IPPB Mock Test Series, IPPB Mains Online Test Series 4, IPPB Quiz
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IPPB Mock Test Series, IPPB Mains Online Test Series 4, IPPB Quiz. IPPB Mains Free Mock Test Exam 2021. IPPB Mains Exam Free Online Quiz 2021. IPPB Mains Full Online Mock Test Series 4th in English.candidates who want to make sure whether their preparation level is up to date then they take this online test series without paying anything. CAknowledge provide IPPB Online Test Series 2021 in order to let you excel in the examination. The IPPB Mains Full online mock test paper is free for all students. IPPB Mains Question and Answers in English and Hindi Series 4. Here we are providing IPPB Mains Full Mock Test Paper in English. IPPB Mains Mock Test Series 4th 2021. Now Test your self for IPPB Mains Exam by using below quiz…
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Question 1 of 50
1. Question
26.98% of 6002 + 45.01% of 4199 = ?
Correct
27% of 6000 + 45% of 4200 = ?
=> ? = 27/100 * 6000 + 45/100 * 4200
=> ? = 1620 + 1890 = 3510 = 3500Incorrect
27% of 6000 + 45% of 4200 = ?
=> ? = 27/100 * 6000 + 45/100 * 4200
=> ? = 1620 + 1890 = 3510 = 3500 
Question 2 of 50
2. Question
A man rows his boat 85 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?
Correct
Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (34 – 18)/2 = 8 kmphIncorrect
Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (34 – 18)/2 = 8 kmph 
Question 3 of 50
3. Question
The average age 9 members of a committee are the same as it was 2 years ago, because an old number has been replaced by a younger number. Find how much younger is the new member than the old number?
Correct
9 * 2 = 18 years
Incorrect
9 * 2 = 18 years

Question 4 of 50
4. Question
If 12 men can reap 120 acres of land in 36 days, how many acres of land can 54 men reap in 54 days?
Correct
12 men 120 acres 36 days
54 men ? 54 days
120 * 54/12 * 54/36
10 * 54 * 3/2
54 * 15 = 810Incorrect
12 men 120 acres 36 days
54 men ? 54 days
120 * 54/12 * 54/36
10 * 54 * 3/2
54 * 15 = 810 
Question 5 of 50
5. Question
The greatest number of four digits that have 144 for their HCF is?
Correct
144) 9999 (69
9936
——–
63
9999 – 63 = 9936Incorrect
144) 9999 (69
9936
——–
63
9999 – 63 = 9936 
Question 6 of 50
6. Question
The sum of three consecutive multiples of 3 is 72. What is the largest number?
Correct
Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27.Incorrect
Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27. 
Question 7 of 50
7. Question
The diameters of two spheres are in the ratio 1:2 what is the ratio of their surface area?
Correct
Incorrect

Question 8 of 50
8. 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 9 of 50
9. Question
A train running at a speed of 36 kmph crosses an electric pole in 12 seconds. In how much time will it cross a 350 m long platform?
Correct
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.Incorrect
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min. 
Question 10 of 50
10. Question
If A:B = 1/2: 1/3 B:C = 1/2:1/3 then A:B:C?
Correct
A:B = 1/2:1/3 = 3:2
B:C = 1/2:1/3 = 3:2
——————–
A:B:C = 9:6:4Incorrect
A:B = 1/2:1/3 = 3:2
B:C = 1/2:1/3 = 3:2
——————–
A:B:C = 9:6:4 
Question 11 of 50
11. Question
A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6AM and they work alternately for one hour each. When will the work be completed?
Correct
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days.Incorrect
Work done by A and B in the first two hours, working alternately = First hour A + Second hour B = 1/4 + 1/12 = 1/3.
Total time required to complete the work = 2 * 3 = 6 days. 
Question 12 of 50
12. Question
Find the greatest number that exactly divides 35, 91 and 840?
Correct
The greatest number that exactly divides 35, 91 and 840 is the HCF of the three numbers. So, calculating HCF we get the answer 7.
Incorrect
The greatest number that exactly divides 35, 91 and 840 is the HCF of the three numbers. So, calculating HCF we get the answer 7.

Question 13 of 50
13. Question
A person purchased a TV set for Rs. 16000 and a DVD player for Rs. 6250. He sold both the items together for Rs. 31150. What percentage of profit did he make?
Correct
The total CP = Rs. 16000 + Rs. 6250 = Rs. 22250 and SP = Rs. 31150
Profit(%) = (31150 – 22250)/22250 * 100 = 40%Incorrect
The total CP = Rs. 16000 + Rs. 6250 = Rs. 22250 and SP = Rs. 31150
Profit(%) = (31150 – 22250)/22250 * 100 = 40% 
Question 14 of 50
14. Question
The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is:
Correct
Let the numbers be a and b. Then,
ab = 17, a = 1 and b = 17So, 1/a^{2} + 1/b^{2} = (a^{2} + b^{2})/a^{2} b^{2}= (1^{2} + 17^{2})/(1 * 17)^{2} = 290/289Incorrect
Let the numbers be a and b. Then,
ab = 17, a = 1 and b = 17So, 1/a^{2} + 1/b^{2} = (a^{2} + b^{2})/a^{2} b^{2}= (1^{2} + 17^{2})/(1 * 17)^{2} = 290/289 
Question 15 of 50
15. Question
Identify the greatest numbers: 4^{50}, 2^{100}, 16^{25}
Correct
4^{50} = (2^{2})^{50} = 2^{100} => 16^{25} = (2^{4})^{25} = 2^{100}
Hence, 4^{50}, 2^{100} and 164^{25} are all equal.Incorrect
4^{50} = (2^{2})^{50} = 2^{100} => 16^{25} = (2^{4})^{25} = 2^{100}
Hence, 4^{50}, 2^{100} and 164^{25} are all equal. 
Question 16 of 50
16. Question
The difference between a twodigit number and the number obtained by interchanging the two digits is 63. Which is the smaller of the two numbers?
Correct
Let the ten’s digit be x and units digit by y.
Then,
(10x + y) – (10y + x) = 63
9(x – y) = 63
x – y = 7
Thus, none of the numbers can be determined.Incorrect
Let the ten’s digit be x and units digit by y.
Then,
(10x + y) – (10y + x) = 63
9(x – y) = 63
x – y = 7
Thus, none of the numbers can be determined. 
Question 17 of 50
17. Question
The value of (2.75)^{3} – (2.00)^{3} – (0.75)^{3} is
Correct
(2.75)^{3} – (2.00)^{3} – (0.75)^{3}
a^{3} + b^{3} + c^{3} = (a + b + c)(a^{3} + b^{3} + c^{3} – ab – bc – ca)
If a + b + c = 0 then a^{3} + b^{3} + c^{3} = 3abc
Hence 2.75 – 2.00 – 0.75 = 0
So, (2.75)^{3} – (2.00)^{3} – (0.75)^{3}
= 3 * 2.75 * 2 * 0.75 = 3 * 1.5 * 2.75
= 4.5 * 2.75Incorrect
(2.75)^{3} – (2.00)^{3} – (0.75)^{3}
a^{3} + b^{3} + c^{3} = (a + b + c)(a^{3} + b^{3} + c^{3} – ab – bc – ca)
If a + b + c = 0 then a^{3} + b^{3} + c^{3} = 3abc
Hence 2.75 – 2.00 – 0.75 = 0
So, (2.75)^{3} – (2.00)^{3} – (0.75)^{3}
= 3 * 2.75 * 2 * 0.75 = 3 * 1.5 * 2.75
= 4.5 * 2.75 
Question 18 of 50
18. Question
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8) = ?
Correct
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125Incorrect
(0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / (1.6 * 1.6 * 1.6 + 1.8 * 1.8 * 1.8)
= (0.9 * 0.9 * 0.9 + 0.8 * 0.8 * 0.8) / 2^{3}(0.8 * 0.8 * 0.8 + 0.9 * 0.9 * 0.9) = 1/8 = 0.125 
Question 19 of 50
19. Question
The average height of 50 pupils in a class is 150 cm. Five of them whose height is 146 cm, leave the class and five others whose average height is 156 cm, join. The new average height of the pupils of the class (in cm) is __________ .
Correct
Total height = 150 * 50 = 7500 cm.
New average = [7500 – 5 * 146 + 5 * 156 ] / 50 = 151 cm.Incorrect
Total height = 150 * 50 = 7500 cm.
New average = [7500 – 5 * 146 + 5 * 156 ] / 50 = 151 cm. 
Question 20 of 50
20. Question
36 * 48 ÷ 64 + 36 ÷ 12 = ?
Correct
36 * 48 / 64 + 36/12 = 27 + 3 = 30
Incorrect
36 * 48 / 64 + 36/12 = 27 + 3 = 30

Question 21 of 50
21. Question
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
Correct
Let the numbers be a, b and C. Then,
a^{2} + b^{2} + c^{2} = 138 and (ab + bc + ca) = 131
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
138 + 2 * 131 = 400
(a + b + c) = √400 = 20Incorrect
Let the numbers be a, b and C. Then,
a^{2} + b^{2} + c^{2} = 138 and (ab + bc + ca) = 131
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)
138 + 2 * 131 = 400
(a + b + c) = √400 = 20 
Question 22 of 50
22. Question
A man walked 20 m to cross a rectangular field diagonally. If the length of the field is 16 cm. Find the breadth of the field is?
Correct
Incorrect

Question 23 of 50
23. Question
Nitin’s salary is reduced by 10% and then reduced salary is increased by 10%. Find ,how many percentage his present salary is less as compared to his previous salary?
Correct
10*10
——– = 1%
100Incorrect
10*10
——– = 1%
100 
Question 24 of 50
24. Question
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x^{2} + 8x + 4 = 0?
Correct
a/b + b/a = (a^{2} + b^{2})/ab = (a^{2} + b^{2} + a + b)/ab
= [(a + b)^{2} – 2ab]/ab
a + b = 8/1 = 8
ab = 4/1 = 4
Hence a/b + b/a = [(8)^{2} – 2(4)]/4 = 56/4 = 14.Incorrect
a/b + b/a = (a^{2} + b^{2})/ab = (a^{2} + b^{2} + a + b)/ab
= [(a + b)^{2} – 2ab]/ab
a + b = 8/1 = 8
ab = 4/1 = 4
Hence a/b + b/a = [(8)^{2} – 2(4)]/4 = 56/4 = 14. 
Question 25 of 50
25. Question
In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?
Correct
Incorrect

Question 26 of 50
26. Question
In digging a pond 20 m * 10 m * 5 m the volumes of the soil extracted will be?
Correct
20 * 10 * 5 = 1000
Incorrect
20 * 10 * 5 = 1000

Question 27 of 50
27. Question
The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?
Correct
Sum of the 10 numbers = 230
If each number is increased by 4, the total increase =
4 * 10 = 40
The new sum = 230 + 40 = 270 The new average = 270/10 = 27.Incorrect
Sum of the 10 numbers = 230
If each number is increased by 4, the total increase =
4 * 10 = 40
The new sum = 230 + 40 = 270 The new average = 270/10 = 27. 
Question 28 of 50
28. Question
Find the roots of the quadratic equation: 2x^{2} + 3x – 9 = 0?
Correct
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2.Incorrect
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2. 
Question 29 of 50
29. Question
If the numerator of a fraction is increased by 20% and its denominator is diminished by 25% value of the fraction is 2/15. Find the original fraction.
Correct
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12Incorrect
X * (120/100)
————— = 2/15
Y * (75/100)
X/Y = 1/12 
Question 30 of 50
30. Question
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 2:3 and B:C = 2:5. If the total runs scored by all of them are 75, the runs scored by B are?
Correct
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18Incorrect
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18 
Question 31 of 50
31. Question
The number of permutations of the letters of the word ‘MESMERISE’ is_.
Correct
n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern ‘MESMERISE’ consists of 10 letters of which there are 2M’s, 3E’s, 2S’s and 1I and 1R.
Number of arrangements = 9!/(2!)^{2} 3!Incorrect
n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern ‘MESMERISE’ consists of 10 letters of which there are 2M’s, 3E’s, 2S’s and 1I and 1R.
Number of arrangements = 9!/(2!)^{2} 3! 
Question 32 of 50
32. Question
Ratio between two numbers is 3: 4 and their sum is 420. Find the smaller number?
Correct
3x + 4x = 420
x = 60 => 3x = 180Incorrect
3x + 4x = 420
x = 60 => 3x = 180 
Question 33 of 50
33. Question
The ratio between the radii of two spheres is 1:3. Find the ratio between their volumes?
Correct
r_{1} : r_{2} = 1:3
r_{1}^{3} : r_{2}^{3} = 1:27Incorrect
r_{1} : r_{2} = 1:3
r_{1}^{3} : r_{2}^{3} = 1:27 
Question 34 of 50
34. Question
If the cost price of 50 articles is equal to the selling price of 40 articles, then the gain or loss percent is?
Correct
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%Incorrect
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25% 
Question 35 of 50
35. Question
Product of two coprime numbers is 117. Their L.C.M should be:
Correct
H.C.F of coprime numbers is 1.
So, L.C.M = 117/1 = 117.Incorrect
H.C.F of coprime numbers is 1.
So, L.C.M = 117/1 = 117. 
Question 36 of 50
36. Question
Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
Correct
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15Incorrect
T = (121 + 165)/ (80 + 65) * 18/5
T = 7.15 
Question 37 of 50
37. Question
If the sides of a cube are in the ratio 4:3. What is the ratio of their diagonals?
Correct
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3Incorrect
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3 
Question 38 of 50
38. Question
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
Correct
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10.Incorrect
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y – 10) = 300 => xy + 5y – 10x – 50 = xy
=>5(300/x) – 10x – 50 = 0 => 150 + x^{2} + 5x = 0
multiplying both sides by 1/10x
=> x^{2} + 15x – 10x – 150 = 0
=> x(x + 15) – 10(x + 15) = 0
=> x = 10 or 15
As x>0, x = 10. 
Question 39 of 50
39. Question
A lends Rs. 2500 to B and a certain to C at the same time at 7% p.a. simple interest. If after 4 years, A altogether receives Rs. 1120 as interest from B and C, then the sum lent to C is?
Correct
Let the sum lent to C be Rs. x. Then,
(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120
7/25 x = (1120 – 700) => x = 1500Incorrect
Let the sum lent to C be Rs. x. Then,
(2500 * 7 * 4) / 100 + (x * 7 * 4) / 100 = 1120
7/25 x = (1120 – 700) => x = 1500 
Question 40 of 50
40. Question
The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas?
Correct
Ratio of the sides = ³√729 : ³√1331 = 9 : 11
Ratio of surface areas = 9^{2} : 11^{2} = 81 : 121Incorrect
Ratio of the sides = ³√729 : ³√1331 = 9 : 11
Ratio of surface areas = 9^{2} : 11^{2} = 81 : 121 
Question 41 of 50
41. Question
In an office, totally there are 6400 employees and 65% of the total employees are males. 25% of the males in the office are atleast 50 years old. Find the number of males aged below 50 years?
Correct
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.Incorrect
Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120. 
Question 42 of 50
42. Question
Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
Correct
Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.Incorrect
Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec. 
Question 43 of 50
43. Question
A man buys milk at a certain rate per liter and after adding water to it sells the mixture at the same rate. Find in what ration he much mix water to milk so as to gain 20% on his outlay?
Correct
Incorrect

Question 44 of 50
44. Question
A bank offers 5% C.I. calculated on halfyearly basis . A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is?
Correct
Amount = [1600 * (1 + 5/(2 * 100)^{2} + 1600 * (1 + 5/(2 * 100)]
= [1600 * 41/40(41/40 + 1)
= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.
C.I. = 3321 – 3200 = Rs. 121.Incorrect
Amount = [1600 * (1 + 5/(2 * 100)^{2} + 1600 * (1 + 5/(2 * 100)]
= [1600 * 41/40(41/40 + 1)
= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.
C.I. = 3321 – 3200 = Rs. 121. 
Question 45 of 50
45. Question
The ratio of two natural numbers is 5:6. If a certain number is added to both the numbers, the ratio becomes 7:8. If the larger number exceeds the smaller number by 10, find the number added?
Correct
Let the two numbers be 5x and 6x.
Let the number added to both so that their ratio becomes 7:8 be k.
(5x + k)/(6x + k) = 7/8
42x = 7k => k = 2x.
6x – 5x = 10 => x = 10
k = 2x = 20.Incorrect
Let the two numbers be 5x and 6x.
Let the number added to both so that their ratio becomes 7:8 be k.
(5x + k)/(6x + k) = 7/8
42x = 7k => k = 2x.
6x – 5x = 10 => x = 10
k = 2x = 20. 
Question 46 of 50
46. Question
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
Correct
C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24
A’s wages : B’s wages : C’s wages
1/6 : 1/8 : 1/24 = 4:3:1
C’s share = 1/8 * 3200 = Rs. 400Incorrect
C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24
A’s wages : B’s wages : C’s wages
1/6 : 1/8 : 1/24 = 4:3:1
C’s share = 1/8 * 3200 = Rs. 400 
Question 47 of 50
47. Question
What is the least number to be subtracted from 11, 15, 21 and 30 each so that 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)
=> (11 – x)(30 – x) = (15 – x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18Incorrect
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)
=> (11 – x)(30 – x) = (15 – x)(21 – x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 
Question 48 of 50
48. Question
The ratio of numbers is 3:4 and their H.C.F is 4. Their L.C.M is:
Correct
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48.Incorrect
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48. 
Question 49 of 50
49. Question
There is 60% increase in an amount in 6 years at S.I. What will be the C.I. of Rs. 12,000 after 3 years at the same rate?
Correct
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 * 60)/(100 * 6) = 10% p.a.
Now, P = Rs. 12000, T = 3 years and R = 10% p.a.
C.I. = [12000 * {(1 + 10/100)3 – 1}]
= 12000 * 331/1000 = Rs. 3972Incorrect
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 * 60)/(100 * 6) = 10% p.a.
Now, P = Rs. 12000, T = 3 years and R = 10% p.a.
C.I. = [12000 * {(1 + 10/100)3 – 1}]
= 12000 * 331/1000 = Rs. 3972 
Question 50 of 50
50. Question
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves onefourth of his income, find the ratio of their monthly savings?
Correct
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.Incorrect
Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) – (His expenditure) = 5x – 2y.
=> 5x = 20x – 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x – 3y : 1/4(5x) = 6x – 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.