBI Assistant Prelims Online Test Series 4th

## RBI Assistant Online Test, RBI Assistant Prelims Online Test Series 4th

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RBI Assistant Online Test, RBI Assistant Prelims Online Test Series 4th. RBI Exam Online Test 2020, RBI Assistant Pre Free Mock Test Exam 2020. RBI Assistant Pre. Exam Free Online Quiz 2020, RBI Assistant Pre. Full Online Mock Test **Series 4th** in English.The **RBI Assistant Exam pattern** is same as other bank clerk exams followed by Preliminary, Main and Language Proficiency Test. Last year new changes were introduced in the exam pattern. RBI Assistant Pre Question and Answers in English and Hindi **Series 4**. Here we are providing** RBI Assistant Pre. Full Mock Test Paper in English. RBI Assistant Pre. **Mock Test **Series 4th** 2020. Now Test your self for RBI Assistant Pre. Exam by using below quiz…

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

Find the lowest common multiple of 24, 36 and 40.

Correct2 24 – 36 – 40

——————–

2 12 – 18 – 20

——————–

2 6 – 9 – 10

——————-

3 3 – 9 – 5

——————–

1 – 3 – 5

L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360.Incorrect2 24 – 36 – 40

——————–

2 12 – 18 – 20

——————–

2 6 – 9 – 10

——————-

3 3 – 9 – 5

——————–

1 – 3 – 5

L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360. - Question 2 of 50
##### 2. Question

Three persons A, B and C divide a certain amount of money such that A’s share is Rs. 4 less than half of the total amount, B’s share is Rs. 8 more than half of what is left and finally C takes the rest which is Rs. 14. Find the total amount they initially had with them?

CorrectLet the total amount be Rs. p.

Let shares of A and B be Rs. x and Rs. y respectively.

C’s share was Rs. 14

we have, x + y + 14 = p —– (1)

From the given data, x = (p/2) – 4 —– (2)

Remaining amount = p – (p/2 – 4) => p/2 + 4.

y = 1/2(p/2 + 4) + 8 => p/4 + 10 —– (3)

From (1), (2) and (3)

p/2 – 4 + p/4 + 10 + 14 = p

3p/4 + 20 = p

p/4 = 20 => p = Rs. 80.IncorrectLet the total amount be Rs. p.

Let shares of A and B be Rs. x and Rs. y respectively.

C’s share was Rs. 14

we have, x + y + 14 = p —– (1)

From the given data, x = (p/2) – 4 —– (2)

Remaining amount = p – (p/2 – 4) => p/2 + 4.

y = 1/2(p/2 + 4) + 8 => p/4 + 10 —– (3)

From (1), (2) and (3)

p/2 – 4 + p/4 + 10 + 14 = p

3p/4 + 20 = p

p/4 = 20 => p = Rs. 80. - Question 3 of 50
##### 3. Question

A is a working partner and B is a sleeping partner in the business. A puts in Rs.15000 and B Rs.25000, A receives 10% of the profit for managing the business the rest being divided in proportion of their capitals. Out of a total profit of Rs.9600, money received by A is?

Correct15:25 => 3:5

9600*10/100 = 960

9600 – 960 = 8640

8640*3/8 = 3240 + 960

= 4200Incorrect15:25 => 3:5

9600*10/100 = 960

9600 – 960 = 8640

8640*3/8 = 3240 + 960

= 4200 - Question 4 of 50
##### 4. Question

A person travels equal distances with speeds of 3 km/hr, 4 km/hr and 5 km/hr and takes a total time of 47 minutes. The total distance is?

CorrectLet the total distance be 3x km.

Then, x/3 + x/4 + x/5 = 47/60

47x/60 = 47/60 => x = 1.

Total distance = 3 * 1 = 3 km.IncorrectLet the total distance be 3x km.

Then, x/3 + x/4 + x/5 = 47/60

47x/60 = 47/60 => x = 1.

Total distance = 3 * 1 = 3 km. - Question 5 of 50
##### 5. 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:

CorrectLet 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%.IncorrectLet 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%. - Question 6 of 50
##### 6. Question

Sreedhar and Sravan together can do a work in 25 days. With the help of Pavan, they completed the work in 8 days and earned Rs. 225. What is the share of Sravan, if Sreedhar alone can do the work in 75 days?

CorrectSravan’s one day’s work = 1/25 – 1/75 = 2/75

Sravan worked for 8 days. So, his 8 days work = 8 * 2/75 = 16/75

Sravan completed 16/75th of total work.

So, his share is 16/75 * 225 = Rs. 48.IncorrectSravan’s one day’s work = 1/25 – 1/75 = 2/75

Sravan worked for 8 days. So, his 8 days work = 8 * 2/75 = 16/75

Sravan completed 16/75th of total work.

So, his share is 16/75 * 225 = Rs. 48. - Question 7 of 50
##### 7. 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 8 of 50
##### 8. Question

If cost of sugar increases by 25%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed?

Correct100

125

—–

125 —– 25

100 —— ? => 20%Incorrect100

125

—–

125 —– 25

100 —— ? => 20% - Question 9 of 50
##### 9. Question

What is the area of square field whose side of length 15 m?

Correct15 * 15 = 225 sq m

Incorrect15 * 15 = 225 sq m

- Question 10 of 50
##### 10. 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 11 of 50
##### 11. Question

108 * 107 * 96 = ? (to the nearest hundred)

Correct108 * 107 * 96 ≡ 1109400

Incorrect108 * 107 * 96 ≡ 1109400

- Question 12 of 50
##### 12. Question

Rs.1200 divided among P, Q and R. P gets half of the total amount received by Q and R. Q gets one-third of the total amount received by P and R. Find the amount received by R?

CorrectLet the amounts to be received by P, Q and R be p, q and r.

p + q + r = 1200

p = 1/2 (q + r) => 2p = q + r

Adding ‘p’ both sides, 3p = p + q + r = 1200

=> p = Rs.400

q = 1/3 (p + r) => 3q = p + r

Adding ‘q’ both sides, 4q = p + q + r = 1200

=> q = Rs.300

r = 1200 – (p + q) => r = Rs.500.IncorrectLet the amounts to be received by P, Q and R be p, q and r.

p + q + r = 1200

p = 1/2 (q + r) => 2p = q + r

Adding ‘p’ both sides, 3p = p + q + r = 1200

=> p = Rs.400

q = 1/3 (p + r) => 3q = p + r

Adding ‘q’ both sides, 4q = p + q + r = 1200

=> q = Rs.300

r = 1200 – (p + q) => r = Rs.500. - Question 13 of 50
##### 13. Question

Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.85. How much was lent at 5%?

Correct(x*5*1)/100 + [(1500 – x)*6*1]/100 = 85

5x/100 + 90 – 6x/100 = 85

x/100 = 5

=> x = 500Incorrect(x*5*1)/100 + [(1500 – x)*6*1]/100 = 85

5x/100 + 90 – 6x/100 = 85

x/100 = 5

=> x = 500 - Question 14 of 50
##### 14. Question

Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?

CorrectC.P. of 1 toy = 375/12 = Rs. 31.25

S.P of 1 toy = Rs. 33

Profit = 1.75/31.25 * 100 = 28/5 = 5.6%IncorrectC.P. of 1 toy = 375/12 = Rs. 31.25

S.P of 1 toy = Rs. 33

Profit = 1.75/31.25 * 100 = 28/5 = 5.6% - Question 15 of 50
##### 15. Question

60 + 5 * 12 / (180/3) = ?

Correct60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60)

= 60 + (5 * 12)/60 = 60 + 1 = 61.Incorrect60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60)

= 60 + (5 * 12)/60 = 60 + 1 = 61. - Question 16 of 50
##### 16. Question

Find the compound interest accrued on an amount of Rs.14,800 at 13.5% p.a at the end of two years. (Round off your answer to nearest integer)

CorrectCI = 14800{ [ 1 + 13.5/100]

^{2}– 1 }

= 14800 { [1 + 27/200]^{2}– 1

= 14800 { 2 + 27/200}{27/200}

= (74)[2 + 27/200](27) =

1998[2 + 27/200] = 3996 + 269.73 = Rs. 4266IncorrectCI = 14800{ [ 1 + 13.5/100]

^{2}– 1 }

= 14800 { [1 + 27/200]^{2}– 1

= 14800 { 2 + 27/200}{27/200}

= (74)[2 + 27/200](27) =

1998[2 + 27/200] = 3996 + 269.73 = Rs. 4266 - Question 17 of 50
##### 17. Question

A batsman in his 17

^{th}innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17^{th}innings?Correct16x + 85 = 17(x + 3)

x = 34 + 3 = 37Incorrect16x + 85 = 17(x + 3)

x = 34 + 3 = 37 - Question 18 of 50
##### 18. Question

A and B can do a piece of work in 12 days and 16 days respectively. Both work for 3 days and then A goes away. Find how long will B take to complete the remaining work?

Correct3/12 + (3 + x)/16 = 1

x = 9 daysIncorrect3/12 + (3 + x)/16 = 1

x = 9 days - Question 19 of 50
##### 19. Question

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

CorrectL.C.M of 12, 18, 21, 30

= 2 * 3 * 2 * 3 * 7 * 5 = 1260

Required number = 1260/2 = 630.IncorrectL.C.M of 12, 18, 21, 30

= 2 * 3 * 2 * 3 * 7 * 5 = 1260

Required number = 1260/2 = 630. - Question 20 of 50
##### 20. Question

A trader marked the price of the T.V. 30% above the cost price of the T.V. and gave the purchaser 10% discount on the marked price, thereby gaining Rs.340. Find the cost price of the T.V?

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

A shopkeeper sold an article offering a discount of 5% and earned a profit of 23.5%. What would have been the percentage of profit earned if no discount was offered?

CorrectLet C.P. be Rs. 100.

Then, S.P. = Rs. 123.50

Let marked price be Rs. x. Then, 95/100 x = 123.50

x = 12350/95 = Rs. 130

Now, S.P. = Rs. 130, C.P. = Rs. 100

Profit % = 30%.IncorrectLet C.P. be Rs. 100.

Then, S.P. = Rs. 123.50

Let marked price be Rs. x. Then, 95/100 x = 123.50

x = 12350/95 = Rs. 130

Now, S.P. = Rs. 130, C.P. = Rs. 100

Profit % = 30%. - Question 22 of 50
##### 22. Question

10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

Correct1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days.Incorrect1 man’s 1 day work = 1/100

(10 men + 15 women)’s 1 day work = 1/6

15 women’s 1 day work = (1/6 – 10/100) = 1/15

1 woman’s 1 day work = 1/225

1 woman alone can complete the work in 225 days. - Question 23 of 50
##### 23. Question

(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = ?

Correct(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = (22 * 33 * 15 + 9 * 99 * 2 * 22) / (33 * 22 * 8 + 5 * 11 * 22 * 12)

= (22 * 33 * 15 + 22 * 33 * 54) / (33 * 22 * 8 + 33 * 22 * 20) = [22 * 33(15 + 54)] / [33 * 22(8 + 20)] = 69/28Incorrect(22 * 495 + 891 * 44) / (33 * 176 + 55 * 264) = (22 * 33 * 15 + 9 * 99 * 2 * 22) / (33 * 22 * 8 + 5 * 11 * 22 * 12)

= (22 * 33 * 15 + 22 * 33 * 54) / (33 * 22 * 8 + 33 * 22 * 20) = [22 * 33(15 + 54)] / [33 * 22(8 + 20)] = 69/28 - 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

Three pipes of same capacity can fill a tank in 8 hours. If there are only two pipes of same capacity, the tank can be filled in.

CorrectThe part of the tank filled by three pipes in one hour = 1/8

=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.

The tank can be filled in 12 hours.IncorrectThe part of the tank filled by three pipes in one hour = 1/8

=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.

The tank can be filled in 12 hours. - Question 26 of 50
##### 26. Question

The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?

Correctr

_{1}: r_{2}= 1: 3

Πr_{1}^{2}: Πr_{2}^{2}

r_{1}^{2}: r_{2}^{2 }= 1: 9Incorrectr

_{1}: r_{2}= 1: 3

Πr_{1}^{2}: Πr_{2}^{2}

r_{1}^{2}: r_{2}^{2 }= 1: 9 - Question 27 of 50
##### 27. Question

A shopkeeper sells 20% of his stock at 10% profit ans sells the remaining at a loss of 5%. He incurred an overall loss of Rs. 400. Find the total worth of the stock?

CorrectLet the total worth of the stock be Rs. x.

The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50

The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50

Total SP = 11x/50 + 38x/50 = 49x/50

Overall loss = x – 49x/50 = x/50

x/50 = 400 => x = 20000IncorrectLet the total worth of the stock be Rs. x.

The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50

The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50

Total SP = 11x/50 + 38x/50 = 49x/50

Overall loss = x – 49x/50 = x/50

x/50 = 400 => x = 20000 - Question 28 of 50
##### 28. Question

Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.

CorrectArea of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm

^{2}IncorrectArea of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm

^{2} - Question 29 of 50
##### 29. Question

The L.C.M of two numbers is 48. The numbers are in the ratio 2:3. The sum of numbers is:

CorrectLet the numbers be 2x and 3x.

Then, their L.C.M = 6x. So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.IncorrectLet the numbers be 2x and 3x.

Then, their L.C.M = 6x. So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40. - Question 30 of 50
##### 30. Question

(50 – ?/29)% of 4200 = 3√196

Correct(50 – ?/29)% of 4200 = 3√196

Incorrect(50 – ?/29)% of 4200 = 3√196

- Question 31 of 50
##### 31. Question

An amount of Rs.1560 was divided among A, B and C in the ratio 1/2:1/3:1/4. Find the share of C?

CorrectLet the shares of A, B and C be a, b and c respectively.

a:b:c = 1/2:1/3:1/4

a:b:c = 6/12:4/12:3:12 = 6:4:3

Share of C = 3/13 * 1560 = Rs.360.IncorrectLet the shares of A, B and C be a, b and c respectively.

a:b:c = 1/2:1/3:1/4

a:b:c = 6/12:4/12:3:12 = 6:4:3

Share of C = 3/13 * 1560 = Rs.360. - Question 32 of 50
##### 32. Question

A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?

CorrectSpeed = 78 * 5/18 = 65/3 m/sec.

Time = 1 min = 60 sec.

Let the length of the train be x meters.

Then, (800 + x)/60 = 65/3

x = 500 m.IncorrectSpeed = 78 * 5/18 = 65/3 m/sec.

Time = 1 min = 60 sec.

Let the length of the train be x meters.

Then, (800 + x)/60 = 65/3

x = 500 m. - Question 33 of 50
##### 33. Question

The sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself?

Correct100 —- 300 — 3

900 — 3

—-

6 yearsIncorrect100 —- 300 — 3

900 — 3

—-

6 years - Question 34 of 50
##### 34. Question

In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes?

CorrectLet the total number of valid votes be x.

70% of x = 70/100 * x = 7x/10

Number of votes secured by the other candidate = x – 7x/100 = 3x/10

Given, 7x/10 – 3x/10 = 172 => 4x/10 = 172

=> 4x = 1720 => x = 430.IncorrectLet the total number of valid votes be x.

70% of x = 70/100 * x = 7x/10

Number of votes secured by the other candidate = x – 7x/100 = 3x/10

Given, 7x/10 – 3x/10 = 172 => 4x/10 = 172

=> 4x = 1720 => x = 430. - Question 35 of 50
##### 35. Question

There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 580. The number of hares in the zoo is?

Correct200*2 = 400

580

—–

180

1—-2

?—-180 = 90Incorrect200*2 = 400

580

—–

180

1—-2

?—-180 = 90 - Question 36 of 50
##### 36. Question

In a three digit number, the hundred digit is 2 more than the tens digit and the units digit is 2 less than the tens digit. If the sum of the digits is 18, find the number?

CorrectLet the three digit numbers be 100a + 10b + c

a = b + 2

c = b – 2

a + b + c = 3b = 18 => b = 6

So a = 8 and b = 4

Hence the three digit number is: 864IncorrectLet the three digit numbers be 100a + 10b + c

a = b + 2

c = b – 2

a + b + c = 3b = 18 => b = 6

So a = 8 and b = 4

Hence the three digit number is: 864 - Question 37 of 50
##### 37. Question

Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

CorrectSI = 40 + 40

CI = 40 + 40 + 1.6 = 81.6IncorrectSI = 40 + 40

CI = 40 + 40 + 1.6 = 81.6 - Question 38 of 50
##### 38. Question

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are picked at random, what is the probability that they are all blue?

CorrectGiven that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455IncorrectGiven that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455 - Question 39 of 50
##### 39. Question

A, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?

Correct1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days

Incorrect1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days

- Question 40 of 50
##### 40. Question

Three years ago the average age of a family of six members was 19 years. A boy have been born, the average age of the family is the same today. What is the age of the boy?

Correct6 * 22 = 132

7 * 19 = 133

————–

1Incorrect6 * 22 = 132

7 * 19 = 133

————–

1 - Question 41 of 50
##### 41. 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?

CorrectM = 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 = 1.5IncorrectM = 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 42 of 50
##### 42. Question

The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?

CorrectLCM = 1400

1400 – 6 = 1394IncorrectLCM = 1400

1400 – 6 = 1394 - Question 43 of 50
##### 43. Question

The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:

CorrectSpeed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.

IncorrectSpeed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.

- Question 44 of 50
##### 44. Question

Find the one which does not belong to that group ?

CorrectAll are blood relation except Daughter-in-law.

IncorrectAll are blood relation except Daughter-in-law.

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

With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 4 km/hr, the same distance could have been covered in 7 1/2 hours. What is the distance covered?

CorrectLet the distance be x km. Then,

x/(7 1/2) – x/8 = 4

2x/15 – x/8 = 4 => x = 480 km.IncorrectLet the distance be x km. Then,

x/(7 1/2) – x/8 = 4

2x/15 – x/8 = 4 => x = 480 km. - Question 46 of 50
##### 46. Question

? % of 400 + 40% of 160 = 17% of 400

Correct? % of 400 + 40% of 160 = 17% of 400

?/100 of 400 + 40/100 of 160 = 17/100 of 400

?(4) + 64 = 68 => ? = 1Incorrect? % of 400 + 40% of 160 = 17% of 400

?/100 of 400 + 40/100 of 160 = 17/100 of 400

?(4) + 64 = 68 => ? = 1 - Question 47 of 50
##### 47. Question

Two pipes can separately fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is full, a leak develops in the tank through which one-third of water supplied by both the pipes goes out. What is the total time taken to fill the tank?

Correct1/20 + 1/30 = 1/12

1 + 1/3 = 4/3

1 — 12

4/3 — ?

4/3 * 12 = 16 hrsIncorrect1/20 + 1/30 = 1/12

1 + 1/3 = 4/3

1 — 12

4/3 — ?

4/3 * 12 = 16 hrs - Question 48 of 50
##### 48. Question

In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case?

CorrectAssume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)

= 1200(30)(3) kg

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg

As the same provisions are available

=> 1200(30)(3) = (1200 + x)(25)(2.5)

x = [1200(30)(3)]/[(25)(2.5)] – 1200

x = 528IncorrectAssume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)

= 1200(30)(3) kg

Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg

As the same provisions are available

=> 1200(30)(3) = (1200 + x)(25)(2.5)

x = [1200(30)(3)]/[(25)(2.5)] – 1200

x = 528 - Question 49 of 50
##### 49. Question

Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?

CorrectRelative speed = (42 + 30) * 5/18 = 4 * 5 = 20 mps.

Distance covered in passing each other = 120 + 280 = 400 m.

The time required = d/s = 400/20 = 20 sec.IncorrectRelative speed = (42 + 30) * 5/18 = 4 * 5 = 20 mps.

Distance covered in passing each other = 120 + 280 = 400 m.

The time required = d/s = 400/20 = 20 sec. - Question 50 of 50
##### 50. 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