## Indian Bank PO Pre Online Test Series 1, Indian Bank Mock Test

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Indian Bank PO Pre Online Test Series 1, Indian Bank Mock Test for pre exam Series 1st. Indian Bank PO Pre Free Online Test Series 1. Indian Bank PO Pre Free quiz Exam 2018 for all students. Indian Bank PO Pre Exam Free Online Quiz 2018, Indian Bank PO Pre. Full Online Mock Test **Series 1st** in English. Indian Bank PO Pre. Free Mock Test Series in English. Indian Bank PO Pre. Free Mock Test **Series 1.** Indian Bank PO Pre. Question and Answers in English and Hindi **Series 1**. Here we are providing** Indian Bank PO Pre. Full Mock Test Paper in English. Indian Bank PO Pre. **Mock Test **Series 1st** 2018. Now Test your self for Indian Bank PO Pre. Exam by using below quiz…

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

The sum of three consecutive numbers is 87. The greatest among these three number is:

CorrectLet the numbers be x, x + 1 and x + 2

Then,

x + (x + 1) + (x + 2) = 87

3x = 84

x = 28

Greatest number, (x + 2) = 30.IncorrectLet the numbers be x, x + 1 and x + 2

Then,

x + (x + 1) + (x + 2) = 87

3x = 84

x = 28

Greatest number, (x + 2) = 30. - Question 2 of 50
##### 2. Question

Twenty times a positive integer is less than its square by 96. What is the integer?

CorrectLet the integer be x. Then,

x^{2}– 20x = 96

(x + 4)(x – 24) = 0

x = 24IncorrectLet the integer be x. Then,

x^{2}– 20x = 96

(x + 4)(x – 24) = 0

x = 24 - Question 3 of 50
##### 3. Question

A rectangular courtyard, the sides of which are in the ratio of 4:3, cost Rs.600 for paving at 50 p per m

^{2}; find the length of the diagonal of the courtyard?Correct1 m

^{2}—- 1/2

? —– 600 => 1200 m^{2}

4x * 3x = 1200 => x = 10

Incorrect1 m

^{2}—- 1/2

? —– 600 => 1200 m^{2}

4x * 3x = 1200 => x = 10

- Question 4 of 50
##### 4. Question

A man on tour travels first 160 km at 64 km/he and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is?

CorrectTotal time taken = (160/64 + 160/8) = 9/2 hrs.

Average speed = 320 * 2/9 = 71.11 km/hr.IncorrectTotal time taken = (160/64 + 160/8) = 9/2 hrs.

Average speed = 320 * 2/9 = 71.11 km/hr. - Question 5 of 50
##### 5. Question

If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is?

CorrectA = 7500(26/25)

^{2}= 8112IncorrectA = 7500(26/25)

^{2}= 8112 - Question 6 of 50
##### 6. Question

If x = √8 – √7, y = √6 – √5 and z = √10 – 3, which of the following is true?

Correctx = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)

As (√8 + √7) > (√6 + √5), (√8 – √7) < (√6 - √5) As (√10 + 3) > (√8 + √7) > (√6 + √5)

z < x < y.Incorrectx = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)

As (√8 + √7) > (√6 + √5), (√8 – √7) < (√6 - √5) As (√10 + 3) > (√8 + √7) > (√6 + √5)

z < x < y. - Question 7 of 50
##### 7. Question

Find the greatest number which divides 83, 125 and 209 leaving the same remainder in each case.

CorrectThe greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q – p), (r – p) and (r – q)).

For the given problem, p = 83, q = 125 and r = 209.

Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 – 83, 209 – 83, 209 – 125) = 42.IncorrectThe greatest number which divides three dividends p, q and r leaving the same remainder in each case is given by H.C.F(any two of (q – p), (r – p) and (r – q)).

For the given problem, p = 83, q = 125 and r = 209.

Hence the greatest number which divides these three leaving the same remainder in each case is H.C.F(any two of (125 – 83, 209 – 83, 209 – 125) = 42. - Question 8 of 50
##### 8. 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:

CorrectLet C.P. be Rs. x.

Then, 5% of x = 350 – 340 = 10

x/20 = 10 => x = 200IncorrectLet C.P. be Rs. x.

Then, 5% of x = 350 – 340 = 10

x/20 = 10 => x = 200 - Question 9 of 50
##### 9. Question

A shopkeeper sells 400 metres of cloth for Rs. 18000 at a loss of Rs.5 per metre. Find his cost price for one metre of cloth?

CorrectSP per metre = 18000/400 = Rs. 45 Loss per metre = Rs. 5 CP per metre = 45 + 5 = Rs. 50

IncorrectSP per metre = 18000/400 = Rs. 45 Loss per metre = Rs. 5 CP per metre = 45 + 5 = Rs. 50

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

10 books are placed at random in a shelf. The probability that a pair of books will always be together is -.

Correct10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5Incorrect10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5 - Question 11 of 50
##### 11. Question

2/3 * 5 1/2 * 2 * 7 5/6 * 9/11 = ?

Correct2/3 * 5 1/2 * 2 * 7 5/6 * 9/11

2/3 * 11/2 * 2 * 47/6 * 9/11 = 47Incorrect2/3 * 5 1/2 * 2 * 7 5/6 * 9/11

2/3 * 11/2 * 2 * 47/6 * 9/11 = 47 - Question 12 of 50
##### 12. Question

Amar, Bhavan and Chetan divide an amount of Rs.5600 among themselves in the ratio 3:6:5. If an amount of Rs.400 is deducted from each of their shares, what will be the new ratio of their shares of the amount?

CorrectLet the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.

3x + 6x + 5x = 5600

14x = 5600 => x = 400

Required ratio = 3x – 400 : 6x – 400 : 5x – 400

= 3x – x : 6x – x : 5x – x

= 2x : 5x : 4x => 2:5:4IncorrectLet the shares of Amar, Bhavan and Chetan be 3x, 6x and 5x respectively.

3x + 6x + 5x = 5600

14x = 5600 => x = 400

Required ratio = 3x – 400 : 6x – 400 : 5x – 400

= 3x – x : 6x – x : 5x – x

= 2x : 5x : 4x => 2:5:4 - Question 13 of 50
##### 13. Question

If x % of 80 = 20% of y, then x = ? and y = ?

Correctx % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)

=> 4x = y

i.e x / y = ¼Incorrectx % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)

=> 4x = y

i.e x / y = ¼ - Question 14 of 50
##### 14. Question

Ronald and Elan are working on an assignment. Ronald takes 6 hrs to type 32 pages on a computer, while Elan takes 5 hrs to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

CorrectNumber of pages typed by Ronald in 1 hour = 32/6 = 16/3

Number of pages typed by Elan in 1 hour = 40/5 = 8

Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3

Time taken by both to type 110 pages = (110 * 3/40) = 8 1/4 = 8 hrs 15 minIncorrectNumber of pages typed by Ronald in 1 hour = 32/6 = 16/3

Number of pages typed by Elan in 1 hour = 40/5 = 8

Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3

Time taken by both to type 110 pages = (110 * 3/40) = 8 1/4 = 8 hrs 15 min - Question 15 of 50
##### 15. Question

A man said to his son, “I was two-third of your present age when you were born”. If the present age of the man is 48 years, find the present age of the son?

CorrectPresent age of the son be P, he was born P years ago.

The age of the man was: (48 – P).

His age when the son was born should be equal to 2/3 of P.

(48 – P) = 2/3 P

5P = 144 => P = 28.8IncorrectPresent age of the son be P, he was born P years ago.

The age of the man was: (48 – P).

His age when the son was born should be equal to 2/3 of P.

(48 – P) = 2/3 P

5P = 144 => P = 28.8 - Question 16 of 50
##### 16. Question

A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5?

Correct30 * 20 * x = (8 * 5.5 * 1.5)/2

Incorrect30 * 20 * x = (8 * 5.5 * 1.5)/2

- Question 17 of 50
##### 17. Question

A sun is divided among X, Y and Z in such a way that for each rupee X gets, Y gets 45 paisa and Z gets 30 paisa. If the share of Y is RS. 27, what is the total amount?

Correctx:y:z = 100:45:30

20:9:6

9 — 27

35 — ? => 105Incorrectx:y:z = 100:45:30

20:9:6

9 — 27

35 — ? => 105 - Question 18 of 50
##### 18. Question

33/4 of 4500 + ? = 118500

Correct? = 118500 – 33/4 * 4500 = 8.1375 * 10

^{4}Incorrect? = 118500 – 33/4 * 4500 = 8.1375 * 10

^{4} - Question 19 of 50
##### 19. Question

Find the number of factors of 300 excluding 1 and itself.

Correct300 = 4.75 = 4.25.3 = 2

^{2}.5^{2}.3^{1}

Number of factors of 300 = (2 + 1)(2 + 1)(1 + 1) = 18.

There are 16 factors of 300 excluding 1 and itself.Incorrect300 = 4.75 = 4.25.3 = 2

^{2}.5^{2}.3^{1}

Number of factors of 300 = (2 + 1)(2 + 1)(1 + 1) = 18.

There are 16 factors of 300 excluding 1 and itself. - Question 20 of 50
##### 20. Question

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is -.

CorrectThe number of exhaustive outcomes is 36.

Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then P(E) = 18/36 = 1/2

P(E) = 1 – 1/2 = 1/2.IncorrectThe number of exhaustive outcomes is 36.

Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then P(E) = 18/36 = 1/2

P(E) = 1 – 1/2 = 1/2. - Question 21 of 50
##### 21. Question

The successive discounts 20% and 15% are equal to a single discount of?

CorrectLet the CP of an article be Rs. 100

Given that successive discounts are 20% and 15%.

SP = 85% of 80% of 100 = (85/100)(80/100)(100)

=> SP = Rs. 68

Clearly, single discount is 32%.IncorrectLet the CP of an article be Rs. 100

Given that successive discounts are 20% and 15%.

SP = 85% of 80% of 100 = (85/100)(80/100)(100)

=> SP = Rs. 68

Clearly, single discount is 32%. - Question 22 of 50
##### 22. Question

Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?

CorrectD = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = 125/3

T = 500 * 3/125 = 12 secIncorrectD = 250 m + 250 m = 500 m

RS = 80 + 70 = 150 * 5/18 = 125/3

T = 500 * 3/125 = 12 sec - Question 23 of 50
##### 23. Question

A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?

Correct2/5 + (2 + x)/10 = 1 => x = 4 days

Incorrect2/5 + (2 + x)/10 = 1 => x = 4 days

- Question 24 of 50
##### 24. Question

A boat goes 100 km downstream in 10 hours, and 75 m upstream in 15 hours. The speed of the stream is?

Correct100 — 10 DS = 10

? —- 1

75 —- 15 US = 5

? —– 1 S = (10 – 5)/2

= 2 2 ½ kmphIncorrect100 — 10 DS = 10

? —- 1

75 —- 15 US = 5

? —– 1 S = (10 – 5)/2

= 2 2 ½ kmph - Question 25 of 50
##### 25. Question

There are two circles of different radii. The are of a square is 784 sq cm and its side is twice the radius of the larger circle. The radius of the larger circle is seven – third that of the smaller circle. Find the circumference of the smaller circle.

CorrectLet the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.

a^{2}= 784 = (4)(196) = (2^{2}).(14^{2})

a = (2)(14) = 28

a = 2l, l = a/2 = 14

l = (7/3)s

Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s = 12∏ cm.IncorrectLet the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.

a^{2}= 784 = (4)(196) = (2^{2}).(14^{2})

a = (2)(14) = 28

a = 2l, l = a/2 = 14

l = (7/3)s

Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s = 12∏ cm. - Question 26 of 50
##### 26. Question

A train crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train?

CorrectLength of the train be ‘X’

X + 120/15 = X + 180/18

6X + 720 = 5X + 900

X = 180mIncorrectLength of the train be ‘X’

X + 120/15 = X + 180/18

6X + 720 = 5X + 900

X = 180m - Question 27 of 50
##### 27. Question

The smallest fraction, which each of 6/7, 5/14, 10/21 will divide exactly is:

CorrectRequired 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) = 30/7IncorrectRequired 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 28 of 50
##### 28. Question

The food in a camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?

Correctone man can consume the same food in 30*40 = 1200 days.

10 more men join, the total number of men = 40

The number of days the food will last = 1200/40 = 30 days.Incorrectone man can consume the same food in 30*40 = 1200 days.

10 more men join, the total number of men = 40

The number of days the food will last = 1200/40 = 30 days. - Question 29 of 50
##### 29. Question

Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?

CorrectLet the speed of the slower train be x m/sec.

Then, speed of the train = 2x m/sec.

Relative speed = ( x + 2x) = 3x m/sec.

(100 + 100)/8 = 3x => x = 25/3.

So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.IncorrectLet the speed of the slower train be x m/sec.

Then, speed of the train = 2x m/sec.

Relative speed = ( x + 2x) = 3x m/sec.

(100 + 100)/8 = 3x => x = 25/3.

So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr. - Question 30 of 50
##### 30. Question

An outlet pipe can empty 2/3 rd of a cistern in 12 minutes. In 8 minutes, what part of the cistern will be emptied?

Correct2/3 —- 12

? —– 8 ==> 4/9Incorrect2/3 —- 12

? —– 8 ==> 4/9 - Question 31 of 50
##### 31. Question

A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?

CorrectLet the length of the train be x m and its speed be y m/sec.

Then, x/y = 15 => y = x/15

(x + 100)/25 = x/15 => x = 150 m.IncorrectLet the length of the train be x m and its speed be y m/sec.

Then, x/y = 15 => y = x/15

(x + 100)/25 = x/15 => x = 150 m. - Question 32 of 50
##### 32. Question

The average weight of a group of boys is 30 kg. After a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?

CorrectLet the number off boys in the group originally be x.

Total weight of the boys = 30x

After the boy weighing 35 kg joins the group, total weight of boys = 30x + 35

So 30x + 35 + 31(x + 1) = > x = 4.IncorrectLet the number off boys in the group originally be x.

Total weight of the boys = 30x

After the boy weighing 35 kg joins the group, total weight of boys = 30x + 35

So 30x + 35 + 31(x + 1) = > x = 4. - Question 33 of 50
##### 33. Question

Simplify the following:

(169/121)^{-3/2}* 27/2 * (13/22)^{-1}Correct(169/121)

^{-3/2}* 27/2 * (13/22)^{-1}

(13^{2}/11^{2})^{-3/2}* 27/2 * 22/13 = 13^{-3}/11^{-3}* 27/2 * 22/13

= 13^{-4}11^{4}3^{3}Incorrect(169/121)

^{-3/2}* 27/2 * (13/22)^{-1}

(13^{2}/11^{2})^{-3/2}* 27/2 * 22/13 = 13^{-3}/11^{-3}* 27/2 * 22/13

= 13^{-4}11^{4}3^{3} - Question 34 of 50
##### 34. Question

(332% of 2113) / 42 = ?

Correct(3.32/100 * 2113) * 1/42 = (70.1516) * 1/42 = 167

Incorrect(3.32/100 * 2113) * 1/42 = (70.1516) * 1/42 = 167

- Question 35 of 50
##### 35. Question

How many minimum number’s of whole square slabs are required for paving the floor 12.96 meters long and 3.84 meters side?

CorrectHCF of 384, 1296 = 48

48 * 48 * x = 384 * 1296

x = 216IncorrectHCF of 384, 1296 = 48

48 * 48 * x = 384 * 1296

x = 216 - Question 36 of 50
##### 36. Question

LCM of 1/3, 5/6, 5/4, 10/7 is:

CorrectLCM of numerators = 10

HCF of denominators = 1

=> 10/1 = 10IncorrectLCM of numerators = 10

HCF of denominators = 1

=> 10/1 = 10 - Question 37 of 50
##### 37. Question

All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?

CorrectB has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.

Quantity of water in B = 8k – 5k = 3k.

Quantity of water in container C = 8k – 3k = 5k

Container: A B C

Quantity of water: 8k 3k 5k

It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.

5k – 148 = 3k + 148 => 2k = 296 => k = 148

The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.IncorrectB has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.

Quantity of water in B = 8k – 5k = 3k.

Quantity of water in container C = 8k – 3k = 5k

Container: A B C

Quantity of water: 8k 3k 5k

It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.

5k – 148 = 3k + 148 => 2k = 296 => k = 148

The initial quantity of water in A = 8k = 8 * 148 = 1184 liters. - Question 38 of 50
##### 38. Question

[(523 + 27) * (187 – 35) / (424 + 16)] / [110/22 of (2 * 38)] = ?

Correct[(523 + 27) * (187 – 35) / (424 + 16)] / [110/22 of (2 * 38)]

= [(550 * 152) / 440] / [(110/44) * 38]

= (5 * 38)/(5/2 * 38) = 2Incorrect[(523 + 27) * (187 – 35) / (424 + 16)] / [110/22 of (2 * 38)]

= [(550 * 152) / 440] / [(110/44) * 38]

= (5 * 38)/(5/2 * 38) = 2 - Question 39 of 50
##### 39. Question

A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good.

CorrectRequired probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

IncorrectRequired probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

- Question 40 of 50
##### 40. Question

The average of 11 results is 50, if the average of first six results is 49 and that of the last six is 52. Find the sixth result?

Correct1 to 11 = 11 * 50 = 550

1 to 6 = 6 * 49 = 294

6 to 11 = 6 * 52 = 312

6th = 294 + 312 – 550 = 56Incorrect1 to 11 = 11 * 50 = 550

1 to 6 = 6 * 49 = 294

6 to 11 = 6 * 52 = 312

6th = 294 + 312 – 550 = 56 - Question 41 of 50
##### 41. 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 42 of 50
##### 42. Question

3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days?

CorrectLet 1 woman’s 1 day work = x.

Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.

So, (3x/2 + 4x + + 6x/4) = 1/7

28x/4 = 1/7 => x = 1/49

1 woman alone can complete the work in 49 days.

So, to complete the work in 7 days, number of women required = 49/7 = 7.IncorrectLet 1 woman’s 1 day work = x.

Then, 1 man’s 1 day work = x/2 and 1 child’s 1 day work x/4.

So, (3x/2 + 4x + + 6x/4) = 1/7

28x/4 = 1/7 => x = 1/49

1 woman alone can complete the work in 49 days.

So, to complete the work in 7 days, number of women required = 49/7 = 7. - Question 43 of 50
##### 43. Question

Two trains are moving at 50 kmph and 70 kmph in opposite directions. Their lengths are 150 m and 100 m respectively. The time they will take to pass each other completely is?

Correct70 + 50 = 120 * 5/18 = 100/3 mps

D = 150 + 100 = 250 m

T = 250 * 3/100 = 15/2 = 7 ½ secIncorrect70 + 50 = 120 * 5/18 = 100/3 mps

D = 150 + 100 = 250 m

T = 250 * 3/100 = 15/2 = 7 ½ sec - Question 44 of 50
##### 44. Question

What is the difference between the largest number and the least number written with the figures 3, 4, 7, 0, 3?

Correct74330 Largest

30347 Smallest

————-

43983Incorrect74330 Largest

30347 Smallest

————-

43983 - Question 45 of 50
##### 45. Question

A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m. ?

CorrectRelative speed = 2 + 3 = 5 rounds per hour.

So, they cross each other 5 times in an hour and 2 times in half an hour.

Hence, they cross each other 7 times before 9.30 a.m.IncorrectRelative speed = 2 + 3 = 5 rounds per hour.

So, they cross each other 5 times in an hour and 2 times in half an hour.

Hence, they cross each other 7 times before 9.30 a.m. - Question 46 of 50
##### 46. 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 47 of 50
##### 47. Question

The population of a town is 45000; 5/9th of them are males and the rest females 40% of the males are married. What is the percentage of married females?

CorrectMale = 45,000* 5/9 = 25,000

Female = 45,000* 4/9 = 20,000

Married Male = 25,000* 40/100 = 10,000

Married Female = 10,000

20,000 ———— 10,000

100 ————- ? => 50%IncorrectMale = 45,000* 5/9 = 25,000

Female = 45,000* 4/9 = 20,000

Married Male = 25,000* 40/100 = 10,000

Married Female = 10,000

20,000 ———— 10,000

100 ————- ? => 50% - Question 48 of 50
##### 48. Question

If 25% of x is 15 less than 15% of 1500, then x is?

Correct25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225

Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.Incorrect25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225

Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840. - Question 49 of 50
##### 49. Question

The triplicate ratio of 1:2 is?

Correct1

^{3}: 2^{3}= 1:8Incorrect1

^{3}: 2^{3}= 1:8 - Question 50 of 50
##### 50. Question

Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is -.

CorrectUsing question number 11 and 12, we get the probability as

1 – (1/36 + 5/9) = 5/12IncorrectUsing question number 11 and 12, we get the probability as

1 – (1/36 + 5/9) = 5/12