Search Previous Year Questions

Shortcut Trick

Let the original fraction be x/y.

New numerator = 180% of x and New denominator = 300% of y.

(x × 180) / (y × 300) = 21 / 50

(x / y) × (180 / 300) = 21 / 50

(x / y) × (3 / 5) = 21 / 50

x / y = (21 / 50) × (5 / 3) = 7 / 10

∴ The correct answer is 7/10.

Alternate Method

Given:

Let the original fraction be x/y.

Numerator increase = 80% ⇒ New Numerator = x + 0.8x = 1.8x

Denominator increase = 200% ⇒ New Denominator = y + 2y = 3y

Resulting ratio = 21/50

Calculations:

⇒ 1.8x / 3y = 21 / 50

⇒ (18x / 10) / 3y = 21 / 50

⇒ 18x / 30y = 21 / 50

⇒ 3x / 5y = 21 / 50

⇒ x / y = (21 × 5) / (50 × 3)

⇒ x / y = 105 / 150

⇒ x / y = 7 / 10

∴ The correct answer is 7/10.

Additional Information

Percentage Change in Fractions

If numerator increases by a% and denominator by b%, the new fraction is [x(100+a)] / [y(100+b)].

Base Value Property

An increase of n% is equivalent to multiplying the original value by (1 + n/100).

Ratio to Fraction Conversion

The ratio a:b is identical to the fraction a/b and can be simplified by dividing both terms by their HCF.

Shortcut Trick

Total Work (LCM of 9, 12, 15) = 180 units.

Efficiency of A = 20, B = 15, C = 12.

Work done by (A + C) in 4 days = (20 + 12) × 4 = 128 units.

Remaining work = 180 − 128 = 52 units.

Time taken by B = 52 ÷ 15 = 3(7/15) days.

∴ The correct answer is 3(7/15) days.

Alternate Method

Given: A completes work in 9 days, B in 12 days, and C in 15 days.

Formula Used: Efficiency = Total Work ÷ Time taken

⇒ Efficiency of (A + C) = 20 + 12 = 32 units/day

⇒ Work completed by A and C in 4 days = 32 × 4 = 128 units

⇒ Remaining work = Total Work − Completed Work = 180 − 128 = 52 units

⇒ Time taken by B to finish remaining work = 52 ÷ 15

⇒ Time = 3(7/15) days

∴ The correct answer is 3(7/15) days.

Additional Information

Total Work Concept

Total work is always considered as the Least Common Multiple (LCM) of the individual days to avoid fractional calculations.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken. If the efficiency increases, the time required to complete the work decreases proportionally.

Pipes and Cisterns

Similar logic applies where filling pipes have positive efficiency and leakage pipes have negative efficiency.

Given:

I. 5x2 + 26x + 33 = 0

II. y2 + 18y + 65 = 0

Calculation:

From I,

5x2 + 26x + 33 = 0

⇒ 5x2 + 15x + 11x + 33 = 0

⇒ 5x(x + 3) + 11(x + 3) = 0

⇒ (x + 3)(5x + 11) = 0

Taking,

⇒ (x + 3) = 0 or (5x + 11) = 0

⇒ x = -3 or (-11/5)

From II,

y2 + 18y + 65 = 0

⇒ y2 + 13y + 5y + 65 = 0

⇒ y(y + 13) + 5(y + 13) = 0

⇒ (y + 13)(y + 5) = 0

Taking,

⇒ (y + 13) = 0 or (y + 5) = 0

⇒ y = -13 or -5

Comparison between x and y (via Tabulation):

 

∴ x > y.

Given:

I. 6x2 - 23x + 20 = 0

II. y2 - 5y + 4 = 0

Calculation:

From I,

6x2 - 23x + 20 = 0

⇒ 6x2 - 15x - 8x + 20 = 0

⇒ 3x(2x - 5) - 4(2x - 5) = 0

⇒ (2x - 5)(3x - 4) = 0

Taking,

⇒ (2x - 5) = 0 or (3x - 4) = 0

⇒ x = (5/2) or (4/3)

From II,

y2 - 5y + 4 = 0

⇒ y2 - 4y - y + 4 = 0

⇒ y(y - 4) - 1(y - 4) = 0

⇒ (y - 4)(y - 1) = 0

Taking,

⇒ (y - 4) = 0 or (y - 1) = 0

⇒ y = 4 or 1

Comparison between x and y (via Tabulation):

 

∴ x = y or relation between x and y can not be established.

Calculations:

From I,

x2 + 30x + 81 = 0

⇒ x2 + 27x + 3x + 81 = 0

⇒ x(x + 27) + 3(x + 27) = 0

⇒ (x + 3)(x + 27) = 0

Taking,

⇒ x + 3 = 0 or x + 27 = 0

⇒ x = – 3 or x = – 27

From II,

y2 + 26y + 48 = 0

⇒ y2 + 24y + 2y + 48 = 0

⇒ y(y + 24) + 2(y + 24) = 0

⇒ (y + 2)(y + 24) = 0

Taking,

⇒ y + 2 = 0 or y + 24 = 0

⇒ y = – 2 or y = – 24

Comparison between x and y (via Tabulation):

∴ After comparing x and y both, No relation in x and y or x = y.

Calculations:

From I,

x2 - 14x + 45  = 0

⇒ x2 - 9x - 5x + 45  = 0

⇒ x(x - 9) - 5(x - 9) = 0

⇒ (x - 5)(x - 9) = 0

⇒ x = 5, 9

From II,

y2 - 11y + 30 = 0

⇒ y2 - 6y  - 5y + 30 = 0

⇒ y(y - 6) - 5(y - 6) = 0

⇒ (y - 5)(y - 6) = 0

⇒ y = 5, 6

Comparison between x and y (via Tabulation):

​

∴ After comparing x and y both, No relation in x and y or x = y.

Given:

I. x2 + 13x + 42 = 0

II. y2 + 21y + 104 = 0

Calculation:

I. x2 + 13x + 42 = 0

⇒ x2 + 6x + 7x + 42 = 0

⇒ x(x + 6) + 7(x + 6)= 0

⇒ (x + 6)(x + 7) = 0

⇒ x = (-6), (-7)

II. y2 + 21y + 104 = 0

⇒ y2 + 13y + 8y + 104 = 0

⇒ y(y + 13) + 8(y + 13) =0

⇒ (y + 13)(y + 8) = 0

⇒ y = (-13), (-8)

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”.

[(15.98 + 9.99) ÷ 1.99] × 4.98 = ?3 + 1.09

⇒ [(16 + 10) ÷ 2] × 5 = ?3 + 1

⇒ (26 ÷ 2) × 5 = ?3 + 1

⇒ 13 × 5 = ?3 + 1

⇒ ?3 = 65 - 1

⇒ ? = 4

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

(222.22 + 71.98) ÷ 2.99 = 8.99% of 299.99 + 29.99 + ?

⇒ (222 + 72) ÷ 3 = 9% of 300 + 30 + ?

⇒ (294 ÷ 3) = (9/100) × 300 + 30 + ?

⇒ 98 = (9 × 3) + 30 + ?

⇒ 98 = 27 + 30 + ?

⇒ ? = 98 – 57

⇒ ? = 41

220.89 + 2.12³ + 205.96 =  1.99 × ? + 214.98 

⇒ 221 + 2³ + 206 =  2 × ? + 215 

⇒ 221 + 8 + 206 =  2 × ? + 215 

⇒ 435 = 2 × ? + 215

⇒ 220 = 2 × ?

⇒ ? = 110