Indian Bank PO Pre Online Test Series 4th, Indian Bank Online Test
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Indian Bank PO Pre Online Test Series 4th, Indian Bank Online Test Series for pre and mains exam. Indian Bank PO Pre Free Online quiz Series 4th. Take CAknowledge Free Indian Bank PO Prelims Mock Test Exam 2024. Indian Bank conducts online recruitment to admission to postgraduate diploma in banking & finance course offered through Manipal global education services 2024. 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 4. Here we are providing Indian Bank PO Pre. Full Mock Test Paper in English. Indian Bank PO Pre. Mock Test Series 4th 2024. 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 radius of a circle is increased by 1%. Find how much % does its area increases?
Correct
r = 100 r = 101
r2 = 10000 r2 = 10201
10000 —- 201
100 —- ? => 2.01%Incorrect
r = 100 r = 101
r2 = 10000 r2 = 10201
10000 —- 201
100 —- ? => 2.01% -
Question 2 of 50
2. Question
A dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent?
Correct
800 — 200
100 — ? => 25%Incorrect
800 — 200
100 — ? => 25% -
Question 3 of 50
3. Question
What is the difference between the largest number and the least number written with the digits 7, 3, 1, 4?
Correct
1347
7431
————
6084Incorrect
1347
7431
————
6084 -
Question 4 of 50
4. 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.
a2 = 784 = (4)(196) = (22).(142)
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.
a2 = 784 = (4)(196) = (22).(142)
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 5 of 50
5. Question
(272/4-3)-5/6 = ?
Correct
(272/4-3)-5/6 = [(33)2/(22)-3]-5/6
= (36 * 26)-5/6
= (66)-5/6 = 1/65 = 1/ 7776.Incorrect
(272/4-3)-5/6 = [(33)2/(22)-3]-5/6
= (36 * 26)-5/6
= (66)-5/6 = 1/65 = 1/ 7776. -
Question 6 of 50
6. Question
In an exam, a candidate secured 504 marks our of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 384 marks. What is the value of M?
Correct
504/M = 384/800
(504 * 800) / 384 = M
M = 1050Incorrect
504/M = 384/800
(504 * 800) / 384 = M
M = 1050 -
Question 7 of 50
7. Question
A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming. The distance between his house and school is?
Correct
Average speed = (2 * 3 * 2) / (3 + 2) = 12/5 km/hr.
Distance traveled = 12/5 * 5 = 12 km.
Distance between house and school = 12/2 = 6 km.Incorrect
Average speed = (2 * 3 * 2) / (3 + 2) = 12/5 km/hr.
Distance traveled = 12/5 * 5 = 12 km.
Distance between house and school = 12/2 = 6 km. -
Question 8 of 50
8. Question
The compound and the simple interests on a certain sum at the same rate of interest for two years are Rs.11730 and Rs.10200 respectively. Find the sum.
Correct
The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year
So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a.
So P = (100 * 10200)/ (30 * 2) = Rs.17000Incorrect
The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year
So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a.
So P = (100 * 10200)/ (30 * 2) = Rs.17000 -
Question 9 of 50
9. Question
I. a3 – 988 = 343,
II. b2 – 72 = 49 to solve both the equations to find the values of a and b?Correct
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ bIncorrect
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ b -
Question 10 of 50
10. Question
The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above conditions is:
Correct
Let the required numbers be 33a and 33b.
Then, 33a + 33b = 528 => a + b = 16.
Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).
Required numbers are (33 * 1, 33 * 15), (33 * 3, 33 * 13), (33 * 5, 33 * 11), (33 * 7, 33 * 9).
The number of such pairs is 4.Incorrect
Let the required numbers be 33a and 33b.
Then, 33a + 33b = 528 => a + b = 16.
Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).
Required numbers are (33 * 1, 33 * 15), (33 * 3, 33 * 13), (33 * 5, 33 * 11), (33 * 7, 33 * 9).
The number of such pairs is 4. -
Question 11 of 50
11. Question
The ratio of the cost price and the selling price is 4:5. The profit percent is:
Correct
Let C.P. = Rs. 4x. Then, S.P. = Rs. 5x
Gain = (5x – 4x) = Rs. x
Gain % = (x * 100)/ 4x = 25%.Incorrect
Let C.P. = Rs. 4x. Then, S.P. = Rs. 5x
Gain = (5x – 4x) = Rs. x
Gain % = (x * 100)/ 4x = 25%. -
Question 12 of 50
12. Question
The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is:
Correct
Correct sum = (36 * 50 + 48 – 23) = 1825.
Correct mean = 1825/50 = 36.5Incorrect
Correct sum = (36 * 50 + 48 – 23) = 1825.
Correct mean = 1825/50 = 36.5 -
Question 13 of 50
13. Question
In a class of 140 students, 60% of them passed. By what percent is the number of students who passed more than the number of failed students?
Correct
Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%.Incorrect
Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%. -
Question 14 of 50
14. Question
The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square.
Correct
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.Incorrect
Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found. -
Question 15 of 50
15. Question
A tradesman by means of his false balance defrauds to the extent of 20%? in buying goods as well as by selling the goods. What percent does he gain on his outlay?
Correct
g% = 20 + 20 + (20*20)/100
= 44%Incorrect
g% = 20 + 20 + (20*20)/100
= 44% -
Question 16 of 50
16. Question
A number when divided by 221 gives a remainder 43, what remainder will be obtained by dividing the same number 17?
Correct
221 + 43 = 264/17 = 9 (Remainder)
Incorrect
221 + 43 = 264/17 = 9 (Remainder)
-
Question 17 of 50
17. Question
Ram’s age and Shyam’s age are in the ratio 3:4. Seven years ago the ratio of their ages was 2:3. Find the ratio of their ages five years hence?
Correct
Let ages of Ramu and Shyam be x and y respectively.
x/y = 3/4 => x = 3/4 y
(x- 7)/(y – 7) = 2/3
=> 3x – 21 = 2y – 14 => 3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28years
=> x = 21 years
Ratio of their ages five years hence, (21 + 5)/(28 + 5) = 26/33.Incorrect
Let ages of Ramu and Shyam be x and y respectively.
x/y = 3/4 => x = 3/4 y
(x- 7)/(y – 7) = 2/3
=> 3x – 21 = 2y – 14 => 3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28years
=> x = 21 years
Ratio of their ages five years hence, (21 + 5)/(28 + 5) = 26/33. -
Question 18 of 50
18. Question
How many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}
Correct
The given digits are 1, 2, 3, 5, 7, 9
A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only ‘2’ and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.Incorrect
The given digits are 1, 2, 3, 5, 7, 9
A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only ‘2’ and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60. -
Question 19 of 50
19. 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?
Correct
2000 —- 50
2000 —- 40
x —– 50
x*50 = 2000*40
x=1600
2000
——-
400Incorrect
2000 —- 50
2000 —- 40
x —– 50
x*50 = 2000*40
x=1600
2000
——-
400 -
Question 20 of 50
20. Question
If 34m+1 = 37m-5, solve for m.
Correct
34m+1 = 37m-5 equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.
Incorrect
34m+1 = 37m-5 equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.
-
Question 21 of 50
21. Question
A man buys an article for 10% less than its value and sells it for 10% more than its value. His gain or loss percent is:
Correct
Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 – 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%Incorrect
Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 – 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20% -
Question 22 of 50
22. Question
3/7 of 4/9 of 2/5 of 7560 = (?)2
Correct
?2 = 3/7 * 4/9 * 2/5 * 7560
=> ?2 = 576 = 242
? = 24Incorrect
?2 = 3/7 * 4/9 * 2/5 * 7560
=> ?2 = 576 = 242
? = 24 -
Question 23 of 50
23. Question
If a man lost 4% by selling oranges at the rate of 12 a rupee at how many a rupee must he sell them to gain 44%?
Correct
96% —- 12
144% —- ?
96/144 * 12 = 8Incorrect
96% —- 12
144% —- ?
96/144 * 12 = 8 -
Question 24 of 50
24. Question
Two pipes P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first is turned off. How much longer will the cistern take to fill?
Correct
3/12 + x/15 = 1
x= 11 1/4Incorrect
3/12 + x/15 = 1
x= 11 1/4 -
Question 25 of 50
25. Question
Ram sold two bicycles, each for Rs.990. If he made 10% profit on the first and 10% loss on the second, what is the total cost of both bicycles?
Correct
(10*10)/100 = 1%loss
100 — 99
? — 1980 => Rs.2000Incorrect
(10*10)/100 = 1%loss
100 — 99
? — 1980 => Rs.2000 -
Question 26 of 50
26. Question
H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 24 * 35 * 52 * 72. The third number is:
Correct
3240 = 23 * 34 * 5; 3600 = 24 * 32 * 52
H.C.F = 36 = 22 * 32
Since H.C.F is the product of lowest powers of common factors, so the third number must have (22 * 32) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors.
Third number = 22 * 35 * 72Incorrect
3240 = 23 * 34 * 5; 3600 = 24 * 32 * 52
H.C.F = 36 = 22 * 32
Since H.C.F is the product of lowest powers of common factors, so the third number must have (22 * 32) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors.
Third number = 22 * 35 * 72 -
Question 27 of 50
27. Question
Which of the following is the greatest?
Correct
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest.Incorrect
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest. -
Question 28 of 50
28. Question
A can do half the work in one day where as B can do it full. B can also do half the work of C in one day. Ratio in their efficiency will be?
Correct
WC of A: B = 1:2
B: C = 1:2
———————
A: B: C = 1:2:4Incorrect
WC of A: B = 1:2
B: C = 1:2
———————
A: B: C = 1:2:4 -
Question 29 of 50
29. Question
HCF of 3/16, 5/12, 7/8 is:
Correct
HCF of numerators = 1
LCM of denominators = 48
=> 1/48Incorrect
HCF of numerators = 1
LCM of denominators = 48
=> 1/48 -
Question 30 of 50
30. Question
Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when fill?
Correct
1/10 + 1/15 = 1/6 * 4 = 2/3
1 – 2/3 = 1/3
1/10 + 1/15 – 1/x = 1/3
x = 8Incorrect
1/10 + 1/15 = 1/6 * 4 = 2/3
1 – 2/3 = 1/3
1/10 + 1/15 – 1/x = 1/3
x = 8 -
Question 31 of 50
31. Question
Find the area of circle whose radius is 7m?
Correct
22/7 * 7 * 7 = 154
Incorrect
22/7 * 7 * 7 = 154
-
Question 32 of 50
32. Question
The diagonals of a rhombus are 15 cm and 20 cm. Find its area?
Correct
1/2 * 15 * 20 = 150
Incorrect
1/2 * 15 * 20 = 150
-
Question 33 of 50
33. Question
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Correct
Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.
Incorrect
Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.
-
Question 34 of 50
34. Question
The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square.
Correct
Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 212
a = (212)1/2 = 26 = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16Incorrect
Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 212
a = (212)1/2 = 26 = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16 -
Question 35 of 50
35. Question
The present average age of a family of five members is 26 years. If the present age of the youngest member in the family is ten years, then what was the average age of the family at the time of the birth of the youngest member ? (Assume no death occurred in the family since the birth of the youngest)
Correct
Present total age of the members = 26(5) = 130 years.
Present age of the youngest member = 10 years
Present total age of the remaining four members = 130 -10 = 120 years
Their average age at the time of the birth of the youngest member = [120 – (4 * 10)] / 4 = 30 – 10 = 20 years.Incorrect
Present total age of the members = 26(5) = 130 years.
Present age of the youngest member = 10 years
Present total age of the remaining four members = 130 -10 = 120 years
Their average age at the time of the birth of the youngest member = [120 – (4 * 10)] / 4 = 30 – 10 = 20 years. -
Question 36 of 50
36. Question
By selling 50 meters of cloth. I gain the selling price of 15 meters. Find the gain percent?
Correct
SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7%Incorrect
SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7% -
Question 37 of 50
37. Question
Sixty men can stitch 200 shirts in 30 days working 8 hours a day. In how many days can 45 men stitch 300 shirts working 6 hours a day?
Correct
We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D2 * 6) / 300
D2 = (60 * 30 * 8 * 300) / (200 * 45 * 6) => D2 = 80.Incorrect
We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D2 * 6) / 300
D2 = (60 * 30 * 8 * 300) / (200 * 45 * 6) => D2 = 80. -
Question 38 of 50
38. Question
The average of 35 numbers is 25. If each number is multiplied by 5, find the new average?
Correct
Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.Incorrect
Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125. -
Question 39 of 50
39. Question
The average age of 8 men increases by 2 years when two women are included in place of two men of ages 20 and 24 years. Find the average age of the women?
Correct
20 + 24 + 8 * 2 = 60/2 = 30
Incorrect
20 + 24 + 8 * 2 = 60/2 = 30
-
Question 40 of 50
40. Question
4 * 0.4 * 0.04 * 0.004 = ?
Correct
4 * 0.4 * 0.04 * 0.004 => 4 * (4/10) * (4/100) * (4/1000) = (256/1000000) = 0.000256
Incorrect
4 * 0.4 * 0.04 * 0.004 => 4 * (4/10) * (4/100) * (4/1000) = (256/1000000) = 0.000256
-
Question 41 of 50
41. Question
A bag contains equal number of Rs.5, Rs.2 and Re.1 coins. If the total amount in the bag is Rs.1152, find the number of coins of each kind?
Correct
Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144Incorrect
Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144 -
Question 42 of 50
42. Question
Simplify : 1 – {1 + (a2 – 1)-1}-1
Correct
1 – {1 + (a2 – 1)-1}-1 = 1 – {1 + 1/a2 – 1}-1 = 1 – {a2 / a2 – 1}-1 = 1 – (a2 – 1)/a2 = 1/a2.
Incorrect
1 – {1 + (a2 – 1)-1}-1 = 1 – {1 + 1/a2 – 1}-1 = 1 – {a2 / a2 – 1}-1 = 1 – (a2 – 1)/a2 = 1/a2.
-
Question 43 of 50
43. Question
The least number of five digits that have 144 their HCF is?
Correct
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080Incorrect
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080 -
Question 44 of 50
44. 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 45 of 50
45. Question
An amount of Rs.1560 was divided among A, B and C, in the ratio 1/2 : 1/3 : 1/4. Find the share of C?
Correct
Let the shares of A, B and C be a, b and c respectively.
a : b : c = 1/2 : 1/3 : 1/4
Let us express each term with a common denominator which is the last number divisible by the denominators of each term i.e., 12.
a : b : c = 6/12 : 4/12 : 3/12 = 6 : 4 : 3.
Share of C = 3/13 * 1560 = Rs. 360.Incorrect
Let the shares of A, B and C be a, b and c respectively.
a : b : c = 1/2 : 1/3 : 1/4
Let us express each term with a common denominator which is the last number divisible by the denominators of each term i.e., 12.
a : b : c = 6/12 : 4/12 : 3/12 = 6 : 4 : 3.
Share of C = 3/13 * 1560 = Rs. 360. -
Question 46 of 50
46. Question
The number of sequences in which 7 players can throw a ball, so that the youngest player may not be the last is -.
Correct
x Not younger_______ ↑
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.
The required number of ways = 6(6!) = 4320Incorrect
x Not younger_______ ↑
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.
The required number of ways = 6(6!) = 4320 -
Question 47 of 50
47. Question
An automobile financier claims to be lending money at S.I., but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes?
Correct
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%.Incorrect
Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%. -
Question 48 of 50
48. Question
52 * 40.1 = ?
Correct
Since, 40.1 = 40
52 * 40 = 2080Incorrect
Since, 40.1 = 40
52 * 40 = 2080 -
Question 49 of 50
49. Question
(1714 * 1716) / 17-8 = ?
Correct
? = 1714+16-8 = 1722
Incorrect
? = 1714+16-8 = 1722
-
Question 50 of 50
50. Question
A number consists of 3 digit whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:
Correct
Let the middle digit be x. Then, 2x = 10 or x = 5.
So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred’s digit is smaller than the unit’s digit. Hence, required number = 253.Incorrect
Let the middle digit be x. Then, 2x = 10 or x = 5.
So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred’s digit is smaller than the unit’s digit. Hence, required number = 253.