Quantitative Aptitude PYQs

**Shortcut TrickThree articles are bought at ₹200 eachOne article is sold at a loss of 12.5% and the overall profit is 20%**By Mixture & Alligation method

∴ The profit percentage on the other two articles is 36.25%.

Alternate MethodGiven:

CP of each article = ₹200

Loss % on one of the articles = 12.5%

Profit % on whole transaction = 20%

Formula Used:

1. Profit = SP – CP

2. Profit % = (Profit ÷ CP) × 100

3. Loss = CP – SP

4. Loss % = (Loss ÷ CP) × 100

Calculation:

CP of three articles = 200 × 3 = ₹600

Profit on whole transaction = 600 × (20/100) = ₹120

SP of three articles = 600 + 120 = ₹720

Loss on one article = 200 × (1/8) = ₹25

Selling price = 200 – 25 = ₹175

Profit on other two articles = (720 – 175) – 400 = ₹145

Profit % on other two articles = (145 ÷ 400) × 100 = 36.25%

∴ The profit percentage on the other two articles is 36.25%.

Shortcut TrickNew Price = \(\frac{1200}{10}\) × \(\frac{25}{100}\) = ₹30

Old Price = New Price × \(\frac{100}{75}\) = 30 × \(\frac{100}{75}\) = ₹40 

Alternate MethodGiven:

The price of sugar is reduced by 25%

Radha is able to buy 10 kg sugar more for ₹ 1,200

Concept Used:

To maintain the expenditure

Change in price of a commodity ∝ \(\frac{1}{Consumption}\)

Calculation:

Let the original price of sugar be ₹P

Quantity of sugar = \(\frac{1200}{P}\)

Reduced price = \(\frac{100 – 25}{100}\) × P = \(\frac{3}{4}\)P

Increased quantity of sugar = \(\frac{1200}{(3/4)P}\) = \(\frac{1600}{P}\)

According to question

(1600/P) – (1200/P) = 10

⇒ 10P = 400

⇒ P = ₹40

 ∴ The original price of sugar per kg is ₹40.

Shortcut Trick

Side of square = √4900 = 70 m.

Radius of circle = Side = 70 m.

Length of rectangle = (2/5) × 70 = 28 m.

Area of rectangle = 28 × 20 = 560 m².

∴ The correct answer is 560 m².

Alternate Method

Given: Area of square = 4900 m², Breadth of rectangle = 20 m.

Formula used: Area of square = side²; Area of rectangle = Length × Breadth.

⇒ Side of square (s) = √4900

⇒ s = 70 m

⇒ Radius of circle (r) = Side of square = 70 m

⇒ Length of rectangle (L) = (2/5) × 70

⇒ L = 2 × 14 = 28 m

⇒ Area of rectangle = L × Breadth

⇒ Area = 28 × 20 = 560 m²

∴ The correct answer is 560 m².

Additional Information

Properties of a Square

A square has all sides equal. Area = side² and Perimeter = 4 × side.

Properties of a Rectangle

A rectangle has opposite sides equal. Area = Length × Breadth and Perimeter = 2(Length + Breadth).

Circle Relations

The radius (r) is the distance from the center to the edge. Area = πr² and Circumference = 2πr.

Given:

Votes received by two candidates in terms of ratio = 3 : 10

Total votes received by successful candidate = 156200

Calculation:

Let the votes got by two candidates be 3x & 10x

10x = 156200

⇒ x = 15620

Total votes polled = 3x + 10x

⇒ 13 × 15620 = 203060

∴ The total votes polled are 203060.

Given:

The radius of a circle is increased by 10%

Formula Used:

Area of circle = πr²

Percentage change = (\(\frac{Final\ Value\ \pm \ Initial\ Value}{Initial\ Value}\)) × 100

Calculation:

Let the Initial radius be r

Initial area = πr²

Increased radius = \(\frac{100+10}{100}\) × r = 1.1r

New area = π(1.1r)²

Percentage increase =\(\frac{π(1.21)r^2 – πr^2}{πr^2}\)  × 100

Percentage increase = \(\frac{πr^2(1.21-1)}{πr^2}\)  × 100 = 21%

∴ The percentage increase in the area of circle is 21%.

Shortcut TrickApply successive formula

Y + Y + \(\frac{Y\times Y}{100}\) = 10 + 10 + \(\frac{10\times 10}{100}\) = 21%

∴ The percentage increase in its area is 21%.

Given:

Length of rectangular plot is 15 m more than its breadth.

Rate of cost of fencing is Rs.16.50 per m.

Total cost of fencing is Rs.4,950.

Formula Used:

Perimeter of rectangle = 2(L + B)

Calculation:

Length of plot L = B + 15

Perimeter of plot = 2(L + B)

⇒ 2 {(B + 15 ) + B}

⇒ 2 (2B + 15)

⇒ 4B + 30

Now, the cost of fencing the plot at the rate of Rs.16.50 per m is Rs.4,950.

⇒ 16.50 × (4B + 30) = Rs.4,950

⇒ 66B + 495 = 4950

⇒ 66B = 4950 - 495

⇒ 66B = 4455

⇒ B = 4455/66 = 67.5

Length = B + 15 

⇒ 67.5 + 15 = 82.5 m

∴ Length of the rectangular plot = 82.5 m

Given:

Speed of first part of journey = 160 km/h

Speed of second part of journey = 240 km/h

Total distance = 3840 km

Total time = 20 hr.

Formula Used:

Distance = Speed × Time

Calculation:

Let time taken to travel in the first part = x hr.

Time taken to travel the second part = 20 - x hr.

Distance travelled in the first part = 160x

Distance travelled in the second part = 240(20 - x)

According to question:

Distance travelled in the first part = Distance travelled in the second part

⇒ 160x +  240(20 - x) = 3840

⇒ 160x + 4800 - 240x = 3840

⇒ - 80x = - 960

⇒ x = (-960)/(-80)

⇒ x = 12

Time taken in first half = x = 12 hr.

Time taken in second half = 20 - x = 20 - 12 = 8 hr.

∴ Time taken in second half = 8 hr.

Alternate Method 

5 = 20 Hrs

2 = 8 Hrs

First take the LCM of 12, 10 and 11.

The LCM is 660

Now, we can see that in each number there is a difference of 1.

⇒ 12 - 11 = 1

⇒ 10 - 9 = 1

⇒ 11 - 10 = 1

⇒ So, subtract the LCM by 1

⇒ 660 - 1 = 659

∴ The smallest number is 659.

Given:

A and B borrow equal sum of money for 3 years at the rate of 5% for SI and CI respectively

Difference between interest of A and B is 427

Formula Used:

SI = (P × R × T)/100

CI = P (1 + R/100)^t - P

Calculation:

SI of A = (P × 3 × 5)/100 = 3P/20    ____(i)

CI of B = P (1 + 5/100)³ - P 

⇒ P ×  \(\frac{{21}}{{20}} × \frac{{21}}{{20}} × \frac{{21}}{{20}}\) - P

⇒ \(\frac{{9261P}}{{8000}}\) - 1

⇒ 1261P/8000    ____(ii)

According to question,

⇒ \(\frac{{1261P}}{{8000}} - \frac{{3P}}{{20}}\) = 427

⇒ 61P/8000 = 427

⇒ P = (427 × 8000)/61

⇒ P = Rs.56000

Interest Paid by B = 56000 (1 + 5/100)3 - P 

⇒ 56000 × \(\frac{{21}}{{20}} × \frac{{21}}{{20}} × \frac{{21}}{{20}}\)  - P

⇒ 56000 × (9261/8000) - P

⇒ 64827 - P

⇒ 64827 - 56000 = Rs.8827

∴ The sum borrowed is Rs.56,000 and interest paid by B is Rs.8827.

Given:

Brand A costing Rs. 15 per 100 g

Brand B costing Rs. 18 per 100 g

Sold the mixture at Rs. 20 per 100 g, making a profit of 20%

Formula used:

Profit% or Loss% = (Profit or Loss/Cost price) × 100

Calculation:

Let x gram of brand A and y gram of brand B is mixed.

For 100g brand A costs Rs. 15

For x g brand A will cost 15x/100

For 100g brand B costs Rs. 18

For y g brand B will cost 18x/100

So, the total cost price for (x + y) grams is

⇒ x + y = 15x/100 + 18y/100      ----(i)

Sold at Rs. 20 with 120% profit, Cost price will be

⇒ 100% = (20/120) × 100 = Rs. 100/6

Cost price for 100g is Rs. 100/6,

Now find cost price for (x + y) grams

⇒ x + y = (100/6) × {(x + y)/100}

⇒ x + y = (x + y)/6

Put value of equation(i)

⇒ 15x/100 + 18y/100 = (x + y)/6

⇒ 90x + 108y = 100x + 100y

⇒ 8y  = 10x

⇒ x/y = 8/10 = 4/5

∴ The ratio of the mixture of Brand A and B is 4 : 5

Sold at Rs. 20 with 120% profit, Cost price will be

⇒ 100% = (20/120) × 100 = Rs. 100/6

∴ The ratio of the mixture of Brand A and B is 4 : 5

Given:

The length of open cuboidal cistern is 1.6 m.

The breadth of open cuboidal cistern is 1.3 m.

The height of open cuboidal cistern is 53 cm.

The capacity of open cuboidal cistern is 900 litres.

The thickness of wall is 5 cm.

Formula Used:

Volume of Cuboidal  = length × Breadth × Height

Calculation:

The length of open cuboidal cistern = 1.6 m = 160 cm

The thickness of wall is 5 cm.

The resultant length of open cuboidal cistern = 160 - 2 × 5 = 150 cm

The breadth of open cuboidal cistern = 1.3 m = 130 cm

The thickness of wall is 5 cm.

The resultant breadth of open cuboidal cistern = 130 - 2 × 5 = 120 cm

The height of open cuboidal cistern = 53 cm 

The thickness of bottom is x cm.

The resultant height of open cuboidal cistern = (53 - x) cm

The capacity of open cuboidal cistern = 900 litres = 900 × 1000 cm³

⇒ 150 × 120 × (53 - x) = 900 × 1000 

⇒ 180 × (53 - x) = 9 × 1000 

⇒ 20 × (53 - x) = 1 × 1000 

⇒ 53 - x = 50

⇒ x = 3

∴ The thickness of bottom is 3 cm.

Given:

A shopkeeper bought 1000 kg  of rice.

A part sold at a 10% profit.

A remaining part at a 20% loss.

A loss of 11% was incurred on the whole transaction. 

Calculation:

A part sold at a 10% profit.

A remaining part at a 20% loss.

A loss of 11% was incurred on the whole transaction.

 

 

The ratio of mixture = 9 : 21 = 3 : 7

A shopkeeper bought 1000 kg  of rice.

The quantity of rice that was sold at a 20% loss = 7/(3 + 7) × 1000 = 700 kg

∴ The quantity of rice that was sold at a 20% loss is 700 kg.

Given:

A's income is 35% more than B's income.

Concept Used:

Percentage = (A - B)/A × 100

Calculation:

Let the value of B be x.

Then, the value of A = (100 + 35)/100 × x = 1.35x

The percentage of B's income less than A's income = (1.35x - x)/1.35x × 100 = 25.92% ≈ 26%

∴ The percentage of B's income less than A's income is 26%.

Let the value of B be 100. 

Then, the value of A = (100 + 35)/100 × 100 = 135

The percentage of B's income less than A's income = (135 - 100)/135 × 100 = 25.92% ≈ 26%

∴ The percentage of B's income less than A's income is 26%.

Given:

A tank full of petrol in a motorcycle lasts for 30 days.

Concept Used:

Total capacity of tank = Petrol used per day × No. of days

Calculation:

Let the quantity of petrol used per day be x.

The capacity of tank = 30 × x = 30x

According to the question;

A  rider starts using 20% more petrol every day.

The new quantity of petrol used per day = 120/100 × x = 1.2x

Let the number of days be y.

The capacity of tank = 30x

⇒ 30x = 1.2x × y

⇒ 30/1.2 = y

⇒ y = 25

∴ The number of days the tank full of petrol last is 25 days.

  

The number of days = 30/(100 + 20) × 100 = 30/1.2 = 25

∴ The number of days the tank full of petrol last is 25 days.

Given:

A can do a piece of work in 45 days.

A works for 30 days.

B completes the remaining work in 5 days.

Concept Used:

Work done = 1/Time

Calculation:

A can do a piece of work in 45 days.

A's 1 days work = 1/45

A's 30 days work = 1/45 × 30 = 2/3

The remaining work = 1 - 2/3 = 1/3

B's complete 1/3^th of work in 5 days

Then, B's 1 day work = 1/(3 × 5) = 1/15

B can complete the whole work = 15 days 

∴ B can complete the whole work is 15 days.

Given:

The average age of a family of 5 members 2 years ago was 18 years.

The average after a new member add to the family is 18 years.

Concept Used:

The average = Sum of the all numbers/Number of the numbers

Calculation:

The average age of a family of 5 members 2 years ago was 18 years.

The sum of age of all family member 2 years ago = 18 × 5 = 90 

The sum of age of all family member at present = 90 + 2 × 5 = 100

Let the age of new member be x.

The sum of age of a family of 6 member = 6 × 18 = 108

⇒ 100 + x = 108

⇒ x = 8 years

∴ The age of new member of family is 8 years.

Let the age of new member be x.

The sum of age of a family of 6 member = 6 × 18 = 108

⇒ 18 × 5 + 2 × 5 + x = 108

⇒ 90 + 10 + x = 108

⇒ x = 8 years

∴ The age of new member of family is 8 years.

Given:

The principal of each person is Rs.43892.

The first on SI at rate of 30% for 3 years.

The second on CI at rate of 30% for 3 years.

Formula Used:

SI = P × R × T/100

Amount = P[1 + R/100]ⁿ

CI = Amount - Principal

Calculation:

The principal of each person is Rs.43892.

The first on Simple Interest at rate of 30% for 3 years.

Simple Interest = 43892 × 30 × 3/100 = Rs.39502.8

The second on Compound Interest at rate of 30% for 3 years.

Amount = 43892[1 + 30/100]³

⇒ 43892 × 130/100 × 130/100 × 130/100

⇒ Rs.96430.72

Compound Interest = 96430.72 - 43892 ≈ Rs.52538

The difference between Compound Interest and Simple Interest = Rs.13036

The percentage = 13036/52538 × 100 = 24.8%

∴ The percentage simple interest be less than the compound interest is 24.8%.

Given:

The sum of two numbers is 26.

The difference of two number is 14.

Calculation:

Let the larger number be x.

Let the smaller number be y.

According to the question;

The sum of two numbers is 26.

⇒ x + y = 26

⇒ x = 26 - y

The difference of two number is 14.

⇒ x - y = 14

⇒ 26 - y - y = 14

⇒ 26 - 14 = 2y

⇒ 12 = 2y

⇒ y = 6

The larger number x = 26 - 6 = 20

Three times the larger number is divided by the smaller number 

⇒ 20 × 3/6 = 10

∴ The quotient is 10.

Given:

The marked Price of an article is Rs. 5000. Three successive discounts of 15%, 10% and 18%, are given on its marked price.

Formula Used:

Selling price = Marked price × (100 - Discount%)/100

Calculation:

The marked Price of an article is Rs. 5000. The Selling Price = 5000 × (100 - 15)/100 × (100 - 10)/100 × (100 - 18)/100

⇒ 5000 × 85/100 × 90/100 × 82/100

⇒ Rs.3136.5

∴ The Selling price of the article is Rs.3136.5.

Given:

\(\frac{{11}}{{12}},\frac{{13}}{{14}},\frac{{12}}{{13}},\frac{{15}}{{16}},\frac{{19}}{{20}}\)

Calculation:

Prime factor of 12 = 1 × 2² × 3

Prime factor of 14 = 1 × 2 × 7

Prime factor of 13 = 1 × 13

Prime factor of 16 = 1 × 2⁴ 

Prime factor of 20 = 1 × 22 × 5

Now the least common multiple of (12, 14, 13, 16 and 20) = 2⁴ × 3 × 7 × 13 × 5

⇒ 16 × 3 × 7 × 13 × 5

⇒ 21,840 

Now multiply each given fraction with this LCM, we get

(11/12) × 21,840

⇒ 20,020 

(13/14) × 21,840

⇒ 20,280

(12/13) × 21,840

⇒ 20,160

(15/16) × 21,840

⇒ 20,475

(19/20) × 21,840

⇒ 20,748

Now we can see only a fraction (11/12) gives the smallest value in all fractions.

∴ The smallest fraction from all fractions is (11/12)

Alternate Method

Concept used:

If series like n/(n + 1), (n + 1)/(n + 2), (n + 2)/(n + 3) and so on

The first term in the series gives the smallest value and the last term in the series gives the maximum value.

Calculation:

The fractions are 11/12, 13/14, 12/13, 15/16, and 19/20

After arranging we get,

11/12, 12/13, 13/14, 15/16, and 19/20

⇒ 11/(11 + 1), (11 + 1)/(11 + 2), (11 + 2)/(11 + 3), (11 + 4)/(11 + 5) and (11 + 8)/(11 + 9)

So the smallest value is 11/(11 + 1)

⇒ 11/12

∴ The smallest fraction from all fractions is (11/12)   

For this type of problem, we must remember the table of each denominator and then we divide each numerator from the respective denominator.

Now 11/12 = 0.916

13/14 = 0.928

12/13 = 0.923

15/16 = 0.937

19/20 = 0.95

So we can see that only a fraction (11/12) gives the smallest value.

∴ The smallest fraction from all fractions is (11/12)