Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.
What should come in place of (?) in the given series?
17, 37, 77, 137, 217, (?)
347
357
327
317
307
Correct answer
317
Why this answer is correct
The series follows the following pattern
⇒ 17 + 20 = 37
⇒ 37 + 40 = 77
⇒ 77 + 60 = 137
⇒ 137 + 80 = 217
⇒ 217 + 100 = 317
Hence, the correct answer is 317.
What should come in the place of (?) in the given series?
2000, 200, 40, 12, 4.8, ?
2.2
1.8
1.6
1.2
2.4
Correct answer
2.4
Why this answer is correct
Solution :
The series follows the following pattern :
⇒2000 × 0.1 = 200
⇒200 × 0.2 = 40
⇒40 × 0.3 = 12
⇒12 × 0.4 = 4.8
⇒4.8 × 0.5 = 2.4
⇒ The next term in the series will be 2.4.
Hence, 2.4 is the correct answer.
A sum when invested at certain rate of simple interest for 3 years gives interest equals to 3/4th of the sum. Find the amount received when Rs. 1500 is invested at the same rate for 2 years at compound interest, compounded annually.
Rs. 2343.75
Rs. 3245.15
Rs. 2509.52
Rs. 3015.08
Rs. 2287.65
Correct answer
Rs. 2343.75
Why this answer is correct
Shortcut Trick
Simple Interest rate: 3/4 of sum in 3 years → 1/4 of sum per year → 25% per annum.
For Compound Interest at 25% for 2 years, the Multiplying Factor is (1.25)2 = 1.5625.
Amount = 1500 × 1.5625 = 2343.75.
∴ The correct answer is Rs. 2343.75.
Alternate Method
Given: Time (T1) = 3 years, SI = 3/4 × P1, P2 = Rs. 1500, T2 = 2 years.
Formula Used: SI = (P × R × T)/100; Amount (CI) = P(1 + R/100)n
⇒ 3/4 × P = (P × R × 3)/100
⇒ 1/4 = R/100
⇒ Rate (R) = 25%
⇒ Amount = 1500 × (1 + 25/100)2
⇒ Amount = 1500 × (5/4) × (5/4)
⇒ Amount = 1500 × 25/16
⇒ Amount = 37500 / 16 = 2343.75
∴ The correct answer is Rs. 2343.75.
Additional Information
Compound Interest Formula
The total amount A is calculated as A = P(1 + r/n)nt, where n is the number of times interest is compounded per year.
Effective Rate for 2 Years
For a rate R, the effective CI rate for 2 years is R + R + (R×R)/100. For 25%, it is 25 + 25 + 6.25 = 56.25%.
SI vs CI Difference
Difference between CI and SI for 2 years is P(R/100)2.
Mixture P of lime juice and water contains 45% lime juice and rest water. x ml of lime juice and 2x ml of water is added to mixture P such that ratio of lime juice to water in the resultant mixture becomes 15:23. Find the ratio of the initial quantity of mixture P to x.
5:7
6:11
3:7
11:7
10:3
Correct answer
10:3
Why this answer is correct
Shortcut Trick
Let the initial mixture P be 100 units.
Initial Lime Juice = 45 units, Water = 55 units.
Ratio: (45 + x) / (55 + 2x) = 15 / 23
Cross-multiplying: 23(45 + x) = 15(55 + 2x) → 1035 + 23x = 825 + 30x → 7x = 210 → x = 30.
Ratio of P to x = 100 : 30 = 10 : 3.
∴ The correct answer is 10:3.
Alternate Method
Given:
Initial Lime Juice in P = 45%
Initial Water in P = 100% − 45% = 55%
Added Lime Juice = x, Added Water = 2x
Final Ratio = 15 : 23
Calculations:
Let the total initial quantity of Mixture P be Q.
⇒ Lime Juice in P = 0.45Q
⇒ Water in P = 0.55Q
⇒ (0.45Q + x) / (0.55Q + 2x) = 15 / 23
⇒ 23 × (0.45Q + x) = 15 × (0.55Q + 2x)
⇒ 10.35Q + 23x = 8.25Q + 30x
⇒ 10.35Q − 8.25Q = 30x − 23x
⇒ 2.1Q = 7x
⇒ Q / x = 7 / 2.1 = 70 / 21 = 10 / 3
∴ The correct answer is 10:3.
Additional Information
Mixture and Alligation
When two ingredients are mixed, the ratio of quantities is inversely proportional to the difference between their individual concentrations and the mean concentration.
Ratio and Proportion
If a/b = c/d, then by cross-multiplication ad = bc. This property is vital for solving linear equations involving ratios.
Percentage to Ratio Conversion
A concentration of 45% can be written as the fraction 45/100 or the ratio 9:11 (9 parts juice to 11 parts water).
Directions: The below table shows the average number of males and females , difference between males and females, total number of boys and girls having chocolates and number of boys having chocolates in five different schools.
| Schools | Average Number of boys and girls | Difference between number of boys and girls | Total number of boys and girls having chocolate | Number of boys having chocolate |
|---|---|---|---|---|
| A | 275 | 50 | 250 | 150 |
| B | 375 | 50 | 300 | 200 |
| C | 375 | 150 | 400 | 150 |
| D | 475 | 50 | 350 | 200 |
| E | 225 | 50 | 300 | 100 |
Note:
i) If in only school B, the number of boys is less than the number of girls in the same school.
Total number of chocolate having students in school E is how much percent more than the total number of girls present in the school E.
45%
50%
75%
60%
40%
Correct answer
50%
Why this answer is correct
General Solution:
School A:
Total number of boys and girls in school A = 275 × 2 = 550
Difference between Number of boys and girls is 50
B + G = 550
B - G = 50
2B = 600
B = 300
Number of boys in school = 300
Number of girls in school = 550 - 300 = 250
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 250 - 150 = 100
School B:
Total number of boys and girls in school B = 375 × 2 = 750
Difference between Number of boys and girls is 50
G + B = 750
G - B = 50
2G = 800
G = 400
Number of girls in school = 400
Number of boys in school = 750 - 400 = 350
Number of boys and girls having chocolate = 200
Number of girls having chocolate = 300 - 200 = 100
School C:
Total number of boys and girls in school C = 375 × 2 = 750
Difference between Number of boys and girls is 150
B + G = 750
B - G = 150
2B = 900
B = 450
Number of boys in school = 450
Number of girls in school = 750 - 450 = 300
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 400 - 150 = 250
School D:
Total number of boys and girls in school D = 475 × 2 = 950
Difference between Number of boys and girls is 50
B + G = 950
B - G = 50
2B = 1000
B = 500
Number of boys in school = 500
Number of girls in school = 950 - 500 = 450
Number of boys and girls having chocolate = 350
Number of girls having chocolate = 350 - 200 = 150
School E:
Total number of boys and girls in school E = 225 × 2 = 450
Difference between Number of boys and girls is 50
B + G = 450
B - G = 50
2B = 500
B = 250
Number of boys in school = 250
Number of girls in school = 450 - 250 = 200
Number of boys and girls having chocolate = 100
Number of girls having chocolate = 300 - 100 = 200
| Schools | Boys in school | Girls in school | Boys having chocolate | Girls having chocolate |
|---|---|---|---|---|
| A | 300 | 250 | 150 | 100 |
| B | 350 | 400 | 200 | 100 |
| C | 450 | 300 | 150 | 250 |
| D | 500 | 450 | 200 | 150 |
| E | 250 | 200 | 100 | 200 |
Solution:
Number of chocolate having students in school E = (200 + 100) = 300
Number of girls in school E = 200
Required percentage = (300 - 200)/200 × 100 = 50%
Directions: The below table shows the average number of males and females , difference between males and females, total number of boys and girls having chocolates and number of boys having chocolates in five different schools.
| Schools | Average Number of boys and girls | Difference between number of boys and girls | Total number of boys and girls having chocolate | Number of boys having chocolate |
|---|---|---|---|---|
| A | 275 | 50 | 250 | 150 |
| B | 375 | 50 | 300 | 200 |
| C | 375 | 150 | 400 | 150 |
| D | 475 | 50 | 350 | 200 |
| E | 225 | 50 | 300 | 100 |
Note:
i) If in only school B, the number of boys is less than the number of girls in the same school.
If the number of boys and girls in school D is increased by 40% and 20% respectively. Find the difference between the chocolate required to all boys and girls in school D.
170
140
110
130
150
Correct answer
110
Why this answer is correct
General Solution:
School A:
Total number of boys and girls in school A = 275 × 2 = 550
Difference between Number of boys and girls is 50
B + G = 550
B - G = 50
2B = 600
B = 300
Number of boys in school = 300
Number of girls in school = 550 - 300 = 250
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 250 - 150 = 100
School B:
Total number of boys and girls in school B = 375 × 2 = 750
Difference between Number of boys and girls is 50
G + B = 750
G - B = 50
2G = 800
G = 400
Number of girls in school = 400
Number of boys in school = 750 - 400 = 350
Number of boys and girls having chocolate = 200
Number of girls having chocolate = 300 - 200 = 100
School C:
Total number of boys and girls in school C = 375 × 2 = 750
Difference between Number of boys and girls is 150
B + G = 750
B - G = 150
2B = 900
B = 450
Number of boys in school = 450
Number of girls in school = 750 - 450 = 300
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 400 - 150 = 250
School D:
Total number of boys and girls in school D = 475 × 2 = 950
Difference between Number of boys and girls is 50
B + G = 950
B - G = 50
2B = 1000
B = 500
Number of boys in school = 500
Number of girls in school = 950 - 500 = 450
Number of boys and girls having chocolate = 350
Number of girls having chocolate = 350 - 200 = 150
School E:
Total number of boys and girls in school E = 225 × 2 = 450
Difference between Number of boys and girls is 50
B + G = 450
B - G = 50
2B = 500
B = 250
Number of boys in school = 250
Number of girls in school = 450 - 250 = 200
Number of boys and girls having chocolate = 100
Number of girls having chocolate = 300 - 100 = 200
| Schools | Boys in school | Girls in school | Boys having chocolate | Girls having chocolate |
|---|---|---|---|---|
| A | 300 | 250 | 150 | 100 |
| B | 350 | 400 | 200 | 100 |
| C | 450 | 300 | 150 | 250 |
| D | 500 | 450 | 200 | 150 |
| E | 250 | 200 | 100 | 200 |
Solution:
Number of boys in school D = 500 × 140/100 = 700
Number of girls in school D = 450 × 120/100 = 540
Number of chocolate required to boys = 700 - 200 = 500
Number of chocolate required to girls = 540 - 150 = 390
Required difference = (500 - 390) = 110
Directions: The below table shows the average number of males and females , difference between males and females, total number of boys and girls having chocolates and number of boys having chocolates in five different schools.
| Schools | Average Number of boys and girls | Difference between number of boys and girls | Total number of boys and girls having chocolate | Number of boys having chocolate |
|---|---|---|---|---|
| A | 275 | 50 | 250 | 150 |
| B | 375 | 50 | 300 | 200 |
| C | 375 | 150 | 400 | 150 |
| D | 475 | 50 | 350 | 200 |
| E | 225 | 50 | 300 | 100 |
Note:
i) If in only school B, the number of boys is less than the number of girls in the same school.
If new 200 chocolates are distributed to boys and girls according to the ratio present in the school C. Find the total number of boys having chocolates in School C at present.
340
330
270
310
290
Correct answer
270
Why this answer is correct
General Solution:
School A:
Total number of boys and girls in school A = 275 × 2 = 550
Difference between Number of boys and girls is 50
B + G = 550
B - G = 50
2B = 600
B = 300
Number of boys in school = 300
Number of girls in school = 550 - 300 = 250
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 250 - 150 = 100
School B:
Total number of boys and girls in school B = 375 × 2 = 750
Difference between Number of boys and girls is 50
G + B = 750
G - B = 50
2G = 800
G = 400
Number of girls in school = 400
Number of boys in school = 750 - 400 = 350
Number of boys and girls having chocolate = 200
Number of girls having chocolate = 300 - 200 = 100
School C:
Total number of boys and girls in school C = 375 × 2 = 750
Difference between Number of boys and girls is 150
B + G = 750
B - G = 150
2B = 900
B = 450
Number of boys in school = 450
Number of girls in school = 750 - 450 = 300
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 400 - 150 = 250
School D:
Total number of boys and girls in school D = 475 × 2 = 950
Difference between Number of boys and girls is 50
B + G = 950
B - G = 50
2B = 1000
B = 500
Number of boys in school = 500
Number of girls in school = 950 - 500 = 450
Number of boys and girls having chocolate = 350
Number of girls having chocolate = 350 - 200 = 150
School E:
Total number of boys and girls in school E = 225 × 2 = 450
Difference between Number of boys and girls is 50
B + G = 450
B - G = 50
2B = 500
B = 250
Number of boys in school = 250
Number of girls in school = 450 - 250 = 200
Number of boys and girls having chocolate = 100
Number of girls having chocolate = 300 - 100 = 200
| Schools | Boys in school | Girls in school | Boys having chocolate | Girls having chocolate |
|---|---|---|---|---|
| A | 300 | 250 | 150 | 100 |
| B | 350 | 400 | 200 | 100 |
| C | 450 | 300 | 150 | 250 |
| D | 500 | 450 | 200 | 150 |
| E | 250 | 200 | 100 | 200 |
Solution:
Ratio of number of boys to girls present in school C = 450: 300 = 3:2
Number of chocolate distributed to boys = 200/5 × 3 = 120
Number of chocolate distributed to girls = (200 - 120) = 80
Number of boys having chocolate at present is (150 + 120) = 270
Directions: The below table shows the average number of males and females , difference between males and females, total number of boys and girls having chocolates and number of boys having chocolates in five different schools.
| Schools | Average Number of boys and girls | Difference between number of boys and girls | Total number of boys and girls having chocolate | Number of boys having chocolate |
|---|---|---|---|---|
| A | 275 | 50 | 250 | 150 |
| B | 375 | 50 | 300 | 200 |
| C | 375 | 150 | 400 | 150 |
| D | 475 | 50 | 350 | 200 |
| E | 225 | 50 | 300 | 100 |
Note:
i) If in only school B, the number of boys is less than the number of girls in the same school.
Find the ratio between the number of boys having chocolate in school B and C to the number of girls having chocolate in school D and E.
2: 1
1: 3
1: 1
2: 3
3: 2
Correct answer
1: 1
Why this answer is correct
General Solution:
School A:
Total number of boys and girls in school A = 275 × 2 = 550
Difference between Number of boys and girls is 50
B + G = 550
B - G = 50
2B = 600
B = 300
Number of boys in school = 300
Number of girls in school = 550 - 300 = 250
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 250 - 150 = 100
School B:
Total number of boys and girls in school B = 375 × 2 = 750
Difference between Number of boys and girls is 50
G + B = 750
G - B = 50
2G = 800
G = 400
Number of girls in school = 400
Number of boys in school = 750 - 400 = 350
Number of boys and girls having chocolate = 200
Number of girls having chocolate = 300 - 200 = 100
School C:
Total number of boys and girls in school C = 375 × 2 = 750
Difference between Number of boys and girls is 150
B + G = 750
B - G = 150
2B = 900
B = 450
Number of boys in school = 450
Number of girls in school = 750 - 450 = 300
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 400 - 150 = 250
School D:
Total number of boys and girls in school D = 475 × 2 = 950
Difference between Number of boys and girls is 50
B + G = 950
B - G = 50
2B = 1000
B = 500
Number of boys in school = 500
Number of girls in school = 950 - 500 = 450
Number of boys and girls having chocolate = 350
Number of girls having chocolate = 350 - 200 = 150
School E:
Total number of boys and girls in school E = 225 × 2 = 450
Difference between Number of boys and girls is 50
B + G = 450
B - G = 50
2B = 500
B = 250
Number of boys in school = 250
Number of girls in school = 450 - 250 = 200
Number of boys and girls having chocolate = 100
Number of girls having chocolate = 300 - 100 = 200
| Schools | Boys in school | Girls in school | Boys having chocolate | Girls having chocolate |
|---|---|---|---|---|
| A | 300 | 250 | 150 | 100 |
| B | 350 | 400 | 200 | 100 |
| C | 450 | 300 | 150 | 250 |
| D | 500 | 450 | 200 | 150 |
| E | 250 | 200 | 100 | 200 |
Solution:
Number of boys having chocolate from school B = 200
Number of boys having chocolate from school C = 150
Number of girls having chocolate from school D = 150
Number of girls having chocolate from school E = 200
Required ratio = 350: 350 = 1: 1
Directions: The below table shows the average number of males and females , difference between males and females, total number of boys and girls having chocolates and number of boys having chocolates in five different schools.
| Schools | Average Number of boys and girls | Difference between number of boys and girls | Total number of boys and girls having chocolate | Number of boys having chocolate |
|---|---|---|---|---|
| A | 275 | 50 | 250 | 150 |
| B | 375 | 50 | 300 | 200 |
| C | 375 | 150 | 400 | 150 |
| D | 475 | 50 | 350 | 200 |
| E | 225 | 50 | 300 | 100 |
Note:
i) If in only school B, the number of boys is less than the number of girls in the same school.
If the number of boys having chocolate in school A is increased by 20% and the number of girls having chocolate in school A is increased by 25%. Find the difference between the number of boys and girls not having chocolate in school A.
20
5
10
15
25
Correct answer
5
Why this answer is correct
General Solution:
School A:
Total number of boys and girls in school A = 275 × 2 = 550
Difference between Number of boys and girls is 50
B + G = 550
B - G = 50
2B = 600
B = 300
Number of boys in school = 300
Number of girls in school = 550 - 300 = 250
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 250 - 150 = 100
School B:
Total number of boys and girls in school B = 375 × 2 = 750
Difference between Number of boys and girls is 50
G + B = 750
G - B = 50
2G = 800
G = 400
Number of girls in school = 400
Number of boys in school = 750 - 400 = 350
Number of boys and girls having chocolate = 200
Number of girls having chocolate = 300 - 200 = 100
School C:
Total number of boys and girls in school C = 375 × 2 = 750
Difference between Number of boys and girls is 150
B + G = 750
B - G = 150
2B = 900
B = 450
Number of boys in school = 450
Number of girls in school = 750 - 450 = 300
Number of boys and girls having chocolate = 150
Number of girls having chocolate = 400 - 150 = 250
School D:
Total number of boys and girls in school D = 475 × 2 = 950
Difference between Number of boys and girls is 50
B + G = 950
B - G = 50
2B = 1000
B = 500
Number of boys in school = 500
Number of girls in school = 950 - 500 = 450
Number of boys and girls having chocolate = 350
Number of girls having chocolate = 350 - 200 = 150
School E:
Total number of boys and girls in school E = 225 × 2 = 450
Difference between Number of boys and girls is 50
B + G = 450
B - G = 50
2B = 500
B = 250
Number of boys in school = 250
Number of girls in school = 450 - 250 = 200
Number of boys and girls having chocolate = 100
Number of girls having chocolate = 300 - 100 = 200
| Schools | Boys in school | Girls in school | Boys having chocolate | Girls having chocolate |
|---|---|---|---|---|
| A | 300 | 250 | 150 | 100 |
| B | 350 | 400 | 200 | 100 |
| C | 450 | 300 | 150 | 250 |
| D | 500 | 450 | 200 | 150 |
| E | 250 | 200 | 100 | 200 |
Solution:
Number of boys having chocolate in school A = 150 × 120/100 = 180
Number of girls having chocolate in school A= 100 × 125/100 = 125
Number of boys not having chocolate in school A = (300 - 180) = 120
Number of girls not having chocolate in school A = (250 - 125) = 125
Required difference = (125 - 120) = 5
The cost price of 8 apples is equal to the selling price of 5 apples and cost price of 13 oranges is equal to selling price of 10 oranges, If the ratio of cost price of 1 apple to 1 orange is 5 : 2, Find the net profit percent on the sale of 8 apples and 10 oranges.
55%
40%
50%
45%
None of these
Correct answer
50%
Why this answer is correct
Shortcut Trick
Ratio of CP of 1 apple to 1 orange = 5 : 2. Let CP of 1 apple = 5 and 1 orange = 2.
Total CP of 8 apples and 10 oranges = (8 × 5) + (10 × 2) = 40 + 20 = 60.
Profit on 8 apples: CP of 8 = SP of 5 ⇒ SP of 8 = (8 ÷ 5) × CP of 8 = 1.6 × 40 = 64.
Profit on 10 oranges: CP of 13 = SP of 10 ⇒ SP of 10 = CP of 13 = 13 × 2 = 26.
Total SP = 64 + 26 = 90. Profit % = [(90 − 60) ÷ 60] × 100 = 50%.
∴ The correct answer is 50%.
Alternate Method
Given → 8 CPa = 5 SPa, 13 CPo = 10 SPo, CPa : CPo = 5 : 2.
Formulas: Profit = SP − CP | Profit % = (Profit ÷ CP) × 100.
⇒ Let CP of 1 apple = 5x and CP of 1 orange = 2x.
⇒ Total CP of 8 apples and 10 oranges = 8(5x) + 10(2x) = 40x + 20x = 60x.
⇒ For apples: 5 SPa = 8 CPa ⇒ SP of 5 apples = 8(5x) = 40x ⇒ SP of 1 apple = 8x.
⇒ SP of 8 apples = 8 × 8x = 64x.
⇒ For oranges: 10 SPo = 13 CPo ⇒ SP of 10 oranges = 13(2x) = 26x.
⇒ Total SP = 64x + 26x = 90x.
⇒ Net Profit % = [(90x − 60x) ÷ 60x] × 100 = (30 ÷ 60) × 100 = 50%.
∴ The correct answer is 50%.
Additional Information
Profit and Loss on Multiple Articles
When handling different items, find the total cost price and total selling price separately before calculating the overall profit percentage.
Ratio of CP and SP
If n CP = m SP, then the ratio of CP/SP = m/n. This allows for quick calculation of individual item profits.
Average of four numbers is 14. Out of four numbers, first number is two times of second number, second number is three times of third number, third number is one-fourth of fourth number. Find the fourth number.
12
13
16
17
18
Correct answer
16
Why this answer is correct
Shortcut Trick
Assume third number = 1 unit.
Second number = 3 units, First number = 6 units, Fourth number = 4 units.
Total units = 6 + 3 + 1 + 4 = 14 units.
Sum of numbers = 14 × 4 = 56.
14 units = 56 ⇒ 1 unit = 4.
Fourth number = 4 units = 4 × 4 = 16.
∴ The correct answer is 16.
Alternate Method
Given: Average of 4 numbers = 14. Relations between numbers are provided.
Formula: Sum = Average × Count of numbers.
⇒ Let the third number be x.
⇒ Second number = 3x.
⇒ First number = 2 × 3x = 6x.
⇒ Third number = (1/4) × Fourth number ⇒ Fourth number = 4x.
⇒ Average = (6x + 3x + x + 4x) ÷ 4 = 14.
⇒ 14x ÷ 4 = 14.
⇒ 14x = 56.
⇒ x = 4.
⇒ Fourth number = 4x = 4 × 4 = 16.
∴ The correct answer is 16.
Additional Information
Average Concept
Average is the sum of observations divided by the total number of observations. Formula: Sum = Average × Number of terms.
Ratio and Proportion
When multiple variables are related linearly, always assume the most central or smallest variable as a single unit (x) to simplify calculations.
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. x2 + 5x – 104 = 0
II. y2 + 30y + 224 = 0
x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.
Correct answer
x > y
Why this answer is correct
I. x2 + 5x – 104 = 0
⇒ x2 + 13x – 8x – 104 = 0
⇒ x(x + 13) – 8(x + 13) = 0
⇒ (x + 13) (x – 8) = 0
⇒ x = –13, 8
II. y2 + 30y + 224 = 0
⇒ y2 + 16y + 14y + 224 = 0
⇒ y(y + 16) + 14(y + 16) = 0
⇒ (y + 16) (y + 14) = 0
⇒ y = –16, –14
| Value of x | Value of y | Relation |
|---|---|---|
| –13 | –16 | x > y |
| –13 | –14 | x > y |
| 8 | –16 | x > y |
| 8 | –14 | x > y |
∴ x > y
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. 2x2 + 5x – 3 = 0
II. 2y2 – 5y + 2 = 0
x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between y and y cannot be established.
Correct answer
x ≤ y
Why this answer is correct
I. 2x2 + 5x – 3 = 0
⇒ 2x2 + 6x – x – 3 = 0
⇒ 2x (x + 3) – 1 (x + 3) = 0
⇒ (2x – 1) (x + 3) = 0,
⇒ x = 1/2, – 3
II. 2y2 – 5y + 2 = 0
⇒ 2y2 – 4y – y + 2 = 0
⇒ 2y (y – 2) – 1 (y – 2) = 0
⇒ (2y – 1) (y – 2) = 0
⇒ y = 1/2, 2
| Value of x | Value of y | Relation |
|---|---|---|
| 1/2 | 1/2 | x = y |
| 1/2 | 2 | x < y |
| – 3 | 1/2 | x < y |
| – 3 | 2 | x < y |
Hence, x ≤ y
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. x2 – 38x + 360 = 0
II. y2 + 10y + 24 = 0
x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.
Correct answer
x > y
Why this answer is correct
I. x2 – 38x + 360 = 0
⇒ x2– 20x – 18x + 360 = 0
⇒ (x – 18) (x – 20) = 0
⇒ x = 18, 20
II. y2+ 10y + 24 = 0
⇒ y2 + 6y + 4y + 2 = 0
⇒ (y + 4) (y + 6) = 0
⇒ y = –4, –6
| Value of x | Value of y | Relation |
|---|---|---|
| 18 | –4 | x > y |
| 18 | –6 | x > y |
| 20 | –4 | x > y |
| 20 | –6 | x > y |
Hence, x > y.
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. x2 – 20x + 99 = 0
II. y2 – 9y + 20 = 0
x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.
Correct answer
x > y
Why this answer is correct
I. x2 – 20x + 99 = 0
⇒ x2 – 11x – 9x + 99 = 0
⇒ x (x – 11) – 9 (x – 11) = 0
⇒ (x – 11) (x – 9) = 0
⇒ x = 11, 9
II. y2 – 9y + 20 = 0
⇒ y2 – 5y – 4y + 20 = 0
⇒ y (y – 5) – 4 (y – 5) = 0
⇒ (y – 5) (y – 4) = 0
⇒ y = 5, 4
| Value of x | Value of y | Relation |
|---|---|---|
| 11 | 5 | x > y |
| 11 | 4 | x > y |
| 9 | 5 | x > y |
| 9 | 4 | x > y |
Hence x > y
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. 6x2 – 17x + 5 = 0
II. 10y2 + 3y – 1 = 0
x > y
x < y
x ≥ y
x ≤ y
x = y or relationship between x and y cannot be established.
Correct answer
x > y
Why this answer is correct
I. 6x2 – 17x + 5 = 0
⇒ 6x2 – 15x – 2x + 5 = 0
⇒ 3x (2x – 5) – 1 (2x – 5) = 0
⇒ (2x – 5) (3x – 1) = 0
⇒ x = 5/2, 1/3
II. 10y2 + 3y – 1 = 0
⇒ 10y2 + 5y – 2y – 1 = 0
⇒ 5y (2y + 1) – 1 (2y + 1) = 0
⇒ (2y + 1) (5y – 1) = 0
⇒ y = -1/2, 1/5
Comparison between x and y (via Tabulation):
| Value of x | Value of y | Relation |
|---|---|---|
| 5/2 | -1/2 | x > y |
| 5/2 | 1/5 | x > y |
| 1/3 | -1/2 | x > y |
| 1/3 | 1/5 | x > y |
∴ x > y
Directions: Read the below information and answer the following questions.
The total number of Sarees sold in Shop A is 530. The number of Cotton Sarees sold in Shop C is 20% more than the number of Cotton Sarees sold in Shop A. The number of SilkSarees sold in Shop C is 55 less than the number of Cotton Sarees sold in Shop B. The number of Silk Sarees sold in Shop B is 20 more than the number of Cotton Sarees sold in Shop C. The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold in Shop C. Total number of Cotton Sarees sold in all three shops is 825.
Note:
Total number of Sarees sold = Number of Cotton Sarees sold + Number of Silk Sarees sold.
Find the difference between the number of Silk and Cotton Sarees sold in all three shops.
10
7
8
5
9
Correct answer
5
Why this answer is correct
General Solution:
The number of Cotton Sarees sold in Shop A = 5x
The number of Silk Sarees sold in Shop A = 530 – 5x
The number of Cotton Sarees sold in Shop C = 6x
The number of Cotton Sarees sold in Shop B = y
The number of Silk Sarees sold in Shop C = y – 55
The number of Silk Sarees sold in Shop B = 6x + 20
The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold
in Shop C
5x = 30 + y – 55
y = 5x + 25 ----------- (1)
The total number of Cotton Sarees sold in all three shops is 825
5x + y + 6x = 825
5x + 5x + 25 + 6x = 825
16x = 800
x = 50
Substitute x = 50 in equation (1)
y = 5 × 50 + 25
y = 275
Solution:
Total number of Silk Sarees sold = (280 + 320 + 220) = 820
Total number of Cotton Sarees sold = (250 + 275 + 300) = 825
Required difference = (825 – 820) = 5
Directions: Read the below information and answer the following questions.
The total number of Sarees sold in Shop A is 530. The number of Cotton Sarees sold in Shop C is 20% more than the number of Cotton Sarees sold in Shop A. The number of SilkSarees sold in Shop C is 55 less than the number of Cotton Sarees sold in Shop B. The number of Silk Sarees sold in Shop B is 20 more than the number of Cotton Sarees sold in Shop C. The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold in Shop C. Total number of Cotton Sarees sold in all three shops is 825.
Note:
Total number of Sarees sold = Number of Cotton Sarees sold + Number of Silk Sarees sold.
If the number of Cotton and Silk Sarees sold in Shop A is increased by 30% and 25% respectively, then find the total number of Sarees sold in Shop A.
525
675
575
475
355
Correct answer
675
Why this answer is correct
General Solution:
The number of Cotton Sarees sold in Shop A = 5x
The number of Silk Sarees sold in Shop A = 530 – 5x
The number of Cotton Sarees sold in Shop C = 6x
The number of Cotton Sarees sold in Shop B = y
The number of Silk Sarees sold in Shop C = y – 55
The number of Silk Sarees sold in Shop B = 6x + 20
The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold
in Shop C
5x = 30 + y – 55
y = 5x + 25 ----------- (1)
The total number of Cotton Sarees sold in all three shops is 825
5x + y + 6x = 825
5x + 5x + 25 + 6x = 825
16x = 800
x = 50
Substitute x = 50 in equation (1)
y = 5 × 50 + 25
y = 275
Solution:
Number of Cotton Sarees sold in Shop A = 250 × (130/100) = 325
Number of Silk Sarees sold in Shop A = 280 × (125/100) = 350
Required sum = 325 + 350 = 675
Directions: Read the below information and answer the following questions.
The total number of Sarees sold in Shop A is 530. The number of Cotton Sarees sold in Shop C is 20% more than the number of Cotton Sarees sold in Shop A. The number of SilkSarees sold in Shop C is 55 less than the number of Cotton Sarees sold in Shop B. The number of Silk Sarees sold in Shop B is 20 more than the number of Cotton Sarees sold in Shop C. The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold in Shop C. Total number of Cotton Sarees sold in all three shops is 825.
Note:
Total number of Sarees sold = Number of Cotton Sarees sold + Number of Silk Sarees sold.
The sum of the number of Silk Sarees sold in Shop B and C is how much percentage more than the number of Cotton Sarees sold in Shop C.
90%
50%
75%
80%
70%
Correct answer
80%
Why this answer is correct
General Solution:
The number of Cotton Sarees sold in Shop A = 5x
The number of Silk Sarees sold in Shop A = 530 – 5x
The number of Cotton Sarees sold in Shop C = 6x
The number of Cotton Sarees sold in Shop B = y
The number of Silk Sarees sold in Shop C = y – 55
The number of Silk Sarees sold in Shop B = 6x + 20
The number of Cotton Sarees sold in Shop A is 30 more than the number of Silk Sarees sold
in Shop C
5x = 30 + y – 55
y = 5x + 25 ----------- (1)
The total number of Cotton Sarees sold in all three shops is 825
5x + y + 6x = 825
5x + 5x + 25 + 6x = 825
16x = 800
x = 50
Substitute x = 50 in equation (1)
y = 5 × 50 + 25
y = 275
Solution:
The sum of the number of Silk Sarees sold in Shop B and C = 320 + 220 = 540
The number of Cotton Sarees sold in Shop C = 300
Required percentage = [(540 – 300)/300] × 100
= [240/300] × 100
= 80%
A cylindrical bucket filled with sand is empty out on the ground, then a conical heap of height 24 cm of sand is formed. If the height of cylindrical bucket is 32 cm and it has 18 cm base radius, then find the area covered on the ground by sand.
3072.14 cm2
500.14 cm2
1296.14 cm2
4073.14 cm2
None of these
Correct answer
4073.14 cm2
Why this answer is correct
Shortcut Trick
Volume of Cylindrical Bucket = Volume of Conical Heap.
π × rc2 × hc = (1/3) × π × rh2 × hh
182 × 32 = (1/3) × rh2 × 24
rh2 = (324 × 32 × 3) / 24 = 1296
Area covered = πrh2 = (22/7) × 1296 ≈ 4073.14 cm2.
∴ The correct answer is 4073.14 cm2.
Alternate Method
Given: Radius of Cylindrical Bucket (r) = 18 cm, Height (h) = 32 cm. Height of Conical Heap (H) = 24 cm.
Formula Used: Volume of Cylinder = πr2h; Volume of Cone = (1/3)πR2H; Area of base = πR2.
⇒ Volume of sand = π × (18)2 × 32
⇒ Volume of conical heap = (1/3) × π × R2 × 24
⇒ (1/3) × π × R2 × 24 = π × 324 × 32
⇒ 8 × R2 = 324 × 32
⇒ R2 = 324 × 4 = 1296 cm2
⇒ Area covered on ground = Area of circular base of cone = πR2
⇒ Area = (22/7) × 1296 = 28512 / 7 ≈ 4073.1428...
∴ The correct answer is 4073.14 cm2.
Additional Information
Volume Invariance
When a 3D object is reshaped or its contents are poured out to form a new shape, the Total Volume remains constant.
Cone Geometry
The area covered by a conical heap on a flat surface is the area of its Circular Base, calculated as A = πr2.
Slant Height of Cone
Slant height l = √(R2 + H2), which is used for calculating the Curved Surface Area (πRl).
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.