Quantitative Aptitude PYQs

Shortcut Trick

Ratio of Height (h) to Radius (r) = 2 :

1. Let h = 2x and r = x.

Total Surface Area (TSA) = 2πr(r + h) = 2 × (22/7) × x × (x + 2x) = 924.

⇒ (44/7) × 3x² = 924 ⇒ x² = 49 ⇒ x = 7 cm.

Volume = πr²h = (22/7) × 7² × 14 = 22 × 7 × 14 = 2156 cm³.

∴ The correct answer is 2156 cm³.

Alternate Method

Given: Ratio of height to radius = 2 : 1, TSA = 924 cm².

Formula Used: TSA = 2πr(h + r); Volume = πr²h

⇒ Let Radius (r) = x and Height (h) = 2x.

⇒ 2 × (22/7) × x × (2x + x) = 924

⇒ 2 × (22/7) × 3x² = 924

⇒ (132/7) × x² = 924

⇒ x² = (924 × 7) ÷ 132

⇒ x² = 7 × 7 = 49 ⇒ x = 7 cm.

⇒ Radius (r) = 7 cm, Height (h) = 14 cm.

⇒ Volume = (22/7) × 7 × 7 × 14

⇒ Volume = 22 × 7 × 14 = 2156 cm³.

∴ The correct answer is 2156 cm³.

Additional Information

Curved Surface Area of Cylinder

The area of the side surface of the cylinder is given by CSA = 2πrh.

Volume of Hollow Cylinder

If R is outer radius and r is inner radius, Volume = π(R² − r²)h.

Base Area of Cylinder

The base of a right circular cylinder is a circle, so Base Area = πr².

Shortcut Trick

Approximate Total mixture = 80 litres, Water = 20%, Milk = 80%.

Mixture remaining after 25% removal = 80 × 0.75 = 60 litres.

Remaining Milk = 80% of 60 = 48 litres; Remaining Water = 20% of 60 = 12 litres.

Final ratio 12:5 means 12 units = 48 litres ⇒ 1 unit = 4 litres.

Required Water = 5 units = 5 × 4 = 20 litres. Added Water = 20 − 12 = 8 litres.

∴ The correct answer is 8 litres.

Alternate Method

Given: Total Mixture ≈ 80 litres, Water ≈ 20.13% ≈ 20%, Removal ≈ 25%.

Step 1: Calculate initial quantities in litres.

⇒ Total mixture = 80 litres.

⇒ Water = 20% of 80 = 16 litres.

⇒ Milk = 80 − 16 = 64 litres.

Step 2: Calculate remaining quantities after taking out 25% of mixture.

⇒ Milk remaining = 64 − (25% of 64) = 64 − 16 = 48 litres.

⇒ Water remaining = 16 − (25% of 16) = 16 − 4 = 12 litres.

Step 3: Let x be the quantity of water added.

⇒ Milk : (Water + x) = 12 : 5

⇒ 48 / (12 + x) = 12 / 5

⇒ 4 / (12 + x) = 1 / 5

⇒ 20 = 12 + x

⇒ x = 20 − 12 = 8 litres.

∴ The correct answer is 8 litres.

Additional Information

Mixture Removal Property

When a portion of a mixture is removed, the ratio of components in the remaining mixture stays the same as the initial ratio.

Mixtures & Alligations

For replacement problems, Quantity left = Initial Quantity × (1 − Removed/Total)ⁿ, where n is the number of operations.

Approximation Rule

In banking exams, values like 79.99 or 20.13% should be rounded to the nearest whole number (80 and 20%) to speed up calculations.

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

12.98 + 417.98 = 34.98 × 19.98 - ?2 - 99.99

13 + 418 = 35 × 20 - ?2 - 100

⇒ 13 + 418 = 700 - ?2 - 100

⇒ 431 = 700 - ?2 - 100

⇒ ?2 = 700 - 431 - 100

⇒ ? = √169 = 13

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

44.99% of 399.99 + 49.99% of 299.99 = ? + 69.99% of 199.99

⇒ 45% of 400 + 50% of 300 = ? + 70% of 200

⇒ 180 + 150 = ? + 140

⇒ 330 = ? + 140

∴ ? = 190

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

[12.935 ÷ 12.984] × [13.033 ÷ 13.136] = 12.83(? – 4)

⇒ [135 ÷ 134] × [133 ÷ 136] = 13(? – 4)

⇒ 13 × 13-3 = 13(? – 4)

⇒ 13-2 = 13(? – 4)

⇒ ? - 4 = -2

⇒ ? = 2

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(124.95 × 5.98) – (110.02 × 7.96) + 669.95 = ? × 8.99

⇒ (125 × 6) – (110 × 8) + 670 = ? × 9

⇒ 750 – 880 + 670 = ? × 9

⇒ 1420 – 880 = ? × 9

⇒ 540 = ? × 9

∴ ? = 60

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

349.99 – [15.952 – (87.88 + 95.94) ÷ 7.99] = ?

⇒ 350 – [162 – (88 + 96) ÷ 8] = ?

⇒ 350 – [162 – 184 ÷ 8] = ?

⇒ 350 – [162 – 23] = ?

⇒ 350 – [256 – 23] = ?

⇒ 350 – 233 = ?

∴ ? = 117

Shortcut Trick

Effective Compound Interest rate for 2 years at 11% = 11 + 11 + (11 × 11)/100 = 23.21%.

First Interest amount = 23.21% of ₹20000 = ₹4642.

Simple Interest on second investment = (4642 × 11 × 2) / 100 = 4642 × 0.22 = ₹1021.24.

∴ The correct answer is ₹1021.24.

Alternate Method

Given: Principal (P₁) = ₹20000, Rate (R) = 11%, Time (T) = 2 years.

Formula Used: CI = P[(1 + R/100)^T − 1]; SI = (P × R × T) / 100.

⇒ 1^st Interest (CI) = 20000 × [(1 + 11/100)² − 1]

⇒ CI = 20000 × [1.2321 − 1] = 20000 × 0.2321 = ₹4642.

⇒ 2^nd Principal (P₂) = 1^st Interest = ₹4642.

⇒ SI received = (4642 × 11 × 2) / 100

⇒ SI = (4642 × 22) / 100 = 102124 / 100 = ₹1021.24.

∴ The correct answer is ₹1021.24.

Additional Information

Effective Rate of Compound Interest

For 2 years, the net percentage change formula (x + y + xy/100) provides the effective CI rate relative to the principal.

Simple Interest vs Compound Interest

While SI is calculated only on the principal, CI is calculated on the principal plus the accumulated interest of previous periods.

CI − SI Difference for 2 years

The difference between CI and SI for 2 years is given by the formula P(R/100)².

Shortcut Trick

Difference in percentage marks = 45% − 25% = 20%

Difference in actual marks = 50 − (−25) = 75 marks

20% of Maximum Marks = 75 marks

100% of Maximum Marks = 75 × 5 = 375 marks

∴ The correct answer is 375.

Alternate Method

Given: Ram score = 25%, Shyam score = 45%, Marks gap = 25 (fail) and 50 (extra).

Let Maximum Marks be M.

Passing Marks = 25% of M + 25

Passing Marks = 45% of M − 50

⇒ 25% of M + 25 = 45% of M − 50

⇒ 45% of M − 25% of M = 50 + 25

⇒ 20% of M = 75

⇒ (20/100) × M = 75

⇒ M = (75 × 100) ÷ 20

⇒ M = 375

∴ The correct answer is 375.

Additional Information

Passing Percentage

To find the passing percentage, substitute M back into the passing marks equation: (25% × 375 + 25) ÷ 375 × 100 = 31.66%.

Percentage Change Base

In examination problems, the Maximum Marks always acts as the base (100%) for calculating scored or passing percentages.

Successive Percentage

For problems involving repeated changes, use the net effect formula: a + b + (ab/100).

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

 

Solution:

Number of females selected from school A from city P= 150 ×  1/3 = 50

Number of females selected from school A from city Q = (150 - 50) = 100

Number of females selected from school B from city P = 150 ×  1/5 = 30

Number of females selected from school B from city Q = 150 - 30 = 120 

Number of males selected from school A from city Q = 350 ×  1/7 = 50

Number of males selected from school B from city Q = 250 ×  1/10 = 25

Required total = (100 + 120 + 50 + 25) = 295

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

Solution:

Average number of males selected = (350 + 250 + 300 + 325 + 300)/5 = 305

Total females selected from all the schools = (150 + 150 + 150 + 225 + 300) = 975

Required percentage = 305/975 × 100 = 31.28%

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

 

Solution:

Number of males selected for TCS from school C = 300 ×  1/6 = 50

Number of females selected for TCS from school C = 150 ×  1/5 = 30

Required total = (50 + 30) = 80

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

 

Solution:

Average number of unselected females = (130 + 240 + 75 + 135 + 195)/5 = 155

Average number of unselected males = (70 + 10 + 225 + 115 + 105)/5 = 105

Required difference = (155 - 105) = 50

Solution:

 

School A:

Total number of students (Male + Female) attend interview = 700

Number of female attend interview = 700 ×  (40/100) = 280

Number of males attend interview = (700 - 280) = 420

Number of males selected in interview = 350

Number of males unselected in interview = 420 - 350 = 70

Total number students selected in interview = 500

Number of females selected in interview = 500 - 350 =150

Number of females unselected in interview = 280 - 150 = 130

 

School B:

Total number of students (Male + Female) attend interview = 650

Number of female attend interview = 650 ×  (60/100) = 390

Number of males attend interview = (650 - 390) = 260

Number of males selected in interview = 250

Number of males unselected in interview = 260 - 250 = 10

Total number students selected in interview = 400

Number of females selected in interview = 400 - 250 =150

Number of females unselected in interview = 390 - 150 = 240

 

School C:

Total number of students (Male + Female) attend interview = 750

Number of female attend interview = 750 ×  (30/100) = 225

Number of males attend interview = (750 - 225) = 525

Number of males selected in interview = 300

Number of males unselected in interview = 525 - 300 = 225

Total number students selected in interview = 450

Number of females selected in interview = 450 - 300 =150

Number of females unselected in interview = 225 - 150 = 75

 

School D:

Total number of students (Male + Female) attend interview = 800

Number of female attend interview = 800 ×  (45/100) = 360

Number of males attend interview = (800 - 360) = 440

Number of males selected in interview = 325

Number of males unselected in interview = 440 - 325 = 115

Total number students selected in interview = 550

Number of females selected in interview = 550 - 325 =225

Number of females unselected in interview = 360 - 225 = 135

 

School E:

Total number of students (Male + Female) attend interview = 900

Number of female attend interview = 900 ×  (55/100) = 495

Number of males attend interview = (900 - 495) = 405

Number of males selected in interview = 300

Number of males unselected in interview = 405 - 300 = 105

Total number students selected in interview = 600

Number of females selected in interview = 600 - 300 = 300

Number of females unselected in interview = 495 - 300 = 195

 

Solution:

Number of females attending interview from school A from section I = 280 × (2/7) = 80

Number of female selected from school A from section I = 150 × (1/5) = 30

Number of females Unselected from school A from section I = (80 - 30) = 50

Shortcut Trick

Let the number of books read on Tuesday be x.

According to the given deviations: Monday = x + 10 and Wednesday = x − 10.

The Average of these three days is simply the middle value (Tuesday) because they are in an Arithmetic Progression with a common difference of 10.

∴ Tuesday = Average = 20 books.

Monday = 20 + 10 = 30 and Wednesday = 20 − 10 = 10.

∴ The correct answer is 30, 20 and 10.

Alternate Method

Given: Average = 20 books, Days = 3.

Formula Used: Total Sum = Average × Number of entities.

⇒ Total books = 20 × 3 = 60.

⇒ Let books on Tuesday = x.

⇒ Monday = x + 10, Wednesday = x − 10.

⇒ (x + 10) + x + (x − 10) = 60.

⇒ 3x = 60 ⇒ x = 20.

⇒ Monday = 20 + 10 = 30; Tuesday = 20; Wednesday = 20 − 10 = 10.

∴ The correct answer is 30, 20 and 10.

Additional Information

Average in Arithmetic Progression

If terms are in AP, the Average is always the Middle Term (for odd number of terms) or the mean of the two middle terms.

Average Change Rule

If each observation is increased or decreased by a constant k, the Average also increases or decreases by k.

Shortcut Trick

Ratio of Profit Share is determined by the ratio of Investment × Time.

Effective investment of P = (50 × 7) + (30 × 5) = 350 + 150 = 500 units.

Effective investment of Q = (75 × 7) + (80 × 5) = 525 + 400 = 925 units.

Profit ratio P : Q = 500 : 925 = 20 : 37.

Share of P = 20 ÷ (20 + 37) × 1,10,000 = 20 ÷ 57 × 1,10,000 ≈ 38,596.49.

∴ The correct answer is Rs. 38,597.

Alternate Method

Given → P initial = Rs. 50,000; Q initial = Rs. 75,000; Time = 12 months; Total Profit = Rs. 1,10,000.
Formulas: Profit Ratio = (I₁ × T₁) : (I₂ × T₂) | Profit Share = (Ratio Part ÷ Total Parts) × Total Profit.

⇒ P total investment units = (50,000 × 7) + (30,000 × 5) = 3,50,000 + 1,50,000 = 5,00,000.
⇒ Q total investment units = (75,000 × 7) + (80,000 × 5) = 5,25,000 + 4,00,000 = 9,25,000.
⇒ Ratio P : Q = 5,00,000 : 9,25,000 = 500 : 925 = 20 : 37.
⇒ Total units = 20 + 37 = 57 units.
⇒ Share of P = (20 ÷ 57) × 1,10,000 ≈ Rs. 38,596.49.

∴ The correct answer is Rs. 38,597.

Additional Information

Partnership Basics

In a compound partnership, the profit is shared based on the product of the capital and the duration of investment.

Active vs Sleeping Partner

An active partner manages the business and may receive a salary or extra profit, while a sleeping partner only provides capital.

Given:

I. x2 – 10x + 25 = 0

II. y2 + 10y + 24 = 0

Calculation:

I. x2 – 10x + 25 = 0

⇒ x^{2 -} 5x - 5x + 25 = 0

⇒ x(x - 5) - 5(x - 5)= 0

⇒ (x - 5)(x - 5) = 0

⇒ x = 5, 5

II. y2 + 10y + 24 = 0

⇒ y² + 6y + 4y + 24 = 0

⇒ y(y + 6) + 4(y + 6) =0

⇒ (y + 6)(y + 4) = 0

⇒ y = (-6), (-4)

 

When we compared the values of ‘x’ and ‘y’ in the table above, we found that there is only ONE relation between X and Y i.e. >. So, a relation between x and y is “x > y”.

∴ After comparison, all the values of x and y the relation is “x > y”.

Shortcut Trick

I. x2 - 10x + 25 = 0

There are three terms = a, b, and c

Where 'a' = value of x², 'b' = Value of x, 'c' = 25

If value of 'c' is positive value and value of 'b' is negative value, then both the values of x is POSITIVE value.

II. y2 + 10y + 24 = 0

There are also three terms = a, b, and c

Where 'a' = value of y², 'b' = Value of y, 'c' = 24

If the value of 'c' is a positive value and the value of 'b' is also a positive value, then both value of y is NEGATIVE value.

∴ After comparison, all the values of x and y, the relation between x and y is "x > y".

Given:

I. 7x2 - 4x - 51 = 0

II. y2 - 24y + 143 = 0

Calculation:

From I,

7x2 - 4x - 51 = 0

⇒ 7x2 - 21x + 17x - 51 = 0

⇒ 7x(x - 3) + 17(x - 3) = 0

⇒ (x - 3)(7x + 17) = 0

Taking,

⇒ (x - 3) = 0 or (7x + 17) = 0

⇒ x = 3 or (-17/7)

From II,

y2 - 24y + 143 = 0

⇒ y2 - 11y - 13y + 143 = 0

⇒ y(y - 11) - 13(y - 11) = 0

⇒ (y - 11)(y - 13) = 0

Taking,

⇒ (y - 11) = 0 or (y - 13) = 0

⇒ y = 11 or 13

Comparison between x and y (via Tabulation):

 

∴ x < y.

Given:

I. 2x2 - 17x + 36 = 0

II. y2 + 10y + 24 = 0

Calculation:

From I

2x2 - 17x + 36 = 0

⇒ 2x² - 8x - 9x + 36 = 0

⇒ 2x(x - 4) - 9(x - 4) = 0

⇒ (x - 4)(2x - 9) = 0

⇒ x = 4, 9/2

From II

y2 + 10y + 24 = 0

⇒ y² + 6y + 4y + 24 = 0

⇒ y(y + 6) + 4(y + 6) = 0

⇒ (y + 6)(y + 4) = 0

⇒ y = -6, -4

Comparison between x and y (via Tabulation) 

 

∴ x > y

Important Points

 

Shortcut Trick

If ax2 - bx + c = 0  then roots will be always +ve

If ax2 + bx + c = 0  then roots will be always -ve

So, without solving the equations we can say the answer.

I. 3x² + 2x – 8 = 0

⇒ 3x² + 6x – 4x – 8 = 0

⇒ 3x(x + 2) – 4(x + 2) = 0

⇒ (3x – 4) (x + 2) = 0

⇒ x = 4/3 or -2

II. 3y² – 14y + 16 = 0

⇒ 3y² – 6y – 8y + 16 = 0

⇒ 3y(y – 2) – 8(y – 2) = 0

⇒ (3y – 8) (y – 2) = 0

⇒ y = 8/3 or 2

∴ x < y