In a pie chart showing the distribution of 1,800 students across five streams, the angle for Commerce is 80° and the angle for Science is 100°. How many more students are in Science than in Commerce?
Answer: B
Understanding:
We must find the difference in the number of students between Science and Commerce.
•Total students=1800
•Angle for Science=100°
•Angle for Commerce=80°
Formula:
Students in a stream=360°Sector angle×Total students
Step 1: Find students in Science.
Science=360100×1800=185×1800=500
Step 2: Find students in Commerce.
Commerce=36080×1800=92×1800=400
Step 3: Find the difference.
500−400=100
Answer:
Science has 100 more students than Commerce.
100
Quick Tip:
Instead of computing each separately, you can find the difference directly: 360(100−80)×1800=36020×1800=100.
Q.22Medium
A pie chart shows the percentage distribution of a company's sales across five regions. Region A = 30%, Region B = 25%, Region C = 20%, Region D = 15%, Region E = 10%. If Region A's sales amount to ₹9 lakh, what are the total sales of Regions B and C combined?
Answer: B
Understanding:
We must find the combined sales of Regions B and C.
•Region A=30%, sales =₹9 lakh
•Region B=25%, Region C=20%
Formula:
First find the total from Region A, then compute B + C:
Total=Region A percentageRegion A sales×100
Combined sales=100B%+C%×Total
Step 1: Find the total sales.
Total=309×100=30 lakh
Step 2: Find combined sales of B and C.
B + C=10025+20×30=10045×30=13.5 lakh
Answer:
The combined sales of Regions B and C are ₹13.5 lakh.
₹13.5 lakh
Quick Tip:
Always derive the total first from the known sector — then every other value is a simple percentage of that total.
Q.23Medium
A pie chart depicts the number of vehicles in a city. Cars represent 40%, Bikes 35%, Buses 10%, Trucks 8%, and Others 7%. If the number of Bikes is 1,05,000, by how much do Cars exceed Trucks?
Answer: A
Understanding:
We must find how much Cars exceed Trucks.
•Bikes=35%=1,05,000
•Cars=40%, Trucks=8%
Formula:
Total=Known %Known count×100
Difference=100Cars%−Trucks%×Total
Step 1: Find the total number of vehicles.
Total=351,05,000×100=3,00,000
Step 2: Find the difference between Cars and Trucks.
Computing the percentage-point difference (40%−8%=32%) directly saves you from finding each category separately.
Q.24Medium
Two pie charts compare a company's expenses in 2022 and 2023. In 2022 (total = ₹40 lakh), the sector for Salaries was 135°. In 2023 (total = ₹48 lakh), the sector for Salaries was 120°. What is the increase in the actual amount spent on Salaries from 2022 to 2023?
Answer: A
Understanding:
We must find the increase in salary expenditure from 2022 to 2023.
•2022 total=₹40 lakh, Salary angle=135°
•2023 total=₹48 lakh, Salary angle=120°
Formula:
Salary amount=360°Sector angle×Total expenses
Step 1: Salary in 2022.
Salary2022=360135×40=83×40=15 lakh
Step 2: Salary in 2023.
Salary2023=360120×48=31×48=16 lakh
Step 3: Find the increase.
16−15=1 lakh
Answer:
The salary expenditure increased by ₹1 lakh.
₹1 lakh
Quick Tip:
A smaller sector angle in a pie chart does NOT always mean a smaller absolute amount — the total may have grown. Always compute actual values before comparing.
Q.25Medium
A pie chart shows the percentage-wise distribution of monthly household expenses. Rent = 25%, Food = 30%, Education = 20%, Travel = 10%, Miscellaneous = 15%. If the family saves ₹4,500 per month after all expenses and the monthly income is ₹36,000, what is the amount spent on Education and Travel together?
Answer: A
Understanding:
We must find the combined expenditure on Education and Travel.
•Monthly income=₹36,000
•Monthly savings=₹4,500
•Education=20%, Travel=10% of total expenses
Formula:
Total expenses=Income−Savings
Amount for a category=100Percentage×Total expenses
Step 1: Find total monthly expenses.
36,000−4,500=31,500
Step 2: Note that Education + Travel = 20%+10%=30% of total expenses.
The combined amount spent on Education and Travel is ₹10,800.
₹10,800
Quick Tip:
When a pie chart shows percentage distribution of income, apply percentages directly to income. The savings information is a distractor here — check whether the chart refers to income or expenditure.
Q.26Medium
A pie chart shows the market share (in %) of five mobile brands: Alpha = 28%, Beta = 22%, Gamma = 18%, Delta = 20%, Others = 12%. The total number of mobiles sold is 5,00,000. If Alpha's market share increases by 4 percentage points (taken equally from Beta and Delta), what is the new number of Alpha mobiles sold?
Answer: B
Understanding:
We must find the new number of Alpha mobiles after its market share increases.
The redistribution detail (taken equally from Beta and Delta) does not affect Alpha's count — only Alpha's final percentage matters for this calculation.
Q.27Medium
A pie chart shows the annual sales distribution (in %) of a retailer across four quarters: Q1 = 20%, Q2 = 35%, Q3 = 25%, Q4 = 20%. The sales in Q2 are ₹1,05,000 more than the sales in Q1. What are the total annual sales?
Answer: B
Understanding:
We must find the total annual sales using the given difference between Q2 and Q1.
•Q1=20%, Q2=35%
•Q2 sales−Q1 sales=₹1,05,000
Formula:
Let total sales =T. Then:
100Q2%×T−100Q1%×T=Difference
100(Q2%−Q1%)×T=Difference
Step 1: Set up the equation.
10035−20×T=1,05,000
Step 2: Solve for T.
10015×TTT=1,05,000=151,05,000×100=7,00,000
Answer:
The total annual sales are ₹7,00,000.
₹7,00,000
Quick Tip:
Whenever a pie chart question gives the difference between two sectors, directly use the percentage-point difference to set up a single equation — it eliminates the need to find individual sector values first.
Q.28Medium
What is the two-digit number?
Statement I: The sum of the digits is 9. Statement II: The difference between the number and the number formed by reversing its digits is 27.
Answer: C
Understanding:
We need to determine whether a unique two-digit number can be found.
•Statement I: sum of digits =9
•Statement II: difference between the number and its reverse =27
Formula:
Let the two-digit number be ab, i.e., 10a+b, where a is the tens digit and b is the units digit.
10a+b is the number, and 10b+a is its reverse.
Step 1: Analyse Statement I alone
From Statement I: a+b=9.
Possible pairs: (1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0).
Multiple solutions exist, so Statement I alone is not sufficient.
Step 2: Analyse Statement II alone
From Statement II:
(10a+b)−(10b+a)9a−9ba−b=27=27=3
Possible pairs: (4,1),(5,2),(6,3),(7,4),(8,5),(9,6).
Multiple solutions exist, so Statement II alone is not sufficient.
Step 3: Combine both statements
From I: a+b=9
From II: a−b=3
2ab=12⇒a=6=9−6=3
The unique number is 63. Both statements together are sufficient.
Answer:
Both statements together uniquely determine the number as 63, but neither alone is sufficient.
Both statements together are sufficient but neither alone is sufficient.
Quick Tip:
In data sufficiency, always check if Statement I alone gives a unique answer before testing combinations. A system of two equations with two unknowns typically requires both.
Q.29Medium
Is the integer n even?
Statement I: n2−1 is divisible by 4. Statement II: n+5 is odd.
Answer: D
Understanding:
We need to determine if n is even.
•Statement I: 4∣(n2−1)
•Statement II: (n+5) is odd
Formula:
An integer is even if it is divisible by 2. We check each statement for sufficiency.
Step 1: Analyse Statement I alone
n2−1=(n−1)(n+1)
If n is even, say n=2k: (2k−1)(2k+1) is a product of two consecutive odd numbers, which is odd — not divisible by 4.
If n is odd, say n=2k+1: (2k)(2k+2)=4k(k+1), which is always divisible by 4.
So 4∣(n2−1)if and only ifn is odd. Statement I tells us n is odd (i.e., not even). This is a definitive answer — Statement I alone is sufficient.
Step 2: Analyse Statement II alone
n+5 is odd ⇒n is even (since odd − odd = even, and 5 is odd, so n must be even for the sum to be odd).
This directly tells us n is even. Statement II alone is sufficient.
Step 3: Conclusion
Both statements independently give a definitive (though opposite in this case) answer about the parity of n. Each statement alone is sufficient.
Answer:
Either statement alone is sufficient to answer the question.
Either statement alone is sufficient.
Quick Tip:
For parity questions, translating algebraic conditions (like divisibility) into odd/even properties quickly reveals sufficiency.
Q.30Medium
What is the area of a rectangle?
Statement I: The perimeter of the rectangle is 40 cm. Statement II: The length of the rectangle is 4 cm more than its breadth.
Answer: C
Understanding:
We need to find the area of a rectangle.
•Statement I: Perimeter =40 cm
•Statement II: l=b+4
Formula:
Perimeter=2(l+b),Area=l×b
Step 1: Analyse Statement I alone
From 2(l+b)=40, we get l+b=20.
This gives infinitely many (l,b) pairs: (11,9),(12,8), etc. — not sufficient.
Step 2: Analyse Statement II alone
l=b+4 gives one equation in two unknowns — infinitely many solutions. Not sufficient.
Step 3: Combine both statements
l+bl−b=20=4
Adding: 2l=24⇒l=12 cm, b=8 cm.
Area=12×8=96 cm2
A unique area is obtained.
Answer:
Both statements together are sufficient to find the area; neither alone is sufficient.
Area=96cm2
Quick Tip:
Perimeter gives l+b; the extra relation gives l−b. Together they form a solvable system — a classic data sufficiency pattern.
Q.31Medium
In a class of 60 students, how many students play neither Cricket nor Football?
Statement I: 35 students play Cricket and 30 students play Football. Statement II: 15 students play both Cricket and Football.
Answer: C
Understanding:
We need to find the number of students who play neither sport in a class of 60.
•Statement I: ∣C∣=35, ∣F∣=30
•Statement II: ∣C∩F∣=15
Formula:
By the inclusion-exclusion principle:
∣C∪F∣=∣C∣+∣F∣−∣C∩F∣
Neither=Total−∣C∪F∣
Step 1: Analyse Statement I alone
We know ∣C∣+∣F∣=65, but without ∣C∩F∣, we cannot find ∣C∪F∣. Not sufficient.
Step 2: Analyse Statement II alone
Knowing only ∣C∩F∣=15 without ∣C∣ and ∣F∣ is insufficient.
Step 3: Combine both statements
∣C∪F∣Neither=35+30−15=50=60−50=10
100%
Answer:
Both statements together give a unique answer; neither alone is sufficient.
Neither=10students
Quick Tip:
For set-based sufficiency questions, inclusion-exclusion needs three values: ∣A∣, ∣B∣, and ∣A∩B∣. Any two of these three is typically insufficient.
Q.32Medium
What is the value of x?
Statement I: x2=49 Statement II: x>0
Answer: C
Understanding:
We need a unique value of x.
•Statement I: x2=49
•Statement II: x>0
Formula:
x2=49⇒x=±7
Step 1: Analyse Statement I alone
x2=49 gives x=7 or x=−7. Two solutions — not sufficient for a unique value.
Step 2: Analyse Statement II alone
x>0 gives infinitely many positive values. Not sufficient.
Step 3: Combine both statements
From Statement I: x∈{7,−7}.
Applying Statement II (x>0): only x=7 qualifies.
x=7
Answer:
Both statements together uniquely determine x; neither alone is sufficient.
x=7
Quick Tip:
Whenever a quadratic equation gives two roots (positive and negative), a sign condition from a second statement is typically what makes the answer unique — a very common data sufficiency pattern.
Q.33Medium
A shopkeeper sold an article. What was his profit percentage?
Statement I: The cost price of the article was ₹500. Statement II: The shopkeeper earned a profit of ₹125 on the article.
Answer: C
Understanding:
We need to find the profit percentage.
•Statement I: CP=₹500
•Statement II: Profit=₹125
Formula:
Profit%=CPProfit×100
Step 1: Analyse Statement I alone
Knowing only CP=₹500 without the profit amount is insufficient to compute profit percentage.
Step 2: Analyse Statement II alone
Knowing only Profit=₹125 without the cost price is insufficient.
Step 3: Combine both statements
Profit%=500125×100=41×100=25%
Answer:
Both statements together are sufficient; neither alone is sufficient.
Profit%=25%
Quick Tip:
Profit percentage always needs both the profit amount and the cost price. If a statement gives only one of these, it is never independently sufficient.
Q.34Medium
Is qp a terminating decimal, where p and q are positive integers and the fraction is in its lowest terms?
Statement I: q=40 Statement II: p=3
Answer: A
Understanding:
We need to determine if qp (in lowest terms) is a terminating decimal.
•Statement I: q=40
•Statement II: p=3
Formula:
A fraction qp in its lowest terms is a terminating decimal if and only if the denominator q has no prime factors other than 2 and 5:
q=2m×5n(m,n≥0)
Step 1: Analyse Statement I alone
Factorise q=40:
40=23×51
The only prime factors are 2 and 5, so 40p in lowest terms is always a terminating decimal, regardless of the value of p. Statement I alone is sufficient.
Step 2: Analyse Statement II alone
Knowing p=3 tells us nothing about the denominator's prime factors. For example, 43 terminates but 73 does not. Not sufficient.
Answer:
Statement I alone is sufficient to determine that the fraction is a terminating decimal.
Statement I alone is sufficient.
Quick Tip:
For terminating decimal questions, the numerator is irrelevant — only the prime factorisation of the denominator (after full reduction) matters. If q only has factors of 2 and 5, the answer is always yes.
Q.35Medium
What is the average speed of a train for an entire journey?
Statement I: The train covers the first half of the journey at 60 km/h. Statement II: The train covers the second half of the journey at 90 km/h.
Answer: C
Understanding:
We need to find the average speed for the entire journey where the two halves are covered at different speeds.
•Statement I: Speed for first half =60 km/h
•Statement II: Speed for second half =90 km/h
Formula:
When equal distances are covered at two different speeds v1 and v2, the average speed is the harmonic mean:
Average speed=v1+v22v1v2
Step 1: Analyse Statement I alone
Only the speed for the first half is known. Without the second half's speed, average speed cannot be determined. Not sufficient.
Step 2: Analyse Statement II alone
Only the speed for the second half is known. Not sufficient.
Step 3: Combine both statements
Average speed=60+902×60×90=15010800=72 km/h
Answer:
Both statements together uniquely determine the average speed; neither alone is sufficient.
Average speed=72km/h
Quick Tip:
For equal-distance two-speed problems, always use the harmonic mean formula. The arithmetic mean 260+90=75 is a common wrong answer — the correct value is always less.
Q.36Medium
How many days does it take Person A alone to complete a work?
Statement I: A and B together can complete the work in 12 days. Statement II: B alone can complete the work in 30 days.
Answer: C
Understanding:
We need to find the number of days A alone takes to complete the work.
•Statement I: (A+B) together complete in 12 days
•Statement II: B alone completes in 30 days
Formula:
Using the work-rate principle:
A1=A+B1−B1
where A and B represent the number of days each person takes alone.
Step 1: Analyse Statement I alone
A1+B1=121. Without knowing B, we cannot find A. Not sufficient.
Step 2: Analyse Statement II alone
Only B1=301 is known. Without A+B combined rate, we cannot find A. Not sufficient.
Step 3: Combine both statements
A1=121−301=605−602=603=201
So A alone completes the work in 20 days.
Answer:
Both statements together uniquely determine the answer; neither alone is sufficient.
A alone completes in 20 days
Quick Tip:
In work problems, the combined rate minus one person's rate gives the other person's rate. Always convert days to rates (days1) before adding or subtracting.
Q.37Medium
What is the compound interest earned on a principal amount after 2 years?
Statement I: The rate of interest is 10% per annum, compounded annually. Statement II: The simple interest on the same principal at the same rate for 2 years is ₹800.
Answer: C
Understanding:
We need to find the compound interest earned after 2 years.
•Statement I: Rate r=10% p.a. compounded annually
•Statement II: Simple Interest for 2 years =₹800
Formula:
SI=100P×r×t,CI=P(1+100r)t−P
Step 1: Analyse Statement I alone
Knowing only r=10% without the principal P is insufficient to calculate CI. Not sufficient.
Step 2: Analyse Statement II alone
From SI =800: 100P×r×2=800, but without r, we cannot find P. Not sufficient.
Both statements together are sufficient; neither alone is sufficient.
CI=₹840
Quick Tip:
A useful shortcut: for 2 years, CI−SI=4PSI2, or equivalently CI=SI+100r×SI1 where SI1 is the first year's interest. Here, CI=800+110%×400=800+40=840.
Q.38Medium
The table below shows the monthly production (in units) of a factory for 6 months. January = 1200, February = 1350, March = ?, April = 1500, May = 1650, June = 1800. If the average monthly production for all 6 months is 1550 units, what is the production in March?
Answer: C
Understanding:
We need to find the missing value for March given that the average of all six months is 1550 units.
•Known months: January =1200, February =1350, April =1500, May =1650, June =1800
•Average =1550 units, Number of months =6
Formula:
Average=Number of valuesSum of all values
Step 1: Find the total sum using the average.
Total Sum=Average×Number of months=1550×6=9300
Step 2: Find the sum of the known months.
Known Sum=1200+1350+1500+1650+1800=7500
Step 3: Find the missing value for March.
March=Total Sum−Known Sum=9300−7500=1850
Answer:
The production in March is 1850 units.
March=1850 units
Quick Tip:
Whenever one value is missing from an average problem, use: Missing value = (Average ×n) − (Sum of known values). This avoids setting up a full equation with x.
Q.39Medium
The table shows the sales (in ₹ lakhs) of a company over five years. 2018 = 42, 2019 = 48, 2020 = ?, 2021 = 56, 2022 = 62. If the average annual sales over the five years is ₹53 lakhs, what is the missing sales figure for 2020?
Answer: B
Understanding:
We need to determine the missing sales value for 2020 given the average of all five years.
•Known values: 2018=42, 2019=48, 2021=56, 2022=62 (all in ₹ lakhs)
•Average =53 lakhs, n=5
Formula:
Missing Value=(Average×n)−Sum of known values
Step 1: Compute the total required sum.
Total Sum=53×5=265
Step 2: Compute the sum of known values.
Known Sum=42+48+56+62=208
Step 3: Find the missing value.
2020 Sales=265−208=57
Answer:
The missing sales figure for 2020 is ₹57 lakhs.
2020 Sales=₹57 lakhs
Q.40Medium
A bar chart shows the number of students who passed an exam in five sections: A = 45, B = 52, C = ?, D = 60, E = 48. The total number of students who passed across all five sections is 265. How many students passed in Section C?
Answer: B
Understanding:
We need to find the number of students who passed in Section C.
•Section A =45, B =52, D =60, E =48
•Total students passed =265
Formula:
Missing Value=Total−Sum of known values
Step 1: Sum the known sections.
Known Sum=45+52+60+48=205
Step 2: Find the value for Section C.
Section C=265−205=60
Answer:
The number of students who passed in Section C is 60.
Section C=60 students
Quick Tip:
In bar chart missing-data problems, always verify the total by adding all values including the one you found — a quick sanity check that takes only seconds.