# RBI Assistant Mains Test Series, RBI Assistant Mains Online Test Series 1

## RBI Assistant Mains Test Series, RBI Assistant Mains Online Test Series 1

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RBI Assistant Mains Test Series, RBI Assistant Mains Online Test Series 1. RBI Exam Online Test 2019, RBI Assistant Mains Free Mock Test Exam 2019. RBI Assistant Mains Exam Free Online Quiz 2019**. ** To Analyse your preparation, one should attempt the test series on a regular basis to score more in the examination. Start FREE mock test and get top rank among all applicants whoever are participating in the exam. RBI Assistant Mains Question and Answers in English and Hindi **Series 1**. Here we are providing** RBI Assistant Mains Full Mock Test Paper in English. RBI Assistant Mains **Mock Test **Series 1st** 2019. Now Test your self for RBI Assistant Mains Exam by using below quiz…

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

12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it so to be completed in 3 days?

Correct1 man’s 1 day work = 1/48; 1 woman’s 1 day work = 1/60.

6 men’s 2 day’s work = 6/48 * 2 = 1/4.

Remaining work = (1 – 1/4) = 3/4

Now, 1/60 work is done in 1 day by 1 woman.

So, 3/4 work will be done in 3 days by (60 * 3/4 * 1/3) = 15 women.Incorrect1 man’s 1 day work = 1/48; 1 woman’s 1 day work = 1/60.

6 men’s 2 day’s work = 6/48 * 2 = 1/4.

Remaining work = (1 – 1/4) = 3/4

Now, 1/60 work is done in 1 day by 1 woman.

So, 3/4 work will be done in 3 days by (60 * 3/4 * 1/3) = 15 women. - Question 2 of 50
##### 2. Question

A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

CorrectC.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5

For Rs. 6/5, toffees sold = 6. For re. 1.

Toffees sold = 6 * 5/6 = 5IncorrectC.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5

For Rs. 6/5, toffees sold = 6. For re. 1.

Toffees sold = 6 * 5/6 = 5 - Question 3 of 50
##### 3. Question

40% of Ram’s marks is equal to 20% of Rahim’s marks which percent is equal to 30% of Robert’s marks. If Robert’s marks is 80, then find the average marks of Ram and Rahim?

CorrectGiven, 40% of Ram’s marks = 20% of Rahim’s marks = 30% of Robert’s marks.

Given, marks of Robert = 80

30% of 80 = 30/100 * 8 = 24

Given, 40% of Ram’s marks = 24.

=> Ram’s marks = (24 * 100)/40 = 60

Also, 20% of Rahim’s marks = 24

=> Rahim’s marks = (24 * 100)/20 = 120

Average marks of Ram and Rahim = (60 + 120)/2 = 90.IncorrectGiven, 40% of Ram’s marks = 20% of Rahim’s marks = 30% of Robert’s marks.

Given, marks of Robert = 80

30% of 80 = 30/100 * 8 = 24

Given, 40% of Ram’s marks = 24.

=> Ram’s marks = (24 * 100)/40 = 60

Also, 20% of Rahim’s marks = 24

=> Rahim’s marks = (24 * 100)/20 = 120

Average marks of Ram and Rahim = (60 + 120)/2 = 90. - Question 4 of 50
##### 4. Question

What is the are of an equilateral triangle of side 16 cm?

CorrectArea of an equilateral triangle = √3/4 S

^{2}

If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm^{2};IncorrectArea of an equilateral triangle = √3/4 S

^{2}

If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm^{2}; - Question 5 of 50
##### 5. Question

What is the difference between the local values of 3 in the number 53403?

Correct3000 – 3 = 2997

Incorrect3000 – 3 = 2997

- Question 6 of 50
##### 6. Question

The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:

CorrectLet P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 * 2) = 10100 — (i)

Q + R = (6250 * 2) = 12500 — (ii)

P + R = (5200 * 2) = 10400 — (iii)

Adding (i), (ii) and (iii), we get:

2(P + Q + R) = 33000 = P + Q + R = 16500 — (iv)

Subtracting (ii) from (iv), we get, P = 4000.

P’s monthly income = Rs. 4000.IncorrectLet P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 * 2) = 10100 — (i)

Q + R = (6250 * 2) = 12500 — (ii)

P + R = (5200 * 2) = 10400 — (iii)

Adding (i), (ii) and (iii), we get:

2(P + Q + R) = 33000 = P + Q + R = 16500 — (iv)

Subtracting (ii) from (iv), we get, P = 4000.

P’s monthly income = Rs. 4000. - Question 7 of 50
##### 7. Question

If x;Y = 5:2, then (8x + 9y):(8x + 2y) is :

CorrectLet x = 5k and y = 2k.

Then, (8x + 9y)/(8x + 2y)

= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22

(8x + 9y):(8x + 2y) = 29:22IncorrectLet x = 5k and y = 2k.

Then, (8x + 9y)/(8x + 2y)

= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22

(8x + 9y):(8x + 2y) = 29:22 - Question 8 of 50
##### 8. Question

A and B put in Rs.300 and Rs.400 respectively into a business. A reinvests into the business his share of the first year’s profit of Rs.210 where as B does not. In what ratio should they divide the second year’s profit?

Correct3: 4

A = 3/7*210 = 90

390: 400

39:40Incorrect3: 4

A = 3/7*210 = 90

390: 400

39:40 - Question 9 of 50
##### 9. Question

What amount does Kiran get if he invests Rs. 18000 at 15% p.a. simple interest for four years?

CorrectSimple interest = (18000 * 4 * 15)/100 = Rs. 10800

Amount = P + I = 18000 + 10800 = Rs. 28800IncorrectSimple interest = (18000 * 4 * 15)/100 = Rs. 10800

Amount = P + I = 18000 + 10800 = Rs. 28800 - Question 10 of 50
##### 10. Question

A man purchased 3 blankets @ Rs.100 each, 5 blankets @ Rs.150 each and two blankets at a certain rate which is now slipped off from his memory. But he remembers that the average price of the blankets was Rs.150. Find the unknown rate of two blankets?

Correct10 * 150 = 1500

3 * 100 + 5 * 150 = 1050

1500 – 1050 = 450Incorrect10 * 150 = 1500

3 * 100 + 5 * 150 = 1050

1500 – 1050 = 450 - Question 11 of 50
##### 11. Question

Find the greatest number which will divide 25, 73 and 97 as so to leave the same remainder in each case?

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

A can do a piece of work in 12 days. He worked for 15 days and then B completed the remaining work in 10 days. Both of them together will finish it in.

Correct15/25 + 10/x = 1 => x = 25

1/25 + 1/25 = 2/25

25/2 = 12 1/2 daysIncorrect15/25 + 10/x = 1 => x = 25

1/25 + 1/25 = 2/25

25/2 = 12 1/2 days - Question 13 of 50
##### 13. Question

A man gets a simple interest of Rs.500 on a certain principal at the rate of 5% p.a in two years. Find the compound interest the man will get on twice the principal in two years at the same rate.

CorrectLet the principal be Rs.P

S.I at 5% p.a in 8 years on Rs.P = Rs.500

(P)(8)(5)/100 = 500

P = 1250

C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years

=2500{ [1 + 5/100]^{2}– 1} = 2500{ 21^{2}– 20^{2}/20^{2}}

= 2500/400(441 – 400)

= 25/4(41) = 1025/4 = Rs.256.25IncorrectLet the principal be Rs.P

S.I at 5% p.a in 8 years on Rs.P = Rs.500

(P)(8)(5)/100 = 500

P = 1250

C.I on Rs.2P i.e., Rs.2500 at 5% p.a in two years

=2500{ [1 + 5/100]^{2}– 1} = 2500{ 21^{2}– 20^{2}/20^{2}}

= 2500/400(441 – 400)

= 25/4(41) = 1025/4 = Rs.256.25 - Question 14 of 50
##### 14. Question

The smallest ratio out of 1:1, 2:1, 1:3 and 3:1 is?

CorrectIncorrect - Question 15 of 50
##### 15. Question

32% of 425 – ?% of 250 = 36

Correct32/100 * 425 – x/100 * 250 = 36

=> x/100 * 250 = 136 – 36 = 100

=> x = (100 * 100)/250 = 40Incorrect32/100 * 425 – x/100 * 250 = 36

=> x/100 * 250 = 136 – 36 = 100

=> x = (100 * 100)/250 = 40 - Question 16 of 50
##### 16. Question

P, Q and R have Rs.6000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R?

CorrectLet the amount with R be Rs.r

r = 2/3 (total amount with P and Q)

r = 2/3(6000 – r) => 3r = 12000 – 2r

=> 5r = 12000 => r = 2400.IncorrectLet the amount with R be Rs.r

r = 2/3 (total amount with P and Q)

r = 2/3(6000 – r) => 3r = 12000 – 2r

=> 5r = 12000 => r = 2400. - Question 17 of 50
##### 17. Question

64 is what percent of 80?

CorrectLet x percent of 80 be 64.

80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.

80% of 80 is 64.IncorrectLet x percent of 80 be 64.

80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.

80% of 80 is 64. - Question 18 of 50
##### 18. Question

A and B start a business with Rs.6000 and Rs.8000 respectively. Hoe should they share their profits at the end of one year?

CorrectThey should share the profits in the ratio of their investments.

The ratio of the investments made by A and B =

6000 : 8000 => 3:4IncorrectThey should share the profits in the ratio of their investments.

The ratio of the investments made by A and B =

6000 : 8000 => 3:4 - Question 19 of 50
##### 19. Question

H.C.F of 4 * 27 * 3125, 8 * 9 * 25 * 7 and 16 * 81 * 5 * 11 * 49 is:

Correct4 * 27 * 3125 = 2

^{2}* 3^{3}* 5^{5};

8 * 9 * 25 * 7 = 2^{3}* 3^{2}* 5^{2}* 7;

16 * 81 * 5 * 11 * 49 = 2^{4}* 3^{4}* 5 * 7^{2}* 11

H.C.F = 2^{2}* 3^{2}* 5 = 180.Incorrect4 * 27 * 3125 = 2

^{2}* 3^{3}* 5^{5};

8 * 9 * 25 * 7 = 2^{3}* 3^{2}* 5^{2}* 7;

16 * 81 * 5 * 11 * 49 = 2^{4}* 3^{4}* 5 * 7^{2}* 11

H.C.F = 2^{2}* 3^{2}* 5 = 180. - Question 20 of 50
##### 20. Question

A building contractor employs 20 male, 15 female and 5 child workers. To a male worker he pays Rs.25 per day, to a female worker Rs.20 per day and a child worker Rs.8 per day. The average wage per day paid by the contractor is?

Correct20 15 5

25 20 8

500 + 300 + 40 = 840/40 = 21Incorrect20 15 5

25 20 8

500 + 300 + 40 = 840/40 = 21 - Question 21 of 50
##### 21. 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 22 of 50
##### 22. Question

The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is?

CorrectP(1 + 20/100)

^{n}> 2P or (6/5)^{n}> 2

Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years.IncorrectP(1 + 20/100)

^{n}> 2P or (6/5)^{n}> 2

Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years. - Question 23 of 50
##### 23. Question

The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is:

CorrectExcluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.

IncorrectExcluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.

- Question 24 of 50
##### 24. 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?

CorrectTotal 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.IncorrectTotal 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 25 of 50
##### 25. Question

5358 * 51 = ?

Correct5358 * 51 = 5358 * (50 + 1)

= 5358 * 50 + 5358 * 1

= 267900 + 5358 = 273258Incorrect5358 * 51 = 5358 * (50 + 1)

= 5358 * 50 + 5358 * 1

= 267900 + 5358 = 273258 - Question 26 of 50
##### 26. Question

A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.

CorrectSpeed = 60*(5/18) m/sec = 50/3 m/sec

Length of Train (Distance) = Speed * Time

(50/3) * 9 = 150 meterIncorrectSpeed = 60*(5/18) m/sec = 50/3 m/sec

Length of Train (Distance) = Speed * Time

(50/3) * 9 = 150 meter - Question 27 of 50
##### 27. Question

A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then, A’s speed is equal to?

CorrectLet A’s speed = x km/hr. Then, B’s speed = (7 – x) km/ hr.

So, 24/x + 24/(7 – x) = 14

x^{2}– 98x + 168 = 0

(x – 3)(x – 4) = 0 => x = 3 or 4.

Since, A is faster than B, so A’s speed = 4 km/hr and B’s speed = 3 km/hr.IncorrectLet A’s speed = x km/hr. Then, B’s speed = (7 – x) km/ hr.

So, 24/x + 24/(7 – x) = 14

x^{2}– 98x + 168 = 0

(x – 3)(x – 4) = 0 => x = 3 or 4.

Since, A is faster than B, so A’s speed = 4 km/hr and B’s speed = 3 km/hr. - Question 28 of 50
##### 28. Question

Excluding stoppages, the speed of a train is 45 kmph and including stoppages it is 36 kmph. Of how many minutes does the train stop per hour?

CorrectT = 9/45 * 60 = 12

IncorrectT = 9/45 * 60 = 12

- Question 29 of 50
##### 29. Question

A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

CorrectLet the ten’s and unit’s digit be x and 8/x respectively.

Then,

(10x + 8/x) + 18 = 10 * 8/x + x

9x^{2}+ 18x – 72 = 0

x^{2}+ 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2

So, ten’s digit = 2 and unit’s digit = 4.

Hence, required number = 24.IncorrectLet the ten’s and unit’s digit be x and 8/x respectively.

Then,

(10x + 8/x) + 18 = 10 * 8/x + x

9x^{2}+ 18x – 72 = 0

x^{2}+ 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2

So, ten’s digit = 2 and unit’s digit = 4.

Hence, required number = 24. - Question 30 of 50
##### 30. Question

The C.P of 15 books is equal to the S.P of 18 books. Find his gain% or loss%?

Correct15 CP = 18 SP

18 — 3 CP loss

100 — ? => 16 2/3% lossIncorrect15 CP = 18 SP

18 — 3 CP loss

100 — ? => 16 2/3% loss - Question 31 of 50
##### 31. Question

In an exam, a candidate secured 504 marks our of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 384 marks. What is the value of M?

Correct504/M = 384/800

(504 * 800) / 384 = M

M = 1050Incorrect504/M = 384/800

(504 * 800) / 384 = M

M = 1050 - Question 32 of 50
##### 32. Question

An engineering student has to secure 36% marks to pass. He gets 130 marks and fails by 14 marks. The maximum No. of marks obtained by him is?

Correct130

14

——-

361—— 144

100%——? => 400Incorrect130

14

——-

361—— 144

100%——? => 400 - Question 33 of 50
##### 33. Question

Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter?

Correct2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132Incorrect2 * 22/7 * 14 = 88

88 * 1 1/2 = Rs.132 - Question 34 of 50
##### 34. Question

What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected?

CorrectThe number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women

= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5

= 30 ways.IncorrectThe number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women

= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5

= 30 ways. - Question 35 of 50
##### 35. Question

The average age of 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:

CorrectSum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) = 90 years.

Sum of the present age of wife and child = (20 * 2 + 5 * 2) = 50 years.

Husband’s present age = (90 – 50) = 40 years.IncorrectSum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) = 90 years.

Sum of the present age of wife and child = (20 * 2 + 5 * 2) = 50 years.

Husband’s present age = (90 – 50) = 40 years. - Question 36 of 50
##### 36. Question

If 15% of 30% of 50% of a number is 90, then what is the number?

CorrectLet the number be a

Given, 15/100 * 30/100 * 50/100 * a = 90

=> 3/20 * 3/10 * 1/2 * a = 90

=> a = 10 * 20 * 10 * 2 = 4000.IncorrectLet the number be a

Given, 15/100 * 30/100 * 50/100 * a = 90

=> 3/20 * 3/10 * 1/2 * a = 90

=> a = 10 * 20 * 10 * 2 = 4000. - Question 37 of 50
##### 37. Question

The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is -.

Correctx Not younger_______ ↑

The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.

The required number of ways = 6(6!) = 4320Incorrectx Not younger_______ ↑

The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.

The required number of ways = 6(6!) = 4320 - Question 38 of 50
##### 38. Question

The salary of a typist was at first raised by 10% and then the same was reduced by 5%. If he presently draws Rs.1045.What was his original salary?

CorrectX * (110/100) * (95/100) = 1045

X * (11/10) * (1/100) = 11

X = 1000IncorrectX * (110/100) * (95/100) = 1045

X * (11/10) * (1/100) = 11

X = 1000 - Question 39 of 50
##### 39. Question

The smallest fraction, which each of 6/7, 5/14, 10/21 will divide exactly is:

CorrectRequired fraction = L.C.M of 6/7, 5/14, 10/21

= (L.C.M of 6, 5, 10) / (H.C.F of 7, 14, 21) = 30/7IncorrectRequired fraction = L.C.M of 6/7, 5/14, 10/21

= (L.C.M of 6, 5, 10) / (H.C.F of 7, 14, 21) = 30/7 - Question 40 of 50
##### 40. Question

– 84 * 29 + 365 = ?

CorrectGiven Exp. = – 84 * (30 – 1) + 365

= – (84 * 30) + 84 + 365

= – 2520 + 449 = – 2071IncorrectGiven Exp. = – 84 * (30 – 1) + 365

= – (84 * 30) + 84 + 365

= – 2520 + 449 = – 2071 - Question 41 of 50
##### 41. Question

On dividing number by 5, we get 3 as remainder. What will be the remainder when the square of this number is divided by 5?

CorrectIncorrect - Question 42 of 50
##### 42. Question

A shopkeeper purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs. 6.50 per kg. What will be his gain percent?

CorrectC.P. of 1 kg = 420/7 = Rs. 6

S.P. of 1 kg = Rs. 6.50

Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 %IncorrectC.P. of 1 kg = 420/7 = Rs. 6

S.P. of 1 kg = Rs. 6.50

Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 % - Question 43 of 50
##### 43. Question

How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?

CorrectThe distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.IncorrectThe distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 * 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec. - Question 44 of 50
##### 44. Question

A person invested in all Rs. 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is?

CorrectLet the parts be x, y and [2600 – (x + y)].

Then,

(x * 4 * 1)/100 = (y * 6 * 1)/100 = {[2600 – (x + y)] * 8 * 1}/100

y/x = 4/6 = 2/3 or y = 2/3 x

So, (x * 4 * 1)/100 = [(2600 – 5/3 x) * 80/100

52x = (7800 * 8) => x = 1200

Money invested at 4% = Rs. 1200.IncorrectLet the parts be x, y and [2600 – (x + y)].

Then,

(x * 4 * 1)/100 = (y * 6 * 1)/100 = {[2600 – (x + y)] * 8 * 1}/100

y/x = 4/6 = 2/3 or y = 2/3 x

So, (x * 4 * 1)/100 = [(2600 – 5/3 x) * 80/100

52x = (7800 * 8) => x = 1200

Money invested at 4% = Rs. 1200. - Question 45 of 50
##### 45. Question

P alone can complete a piece of work in 6 days. Work done by Q alone in one day is equal to one-third of the work done by P alone in one day. In how many days can the work be completed if P and Q work together?

CorrectWork done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.

Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total

They would take 9/2 days = 4 (1/2) days to complete the work working together.IncorrectWork done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.

Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total

They would take 9/2 days = 4 (1/2) days to complete the work working together. - Question 46 of 50
##### 46. Question

(27

^{2}/4^{-3})^{-5/6}= ?Correct(27

^{2}/4^{-3})^{-5/6}= [(3^{3})^{2}/(2^{2})^{-3}]^{-5/6}

= (3^{6}* 2^{6})^{-5/6 = (6}^{6})^{-5/6}= 1/6^{5}= 1/ 7776.Incorrect(27

^{2}/4^{-3})^{-5/6}= [(3^{3})^{2}/(2^{2})^{-3}]^{-5/6}

= (3^{6}* 2^{6})^{-5/6 = (6}^{6})^{-5/6}= 1/6^{5}= 1/ 7776. - Question 47 of 50
##### 47. Question

If an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price?

CorrectLet the cost price of an article be Rs. x.

(19% of x) – (12% of x) = 105

19x/100 – 12x/100 = 105 => 7x = 105 * 100

=> x = 1500

Cost price = Rs. 1500IncorrectLet the cost price of an article be Rs. x.

(19% of x) – (12% of x) = 105

19x/100 – 12x/100 = 105 => 7x = 105 * 100

=> x = 1500

Cost price = Rs. 1500 - Question 48 of 50
##### 48. Question

The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.

CorrectLet the radii of the smaller and the larger circles be s m and l m respectively.

2∏s = 264 and 2∏l = 352

s = 264/2∏ and l = 352/2∏

Difference between the areas = ∏l^{2}– ∏s^{2}

= ∏{176^{2}/∏^{2}– 132^{2}/∏^{2}}

= 176^{2}/∏ – 132^{2}/∏

= (176 – 132)(176 + 132)/∏

= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq mIncorrectLet the radii of the smaller and the larger circles be s m and l m respectively.

2∏s = 264 and 2∏l = 352

s = 264/2∏ and l = 352/2∏

Difference between the areas = ∏l^{2}– ∏s^{2}

= ∏{176^{2}/∏^{2}– 132^{2}/∏^{2}}

= 176^{2}/∏ – 132^{2}/∏

= (176 – 132)(176 + 132)/∏

= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m - Question 49 of 50
##### 49. Question

A and B together can do a piece of work in 8 days. If A alone can do the same work in 12 days, then B alone can do the same work in?

CorrectB = 1/8 – 1/2 = 1/24 => 24 days

IncorrectB = 1/8 – 1/2 = 1/24 => 24 days

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

A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle?

CorrectThe circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5

=> x = 1

Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2}IncorrectThe circumference of the circle is equal to the permeter of the rectangle.

Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5

=> x = 1

Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm^{2}