DCCB Quiz, DCCB Online Test in English, DCCB Mock Test Series
DCCB Quiz Series 5, DCCB Online Test in English Series 5, DCCB Mock Test
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DCCB Quiz Series 5, DCCB Online Test in English Series 5, DCCB Mock Test. DCCB Free Online Test Series 5. RRB Exam Online Test 2021, DCCB Free Mock Test Exam 2021. DCCB Exam Free Online Quiz 2021, DCCB Full Online Mock Test Series 5th in English. RRB Online Test for All Subjects, DCCB Free Mock Test Series in English. DCCB Free Mock Test Series 5. DCCB English Language Online Test in English Series 5th. DCCB Quantitative Aptitude Quiz 2021, DCCB Reasoning Ability Free Online Test. Take DCCB Online Quiz. The DCCB Full online mock test paper is free for all students. DCCB Question and Answers in English and Hindi Series 5. Here we are providing DCCB Full Mock Test Paper in English. DCCB Mock Test Series 5th 2021. Now Test your self for DCCB Exam by using below quiz…
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Question 1 of 50
1. Question
The simple interest accrued on an amount Rs.10,000 at the end of two years is same as the compound interest on Rs.8,000 at the end of two years. The rate of interest is same in both the cases. What is the rate of interest?
Correct
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50%Incorrect
Given that Rs.10,000 is invested in S.I for two years and Rs.8,000 in C.I for two years
=> C.I – S.I
=> 8,000 { [1 + r/100]^{2 } – 1} = (10,000)2r /100
=> 8{ 1 + 2r/100 + r^{2} / (100)^{2} – 1} = r/5
=> 16r/100 + 8r^{2}/(100)^{2} = 20r/100
=> 4r/10 = 8r^{2}/(100)^{2}
=> 8[r/100]^{2} – 4r/100 = 0
=> r/100 {8r/100 4} = 0
=> r = 0% of 50%
Since r!= 0%, r =50% 
Question 2 of 50
2. Question
The smallest number when increased by ” 1 ” is exactly divisible by 12, 18, 24, 32 and 40 is:
Correct
LCM = 1440
1440 – 1 = 1439Incorrect
LCM = 1440
1440 – 1 = 1439 
Question 3 of 50
3. Question
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is?
Correct
LCM = 1400
1400 – 6 = 1394Incorrect
LCM = 1400
1400 – 6 = 1394 
Question 4 of 50
4. Question
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can?
Correct
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T – 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 – 8 = 4/9(T – 8)
10T = 792 – 352 => T = 44.Incorrect
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T – 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 – 8 = 4/9(T – 8)
10T = 792 – 352 => T = 44. 
Question 5 of 50
5. Question
421 * 0.9 + 130 * 101 + 10000 = ?
Correct
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500Incorrect
421 * 0.9 + 130 * 101 + 10000
= 378.9 + 1313 + 10000 = 23508.9 = 23500 
Question 6 of 50
6. Question
The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?
Correct
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.Incorrect
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially. 
Question 7 of 50
7. Question
The wheels revolve round a common horizontal axis. They make 15, 20 and 48 revolutions in a minute respectively. Starting with a certain point on the circumference down wards. After what interval of time will they come together in the same position?
Correct
Time for one revolution = 60/15 = 4
60/ 20 = 3
60/48 = 5/4
LCM of 4, 3, 5/4
LCM of Numerators/HCF of Denominators =
60/1 = 60Incorrect
Time for one revolution = 60/15 = 4
60/ 20 = 3
60/48 = 5/4
LCM of 4, 3, 5/4
LCM of Numerators/HCF of Denominators =
60/1 = 60 
Question 8 of 50
8. Question
Twofifth of onethird of threeseventh of a number is 15. What is 40% of that number?
Correct
Let the number be x. Then,
2/5 of 1/3 of 3/7 of
x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2
40% of 525/2 = (40/100 * 525/2) = 105Incorrect
Let the number be x. Then,
2/5 of 1/3 of 3/7 of
x = 15 => x = (15 * 5/3 * 3 * 5/2) = 525/2
40% of 525/2 = (40/100 * 525/2) = 105 
Question 9 of 50
9. Question
A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A’s share in the profit is?
Correct
2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400Incorrect
2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400 
Question 10 of 50
10. Question
The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned halfyearly is?
Correct
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)^{2} – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.Incorrect
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)^{2} – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3. 
Question 11 of 50
11. Question
H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2^{4} * 3^{5} * 5^{2} * 7^{2}. The third number is:
Correct
3240 = 2^{3} * 3^{4} * 5; 3600 = 2^{4} * 3^{2} * 5^{2}
H.C.F = 36 = 2^{2} * 3^{2}
Since H.C.F is the product of lowest powers of common factors, so the third number must have (2^{2} * 3^{2}) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 3^{5} and 7^{2} as its factors.
Third number = 2^{2} * 3^{5} * 7^{2}Incorrect
3240 = 2^{3} * 3^{4} * 5; 3600 = 2^{4} * 3^{2} * 5^{2}
H.C.F = 36 = 2^{2} * 3^{2}
Since H.C.F is the product of lowest powers of common factors, so the third number must have (2^{2} * 3^{2}) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 3^{5} and 7^{2} as its factors.
Third number = 2^{2} * 3^{5} * 7^{2} 
Question 12 of 50
12. Question
The ratio of the earnings of P and Q is 9:10. If the earnings of P increases by onefourth and the earnings of Q decreases by onefourth, then find the new ratio of their earnings?
Correct
Let the earnings of P and Q be 9x and 10x respectively.
New ratio = [9x + 1/4 (9x)]/[10x – 1/4 (10x)]
=> 9*(1 + 1/4)/10*(1 – 1/4)
=> 9/10 * (5/4)/(3/4) = 3/2Incorrect
Let the earnings of P and Q be 9x and 10x respectively.
New ratio = [9x + 1/4 (9x)]/[10x – 1/4 (10x)]
=> 9*(1 + 1/4)/10*(1 – 1/4)
=> 9/10 * (5/4)/(3/4) = 3/2 
Question 13 of 50
13. Question
I. x^{2} + 9x + 20 = 0,
II. y^{2} + 5y + 6 = 0 to solve both the equations to find the values of x and y?Correct
I. x^{2} + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = 4, 5
II. y^{2} + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = 3, 2
= x < y.Incorrect
I. x^{2} + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = 4, 5
II. y^{2} + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = 3, 2
= x < y. 
Question 14 of 50
14. Question
LCM of 455, 117, 338 is:
Correct
Incorrect

Question 15 of 50
15. Question
The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream?
Correct
The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(21) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.Incorrect
The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(21) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph. 
Question 16 of 50
16. Question
A person takes 20 minutes more to cover a certain distance by decreasing his speed by 20%. What is the time taken to cover the distance at his original speed?
Correct
Let the distance and original speed be d km and k kmph respectively.
d/0.8k – d/k = 20/60 => 5d/4k – d/k = 1/3
=> (5d – 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes.Incorrect
Let the distance and original speed be d km and k kmph respectively.
d/0.8k – d/k = 20/60 => 5d/4k – d/k = 1/3
=> (5d – 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes. 
Question 17 of 50
17. Question
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
Correct
x – y = 5
4x – 6y = 6
x = 12 y = 7Incorrect
x – y = 5
4x – 6y = 6
x = 12 y = 7 
Question 18 of 50
18. Question
A gardener wants to plant trees in his garden in rows in such away that the number of trees in each row to be the same. If there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row?
Correct
Required number of trees
= 24/36 * 42 = 28.Incorrect
Required number of trees
= 24/36 * 42 = 28. 
Question 19 of 50
19. Question
The inverse ratio of 3: 2: 1 is?
Correct
1/3: 1/2: 1/1 = 2:3:6
Incorrect
1/3: 1/2: 1/1 = 2:3:6

Question 20 of 50
20. Question
A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 A.M. When will the cistern be empty?
Correct
1 to 2 = 1/4
2 to 3 = 1/4 + 1/5 = 9/20
After 3 AM = 1/4 + 1/5 – 1/2 = 1/20
1/4 + 9/20 = 14/20
1 h — 1/20
? —– 14/20
14 hours ==> 5 PMIncorrect
1 to 2 = 1/4
2 to 3 = 1/4 + 1/5 = 9/20
After 3 AM = 1/4 + 1/5 – 1/2 = 1/20
1/4 + 9/20 = 14/20
1 h — 1/20
? —– 14/20
14 hours ==> 5 PM 
Question 21 of 50
21. Question
Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.15of money?
Correct
P = 15(100/10)^{2 }=> P = 1500
Incorrect
P = 15(100/10)^{2 }=> P = 1500

Question 22 of 50
22. Question
A person got Rs.48 more when he invested a certain sum at compound interest instead of simple interest for two years at 8% p.a. Find the sum?
Correct
P = (d * 100^{2}) / R^{2}
=> (48 * 100 * 100) / 8 * 8 = Rs.7500Incorrect
P = (d * 100^{2}) / R^{2}
=> (48 * 100 * 100) / 8 * 8 = Rs.7500 
Question 23 of 50
23. 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 24 of 50
24. Question
A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
Correct
D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmphIncorrect
D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph 
Question 25 of 50
25. Question
The length of the bridge, which a train 130 m long and traveling at 45 km/hr can cross in 30 sec is?
Correct
Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (130 + x)/30 = 25/2
x = 245 m.Incorrect
Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (130 + x)/30 = 25/2
x = 245 m. 
Question 26 of 50
26. Question
Rs.525 among A, B and C such that B may get 2/3 of A and C together get. Find the share of C?
Correct
Incorrect

Question 27 of 50
27. Question
From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is .
Correct
Let E_{1} be the event of drawing a red card.
Let E_{2} be the event of drawing a king .
P(E_{1} ∩ E_{2}) = P(E_{1}) . P(E_{2})
(As E_{1} and E_{2} are independent)
= 1/2 * 1/13 = 1/26Incorrect
Let E_{1} be the event of drawing a red card.
Let E_{2} be the event of drawing a king .
P(E_{1} ∩ E_{2}) = P(E_{1}) . P(E_{2})
(As E_{1} and E_{2} are independent)
= 1/2 * 1/13 = 1/26 
Question 28 of 50
28. 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?
Correct
Let 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.Incorrect
Let 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 29 of 50
29. Question
the present average age of a couple and their daughter is 35 years. Fifteen years from now, the age of the mother will be equal to the sum of present ages of the father and the daughter. Find the present age of mother?
Correct
(f + m + d)/3 = 35
=> f + m + d = 105 — (1)
m + 15 = f + d
Substituting f + d as m + 15 in (1), we get
2m + 15 = 105
2m = 90 => m = 45 years.Incorrect
(f + m + d)/3 = 35
=> f + m + d = 105 — (1)
m + 15 = f + d
Substituting f + d as m + 15 in (1), we get
2m + 15 = 105
2m = 90 => m = 45 years. 
Question 30 of 50
30. Question
A man sells two articles for Rs.3600 each and he gains 30% on the first and loses 30% on the next. Find his total gain or loss?
Correct
(30*30)/100 = 9%loss
Incorrect
(30*30)/100 = 9%loss

Question 31 of 50
31. Question
96 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part?
Correct
x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42Incorrect
x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42 
Question 32 of 50
32. Question
Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Correct
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}
Incorrect
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm^{2}

Question 33 of 50
33. Question
The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years. Tee age of the 15th student is:
Correct
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.Incorrect
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years. 
Question 34 of 50
34. 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?
Correct
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
Incorrect
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days

Question 35 of 50
35. Question
Five bells first begin to toll together and then at intervals of 5, 10, 15, 20 and 25 seconds respectively. After what interval of time will they toll again together?
Correct
LCM = 300/60 = 5 min
Incorrect
LCM = 300/60 = 5 min

Question 36 of 50
36. Question
What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58?
Correct
Let the side of the square plot be a ft.
a^{2} = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944.Incorrect
Let the side of the square plot be a ft.
a^{2} = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944. 
Question 37 of 50
37. Question
I. a^{2} – 13a + 42 = 0,
II. b^{2} – 15b + 56 = 0 to solve both the equations to find the values of a and b?Correct
I. a^{2} – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b^{2} – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ bIncorrect
I. a^{2} – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b^{2} – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ b 
Question 38 of 50
38. Question
Simplify : (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
Correct
(y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
= (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a+b+c})^{3}
= y^{4ba+4cb+4ac} / y^{3(a+b+c)}
= y^{3(a+b+c)} / y^{3(a+b+c)} = 1.Incorrect
(y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a} . y^{b} . y^{c})^{3}
= (y^{4ba} . y^{4cb} . y^{4ac}) / (y^{a+b+c})^{3}
= y^{4ba+4cb+4ac} / y^{3(a+b+c)}
= y^{3(a+b+c)} / y^{3(a+b+c)} = 1. 
Question 39 of 50
39. Question
A man can row 30 km downstream and 20 km upstream in 4 hours. He can row 45 km downstream and 40 km upstream in 7 hours. Find the speed of man in still water?
Correct
Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a – b) = 4 and 45/(a + b) + 40/(a – b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph.Incorrect
Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a – b) = 4 and 45/(a + b) + 40/(a – b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph. 
Question 40 of 50
40. 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 41 of 50
41. Question
Income and expenditure of a person are in the ratio 5:4. If the income of the person is Rs.18000, then find his savings?
Correct
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600.Incorrect
Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income – expenditure = 5x – 4x = x
So, savings = Rs.3600. 
Question 42 of 50
42. Question
What is the least number to be subtracted from 11, 15, 21 and 30 each so that the resultant numbers become proportional?
Correct
Let the least number to be subtracted be ‘x’, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.
<=> (11 – x):(15 – x) = (21 – x):(30 x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 – x)(30 – x) = (15 – x)(21 – x)Incorrect
Let the least number to be subtracted be ‘x’, then 11 – x, 15 – x, 21 – x and 30 – x are in proportion.
<=> (11 – x):(15 – x) = (21 – x):(30 x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 – x)(30 – x) = (15 – x)(21 – x) 
Question 43 of 50
43. Question
Two pipes can fill a tank in 18 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?
Correct
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.Incorrect
Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes. 
Question 44 of 50
44. Question
The simple interest on Rs.12000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for five years at the same rate of interest.
Correct
R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]^{n} – 1}
= 12000 { [ 1 + 12 / 100]^{2} – 1} = Rs.3052.80Incorrect
R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]^{n} – 1}
= 12000 { [ 1 + 12 / 100]^{2} – 1} = Rs.3052.80 
Question 45 of 50
45. Question
The difference between a twodigit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2?
Correct
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8.Incorrect
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8. 
Question 46 of 50
46. Question
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
Correct
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24Incorrect
(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24 
Question 47 of 50
47. Question
X and Y started a business with capitals Rs. 20000 and Rs. 25000. After few months Z joined them with a capital of Rs. 30000. If the share of Z in the annual profit of Rs. 50000 is Rs. 14000, then after how many months from the beginning did Z join?
Correct
Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months.Incorrect
Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 – 7) months. i.e., 5 months. 
Question 48 of 50
48. Question
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in?
Correct
Net part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.Incorrect
Net part filled in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs. 
Question 49 of 50
49. Question
A certain number of men can do a work in 65 days working 6 hours a day. If the number of men are decreased by onefourth, then for how many hours per day should they work in order to complete the work in 40 days?
Correct
Let the number of men initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
So, x * 65 * 6 = (3x)/4 * 40 * h_{2}
=> h_{2} = (65 * 6 * 4)/(3 * 40) = 13.Incorrect
Let the number of men initially be x. we have M_{1} D_{1} H_{1}= M_{2} D_{2} H_{2}
So, x * 65 * 6 = (3x)/4 * 40 * h_{2}
=> h_{2} = (65 * 6 * 4)/(3 * 40) = 13. 
Question 50 of 50
50. Question
A gardener wants to plant trees in his garden in rows in such a way that the number of trees in each row to be the same. If there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row?
Correct
Required number of trees
= 24/36 * 42 = 28.Incorrect
Required number of trees
= 24/36 * 42 = 28.