RRB JE Mock Test series, RRB JE Online Test 2021, RRB JE MCQ
RRB JE Mock Test series, RRB JE Online Test in English Series 2
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RRB JE Mock Test series 2, RRB JE Online Test in English Series 2. RRB Exam Online Test 2021, RRB JE Free Mock Test Exam 2021. RRB JE Exam Free Online Quiz 2021, RRB JE Full Online Mock Test Series 2nd in English. RRB Online Test for All Subjects, RRB JE Free Mock Test Series in English. RRB JE Free Mock Test Series 2. RRB JE English Language Online Test in English Series 2nd. RRB JE Quantitative Aptitude Quiz 2021, RRB JE Reasoning Ability Free Online Test. Take RRB JE Online Quiz. The RRB JE Full online mock test paper is free for all students. RRB JE Question and Answers in English and Hindi Series 2. Here we are providing RRB JE Full Mock Test Paper in English. RRB JE Mock Test Series 2nd 2021. Now Test your self for RRB JE Exam by using below quiz…
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Question 1 of 50
1. 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?
Correct
200 + 130 = 330
800 – 330 = 470
200 — 5000
470 — ? => 11750Incorrect
200 + 130 = 330
800 – 330 = 470
200 — 5000
470 — ? => 11750 
Question 2 of 50
2. Question
A sum of Rs. 125000 amounts to Rs. 15500 in 4 years at the rate of simple interest. What is the rate of interest?
Correct
S.I. = (15500 – 12500) = Rs. 3000\
Rate = (100 * 3000) / (12500 * 4) = 6%Incorrect
S.I. = (15500 – 12500) = Rs. 3000\
Rate = (100 * 3000) / (12500 * 4) = 6% 
Question 3 of 50
3. Question
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
Correct
Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 2/25 = 40 secIncorrect
Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 2/25 = 40 sec 
Question 4 of 50
4. Question
The diameters of two spheres are in the ratio 1:2 what is the ratio of their surface area?
Correct
Incorrect

Question 5 of 50
5. Question
0.003 * ? * 0.0003 = 0.00000027
Correct
0.003 * ? * 0.0003 = 0.00000027
3/1000 * ? * 3/10000 = 3/1000 * 3/1000 * 3/100
? = 3/10 = 0.3Incorrect
0.003 * ? * 0.0003 = 0.00000027
3/1000 * ? * 3/10000 = 3/1000 * 3/1000 * 3/100
? = 3/10 = 0.3 
Question 6 of 50
6. Question
Anand and Deepak started a business investing Rs. 22,500 and Rs. 35,000 respectively. Out of a total profit of Rs. 13,800, Deepak’s share is:
Correct
Ratio of their shares = 22500 : 35000 = 9:14
Deepak’s share = 13800 * 14/23 = Rs. 8400.Incorrect
Ratio of their shares = 22500 : 35000 = 9:14
Deepak’s share = 13800 * 14/23 = Rs. 8400. 
Question 7 of 50
7. Question
Find the least number of five digits which is exactly divisible by 12, 15 and 18?
Correct
LCM = 180
180) 10000 (55
9900
———
100
10000 + 180 – 100 = 10080Incorrect
LCM = 180
180) 10000 (55
9900
———
100
10000 + 180 – 100 = 10080 
Question 8 of 50
8. Question
If onethird of one fourth of a number is 15, then threetenth of that number is:
Correct
Let the number be x. Then,
1/3 of 1/4 of x = 15
x = 15 * 12 = 180
So, required number = (3/10 * 180) = 54.Incorrect
Let the number be x. Then,
1/3 of 1/4 of x = 15
x = 15 * 12 = 180
So, required number = (3/10 * 180) = 54. 
Question 9 of 50
9. Question
The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?
Correct
LCM = 48 + 3 = 51
Incorrect
LCM = 48 + 3 = 51

Question 10 of 50
10. Question
Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?
Correct
Incorrect

Question 11 of 50
11. Question
Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
Correct
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.Incorrect
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec. 
Question 12 of 50
12. Question
The ratio of the volumes of a cube to that of the sphere which will fit inside the cube is?
Correct
a^{3} : a^{3}/8 * 4/3 π => 6: π
Incorrect
a^{3} : a^{3}/8 * 4/3 π => 6: π

Question 13 of 50
13. Question
A mixture of 70 liters of wine and water contains 10% water. How much water must be added to make water 12 ½% of the total mixture?
Correct
70 * (10/100) = 7
Wine Water
87 1/2% 12 1/2%
87 1/2% —— 63
12 1/2% ——? => 97=2Incorrect
70 * (10/100) = 7
Wine Water
87 1/2% 12 1/2%
87 1/2% —— 63
12 1/2% ——? => 97=2 
Question 14 of 50
14. 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 15 of 50
15. Question
A person buys an article at Rs.500. At what price should he sell the article so as to make a profit of 20%?
Correct
Cost price = Rs.500
profit = 20% of 500 = Rs.100
Selling price = Cost price + Profit
= 500 + 100 = 600Incorrect
Cost price = Rs.500
profit = 20% of 500 = Rs.100
Selling price = Cost price + Profit
= 500 + 100 = 600 
Question 16 of 50
16. Question
The current of a stream runs at the rate of 4 kmph. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of the boat in still water?
Correct
S = 4
M = x
DS = x + 4
US = x – 4
6/(x + 4) + 6/(x – 4) = 2
x = 8Incorrect
S = 4
M = x
DS = x + 4
US = x – 4
6/(x + 4) + 6/(x – 4) = 2
x = 8 
Question 17 of 50
17. Question
Visitors to show were charged Rs.15 each on the first day. Rs.7.50 on the second day, Rs.2.50 on the third day and total attendance on the three days were in ratio 2:5:13 respectively. The average charge per person for the whole show is?
Correct
2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5Incorrect
2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5 
Question 18 of 50
18. Question
How many numbers up to 100 are divisible by 7?
Correct
100/7 = 14 2/7 => 14 Numbers
Incorrect
100/7 = 14 2/7 => 14 Numbers

Question 19 of 50
19. Question
A sells his goods 50% cheaper than B but 50% dearer than C. The cheapest is?
Correct
Let B = 100
A = 50
C * (150/100) = 50
3C = 100
C = 33.3 then ‘C’ CheapestIncorrect
Let B = 100
A = 50
C * (150/100) = 50
3C = 100
C = 33.3 then ‘C’ Cheapest 
Question 20 of 50
20. Question
What is the greater of the two numbers whose product is 2560, given that the sum of the two numbers exceeds their difference by 64?
Correct
Let the greater and the smaller number be g and s respectively.
gs = 2560
g + s exceeds g – s by 64 i.e., g + s – (g – s) = 64
i.e., 2s = 64 => s = 32.
g = 2560/s = 80.Incorrect
Let the greater and the smaller number be g and s respectively.
gs = 2560
g + s exceeds g – s by 64 i.e., g + s – (g – s) = 64
i.e., 2s = 64 => s = 32.
g = 2560/s = 80. 
Question 21 of 50
21. Question
A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7.30 a.m. They shall first cross each other at?
Correct
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = 6 – 1 = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min.Incorrect
Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two.
Relative speed of A and B = 6 – 1 = 5 rounds per hour.
Time taken to complete one round at this speed = 1/5 hr = 12 min. 
Question 22 of 50
22. 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 23 of 50
23. 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 24 of 50
24. Question
Find the greatest number which leaves the same remainder when it divides 25, 57 and 105.
Correct
105 – 57 = 48
57 – 25 = 32
105 – 25 = 80
The H.C.F of 32, 48 and 80 is 16.Incorrect
105 – 57 = 48
57 – 25 = 32
105 – 25 = 80
The H.C.F of 32, 48 and 80 is 16. 
Question 25 of 50
25. Question
Find the mean proportional of the following pairs of numbers 6 and 24.
Correct
Incorrect

Question 26 of 50
26. Question
Vijay lent out an amount Rs. 10000 into two parts, one at 8% p.a. and the remaining at 10% p.a. both on simple interest. At the end of the year he received Rs. 890 as total interest. What was the amount he lent out at 8% pa.a?
Correct
Let the amount lent out at 8% p.a. be Rs. A
=> (A * 8)/100 + [(10000 – A) * 10]/100 = 890
=> A = Rs. 5500.Incorrect
Let the amount lent out at 8% p.a. be Rs. A
=> (A * 8)/100 + [(10000 – A) * 10]/100 = 890
=> A = Rs. 5500. 
Question 27 of 50
27. Question
A take twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in?
Correct
Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.Incorrect
Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work. 
Question 28 of 50
28. Question
The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
Correct
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.Incorrect
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l^{2} + b^{2}) = (1 + 108 1/3 %)lb
=> l^{2} + b^{2} = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l^{2} + b^{2})/lb = 25/12
12(l^{2} + b^{2}) = 25lb
Adding 24lb on both sides
12l^{2} + 12b^{2} + 24lb = 49lb
12(l^{2} + b^{2} + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)^{2} = 49lb
=> 12(14)^{2} = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m. 
Question 29 of 50
29. Question
If 10% of x = 20% of y, then x:y is equal to:
Correct
10% of x = 20% of y
10x/100 = 20y/100 => x/10 = y/5
x/y = 10/5 = 2/1
x:y = 2:1.Incorrect
10% of x = 20% of y
10x/100 = 20y/100 => x/10 = y/5
x/y = 10/5 = 2/1
x:y = 2:1. 
Question 30 of 50
30. Question
Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
Correct
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15Incorrect
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15 
Question 31 of 50
31. Question
If the roots of the equation 2x^{2} – 5x + b = 0 are in the ratio of 2:3, then find the value of b?
Correct
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = ( 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3.Incorrect
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = ( 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3. 
Question 32 of 50
32. Question
36 * 48 ÷ 64 + 36 ÷ 12 = ?
Correct
36 * 48 / 64 + 36/12 = 27 + 3 = 30
Incorrect
36 * 48 / 64 + 36/12 = 27 + 3 = 30

Question 33 of 50
33. Question
A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously, the probability that both are of the same colour is .
Correct
Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = ⁷C₂/¹⁵C₂ + ⁸C₂/¹⁵C₂ = 7/15Incorrect
Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = ⁷C₂/¹⁵C₂ + ⁸C₂/¹⁵C₂ = 7/15 
Question 34 of 50
34. Question
The diagonals of a rhombus are 15 cm and 20 cm. Find its area?
Correct
1/2 * 15 * 20 = 150
Incorrect
1/2 * 15 * 20 = 150

Question 35 of 50
35. 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?
Correct
5% – 2% = 3%
3% — 5
100% — ? => 166 2/3Incorrect
5% – 2% = 3%
3% — 5
100% — ? => 166 2/3 
Question 36 of 50
36. Question
A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?
Correct
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.Incorrect
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000. 
Question 37 of 50
37. Question
A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
Correct
We need to know the S.I., principal and time to find the rate. Since the principal is not given, so data is inadequate.
Incorrect
We need to know the S.I., principal and time to find the rate. Since the principal is not given, so data is inadequate.

Question 38 of 50
38. 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?
Correct
They 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.Incorrect
They 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 39 of 50
39. Question
[(3.572)^{3} + (2.428)^{3}] / [(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}] is equal to .
Correct
[(3.572)^{3} + (2.428)^{3}] / [(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}]
= {(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}}{3.752 + 2.428} / [(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}]
= 3.752 + 2.428 = 6Incorrect
[(3.572)^{3} + (2.428)^{3}] / [(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}]
= {(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}}{3.752 + 2.428} / [(3.572)^{2} – (3.572) * (2.428) + (2.428)^{2}]
= 3.752 + 2.428 = 6 
Question 40 of 50
40. Question
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
Correct
Let the two consecutive positive integers be x and x + 1
x^{2} + (x + 1)^{2} – x(x + 1) = 91
x^{2} + x – 90 = 0
(x + 10)(x – 9) = 0 => x = 10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.Incorrect
Let the two consecutive positive integers be x and x + 1
x^{2} + (x + 1)^{2} – x(x + 1) = 91
x^{2} + x – 90 = 0
(x + 10)(x – 9) = 0 => x = 10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10. 
Question 41 of 50
41. Question
Two trains travelling in the same direction at 40 and 22 kmph completely pass off another in 1 minute. If the length of the first train is 125 m, what is the length of the second train?
Correct
RS = 40 – 22 = 18 * 5/18 = 5 mps
T = 60 sec
D = 5 * 60 = 300 m
125
——–
175 mIncorrect
RS = 40 – 22 = 18 * 5/18 = 5 mps
T = 60 sec
D = 5 * 60 = 300 m
125
——–
175 m 
Question 42 of 50
42. 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 43 of 50
43. Question
In how many ways can three consonants and two vowels be selected from the letters of the word “TRIANGLE”?
Correct
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.Incorrect
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways. 
Question 44 of 50
44. 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?
Correct
Assume 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 = 528Incorrect
Assume 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 45 of 50
45. Question
105 * 99 * 299 = ?
Correct
105 * 99 * 299 = 105 * 100 * 300 = 315000
Incorrect
105 * 99 * 299 = 105 * 100 * 300 = 315000

Question 46 of 50
46. Question
Two persons A and B can complete a piece of work in 30 days and 45 days respectively. If they work together, what part of the work will be completed in 3 days?
Correct
A’s one day’s work = 1/30
B’s one day’s work = 1/45
(A + B)’s one day’s work = 1/30 + 1/45 = 1/18
The part of the work completed in 3 days = 3 (1/18) = 1/6.Incorrect
A’s one day’s work = 1/30
B’s one day’s work = 1/45
(A + B)’s one day’s work = 1/30 + 1/45 = 1/18
The part of the work completed in 3 days = 3 (1/18) = 1/6. 
Question 47 of 50
47. Question
Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.441 in 2 years?
Correct
441 = P(21/20)^{2}
P = 400Incorrect
441 = P(21/20)^{2}
P = 400 
Question 48 of 50
48. Question
The duplicate ratio of 3:4 is?
Correct
3^{2}: 4^{2} = 9:16
Incorrect
3^{2}: 4^{2} = 9:16

Question 49 of 50
49. Question
A laborer is engaged for 30 days on the condition that he receives Rs.25 for each day he works and is fined Rs.7.50 for each day is absent. He gets Rs.425 in all. For how many days was he absent?
Correct
30 * 25 = 750
425
———–
325
25 + 7.50 = 32.5
325/32.5 = 10Incorrect
30 * 25 = 750
425
———–
325
25 + 7.50 = 32.5
325/32.5 = 10 
Question 50 of 50
50. Question
A trader marks his articles 20% more than the cost price. If he allows 20% discount, then find his gain or loss percent?
Correct
Let CP of an article = RS. 100
MP= Rs. 120
Discount = 20%
SP = M[(100 – d%)/100] = 120(80/100) = Rs. 96
Clearly, the trader gets 4% loss.Incorrect
Let CP of an article = RS. 100
MP= Rs. 120
Discount = 20%
SP = M[(100 – d%)/100] = 120(80/100) = Rs. 96
Clearly, the trader gets 4% loss.