Search Previous Year Questions

Shortcut Trick

Sum of ratio parts = 2 + 3 + 5 = 10 units.

Each variable appears twice in the sum, so 2 × (a + b + c) = 10 units.

Since a + b + c = 20, then 2 × 20 = 40 = 10 units ⇒ 1 unit = 4.

Therefore, a + b = 2 units = 2 × 4 = 8.

Value of c = (a + b + c) − (a + b) = 20 − 8 = 12.

∴ The correct answer is 12.

Alternate Method

Given: (a + b) : (b + c) : (c + a) = 2 : 3 : 5 and a + b + c = 20.

⇒ Let a + b = 2k, b + c = 3k, and c + a = 5k.

⇒ Adding the ratios: (a + b) + (b + c) + (c + a) = 2k + 3k + 5k.

⇒ 2(a + b + c) = 10k.

⇒ Given a + b + c = 20, substitute: 2(20) = 10k.

⇒ 40 = 10k ⇒ k = 4.

⇒ Now, a + b = 2k = 2 × 4 = 8.

⇒ We know c = (a + b + c) − (a + b).

⇒ c = 20 − 8 = 12.

∴ The correct answer is 12.

Additional Information

Constant of Proportionality

When terms are given in a ratio (x : y = m : n), they can be expressed as x = mk and y = nk, where k is the common multiplier.

Cyclic Ratio Sums

For expressions involving pairs like (a+b), (b+c), and (c+a), the sum of all pairs is always equal to 2(a + b + c).

Finding Individual Variables

To find a single variable like c from the total sum S = a+b+c, subtract the sum of the other two variables: c = S − (a+b).

Given:

A completes work in 15 days.

B completes work in 10 days.

Concept Used:

Share of wages is distributed as the ratio of effciency

Calculations:

We know that amount must be shared as the ratio of their efficiencies.

So efficiencies of A : B = 1/15 : 1/10 

⇒ 10 : 15 = 2 : 3

Sum of efficiencies = 2 + 3 = 5

So A share in the total = (2/5) × 30,000

⇒ Rs.12000

∴ The share of A in the contracted money will be Rs.12000

Shortcut Trick

For finding the HCF of prime factorised numbers, multiply the lowest powers of all common prime factors.

For base 2 → minimum power is 2 → 2²

For bases 3 and 5 → minimum powers are 2 and 1 → 3² and 5

HCF = 2² × 3² × 5

∴ The correct answer is 2² × 3² × 5.

Alternate Method

Given:

First number = 2² × 3⁵ × 5³

Second number = 2⁴ × 3² × 5

Formula Used:

HCF = Product of the lowest power of each common prime factor.

Calculations:

⇒ Compare the powers of the common base 2: the lowest power is 2².

⇒ Compare the powers of the common base 3: the lowest power is 3².

⇒ Compare the powers of the common base 5: the lowest power is 5.

⇒ Multiplying these lowest powers together gives the Highest Common Factor.

⇒ HCF = 2² × 3² × 5

∴ The correct answer is 2² × 3² × 5.

Additional Information

LCM of Prime Factors

To find the LCM of numbers in prime factorised form, multiply the highest powers of all prime factors present in the given numbers.

Co-prime Numbers

Two numbers are considered co-prime if their HCF is 1, meaning they do not share any common prime factors.

HCF and LCM Relation

For any two numbers, the product of the numbers is always equal to the product of their HCF and LCM.

Given:

15 males or 20 females can complete a project in 26 days.

Formula used:

(M1 × D1)/W1 = (M2 × D2)/W2

Calculation:

15 males × 26 days = 20 females × 26 days 

∵ 15 males = 20 females 

∴  30 males = 40 females 

Total work in terms of females  = 20 females × 26 days = 520 units 

No of days taken by 30 males and 12 females:

⇒ (40 females + 12 females) = 52 females 

∴ No of days taken = 520 units /52 females = 10

Given:

The ratio of the radius to the height of the cylinder = 6 : 7

Volume of the cylinder = 792 cm³

Formula used:

Volume of a cylinder = πr²h

Curved surface area of a cylinder = 2πrh

Calculation:

Let the common ratio be x.

Then, the radius r = 6x and the height h = 7x.

Using the volume formula:

π(6x)²(7x) = 792

⇒ π × 36x² × 7x = 792

⇒ 252πx³ = 792

Take π as 22/7:

⇒ 252 × 22/7 × x³ = 792

⇒ 792x³ = 792

⇒ x³ = 1

⇒ x = 1

Thus, radius r = 6x = 6 × 1 = 6 cm

And height h = 7x = 7 × 1 = 7 cm

Now, calculate the curved surface area:

Curved surface area = 2πrh = 2 × 22/7 × 6 × 7

= 2 × 22 × 6

= 264 cm²

∴ The curved surface area of the cylinder is 264 cm².

Side of triangle a = 8 cm

Area of triangle = √3/4 × a²

⇒ Area of triangle ABC = √3/4 × 8 × 8 = 16√3

As we know, If P,Q and R are the mid points of sides AB, BC and CA, then

Area of ΔPQR = 1/4 × Area of ABC

⇒ Area of ΔPQR = 1/4 × 16√3 = 4√3

Shortcut Trick

Difference in Selling Prices = 1,498 − 1,300 = ₹198.

Ratio of Gain to Loss = 125 : 100 = 5 : 4.

Value of 1 unit = 198 ÷ (5 + 4) = 198 ÷ 9 = ₹22.

Cost Price (CP) = 1,300 + (4 × 22) = 1,300 + 88 = ₹1,388.

New SP for 25% gain = 1,388 × 1.25 = ₹1,735.

∴ The correct answer is ₹1,735.

Alternate Method

Given: SP₁ = ₹1,498, SP₂ = ₹1,300, Gain = 1.25 × Loss.

Formula: CP = SP − Gain = SP + Loss

⇒ Let Loss = L

⇒ Gain = 1.25L

⇒ 1,498 − CP = 1.25L and CP − 1,300 = L

⇒ (1,498 − CP) ÷ (CP − 1,300) = 1.25 ÷ 1 = 5 ÷ 4

⇒ 4 × (1,498 − CP) = 5 × (CP − 1,300)

⇒ 5,992 − 4CP = 5CP − 6,500

⇒ 9CP = 12,492

⇒ CP = ₹1,388

⇒ New SP = 1,388 × (125/100) = 1,388 × (5/4)

⇒ New SP = 347 × 5 = ₹1,735

∴ The correct answer is ₹1,735.

Additional Information

Profit and Loss Basic Relations

Profit = SP − CP and Loss = CP − SP. CP is always considered as the 100% base for calculations.

Marked Price and Discount

Discount is always calculated on Marked Price (MP). SP = MP × (100 − Discount%)/100.

Successive Profit/Loss

If there are successive percentage changes a and b, the effective change is [a + b + (ab/100)]%.

Given:

Length of the train = 80 meters

Length of the platform = 120 meters

Speed of the train = 72 km/hr

Formula Used:

Time = Distance / Speed

Calculation:

Convert speed from km/hr to m/s:

Speed = 72 × (1000/3600) = 72 × (10/36) = 72 / 3.6 = 20 m/s

Total distance to be covered = Length of the train + Length of the platform

Total distance = 80 + 120 = 200 meters

Time taken to cross the platform:

Time = Distance / Speed

Time = 200 / 20

Time = 10 seconds

The train will take 10 seconds to cross the platform.

Given:

A = 2C

A + B in 20 days

B + C in 24 days

Concept used:

Total work = LCM of time taken by the workers

Calculation:

LCM of 20 and 24 is 120

So, efficiency of A and B = 120/20 = 6 and efficiency of B and C = 120/24 = 5

Now 2C + B = 6 and B + C = 5

So, C = 1

B = 4

40% of the work = 120 × 2/5 = 48 units

So, B will take 48/4 = 12 days

∴ B alone do 40% of the same work in 12 days

Shortcut Trick

Now 2C + B = 6 and B + C = 5

So, C = 1

B = 4

So,

A, B and C = 2, 4, and 1

40% of the work = 120 × 2/5 = 48 units

So, B will take 48/4 = 12 days

Given:

S.P. of 1 kg of mixture = Rs. 9.24,

Gain = 10%100(100+gain%)×SP

Formula Used: 

C.P = S.P x 100/(100 + P%)

Calculation: 

C.P = Rs 9.24 x 100/(100 + 10%)

C.P= Rs 924 x 100/(110%)

C.P = Rs 8.4

By Rule of Alligation 

Ratio of quantities of 1^st and 2^nd kind = 1.4 : 0.6 = 7 : 3.

Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

Then, 7 : 3 = x : 27

∴ x = (27 × 7)/3 = 63 kg.

∴ The quantity of sugar that should be mixed is 63 Kg

ggggg×9.24=100110×9.24=��8.40[Math Processing Error]