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 2021 for all students. Indian Bank PO Pre Exam Free Online Quiz 2021, 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 2021. 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:
Correct
Let 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.Incorrect
Let 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?
Correct
Let the integer be x. Then,
x^{2} – 20x = 96
(x + 4)(x – 24) = 0
x = 24Incorrect
Let 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?
Correct
1 m^{2} — 1/2
? —– 600 => 1200 m^{2}
4x * 3x = 1200 => x = 10
Incorrect
1 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?
Correct
Total time taken = (160/64 + 160/8) = 9/2 hrs.
Average speed = 320 * 2/9 = 71.11 km/hr.Incorrect
Total 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?
Correct
A = 7500(26/25)^{2} = 8112
Incorrect
A = 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?
Correct
x = √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.Incorrect
x = √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.
Correct
The 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.Incorrect
The 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:
Correct
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200Incorrect
Let 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?
Correct
SP per metre = 18000/400 = Rs. 45 Loss per metre = Rs. 5 CP per metre = 45 + 5 = Rs. 50
Incorrect
SP 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 .
Correct
10 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 = 1/5Incorrect
10 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 = ?
Correct
2/3 * 5 1/2 * 2 * 7 5/6 * 9/11
2/3 * 11/2 * 2 * 47/6 * 9/11 = 47Incorrect
2/3 * 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?
Correct
Let 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 => 2:5:4Incorrect
Let 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 = ?
Correct
x % of 80 = 20 % of y => x / 100 (80) = 20 / 100 (y)
=> 4x = y
i.e x / y = ¼Incorrect
x % 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?
Correct
Number 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 minIncorrect
Number 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 twothird 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?
Correct
Present 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 = 28.8Incorrect
Present 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?
Correct
30 * 20 * x = (8 * 5.5 * 1.5)/2
Incorrect
30 * 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?
Correct
x:y:z = 100:45:30
20:9:6
9 — 27
35 — ? => 105Incorrect
x: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.
Correct
300 = 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.Incorrect
300 = 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 .
Correct
The 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.Incorrect
The 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?
Correct
Let 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%.Incorrect
Let 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?
Correct
D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 secIncorrect
D = 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?
Correct
2/5 + (2 + x)/10 = 1 => x = 4 days
Incorrect
2/5 + (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?
Correct
100 — 10 DS = 10
? — 1
75 — 15 US = 5
? —– 1 S = (10 – 5)/2
= 2 2 ½ kmphIncorrect
100 — 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.
Correct
Let 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.Incorrect
Let 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?
Correct
Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180mIncorrect
Length 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:
Correct
Required 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) = 30/7Incorrect
Required 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?
Correct
one 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.Incorrect
one 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?
Correct
Let 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.Incorrect
Let 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?
Correct
2/3 — 12
? —– 8 ==> 4/9Incorrect
2/3 — 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?
Correct
Let 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.Incorrect
Let 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 ?
Correct
Let 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.Incorrect
Let 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?
Correct
HCF of 384, 1296 = 48
48 * 48 * x = 384 * 1296
x = 216Incorrect
HCF 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:
Correct
LCM of numerators = 10
HCF of denominators = 1
=> 10/1 = 10Incorrect
LCM 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?
Correct
B 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.Incorrect
B 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.
Correct
Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63
Incorrect
Required 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?
Correct
1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 – 550 = 56Incorrect
1 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:
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 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?
Correct
Let 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.Incorrect
Let 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?
Correct
70 + 50 = 120 * 5/18 = 100/3 mps
D = 150 + 100 = 250 m
T = 250 * 3/100 = 15/2 = 7 ½ secIncorrect
70 + 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?
Correct
74330 Largest
30347 Smallest
————
43983Incorrect
74330 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. ?
Correct
Relative 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.Incorrect
Relative 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?
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 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?
Correct
Male = 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%Incorrect
Male = 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?
Correct
25% of x = x/4 ; 15% of 1500 = 15/100 * 1500 = 225
Given that, x/4 = 225 – 15 => x/4 = 210 => x = 840.Incorrect
25% 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?
Correct
1^{3}: 2^{3} = 1:8
Incorrect
1^{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 .
Correct
Using question number 11 and 12, we get the probability as
1 – (1/36 + 5/9) = 5/12Incorrect
Using question number 11 and 12, we get the probability as
1 – (1/36 + 5/9) = 5/12