# 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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- 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?

CorrectGiven 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%IncorrectGiven 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:

CorrectLCM = 1440

1440 – 1 = 1439IncorrectLCM = 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?

CorrectLCM = 1400

1400 – 6 = 1394IncorrectLCM = 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?

CorrectLet 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.IncorrectLet 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 = ?

Correct421 * 0.9 + 130 * 101 + 10000

= 378.9 + 1313 + 10000 = 23508.9 = 23500Incorrect421 * 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?

CorrectLet 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.IncorrectLet 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?

CorrectTime 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 = 60IncorrectTime 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

Two-fifth of one-third of three-seventh of a number is 15. What is 40% of that number?

CorrectLet 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) = 105IncorrectLet 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?

Correct2*12 : 3*12 : 4*24

1: 3: 4

1/8 * 3200 = 400Incorrect2*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 half-yearly is?

CorrectS.I. = (1200 * 10 * 1)/100 = Rs. 120

C.I. = [1200 * (1 + 5/100)^{2}– 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.IncorrectS.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:Correct3240 = 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}Incorrect3240 = 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 one-fourth and the earnings of Q decreases by one-fourth, then find the new ratio of their earnings?

CorrectLet 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) = 3/2IncorrectLet 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?CorrectI. 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.IncorrectI. 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:

CorrectIncorrect - 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?

CorrectThe 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)/(2-1) = 3/1 = 3:1

Speed of the stream = 42/3 = 14 kmph.IncorrectThe 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)/(2-1) = 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?

CorrectLet 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.IncorrectLet 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?

Correctx – y = 5

4x – 6y = 6

x = 12 y = 7Incorrectx – 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?

CorrectRequired number of trees

= 24/36 * 42 = 28.IncorrectRequired number of trees

= 24/36 * 42 = 28. - Question 19 of 50
##### 19. Question

The inverse ratio of 3: 2: 1 is?

Correct1/3: 1/2: 1/1 = 2:3:6

Incorrect1/3: 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?

Correct1 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 PMIncorrect1 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?

CorrectP = 15(100/10)

^{2 }=> P = 1500IncorrectP = 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?

CorrectP = (d * 100

^{2}) / R^{2}

=> (48 * 100 * 100) / 8 * 8 = Rs.7500IncorrectP = (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?

CorrectD = 100 + 125 = 225

T = 15

S = 225/15 * 18/5 = 54 kmphIncorrectD = 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?

CorrectSpeed = 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.IncorrectSpeed = 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?

CorrectIncorrect - 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 -.

CorrectLet 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 = 1/26IncorrectLet 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?

CorrectLet 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.IncorrectLet 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?

Correctx/7 = y/9 => x:y = 7:9

7/16 * 96 = 42Incorrectx/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.

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 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:

CorrectAge of the 15th student=

[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.IncorrectAge 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?

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 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?

CorrectLCM = 300/60 = 5 min

IncorrectLCM = 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?

CorrectLet 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.IncorrectLet 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?CorrectI. 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 ≤ bIncorrectI. 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

^{4b-a}. y^{4c-b}. y^{4a-c}) / (y^{a}. y^{b}. y^{c})^{3}Correct(y

^{4b-a}. y^{4c-b}. y^{4a-c}) / (y^{a}. y^{b}. y^{c})^{3}

= (y^{4b-a}. y^{4c-b}. y^{4a-c}) / (y^{a+b+c})^{3}

= y^{4b-a+4c-b+4a-c}/ y^{3(a+b+c)}

= y^{3(a+b+c)}/ y^{3(a+b+c)}= 1.Incorrect(y

^{4b-a}. y^{4c-b}. y^{4a-c}) / (y^{a}. y^{b}. y^{c})^{3}

= (y^{4b-a}. y^{4c-b}. y^{4a-c}) / (y^{a+b+c})^{3}

= y^{4b-a+4c-b+4a-c}/ 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?

CorrectLet 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.IncorrectLet 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?

CorrectSince 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.IncorrectSince 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?

CorrectLet 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.IncorrectLet 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?

CorrectLet 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)IncorrectLet 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?

CorrectPart 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.IncorrectPart 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.

CorrectR = 100 I / PT

=> R = (100 * 7200)/ (12000 * 5) = 12%

CI = P{ [1 + R /100]^{n}– 1}

= 12000 { [ 1 + 12 / 100]^{2}– 1} = Rs.3052.80IncorrectR = 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 two-digit 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?

CorrectSince 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.IncorrectSince 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?

CorrectInvestments 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.IncorrectInvestments 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?

CorrectNet 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.IncorrectNet 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 one-fourth, then for how many hours per day should they work in order to complete the work in 40 days?

CorrectLet 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.IncorrectLet 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?

CorrectRequired number of trees

= 24/36 * 42 = 28.IncorrectRequired number of trees

= 24/36 * 42 = 28.