Mathematics — Number System (HCF)
Q971EasyBPSC Prelims
What is the HCF of 48 and 64?
A8
B16
C24
D32
Show Answer
✔ B — 16
48 = 2⁴ × 3 = 2 × 2 × 2 × 2 × 3
64 = 2⁶ = 2 × 2 × 2 × 2 × 2 × 2
Common prime factors: 2⁴ = 16
HCF = 16
*Why A is wrong:* 8 = 2³. Both 48 and 64 are divisible by 16, not just 8. The HCF is the highest common factor, not just any common factor.
*Why C is wrong:* 24 divides 48 but does not divide 64 (64 ÷ 24 = 2.67), so 24 cannot be a common factor.
*Why D is wrong:* 32 divides 64 but does not divide 48 (48 ÷ 32 = 1.5), so 32 is not a common factor.
64 = 2⁶ = 2 × 2 × 2 × 2 × 2 × 2
Common prime factors: 2⁴ = 16
HCF = 16
*Why A is wrong:* 8 = 2³. Both 48 and 64 are divisible by 16, not just 8. The HCF is the highest common factor, not just any common factor.
*Why C is wrong:* 24 divides 48 but does not divide 64 (64 ÷ 24 = 2.67), so 24 cannot be a common factor.
*Why D is wrong:* 32 divides 64 but does not divide 48 (48 ÷ 32 = 1.5), so 32 is not a common factor.
Mathematics — Percentage (Basic)
Q972EasyBPSC Prelims
What is 25% of 480?
A100
B110
C120
D140
Show Answer
✔ C — 120
25% = 25/100 = 1/4
25% of 480 = 480 ÷ 4 = 120
Alternatively: 10% of 480 = 48; 25% = 2.5 × 48 = 120 ✓
*Why A is wrong:* 100 would be approximately 20.8% of 480, not 25%.
*Why B is wrong:* 110 is approximately 22.9% of 480.
*Why D is wrong:* 140 is approximately 29.2% of 480.
25% of 480 = 480 ÷ 4 = 120
Alternatively: 10% of 480 = 48; 25% = 2.5 × 48 = 120 ✓
*Why A is wrong:* 100 would be approximately 20.8% of 480, not 25%.
*Why B is wrong:* 110 is approximately 22.9% of 480.
*Why D is wrong:* 140 is approximately 29.2% of 480.
Mathematics — Ratio & Proportion
Q973EasyBPSC Prelims
Two numbers are in the ratio 3:4, and their sum is 280. What is the larger number?
A120
B140
C150
D160
Show Answer
✔ D — 160
Ratio 3:4 → total parts = 3 + 4 = 7 parts
Each part = 280 ÷ 7 = 40
Larger number (4 parts) = 4 × 40 = 160
Smaller number (3 parts) = 3 × 40 = 120
*Why A is wrong:* 120 is the smaller number (3 parts), not the larger.
*Why B is wrong:* 140 would be correct if the ratio were equal (both halves), but ratio 3:4 gives unequal parts.
*Why C is wrong:* 150 does not correspond to either part; 150 would require one part = 37.5, which is inconsistent.
Each part = 280 ÷ 7 = 40
Larger number (4 parts) = 4 × 40 = 160
Smaller number (3 parts) = 3 × 40 = 120
*Why A is wrong:* 120 is the smaller number (3 parts), not the larger.
*Why B is wrong:* 140 would be correct if the ratio were equal (both halves), but ratio 3:4 gives unequal parts.
*Why C is wrong:* 150 does not correspond to either part; 150 would require one part = 37.5, which is inconsistent.
Mathematics — Simple Interest
Q974EasyBPSC Prelims
Find the Simple Interest on ₹5,000 at 8% per annum for 3 years.
A₹900
B₹1,000
C₹1,100
D₹1,200
Show Answer
✔ D — ₹1,200
SI = (P × R × T) / 100
SI = (5000 × 8 × 3) / 100
SI = 120,000 / 100 = ₹1,200
*Why A is wrong:* ₹900 would result from SI = (5000 × 6 × 3)/100, i.e., at 6% rate, not 8%.
*Why B is wrong:* ₹1,000 = SI at approximately 6.67% for 3 years, not 8%.
*Why C is wrong:* ₹1,100 has no direct correspondence to P=5000, R=8%, T=3.
SI = (5000 × 8 × 3) / 100
SI = 120,000 / 100 = ₹1,200
*Why A is wrong:* ₹900 would result from SI = (5000 × 6 × 3)/100, i.e., at 6% rate, not 8%.
*Why B is wrong:* ₹1,000 = SI at approximately 6.67% for 3 years, not 8%.
*Why C is wrong:* ₹1,100 has no direct correspondence to P=5000, R=8%, T=3.
Mathematics — Time & Work
Q975EasyBPSC Prelims
A can complete a work in 10 days and B in 15 days. How many days will they take working together?
A5 days
B6 days
C8 days
D12 days
Show Answer
✔ B — 6 days
A's work rate = 1/10 per day
B's work rate = 1/15 per day
Combined rate = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 per day
Time = 1 ÷ (1/6) = 6 days
*Why A is wrong:* 5 days would require combined rate of 1/5, but actual combined rate is 1/6.
*Why C is wrong:* 8 days would imply combined rate of 1/8, which is less than A's individual rate of 1/10 — impossible.
*Why D is wrong:* 12 days is less efficient than either A or B working alone — working together must be faster than either alone.
B's work rate = 1/15 per day
Combined rate = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 per day
Time = 1 ÷ (1/6) = 6 days
*Why A is wrong:* 5 days would require combined rate of 1/5, but actual combined rate is 1/6.
*Why C is wrong:* 8 days would imply combined rate of 1/8, which is less than A's individual rate of 1/10 — impossible.
*Why D is wrong:* 12 days is less efficient than either A or B working alone — working together must be faster than either alone.
Mathematics — Speed, Distance, Time
Q976EasyBPSC Prelims
A car travels 120 km at a speed of 40 km/h. How long does the journey take?
A2 hours
B2.5 hours
C3 hours
D4 hours
Show Answer
✔ C — 3 hours
Time = Distance ÷ Speed
Time = 120 ÷ 40 = 3 hours
*Why A is wrong:* 2 hours at 40 km/h would cover only 80 km, not 120 km.
*Why B is wrong:* 2.5 hours at 40 km/h = 100 km, not 120 km.
*Why D is wrong:* 4 hours at 40 km/h = 160 km, more than the required distance.
Time = 120 ÷ 40 = 3 hours
*Why A is wrong:* 2 hours at 40 km/h would cover only 80 km, not 120 km.
*Why B is wrong:* 2.5 hours at 40 km/h = 100 km, not 120 km.
*Why D is wrong:* 4 hours at 40 km/h = 160 km, more than the required distance.
Mathematics — Average
Q977EasyBPSC Prelims
Find the average of: 5, 10, 15, 20, 25.
A12
B13
C15
D17
Show Answer
✔ C — 15
Sum = 5 + 10 + 15 + 20 + 25 = 75
Number of terms = 5
Average = 75 ÷ 5 = 15
Note: This is an arithmetic progression with first term 5 and last term 25.
Average of AP = (First + Last) ÷ 2 = (5 + 25) ÷ 2 = 15 ✓
*Why A is wrong:* 12 is the average of a different set; the middle term of this AP is 15, not 12.
*Why B is wrong:* 13 does not correspond to the arithmetic mean of this dataset.
*Why D is wrong:* 17 exceeds the median and mean of this symmetric dataset.
Number of terms = 5
Average = 75 ÷ 5 = 15
Note: This is an arithmetic progression with first term 5 and last term 25.
Average of AP = (First + Last) ÷ 2 = (5 + 25) ÷ 2 = 15 ✓
*Why A is wrong:* 12 is the average of a different set; the middle term of this AP is 15, not 12.
*Why B is wrong:* 13 does not correspond to the arithmetic mean of this dataset.
*Why D is wrong:* 17 exceeds the median and mean of this symmetric dataset.
Mathematics — Algebra (Linear Equation)
Q978EasyBPSC Prelims
If 2x + 5 = 17, what is the value of x?
A5
B6
C7
D8
Show Answer
✔ B — 6
2x + 5 = 17
2x = 17 − 5 = 12
x = 12 ÷ 2 = 6
Verification: 2(6) + 5 = 12 + 5 = 17 ✓
*Why A is wrong:* If x=5, then 2(5)+5 = 15, not 17.
*Why C is wrong:* If x=7, then 2(7)+5 = 19, not 17.
*Why D is wrong:* If x=8, then 2(8)+5 = 21, not 17.
2x = 17 − 5 = 12
x = 12 ÷ 2 = 6
Verification: 2(6) + 5 = 12 + 5 = 17 ✓
*Why A is wrong:* If x=5, then 2(5)+5 = 15, not 17.
*Why C is wrong:* If x=7, then 2(7)+5 = 19, not 17.
*Why D is wrong:* If x=8, then 2(8)+5 = 21, not 17.
Mathematics — Geometry (Pythagoras Theorem)
Q979EasyBPSC Prelims
A right-angled triangle has legs of 3 cm and 4 cm. What is the length of the hypotenuse?
A4 cm
B5 cm
C6 cm
D7 cm
Show Answer
✔ B — 5 cm
By Pythagoras Theorem: Hypotenuse² = Base² + Height²
h² = 3² + 4² = 9 + 16 = 25
h = √25 = 5 cm
This is the classic "3-4-5 Pythagorean triple," the most frequently tested in competitive exams.
*Why A is wrong:* 4 cm is the length of one of the legs, not the hypotenuse; the hypotenuse is always the longest side.
*Why C is wrong:* h=6 would require base²+height²=36; 3²+4²=25≠36.
*Why D is wrong:* h=7 would require 3²+4²=49; but 25≠49.
h² = 3² + 4² = 9 + 16 = 25
h = √25 = 5 cm
This is the classic "3-4-5 Pythagorean triple," the most frequently tested in competitive exams.
*Why A is wrong:* 4 cm is the length of one of the legs, not the hypotenuse; the hypotenuse is always the longest side.
*Why C is wrong:* h=6 would require base²+height²=36; 3²+4²=25≠36.
*Why D is wrong:* h=7 would require 3²+4²=49; but 25≠49.
Mathematics — Number Series
Q980EasyBPSC Prelims
Find the next term: 2, 6, 18, 54, ?
A108
B144
C162
D216
Show Answer
✔ C — 162
Pattern: Each term is multiplied by 3
2 × 3 = 6
6 × 3 = 18
18 × 3 = 54
54 × 3 = 162
This is a geometric series with first term 2 and common ratio 3.
*Why A is wrong:* 108 = 54 × 2; the multiplier here is 2, not 3. The pattern is ×3, not ×2.
*Why B is wrong:* 144 has no direct multiplicative relationship with 54 consistent with the pattern.
*Why D is wrong:* 216 = 54 × 4; overcounts by using multiplier 4 instead of 3.
2 × 3 = 6
6 × 3 = 18
18 × 3 = 54
54 × 3 = 162
This is a geometric series with first term 2 and common ratio 3.
*Why A is wrong:* 108 = 54 × 2; the multiplier here is 2, not 3. The pattern is ×3, not ×2.
*Why B is wrong:* 144 has no direct multiplicative relationship with 54 consistent with the pattern.
*Why D is wrong:* 216 = 54 × 4; overcounts by using multiplier 4 instead of 3.
Mathematics — LCM-HCF Relationship
Q981MediumBPSC Prelims
The LCM of two numbers is 180 and their HCF is 12. If one number is 36, find the other number.
A45
B48
C60
D72
Show Answer
✔ C — 60
Key property: Product of two numbers = LCM × HCF
36 × Other = 180 × 12
36 × Other = 2160
Other = 2160 ÷ 36 = 60
Verification: HCF(36, 60) = 12 ✓; LCM(36, 60) = ?
36 = 2² × 3², 60 = 2² × 3 × 5
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 ✓
*Why A is wrong:* HCF(36, 45) = 9, not 12; and LCM(36, 45) = 180 but HCF fails.
*Why B is wrong:* 36 × 48 = 1728 ≠ 2160; doesn't satisfy LCM × HCF condition.
*Why D is wrong:* 36 × 72 = 2592 ≠ 2160.
36 × Other = 180 × 12
36 × Other = 2160
Other = 2160 ÷ 36 = 60
Verification: HCF(36, 60) = 12 ✓; LCM(36, 60) = ?
36 = 2² × 3², 60 = 2² × 3 × 5
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 ✓
*Why A is wrong:* HCF(36, 45) = 9, not 12; and LCM(36, 45) = 180 but HCF fails.
*Why B is wrong:* 36 × 48 = 1728 ≠ 2160; doesn't satisfy LCM × HCF condition.
*Why D is wrong:* 36 × 72 = 2592 ≠ 2160.
Mathematics — Profit & Loss with Discount
Q982MediumBPSC Prelims
An article is marked at ₹1,000, sold at 10% discount. The cost price is ₹800. Find the profit percentage.
A10%
B12.5%
C15%
D20%
Show Answer
✔ B — 12.5%
Marked Price (MP) = ₹1,000
Discount = 10% of 1000 = ₹100
Selling Price (SP) = 1000 − 100 = ₹900
Cost Price (CP) = ₹800
Profit = SP − CP = 900 − 800 = ₹100
Profit% = (Profit/CP) × 100 = (100/800) × 100 = 12.5%
*Why A is wrong:* 10% profit on CP=800 would give profit=₹80 and SP=₹880, but actual SP=₹900.
*Why C is wrong:* 15% profit on CP=800 would give profit=₹120 and SP=₹920≠₹900.
*Why D is wrong:* 20% profit on CP=800 gives SP=₹960, which requires no discount or higher MP.
Discount = 10% of 1000 = ₹100
Selling Price (SP) = 1000 − 100 = ₹900
Cost Price (CP) = ₹800
Profit = SP − CP = 900 − 800 = ₹100
Profit% = (Profit/CP) × 100 = (100/800) × 100 = 12.5%
*Why A is wrong:* 10% profit on CP=800 would give profit=₹80 and SP=₹880, but actual SP=₹900.
*Why C is wrong:* 15% profit on CP=800 would give profit=₹120 and SP=₹920≠₹900.
*Why D is wrong:* 20% profit on CP=800 gives SP=₹960, which requires no discount or higher MP.
Mathematics — Mixture (Replacement)
Q983MediumBPSC Prelims
A jar contains milk and water in ratio 5:1 (total 120 litres). 12 litres of the mixture are replaced by water. What is the new ratio of milk to water?
A4:1
B3:1
C2:1
D5:2
Show Answer
✔ B — 3:1
Initial: Total = 120 L, Milk = (5/6) × 120 = 100 L, Water = 20 L
12 litres mixture removed: Milk removed = (5/6) × 12 = 10 L, Water removed = (1/6) × 12 = 2 L
After removal: Milk = 100 − 10 = 90 L, Water = 20 − 2 = 18 L
12 litres of water added: Water = 18 + 12 = 30 L
New ratio = 90:30 = 3:1
*Why A is wrong:* 4:1 would mean milk=96, water=24; but actual milk=90, water=30.
*Why C is wrong:* 2:1 (milk=80, water=40) would require removing much more milk than happens here.
*Why D is wrong:* 5:2 would mean 102.86 L milk in 120 L — inconsistent with the calculation.
12 litres mixture removed: Milk removed = (5/6) × 12 = 10 L, Water removed = (1/6) × 12 = 2 L
After removal: Milk = 100 − 10 = 90 L, Water = 20 − 2 = 18 L
12 litres of water added: Water = 18 + 12 = 30 L
New ratio = 90:30 = 3:1
*Why A is wrong:* 4:1 would mean milk=96, water=24; but actual milk=90, water=30.
*Why C is wrong:* 2:1 (milk=80, water=40) would require removing much more milk than happens here.
*Why D is wrong:* 5:2 would mean 102.86 L milk in 120 L — inconsistent with the calculation.
Mathematics — Compound Interest
Q984MediumBPSC Prelims
Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.
A₹1,900
B₹2,000
C₹2,100
D₹2,200
Show Answer
✔ C — ₹2,100
Formula: A = P(1 + R/100)^T
A = 10,000 × (1 + 10/100)² = 10,000 × (1.1)² = 10,000 × 1.21 = ₹12,100
CI = A − P = 12,100 − 10,000 = ₹2,100
Comparison with SI: SI = 10,000 × 10 × 2/100 = ₹2,000
Difference = CI − SI = ₹100 (this equals P × (R/100)² = 10,000 × 0.01 = 100 ✓)
*Why A is wrong:* ₹1,900 is less than the SI (₹2,000); CI is always greater than SI for the same P, R, T (T>1 year).
*Why B is wrong:* ₹2,000 is the Simple Interest, not the compound interest.
*Why D is wrong:* ₹2,200 would mean A = ₹12,200; CI at 10% for 2 years on ₹10,000 is exactly ₹2,100.
A = 10,000 × (1 + 10/100)² = 10,000 × (1.1)² = 10,000 × 1.21 = ₹12,100
CI = A − P = 12,100 − 10,000 = ₹2,100
Comparison with SI: SI = 10,000 × 10 × 2/100 = ₹2,000
Difference = CI − SI = ₹100 (this equals P × (R/100)² = 10,000 × 0.01 = 100 ✓)
*Why A is wrong:* ₹1,900 is less than the SI (₹2,000); CI is always greater than SI for the same P, R, T (T>1 year).
*Why B is wrong:* ₹2,000 is the Simple Interest, not the compound interest.
*Why D is wrong:* ₹2,200 would mean A = ₹12,200; CI at 10% for 2 years on ₹10,000 is exactly ₹2,100.
Mathematics — Time & Work (Multiple Workers)
Q985MediumBPSC Prelims
A, B, and C can do a work in 20, 30, and 60 days respectively. All three start together; A leaves after 4 days. How many more days do B and C take to finish the remaining work?
A10 days
B12 days
C14 days
D16 days
Show Answer
✔ B — 12 days
Combined rate of A+B+C = 1/20 + 1/30 + 1/60 = 3/60 + 2/60 + 1/60 = 6/60 = 1/10 per day
Work done in 4 days = 4 × 1/10 = 2/5
Remaining work = 1 − 2/5 = 3/5
Rate of B+C = 1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20 per day
Days for B+C to finish remaining = (3/5) ÷ (1/20) = (3/5) × 20 = 12 days
*Why A is wrong:* 10 days would require B+C to finish 3/5 work at rate 1/16.67, which is not their actual combined rate.
*Why C is wrong:* 14 days × 1/20 = 14/20 = 7/10 ≠ 3/5.
*Why D is wrong:* 16 days × 1/20 = 4/5 ≠ 3/5.
Work done in 4 days = 4 × 1/10 = 2/5
Remaining work = 1 − 2/5 = 3/5
Rate of B+C = 1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20 per day
Days for B+C to finish remaining = (3/5) ÷ (1/20) = (3/5) × 20 = 12 days
*Why A is wrong:* 10 days would require B+C to finish 3/5 work at rate 1/16.67, which is not their actual combined rate.
*Why C is wrong:* 14 days × 1/20 = 14/20 = 7/10 ≠ 3/5.
*Why D is wrong:* 16 days × 1/20 = 4/5 ≠ 3/5.
Mathematics — Pipes & Cisterns
Q986MediumBPSC Prelims
Pipe A fills a tank in 12 hours, Pipe B fills it in 16 hours, and Pipe C (drain) empties it in 8 hours. If all three are opened simultaneously, how many hours will it take to fill the tank?
A24 hours
B36 hours
C48 hours
DCannot be filled
Show Answer
✔ C — 48 hours
Net rate per hour = (Filling rate A + Filling rate B) − Drain rate C
= 1/12 + 1/16 − 1/8
= 4/48 + 3/48 − 6/48
= 1/48 per hour (positive → tank fills)
Time = 48 hours
*Why A is wrong:* 24 hours would require net rate = 1/24; actual net rate = 1/48.
*Why B is wrong:* 36 hours requires net rate = 1/36; not matching our calculation.
*Why D is wrong:* Since net rate = +1/48 (positive), the tank does fill — it's just slow because C drains faster than either A or B alone.
= 1/12 + 1/16 − 1/8
= 4/48 + 3/48 − 6/48
= 1/48 per hour (positive → tank fills)
Time = 48 hours
*Why A is wrong:* 24 hours would require net rate = 1/24; actual net rate = 1/48.
*Why B is wrong:* 36 hours requires net rate = 1/36; not matching our calculation.
*Why D is wrong:* Since net rate = +1/48 (positive), the tank does fill — it's just slow because C drains faster than either A or B alone.
Mathematics — Speed (Trains)
Q987MediumBPSC Prelims
A train 200 m long passes a stationary pole in 10 seconds. What is the train's speed in km/h?
A60 km/h
B66 km/h
C72 km/h
D80 km/h
Show Answer
✔ C — 72 km/h
Speed = Distance ÷ Time = 200 m ÷ 10 s = 20 m/s
Convert to km/h: 20 × (18/5) = 20 × 3.6 = 72 km/h
Conversion formula: m/s × 18/5 = km/h
*Why A is wrong:* 60 km/h = 60 × 5/18 = 16.67 m/s; at this speed the train would take 200/16.67 = 12 seconds, not 10 seconds.
*Why B is wrong:* 66 km/h = 18.33 m/s; time = 200/18.33 ≈ 10.9 seconds, not 10 seconds.
*Why D is wrong:* 80 km/h = 22.22 m/s; time = 200/22.22 = 9 seconds, not 10 seconds.
Convert to km/h: 20 × (18/5) = 20 × 3.6 = 72 km/h
Conversion formula: m/s × 18/5 = km/h
*Why A is wrong:* 60 km/h = 60 × 5/18 = 16.67 m/s; at this speed the train would take 200/16.67 = 12 seconds, not 10 seconds.
*Why B is wrong:* 66 km/h = 18.33 m/s; time = 200/18.33 ≈ 10.9 seconds, not 10 seconds.
*Why D is wrong:* 80 km/h = 22.22 m/s; time = 200/22.22 = 9 seconds, not 10 seconds.
Mathematics — Average (Missing Number)
Q988MediumBPSC Prelims
The average of 7 numbers is 40. The average of the first 4 numbers is 36, and the average of the last 4 numbers is 45. Find the 4th number (the middle one).
A40
B42
C44
D46
Show Answer
✔ C — 44
Sum of all 7 numbers = 7 × 40 = 280
Sum of first 4 numbers = 4 × 36 = 144
Sum of last 4 numbers = 4 × 45 = 180
The 4th number is counted in both groups:
Sum(first 4) + Sum(last 4) = Sum(all 7) + 4th number
144 + 180 = 280 + 4th number
324 = 280 + 4th number
4th number = 44
*Why A is wrong:* 40 is the overall average, not the 4th number's value.
*Why B is wrong:* 42 does not satisfy 144 + 180 = 280 + x → x = 44 ≠ 42.
*Why D is wrong:* 46 gives 280 + 46 = 326 ≠ 324.
Sum of first 4 numbers = 4 × 36 = 144
Sum of last 4 numbers = 4 × 45 = 180
The 4th number is counted in both groups:
Sum(first 4) + Sum(last 4) = Sum(all 7) + 4th number
144 + 180 = 280 + 4th number
324 = 280 + 4th number
4th number = 44
*Why A is wrong:* 40 is the overall average, not the 4th number's value.
*Why B is wrong:* 42 does not satisfy 144 + 180 = 280 + x → x = 44 ≠ 42.
*Why D is wrong:* 46 gives 280 + 46 = 326 ≠ 324.
Mathematics — Mixture & Alligation
Q989MediumBPSC Prelims
In what ratio should water be mixed with milk (costing ₹60/litre) so that the mixture costs ₹48/litre? (Water costs nothing)
A1:2
B1:3
C1:4
D2:3
Show Answer
✔ C — 1:4
Using the Alligation rule:
- Milk price: ₹60/litre
- Water price: ₹0/litre
- Desired mixture price: ₹48/litre
Water : Milk = (Milk price − Desired price) : (Desired price − Water price)
= (60 − 48) : (48 − 0)
= 12 : 48
= 1 : 4
So 1 part water mixed with 4 parts milk gives mixture at ₹48/litre.
Verification: (1×0 + 4×60)/(1+4) = 240/5 = 48 ✓
*Why A is wrong:* Ratio 1:2 gives mixture price = (0+2×60)/3 = 40/litre ≠ 48.
*Why B is wrong:* Ratio 1:3 gives mixture price = (0+3×60)/4 = 45/litre ≠ 48.
*Why D is wrong:* Ratio 2:3 gives mixture price = (0+3×60)/5 = 36/litre ≠ 48.
- Milk price: ₹60/litre
- Water price: ₹0/litre
- Desired mixture price: ₹48/litre
Water : Milk = (Milk price − Desired price) : (Desired price − Water price)
= (60 − 48) : (48 − 0)
= 12 : 48
= 1 : 4
So 1 part water mixed with 4 parts milk gives mixture at ₹48/litre.
Verification: (1×0 + 4×60)/(1+4) = 240/5 = 48 ✓
*Why A is wrong:* Ratio 1:2 gives mixture price = (0+2×60)/3 = 40/litre ≠ 48.
*Why B is wrong:* Ratio 1:3 gives mixture price = (0+3×60)/4 = 45/litre ≠ 48.
*Why D is wrong:* Ratio 2:3 gives mixture price = (0+3×60)/5 = 36/litre ≠ 48.
Mathematics — Algebra (Quadratic Equation)
Q990MediumBPSC Prelims
Solve: x² − 5x + 6 = 0. The values of x are:
A1 and 6
B2 and 3
C3 and 4
D−2 and −3
Show Answer
✔ B — 2 and 3
x² − 5x + 6 = 0
Factor: find two numbers that multiply to +6 and add to −5
Those numbers: −2 and −3
(x − 2)(x − 3) = 0
x = 2 or x = 3
Verification: (2)²−5(2)+6 = 4−10+6 = 0 ✓; (3)²−5(3)+6 = 9−15+6 = 0 ✓
*Why A is wrong:* x=1 gives 1−5+6=2≠0; x=6 gives 36−30+6=12≠0.
*Why C is wrong:* x=3 works ✓ but x=4 gives 16−20+6=2≠0.
*Why D is wrong:* x=−2 gives 4+10+6=20≠0; these would be roots of x²+5x+6=0, not x²−5x+6=0.
Factor: find two numbers that multiply to +6 and add to −5
Those numbers: −2 and −3
(x − 2)(x − 3) = 0
x = 2 or x = 3
Verification: (2)²−5(2)+6 = 4−10+6 = 0 ✓; (3)²−5(3)+6 = 9−15+6 = 0 ✓
*Why A is wrong:* x=1 gives 1−5+6=2≠0; x=6 gives 36−30+6=12≠0.
*Why C is wrong:* x=3 works ✓ but x=4 gives 16−20+6=2≠0.
*Why D is wrong:* x=−2 gives 4+10+6=20≠0; these would be roots of x²+5x+6=0, not x²−5x+6=0.
Mathematics — Geometry (Triangle)
Q991MediumBPSC Prelims
A triangle has sides 5 cm, 12 cm, and 13 cm. Is it a right triangle? If yes, find its area.
ANot a right triangle
BArea = 24 cm²
CArea = 30 cm²
DArea = 36 cm²
Show Answer
✔ C — Area = 30 cm²
Check: 5² + 12² = 25 + 144 = 169 = 13² ✓
So it IS a right-angled triangle (5-12-13 Pythagorean triple).
For a right triangle: Area = ½ × base × height = ½ × 5 × 12 = 30 cm²
(The two legs serve as base and height.)
*Why A is wrong:* 5² + 12² = 169 = 13², so it is indeed a right triangle by the converse of Pythagoras theorem.
*Why B is wrong:* Area = 24 cm² would require base × height = 48; for this triangle, legs are 5 and 12, giving 60.
*Why D is wrong:* Area = 36 requires base × height = 72; not consistent with the given side lengths.
So it IS a right-angled triangle (5-12-13 Pythagorean triple).
For a right triangle: Area = ½ × base × height = ½ × 5 × 12 = 30 cm²
(The two legs serve as base and height.)
*Why A is wrong:* 5² + 12² = 169 = 13², so it is indeed a right triangle by the converse of Pythagoras theorem.
*Why B is wrong:* Area = 24 cm² would require base × height = 48; for this triangle, legs are 5 and 12, giving 60.
*Why D is wrong:* Area = 36 requires base × height = 72; not consistent with the given side lengths.
Mathematics — Mensuration (Circle)
Q992MediumBPSC Prelims
Find the area of a circle with radius 7 cm. (Use π = 22/7)
A132 cm²
B144 cm²
C154 cm²
D176 cm²
Show Answer
✔ C — 154 cm²
Area of circle = πr²
= (22/7) × 7 × 7
= (22/7) × 49
= 22 × 7
= 154 cm²
Circumference = 2πr = 2 × (22/7) × 7 = 44 cm (for reference)
*Why A is wrong:* 132 = 22/7 × 6² × π adjustment; corresponds to r=6.5 approximately, not r=7.
*Why B is wrong:* 144 = 12² (perfect square), but Area = π × 49 = 154, not 144.
*Why D is wrong:* 176 ÷ 22 = 8 = r² only if r = 2√2 ≈ 2.83 cm, not r = 7 cm.
= (22/7) × 7 × 7
= (22/7) × 49
= 22 × 7
= 154 cm²
Circumference = 2πr = 2 × (22/7) × 7 = 44 cm (for reference)
*Why A is wrong:* 132 = 22/7 × 6² × π adjustment; corresponds to r=6.5 approximately, not r=7.
*Why B is wrong:* 144 = 12² (perfect square), but Area = π × 49 = 154, not 144.
*Why D is wrong:* 176 ÷ 22 = 8 = r² only if r = 2√2 ≈ 2.83 cm, not r = 7 cm.
Mathematics — Mensuration (Cylinder)
Q993MediumBPSC Prelims
Find the volume of a cylinder with radius 7 cm and height 10 cm. (Use π = 22/7)
A1,100 cm³
B1,320 cm³
C1,540 cm³
D1,760 cm³
Show Answer
✔ C — 1,540 cm³
Volume of cylinder = πr²h
= (22/7) × 7² × 10
= (22/7) × 49 × 10
= 22 × 7 × 10
= 1,540 cm³
*Why A is wrong:* 1,100 = 22/7 × 5² × 10; corresponds to r=5 cm, not r=7 cm.
*Why B is wrong:* 1,320 ÷ (22 × 10/7) = 1320 × 7/220 = 42 ≠ r²=49; doesn't match r=7 cm.
*Why D is wrong:* 1,760 corresponds to r=8 cm (22/7 × 64 × 10 = 2011.4... not 1760); inconsistent.
= (22/7) × 7² × 10
= (22/7) × 49 × 10
= 22 × 7 × 10
= 1,540 cm³
*Why A is wrong:* 1,100 = 22/7 × 5² × 10; corresponds to r=5 cm, not r=7 cm.
*Why B is wrong:* 1,320 ÷ (22 × 10/7) = 1320 × 7/220 = 42 ≠ r²=49; doesn't match r=7 cm.
*Why D is wrong:* 1,760 corresponds to r=8 cm (22/7 × 64 × 10 = 2011.4... not 1760); inconsistent.
Mathematics — Data Interpretation (Table)
Q994MediumBPSC Prelims
The table below shows wheat production (in thousand tonnes) across 5 districts of Bihar:
| District | Production |
|----------|-----------|
| Rohtas | 200 |
| Bhojpur | 300 |
| Kaimur | 150 |
| Gaya | 250 |
| Aurangabad | 100 |
| **Total** | **1,000** |
What percentage of total production comes from Bhojpur?
A25%
B28%
C30%
D35%
Show Answer
✔ C — 30%
Bhojpur's production = 300 thousand tonnes
Total production = 1,000 thousand tonnes
Percentage = (300/1000) × 100 = 30%
*Why A is wrong:* 25% of 1000 = 250 = Gaya's production, not Bhojpur's.
*Why B is wrong:* 28% of 1000 = 280 ≠ 300.
*Why D is wrong:* 35% of 1000 = 350 ≠ 300.
Total production = 1,000 thousand tonnes
Percentage = (300/1000) × 100 = 30%
*Why A is wrong:* 25% of 1000 = 250 = Gaya's production, not Bhojpur's.
*Why B is wrong:* 28% of 1000 = 280 ≠ 300.
*Why D is wrong:* 35% of 1000 = 350 ≠ 300.
Mathematics — Statistics (Median & Mode)
Q995MediumBPSC Prelims
For the dataset: 3, 7, 7, 5, 9, 7, 11 — find the Median and Mode.
AMedian = 7, Mode = 7
BMedian = 8, Mode = 7
CMedian = 7, Mode = 9
DMedian = 6, Mode = 7
Show Answer
✔ A — Median = 7, Mode = 7
Step 1 — Arrange in ascending order: 3, 5, 7, 7, 7, 9, 11
Step 2 — Median (middle value, 4th of 7): 7
Step 3 — Mode (most frequent): 7 appears 3 times → Mode = 7
Both Median = 7 and Mode = 7 coincide here. The Mean = (3+5+7+7+7+9+11)/7 = 49/7 = 7 (all three equal!)
*Why B is wrong:* Median = 8 would require the middle value to be 8; the 4th value is clearly 7.
*Why C is wrong:* Mode = 9 is wrong; 9 appears only once; 7 appears 3 times → Mode = 7.
*Why D is wrong:* Median = 6 is wrong; the arranged dataset's 4th value (middle) is 7, not 6.
Step 2 — Median (middle value, 4th of 7): 7
Step 3 — Mode (most frequent): 7 appears 3 times → Mode = 7
Both Median = 7 and Mode = 7 coincide here. The Mean = (3+5+7+7+7+9+11)/7 = 49/7 = 7 (all three equal!)
*Why B is wrong:* Median = 8 would require the middle value to be 8; the 4th value is clearly 7.
*Why C is wrong:* Mode = 9 is wrong; 9 appears only once; 7 appears 3 times → Mode = 7.
*Why D is wrong:* Median = 6 is wrong; the arranged dataset's 4th value (middle) is 7, not 6.
Mathematics — Trigonometry (Basic Ratio)
Q996MediumBPSC Prelims
If sin θ = 3/5, find the value of cos θ (assuming θ is in the first quadrant).
A3/4
B4/5
C5/3
D4/3
Show Answer
✔ B — 4/5
Using the identity: sin²θ + cos²θ = 1
cos²θ = 1 − sin²θ = 1 − (3/5)² = 1 − 9/25 = 16/25
cos θ = √(16/25) = 4/5 (positive in first quadrant)
This corresponds to the 3-4-5 right triangle: opposite=3, hypotenuse=5, adjacent=4.
sin θ = 3/5, cos θ = 4/5, tan θ = 3/4
*Why A is wrong:* 3/4 = tan θ, not cos θ.
*Why C is wrong:* 5/3 = cosec θ (reciprocal of sin θ), not cos θ.
*Why D is wrong:* 4/3 = cot θ (reciprocal of tan θ), not cos θ.
cos²θ = 1 − sin²θ = 1 − (3/5)² = 1 − 9/25 = 16/25
cos θ = √(16/25) = 4/5 (positive in first quadrant)
This corresponds to the 3-4-5 right triangle: opposite=3, hypotenuse=5, adjacent=4.
sin θ = 3/5, cos θ = 4/5, tan θ = 3/4
*Why A is wrong:* 3/4 = tan θ, not cos θ.
*Why C is wrong:* 5/3 = cosec θ (reciprocal of sin θ), not cos θ.
*Why D is wrong:* 4/3 = cot θ (reciprocal of tan θ), not cos θ.
Mathematics — Divisibility Rules
Q997MediumBPSC Prelims
Which of the following numbers is divisible by 11?
A10,234: 1 − 0 + 2 − 3 + 4 = 4 → NOT divisible by 11
B121: 1 − 2 + 1 = 0 → **Divisible by 11** ✓ (121 = 11 × 11)
C1,234: 1 − 2 + 3 − 4 = −2 → NOT divisible by 11
D5,678: 5 − 6 + 7 − 8 = −2 → NOT divisible by 11
Show Answer
✔ B — 121: 1 − 2 + 1 = 0 → **Divisible by 11** ✓ (121 = 11 × 11)
Divisibility rule for 11: Alternating sum of digits (from left, alternating +/−) must be 0 or a multiple of 11.
A) 10,234: 1 − 0 + 2 − 3 + 4 = 4 → NOT divisible by 11
B) 121: 1 − 2 + 1 = 0 → Divisible by 11 ✓ (121 = 11 × 11)
C) 1,234: 1 − 2 + 3 − 4 = −2 → NOT divisible by 11
D) 5,678: 5 − 6 + 7 − 8 = −2 → NOT divisible by 11
*Why A is wrong:* Alternating digit sum of 10234 = 4, not 0 or 11.
*Why C is wrong:* Alternating digit sum = −2, not divisible by 11.
*Why D is wrong:* Alternating digit sum = −2, not divisible by 11.
A) 10,234: 1 − 0 + 2 − 3 + 4 = 4 → NOT divisible by 11
B) 121: 1 − 2 + 1 = 0 → Divisible by 11 ✓ (121 = 11 × 11)
C) 1,234: 1 − 2 + 3 − 4 = −2 → NOT divisible by 11
D) 5,678: 5 − 6 + 7 − 8 = −2 → NOT divisible by 11
*Why A is wrong:* Alternating digit sum of 10234 = 4, not 0 or 11.
*Why C is wrong:* Alternating digit sum = −2, not divisible by 11.
*Why D is wrong:* Alternating digit sum = −2, not divisible by 11.
Mathematics — Fractions (Comparison)
Q998MediumBPSC Prelims
Arrange in descending order and identify the largest: 3/4, 5/7, 7/9, 9/11.
A3/4
B5/7
C7/9
D9/11
Show Answer
✔ D — 9/11
Convert to decimals:
3/4 = 0.7500
5/7 = 0.7143
7/9 = 0.7778
9/11 = 0.8182
Descending order: 9/11 > 7/9 > 3/4 > 5/7
Largest fraction = 9/11
Pattern observation: These fractions are of the form (2n−1)/(2n+1): 3/4 doesn't fit exactly, but note 9/11 > 7/9 > 3/4 > 5/7.
*Why A is wrong:* 3/4 = 0.75 < 9/11 = 0.818.
*Why B is wrong:* 5/7 ≈ 0.714 is actually the smallest of the four.
*Why C is wrong:* 7/9 ≈ 0.778 < 9/11 ≈ 0.818.
3/4 = 0.7500
5/7 = 0.7143
7/9 = 0.7778
9/11 = 0.8182
Descending order: 9/11 > 7/9 > 3/4 > 5/7
Largest fraction = 9/11
Pattern observation: These fractions are of the form (2n−1)/(2n+1): 3/4 doesn't fit exactly, but note 9/11 > 7/9 > 3/4 > 5/7.
*Why A is wrong:* 3/4 = 0.75 < 9/11 = 0.818.
*Why B is wrong:* 5/7 ≈ 0.714 is actually the smallest of the four.
*Why C is wrong:* 7/9 ≈ 0.778 < 9/11 ≈ 0.818.
Mathematics — Partnership
Q999MediumBPSC Prelims
A invests ₹3,000 for 4 months, B invests ₹4,000 for 6 months, and C invests ₹5,000 for 3 months in a business. Total profit is ₹5,100. Find A's share of profit.
A₹1,000
B₹1,200
C₹1,500
D₹2,000
Show Answer
✔ B — ₹1,200
Profit sharing is proportional to Capital × Time:
A = 3,000 × 4 = 12,000
B = 4,000 × 6 = 24,000
C = 5,000 × 3 = 15,000
Total = 12,000 + 24,000 + 15,000 = 51,000
A's share = (12,000/51,000) × 5,100 = (12/51) × 5,100 = 12 × 100 = ₹1,200
B's share = (24/51) × 5,100 = ₹2,400
C's share = (15/51) × 5,100 = ₹1,500
Check: 1200 + 2400 + 1500 = 5100 ✓
*Why A is wrong:* ₹1,000 = 5100 × (10/51); A's ratio is 12/51, not 10/51.
*Why C is wrong:* ₹1,500 is C's share (15/51 × 5100), not A's share.
*Why D is wrong:* ₹2,000 > A's actual share; A has the smallest investment×time product.
A = 3,000 × 4 = 12,000
B = 4,000 × 6 = 24,000
C = 5,000 × 3 = 15,000
Total = 12,000 + 24,000 + 15,000 = 51,000
A's share = (12,000/51,000) × 5,100 = (12/51) × 5,100 = 12 × 100 = ₹1,200
B's share = (24/51) × 5,100 = ₹2,400
C's share = (15/51) × 5,100 = ₹1,500
Check: 1200 + 2400 + 1500 = 5100 ✓
*Why A is wrong:* ₹1,000 = 5100 × (10/51); A's ratio is 12/51, not 10/51.
*Why C is wrong:* ₹1,500 is C's share (15/51 × 5100), not A's share.
*Why D is wrong:* ₹2,000 > A's actual share; A has the smallest investment×time product.
Mathematics — Discount & Selling Price
Q1000MediumBPSC Prelims
An article is marked at ₹1,200 and sold for ₹1,020. What is the discount percentage?
A12%
B13%
C15%
D18%
Show Answer
✔ C — 15%
Discount = Marked Price − Selling Price = 1,200 − 1,020 = ₹180
Discount% = (Discount/Marked Price) × 100 = (180/1,200) × 100 = 15%
*Why A is wrong:* 12% of 1200 = ₹144; SP would be ₹1,056 ≠ ₹1,020.
*Why B is wrong:* 13% of 1200 = ₹156; SP = ₹1,044 ≠ ₹1,020.
*Why D is wrong:* 18% of 1200 = ₹216; SP = ₹984 ≠ ₹1,020.
Discount% = (Discount/Marked Price) × 100 = (180/1,200) × 100 = 15%
*Why A is wrong:* 12% of 1200 = ₹144; SP would be ₹1,056 ≠ ₹1,020.
*Why B is wrong:* 13% of 1200 = ₹156; SP = ₹1,044 ≠ ₹1,020.
*Why D is wrong:* 18% of 1200 = ₹216; SP = ₹984 ≠ ₹1,020.
Mathematics — LCM Application (Bell Problem)
Q1001HardBPSC Prelims
Three bells ring at intervals of 12, 15, and 20 minutes. If they all ring together at 8:00 AM, at what time will they next ring together?
A8:45 AM
B9:00 AM
C9:30 AM
D10:00 AM
Show Answer
✔ B — 9:00 AM
The bells ring together after LCM(12, 15, 20) minutes.
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5
LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60 minutes
60 minutes after 8:00 AM = 9:00 AM
*Why A is wrong:* 45 minutes = LCM(9,15)... not LCM(12,15,20). At 8:45 AM, bell 1 (12-min interval) rings at 8:48 AM, not 8:45.
*Why C is wrong:* 90 minutes is not the LCM of 12, 15, and 20. LCM = 60.
*Why D is wrong:* 120 minutes (2 hours) is 2×LCM; the first common ring is at 60 minutes = 9:00 AM.
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5
LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60 minutes
60 minutes after 8:00 AM = 9:00 AM
*Why A is wrong:* 45 minutes = LCM(9,15)... not LCM(12,15,20). At 8:45 AM, bell 1 (12-min interval) rings at 8:48 AM, not 8:45.
*Why C is wrong:* 90 minutes is not the LCM of 12, 15, and 20. LCM = 60.
*Why D is wrong:* 120 minutes (2 hours) is 2×LCM; the first common ring is at 60 minutes = 9:00 AM.
Mathematics — Percentage (Successive Changes)
Q1002HardBPSC Prelims
A number is first increased by 20%, then decreased by 25%. What is the net percentage change?
A5% decrease
B10% decrease
C10% increase
D5% increase
Show Answer
✔ B — 10% decrease
Let original number = 100
After 20% increase: 100 × 1.20 = 120
After 25% decrease: 120 × 0.75 = 90
Net change = 90 − 100 = −10
Net percentage change = 10% decrease
Formula shortcut: Net% = a + b + (ab/100) where a=+20, b=−25
= 20 + (−25) + (20×(−25)/100) = 20 − 25 − 5 = −10% ✓
*Why A is wrong:* 5% decrease would give final value 95; but actual final value = 90.
*Why C is wrong:* The net result is a decrease, not an increase; the 25% decrease is applied to a larger number (120), making the absolute loss bigger than the absolute gain.
*Why D is wrong:* 5% increase means final value = 105; but 120 × 0.75 = 90 ≠ 105.
After 20% increase: 100 × 1.20 = 120
After 25% decrease: 120 × 0.75 = 90
Net change = 90 − 100 = −10
Net percentage change = 10% decrease
Formula shortcut: Net% = a + b + (ab/100) where a=+20, b=−25
= 20 + (−25) + (20×(−25)/100) = 20 − 25 − 5 = −10% ✓
*Why A is wrong:* 5% decrease would give final value 95; but actual final value = 90.
*Why C is wrong:* The net result is a decrease, not an increase; the 25% decrease is applied to a larger number (120), making the absolute loss bigger than the absolute gain.
*Why D is wrong:* 5% increase means final value = 105; but 120 × 0.75 = 90 ≠ 105.
Mathematics — Profit & Loss (Marked Price & Cost Price)
Q1003HardBPSC Prelims
A trader buys goods at 20% below the list price and sells at 10% above the list price. What is the profit percentage on cost price?
A30%
B33.33%
C37.5%
D40%
Show Answer
✔ C — 37.5%
Let List Price = ₹100
Cost Price = 100 − 20% of 100 = ₹80 (bought at 20% discount)
Selling Price = 100 + 10% of 100 = ₹110 (sold at 10% above list price)
Profit = SP − CP = 110 − 80 = ₹30
Profit% = (30/80) × 100 = 37.5%
*Why A is wrong:* 30% profit on CP=80 gives profit=₹24 and SP=₹104, not ₹110.
*Why B is wrong:* 33.33% = 1/3; profit would be 80/3 ≈ ₹26.67, giving SP≈₹106.67 ≠ ₹110.
*Why D is wrong:* 40% profit gives SP = 80 × 1.4 = ₹112 ≠ ₹110.
Cost Price = 100 − 20% of 100 = ₹80 (bought at 20% discount)
Selling Price = 100 + 10% of 100 = ₹110 (sold at 10% above list price)
Profit = SP − CP = 110 − 80 = ₹30
Profit% = (30/80) × 100 = 37.5%
*Why A is wrong:* 30% profit on CP=80 gives profit=₹24 and SP=₹104, not ₹110.
*Why B is wrong:* 33.33% = 1/3; profit would be 80/3 ≈ ₹26.67, giving SP≈₹106.67 ≠ ₹110.
*Why D is wrong:* 40% profit gives SP = 80 × 1.4 = ₹112 ≠ ₹110.
Mathematics — Compound Interest (Half-Yearly)
Q1004HardBPSC Prelims
Find the compound interest on ₹16,000 at 20% per annum, compounded half-yearly, for 1 year.
A₹3,000
B₹3,200
C₹3,360
D₹3,400
Show Answer
✔ C — ₹3,360
When compounded half-yearly:
Rate per half year = 20%/2 = 10%
Number of periods in 1 year = 2
Amount = P × (1 + R/100)^n = 16,000 × (1 + 10/100)² = 16,000 × (1.1)² = 16,000 × 1.21 = ₹19,360
CI = 19,360 − 16,000 = ₹3,360
Compare with annual compounding: CI = 16,000 × (1.2)¹ − 16,000 = ₹3,200
Half-yearly compounding gives ₹160 more than annual compounding.
*Why A is wrong:* ₹3,000 = 16,000 × 20% × 1 year = Simple Interest; CI > SI.
*Why B is wrong:* ₹3,200 is CI with annual compounding (16,000 × 0.20 = 3,200), not half-yearly.
*Why D is wrong:* ₹3,400 would correspond to a rate slightly above 20%; not consistent with 10% per half-year calculation.
Rate per half year = 20%/2 = 10%
Number of periods in 1 year = 2
Amount = P × (1 + R/100)^n = 16,000 × (1 + 10/100)² = 16,000 × (1.1)² = 16,000 × 1.21 = ₹19,360
CI = 19,360 − 16,000 = ₹3,360
Compare with annual compounding: CI = 16,000 × (1.2)¹ − 16,000 = ₹3,200
Half-yearly compounding gives ₹160 more than annual compounding.
*Why A is wrong:* ₹3,000 = 16,000 × 20% × 1 year = Simple Interest; CI > SI.
*Why B is wrong:* ₹3,200 is CI with annual compounding (16,000 × 0.20 = 3,200), not half-yearly.
*Why D is wrong:* ₹3,400 would correspond to a rate slightly above 20%; not consistent with 10% per half-year calculation.
Mathematics — Time & Work (Efficiency)
Q1005HardBPSC Prelims
A is twice as efficient as B. Together they complete a work in 18 days. In how many days can A alone complete the work?
A24 days
B27 days
C30 days
D36 days
Show Answer
✔ B — 27 days
Let B's efficiency = 1 unit/day → A's efficiency = 2 units/day
Combined efficiency = 3 units/day
Total work = Combined efficiency × Time = 3 × 18 = 54 units
A alone: Time = Total work / A's efficiency = 54/2 = 27 days
B alone: Time = 54/1 = 54 days
*Why A is wrong:* 24 days requires total work = 2 × 24 = 48 units; but combined rate × 18 days = 3×18 = 54 ≠ 48.
*Why C is wrong:* 30 days requires total work = 2 × 30 = 60; but total work = 54.
*Why D is wrong:* 36 days is B's time (if efficiency ratio is 1:2); but actually B takes 54 days, not 36.
Combined efficiency = 3 units/day
Total work = Combined efficiency × Time = 3 × 18 = 54 units
A alone: Time = Total work / A's efficiency = 54/2 = 27 days
B alone: Time = 54/1 = 54 days
*Why A is wrong:* 24 days requires total work = 2 × 24 = 48 units; but combined rate × 18 days = 3×18 = 54 ≠ 48.
*Why C is wrong:* 30 days requires total work = 2 × 30 = 60; but total work = 54.
*Why D is wrong:* 36 days is B's time (if efficiency ratio is 1:2); but actually B takes 54 days, not 36.
Mathematics — Boats & Streams
Q1006HardBPSC Prelims
A boat covers 36 km downstream in 3 hours and 24 km upstream in 4 hours. Find the speed of the stream.
A2 km/h
B3 km/h
C4 km/h
D5 km/h
Show Answer
✔ B — 3 km/h
Downstream speed = 36/3 = 12 km/h
Upstream speed = 24/4 = 6 km/h
Speed of boat in still water = (Downstream + Upstream)/2 = (12 + 6)/2 = 9 km/h
Speed of stream = (Downstream − Upstream)/2 = (12 − 6)/2 = 6/2 = 3 km/h
*Why A is wrong:* Speed of stream = 2 km/h would give downstream = 9+2 = 11 km/h → 36 km in 36/11 ≈ 3.27 hrs ≠ 3 hrs.
*Why C is wrong:* If stream = 4 km/h, then boat speed = 8 km/h; upstream = 4 km/h, 24/4=6 hrs ≠ 4 hrs.
*Why D is wrong:* Stream = 5 km/h would give boat speed = 7 km/h; upstream = 2 km/h, 24/2=12 hrs ≠ 4 hrs.
Upstream speed = 24/4 = 6 km/h
Speed of boat in still water = (Downstream + Upstream)/2 = (12 + 6)/2 = 9 km/h
Speed of stream = (Downstream − Upstream)/2 = (12 − 6)/2 = 6/2 = 3 km/h
*Why A is wrong:* Speed of stream = 2 km/h would give downstream = 9+2 = 11 km/h → 36 km in 36/11 ≈ 3.27 hrs ≠ 3 hrs.
*Why C is wrong:* If stream = 4 km/h, then boat speed = 8 km/h; upstream = 4 km/h, 24/4=6 hrs ≠ 4 hrs.
*Why D is wrong:* Stream = 5 km/h would give boat speed = 7 km/h; upstream = 2 km/h, 24/2=12 hrs ≠ 4 hrs.
Mathematics — Trains (Same Direction)
Q1007HardBPSC Prelims
Two trains, each 100 m long, travel in the same direction at speeds of 90 km/h and 72 km/h. How long does the faster train take to completely pass the slower train?
A20 seconds
B30 seconds
C40 seconds
D50 seconds
Show Answer
✔ C — 40 seconds
Relative speed (same direction) = 90 − 72 = 18 km/h
Convert to m/s: 18 × (5/18) = 5 m/s
Total distance to be covered = Length of Train 1 + Length of Train 2 = 100 + 100 = 200 m
Time = Distance/Speed = 200/5 = 40 seconds
*Why A is wrong:* 20 seconds would require relative speed = 200/20 = 10 m/s = 36 km/h relative difference; but actual difference = 18 km/h = 5 m/s.
*Why B is wrong:* 30 seconds requires 200/30 = 6.67 m/s relative speed = 24 km/h difference; not consistent.
*Why D is wrong:* 50 seconds requires relative speed = 200/50 = 4 m/s = 14.4 km/h; actual = 5 m/s.
Convert to m/s: 18 × (5/18) = 5 m/s
Total distance to be covered = Length of Train 1 + Length of Train 2 = 100 + 100 = 200 m
Time = Distance/Speed = 200/5 = 40 seconds
*Why A is wrong:* 20 seconds would require relative speed = 200/20 = 10 m/s = 36 km/h relative difference; but actual difference = 18 km/h = 5 m/s.
*Why B is wrong:* 30 seconds requires 200/30 = 6.67 m/s relative speed = 24 km/h difference; not consistent.
*Why D is wrong:* 50 seconds requires relative speed = 200/50 = 4 m/s = 14.4 km/h; actual = 5 m/s.
Mathematics — Successive Replacement (Milk & Water)
Q1008HardBPSC Prelims
A vessel contains 80 litres of pure milk. 20 litres are removed and replaced with water. Then, 20 litres of the mixture are again removed and replaced with water. How many litres of milk remain?
A40 litres
B42 litres
C45 litres
D48 litres
Show Answer
✔ C — 45 litres
After 1st replacement:
Milk = 80 − 20 = 60 litres; Water = 20 litres
After 2nd replacement (removing 20 litres of 80-litre mixture):
Milk removed = (60/80) × 20 = 15 litres
Milk remaining = 60 − 15 = 45 litres
General formula: Final milk = Initial × ((Total − Removed)/Total)^n
= 80 × (60/80)² = 80 × (3/4)² = 80 × 9/16 = 45 litres ✓
*Why A is wrong:* 40 litres would result from removing half the milk each time (50% replacement), not 25%.
*Why B is wrong:* 42 has no mathematical basis from the formula; 80 × (3/4)² = 45, not 42.
*Why D is wrong:* 48 = 80 × 3/5; this doesn't correspond to (3/4)² = 9/16.
Milk = 80 − 20 = 60 litres; Water = 20 litres
After 2nd replacement (removing 20 litres of 80-litre mixture):
Milk removed = (60/80) × 20 = 15 litres
Milk remaining = 60 − 15 = 45 litres
General formula: Final milk = Initial × ((Total − Removed)/Total)^n
= 80 × (60/80)² = 80 × (3/4)² = 80 × 9/16 = 45 litres ✓
*Why A is wrong:* 40 litres would result from removing half the milk each time (50% replacement), not 25%.
*Why B is wrong:* 42 has no mathematical basis from the formula; 80 × (3/4)² = 45, not 42.
*Why D is wrong:* 48 = 80 × 3/5; this doesn't correspond to (3/4)² = 9/16.
Mathematics — Algebra (Algebraic Identity)
Q1009HardBPSC Prelims
If (x + 1/x) = 5, find the value of (x² + 1/x²).
A21
B23
C25
D27
Show Answer
✔ B — 23
Squaring both sides of (x + 1/x) = 5:
(x + 1/x)² = 25
x² + 2·x·(1/x) + 1/x² = 25
x² + 2 + 1/x² = 25
x² + 1/x² = 25 − 2 = 23
Identity used: (a + b)² = a² + 2ab + b²; here a=x, b=1/x, so ab = x·(1/x) = 1.
*Why A is wrong:* 21 would result from (x+1/x)² − 4 = 25 − 4; but the correct subtraction is 2, not 4.
*Why C is wrong:* 25 = (x+1/x)²; that's before subtracting the 2 from the middle term.
*Why D is wrong:* 27 = 25 + 2; this would mean adding 2 rather than subtracting it.
(x + 1/x)² = 25
x² + 2·x·(1/x) + 1/x² = 25
x² + 2 + 1/x² = 25
x² + 1/x² = 25 − 2 = 23
Identity used: (a + b)² = a² + 2ab + b²; here a=x, b=1/x, so ab = x·(1/x) = 1.
*Why A is wrong:* 21 would result from (x+1/x)² − 4 = 25 − 4; but the correct subtraction is 2, not 4.
*Why C is wrong:* 25 = (x+1/x)²; that's before subtracting the 2 from the middle term.
*Why D is wrong:* 27 = 25 + 2; this would mean adding 2 rather than subtracting it.
Mathematics — Geometry (Angles in Triangle)
Q1010HardBPSC Prelims
The angles of a triangle are in the ratio 2:3:5. Find the measure of the largest angle.
A72°
B80°
C90°
D108°
Show Answer
✔ C — 90°
Sum of angles in a triangle = 180°
Let angles be 2k, 3k, 5k
2k + 3k + 5k = 180°
10k = 180°
k = 18°
Largest angle = 5k = 5 × 18° = 90°
This means the triangle is a right-angled triangle.
Other angles: 2k = 36°, 3k = 54°
*Why A is wrong:* 72° = 4k; but the largest angle = 5k = 90°.
*Why B is wrong:* If largest angle = 80°, then 5k = 80° → k=16°; sum would be 10×16=160° ≠ 180°.
*Why D is wrong:* If largest angle = 108°, the triangle would be obtuse; 5k=108° → k=21.6°; sum = 10×21.6=216° ≠ 180°.
Let angles be 2k, 3k, 5k
2k + 3k + 5k = 180°
10k = 180°
k = 18°
Largest angle = 5k = 5 × 18° = 90°
This means the triangle is a right-angled triangle.
Other angles: 2k = 36°, 3k = 54°
*Why A is wrong:* 72° = 4k; but the largest angle = 5k = 90°.
*Why B is wrong:* If largest angle = 80°, then 5k = 80° → k=16°; sum would be 10×16=160° ≠ 180°.
*Why D is wrong:* If largest angle = 108°, the triangle would be obtuse; 5k=108° → k=21.6°; sum = 10×21.6=216° ≠ 180°.
Mathematics — Mensuration (Cone — Slant Height)
Q1011HardBPSC Prelims
A cone has a base radius of 7 cm and height of 24 cm. Find the slant height.
A20 cm
B23 cm
C25 cm
D26 cm
Show Answer
✔ C — 25 cm
Slant height (l) = √(r² + h²)
l = √(7² + 24²)
l = √(49 + 576)
l = √625
l = 25 cm
This is another Pythagorean triple: 7-24-25.
Volume = ⅓πr²h = ⅓ × (22/7) × 49 × 24 = 1,232 cm³
Curved Surface Area = πrl = (22/7) × 7 × 25 = 550 cm²
*Why A is wrong:* 20 cm → l²=400; r²+h²=49+576=625≠400.
*Why B is wrong:* 23 cm → l²=529; 49+576=625≠529.
*Why D is wrong:* 26 cm → l²=676; 49+576=625≠676.
l = √(7² + 24²)
l = √(49 + 576)
l = √625
l = 25 cm
This is another Pythagorean triple: 7-24-25.
Volume = ⅓πr²h = ⅓ × (22/7) × 49 × 24 = 1,232 cm³
Curved Surface Area = πrl = (22/7) × 7 × 25 = 550 cm²
*Why A is wrong:* 20 cm → l²=400; r²+h²=49+576=625≠400.
*Why B is wrong:* 23 cm → l²=529; 49+576=625≠529.
*Why D is wrong:* 26 cm → l²=676; 49+576=625≠676.
Mathematics — Mensuration (Sphere — Surface Area)
Q1012HardBPSC Prelims
Find the total surface area of a sphere with radius 7 cm. (Use π = 22/7)
A492 cm²
B528 cm²
C616 cm²
D726 cm²
Show Answer
✔ C — 616 cm²
Total Surface Area of sphere = 4πr²
= 4 × (22/7) × 7²
= 4 × (22/7) × 49
= 4 × 22 × 7
= 4 × 154
= 616 cm²
Volume = (4/3)πr³ = (4/3) × (22/7) × 343 = (4/3) × 22 × 49 = 4312/3 ≈ 1,437.3 cm³
*Why A is wrong:* 492 ≠ 4 × 22/7 × 49; there's no combination giving 492 with r=7.
*Why B is wrong:* 528 = 4 × 22 × 6 = 4πr² for r=6, not r=7.
*Why D is wrong:* 726 would require r² = 726/(4×22/7) = 726×7/88 = 57.75; √57.75 ≈ 7.6 ≠ 7.
= 4 × (22/7) × 7²
= 4 × (22/7) × 49
= 4 × 22 × 7
= 4 × 154
= 616 cm²
Volume = (4/3)πr³ = (4/3) × (22/7) × 343 = (4/3) × 22 × 49 = 4312/3 ≈ 1,437.3 cm³
*Why A is wrong:* 492 ≠ 4 × 22/7 × 49; there's no combination giving 492 with r=7.
*Why B is wrong:* 528 = 4 × 22 × 6 = 4πr² for r=6, not r=7.
*Why D is wrong:* 726 would require r² = 726/(4×22/7) = 726×7/88 = 57.75; √57.75 ≈ 7.6 ≠ 7.
Mathematics — Data Interpretation (Bar Graph)
Q1013HardBPSC Prelims
A bar graph shows agricultural output (in lakh tonnes) in Bihar for 5 crops:
Wheat: 40, Rice: 60, Maize: 30, Pulses: 20, Vegetables: 50.
What is the ratio of Rice production to Pulses production? Also, what percentage of total output is from Rice and Vegetables combined?
ARatio 3:1; Combined 46%
BRatio 3:1; Combined 55%
CRatio 3:1; Combined 50%
DRatio 4:1; Combined 50%
Show Answer
✔ B — Ratio 3:1; Combined 55%
Rice : Pulses = 60 : 20 = 3 : 1
Total output = 40 + 60 + 30 + 20 + 50 = 200 lakh tonnes
Rice + Vegetables = 60 + 50 = 110 lakh tonnes
(110/200) × 100 = 55%
*Why A is wrong:* Ratio 3:1 is correct but 46% is incorrect; (60+50)/200 = 55%, not 46%.
*Why C is wrong:* 50% requires Rice+Veg=100 lakh tonnes; but 60+50=110.
*Why D is wrong:* Ratio 4:1 means Rice=4×Pulses=4×20=80; but Rice=60, not 80.
Total output = 40 + 60 + 30 + 20 + 50 = 200 lakh tonnes
Rice + Vegetables = 60 + 50 = 110 lakh tonnes
(110/200) × 100 = 55%
*Why A is wrong:* Ratio 3:1 is correct but 46% is incorrect; (60+50)/200 = 55%, not 46%.
*Why C is wrong:* 50% requires Rice+Veg=100 lakh tonnes; but 60+50=110.
*Why D is wrong:* Ratio 4:1 means Rice=4×Pulses=4×20=80; but Rice=60, not 80.
Mathematics — Data Interpretation (Pie Chart)
Q1014HardBPSC Prelims
A family's monthly expenditure is distributed as: Food=30%, Rent=25%, Education=20%, Transport=15%, Others=10%. Total monthly income = ₹40,000.
If the family saves 10% of income and spends the rest, how much is spent on Education?
A₹6,000
B₹7,200
C₹8,000
D₹10,000
Show Answer
✔ B — ₹7,200
Monthly income = ₹40,000
Savings = 10% of ₹40,000 = ₹4,000
Total expenditure = ₹40,000 − ₹4,000 = ₹36,000
The pie chart percentages apply to total expenditure (₹36,000):
Education = 20% of ₹36,000 = 0.20 × 36,000 = ₹7,200
*Why A is wrong:* ₹6,000 = 20% of ₹30,000; this would imply total expenditure of ₹30,000, which requires savings=₹10,000 (25%), not 10%.
*Why C is wrong:* ₹8,000 = 20% of ₹40,000; this treats 20% of total income, not of expenditure after saving.
*Why D is wrong:* ₹10,000 = 25% of ₹40,000; this incorrectly applies a different percentage to total income.
Savings = 10% of ₹40,000 = ₹4,000
Total expenditure = ₹40,000 − ₹4,000 = ₹36,000
The pie chart percentages apply to total expenditure (₹36,000):
Education = 20% of ₹36,000 = 0.20 × 36,000 = ₹7,200
*Why A is wrong:* ₹6,000 = 20% of ₹30,000; this would imply total expenditure of ₹30,000, which requires savings=₹10,000 (25%), not 10%.
*Why C is wrong:* ₹8,000 = 20% of ₹40,000; this treats 20% of total income, not of expenditure after saving.
*Why D is wrong:* ₹10,000 = 25% of ₹40,000; this incorrectly applies a different percentage to total income.
Mathematics — Statistics (Combined Mean)
Q1015HardBPSC Prelims
Class A has 30 students with average marks of 60. Class B has 20 students with average marks of 70. What is the combined average of both classes?
A63
B64
C65
D66
Show Answer
✔ B — 64
Total marks of Class A = 30 × 60 = 1,800
Total marks of Class B = 20 × 70 = 1,400
Combined total marks = 1,800 + 1,400 = 3,200
Total students = 30 + 20 = 50
Combined average = 3,200 ÷ 50 = 64
Note: Simple average of 60 and 70 = 65, but this is WRONG because the classes have different sizes. Class A (larger) pulls the average toward 60.
*Why A is wrong:* 63 = 60 + 3; this doesn't follow the weighted formula.
*Why C is wrong:* 65 = (60+70)/2; this is the unweighted average, which is wrong when class sizes differ.
*Why D is wrong:* 66 would mean 66×50=3300 total marks; but actual total=3200.
Total marks of Class B = 20 × 70 = 1,400
Combined total marks = 1,800 + 1,400 = 3,200
Total students = 30 + 20 = 50
Combined average = 3,200 ÷ 50 = 64
Note: Simple average of 60 and 70 = 65, but this is WRONG because the classes have different sizes. Class A (larger) pulls the average toward 60.
*Why A is wrong:* 63 = 60 + 3; this doesn't follow the weighted formula.
*Why C is wrong:* 65 = (60+70)/2; this is the unweighted average, which is wrong when class sizes differ.
*Why D is wrong:* 66 would mean 66×50=3300 total marks; but actual total=3200.
Mathematics — Trigonometry (Identity Simplification)
Q1016HardBPSC Prelims
Simplify: sin²θ + cos²θ + tan²θ − sec²θ
A0
B1
C−1
D2
Show Answer
✔ A — 0
Using trigonometric identities:
sin²θ + cos²θ = 1 ... (Identity 1)
sec²θ − tan²θ = 1 ... (Identity 2) → tan²θ − sec²θ = −1
Therefore:
sin²θ + cos²θ + tan²θ − sec²θ
= (sin²θ + cos²θ) + (tan²θ − sec²θ)
= 1 + (−1)
= 0
*Why B is wrong:* 1 = sin²θ + cos²θ alone; adding tan²θ − sec²θ = −1 gives 1 + (−1) = 0.
*Why C is wrong:* −1 = tan²θ − sec²θ alone; adding sin²θ + cos²θ = 1 gives −1 + 1 = 0.
*Why D is wrong:* 2 would require both partial sums to be +1, but tan²θ − sec²θ = −1, not +1.
sin²θ + cos²θ = 1 ... (Identity 1)
sec²θ − tan²θ = 1 ... (Identity 2) → tan²θ − sec²θ = −1
Therefore:
sin²θ + cos²θ + tan²θ − sec²θ
= (sin²θ + cos²θ) + (tan²θ − sec²θ)
= 1 + (−1)
= 0
*Why B is wrong:* 1 = sin²θ + cos²θ alone; adding tan²θ − sec²θ = −1 gives 1 + (−1) = 0.
*Why C is wrong:* −1 = tan²θ − sec²θ alone; adding sin²θ + cos²θ = 1 gives −1 + 1 = 0.
*Why D is wrong:* 2 would require both partial sums to be +1, but tan²θ − sec²θ = −1, not +1.
Mathematics — Number Series (Complex Pattern)
Q1017HardBPSC Prelims
Find the next term: 3, 7, 15, 31, 63, ?
A105
B120
C127
D135
Show Answer
✔ C — 127
Pattern: Each term = 2 × (previous term) + 1
3 × 2 + 1 = 7 ✓
7 × 2 + 1 = 15 ✓
15 × 2 + 1 = 31 ✓
31 × 2 + 1 = 63 ✓
63 × 2 + 1 = 127 ✓
Alternative: Terms are 4−1, 8−1, 16−1, 32−1, 64−1, 128−1 = 127 (powers of 2 minus 1)
*Why A is wrong:* 105 has no multiplicative pattern with 63 (63×2+1=127≠105).
*Why B is wrong:* 120 = 63×1.9+...; doesn't fit the ×2+1 rule.
*Why D is wrong:* 135 = 63×2+9; the +1 increment is fixed, not variable.
3 × 2 + 1 = 7 ✓
7 × 2 + 1 = 15 ✓
15 × 2 + 1 = 31 ✓
31 × 2 + 1 = 63 ✓
63 × 2 + 1 = 127 ✓
Alternative: Terms are 4−1, 8−1, 16−1, 32−1, 64−1, 128−1 = 127 (powers of 2 minus 1)
*Why A is wrong:* 105 has no multiplicative pattern with 63 (63×2+1=127≠105).
*Why B is wrong:* 120 = 63×1.9+...; doesn't fit the ×2+1 rule.
*Why D is wrong:* 135 = 63×2+9; the +1 increment is fixed, not variable.
Mathematics — Partnership (Time-Based)
Q1018HardBPSC Prelims
A starts a business with ₹50,000. After 3 months, B joins with ₹40,000. After 6 more months (i.e., 9 months from A's start), C joins with ₹30,000. Find the ratio of profits at the end of 12 months.
A10:6:3
B20:12:3
C15:9:4
D25:15:5
Show Answer
✔ B — 20:12:3
A invested for 12 months: 50,000 × 12 = 6,00,000
B invested for 9 months (joined at month 3): 40,000 × 9 = 3,60,000
C invested for 3 months (joined at month 9): 30,000 × 3 = 90,000
Ratio = 6,00,000 : 3,60,000 : 90,000
= 600 : 360 : 90
= 20 : 12 : 3 (dividing by 30)
*Why A is wrong:* 10:6:3 = 20:12:6 scaled down by 2; but C's share should be 90, not 180 (3 months, not 6).
*Why C is wrong:* 15:9:4 = 150:90:40 scaled; 40 would correspond to 30,000 × 40/10 months, not 3 months.
*Why D is wrong:* 25:15:5 = 5:3:1; A:B ratio = 5:3 but actual is 20:12 = 5:3 ✓ (matches), but A:C should be 20:3, not 5:1.
B invested for 9 months (joined at month 3): 40,000 × 9 = 3,60,000
C invested for 3 months (joined at month 9): 30,000 × 3 = 90,000
Ratio = 6,00,000 : 3,60,000 : 90,000
= 600 : 360 : 90
= 20 : 12 : 3 (dividing by 30)
*Why A is wrong:* 10:6:3 = 20:12:6 scaled down by 2; but C's share should be 90, not 180 (3 months, not 6).
*Why C is wrong:* 15:9:4 = 150:90:40 scaled; 40 would correspond to 30,000 × 40/10 months, not 3 months.
*Why D is wrong:* 25:15:5 = 5:3:1; A:B ratio = 5:3 but actual is 20:12 = 5:3 ✓ (matches), but A:C should be 20:3, not 5:1.
Mathematics — Simplification (BODMAS)
Q1019HardBPSC Prelims
Simplify: 18 + 6 ÷ 3 × 4 − 5 × 2
A12
B14
C16
D18
Show Answer
✔ C — 16
Apply BODMAS (Brackets → Orders → Division → Multiplication → Addition → Subtraction):
= 18 + 6 ÷ 3 × 4 − 5 × 2 [Division first]
= 18 + 2 × 4 − 5 × 2 [Multiplication, left to right]
= 18 + 8 − 10 [All multiplications done]
= 26 − 10 [Addition, then subtraction]
= 16
*Why A is wrong:* 12 results from incorrect order — if one adds/subtracts sequentially from left (18+6=24, 24÷3=8, 8×4=32, 32−5=27, 27×2=54 — completely wrong order).
*Why B is wrong:* 14 = 16 − 2; off by 2, suggesting an arithmetic error in one of the multiplication steps.
*Why D is wrong:* 18 = original first term; suggests the rest evaluated to 0, which is incorrect.
= 18 + 6 ÷ 3 × 4 − 5 × 2 [Division first]
= 18 + 2 × 4 − 5 × 2 [Multiplication, left to right]
= 18 + 8 − 10 [All multiplications done]
= 26 − 10 [Addition, then subtraction]
= 16
*Why A is wrong:* 12 results from incorrect order — if one adds/subtracts sequentially from left (18+6=24, 24÷3=8, 8×4=32, 32−5=27, 27×2=54 — completely wrong order).
*Why B is wrong:* 14 = 16 − 2; off by 2, suggesting an arithmetic error in one of the multiplication steps.
*Why D is wrong:* 18 = original first term; suggests the rest evaluated to 0, which is incorrect.
Mathematics — Number System (Conditional Sum)
Q1020HardBPSC Prelims
Find the sum of all natural numbers from 1 to 100 that are divisible by 3 but NOT divisible by 9.
A972
B1,000
C1,089
D1,200
Show Answer
✔ C — 1,089
Step 1 — Sum of all multiples of 3 from 1 to 100:
Multiples: 3, 6, 9, ..., 99 → n = 33 terms
Sum = 3 + 6 + ... + 99 = 3(1 + 2 + ... + 33) = 3 × (33 × 34/2) = 3 × 561 = 1,683
Step 2 — Sum of all multiples of 9 from 1 to 100:
Multiples: 9, 18, 27, ..., 99 → n = 11 terms
Sum = 9(1 + 2 + ... + 11) = 9 × (11 × 12/2) = 9 × 66 = 594
Step 3 — Sum of numbers divisible by 3 but NOT by 9:
= Sum(div by 3) − Sum(div by 9) = 1,683 − 594 = 1,089
Verification: In each set of 3 consecutive multiples of 3 (e.g., 3,6,9), two are not divisible by 9.
Sum of qualifying numbers from 1–9: 3+6=9; from 1–18: 9+12+15=36; pattern confirms.
*Why A is wrong:* 972 = sum of multiples of 3 only up to ~87; the full sum to 99 is 1,683 and after removing multiples of 9 (594), we get 1,089, not 972.
*Why B is wrong:* 1,000 is a round number with no mathematical basis from the correct calculation.
*Why D is wrong:* 1,200 > 1,089; would require including some numbers that are multiples of 9, which should be excluded.
Multiples: 3, 6, 9, ..., 99 → n = 33 terms
Sum = 3 + 6 + ... + 99 = 3(1 + 2 + ... + 33) = 3 × (33 × 34/2) = 3 × 561 = 1,683
Step 2 — Sum of all multiples of 9 from 1 to 100:
Multiples: 9, 18, 27, ..., 99 → n = 11 terms
Sum = 9(1 + 2 + ... + 11) = 9 × (11 × 12/2) = 9 × 66 = 594
Step 3 — Sum of numbers divisible by 3 but NOT by 9:
= Sum(div by 3) − Sum(div by 9) = 1,683 − 594 = 1,089
Verification: In each set of 3 consecutive multiples of 3 (e.g., 3,6,9), two are not divisible by 9.
Sum of qualifying numbers from 1–9: 3+6=9; from 1–18: 9+12+15=36; pattern confirms.
*Why A is wrong:* 972 = sum of multiples of 3 only up to ~87; the full sum to 99 is 1,683 and after removing multiples of 9 (594), we get 1,089, not 972.
*Why B is wrong:* 1,000 is a round number with no mathematical basis from the correct calculation.
*Why D is wrong:* 1,200 > 1,089; would require including some numbers that are multiples of 9, which should be excluded.