Profit and Loss:

1. kavita sold a mobile phone at a price of rs.1950 and made a loss of 25%. At what price will she have to sell it against to get a profit of 30%?
1. rs.3380
1. SP=rs.1950
2. 75/100*x = 1950
1. x = rs.2600
2. 130/100 * 2600 = rs.3380
2. What is the actual profit when the profit on selling price is 25%? – 33.33%
1. P% = SP-CP/CP * 100
2. Let SP be 100x
3. Profit = 25% of 100x = 25x
4. 25x = 100x – CP
5. CP = 75x
6. P% = 25x/75x * 100 = 100/3 = 33.33%
1. Reference:
3. A trader marks up the price of goods he sells by 50% and gives a discount of 20%. Besides this, he also weighs the goods by 20% less. What is the net profit he gets?
1. 50%
1. Let the CP of the goods be 100x
2. then, MP = CP + 50% of CP
3. 100x+50x = 150x
4. SP = MP*(100-D%)/100
5. 150x(100-20)/100 = 120x
6. Profit = SP – CP = 120x-100x = 20x
7. P% = 20x/100x * 100 = 20%
1. When the trader weighs the goods 20% less than it means, instead of 1000gms he only weighed 800 gms
2. trader earns profit of 200gms in 800gms then profit% = 200/800 * 100 = 25%
3. Overall profit = 20+25+(20*25)/100 = 50%
4. Net profit earned by trader is 50%
5. Reference:
4. A shopkeeper sells an item at a discount of 40% on the items listed price of rs.1500. He still makes a profit of 20% on the original cost of the item. What is the original cost of the item?
1. rs.750
1. D=40%, LP = 1500, P=20%
1. SP = MP(100-D%)/100 and CP = (SP*100)/100+P%
2. Solving SP = 60/100*1500 = rs.900
3. CP = 100/120 *900 = rs.750
1. Reference:
5. S-1: If an article is sold at a gain of 19% then Selling price is 109% of CP
1. S-2: If an article is sold at a loss of 9% then SP is 109% of CP
1. Both S-1 and S-2 incorrect

Interest and Discounting:

1. A sum of money with C.I becomes rs.400 in 2yrs and rs.500 in 3 yrs. Find the sum
1. 256
1. Formuale required:
1.  Amount = sum + interest
2. Interest = PTR/100     (∵ for simple and compound interest it is same first year)
3. Compound Interest = (P[(1+ {R}/{100})^n-1])
4. Amount = (P(1 + {R}/{100})^n)
1. Calculations:
5. Interest for 1 year = 500 – 400⇒ 100
1. Interest for 1 year = PRT/100⇒ 100 = PRT/100
1. ⇒ 100 = 400 × R × 1/100⇒ 100 = 4R⇒ R = 25%
2. Hence, rate of interest = 25%
3. Amount = (P(1 + {R}/{100})^n)
2. ⇒ 400 =(P(1 +{25}/{100})^2) ⇒ 400 = P(5/4)⇒ 400 × (4/5) × (4/5) = P
3. ⇒ 16 × 16 = P= 256
6. ∴ The sum is 256
7. Reference
2. A certain principal is invested at the annual rate of interest of 5% for 2 yrs. The difference between CI and SI is rs.1What is the principal?
1. rs.400
1. SI = PRT/100
2. P*5*2/100 = 10P/100 = P/10
3. A = P(1+r/100)^n = P(1+5/100)^2
4. P(21/20)^2 = 441P/200
5. CI = A-P
6. 441P/400 – P = 41P/400
7. Now, A/Q = 41P/400 – P/10 = 1, P/400 = 1 = P=rs.400
1. Short trick
1. (CI-SI)for 2yrs = P*R^2/100^2
2. 1 =P*5^2/100^2 =(100^2)/50^2 =400
3. A sum of rs.50,000 is invested in a bank deposit for 1 year at an interest rate of 10%p.a compounded on a half yearly basis. What is the amount of interest at the end of 1 year?
1. 5125
1. A=P(1+r/100)^n and A =P+I
2. A = 50000*(1+5/100)^2
3. A= rs. 55,125
4. Amount of Interest = P+I = 50,000+Interst = 55,125
5. I = rs.5125
4. A sum of money earning compound interest annually doubles itself after 4 years. What is the rate of interest?(Take 2^1/4 = 1.19)
1. 19%
1. t=4yrs, A=2P
2. A = P(1+r/100)^n
3. 2P =P(1+r/100)^4
4. 2^1/4 = 1+r/100
5. 2^1/4 = (100+r)/100
6. 1.19*100 = 100+r
7. 119 = 100+r so r=19%
1. Reference:
5. A household incurs a certain expenditure on the consumption of a commodity in a given month. If the price of the commodity is decreased by 20% the net % increase or decrease in the expenditure of the household?
1. Given:
2. The decrease in the price of the commodity = 20%
3. The increase in consumption = 20%
4. Formula used:
5. Percentage change = (Final value – Initial value)/Initial value × 100
6. Calculation:
7. Let the initial expenditure on the commodity be 100x
8. Price of the commodity after decrease = 100x – 20% of 100x = 80x
9. The final expenditure on the commodity after 20% increase in consumption = 80x + 20% of 80x = 96x
10. The percentage decrease in the expenditure of the household = (100x – 96x)/100x × 100 = 4%
1. ∴ The percentage decrease in the expenditure of the household is 4%
1. Shortcut
1. Net increase or decrease = + a + b + ab/100
2. Where, a = first percentage change and, b = second percentage change.
3. Calculation:
4. Net percentage change = + 20 – 20 – (20 ×20)/100 = – 4%
5. ∴ The percentage decrease in the expenditure of the household is 4%
1. Note – while using this formula as here we need to take the signs while choosing the values of a and b as well.
2. In this case, a = + 20 and b = – 20
2. Reference:
6. S-1: CI involves the reinvestment of the earned interest which in future years also earns interest
1. S-2: SI is the amount earned or paid on a sum of compound interest and principal amount
2. S-1 is correct s-2 is false.
7. What is the actual (equivalent) discount of three successive discounts of 8%, 10%, and 12% by sale of an article?
1. 27.136%
1. Equivalent discount = (-d1+(-d2)+(d1*d2)/100)
2. -8+-10+8*10/100 = -17.2%
3. equivalent discount of -17.2% and 12% = -17.2 + (-12) + (17.2*12)/100 = 27.136
2. Short trick: ratio method
1. 8% = 2/25, 10%=1/10, 12% = 3/25
1. D% = D/MP *100
1. D =MP – SP
1. D – 8% MP – 25 SP- 23
2. D- 10% MP-10 SP-9
3. D- 12% MP-25 SP-22
1. 6250 4554
2. D = 6250-4554 = 1696
3. equivalent discount= 1696 / 6250 * 100 = 27.136%
8. The volume of a cube of a side is 10 cm is changed by increasing its side by 1%. The % increase in the volume of the cube is
1. 3%
1. The volume of the cube = side3
2. Change in percentage = (Final value – Initial value)/Initial value × 100
3. Calculation:
4. The initial volume of the cube = 10^3(a^3) = 1000 cm3
5. The side of the cube after increase = 10 + 1% of 10 = 10.1
6. The final volume of the cube = 10.1^3 = 1030.3
7. Increase in the volume of the cube = Final volume – Initial volume
8. ⇒ 1030.3 – 1000 = 30.3
9. Percentage Increase = (30.3/1000) × 100 = 3.03%
10. ∴ The percentage increase in the volume of the cube is 3.03%
1. reference:

Averages:

1. S-1: the mean is the numerical average and is found by adding up all the values and dividing the sum by the number of values
1. S-1 mean of the following wages is 16000
1. 12000, 10000, 23000, 15000
1. Mean = Average ⇒ Average = sum of numbers/number of terms
2. Statement I is true
3. Statement II :
4. Mean of wages = (12000 + 10000 + 23000 + 15000)/4
5. ⇒ 60000/4 = 15000
6. Statement II is false
7. ∴ Statement I is correct and statement II is false
1. Logic:
2. A man is 30 yrs older than his son. five years ago his age was 6 times the age of his son. The age of the son is
1. Man’s age = x+30
2. 5 yrs ago sons age = x-5
3. 5yrs ago mans age = x+30-5 = x+25
4. 5yrs ago ratio of age of man and son = 6:1
5. x-5/x+25 = 1/6
6. solving for x = 11yrs
3. S-1: the mean is the numerical average and is found by adding up all the values and dividing the sum by the number of values
1. S-1 mean of the following wages is 16000
1. 12000, 10000, 23000, 15000
4. Consider the following 3 sets of numbers
1. A. 2,5,9
2. B. 1,6,10
3. C. 3,4,8 – The correct ascending order of the averages of 3 sets A, B and C?
1. C<A<B
2. Average = Sum of all the terms/Total number of terms
3. Concept used:
1. Ascending Order means number should be arranged in the increasing sequence
4. Calculation:
1. Average of the set A = (2 + 5 + 9)/3 = 5.33
2. Average of the set B = (1 + 6 + 10)/3 = 5.66
3. Average of the set C = (3 + 4 + 8)/3 = 5
4. Arranging in the ascending order, 5 < 5.33 < 5.66
5. ∴ The correct sequence is C < A < B
6. Logic:
5. If 30% of A = 0.25 of B=1/5 of C which of the following is true about A:B:C?
1. 10:12:15
1. Calulation:
2. 30% of A = 0.25 of B
3. ⇒ 3A/10 = B/4
4. ⇒ A/B = 5/6
5. ⇒ A : B = 5 : 6      —-(1)
6. 0.25 of B = 1/5h part of C
7. ⇒ B/4 = C/5
8. ⇒ B/C = 4/5
9. ⇒ B : C = 4 : 5      —-(2)
10. Multipling (1) by 4 and (2) by 6 get,
11. A : B = 20 : 24
12. B : C = 24 : 30
13. ⇒ A : B : C = 20 : 24 : 30
14. ⇒ A : B : C = 10 : 12 : 15
15. ∴ A : B : C = 10 : 12 : 15
1. Logic:
6. The product of the ages of sagun and srishti is 240. If twice the age of srishti is more than sagun’s age by 4 years, then what is srishti’s age
1. 12
1. let the present age of sagun and srishti be X and Y yrs respectively
1. xy = 240
2. and 2y-x=4, x=2y-4
3. (2y-4)y = 240
4. (y-2)y = 240
5. on solving y=12 and y= -10 as age cannot be negative present age of srishti be 12yrs
7. average of seven consecutive numbers is 33. The largest of these numbers is
1. 36
1. Average of seven consecutive numbers = 33
2. Calculations:
3. Let the seven consecutive numbers be ‘(x – 3)’,’ (x – 2)’, ‘(x – 1)’, ‘x’, ‘(x + 1)’,'(x + 2)’, ‘(x + 3)’
4. Sum of the seven numbers = x – 3 + x – 2 + x – 1 + x + x + 1 + x + 2 + x + 3
5. ⇒ 7x
6. Average = sum of numbers / number of terms
7.  ⇒ 33 = 7x/7
8. ⇒ x = 33
9. Hence, largest number = x + 3
10. ⇒ Largest number = 33 + 3
11. ⇒ 36
12. ∴ The largest number in the consecutive series is 36
1. Reference:
8. The ratio of ages of two persons A and B is 3:4 and that of ages B and C is 4:5. If the average of all 3 is 40 yrs what is B’s age?
1. 40 yrs
1. Average age = 40 years
2. Formula used:
3. Average = Sum of all the terms/Total number of terms
4. Calculation:
5. A ∶ B = 3 ∶ 4
6. And, B ∶ C = 4 ∶ 5
7. Then, A ∶ B ∶ C = 3 ∶ 4 ∶ 5
8. Let the age of A, B and C be 3x, 4x and 5x
9. Sum of ages of all the persons = 3x + 4x + 5x = 12x
10. Average = 12x/3 = 4x
11. ⇒ 4x = 40
12. ⇒ x = 10
13. ∴ The age of B is 40
1. Reference:
9. Average of 3 numbers is 22. Sum of two of the numbers is 24 and these two numbers are in the ratio 7:5. Find three numbers
1. 14,10,42
1. Let the numbers be ‘a’, ‘b’ and ‘c’
2. If sum of two numbers  = a + b ⇒ a + b = 24
3. The ratio of these two numbers = a ∶ b
4. a ∶ b = 7 ∶ 5
5. Sum of the ratio (a + b) = (7x + 5x)
6. ⇒ 12x = 24 ⇒ x = 2
7. Hence, the value of a = 7 × 2= 14
8. Hence, the ratio of b = 5 × 2 = 10
9. Sum of (a + b) = 10 + 14 ⇒ (a + b) = 24
10. Average of three numbers = (a + b + c)/3
1. ⇒ (a + b + c)/3 = 22
2. ⇒ (a + b + c) = 22 × 3
3. ⇒ (a + b + c) = 66
4. ⇒ (24 + c) = 66
5. ⇒ c = 66 – 24 ⇒ c = 42
6. ∴ Three numbers a, b, and c are 10, 14 and 42 respectively
1. Reference:
10. The average of 70 numbers is 32. If each number is increased by 3 , then the new average will
1. 35
1. Total number = 70
2. Average = 32
3. Formula used:
4. Average = Sum of all the terms/Total number of terms
5. Calculation:
6. Sum of all the numbers = 32 × 70 = 2240
7. Sum of all the number after increase = 2240 + 3 × 70 = 2450
8. New average = 2450/70 = 35
9. ∴ The new average is 35
1. Method-2 short
2. When every number is increased by the same number then the average is also increased by the same number
3. New average = 32 + 3 = 35
4. ∴ The new average is 35
5. Reference:
11. A student scores 55,60,65, and 70 respectively in english , physics, chemistry and math. If the credits assigned to E,P, C, M are 2,3,3 and 4what would be the average marks of students?
1. 63.75
1. The marks of students in English, Physics, Chemistry and Mathematics is 55, 60, 65 and 70 respectively
2. Formula used:
3. Average = Sum of all the terms/Total number of terms
4. Concept used:
5. Here the credit means the weightage of the subjects
6. Ex- 2 credit of English means English is considered as the two subjects
7. Calculation:
8. Sum of marks of all the subjects according to the credits = 55 × 2 + 60 × 3 + 65 × 3 + 70 × 4 = 765
9. Average marks = 765/(2 + 3 + 3 + 4) = 63.75
10. ∴ The average mark of the students is 63.75
1. Reference:
12. Given 3numbers 1st is twice the second and is half of third. If the average of 3 numbers is 56 then the difference of 1st and third number is
1. 48
1. Let the first number be x
2. Second number = x/2
3. Third number = 2x
4. Average of three number x, x/2 and 2x = 56
5. ⇒ (x + x/2 + 2x)/3 = 56
6. ⇒ (7x/2)/3 = 56
7. ⇒ x = 336/7 = 48
8. Difference between first and third number = 2x – x = x
9. ∴ Difference between first and third number = 48
1. Reference:
13. Mr.X is 40 yrs old and Mr.Y is 60 yrs old. How many years ago was the ratio of their ages 3:5?
1. 10 yrs
1. The age of Mr X = 40 years
2. The age of Mr Y = 60 years
3. Calculation:
1. Let t years ago their ratio of age was 3 ∶ 5
2. Now, t years ago
3. The age of Mr X = (40 – t) years
4. The age of Mr Y = (60 – t) years
5. Now, A/Q
1. (40 – t)/(60 – t) = 3/5
2. ⇒ 5(40 – t) = 3(60 – t)
3. ⇒ 200 – 5t = 180 – 3t
4. ⇒ 2t = 20⇒ t = 10 years
6. ∴ The required ratio of their age was 10 years ago
14. A man is 30yrs older than his son. 5 yrs ago his age wasa 6 times the age of his son the age of the son is
1. 11 yrs
1. Let the age of son be ‘x’
1. Hence, The man age = x + 30
2. Five years ago, son age = x – 5
3. Five years ago, man age = x + 30 –  5
4. Hence, age of man five years ago = x + 25
5. According to the question,
6. ⇒ x – 5/(x + 25) = 1/6
7. ⇒ 6 × x = 1 × (x +25)
8. ⇒ 6x – 30 = x + 25
9. ⇒ 6x – x = 25 + 30
10. ⇒ 5x = 55
11. ⇒ x = 11
12. ∴ The age of son is 11 years
1. Reference:
15. The difference between the ages of two men is 10yrs. 15 yrs ago the elder one was twice as old as the younger one. The present age of the elder man is
1. 35 yrs
1. The difference between the ages of two men = 10 years
2. 15 years ago, the ratio of elder man to younger man = 2 ∶ 1
3. Calculations :
4. Let the present age of elder man be ‘a’
5. The present age of a younger man be ‘b’
6. According to the question,
7. 15 years ago the age of elder man = a – 15
8. 15 years ago the age of younger man = b – 15
9. a – b = 10
10. ⇒ a = b + 10
11. The ratio of ages of the elder man and younger man = (a -15) ∶ (b – 15)
12. (a -15) / (b – 15) = 2 / 1
13. ⇒ 1 × (a – 15) = 2 × (b – 15)
14. ⇒ a – 15 = 2b – 30
15. ⇒ a – 2b = -30 + 15
16. ⇒ b + 10 – 2b = -15
17. ⇒ -b = -15 -10
18. ⇒ b = 25
19. a = b + 10
20. ⇒ a = 25 + 10                   (∵ a = b + 10 )
21. Hence a = 35 years
22. ∴The elder man age is 35 years
1. Reference:

### Time & Distance:

1. A man covers a distance of 22km in 4hrs initially walking and then jogging at speed of 4km/hr and 10km/hr. The distance covered by him during walking is
1. Total distance covered = 22 km
2. Time taken = 4 hours
3. Speed of man when walking = 4 km/hr
4. Speed of man when jogging = 10 km/hr
5. Formula used:
6. Distance = Speed × Time
7. Calculation:
8. Let the walking time of the man be x hours then jogging time be (4 – x) hours
9. Now, A/Q
10. Speed × Time = Distance
11. ⇒ 4x + 10(4 – x) = 22
12. ⇒ 4x + 40 – 10x = 22
13. ⇒ 6x = 18
14. ⇒ x = 3
15. Distance covered by man during walking = 4 × 3 = 12
16. ∴ The distance covered by man during walking is 12 km
2. A motorist travels to a place 150km away, at an average speed of 50 km/hr and returns at a speed of 30km/hr. His average speed for the whole journey is
1. 37.5 km/hr
1. Average speed = 2S1S2/(S1 + S2)
2. Calculation:
3. According to the question,
4. Average speed = 2S1S2/(S1 + S2)
5. S1 = 50 km/hr & S2 = 30 km/hr
6. ⇒ 2 × 50 × 30/(50 + 30)
7. ⇒ 3000/80
8. ⇒ 37.5 km/hr
9. ∴ His average speed for the whole journey in km per h is 37.5.
3. A man walks at a speed of 3km/hr from location A towards east for 1hr and then turning left walks at a speed of 1.5km/hr for 6hrs towards north to a location B. The shortest distance between locations A and B is
1. 9.48km
1. The speed of man in east direction = 3 km/hr
2. The speed of man in north direction = 1.5 km/hr
3. Time taken is taken by a man in east direction = 1 hour
4. Time taken is taken by a man in north direction = 6 hour
5. Formula used:
6. Distance = Speed × Time
7. Concept used:
8. The Hypotenuse is the shortest distance while travelling through the right angle
1. The distance covered by man in east direction = 3 × 1 = 3 km
2. The distance covered by man in North direction = 1.5 × 6 = 9 km
3. The shortest distance between point A and B = √(32 + 92) = 9.48
4. ∴ The shortest distance between point A and B is 9.48 km
9. Reference:
4. A man walks 400 mts in north direction, took a left turn and walks 600 mts. He again take left turn and walks for 400 mts . How far and in which direction he is from original direction?
1. 600 mts west
1. Logic:
5. A person in a car drives along a straight road from point A to point B at a speed of 60km/hr for 15min and then from point B to C at a speed of 90km/hr for 10minutes. The average speed of his travel from point A to C is
1. 72km/hr
2. Average speed = Total distance/Total time
3. Distance = Speed × Time
4. Calculation:
5. Distance from A to B = 60 × (15/60) = 15 km
6. Distance from B to C = 90 × (10/60) = 15 km
7. Average speed = (15 + 15)/(15/60 + 10/60) = 72 km/hr
8. ∴ The average speed of his travel from point A to C is 72 km/hr
1. Reference:
6. A man reaches his office 30 minutes late if he walks by 2/3rd of his routine speed. What is the time he usually takes to reach office?
1. 60 minutes
1. The man reaches the office late by 30 minutes
2. Formula used:
3. Distance = Speed × Time
4. Concept used:
5. Speed is inversely proportional to the time taken
6. Calculation:
7. Let the distance from the office and home be x and original speed be y
8. Then, the Original time is taken = x/y
9. Decreased speed = 2y/3
10. Time is taken when speed is decreased = Distance/Speed = x/(2y/3)
11. Now, A/Q
12. x/(2y/3) – x/y = 30
13. ⇒ 3x/2y – x/y = 30
14. ⇒ x/2y = 30
15. ⇒ x/y = 30 × 2 = 60 minutes
16. ∴ The actual time is taken by the man to reach the office is 60 minutes
1. Reference:
7. In 100 mt race A beat B by 10mts or 2 sec what is the speed of A?
1. 20 km/hr
1. Distance to cover = 100 m
2. Formula used:
3. Distance = Speed × Time
4. Concept used:
5. A beats B by 10 meters or 2 seconds means B covers 10 meters distance in 2 seconds
6. Calculation:
1. Speed of B = 10/2 = 5 m/s
2. Total time is taken by B to complete 100 m race = 100/5 = 20 seconds
3. A takes 2 seconds less than B to complete the race
4. So, time taken by A to complete the race = 20 – 2 = 18 sec
5. Speed of A = 100/18 m/s = (100/18) × (18/5) = 20 km/h
6. ∴ The speed of A is 20 km/h
1. reference
8. 3 pipes fill a tank in 4hrs. If two of them take 8 and 12 hrs each to fill the tank, how many hours will third pipe take to fill the tank?
1. 24 hrs
1. Shortcut: To find P3’s time
2. Time taken= 4,8,12,
3. Efficiency of all pipes, P1,P2 = 6,3,2
4. Total work = 24(LCM of 4,8,12)
5. Efficiency of the third pipe = 6 – (3 + 2) = 1
1. Time is taken by pipe 3 = 24/1 = 24 hours
2. ∴ Total time is taken by pipe 3 to fill the tank is 24 hours
2. Method-2
1. Time taken by all the pipes to fill the tank = 4 hours
2. Time taken by pipe 1 to fill the tank = 8 hours
3. Time taken by pipe 2 to fill the tank = 12 hours
4. Formula used:
1. Time is taken = Total work/Efficiency
5. Calculation:
1. In 1 hour part of the tank filled by all the pipes = 1/4
2. In 1 hour part of the tank filled by pipe 1 = 1/8
3. In 1 hour part of the tank filled by pipe 2 = 1/12
4. In 1 hour part of the tank filled by pipe 3 = 1/4 – (1/8 + 1/12) = 1/24
5. ∴ Total time is taken by pipe 3 to fill the tank is 24 hours
6. Reference
9. The LCM of 2 numbers is 144 and their GCD is 2. What are the numbers given their sum is 34.
1. 18,16
1. Given:
2. L.C.M = 144
3. G.C.D or H.C.F = 2
4. Sum of the two number = 34
5. Formula used:
6. Product of two numbers = L.C.M × H.C.F
7. (x + y)2 = x2 + y2 + 2xy
8. (x – y)2 = x2 + y2 – 2xy
9. Calculation:
10. Let the two number be x and y
11. Now, Product of two numbers = L.C.M × H.C.F
12. ⇒ xy = 144 × 2 = 288      —-(1)
13. And, x + y = 34      —-(2)
14. (x + y)2 = 342
15. ⇒ x2 + y2 + 2xy = 1156
16. ⇒ x2 + y2 + 2 × 288 = 1156
17. ⇒ x2 + y2 = 1156 – 576 = 580
18. ⇒ x2 + y2 = 580      —-(3)
19. Now, (x – y)2 = x2 + y2 – 2xy
20. ⇒ (x – y)2 = 580 – 2 × 288 = 4
21. ⇒ (x – y) = ±2      —-(4)
22. From eq (2) and eq (4), Taking positive value
23. x + y + x – y = 34 + 2
24. ⇒ 2x = 36
25. x = 18 and y = 16
26. When we take negative value then,
27. x= 16 and y = 18
28. ∴ The two numbers are 18 and 16
1. Reference
10. Two water pipes can fill a tank in 10 minutes and 15 minutes, respectively, if each of them is used separately. If both the pipes are used to fill the tank simultaneously how long will it take to fill the tank –
1. 6 minutes
1. Time is taken by pipe 1 to fill the tank = 10 min
2. Time is taken by pipe 2 to fill the tank = 15 min
3. Formula used:
4. Time to do work = Total work/Efficiency
5. Calculation:
6. Work is done by pipe 1 in 1 min = 1/10
7. Work is done by pipe 2 in 1 min = 1/15
8. Work is done by pipe 1 and pipe 2 in 1 min = 1/10 + 1/15
9. = (3 + 2)/30
10. = 5/30 = 1/6 min
11. Time is taken by both the pipe to do work = 6 min
12. ∴ The time is taken by both the pipe to fill the tank simultaneously is 6 min
1. Method-2:
1. Time is taken to do the work together = 30/(3 + 2) = 6
2. ∴ The time is taken by both the pipe to fill the tank simultaneously is 6 min
1. efficiency Time total work
2. 3,2 (P1,P2) 10,15 (LCM of 10,15)
11. 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?
1. Time taken by 3 men, 4 women, 6 children to complete the work = 7 days.
2. A woman does double work compare to a man and a child does half work compare to a man.
3. Ratio of efficiency of man and woman = 1 ∶ 2,
4. Ratio of efficiency of child and man = 1 ∶ 2
5. Calculation∶
1. Ratio of efficiency of man, woman and child = 2 ∶ 4 ∶ 1
2. Total work = (3 man + 4 woman + 6 child) × 7 days = (3 × 2 + 4 × 4 + 6 × 1) × 7 = 28 × 7 = 196
3. Number of women necessary to complete to the work in 7 days = 196/(4 × 7) = 7
1. Reference:
12. 3 persons are walking around a ground. They take 20,30, and 40 minutes to complete one round. If they start from a single point together they will meet again at that point after
1. 2 hrs
1. Logic:
2. Three persons take 20, 30 and 40 minutes to complete one round.
3. L.C.M of 20, 30 and 40 minutes is 120 minutes.
4. 120 minutes = 2 hours
5. So, again after 120 minutes or 2 hours, they will meet again at that point.
6. Hence, ‘2 hours’ is the correct answer.
13. In a river from a fixed location two identical boats start moving upstream and down stream respectively. The speed of the water current in the river is 3km/hr. If the speed of each of the boats in still water is 15km/hr the separation between the boats after 12 min will be
1. 6 km
1. Concept used:
2. Speed of the boat in up-stream = u – v units
3. Speed of the boat in downstream = u + v units
4. Here, u and v is the speed of the boat in still water and speed of current respectively
5. Calculation:
6. Speed of the boat in up-stream = 15 – 3 = 12 km/hr
7. Speed of the boat in downstream = 15 + 3 = 18 km/hr
8. The distance covered by boat in upstream = 12 × 1/5 = 12/5 km
9. The distance covered by the boat in downstream = 18 × 1/5 = 18/5 km
10. The separation between both the boat = 12/5 + 18/5 = 30/5 = 6 km
11. ∴ The separation between the boats is 6 km
1. Note: The boat will move in a different direction because the boat is moving in the downstream and upstream so, to find the separation distance we’ll add both the distances.
14. A man in a train notices that he can count 31 telephone poles in a minute. If the poles are at equal intervals of 50m what is the speed of the train?
1. 90km/hr
1. According to the question, there is a total of 31 pole. Hence,
2. Number of gaps between 31 telephone posts = 30
3. Since poles are at equal intervals of 50 m. Therefore,
4. Distance between the 31 telephone = 30 × 50 m = 1500 m
5. Time taken to cross these posts = 1 minute = 60 seconds
6. Now, speed of the train = {1500}/{60}*18/5 =90 Km/hr)
1. Hence, the speed of the train is 90 km/hour.
15. A train is travelling at a speed of 96kms /hr. It takes 3seconds to enter a tunnel and 30 seconds more to pass through it completely. Correct statements are
1. Tunnel is lengthier than train
2. Length of train is 80 mtrs
3. Length of tunnel is 800 mtrs
1. Speed of train = 96 km/h
2. time taken to enter a tunnel = 3 sec
3. Total time taken to cross the tunnel = 33 seconds
4. Formulae required:
5. Distance = speed × time
6. a km/hr = a × 5/18 m/sec
7. Calculations:
8. convert 96km/hr in to meter/sec = 96 × 5/18 ⇒ 80/3 m/sec
9. Length of train(dis) = 80/3 × 3 ⇒ 80 meters
1. Distance covered in 33 seconds  = 80/3 × 33 ⇒ 880 meters
10. Length of tunnel = Distance covered in 33 seconds – length of train
1. ⇒ 880 – 80 = 800 meters
2. ∴ Statement B, C and E are correct.

Number series:

1. The sum of four consecutive odd integers is 2160. The greatest of them is
1. Let the first integer be ‘n’
2. The next three odd integers be ‘n + 2’, ‘n + 4’ , ‘n + 6’
3. The sum number = n + n + 2 + n + 4 + n + 6
4. ⇒ 4n + 12, 4n + 12 = 2160
1. ⇒ 4(n + 3) = 2160
2. ⇒ (n + 3) = 2160/4
3. ⇒ n = 540 – 3 = 537
4. The greatest number n + 6 = 537 + 6
5. ⇒ 543
6. ∴ The greatest of them is 543
7. Logic:
2. Identify the next term in the series
3. 0,8,24,48.__
1. 80
1. 0+8=8+16=24+24=48+32=80
1. Identify the next term in the series 2,10,26,50,?
1. 82
1. 1^2+1=2
2. 3^2+1=10
3. 5^2+1=26
2. Consider the following number series 2,10,30,68,X
1. what is the term X?
2. 130
1. The logic is:
2. 13 + 1 = 2
3. 23 + 2 = 10
4. 33 + 3 = 30
5. 43 + 4 = 68
6. 53 + 5 = 130 = X
1. Reference:
3. Find odd one out from the following
1. 5,7,12,17,19,23
1. All options except ‘12’ are prime numbers.Hence, ‘12’ is the odd one out.
2. Reference
4. Find the wrong number in the series:
1. 7,8, 18, 57, 228, 1165, 6996
1. 6996-1165=5831/7 = 833 rest of the numbers not divisible so ans is 228
2. 7*1+1 = 8, 8*2+2 =18, 18*3+3=57, 57*4+4=232, 232*5+5=1165,1165*6+6=6996
5. Next term in the following series 4,12, 28,52
1. gap between numbers 52-28 = 24
2. 28-12=16
3. 12-4=8 so the next number is 52+32 = 84
1. 4, 7, 12, 19, 28, ?
1. The pattern is + 3, + 5, + 7, + 9, …..So, missing term = 28 + 11 = 39
2. logic
6. Find the odd one out in the following
1. 1,5,14,30,50,55,91
1. 50
1. 1 + 22 = 5
2. 5 + 32 = 14
3. 14 + 42 = 30
4. 30 + 52 = 55 ≠ 50
5. 55 + 62 = 91
6. Hence, 50 is the odd one out in the following sequence.
1. Logic:
7. The next number in the given series
1. 196,384, 736,-
1. 1344
1. 196=197
2. 384=385
3. 736=737
8. A rational number has its denominator greater than its numerator by 6. If the numerator is increased by 4 and the denominator is decreased by 8, the number becomes 5/3.What is the original rational number?
1. Let the numerator of the rational number be x then its denominator be x + 6
2. Rational number = Numerator/Denominator
3. The original rational number = x/(x + 6)
4. Now, A/Q
5. (x + 4)/(x + 6 – 8) = 5/3
6. ⇒ (x + 4)/(x – 2) = 5/3
7. ⇒ 3(x + 4) = 5(x – 2)
8. ⇒ 3x + 12 = 5x – 10
9. ⇒ 2x = 22
10. ⇒ x = 11
11. Numerator = x and Denominator = 11 + 6 = 17
12. ∴ The original rational number is 11/17
1. Reference:
9. What is the remainder o f 4^5 is divided by 15?
1. 4
1. The remainder can be multiplied in the same way as the actual number when divided by any other number
2. Ex- 52/4 = (5 × 5)/4 = (1 × 1)/4 = 1/4, Remainder = 1
3. Calculation:
4. 45 is divided by 15
5. ⇒ 45/15
6. ⇒ (42 × 42 × 4)/15
7. ⇒ (16 × 16 × 4)/15
8. ⇒ (1 × 1× 4)/15      —-(When 16 and 4 is divided by 15 then it gives 1 and 4 as remainder respectively)
9. ⇒ 4/15
10. ∴ The remainder is 4
1. logic
10. What is the missing number?
1. 4,-8,16,?,64,-128
1. -32
1. Each number is the proceeding number multiplied by -2.So, the required number is -128.
11. The missing two number in the following series
1. 1,4,3,16,5,36,?,?, 9,100
1. 7,64
1. Square the even numbers each time but leave the odd numbers alone.
12. The difference of the square of two consecutive even numbers is 84. What is the sum of these consecutive even numbers?
1. 42
1. Let the two consecutive even number be x and x + 2
2. Now, A/Q
3. (x + 2)2 – x2 = 84
4. ⇒ x2 + 4 + 4x – x2 = 84
5. ⇒ 4x = 80
6. ⇒ x = 20
7. The two consecutive number are 20 and 22
8. Sum = 20 + 22 = 42
9. ∴ The sum of two consecutive even number is 42
1. Logic:
13. Consider the following number series
1. 0,6,24,60,120,X What is X
1. 0+6=6+18=24+36=60+60=120+90=210+90=316
2. 18-12=6, 36-18=18, 60-24=36
3. Reference:
14. Complete the following series
1. 9,11,15,23,39,_?
2. 71
1. 9(+2), 11(+4), 15(+6), 23(+8), 39(+16),
1. +2*2,+4*2,+
2. Logic:
15. Complete the series
1. 2,5,9,19,37,?
2. 75
1. The logic is:
2. 2 × 2 + 1 = 5
3. 5 × 2 – 1 = 9
4. 9 × 2 + 1 = 19
5. 19 × 2 – 1 = 37
6. 37 × 2 + 1 = 75
7. Hence, 75 is the correct answer.
1. Logic:
16. next term of the series
1. 4,12,28,52,?
1. 84
1. 4+8=12+16=28+24=52+32=84
1. Reference:
17. Consider the following number series 2,10, 30,68,X
1. what is the term X?
1. The logic is:
2. 13 + 1 = 2
3. 23 + 2 = 10
4. 33 + 3 = 30
5. 43 + 4 = 68
6. 53 + 5 = 130 = X
7. Hence, ‘130’ is the correct answer.
1. Logic
18. what part of 1/12 is 3/8?
1. 9/2
1. Let the number be ‘x’
2. 1/12 × x = 3/8
3. ⇒ x = 3/8 ÷ 1/12
4. ⇒ x = 3/8 × 12
5. ⇒ x = 9/2
6. ∴ 3/8 is 9/2 part of 1/12
1. Logic
19. Consider the following fractions
1. 6/11, 4/9, 3/8, 2/5, 5/13
2. For the above fraction s, which one of the following is in the correct descending order?
3. A>B>D>E>C
1. 6/11, 4/9, 3/8, 2/5, 5/13
2. Calculations:
3. LCM of the given fractions = 28080/51480, 22880/51480, 19305/51480, 20592/51480, 19800/51480
4. Now, we can arrange in descending order
5. 28080/51480  > 22880/51480 >  20592/51480 >  19800/51480 > 19305/51480
1. ∴ A > B > D > E > C
4. Alternate method
1. 6/11 = 0.545
2. 4/9 = 0.44
3. 3/8 = 0.35
4. 2/5 = 0.4
5. 5/13 = 0.38
6. When we write in descending order, (highest value to lowest value)
7. 6/11 > 4/9 > 2/5 > 5/13 > 3/8
8. ∴ The correct descending order is 6/11 > 4/9 > 2/5 > 5/13 > 3/8
20. If 20% of (a+b) =50% of (ab) and 40%(a-b)=70% of ab , then what fraction of a is b?
1. 3/17
1. Given
2. 20% of (a + b) = 50% of ab       —-(i)
3. 40% of (a – b) = 70% of ab      —-(ii)
4. Calculations:
5. What fraction of a is b means b/a
6. 20% = 20/100 ⇒ 1/5
7. 40% = 40/100 ⇒ 2/5
8. 50% = 50/100 ⇒ 1/2
9. 70% = 70/100 ⇒ 7/10
10. Substitute the values in equation (i) and (ii)
11. 20% of (a + b) = 50% of ab
12. ⇒ (1/5) × (a + b) = (1/2) × ab
13. ⇒ 2 × (a + b) = 5 × ab
14. ⇒ 2a + 2b = 5ab       —-(iii)
15. 40% of (a – b) = 70% of ab
16. (2/5) × (a – b) = (7/10) × ab
17. 4 × (a – b) =  7ab
18. ⇒ 4a – 4b = 7ab       —-(iv)
19. Multiply equation iii with 2
20. ⇒ 2 × ( 2a + 2b) = 5ab × 2
21. ⇒ 4a + 4b = 10ab       —-(v)
22. Add equation iv and v, we get
23. 8a = 17ab
24. ⇒ b = 8/17
25. Substitute  value of b in equation iv
26. 4a – 4 × 8/17 = 7 × a × 8/17
27. ⇒ (68a – 32)/17 = 56a/17
28. ⇒ 68a – 32 = 56a
29. ⇒ 68a – 56a = 32
30. ⇒ 12a = 32
31. ⇒ a = 8/3
32. Fraction of a is b = (8/17)/(8/3)
33. ⇒ 3/17
34. ∴ Fraction of a is b = 3/17
1. Logic:
21. A two digit number is 4 times the sum of its 2 digits. the number that is formed by reversing its digits is 27 more than the original number. What is the number?
1. 36
1. Let the tens and units digit of number are ‘x’, ‘y’ respectively
2. First condition:
3. 10x + y = 4(x + y)
4. ⇒ 10x + y = 4x + 4y
5. ⇒ 10x – 4x = 4y – y
6. ⇒ 6x = 3y
7. ⇒ 2x = y        —-(1)
8. Second condition:
9. Reversed Number = (10y + x)
10. (10x + y) + 27 = 10y + x
11. ⇒ 10x – x + y – 10y = -27
12. ⇒ 9x – 9y = -27
13. ⇒  9(x – y) = -27
14. ⇒ (x – y) = -27/3
15. ⇒ (x – y) = -3      —-(2)
16. Substitute the value of eq.1 in equation.2, we can get the value of x
17. x – 2x = -3
18. ⇒ x = 3
19. Substitute the value of x in in eq.1, we can get the value of y
20. 2 × 3 = y
21. ⇒ y = 6
22. According to first condition the original number = 10 × 3 + 6
23. ⇒ 36
24. Adding 27 to the first condition = 36 + 27
25. ⇒ 63
26. According to the second condition the reversed number = 10 × 6 + 3
27. ⇒ 6
28. ∴ The original number is 36
1. logic:
22. The sum of squares of 3 numbers is 83, while sum of their products taken two at a time is 71. Their sum will be
1. 15
1. Let three numbers be ‘a’, ‘b’, and ‘c’
2. The sum of the squares of a, b, and c = a+ b+ c2
3. ⇒ a2 + b2 + c2
4. The sum of their products taken two at a time = ab + bc + ca
5. ⇒ ab + bc + ca = 71
6. We know that,
7. (a + b + c)= a2 + b2 + c+ 2(ab + bc + ca)
8. ⇒ (a + b + c)2 = 83 + 2 × 71
9. ⇒ (a + b + c)2 = 83 + 142
10. ⇒ (a + b + c)2 = 225
11. ⇒ (a + b + c)2 = 152
12. a + b + c = 15     (∵ a= bm  then, a = b)
13. ∴ The sum of three number is 15
1. Logic:
23. root(4)^n=256 then root n is equal to
1. 2root 2
1. Given:
2. \(√ {{{(4)}^n}}\) = 256
3. Concept used:
4. ab = ac then b = c
5. Formula used:
6. (ab)c = a(b × c)
7. Calculation:
8. Our equation:
9. \(√ {{{(4)}^n}}\) = 256
10. ⇒ 4n = (256)2      —-(By squaring both the side)
11. ⇒ 4n = (44)2      —-(∵ 256 = 4 × 4 × 4 × 4 = 44)
12. ⇒ 4n = 48
13. ⇒ n = 8
14. Now, √n = √8 = 2√2
15. ∴ The answer of the given expression is 2√2
1. Logic:
24. 20 times a positive integer is less than its square is by 96. what is the integer
1. 24
1. Let the positive integer be ‘x’
2. Then, 20 times integer = 20x
3. Square of integer = x2
4. Twenty times a positive integer is less than its square by 96
5. x2  = 20x + 96
6. ⇒ x2 – 20x – 96 = 0
7. ⇒ x– 24x + 4x – 96 = 0
8. ⇒ x ( x -24) + 4( x – 24) = 0
9. ⇒ (x + 4) × (x -24) = 0
10. ⇒ x = -4,  x = 24
11. ∴ The positive integer is 24
1. Logic
25. Consider the following 3 expressions
1. 35/0.07
2. 2.5/0.0005
3. 136.09/43.9
1. The correct descending order for the simplified values of A.B and C?
2. B>A>C
26. The following numbers are non-prime?
1. 21 and 15
1. other are11,13 and 19
27. Find the odd umber in the following series
1. 544, 328, 164, 84, 44, 24,14
2. 328
1. List-1 and list-2
1. for two numbers 84 and 86 and result
2. HCF – 12
3. LCM – 672
4. Product – 8064
5. SUm -180
2. S-1 A number is divisible by 25 if the number formed by its last two digit is either 00 or divisible by 25
1. s-2: a number is divisible 16 if the number formed by its last 4 digits is divisible by 16
2. Both S-1 and S-2 are true
1. Divisibility rule  for 25:
2. To determine if a number is divisible by 25, the number formed by the last two digits of the number must be 00, 25, 50, or 75.
3. Divisibility rule for 16:
4.  A number is divisible by 16 if the number formed by its last 4 digits is divisible by 16.
5. From the above,
1. ∴ Statement II is appropriate, hence statement I is not appropriate
3. List-1 and List-2
1. triangle area – 1/2 bh
2. square area – side ^2
3. rhombus area -1/2*(product of diagonals)
4. parallelogram area – base*ht
4. S-1: sometime ratio has units
1. S-2: When two ratios are equal a proportion is formed
2. S-1 is incorrect but s-2 is true
5. S-1: Co-prime numbers are the numbers  which have only 1 as their commn factor
1. S-2: factorisation is the decomposition of a number into the other numbers which when added together will give the original number
1. S-1 is correct but S-2 is false
1. 3 and 4 are Co prime because their HCF is 1 i.e they have only 1 as their common factor.
6. S-1 sum of the angles of a triangle is 180*
1. S-2: The sum of any 2 sides of a triangle is greater than the third side
1. Both s-1 and S-2 are true
7. S-1 If you know length of 2 sides of a right angled triangle then the length of the third side can not be found
1. S-2 the internal angles of a triangle add upto 180*
2. S-1 is incorrect but S-2 is true
8. The given equation becomes correct after interchanging 2 signs, One fo the four alternatives under it specifies the interchange of signs in the equation which when made make the equation correct
1. Find the correct alternative – 9+5+4*3-6=12
1. + and –
9. S-1: A percentile divides the distribution into 100 equal parts
1. S-2: range is the distribution between zero and highest value
2. S-1 is correct s-2 is false
10. S-1: If the ratio of the speeds of A -and B is a:b then the ratio of the times taken by them to cover the same distance is b:a
1. S-2: Suppose two men are moving in the same direction at u m/s and v m/s respectively where u>v then their relative speed is (u_v) m/s
1. S-1 is correct s-2 is false

Clocks and Calendars:

1. The time is 2:15P.M. What is the angle between hour and minute hand
1. 22 1/2*
1. Logic:
2. A clock takes 5seconds to strike 5 times at 5Oclock How long will it take to strike 9 times at 9O clock?
1. 10 seconds
1. But we know that, there should be an interval of time between two successive strikes.
2. Then, between 5 strikes, there must have 4 intervals of time.
3. And it is given that it takes 5 seconds for that 4 intervals.
4. Hence, 5 strikes → 4 intervals → 5 seconds
1. So, each interval is 5/4 seconds long.
2. Now we have to find out the time taken to strike 9 times at 9 O’clock.
3. As per the given concept, between 9 strikes, there must have 8 intervals of time.
4. That is : 9 strikes → 8 intervals.
5. Then the time taken for 9 strikes = Number of intervals between them × Time taken for one interval = 8 × 5/ 4 = 10 seconds
6. Hence, 10 seconds is the correct answer.
1. Note:
1. As 5 strikes take 5 seconds, do not count like each strike takes 1 second. Then the answer will be marked as 9 seconds for 9 strikes.
2. reference:
3. At what angle the hands of clocks are inclined at 15minutes past 5?
1. 67 1/2*
1. Angle traced by hour hand in 21/4 hrs = (360/12*21/4)*=157 1/2*
2. Angle traced by minute hand in 15min = (360/60 *15) = 90*
3. Reqd angle = 157 1/2*- 90* = 671/2*
1. Method-2:
2. By formula,
1. theta = 11*minutes/2- 30*hour
2. 11/2*15 – 30*5 = 67.5*
3. Reference:
4. If jan1, 2007 was a monday then jan1 2008 is a
1. Tuesday
1. January 1, 2007 → Monday
2. So, January 1, 2008 will be a day beyond Monday i.e, Tuesday as 2007 is an ordinary year.
3. Hence, Tuesday is the correct answer.
1. Logic:
5. Today is what day if day before yesterday it was 13th of the present month and 3rd day of the last month, with 31 days was a tuesday?
1. Wednesday
1. Day before yesterday = 13th day of the month
2. Today is 15th day of the month.
3. 3rd day of the last month with 31 days = Tuesday
4. 1st day of last month = Sunday
5. 31st day of the last month = Tuesday
6. 1st day of this month is Wednesday
7. 15th day of this month is Wednesday.
8. Hence, “Wednesday” is the correct answer.
1. logic:
6. S-1: It is possible that there are 53 tuesday in a non-leap year
1. S-2: In a non-leap year there are 54 of each day plus one extra day
2. S-1 is correct s-2 is false
1. Logic:
2. Statement I: It is possible that there are 53 Tuesdays in a non-leap year.
1. In a non-leap year, there are 365 days i.e, 52 weeks and 1 odd day.
2. So, there are 52 Tuesdays and if the 1 odd day is Tuesday so it is possible that there are 53 Tuesdays in a non-leap year.
3. Statement II: In a non-leap year, there are 54 of each day plus one extra day.
1. In a non-leap year, there can be maximum of 53 days of a particular day.
2. Hence, Statement I is correct but Statement II is false.
7. At what time between 2 and 3 o clock will the hour hand and minute hand of a clock together?
1. 10 10/11min past 2
1. The hour hand and minute hand of a clock together means the angle between angle and minute hand will be 0°.
2. Lower value of time (H) = 2 and Angle (θ) = 0° (Ashands of a clock will be together)
3. By formula,
4. Time = H : (H × 5 ± θ ÷ 60) × {12}/{11}
1. 2 : (2 × 5 ± 0 ÷ 60) × {12}/{11})
2. 2 : (10 ± 0) × {12}/{11}
3. = 2 : 10 ×{12}/{11
4. = 2 : {120}/{11}
5. = 2 : 10 {10}/{11}
1. Hence, 10 {10}/{11} min. past 2 is the correct answer.
5. Logic:
8. If the day before yesterday was friday what will be the third day after the day after tomorrow?
1. Friday
1. The day before yesterday was Friday.
2. Therefore, today is Sunday.
3. The day after tomorrow will be Tuesday.
4. Tuesday + 3 = Friday
5. Hence, Friday is the correct answer.
1. Logic:

Letter series:

1. Consider the following letter series
1. what is the next of acj, egj, ikn, oqt,
1. a+4=e+4 =i+4=q
1. Next term in the following letter series
1. aceg, cegi, egik, ,____
1. a+2=c+2=e+2=g+2=i
2. Ikmo
1. Reference:
2. Next term in the letter series
1. abcf, bcdg, cdeh, defi,____
1. efgj
1. a+1=b+1=c+1=d+1=e
1. In the following letter series the next term is
1. bdf, egi, hjl,
1. kmo
1. Identify the next term in the letter series
1. ace, fhj, kmo, ___PRT
1. A+5=F+5=K+5=P+5=U
2. Consider the following letter series
1. what is next of acf, egj, ikn, oqt,___
1. uwz
1. a+4=e+4=i+4=m+4=q
3. Missing letter in the series
1. BCA, YZX, EFD, ___, HIG
1. VWU
1. reasoning:
4. The missing letter in the series
1. BCE, CDG, DEI, ___, FGM
1. EFK
1. A series is given with one term missing. Select the correct alternative from the given ones that will complete the series.BCF, CDG, DEH, ?
2. he pattern followed here is,
3. → B + 1 = C; C + 1 = D; D + 1 = E
4. → C + 1 = D; D + 1 = E; E + 1 = F
5. → F + 1 = G; G + 1 = H; H + 1 = I
6. Hence, “EFI” is a term that will complete the series.
5. In a certain code, FISH is written as EHRG, The code for JUNGLE is
1. ITMFKD
1. Logic:
6. In a certain code, ‘Universal’ is written as ‘LASREVINU’. In the same coding language ‘COMMERCIALS’ will be written as
1. SLAICREMMOC
1. Reference:
7. The next term in the following series is ABB, BCF, CDL?
1. DET
1. The logic is as shown below,
2. A + 1 → B, B + 1 → C, C + 1 → D, D + 1 → E
3. B + 1 → C, C + 1 → D, D + 1 → E, E + 1 → F
4. B + 4 → F, F + 6 → L, L + 8 → T, T + 10 → D
5. Hence, the required term is DET.
6. Logic:
8. In a certain code language ‘CHEMISTRY’ is written as ‘DGFLJRUQZ’ then in the same code ‘GEOMETRY’ will be written as
1. HDPLFSSX
9. If UGCNET = 35, then CSIRNET =
1. 44
1. UGC – NET = 35
2. U = 21
3. G = 7
4. C = 3
5. N = 14
6. E = 5
7. T = 20
8. (21 + 7 + 3 + 14  + 5 + 20)/2  = 35
9. CSIR-NET
10. C  = 3
11. S   = 19
12. I   = 9
13. R  = 18
14. N = 14
15. E = 5
16. T = 20
17. (3 +19 +9 + 18 + 14  + 5 + 20)/2  = 44
10. Next letter in the logical pattern O,T,T,F,F,S,S,E,N_ is
1. T
1. O = One
2. T = Two
3. T = Three
4. F = Four
5. F = Five
6. S = Six
7. S = Seven
8. E = Eight
9. N = Nine
10. So, the next letter is
11. T = Ten
11. A’s father has  five sons. Name of four sons are M,O,Q, and S . What is the name of the fifth
1. A
1. A’s father has five sons.
2. Four are: M, O, Q, and S.
3. Fifth son is A himself.
4. Hence, “A” is the correct answer.
1. Logic
12. S-1: If A,B and C are standing in a circle looking at the centre and B is just left of A then C is just right of B
1. If X,Y and Z are standing in a circle looking at the centre and X is just right of Y, then Z is just right of X
2. S-1 is false s-2 is true
1. Logic:
1. A,B,C,D,E,F are standing in a circle and looking at the centre. Only D is between A and B. Only F is between A and C. E is just left of B. Who is the only one standing between E and F?
1. C
1. Logic:
2. Kavitha collected 8 spiders (with 8leags each) and beetles (with 6 legs each). When she counted the legs, she found that there are altogether 54. Which of the following are true?
1. She collected 5S, 3S, 5B, 3B
1. Logic:
1. Kavitha collected 8 spiders (with 8 legs each) and beetles (with 6 legs each). When she counted the legs, she found that there were altogether 54.
2. Let us assume that she collected ‘a’ number of spiders and ‘b’ number of beetles.
3. According to question:
4. 8a + 6b = 54
5. => 2 (4a + 3b) = 54
6. => 4a + 3b = 54 ÷ 2
7. => 4a + 3b = 27
8. 1) a and c only
9. a) She collected 5 spiders
10. c) She collected 5 beetles
11. 4a + 3b = 27
12. L.H.S = 4 × 5 + 3 × 5
13. = 20 + 15
14. = 35 ≠ 27
15. 2) a and d only
16. a) She collected 5 spiders
17. d) She collected 3 beetles
18. 4a + 3b = 27
19. L.H.S = 4 × 5 + 3 × 3
20. = 20 + 9
21. = 29 ≠ 27
22. 3) b and d only
23. b) She collected 3 spiders
24. d) She collected 3 beetles
25. 4a + 3b = 27
26. L.H.S = 4 × 3 + 3 × 3
27. = 12 + 9
28. = 21 ≠ 27
29. 4) b and c only
30. b) She collected 3 spiders
31. c) She collected 5 beetles
32. 4a + 3b = 27
33. L.H.S = 4 × 3 + 3 × 5
34. = 12 + 15
35. => 27 = 27
36. Hence, ‘b and c only’ is the correct answer.
1. She collected 3 S and 5B
2. Logic:
3. In a class of 65 students 27 speak hindi, 21 speak english, and 17 speak bangla, 7 speak only english and hindi, 6 speak only hindi and bangla and 5 speak only english and bangla . If 12 speak only hindi then how many speak all three language
1. 4
1. Students who speak all the three languages = Students who can speak Hindi – (Students who can speak only Hindi + Students who speak only English and Hindi + Students who speak only Hindi and Bangla)
2. = 27 – (12 + 7 + 6)
3. = 27 – 25
4. = 2
5. Hence, ‘2’ is the correct answer.
1. The Venn diagram for the following question is given below:
1. Logic:
4. R,S,V join a running race. The distance is 1500mts. Rajiv beats Sanjiv by 30mts and vijay by 100mts, By how much could sanjiv beat vijay over the full distance if they both ran as before?
1. 71.4
1. Distance = 1500 meters
2. Rajiv beats Sanjay by 30 meters
3. Rajiv beats Vijay by 100 meters
4. Calculations:
5. Rajiv runs distance = 1500 meters
6. Sanjiv runs distance = 1500 – 30
7. ⇒ 1470 metes
8. Vijay runs distance = 1500 – 100
9. ⇒ 1400 meters
10. At the same rate, Sanjiv runs 1500 meters while Vijay runs = 1500 × 1400/1470
11. ⇒ 1428.57
12. Sanjay beats Vijay by =1500 – 1428.57
13. ⇒ 71.42 meters
14.  ∴ Sanjiv ought to beat Vijay by 71.4 meters.
1. Logic:
5. ‘A’ is the only sister of a man ‘B’ and ‘C’ is the maternal grandson of B’s father. How is A’ related to ‘C’?
1. Mother
1. Reference: