RBI Assistant Mains Test Series, RBI Assistant Mains Online Test Series
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 2020, RBI Assistant Mains Free Mock Test Exam 2020. RBI Assistant Mains Exam Free Online Quiz 2020. 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 2020. 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?
Correct
1 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.Incorrect
1 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%?
Correct
C.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 = 5Incorrect
C.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?
Correct
Given, 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.Incorrect
Given, 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?
Correct
Area of an equilateral triangle = √3/4 S^{2}
If S = 16, Area of triangle = √3/4 * 16 * 16 = 64√3 cm^{2};Incorrect
Area 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?
Correct
3000 – 3 = 2997
Incorrect
3000 – 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:
Correct
Let 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.Incorrect
Let 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 :
Correct
Let 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) = 29:22Incorrect
Let 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?
Correct
3: 4
A = 3/7*210 = 90
390: 400
39:40Incorrect
3: 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?
Correct
Simple interest = (18000 * 4 * 15)/100 = Rs. 10800
Amount = P + I = 18000 + 10800 = Rs. 28800Incorrect
Simple 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?
Correct
10 * 150 = 1500
3 * 100 + 5 * 150 = 1050
1500 – 1050 = 450Incorrect
10 * 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?
Correct
Incorrect

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.
Correct
15/25 + 10/x = 1 => x = 25
1/25 + 1/25 = 2/25
25/2 = 12 1/2 daysIncorrect
15/25 + 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.
Correct
Let 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.256.25Incorrect
Let 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?
Correct
Incorrect

Question 15 of 50
15. Question
32% of 425 – ?% of 250 = 36
Correct
32/100 * 425 – x/100 * 250 = 36
=> x/100 * 250 = 136 – 36 = 100
=> x = (100 * 100)/250 = 40Incorrect
32/100 * 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 twothirds of the total amount with P and Q. Find the amount with R?
Correct
Let 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.Incorrect
Let 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?
Correct
Let x percent of 80 be 64.
80 * x/100 = 64 => x = (64 * 100)/80 => x = 80.
80% of 80 is 64.Incorrect
Let 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?
Correct
They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
6000 : 8000 => 3:4Incorrect
They 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:
Correct
4 * 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.Incorrect
4 * 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?
Correct
20 15 5
25 20 8
500 + 300 + 40 = 840/40 = 21Incorrect
20 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:
Correct
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200Incorrect
Let 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?
Correct
P(1 + 20/100)^{n} > 2P or (6/5)^{n} > 2
Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years.Incorrect
P(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:
Correct
Excluded number = (27 * 5) – (25 * 4) = 135 – 100 = 35.
Incorrect
Excluded 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?
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 25 of 50
25. Question
5358 * 51 = ?
Correct
5358 * 51 = 5358 * (50 + 1)
= 5358 * 50 + 5358 * 1
= 267900 + 5358 = 273258Incorrect
5358 * 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.
Correct
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meterIncorrect
Speed = 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?
Correct
Let 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.Incorrect
Let 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?
Correct
T = 9/45 * 60 = 12
Incorrect
T = 9/45 * 60 = 12

Question 29 of 50
29. Question
A twodigit 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:
Correct
Let 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.Incorrect
Let 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%?
Correct
15 CP = 18 SP
18 — 3 CP loss
100 — ? => 16 2/3% lossIncorrect
15 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?
Correct
504/M = 384/800
(504 * 800) / 384 = M
M = 1050Incorrect
504/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?
Correct
130
14
——
361—— 144
100%——? => 400Incorrect
130
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?
Correct
2 * 22/7 * 14 = 88
88 * 1 1/2 = Rs.132Incorrect
2 * 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?
Correct
The 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.Incorrect
The 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:
Correct
Sum 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.Incorrect
Sum 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?
Correct
Let 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.Incorrect
Let 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 .
Correct
x 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!) = 4320Incorrect
x 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?
Correct
X * (110/100) * (95/100) = 1045
X * (11/10) * (1/100) = 11
X = 1000Incorrect
X * (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:
Correct
Required 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) = 30/7Incorrect
Required 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 = ?
Correct
Given Exp. = – 84 * (30 – 1) + 365
= – (84 * 30) + 84 + 365
= – 2520 + 449 = – 2071Incorrect
Given 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?
Correct
Incorrect

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?
Correct
C.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 %Incorrect
C.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?
Correct
The 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.Incorrect
The 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?
Correct
Let 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.Incorrect
Let 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 onethird 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?
Correct
Work 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.Incorrect
Work 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?
Correct
Let 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. 1500Incorrect
Let 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.
Correct
Let 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 mIncorrect
Let 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?
Correct
B = 1/8 – 1/2 = 1/24 => 24 days
Incorrect
B = 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?
Correct
The 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}Incorrect
The 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}