Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.
What is the value of \(\rm\frac{\sin (A+B)}{\sin A \cos B}\)?
1 + cot A tan B
1 + tan A cot B
1 – sin A cos B
1 − cot A tan B
Correct answer
1 + cot A tan B
In the given figure, the circle with centre O has radius 10 cm. The radius of the circle with centre P is x. STR is a common tangent to the two circles at points R and S as shown in the figure. RT = 16 cm and TS = 24 cm. What is the value of x (in cm)?
15
16
12
18
Correct answer
15
The following pie chart shows the number of workers of different categories A, B, C, D, E, F, G and H of a factory in 1995.
If the number of workers in category B is increased by 10% and the number of workers in category D is decreased by 5% in 1996, then what is the total number of workers in categories B and D in 1996, if no. of workers in B was 1500 in the year 1995?
4025
4500
5000
4000
Correct answer
4025
In an election between two candidates, 65% of the voters cast their votes, out of which 3% of the votes were decided to be invalid. A candidate got 81965 votes which are 65% of the total valid votes. What is the total number of votes enrolled in that election?
200000
190000
215000
185000
Correct answer
200000
A man goes from place A to B at a speed of 12 km/h and returns from B to A at a speed of 18 km/h. His average speed for the entire journey is:
14\(\frac{2}{5}\) km/h
15 km/h
6 km/h
30 km/h
Correct answer
14\(\frac{2}{5}\) km/h
After deducting 20% from a certain sum, and then 10% from the remainder, there is Rs. 7,200 left. Find the original sum.
Rs. 9,000
Rs. 10,000
Rs. 7,500
Rs. 7,800
Correct answer
Rs. 9,000
Study the given bar-graph and answer the question that follows.
The bar-graph shows the number of employees recruited (in lakhs) by three different companies in five different years.
The number of employees recruited in company B in the year 2019 was what percentage of the number of employees recruited in company C in the year 2021 (correct up to 2 decimal places)?
88.85%
85.25%
105.88%
102.35%
Correct answer
88.85%
Find the largest number which completely divides 24, 56 and 96.
4
1
8
2
Correct answer
4
The perimeter of a rectangle is equal to the perimeter of a square. If the length and the breadth of the rectangle are 10 cm and 8 cm, respectively, then what will be the area of the square?
72 cm2
98 cm2
81 cm2
64 cm2
Correct answer
72 cm2
The value of 23 cosec2A − 23 cot2A is:
23
1
27
0
Correct answer
23
The given pie chart represents the popularity of ice-cream flavours in the year 2015.
In 2015, if the total sale of vanila flavour is for Rs. 3,300 then the total sale (in Rs.) for chocolate flavour is:
Rs. 5,160
Rs. 4,100
Rs. 4,120
Rs. 3,320
Correct answer
Rs. 5,160
Why this answer is correct
Given:
The total sale of vanilla flavour is for Rs. 3,300
Percentage of Vanilla is 16.50%
Percentage of chocolate flavour is 25.80%
Concept Used:
Concept of the unitary method.
Calculation:
16.50% → Rs. 3300
⇒ 1% → (3300/16.50)
⇒ 25.80% → (3300/16.50) × 25.80
⇒ 25.80% → 5160
∴ The total sale (in Rs.) for chocolate flavour is Rs. 5160
If \(a^2 + b^2 + c^2+ 216 = 12(a+b-2c)\), then \(\sqrt{ab-bc+ca}\) is:
6
4
8
3
Correct answer
6
Why this answer is correct
Given:
a2 + b2 + c2 + 216 = 12(a + b – 2c)
Identity used:
(a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab
Calculation:
a2 + b2 + c2 + 216 = 12(a + b – 2c)
⇒ a2 – 12a + b2 – 12b + c2 + 24c + 216 = 0
⇒ a2 – 12a + 36 + b2 – 12b + 36 + c2 + 24c + 144 = 0
⇒ (a – 6)2 + (b – 6)2 + (c + 12)2 = 0
⇒ a = 6, b = 6, c = –12
\(\sqrt{ab-bc+ca}\)
⇒ \(\sqrt{6\times 6-(6 \times (-12)+(-12 \times 6)}\)
⇒ √36
⇒ 6
∴ \(\sqrt{ab-bc+ca}\) is 6
The value of 8 - 3 ÷ 6 of \(2+\left(4 \div 4 \ of \ \frac{1}{4}\right)\div 8 + \left(4\times 8 \div \frac{1}{4}\right)\times \frac{1}{8}\) is:
\(-\dfrac{97}{4}\)
\(\dfrac{7}{4}\)
\(-\dfrac{7}{4}\)
\(\dfrac{97}{4}\)
Correct answer
\(\dfrac{97}{4}\)
Why this answer is correct
Given:
8 - 3 ÷ 6 of \(2+\left(4 \div 4 \ of \ \frac{1}{4}\right)\div 8 + \left(4\times 8 \div \frac{1}{4}\right)\times \frac{1}{8}\)
Concept Used:
Follow the BODMAS rule to solve this question, as per the order given below,
Calculation:
8 - 3 ÷ 6 of 2 + (4 ÷ 4 of 1/4) ÷ 8 + (4 × 8 ÷ 1/4) × 1/8
⇒ 8 - 3 ÷ 12 + (4 ÷ 1) ÷ 8 + (4 × 8 × 4) × 1/8
⇒ 8 - 1/4 + 4 ÷ 8 + 16
⇒ 8 - 1/4 + 1/2 + 16
⇒ (32 - 1 + 2 + 64)/4
⇒ 97/4
∴ The Required value is 97/4
The given bar graph shows the number of students of two schools over a period of six years.
In the bar graph, in which year is the sum of the students from schools A and B taken together, the minimum?
2010
2011
2012
2013
Correct answer
2010
Why this answer is correct
Given:
| 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | |
|---|---|---|---|---|---|---|
| A | 640 | 800 | 500 | 700 | 900 | 750 |
| B | 550 | 820 | 600 | 750 | 500 | 480 |
Calculation:
Total students in 2008 = 640 + 550 = 1190
Total students in 2009 = 800 + 820 = 1620
Total students in 2010 = 500 + 600 = 1100
Total students in 2011 = 700 + 750 = 1450
Total students in 2012 = 900 + 500 = 1400
Total students in 2013 = 750 + 480 = 1230
Minimum number of students together in A and B is 1100 in year 2010
∴ Minimum number of students is in year 2010
Two pipes A and B can fill a tank in 15 hours and 18 hours, respectively. Both pipes are opened simultaneously to fill the tank. In how many hours will the empty tank be filled?
\(8 \frac{2}{11}\)
\(7 \frac{2}{11}\)
\(9 \frac{2}{11}\)
\(10 \frac{2}{11}\)
Correct answer
\(8 \frac{2}{11}\)
Why this answer is correct
Given:
Pipe A can fill in 15 hours.
Pipe B can fill in 18 hours.
Calculation:
Tank filled by Pipe A in 1 hour
⇒ 1/15
Tank filled by Pipe B in 1 hour
⇒ 1/18
Tank filled by both Pipes in 1 hour
⇒ (1/15) + (1/18)
⇒ 11/90
Time taken By both pipes to fill the tank
⇒ 90/11
\(\Rightarrow 8\frac{2}{{11}};{\rm{hours}}\)
∴ Time taken by both pipes is \(8\frac{2}{{11}};{\bf{hours}}\).
In ΔABC, BD ⊥ AC at D, E is a point on BC such that ∠BEA = x°. If ∠EAC = 46° and ∠EBD = 60°, then the value of x is:
76°
72°
78°
68°
Correct answer
76°
Why this answer is correct
Given:
In ΔABC, BD ⊥ AC at D, E is a point on BC such that ∠BEA = x°. ∠EAC = 46° and ∠EBD = 60°
Concept Used:
The Sum of all three angles of a triangle is 180°
Calculation:
Given BD ⊥ AC at D
∠BDC = 90°
In ΔBDC,
∠EBD = 60° and ∠BDC = 90°
∠DCB = 180° - (60° + 90°)
⇒ ∠DCB = 30°
In ΔAEC,
∠EAC = 46° and ∠ACE = 30°
∠AEC = 180° - (46° + 30°)
⇒ ∠AEC = 104°
∠BEA = 180° - ∠AEC
⇒ ∠BEA = 180° - 104°
⇒ ∠BEA = 76°
∴ The value of x is 76°
The average weight of A, B and C is 65 kg. If the average weight of A and B is 63.5 kg, and the average weight of A and C is 67.5 kg, then the weight of A (in kg) is:
67
65
68
60
Correct answer
67
Why this answer is correct
Shortcut Trick
Using the deviation method relative to the total average of 65 kg:
Net deviation for (A + B) = 2 × (63.5 − 65) = −3 kg.
Net deviation for (A + C) = 2 × (67.5 − 65) = +5 kg.
Since A is the common element in both subsets, its weight is calculated by adding these deviations to the original average.
Weight of A = 65 + (−3 + 5) = 65 + 2 = 67 kg.
∴ The correct answer is 67 kg.
Alternate Method
Given:
Average weight of A, B, and C = 65 kg
Average weight of A and B = 63.5 kg
Average weight of A and C = 67.5 kg
Formula Used:
Sum of values = Average × Number of values
Calculations:
⇒ Total weight of (A + B + C) = 65 × 3 = 195 kg
⇒ Total weight of (A + B) = 63.5 × 2 = 127 kg
⇒ Total weight of (A + C) = 67.5 × 2 = 135 kg
⇒ Sum of (A + B) and (A + C) = 127 + 135 = 262 kg
⇒ This sum represents (A + B + C + A). Therefore:
⇒ Weight of A = (A + B + A + C) − (A + B + C)
⇒ Weight of A = 262 − 195 = 67 kg
∴ The correct answer is 67 kg.
Additional Information
Average Concept
The Average is the sum of observations divided by the total number of observations. Formula: Avg = Σx ÷ n.
Deviation Principle
The algebraic sum of deviations of all observations from their arithmetic mean is always zero. This is used to find missing values in datasets.
Weighted Average
When combining two groups, the combined average = (n1x1 + n2x2) ÷ (n1 + n2), where n is count and x is the average.
A person sold an article at a loss of 16%. Had he sold it for Rs. 660 more, he would have gained 8%. What should be the selling price (in Rs.) to gain a profit of 12%?
2,970
3,200
2,750
3,080
Correct answer
3,080
Why this answer is correct
Given:
The person sold the article at a loss of 16% If he sold it for Rs. 660 more he have gained 8%
Concept Used:
Selling price = Cost Price + Profit
Calculation:
Accordingly,
(16 + 8)% → Rs. 660
⇒ 100% → (660/24) × 100
⇒ 100% → 2750
The cost price of the article is Rs. 2750
To get a profit of 12% net profit will be (12/100) × 2750
⇒ 330
The selling price should be (2750 + 330) = 3080
Rs. 3080 should be the selling price to gain a profit of 12%
The given histogram shows the height of the students.
The difference between the number of students whose height is between 150-155 cm and the number of students whose height lies between 175-180 cm is:
8
9
3
7
Correct answer
9
Why this answer is correct
Given:
The number of students whose height is between 150-155 cm is 15
The number of students whose height is between 175-180 cm is 6
Calculation:
The required difference is (15 - 6)
⇒ 9
∴ The difference between the number of students whose height is between 150-155 cm and the number of students whose height lies between 175-180 cm is 9
The value of \(\dfrac{40-\dfrac{3}{4} \ of \ 32}{37 - \dfrac{3}{4} \ of \ (34-6)}\) is :
0
–1/2
1
1/2
Correct answer
1
Why this answer is correct
Concept used:
Follow BODMAS rule to solve this question, as per the order given below,
Now,
Calculation:
\(\dfrac{40-\dfrac{3}{4} \ of \ 32}{37 - \dfrac{3}{4} \ of \ (34-6)}\)
⇒ \(\dfrac{40-\dfrac{3}{4} \times 32}{37 - \dfrac{3}{4} \ of \ 28}\)
⇒ (40 – 24)/(37 – 21)
⇒ 16/16
⇒ 1
∴ The required value is 1
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.