Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.
What will come in the place of the question mark ‘?’ in the following question?
(√121 × √169 × √324) ÷ 78 + 24 = ?
63
27
57
67
43
Correct answer
57
Why this answer is correct
Concept used:
Calculation:
(√121 × √169 × √324) ÷ 78 + 24 = ?
(11 × 13 × 18) ÷ 78 + 24 = ?
2574 ÷ 78 + 24 = ?
33 + 24 = ?
57 = ?
∴ 57 will come in place of the question mark.
What will come in place of the question mark (?) in the following question?
(1/4) + (22/7) + (32/15) + ? = 6
299/380
199/420
363/480
119/299
549/1099
Correct answer
199/420
Why this answer is correct
Concept used:
Follow BODMAS rule to solve this question, as per the order given below:
Calculation:
(1/4) + (22/7) + (32/15) + ? = 6
⇒ 6 – [(1/4) + (22/7) + (32/15)] = ?
⇒ 6 – [(105 + 1320 + 896)/420] = ?
⇒ 6 – (2321/420) = ?
⇒ (2520 – 2321)/420 = ?
⇒ 199/420 = ?
∴ The value of ? is 199/420
What will come in place of the question mark (?) in the following question?
18% of 27 + 2% of 207 = (?)2
4
6
5
3
2
Correct answer
3
Why this answer is correct
Concept Used:
Follow BODMAS rule to solve this question, as per the order given below,
Calculation:
18% of 27 + 2% of 207 = (?)2
⇒ 4.86 + 4.14 = (?)2
⇒ 9 = (?)2
⇒ 3 = ?
∴ 3 will come in place of question mark
What will come in place of question mark(?) in the following questions?
? = (7/5) of 55 + (3/8) of 1376 − 52
568
506
554
550
571
Correct answer
568
Why this answer is correct
Concept Used:
Follow BODMAS rule to solve this question, as per the order given below,
Calculation:
? = (7/5) of 55 + (3/8) of 1376 − 52
⇒ ? = (7) × 11 + (3) × 172 − 52
⇒ ? = 77 + 516 − 25
⇒ ? = 52 + 516
⇒ ? = 568
Hence, option 1 is correct.
What will come in place of the question mark (?) in the following question?
(65 ÷ 26) ÷ 5 of 10 × (47 + 78) = ?
2.25
50
6.25
45
2.75
Correct answer
6.25
Why this answer is correct
Concept Used:
Follow BODMAS rule to solve this question, as per the order given below,
Calculation:
(65 ÷ 26) ÷ 5 of 10 × (47 + 78) = ?
2.5 ÷ 5 of 10 × 125 = ?
2.5 ÷ 50 × 125 = ?
0.05 × 125 = ?
6.25 = ?
∴ 6.25 will come in place of the question mark
A train running at a speed of 20 m/s takes 22 seconds to pass a bridge and it takes 10 seconds to pass a man who is running at a speed of 5 m/s in the same direction. Find the length of the bridge.
150 m
200 m
290 m
300 m
350 m
Correct answer
290 m
Why this answer is correct
Shortcut Trick
Relative Speed with man = 20 − 5 = 15 m/s.
Length of Train (LT) = Relative Speed × Time = 15 × 10 = 150 m.
Total distance to cross bridge = LT + LB = Train Speed × Time = 20 × 22 = 440 m.
Length of Bridge (LB) = 440 − 150 = 290 m.
∴ The correct answer is 290 m.
Alternate Method
Given:
Speed of train (VT) = 20 m/s
Speed of man (VM) = 5 m/s (Same direction)
Time to pass man (t1) = 10 s
Time to pass bridge (t2) = 22 s
Formula Used:
Distance = Speed × Time
Relative Speed (Same direction) = V1 − V2
Calculations:
⇒ Relative speed of train w.r.t man = 20 − 5 = 15 m/s
⇒ Length of train (LT) = Relative Speed × t1 = 15 × 10 = 150 m
⇒ Total distance to cross bridge = LT + Length of Bridge (LB)
⇒ LT + LB = VT × t2
⇒ 150 + LB = 20 × 22
⇒ 150 + LB = 440
⇒ LB = 440 − 150 = 290 m
∴ The correct answer is 290 m.
Additional Information
Relative Speed
When two objects move in the same direction, relative speed = S1 − S2. In opposite directions, relative speed = S1 + S2.
Crossing Stationary Objects
To cross a point object (pole, man), distance = LTrain. To cross an object with length (bridge, platform), distance = LTrain + LObject.
Unit Conversion
To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.
A container contains 100 litres of milk. If 10 % of milk was replaced by water and this process was repeated two times more, then find the quantity of the milk in the final mixture.
82.9 litres
72.9 litres
62.9 litres
72 litres
None of these
Correct answer
72.9 litres
Why this answer is correct
Shortcut Trick
Initial quantity = 100 litres
Replacement fraction = 10% = 1/10
Remaining fraction after 1 process = 1 − 1/10 = 9/10
Since the process is repeated two times more (total 3 times):
Final Milk = 100 × (9/10)3 = 100 × 729/1000 = 72.9 litres
∴ The correct answer is 72.9 litres.
Alternate Method
Given:
Initial quantity of milk (x) = 100 litres
Quantity replaced (y) = 10% of 100 = 10 litres
Number of times the process is performed (n) = 3
Formula Used:
Final quantity = x × (1 − y/x)n
Calculations:
⇒ Final Quantity = 100 × (1 − 10/100)3
⇒ Final Quantity = 100 × (1 − 1/10)3
⇒ Final Quantity = 100 × (9/10) × (9/10) × (9/10)
⇒ Final Quantity = 100 × 0.9 × 0.9 × 0.9
⇒ Final Quantity = 100 × 0.729
⇒ Final Quantity = 72.9 litres
∴ The correct answer is 72.9 litres.
Additional Information
Successive Replacement Formula
The formula Final Quantity = Initial × (1 − r/100)n is used when a fixed percentage r is replaced n times.
Mixture Replacement with Different Quantities
If different amounts a, b, c are replaced each time, the final quantity is x × (1 − a/x) × (1 − b/x) × (1 − c/x).
Ratio of Milk to Water
After the process, the quantity of water = Total Capacity − Final Milk. In this case, 100 − 72.9 = 27.1 litres.
The average of 30 numbers is 9, the average of first 15 numbers is 8 and that of next 13 numbers is also 8. What will be the average of last two numbers?
46
20
23
12
9
Correct answer
23
Why this answer is correct
Shortcut Trick
Total deviation from average (9) for 28 numbers = 28 × (8 − 9) = −28.
To maintain the average, the last two numbers must compensate for this deficit.
Sum of last two numbers = (2 × 9) + 28 = 18 + 28 = 46.
Average of last two numbers = 46 ÷ 2 = 23.
∴ The correct answer is 23.
Alternate Method
Given: Total numbers = 30, Overall average = 9, First 15 average = 8, Next 13 average = 8.
⇒ Total sum of 30 numbers = 30 × 9 = 270
⇒ Sum of first 15 numbers = 15 × 8 = 120
⇒ Sum of next 13 numbers = 13 × 8 = 104
⇒ Sum of last 2 numbers = Total sum − (Sum of 15 + Sum of 13)
⇒ Sum of last 2 numbers = 270 − (120 + 104) = 270 − 224 = 46
⇒ Average of last 2 numbers = 46 ÷ 2 = 23
∴ The correct answer is 23.
Additional Information
Basic Average Formula
Average = (Sum of all observations) ÷ (Total number of observations).
Net Deviation Method
The sum of deviations of all observations from the arithmetic mean is always zero. This helps in finding missing values quickly.
Weighted Average
When groups have different sizes, Combined Average = (n1x1 + n2x2 + ...) ÷ (n1 + n2 + ...).
A can complete a work in 12 days, B can complete the same work in 7(1/2) days, and B along with C together can complete the work in 7(1/5) days. How many days will C alone take to complete the work?
100 days
120 days
150 days
180 days
200 days
Correct answer
180 days
Why this answer is correct
Shortcut Trick
Convert mixed fractions: B = 15/2 days, B+C = 36/5 days.
Total Work (LCM of 12, 15, 36) = 180 units.
Efficiency: B = 180 ÷ (15/2) = 24; B + C = 180 ÷ (36/5) = 25.
Efficiency of C = (B + C) − B = 25 − 24 = 1 unit/day.
Time taken by C = 180 ÷ 1 = 180 days.
∴ The correct answer is 180 days.
Alternate Method
Given:
Time taken by A = 12 days
Time taken by B = 7(1/2) days = 15/2 days
Time taken by B + C = 7(1/5) days = 36/5 days
Formula Used:
Total Work = Time × Efficiency
Efficiency = Total Work ÷ Time
| Person | Days | Total Work | Efficiency |
|---|---|---|---|
| A | 12 | 180 | 15 |
| B | 15/2 | 24 | |
| B + C | 36/5 | 25 |
Calculations:
⇒ Total work = LCM of numerators (12, 15, 36) = 180 units
⇒ Efficiency of B = 180 ÷ (15/2) = (180 × 2) ÷ 15 = 24
⇒ Efficiency of B + C = 180 ÷ (36/5) = (180 × 5) ÷ 36 = 25
⇒ Efficiency of C = (Efficiency of B + C) − (Efficiency of B)
⇒ Efficiency of C = 25 − 24 = 1
⇒ Time taken by C alone = Total Work ÷ Efficiency of C
⇒ Time taken by C alone = 180 ÷ 1 = 180 days
∴ The correct answer is 180 days.
Additional Information
LCM of Fractions
To find the LCM of fractions, calculate (LCM of Numerators) ÷ (HCF of Denominators).
Work and Wages
Wages are always distributed in the ratio of the efficiency of the workers or the work done by them.
Inverse Relation
Efficiency is inversely proportional to the time taken to complete a specific task. If Time ratio is x:y, Efficiency ratio is y:x.
Directions: Read the data carefully and answer the following questions.
The following table shows the number of candidates visiting three colleges (A, B, and C) as exam centers over three years (2020, 2021, and 2022):
| College | 2020 | 2021 | 2022 |
|---|---|---|---|
| A | 45 | 52 | 60 |
| B | 38 | 47 | 55 |
| C | 42 | 50 | 58 |
The average number of candidates who visited College C over the three years is what percentage of the total number of candidates who visited all three colleges in 2022?
28.90%
32.50%
34.80%
37.30%
41.20%
Correct answer
28.90%
Why this answer is correct
College C total = 42 (2020) + 50 (2021) + 58 (2022) = 42 + 50 + 58 = 150
Average = 150 / 3 = 50
Total candidates in 2022 = College A (60) + College B (55) + College C (58) = 60 + 55 + 58 = 173
Percentage = 50 / 173 × 100 = 28.90%
Thus, the correct answer is 28.90%
Directions: Read the data carefully and answer the following questions.
The following table shows the number of candidates visiting three colleges (A, B, and C) as exam centers over three years (2020, 2021, and 2022):
| College | 2020 | 2021 | 2022 |
|---|---|---|---|
| A | 45 | 52 | 60 |
| B | 38 | 47 | 55 |
| C | 42 | 50 | 58 |
What is the difference between the number of candidates who visited College C in 2021 and College B in 2020, and what is the ratio of this difference to the number of candidates who visited College A in 2021?
7, 7 : 52
8, 2 : 13
9, 9 : 52
10, 5 : 26
12, 3 : 13
Correct answer
12, 3 : 13
Why this answer is correct
College C in 2021 = 50
College B in 2020 = 38
Difference = 50 − 38 = 12
College A in 2021 = 52
Ratio = 12 / 52 or 12:52 = 3:13
Hence the correct answer is 12, 3:13
Directions: Read the data carefully and answer the following questions.
The following table shows the number of candidates visiting three colleges (A, B, and C) as exam centers over three years (2020, 2021, and 2022):
| College | 2020 | 2021 | 2022 |
|---|---|---|---|
| A | 45 | 52 | 60 |
| B | 38 | 47 | 55 |
| C | 42 | 50 | 58 |
Which college had the highest total number of candidates visiting over the three years?
College A
College B
College C
Both A and C
Both B and C
Correct answer
College A
Why this answer is correct
College with the highest total number of candidates visiting over the three years:
College A:
2020 = 45
2021 = 52
2022 = 60
Total = 45 + 52 + 60 = 157
College B:
2020 = 38
2021 = 47
2022 = 55
Total = 38 + 47 + 55 = 140
College C:
2020 = 42
2021 = 50
2022 = 58
Total = 42 + 50 + 58 = 150
College A (157 candidates) had the highest total number of candidates visiting over the three years.
Directions: Read the data carefully and answer the following questions.
The following table shows the number of candidates visiting three colleges (A, B, and C) as exam centers over three years (2020, 2021, and 2022):
| College | 2020 | 2021 | 2022 |
|---|---|---|---|
| A | 45 | 52 | 60 |
| B | 38 | 47 | 55 |
| C | 42 | 50 | 58 |
The number of candidates who visited College B in 2020 is what percentage less than the number of candidates who visited College A in 2022?
33.33%
36.67%
40.00%
43.33%
46.67%
Correct answer
36.67%
Why this answer is correct
College B in 2020 = 38
College A in 2022 = 60
Difference = 60 − 38 = 22
Percentage less = 22 / 60 × 100 = 36.67%
Thus, the correct answer is 36.67%
Directions: Read the data carefully and answer the following questions.
The following table shows the number of candidates visiting three colleges (A, B, and C) as exam centers over three years (2020, 2021, and 2022):
| College | 2020 | 2021 | 2022 |
|---|---|---|---|
| A | 45 | 52 | 60 |
| B | 38 | 47 | 55 |
| C | 42 | 50 | 58 |
What is the ratio of the total number of candidates who visited College A over the three years to the total number of candidates who visited College C over the three years?
150:157
157:150
14:15
157:140
15:14
Correct answer
157:150
Why this answer is correct
Total for College A = 45 + 52 + 60 = 157
Total for College C = 42 + 50 + 58 = 150
Ratio = 157:150
Thus, the correct answer is 157:150.
Find the wrong number in the following series:
2880, 480, 96, 24, 6, 4
480
96
24
6
4
Correct answer
6
Why this answer is correct
Solution:
Given Series: 2880, 480, 96, 24, 6, 2
Logic Used:
2880 ÷ 6 = 480
480 ÷ 5 = 96
96 ÷ 4 = 24
24 ÷ 3 = 8 (given 6 instead of 8)
8 ÷ 2 = 4
∴ The wrong number in the series is 6.
Find the wrong term in the given series.
69, 117, 161, 253, 299, 391
161
117
253
299
69
Correct answer
117
Why this answer is correct
Solution :
The series follows the following pattern :
⇒ 23 × 3 = 69
⇒ 23 × 5 = 115
⇒ 23 × 7 = 161
⇒ 23 × 11 = 253
⇒ 23 × 13 = 299
⇒ 23 × 17 = 391
⇒ The wrong term in the series will be 117.
Hence 117 is the correct answer.
In the following number series, a wrong number is given. Find out the wrong number.
6, 14, 22, 36, 50, 66, 84
36
66
22
6
50
Correct answer
22
Why this answer is correct
Calculation:
The series follows the following pattern
6 + 8 = 14
14 + 10 = 24 ≠ 22
24 + 12 = 36
36 + 14 = 50
50 + 16 = 66
66 + 18 = 84
∴ The wrong value is 22
An amount is invested at 10% per annum compound interest for 2 years, and the interest earned is ₹945. What is the principal amount?
₹4,000
₹4,500
₹4,800
₹5,000
₹5,200
Correct answer
₹4,500
Why this answer is correct
Shortcut Trick
Effective Successive Interest Rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%.
Given Compound Interest (21%) = ₹945.
1% = 945 ÷ 21 = ₹45.
Principal (100%) = 45 × 100 = ₹4,500.
∴ The correct answer is ₹4,500.
Alternate Method
Given: Rate (R) = 10%, Time (T) = 2 years, Compound Interest (CI) = ₹945.
Formula Used: CI = P[(1 + R/100)T − 1]
⇒ 945 = P[(1 + 10/100)2 − 1]
⇒ 945 = P[(11/10)2 − 1]
⇒ 945 = P[121/100 − 1]
⇒ 945 = P[21/100]
⇒ P = (945 × 100) ÷ 21
⇒ P = 45 × 100 = 4,500
∴ The correct answer is ₹4,500.
Additional Information
Effective Rate for Compound Interest
For 2 years, the effective rate is x + y + (xy/100)%. For 10%, it is 10 + 10 + 1 = 21%.
Difference between CI and SI
The difference for 2 years is P(R/100)2. For 3 years, it is P(R/100)2 × (3 + R/100).
Amount Formula
Total Amount A = P(1 + R/100)n, where n is the number of compounding periods.
The speed of a boat in still water is 24 km/hr, and the speed of the stream is 6 km/hr. If the speed of the stream decreases by 50% and the speed of the boat decreases by 25%, what will be the time taken by the boat to travel 63 km downstream?
2 hours
3 hours
4 hours
5 hours
6 hours
Correct answer
3 hours
Why this answer is correct
Shortcut Trick
New Boat Speed = 24 − 25% of 24 = 18 km/hr.
New Stream Speed = 6 − 50% of 6 = 3 km/hr.
New Downstream Speed = 18 + 3 = 21 km/hr.
Time taken = 63 ÷ 21 = 3 hr.
∴ The correct answer is 3 hr.
Alternate Method
Given: Initial boat speed (u) = 24 km/hr, Initial stream speed (v) = 6 km/hr, Distance (D) = 63 km.
Formula Used: Downstream Speed = New Boat Speed + New Stream Speed; Time = Distance ÷ Speed
⇒ New Boat Speed (unew) = 24 × (1 − 25/100) = 24 × 0.75 = 18 km/hr
⇒ New Stream Speed (vnew) = 6 × (1 − 50/100) = 6 × 0.50 = 3 km/hr
⇒ Downstream Speed = unew + vnew = 18 + 3 = 21 km/hr
⇒ Time Required = Distance ÷ Downstream Speed = 63 ÷ 21 = 3 hr
∴ The correct answer is 3 hr.
Additional Information
Downstream Speed
When the boat travels in the direction of the stream, its effective speed is (u + v) km/hr.
Upstream Speed
When the boat travels against the direction of the stream, its effective speed is (u − v) km/hr.
Speed in Still Water
Calculated as (Downstream Speed + Upstream Speed) ÷ 2.
Stream Speed
Calculated as (Downstream Speed − Upstream Speed) ÷ 2.
A can fill a tank in 30 hours and B is 120% as efficient as A. If they fill the tank on alternate hours starting with A, find time to fill the tank completely (in hours).
27.33
28.4
26.6
26.33
None of these
Correct answer
27.33
Why this answer is correct
Shortcut Trick
Efficiency ratio A : B = 100 : 120 = 5 : 6.
Total Work = Efficiency of A × Time of A = 5 × 30 = 150 units.
Alternate hours (A then B): 2-hour cycle work = 5 + 6 = 11 units.
Number of cycles = 150 ÷ 11 = 13 full cycles (26 hours) + 7 units remaining.
Next hour (A's turn): A fills 5 units in 1 hour. Remaining 2 units filled by B in 2/6 = 0.33 hours.
Total time = 26 + 1 + 0.33 = 27.33 hours.
∴ The correct answer is 27.33 hours.
Alternate Method
Given: Time taken by A = 30 hours. Efficiency of B = 120% of A.
Formula: Total Work = Efficiency × Time; Time = Work ÷ Efficiency.
| Person | Efficiency | Total Work | Time Taken |
|---|---|---|---|
| A | 5 | 150 | 30 hours |
| B | 6 | 25 hours |
⇒ Efficiency of A = 5 units/hr (assumed from 100%).
⇒ Efficiency of B = 1.20 × 5 = 6 units/hr.
⇒ Total capacity of tank = 30 × 5 = 150 units.
⇒ In 1st hour (A) = 5 units, 2nd hour (B) = 6 units. Total in 2 hours = 11 units.
⇒ In 13 cycles (26 hours), work done = 13 × 11 = 143 units.
⇒ Remaining work = 150 − 143 = 7 units.
⇒ 27th hour: A fills 5 units. Remaining = 7 − 5 = 2 units.
⇒ Time for B to fill remaining = 2 ÷ 6 = 1/3 hour = 0.33 hours.
⇒ Total time = 26 + 1 + 0.33 = 27.33 hours.
∴ The correct answer is 27.33 hours.
Additional Information
Efficiency and Time Relation
Efficiency is inversely proportional to the time taken to complete a task. If efficiency ratio is x:y, time ratio is y:x.
Alternate Work Pattern
In alternate work, always calculate the work done in one complete cycle (sum of efficiencies of all persons involved) before finding the remaining work.
Pipes and Cisterns
Filling pipes have positive efficiency while outlet/leakage pipes have negative efficiency. Total work is the tank capacity.
Solving previous year questions by topic helps identify recurring concepts, question patterns, and frequently asked sub-topics. Attempt each question, review the step-by-step explanation, and note down key formulas or facts for quick revision before the exam.