Quantitative Aptitude PYQs

Shortcut Trick

For faulty weight problems where the article is sold at cost price:

Profit % = [(Stated Weight − Actual Weight) ÷ Actual Weight] × 100

Here, Stated Weight = 1000 gm and Actual Weight = 800 gm.

Profit % = [(1000 − 800) ÷ 800] × 100 = [200 ÷ 800] × 100 = 1/4 × 100 = 25%

∴ The correct answer is 25%.

Alternate Method

Given:

Reading on balance = 1000 gm (Quantity for which customer pays)

Actual weight delivered = 800 gm (Quantity that actually goes out of stock)

Selling price = Cost price (per unit weight)

Formula Used:

Profit = Selling Price (SP) − Cost Price (CP)

Profit % = (Profit ÷ CP) × 100

Let the Cost Price (CP) of 1 gm of the article be &inr;1.

⇒ Shopkeeper pays for 800 gm = &inr;800 (This is the effective CP).

⇒ Shopkeeper charges the customer for 1000 gm = &inr;1000 (Since SP = CP, this is the effective SP).

⇒ Total Profit = &inr;1000 − &inr;800 = &inr;200.

⇒ Profit % = (Profit ÷ Effective CP) × 100

⇒ Profit % = (200 ÷ 800) × 100

⇒ Profit % = 1/4 × 100 = 25%

∴ The correct answer is 25%.

Additional Information

Dishonest Dealer Formula

Profit % = [Error ÷ (True Value − Error)] × 100, where Error is the difference between the weight shown and the actual weight used.

Profit with Mark-up

If a dealer marks up price by x% and uses y% less weight, Net Profit % = [(x + y) ÷ (100 − y)] × 100.

Effective Ratio Method

Profit ratio can be found by multiplying individual ratios: (Stated Weight/Actual Weight) × (SP/CP). If SP=CP, the ratio is simply Stated Weight : Actual Weight.

Shortcut Trick

Area of river = depth × width

⇒ Area = 3 × 40 = 120 m²

Amount of water that falls into the sea = Area of river × flow rate

⇒ 120 m² × 2 km/h × 1000           (∵ 1 m³ = 1000 l)

⇒ 120 × (2000/60) m/min × 1000

⇒ 40,00,000 l

Given:

Depth of a river = 3 m

Width of a river = 40 m

Water flowing at the rate of 2 km/hr

Concept used:

Area of cross section of the river = L × B

Volume = Area × Rate of flow

1 m³ = 1000 litre

Calculation:

Depth of a river = 3 m

Width of a river = 40 m

Water flowing at the rate of 2 km/hr

2 km/hr =  (2000/60) m/min

Area of cross section of the river = L × B

Area of cross section of the river = 3 × 40 = 120 m

Volume of water flowing through this cross section in every minute = Area × Rate of flow

⇒ 120 × (2000/60)

⇒ 4000 m³

Now,

1 m3 = 1000 litres

4000 × 1000 = 4000000 litres

∴ The amount of water flow in one minute is 4000000 litres.

Shortcut TrickLet the cost price = C

y = (C × 112)/100

x = (C × 72)/100

y/x = 112/72 = 14/9

∴ The correct answer is 14:9.

Solution:

Given:

SP = Rs. x

Loss = 28%

SP = Rs. y

Profit = 12%

Concept used:

Loss% = (Loss/CP) × 100 

Calculation:

Let the cost price be Rs. 100

If television is sold at Rs. x, then

Loss % = 28%

Loss % = (Loss/CP) × 100 

28 = (Loss/100) × 100

⇒ Loss = 28

SP = CP - Loss

SP = 100 - 28 = Rs. 72

⇒ x = Rs. 72

Now,

If television is sold at Rs. y, then

Profit % = 12%

12 = (Profit/100) × 100

Profit = Rs. 12

SP = CP + Profit

SP = 100 + 12 = Rs. 112

⇒ y = Rs. 112

We have to find y/x 

y/x = 112/72 = 14/9

∴ The ratio of y to x is 14 : 9.

Alternate Method

Let CP be Rs. 100

Loss = 28%

SP(x) = 100 - 28 = Rs. 72

Profit = 12%

SP(y) = 100 + 12 = Rs. 112

y/x = 112/72 = 14/9

∴ The ratio of y to x is 14 : 9.

Shortcut TrickWe know that Time = Distance / Speed

Let the original speed be x km/hr

As per the question,

⇒ (300/x) - [300/(x + 15)] = 1

Now, we will check through the options.

Check for those values which will divide 300 completely and give us an integral value.

As we can understand easily when x = 60 km/hr,

⇒ (300/60) - [300/(60 + 15)] = 1

∴ The original speed of the car is 60 km/hr.

Given:

Distance = 300 km

Speed increased by car = 15 km/hr

Time difference = 1 hour

Formula Used:

Distance = {(S₁ ×  S₂)/(S₂ - S₁)} × Time difference

Where,

S₁ = Original speed

S₂ = Increased speed

Calculation:

Let the original speed of the car be x km/hr

⇒ Increased speed of car = (x + 15) km/hr

According to the question, we have

300 = [{x × (x + 15)}/(x + 15 - x)} × 1

⇒ 300 = (x² + 15x)/15

⇒ 4500 = x2 + 15x

⇒ x2 + 15x - 4500 = 0

⇒ x² + 75x - 60x - 4500 = 0

⇒ x(x + 75) - 60(x + 75) = 0

⇒ (x + 75)(x + 60) = 0

⇒ x = 60 and (- 75)

Now, the negative distance is not possible

 x = 60 km/hr

∴ The original speed of car is 60 km/hr.

Shortcut Trick 

                   A                :          B              :          C

{(20 × 4) + (10 × 8)}     :      (15 × 12)       :     (12 × 12)

                  40              :          45              :         36

⇒ A's share = (96800 / 121) × 40 = 32000

∴ The share of the first person in profit is Rs. 32,000.

Alternate Method 

Given:

The ratio of capital invested by three persons = (1/3) ∶ (1/4) ∶ (1/5)

Total profit = Rs. 96,800

Calculation:

Let the three persons be A, B and C

According to the question, we have

A ∶ B ∶ C = (1/3) ∶ (1/4) ∶ (1/5)      ----(1)

LCM of 3, 4 and 5 is 60

Multiply of 60 in equation (1)

⇒ A ∶ B ∶ C = 20 ∶ 15 ∶ 12

⇒ So, the investment of A, B and C are 20x, 15x and 12x

Now, according to the question

A withdraws half of his capital after 4 months

⇒ A's investment after 12 months =(20x × 4) + (10x × 8)

⇒ 80x + 80x

⇒ 160x

B's investment after 12 months = 15x × 12

⇒ 180x

⇒ C's investment after 12 months = 12x × 12

⇒ 144x

Total investment = 160x + 180x + 144x

⇒ 484x

Now,

Total profit  = Rs. 96,800

⇒ Profit of A's = (96800 × 160x)/(484x)

⇒ 32000

∴ The share of the first person in profit is Rs. 32,000.

Given:

1. x2 + y2 varies as x2 - y2 2. \(\frac{x}{y^2}\) varies inversely as y 3. \(\sqrt[n]{x^2y}\) varies as \(\sqrt[2n]{x^4y^2}\)

Concept:

If x varies as y 

then x = ky

Calculation:

Check the options:

1. x2 + y2 varies as x2 - y2

Put x = ky

x² + y² = (ky)² + y² = y²(k² + 1)

And x² - y² = (ky)² - y² = y²(k² - 1)

So, \(\rm {{x^2\ +\ y^2}\over{x^2\ -\ y^2}} = {{y^2(k^2\ +\ 1)}\over {y^2(k^2\ -\ 1)}} = {{(k^2\ +\ 1)}\over {(k^2\ -\ 1)}}\)

\(\rm x^2 + y^2 = {{(k^2\ +\ 1)}\over {(k^2\ -\ 1)}}(x^2 - y^2)\)

Then x² + y² varies as x² - y²

2. \(\frac{x}{y^2}\) varies inversely as y

Put x = ky

\(\rm {x\over y^2} = {ky\over y^2} = {k\over y}\)

So, \(\rm x\over y^2\) is inversely proportional to y

3. \(\rm \sqrt[n]{x^2y}\) varies as \(\rm \sqrt[2n]{x^4y^2}\) 

Put x = ky

\(\rm \sqrt[2n]{x^4y^2} = ​​\sqrt[2n]{(x^2y)^2} = \sqrt[n]{(x^2y)}\)

So,  \(\rm \sqrt[n]{x^2y}\) varies with \(\rm \sqrt[2n]{x^4y^2}\) 

∴ 1, 2 and 3 all are correct.

Shortcut TrickEna's age = 13 years

Ena' mother age = 13 + 24 = 37 years

Ena' father age = 37 + 3 = 40 years

The age of Ena's father at the time of marriage = 40 - (13 + 4)

⇒ (40 - 17) years = 23 years

Given:

Ena born after the parent's marriage = 4 years

Age of father - Age of mother = 3 years

Age of mother - Age of Ena = 24 years

The present age of Ena = 13 years

Calculation:

According to the question, we have

Age of mother - Age of Ena = 24 years

⇒ Age of mother - 13 = 24

⇒ Age of mother = 24 + 13

⇒ Age of mother = 37 years

Now, according to the question

Age of father - Age of mother = 3 years

⇒ Age of father - 37 = 3

⇒ Age of father = 3 + 37

⇒ Age of father = 40 years

Now,

The age of Ena's father at the time of marriage = 40 - (13 + 4)

⇒ 40 - 17

⇒ 23 years

∴ The age of Ena's father at the time of marriage is 23 years.

Shortcut TrickMahesh's age = 60 years

Ram's age = Mahesh's age - 6 years = 55 years

Raju's age = Ram's age - 4 years = 51 years

Babu's age = Raju's age - 6 years = 45 years

Difference between Mahesh and Babu age = (60 - 45) years

⇒ 15 years

Given:

Mahesh = 60 years

Mahesh - Ram = 5 years

Ram - Raju = 4 years

Raju - Babu = 6 years

Calculation:

According to the question, we have 

Mahesh - Ram = 5 years

⇒ 60 - Ram = 5

⇒ Ram = 60 - 5

⇒ Ram = 55 years

Now, according to the question

Ram - Raju = 4 years

⇒ 55 - Raju = 4

⇒ Raju = 55 - 4

⇒ Raju = 51 years

Now, according to the question

Raju - Babu = 6 years

⇒ 51 - Babu = 6

⇒ Babu = 51 - 6

⇒ Babu = 45 years

Now, the age difference between Mahesh and Babu 

Mahesh - Babu = 60 - 45

⇒ 15 years

∴ The age difference between Mahesh and Babu is 15 years.

Shortcut TrickThe formula for Successive Percentage change for 2 years

⇒ x + y + (xy/100)

As per question,

⇒ Effective change in no. of items for 2 yr = x - x - (x²/100)

⇒ Effective change in no. of items for 2 yr = -x2/100

⇒ Negative sign shows decrease.

So, after 2 years the number of items = N - x2/100

∴ The number of items will be less than N.

Concept used :

If there are N number of items and there is an increment and decrement of x% respectively in the amount then final value of N 

N = N.(100 + x)/100).(100 - x)/100) 

Calculations :

Let the value of N be 100 

Let the value of x be 10 

Final number of products = 100 × (100 + 10)/100) × (100 - 10)/100) 

⇒ 100 × (110/100) × (90/100) 

⇒ 99 

Final amount is less than the initial amount(N)

∴  Option 1 will be the correct choice.

Given :

ab + xy - xb = 0 and bc + yz - cy = 0

Calculations :

According to the question 

ab + xy - xb = 0 

⇒ ab = xb - xy 

⇒ ab = x(b -y) 

⇒ x/a = b/(b - y) 

Now 

bc + yz - cy = 0 

⇒ yz = cy - bc 

⇒ yz = c(y - b) 

⇒ c/z = y/(y - b) 

Now, 

(x/a) + (c/z) =  (b/(b - y) + (y/(y - b)

⇒ (b/(b - y) - (y/(b - y)

⇒ (b - y)/(b - y) 

⇒ 1 

∴ Option 3 will be the correct answer.

Given:

Polynomial f(x) = x6 - 3x4 + 3x2 - 1 

And polynomial g(x) = x3 + 3x2 + 3x + 1

Formula Used:

(x² - y²) = (x + 1)(x - y)

(x + y)³ = x³ + 3x²y + 3xy² + y³

(x - y)3 = x3 - 3x2y + 3xy2 - y3

Shortcut Trick

f(x) = x6 - 3x4 + 3x2 - 1

⇒ f(x) = (x²)³ - 3(x²)².1 + 3.1.(x²) - 1

⇒ f(x) = (x² - 1)³

⇒ f(x) = (x - 1)3 (x + 1)3

g(x) = x3 + 3x2 + 3x + 1

⇒ g(x) = (x + 1)3

Thus,HCF of f(x) and g(x) = (x + 1)3

∴ The correct answer is option (4).

Alternate Method

Calculation:

According to the question, we have

f(x) = x6 - 3x4 + 3x2 - 1 

⇒ x⁶ - x⁴ - 2x⁴ + 2x² + x² - 1

⇒ x⁴(x² - 1) - 2x²(x² - 1) +1(x² + 1)

⇒ (x² - 1)(x⁴ - 2x² + 1)

⇒ (x + 1)(x - 1)(x4 - 2x2 + 1)

⇒ (x + 1)(x - 1)(x² - 1)²

⇒ (x + 1) (x - 1) (x + 1)² (x - 1)²

⇒ (x + 1)³(x - 1)³      ----(1)

And,

g(x) = x3 + 3x2 + 3x + 1

⇒ x³ + x² + 2x² + 2x + x + 1

⇒ x²(x + 1) + 2x(x + 1) + 1(x + 1)

⇒ (x + 1)(x² + 2x + 1)

⇒ (x + 1) (x + 1)²

⇒ (x + 1)³      ----(2)

Now,

The HCF of f(x) and g(x) is the common factor of equations (1) and (2)

⇒ (x + 1)³

∴ The HCF of the polynomials x6 - 3x4 + 3x2 - 1 and x3 + 3x2 + 3x + 1 is (x + 1)³.

Given:

The HCF of two polynomials = 3x + 1

The LCM of two polynomials = 30x3 + 7x2 - 10x - 3

One polynomial = 6x2 + 5x + 1

Formula used:

The product of two polynomials = Product of their HCF and LCM

Calculation:

Let the two polynomials be f(x) and g(x).

Here, f(x) = 6x2 + 5x + 1

According to the question,

f(x) × g(x) = (3x + 1) × (30x3 + 7x2 - 10x - 3)

⇒ (6x2 + 5x + 1) × g(x) = (3x + 1) × (30x3 + 7x2 - 10x - 3)

⇒ (6x2 + 2x + 3x + 1) × g(x) = (3x + 1) × (30x3 + 15x2 - 8x2 - 4x - 6x - 3)

⇒ [2x(3x + 1) + 1(3x + 1)] × g(x) = (3x + 1)[15x²(2x + 1) - 4x(2x + 1) - 3(2x + 1)] 

⇒ (3x + 1) × (2x + 1) × g(x) = (3x + 1) × (2x + 1) × (15x² - 4x - 3)

⇒ g(x) = (15x2 - 4x - 3)

∴ The other polynomial is 15x2 - 4x - 3.

Given:

(p + 2)(2q - 1) = 2pq - 10 

(p - 2)(2q - 1) = 2pq - 10

Calculation:

(p + 2)(2q - 1) = 2pq - 10

⇒ 2pq - p + 4q - 2 = 2pq - 10

⇒ 4q - p + 10 - 2 = 0     

⇒ 4q - p + 8 = 0      ----(1)

(p - 2)(2q - 1) = 2pq - 10

⇒ 2pq - p - 4q + 2 = 2pq - 10

⇒ -4q - p + 10 + 2 = 0

⇒ -4q - p + 12 = 0     

⇒ 4q + p -12 = 0      ----(2)

Now, add equation (1) and (2)

4q - p + 8 + 4q + p -12 = 0

⇒ 8q - 4 = 0

⇒ 8q = 4

⇒ q = 1/2

By putting the value of q in equation (1), we get

4(1/2) - p + 8 = 0

⇒ 2 - p + 8 = 0

⇒ 10 - p = 0

⇒ p = 10

Now, pq = 10 × (1/2)

⇒ 5

∴ The value of pq is 5.

Mistake Points

We can't compare p + 2 = p - 2 as these two are the terms of the two

given equations. When we divide the given equation we get 0/0 form.

So we can't do this, and to satisfy the equations you have to put the

value of p and q separately in both the equations.

Given:

The given expression is \(\frac{a^{2}~+~ac}{a^{2}c~-~c^{3}}-\frac{a^{2}~-~c^{2}}{a^{2}c~+~2ac^2~+~c^3}-\frac{2c}{a^2~-~c^2}+\frac{3}{a~+~c}\)

Shortcut TrickLet, a  =0, c = 1

Put these values on a given expression, and we get

⇒  \(0-\frac{-1}{1}-\frac{2}{-1}+\frac{3}{1}\)

⇒  1 + 2 + 3 = 6

From option 4

6/(0 + 1) = 6

∴ Option 4 is correct.

 Alternate Method

\(\frac{a^{2}~+~ac}{a^{2}c~-~c^{3}}-\frac{a^{2}~-~c^{2}}{a^{2}c~+~2ac^2~+~c^3}-\frac{2c}{a^2~-~c^2}+\frac{3}{a~+~c}\)

⇒ \(\frac{a(a~+~c)}{c(a^2~-~c^2)}-\frac{a^2~-~c^2}{c(a^2~+~2ac~+~c^2)}-\frac{2c}{a^2~-~c^2}+\frac{3}{a~+~c}\)

⇒ \(\frac{a(a~+~c)}{c(a~-~c)(a~+~c)}-\frac{(a~-~c)(a~+~c)}{c(a~+~c)^2}-\frac{2c}{a^2~-~c^2}+\frac{3}{a~+~c}\)

⇒ \(\frac{a}{c(a~-~c)}-\frac{(a~-~c)}{c(a~+~c)}-\frac{2c}{(a~-~c)(a~+~c)}+\frac{3}{a~+~c}\)

⇒ \(\frac{a(a~+~c)~-~(a~-~c)^2~-~2c^2~+~3c(a~-~c)}{c(a~-~c)(a~+~c)}\)

⇒ \(\frac{a^2~+~ac~-~(a^2~-~2ac~+~c^2)~-~2c^2~+~3ca~-~3c^2}{c(a~-~c)(a~+~c)}\)

⇒ \(\frac{a^2~+~ac~-~a^2~+~2ac~-~c^2~-~2c^2~+~3ca~-~3c^2}{c(a~-~c)(a~+~c)}\)

⇒ \(\frac{6ac~-~6c^2}{c(a~-~c)(a~+~c)}\)

⇒ \(\frac{6c(a~-~c)}{c(a~-~c)(a~+~c)}\)

⇒ \(\frac{6}{a~+~c}\)

∴ The value of \(\frac{a^{2}~+~ac}{a^{2}c~-~c^{3}}-\frac{a^{2}~-~c^{2}}{a^{2}c~+~2ac^2~+~c^3}-\frac{2c}{a^2~-~c^2}+\frac{3}{a~+~c}\) is \(\frac{6}{a~+~c}\).

Given:

The given expression is 4x⁴ + 8x³ - 4x + 1

Formula used:

(a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ca

Calculation:

4x4 + 8x3 - 4x + 1

⇒ 4x4 + 8x3 - 4x + 1 + 4x² - 4x²

⇒ 4x4 + 4x2 + 1 + 8x3 - 4x - 4x2

⇒ (2x²)² + (2x)² + (1)² + 2 × 2x2 × 2x - 2 × 2x × 1 - 2 × 2x2 × 1

⇒ (2x² + 2x - 1)²

So, \(\sqrt{4x^4 + 8x^3 - 4x + 1}\)

⇒ \(\sqrt{(2x^2 + 2x - 1)^2}\)

⇒ 2x2 + 2x - 1

∴ The square root of 4x4 + 8x3 - 4x + 1 is 2x2 + 2x - 1.

Alternate Method x = 1

4x4 + 8x3 - 4x + 1

4 + 8 - 4 + 1 = 9

From option (4)

2x2​ + 2x - 1

2 + 2 - 1 = 3

3² = 9

Shortcut Trick 

⇒ x + y = 13    ......(1)

⇒ (10y + x) - (10x + y) = 27

⇒ y - x = 3        .....(2)

⇒ x × y = 5 × 8 = 40

∴ The product of the digits of the number is 40.

Alternate Method 

Given:

The sum of the digits of a two-digit number = 13

The difference between the number and that formed by reversing the digits = 27

Calculation:

Let the digit in the 10^th place and unit place are a and b respectively.

So, the two-digit number is 10a + b.

According to the question,

a + b = 13      ----(1)

Again, according to the question,

(10b + a) - (10a + b) = 27

⇒ 10b + a - 10a - b = 27

⇒ 9b - 9a = 27

⇒ 9(a - b) = 27

⇒ a - b = 3      ----(2)

By adding equ(1) and equ(2), we get

(a + b) + (a - b) = 13 + 3

⇒ 2a = 16

⇒ a = 8

From equ(1), b = 13 - a

⇒ 13 - 8 = 5

Now, the product of the digits = 8 × 5 = 40

∴ The product of the digits of the number is 40.

Given :

x/(b + c) = y/(c + a) = z/(b - a) 

Calculations :

Let, x/(b + c) = y/(c + a) = z/(b - a) = k 

⇒ x = (b + c)k = bk + ck 

⇒ y = (c + a)k = ck + ak 

⇒ z = (b - a)k = bk - ak 

Now check for each option given 

1. x + y + z = bk + ck + ck + ak + bk - ak 

⇒ 2(bk + ck) 

2. x - y - z = bk + ck - (ck + ak ) - (bk - ak) 

⇒ bk + ck - ck - ak - bk + ak

⇒  0 

3. x + y - z = bk + ck + ck + ak - (bk - ak) 

⇒ bk + ck + ck + ak - bk + ak

⇒ 2(ak + ck)

4. x + 2y + 3z = bk + ck + 2(ck + ak) + 3(bk - ak) 

⇒ bk + ck + 2ck + 2ak + 3bk - 3ak 

⇒ 4bk + 3ck - ak 

∴ Option 2 will be the correct choice.

Given:

Speed of X, Y and Z = 4 km/h, 5 km/h and 6 km/h.

Formula used:

Distance = Speed × Time

The relative speed of Y with respect to X = (S_x - S_y)

Where S_x and S_y are the speed of X and Y respectively.

Calculation:

Distance covered by X in 2 hours = 4 × 2 = 8 km

Time required for Y to cover this distance = 8/(5 - 4) = 8 hours

Let Z starts t hours after Y.

According to the question,

Distance covered by Y = Distance covered by Z 

⇒ 8 × 5 = (8 - t) × 6

⇒ 40 = 48 - 6t

⇒ 6t = 8

⇒ t = 4/3 

∴ Z starts travelling 4/3 hours after Y.

Given:

\(1-x-x^{n}+x^{n+1}\)

Shortcut Trick

Concept used:

aⁿ - bⁿ is divisible by (a - b); if n = odd

Calculations:

1 - x - xⁿ + xⁿ⁺¹

⇒ (1 - x) - xn​ (1 - x)

⇒ (1 - x)(1 - xn)

Now, (1 - xn) = (1ⁿ - xn) 

Here, n = odd, so it is divisible by (1 -x)

Thus, expression is divisible by 

⇒ (1 -x)(1 -x) = (1 -x)²

∴ The correct answer is option (2).

Alternate Method

Formula Used:

(1 - xn) = (1 - x)(1 + x + x2 + x3 + ...... + xn - 1)

Calculation:

Let, f(x) = \(1-x-x^{n}+x^{n+1}\)

= (1 - x) - xⁿ(1 - x)

= (1 - x)(1 - xⁿ)

= (1 - x)(1 - x)(1 + x + x² + x³ + .....+ x^{n - 1})             [∵ (1 - xⁿ) = (1 - x)(1 + x + x² + x³ + ...... + x^{n - 1})]

= (1 - x)²(1 + x + x2 + x3 + .....+ xn - 1)

∴  \(1-x-x^{n}+x^{n+1}\) is divisible by (1 - x)²

Given:

Cost price of the article = Rs. x

Selling price of the article = Rs. 75

Profit% = x%

Formula used:

Profit% = [(S.P - C.P)/C.P] × 100

Calculation:

According to the question,

x = [(75 - x)/x] × 100

⇒ x² = 7500 - 100x 

⇒ x(x + 100) = 7500

⇒ x(x + 100) = 50(50 + 100)

On comparing both sides,

x = 50

∴ The value of x is 50.