Search Previous Year Questions

We know that, for a year

SI = CI = pr/100

Where, p = principal, r = % rate of interest per annum

Amount after first year \(= ;15000 + \frac{{15000; \times ;10;}}{{100}}\)  = Rs. 16500

Amount After second year  \(= ;16500; + ;\frac{{16500; \times ;12; }}{{100}}\)

= Rs. (16500 + 1980)

= Rs. 18480

SI = P × R × T/100

⇒ 5264 = P × 7 × 8/100

⇒ P = 5264 × 100/56

∴ P = Rs. 9400

The perimeter of a rhombus = 20 cm

⇒ Perimeter = 4 × side = 20

⇒ Side = 5 cm

\(\Rightarrow {\rm{}}Side = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \times \sqrt {{{\left( {{d_1}} \right)}^2} + {{\left( {{d_2}} \right)}^2}}\)

\(\Rightarrow {\rm{}}5 = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} \times \sqrt {{{\left( 6 \right)}^2} + {{\left( {{d_2}} \right)}^2}} \)

⇒ (d₂)² = 100 – 36 = 64

⇒ d₂ = 8 cm

∴ The length of the other diagonal is 8 cm.

Shortcut Trick

Total average rate of interest = (Total Interest ÷ Total Principal) × 100 = (100 ÷ 3000) × 100 = 10/3%

Using Alligation ratio (Amount at 3% : Amount at 5%):

(5 − 10/3) : (10/3 − 3) = 5/3 : 1/3 = 5 : 1

Amount lent at 3% = (5 ÷ (5+1)) × 3000 = 5/6 × 3000 = 2500

∴ The correct answer is Rs. 2500.

Alternate Method

Given: Total Principal (P) = Rs. 3000, Total Yearly Interest = Rs. 100, Rate 1 (R₁) = 3%, Rate 2 (R₂) = 5%.

Formula: Simple Interest (SI) = (P × R × T) ÷ 100

Let the amount lent at 3% be x and the amount at 5% be (3000 − x).

⇒ Total Interest = (x × 3 × 1)/100 + ((3000 − x) × 5 × 1)/100 = 100

⇒ 3x + 5(3000 − x) = 100 × 100

⇒ 3x + 15000 − 5x = 10000

⇒ −2x = 10000 − 15000

⇒ −2x = −5000

⇒ x = 5000 ÷ 2 = 2500

∴ The correct answer is Rs. 2500.

Additional Information

Simple Interest Rule

Interest is calculated only on the initial principal. Formula: SI = P×R×T/100, where T is in years.

Alligation in Finance

It is used to find the ratio in which two components with different rates are mixed to get an average rate. Ratio = (Higher Rate − Average Rate) : (Average Rate − Lower Rate).

Effective Interest Rate

The actual rate of return when interest is compounded or when multiple principals are combined: R_eff = (Total Interest / Total Principal) × 100.

Given:

John travels a distance by bike =\(\frac{2}{3}^{rd} \) of the total distance 

John travels a distance by foot = 6 km

Calculation:

Let us take the total distance traveled be x

John travels a distance by foot  = x - \(\frac{2}{3} \)x ⇒ \(\frac{1}{3} \) x          -----------------------(I)

John travels a distance by foot = 6 km                   -----------------------(ii)

Equating (I) and (ii)

6 = \(\frac{1}{3} \)x

x = 18 km 

∴ The total distance traveled = 18 km

Alternate Method Unitary Method Approach:

Distance travelled by Bike = 2/3

Total Distance = 3 units

That means, 2 units out of 3 units travelled by Bike

Remaining Distance travelled by Foot = 1 unit (given as 6 km)

1 unit = 6 km

3 units = 6 × 3 = 18 km

Given:

Radius ∶ Height = 77 ∶ 36

Curved surface area = 20570 cm2

Formula used:

Curved surface area of cone = π × R × l

Volume of cone = (1/3) × π × R² × h

l2 = R2 + h2  

Where,

R = Radius of cone

l = Slant height of cone 

h = Height of cone

Calculations:

Let the radius and height of the cone be 77x and 36x respectively

⇒ l2 = (77x)2 + (36x)2

⇒ l2 = 5929x2 + 1296x2

⇒ l2 = 7225x2

⇒ l = 85x

Curved surface area = (22/7) × 77x × 85x

⇒ 20570 = 20570x2

⇒ x = 1 

Radius = 77 cm, Height = 36 cm, Slant height = 85 cm

Volume of the cone = (1/3) × (22/7) × 772 × 36

⇒ 223608 cm³

∴ The volume of the cone is 223608 cm3

Given:

Side of old cube = 4 cm

Side of new cube = 1 cm

Formula Used:

The volume of cube = (Side)³

Total surface area of cube = 6 × (Side)2

Calculation:       

The volume of old cube = 4 × 4 × 4 cm³

The volume of new cube = 1 × 1 × 1 cm3

Number of cubes = \(\frac{4 × 4 × 4}{1 × 1 × 1}\) = 64 cubes

⇒Total surface area of old cube = 6 × (4)² = 96 cm²

Total surface area of new cube = 6 × (1)² = 6 cm²

⇒ Total surface area of 64 new cubes = 64 × 6 = 384 cm²

Difference in area = 384 - 96 = 288 cm²

⇒ % increase = \(\frac{288}{96}\times 100 = 300\)%

∴ 300% increase in the surface area after such cutting.

The correct option is 2 i.e. 300%

Shortcut Trick

Volume of hollow cylinder = π × (R² − r²) × h

Inner radius (r) = 7 cm, Outer radius (R) = 8 cm, height (h) = 7 cm

Volume = (22/7) × (8² − 7²) × 7 = 22 × (64 − 49) = 22 × 15 = 330 cm³

Weight = Volume × Density = 330 × 7.8 = 2574 gm = 2.574 kg

∴ The correct answer is 2.574 kg.

Alternate Method

Given:

Inner diameter = 14 cm ⇒ Inner radius (r) = 7 cm

Outer diameter = 16 cm ⇒ Outer radius (R) = 8 cm

Length (h) = 7 cm

Density = 7.8 gm/cm³

Formula Used:

Volume of material = π × (R² − r²) × h

Weight = Volume × Density

Calculations:

⇒ Volume = (22/7) × (8² − 7²) × 7

⇒ Volume = 22 × (64 − 49)

⇒ Volume = 22 × 15 = 330 cm³

⇒ Weight in gm = 330 × 7.8

⇒ Weight in gm = 2574 gm

⇒ Weight in kg = 2574 / 1000 = 2.574 kg

∴ The correct answer is 2.574 kg.

Additional Information

Hollow Cylinder Surface Area

Total Surface Area = 2πR h + 2πr h + 2π(R² − r²). This includes inner, outer, and base areas.

Density Conversion

1 gm/cm³ = 1000 kg/m³. To convert grams to kilograms, divide the value by 1000.

Volume of Solid Cylinder

The volume of a solid cylinder is given by πr²h, where r is the radius and h is the height.

Given:

Area of triangle = 1,815 cm².

the ratio of its base and the corresponding altitude in 5 : 6.

Formula used:

Area of triangle = (1/2) × base × altitude

Calculation:

Let the base and altitude be 5x and 6x.

(1/2) × 5x × 6x = 1815

⇒ 15x² = 1815

⇒ x² = 121

⇒ x = 11

Altitude = 6(11)

⇒ 66 cm

∴ Altitude is 66 cm.

Shortcut Trick

Work done by Pipe C in 3 hours = 3/4 of total work.

Remaining work to be filled by Pipe D = 1 − 3/4 = 1/4.

Time taken by Pipe D = (Remaining Work) × (Total time for D) = 1/4 × 5 = 1.25 hours.

∴ The correct answer is 1.25 hours.

Alternate Method

Given: Time for Pipe C = 4 hours, Time for Pipe D = 5 hours.

Formula Used: Total Work = LCM of time; Efficiency = Total Work ÷ Time.

⇒ Work done by Pipe C in 3 hours = 3 × 5 = 15 units.

⇒ Remaining Work = 20 − 15 = 5 units.

⇒ Time taken by Pipe D to fill remaining part = 5 ÷ 4 = 1.25 hours.

∴ The correct answer is 1.25 hours.

Additional Information

Pipes & Cisterns Efficiency

Efficiency is inversely proportional to time. If a pipe fills a tank in 'x' hours, its 1-hour work is 1/x.

Inlet and Outlet Pipes

Inlet pipes (filling) have positive efficiency, while outlet pipes (emptying/leakage) have negative efficiency.