## Indian Bank PO Pre mock test series 2nd, Indian bank po prelims online test series 2nd

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Indian Bank PO Pre mock test series 2nd, Indian bank po test series 2nd. Indian Bank PO Pre Free Mock Test Exam 2019. Indian Bank PO Pre Quiz 2019. Indian Bank PO Pre. Full Online Mock Test **Series 2nd** in English. The Indian Bank PO Pre. Full online mock test paper is free for all students. Indian Bank PO Pre. Question and Answers in English and Hindi **Series 2**. Here we are providing** Indian Bank PO Pre. Full Mock Test Paper in English. Indian Bank PO Pre. **Mock Test **Series 2nd** 2019. Now Test your self for Indian Bank PO Pre. Exam by using below quiz…

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

(743.30)

^{2}= ?Correct(743.30)

^{2}= 552500Incorrect(743.30)

^{2}= 552500 - Question 2 of 50
##### 2. Question

A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man?

CorrectM = 6

S = 1.2

DS = 7.2

US = 4.8

x/7.2 + x/4.8 = 1

x = 2.88

D = 2.88 * 2 = 5.76IncorrectM = 6

S = 1.2

DS = 7.2

US = 4.8

x/7.2 + x/4.8 = 1

x = 2.88

D = 2.88 * 2 = 5.76 - Question 3 of 50
##### 3. Question

If a + b + c = 0 and abc ≠ 0, then [X

^{a3 / abc}][X^{b3 / abc}][X^{c3 / abc}] equalsCorrectX

^{a3 / abc}* X^{b3 / abc}* X^{c3 / abc = X}^{(a3 + b3 + c3) / abc}

As a + b + c = 0, (a^{3}+ b^{3}+ c^{3}) / abc = 3

The given expression equals X^{3}IncorrectX

^{a3 / abc}* X^{b3 / abc}* X^{c3 / abc = X}^{(a3 + b3 + c3) / abc}

As a + b + c = 0, (a^{3}+ b^{3}+ c^{3}) / abc = 3

The given expression equals X^{3} - Question 4 of 50
##### 4. Question

A man buys two articles for Rs.1980 each and he gains 10% on the first and loses 10% on the next. Find his total gain or loss percent?

CorrectIncorrect - Question 5 of 50
##### 5. Question

If x = √196 + √200, then x/2 + 2/x is?

Correctx = 14 + 10√2

x/2 = 7 + 5√2

2/x = (7 – 5√2) / (7 + 5√2)(7 – 5√2) = 5√2 – 7

x/2 + 2/x = 10√2 = 2√50Incorrectx = 14 + 10√2

x/2 = 7 + 5√2

2/x = (7 – 5√2) / (7 + 5√2)(7 – 5√2) = 5√2 – 7

x/2 + 2/x = 10√2 = 2√50 - Question 6 of 50
##### 6. Question

Sum of two consecutive even terms lacks by 98 from their product. Find the sum of these numbers?

Correctx(x + 2) – (x + x + 2) = 98

Incorrectx(x + 2) – (x + x + 2) = 98

- Question 7 of 50
##### 7. Question

A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs.18000. A’s share is?

Correctx* 12 : 2x* 6: 3x* 4

1:1:1

1/3 * 18000 = 6000Incorrectx* 12 : 2x* 6: 3x* 4

1:1:1

1/3 * 18000 = 6000 - Question 8 of 50
##### 8. Question

One half of a two digit number exceeds its one third by 6. What is the sum of the digits of the number?

Correctx/2 – x/3 = 6 => x =6

3 + 6 = 9Incorrectx/2 – x/3 = 6 => x =6

3 + 6 = 9 - Question 9 of 50
##### 9. Question

A 180 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?

CorrectS = 180/6 * 18/5 = 108 kmph

IncorrectS = 180/6 * 18/5 = 108 kmph

- Question 10 of 50
##### 10. Question

Arun purchased 30 kg of wheat at the rate of Rs. 11.50 per kg and 20 kg of wheat at the rate of 14.25 per kg. He mixed the two and sold the mixture. Approximately what price per kg should be sell the mixture to make 30% profit?

CorrectC.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.

S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.

S.P. per kg = 819/50 = Rs. 16.38 = 16.30.IncorrectC.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.

S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.

S.P. per kg = 819/50 = Rs. 16.38 = 16.30. - Question 11 of 50
##### 11. Question

There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and of the last three is 1001. Find the fourth number?

Correct(385/1001) = 5/13 First Number is 5

Incorrect(385/1001) = 5/13 First Number is 5

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

A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is:

CorrectLet the son’s present age be x years.

Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 => x = 22.IncorrectLet the son’s present age be x years.

Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 => x = 22. - Question 13 of 50
##### 13. Question

A fair price shopkeeper takes 10% profit on his goods. He lost 20% goods during theft. His loss percent is:

CorrectSuppose he has 100 items. Let C.P. of each item be Re. 1.

Total cost = Rs. 100. Number of items left after theft = 80.

S.P. of each item = Rs. 1.10

Total sale = 1.10 * 80 = Rs. 88

Hence, loss % = 12/100 * 100 = 12%IncorrectSuppose he has 100 items. Let C.P. of each item be Re. 1.

Total cost = Rs. 100. Number of items left after theft = 80.

S.P. of each item = Rs. 1.10

Total sale = 1.10 * 80 = Rs. 88

Hence, loss % = 12/100 * 100 = 12% - Question 14 of 50
##### 14. Question

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are picked at random, what is the probability that they are all blue?

CorrectGiven that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455IncorrectGiven that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455 - Question 15 of 50
##### 15. Question

Divide Rs. 1500 among A, B and C so that A receives 1/3 as much as B and C together and B receives 2/3 as A and C together. A’s share is?

CorrectA+B+C = 1500

A = 1/3(B+C); B = 2/3(A+C)

A/(B+C) = 1/3

A = 1/4 * 1500 => 375IncorrectA+B+C = 1500

A = 1/3(B+C); B = 2/3(A+C)

A/(B+C) = 1/3

A = 1/4 * 1500 => 375 - Question 16 of 50
##### 16. Question

109.13 * 34.14 – 108.12 * 33 = ?

Correct109 * 34 – 108 * 33

= 3706 – 3564 = 142Incorrect109 * 34 – 108 * 33

= 3706 – 3564 = 142 - Question 17 of 50
##### 17. Question

A man saves Rs. 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years?

CorrectAmount = [200(1 + 5/100)

^{3}+ 200(1 + 5/100)^{2}+ 200(1 + 5/100)]

= [200 * 21/20(21/20 * 21/20 + 21/20 + 1)] = Rs. 662.02IncorrectAmount = [200(1 + 5/100)

^{3}+ 200(1 + 5/100)^{2}+ 200(1 + 5/100)]

= [200 * 21/20(21/20 * 21/20 + 21/20 + 1)] = Rs. 662.02 - Question 18 of 50
##### 18. Question

In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.42, the number of 25 paisa coins is?

Correct5x 6x 2x

50 25 100

250x + 150x + 200x = 4200

600x = 4200

x = 7 => 6x = 42Incorrect5x 6x 2x

50 25 100

250x + 150x + 200x = 4200

600x = 4200

x = 7 => 6x = 42 - Question 19 of 50
##### 19. Question

Find the one which does not belong to that group ?

Correct343 = 7

^{3}, 121 = 11^{2}, 1331 = 11^{3}, 2197 = 13^{3}and 125 = 5^{3}.

343, 1331, 2197 and 125 are perfect cubes, but not 121.Incorrect343 = 7

^{3}, 121 = 11^{2}, 1331 = 11^{3}, 2197 = 13^{3}and 125 = 5^{3}.

343, 1331, 2197 and 125 are perfect cubes, but not 121. - Question 20 of 50
##### 20. Question

A 70 cm long wire is to be cut into two pieces so that one piece will be 2/5th of the other, how many centimeters will the shorter piece be?

Correct1: 2/5 = 5: 2

2/7 * 70 = 20Incorrect1: 2/5 = 5: 2

2/7 * 70 = 20 - Question 21 of 50
##### 21. Question

– 84 * 29 + 365 = ?

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

= – (84 * 30) + 84 + 365

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

= – (84 * 30) + 84 + 365

= – 2520 + 449 = – 2071 - Question 22 of 50
##### 22. Question

The edges of cuboid are 4 cm; 5 cm and 6 cm. Find its surface area?

Correct2(4*5 + 5*6 + 4*6) = 148

Incorrect2(4*5 + 5*6 + 4*6) = 148

- Question 23 of 50
##### 23. Question

Rs.160 contained in a box consists of one rupee, 50 paisa and 25 paisa coins in the ratio 4:5:6. What is the number of 25 paisa coins?

Correct4x 5x 6x

100 50 25

400x + 350x + 150x = 16000

x = 20

6x = 120Incorrect4x 5x 6x

100 50 25

400x + 350x + 150x = 16000

x = 20

6x = 120 - Question 24 of 50
##### 24. Question

In an examination, 47% failed in English and 54% failed in Mathematics. Find the pass percentage in both the subjects if 31% failed in both the subjects?

CorrectIncorrect - Question 25 of 50
##### 25. Question

Find the one which does not belong to that group ?

Correct20 = 4

^{2}+ 4, 42 = 6^{2}+ 6, 58 = 7^{2}+ 9, 72 = 8^{2}+ 8 and 90 = 9^{2}+ 9.

20, 42, 72 and 90 can be expressed in n^{2}+ n form but not 58.Incorrect20 = 4

^{2}+ 4, 42 = 6^{2}+ 6, 58 = 7^{2}+ 9, 72 = 8^{2}+ 8 and 90 = 9^{2}+ 9.

20, 42, 72 and 90 can be expressed in n^{2}+ n form but not 58. - Question 26 of 50
##### 26. Question

On an order of 5 dozen boxes of a consumer product, a retailer receives an extra dozen free. This is equivalent to allowing him a discount of:

CorrectClearly, the retailer gets 1 dozen out of 6 dozens free.

Equivalent discount = 1/6 * 100 = 16 2/3%.IncorrectClearly, the retailer gets 1 dozen out of 6 dozens free.

Equivalent discount = 1/6 * 100 = 16 2/3%. - Question 27 of 50
##### 27. Question

If A got 80 marks and B got 60 marks, then what percent of A’s mark is B’s mark?

CorrectA’s marks = 80 ; B’s marks = 60.

Let x% of A = B => x/100 * 80 = 60

=> x = (60 * 100)/80 = 75

B’s marks is 75% of A’s marks.IncorrectA’s marks = 80 ; B’s marks = 60.

Let x% of A = B => x/100 * 80 = 60

=> x = (60 * 100)/80 = 75

B’s marks is 75% of A’s marks. - Question 28 of 50
##### 28. Question

The cost price of 13 articles is equal to the selling price of 11 articles. Find the profit percent?

Correct13 CP = 11 SP

11 — 2 CP

100 — ? =>18 2/11%Incorrect13 CP = 11 SP

11 — 2 CP

100 — ? =>18 2/11% - Question 29 of 50
##### 29. Question

Ten years ago, the age of Anand was one-third the age of Bala at that time. The present age of Bala is 12 years more than the present age of Anand. Find the present age of Anand?

CorrectLet the present ages of Anand and Bala be ‘a’ and ‘b’ respectively.

a – 10 = 1/3 (b – 10) — (1)

b = a + 12

Substituting b = a + 12 in first equation,

a – 10 = 1/3 (a + 2) => 3a – 30 = a + 2

=> 2a = 32 => a = 16.IncorrectLet the present ages of Anand and Bala be ‘a’ and ‘b’ respectively.

a – 10 = 1/3 (b – 10) — (1)

b = a + 12

Substituting b = a + 12 in first equation,

a – 10 = 1/3 (a + 2) => 3a – 30 = a + 2

=> 2a = 32 => a = 16. - Question 30 of 50
##### 30. Question

If a:b = 4:1, then find (a – 3b)/(2a – b)?

Correcta/b = 4/1 => a = 4b

(a – 3b)/(2a – b) = (4b – 3b)/(8b – b)

= b/7b => 1/7Incorrecta/b = 4/1 => a = 4b

(a – 3b)/(2a – b) = (4b – 3b)/(8b – b)

= b/7b => 1/7 - Question 31 of 50
##### 31. Question

A train 400 m long can cross an electric pole in 20 sec and then find the speed of the train?

CorrectLength = Speed * time

Speed = L/T

S = 400/20

S = 20 M/Sec

Speed= 20*18/5 (To convert M/Sec in to Kmph multiply by 18/5)

Speed = 72 KmphIncorrectLength = Speed * time

Speed = L/T

S = 400/20

S = 20 M/Sec

Speed= 20*18/5 (To convert M/Sec in to Kmph multiply by 18/5)

Speed = 72 Kmph - Question 32 of 50
##### 32. Question

An amount of Rs. 3000 becomes Rs. 3600 in four years at simple interest. If the rate of interest was 1% more, then what was be the total amount?

CorrectA = P(1 + TR/100)

=> 3600 = 3000[1 + (4 * R)/100] => R = 5%

Now R = 6%

=> A = 3000[1 + (4 * 6)/100] = Rs. 3720.IncorrectA = P(1 + TR/100)

=> 3600 = 3000[1 + (4 * R)/100] => R = 5%

Now R = 6%

=> A = 3000[1 + (4 * 6)/100] = Rs. 3720. - Question 33 of 50
##### 33. Question

The average of the two-digit numbers, which remain the same when the digits interchange their positions, is:

CorrectAverage = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99)/9

= [(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55]/9

= [(4 * 110) + 55]/9 = 495/9 = 55.IncorrectAverage = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99)/9

= [(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55]/9

= [(4 * 110) + 55]/9 = 495/9 = 55. - Question 34 of 50
##### 34. Question

At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.20000?

CorrectA = 20000(11/10)

^{3}

= 26620

= 20000

———-

6620IncorrectA = 20000(11/10)

^{3}

= 26620

= 20000

———-

6620 - Question 35 of 50
##### 35. 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 36 of 50
##### 36. Question

The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?

CorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560IncorrectLet the rate of interest be R% p.a.

4400{[1 + R/100]^{2}– 1} = 11193.60

[1 + R/100]^{2}= (44000 + 11193.60)/44000

[1 + R/100]^{2}= 1 + 2544/1000 = 1 + 159/625

[1 + R/100]^{2}= 784/625 = (28/25)^{2}

1 + R/100 = 28/25

R/100 = 3/25

Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100

=Rs.10560 - Question 37 of 50
##### 37. Question

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

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

Then,

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

9x^{2}+ 18x – 72 = 0

x^{2}+ 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2

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

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

Then,

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

9x^{2}+ 18x – 72 = 0

x^{2}+ 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2

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

Hence, required number = 24. - Question 38 of 50
##### 38. Question

Which one of the following represents the sum of the first n even numbers?

CorrectN (N + 1)

IncorrectN (N + 1)

- Question 39 of 50
##### 39. Question

Find the roots of the quadratic equation: 2x

^{2}+ 3x – 9 = 0?Correct2x

^{2}+ 6x – 3x – 9 = 0

2x(x + 3) – 3(x + 3) = 0

(x + 3)(2x – 3) = 0

=> x = -3 or x = 3/2.Incorrect2x

^{2}+ 6x – 3x – 9 = 0

2x(x + 3) – 3(x + 3) = 0

(x + 3)(2x – 3) = 0

=> x = -3 or x = 3/2. - Question 40 of 50
##### 40. Question

What is the least number to be subtracted from 696 to make it a perfect square?

CorrectThe numbers less than 696 and are squares of certain numbers are 676, 625.

The least number that should be subtracted from 696 to make it perfect square = 696 – 676 = 20.IncorrectThe numbers less than 696 and are squares of certain numbers are 676, 625.

The least number that should be subtracted from 696 to make it perfect square = 696 – 676 = 20. - Question 41 of 50
##### 41. Question

A 25 cm wide path is to be made around a circular garden having a diameter of 4 meters. Approximate area of the path is square meters is

CorrectArea of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}

^{2}– ∏[4/2]^{2}

= ∏[2.25^{2}– 2^{2}] = ∏(0.25)(4.25) { (a^{2}– b^{2}= (a – b)(a + b) }

= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq mIncorrectArea of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}

^{2}– ∏[4/2]^{2}

= ∏[2.25^{2}– 2^{2}] = ∏(0.25)(4.25) { (a^{2}– b^{2}= (a – b)(a + b) }

= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m - Question 42 of 50
##### 42. Question

Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 15 km/hr and 25 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?

CorrectTime taken to meet for the first time anywhere on the track

= length of the track / relative speed

= 300 / (15 + 25)5/18 = 300* 18 / 40 * 5 = 27 seconds.IncorrectTime taken to meet for the first time anywhere on the track

= length of the track / relative speed

= 300 / (15 + 25)5/18 = 300* 18 / 40 * 5 = 27 seconds. - Question 43 of 50
##### 43. Question

Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?

CorrectSI = 40 + 40

CI = 40 + 40 + 1.6 = 81.6IncorrectSI = 40 + 40

CI = 40 + 40 + 1.6 = 81.6 - Question 44 of 50
##### 44. Question

A fort of 2000 soldiers has provisions for 50 days. After 10 days some of them left and the food was now enough for the same period of 50 days as before. How many of them left?

Correct2000 —- 50

2000 —- 40

x —– 50

x*50 = 2000*40

x=1600

2000

——-

400Incorrect2000 —- 50

2000 —- 40

x —– 50

x*50 = 2000*40

x=1600

2000

——-

400 - Question 45 of 50
##### 45. Question

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately?

CorrectLet the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

1/x + 1/(x + 6) = 1/4

x^{2}– 2x – 24 = 0

(x – 6)(x + 4) = 0 => x = 6.IncorrectLet the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

1/x + 1/(x + 6) = 1/4

x^{2}– 2x – 24 = 0

(x – 6)(x + 4) = 0 => x = 6. - Question 46 of 50
##### 46. Question

The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream?

CorrectSpeed downstream = 60 + 20 = 80 kmph

Speed upstream = 60 – 20 = 40 kmphIncorrectSpeed downstream = 60 + 20 = 80 kmph

Speed upstream = 60 – 20 = 40 kmph - Question 47 of 50
##### 47. Question

A and B can finish a work in 16 days while A alone can do the same work in 24 days. In how many days B alone will complete the work?

CorrectB = 1/16 – 1/24 = 1/48 => 48 days

IncorrectB = 1/16 – 1/24 = 1/48 => 48 days

- Question 48 of 50
##### 48. 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 49 of 50
##### 49. Question

There were two candidates in an election. Winner candidate received 62% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate?

CorrectW = 62% L = 38%

62% – 38% = 24%

24% ——– 288

62% ——– ? => 744IncorrectW = 62% L = 38%

62% – 38% = 24%

24% ——– 288

62% ——– ? => 744 - Question 50 of 50
##### 50. Question

In how many ways can a committee consisting of three men and four women be formed from a group of six men and seven women?

CorrectThe group contain six men and seven women

Three men can be selected from six men in ⁶C₃ ways.

Four women can be selected from seven women in ⁷C₄ ways.

Total number of ways = (⁷C₄)(⁶C₃).IncorrectThe group contain six men and seven women

Three men can be selected from six men in ⁶C₃ ways.

Four women can be selected from seven women in ⁷C₄ ways.

Total number of ways = (⁷C₄)(⁶C₃).