Quantitative Aptitude PYQs

Given:

\(\left[\frac{{40 + \frac{3}{4}\;of\;32}}{{37 + \frac{3}{4}\;of\;\left( {34 - 6} \right)}}\right]\)

Concept used:

Calculation:

\(\left[\frac{{40 + \frac{3}{4}\;of\;32}}{{37 + \frac{3}{4}\;of\;\left( {34 - 6} \right)}}\right]\)

⇒ [40 + (3/4) × 32]/[37 + (3/4) × 28]

⇒ [40 + 24]/[37 + 21]

⇒ 64/58

⇒ \(1 \frac{3}{29}\)

∴ The value is \(1 \frac{3}{29}\).

Given:

The number of students weigh less than 55 kg and the number of students weight 55 kg to 65 kg

Concept Used:

Concept of percentage, it is calculated on the base of 100

For example, x% means x out of 100

Calculation:

The number of students weight less than 55 kg is (20 + 45 + 30)

⇒ 95

The number of students weight from 55 kg to 65 kg is (60 + 40)

⇒ 100

The required percentage of students {(100 - 95)/100} × 100

⇒ 5%

∴ The number of students weighing less than 55 kg is 5% percent less than the number of students weighing 55 kg to 65 kg.

Given:

40% of other category is mix flavour

People preferred mix fruits = 1570

Calculation:

Let the total number of people be x

People eating other flavoured ice cream = 15.7% of x

⇒ People eating other flavoured ice cream = 0.157x

People having mix flavoured ice cream = 40% of 0.157x

⇒ People having mix flavoured ice cream = 0.0628x

According to question 0.0628x = 1570

⇒ x = 1570 ÷ 0.0628

⇒ x = 25000

∴ Total people surveyed are 25,000

Shortcut Trick

Ratio of the two numbers = 7 : 11

Sum of the ratio parts = 7 + 11 = 18 units

Since HCF is the common factor, 1 unit = HCF = 28

Total sum of the numbers = 18 × 28 = 504

∴ The correct answer is 504.

Alternate Method

Given:

Ratio of numbers = 7 : 11

HCF of numbers = 28

Formula Used:

If two numbers are in the ratio a : b and their HCF is x, then the numbers are ax and bx.

7 Units (Number 1)
11 Units (Number 2)

Sum = 7 + 11 = 18 Units
Common Factor (HCF) = 28

Calculations:

⇒ Let the first number be 7x and the second number be 11x.

⇒ HCF of 7x and 11x is x (since 7 and 11 are co-prime).

⇒ x = 28

⇒ Sum of the numbers = 7x + 11x = 18x

⇒ Sum = 18 × 28

⇒ Sum = 504

∴ The correct answer is 504.

Additional Information

Relationship between Numbers, HCF and Ratio

If HCF of two numbers is H and they are in ratio a:b, then the numbers are aH and bH, and their LCM = a × b × H.

Product of Numbers

Product of two numbers = HCF × LCM. For the given problem, Product = (7 × 28) × (11 × 28).

Co-prime Numbers

Numbers that have only 1 as their common factor (HCF = 1) are called co-prime numbers. In a reduced ratio a:b, a and b are always co-prime.

Given:

Decrease percent = 17%

Concept used:

Using the concept of percentage

Area of square = (Side)²

Calculation:

Let the original side be 10.

Area of original square = (Side)²

⇒ Original area = 10²

New side = (100 – 17)% of 10

⇒ New side = 83/100 × 10

⇒ New side = 8.3

New area = (New side)²

⇒ New area = (8.3)²

⇒ New area = 68.89

Percent decrease in area = (Original area – New area)/(Original area) × 100

⇒ Percent decrease in area = (100 – 68.89)/100 × 100

⇒ Percent decrease in area = 31.11%

∴ The percentage decrease in area is 31.11%.

Shortcut TrickDecreasing(%) in area = - x - x + (- x × - x)/100

Decreasing(%) in area = - 17 - 17 + (- 17 × - 17)/100

⇒ - 34 + 2.89 = - 31.11

∴ The percentage decrease in area is 31.11%

Given:

A can do work in 10 days

B can do work in 5 days

A and B worked for 2 days

A works for 5 days

B works for 2 days

C works for 3 days

Concept used:

If a person can do a piece of work in x days, then the person will complete 1/x work in 1 day.

Calculation:

Total work = 10 unit

 

Total efficiency = 3

Two days of work of A and B = 3 × 2 = 6 unit

Remaining work = 10 - 6 = 4 unit

According to question

\(\frac{4}{1 + c}\) = 3

⇒ 1 + c = 4/3

⇒ c = 1/3

∴ C can complete 40% work =\(\frac{10 \times 40}{100\times 1/3}\)  = 12 days

Given:

Amount after 4 years (A₄) = Rs. 7,562

Amount after 5 years (A₅) = Rs. 8,469.44

Principal (P) = Rs. 10,000

Formula used:

CI = P × [(1 + R/100)ⁿ – 1]

Where CI is compound interest, P is principal, R is rate percent, and n is time in years

Calculation:

Let the rate of interest be R%

CI for fifth year = A₅ – A₄ = 8,469.44 – 7562

⇒ CI for fifth year = 907.44

CI for fifth year = R% of A₄ 

⇒ 907.44 = (R/100) × 7562

⇒ R = 90744/7562

⇒ R = 12%

CI = 10,000 × [(1 + 12/100)² – 1]

⇒ CI = 10000 × 0.2544

∴ CI is Rs. 2,544

Given:

Three successive discounts are 12%, 13% and 11%

Concept Used:

Concept of percentage. It is calculated on the base of 100

For example, x% means x out of 100

Calculation:

Let, the actual value is 100

After 1^st discount value becomes (100 - 12) = 88

In 2^nd discount given 13%

Discount = (13 × 88)/100

⇒ 11.4

After 2^nd discount value becomes (88 - 11.4) = 76.6

3^rd discount is 11%

Discount = (11 × 76.6)/100

⇒ 8.4

Overall single discount (12 + 11.4 + 8.4) = 32.8% = 32% (approximate)

∴ Three successive discounts of 12%, 13% and 11% are equivalent to an approximate single discount of 32%

Shortcut TrickSuccessive discount = {x + y - (x × y)/100}%

Successive discount = {12 + 13 - (12 × 13)/100}%

⇒ (25 - 1.56)% = 23.44%

Again,

Successive discount = {23.44 + 11 - (23.44 × 11)/100}%

⇒ (34.44 - 2.5784)%

⇒ 31.8616% = 32% (approximate)

∴ Three successive discounts of 12%, 13% and 11% are equivalent to an approximate single discount of 32%

Calculation:

Pipe A, B, and C can fill in 1 hour

⇒ 15 + 10 - 6

⇒ 19

(A + B + C) can fill in 10 hours

⇒ 19 × 10

⇒ 190

Remaining empty tank

⇒ 240 - 190

⇒ 50

Remaining empty tank filled by A and B

⇒ 50/25

⇒ 2 hours

∴ Time taken to fill remaining tank is 2 hours.

Given:

The perimeter of a square is half the perimeter of a rectangle. The perimeter of the square is 40 m. If its breadth is two­-thirds of its length.

Concept Used:

The perimeter of a square = 4 × length of the sides

The perimeter of a rectangle = 2 × (Length + Breadth)

Area of rectangle = Length × Breadth

Calculation:

The perimeter of the square is 40 m

The perimeter of a square is half the perimeter of a rectangle.

⇒ The perimeter of the rectangle is 2 × 40 = 80 m

Let, the length of the rectangle be x

⇒ The breadth of the rectangle is 2x/3

Accordingly,

2 × ( x + 2x/3) = 80

⇒ 5x/3 = 40

⇒ x = 24

Length of the rectangle is 24 m and breadth is (2/3) × 24 = 16 m

Area of the rectangle is (24 × 16) = 384 m²

∴ The area of the rectangle is 384 m²

Given:

Total sale of chocolate flavour = 25.8% of the total sale

Total sale of vanilla flavour = 16.5% of the total sale

Total sale of chocolate flavours is for Rs. 5,160

Calculation:

Let the total sale of all flavours be Rs.

x. Total sale of chocolate flavours = Rs. 5160

⇒ 25.8% of x = 5160

⇒ x = (5160/25.8) × 100

Total sale of vanilla flavor = 16.5% of x

⇒ Total sale of vanilla = (16.5/100) × (5160/25.8) × 100

⇒ Total sale of vanilla = Rs. 3300

∴ The sale of vanilla is Rs. 3300.

Given:

BO and CO are bisectors of ∠B and ∠C

∠BOC = 134°

Concept used:

If in a given ΔABC, If BO and CO are angle bisectors then ∠BOC = 90° + ∠A/2

The angle formed by the 

Calculation:

As, BO and CO are angle bisectors of ∠B and ∠C

∠BOC = 90° + ∠A/2

⇒ 134° = 90° + ∠A/2

⇒ ∠A = 2 × 44° 

∴ ∠A is 88°

Given:

x + y + z = 19, x2 + y2 + z2 = 133 and xz = y2, 

Formula used:

(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)

Calculation:

(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)

⇒ (19)2 = 133 + 2(xy + yz + y²)

⇒ 133 + 2[y(x + y + z)] = 361

⇒ 2y(19) = 361 - 133

⇒ y = 6

x + y + z = 19

⇒ x + z = 13

The possible value of x and z is 9 and 4.

x - z

⇒ 9 - 4

⇒ 5

∴ The value is 5.

Given:

4(cosec2 57 - tan2 33) - cos 90 + y × tan2 66 × tan2 24 = y/2

Formula used:

(i) cosec(90 - θ) = secθ

(ii) tan(90 - θ) = cotθ

(iii) sec2θ - tan2θ = 1

(iv) tanθ = 1/cotθ 

Calculations:

4(cosec2 (90 - 33) - tan2 33) - cos 90 + y × tan2 66 × tan2 (90 - 66) = y/2

⇒ 4(sec2 33 - tan2 33) - cos 90 + y × tan2 66 × cot2 66 = y/2

⇒ 4 × 1 - 0 + y × tan2 66 × 1/tan2 66 = y/2

⇒ 4 + y = y/2

⇒ y - y/2 = -4

⇒ y/2 = -4

⇒ y = -8

∴ The value of y is -8.

Given:

Salary of A is 30% more than salary of B.

Calculations:

Let the salary of B be 100x.

Salary of A  = 100x + 100x × 30%

⇒ 130x

Percentage Difference = {(130x - 100x)/130x} × 100

⇒ (30x/130x) × 100

⇒ 23.07% ≈ 23.1%

∴ B's salary is 23.1% less than that of A

Shortcut Trick

 

Percentage Difference = \(\frac{130-100}{130}\) × 100

⇒ 300/100

⇒ 23.07% ≈ 23.1%

∴ B's salary is 23.1% less than that of A

Given:

The number when divided by 15, 18 and 36 and leaves the same remainder 9.

The number is also divisible by 11.

Calculations:

The least number which will be completely divisible by the 15, 18 and 36 will be the LCM of 15, 18 and 36.

⇒ LCM of 15, 18 and 36 = 180

The least number will be completely divisible by the 15, 18 and 36 and leaves remainder 9 = 180k + 9

Where,

k = natural number

when k = 1

⇒ 189

Here 189 is not divisible by 11.

When k = 1, 2, 3, 4, 5. The number is not divisible by 11.

Now, k = 6

Number = 180 × 6 + 9

⇒ Number = 1089

The number 1089 is divisible by 11.

Sum of digits = 1 + 0 + 8 + 9

⇒ 18

∴ The sum of the digits of the least number which when divided by 15, 18 and 36 leaves the same remainder 9 in each case and is divisible by 11 is 18.

Given:

Side of rhombus = 13 cm

D1 (Diagonal) = 24 cm

Concept and Formula used:

Diagonals of a rhombus bisect each other at 90° 

Area of rhombus = 1/2 × (D1 × D2)

Where D1 and D2 are diagonals of the rhombus

Calculation:

Let the length of the other diagonal (D2) be 2x cm

As, diagonals of rhombus bisect each other at 90° 

AB = 24/2 = 12 cm

BC = 2x/2 = x cm

In ΔABC, ∠ABC = 90° 

AC2 = AB2 + BC2 

⇒ 132 = 122 + x2 

⇒ 169 – 144 = x2 

⇒ x = 5 cm

D2 = 2x = 10 cm

Area of rhombus = 1/2 × (D1 × D2) = 1/2 × (10 × 24)

∴ Area of rhombus is 120 cm2

Given:

Nine digit number = 785x3678y

Concept used:

The Law of Divisibility by 8 is the last three digits must be divisible by 8.

The Sum of digits should be divisible by 9.

Calculation:

785x3678y

If the number is divisible by 72 (8 × 9) then it must be divisible by 8 and 9.

Divisible by 8

Law of Divisibility by 8 is last three digits must be divisible by 8

⇒ 78y = 780 + y

⇒ 97 × 8 + ( 4 + y ) is divisible by 8

So value of 4 + y = 8

⇒ y = 4

Divisible by 9

Sum of digits should be divisible by 9

⇒ 7 + 8 + 5 + x + 3 + 6 + 7 + 8 + 4

⇒ 48 + x

⇒ x = 6

x + y

⇒ 6 + 4

⇒ 10

∴ The value of (x + y) is 10.

Given: 

The sides of the triangle are 24 cm, 26 cm, and 10 cm. The radius of those circle at the vertex is 4.2 cm

Concept Used:

The area made by three circles in three vertices of a triangle of the same radius will be half the area of the circle.

Calculation:

The radius of the circle in three vertices is 4.2 cm

The total angle created in three circles in the vertices is 180°  [∵ Sum of three angles of a triangle is 180°]

The area of the triangle which are covered by three sectors of the circle is (1/2) π (4.2)2

⇒ 27.72 cm2

∴ The area of the portion covered by the three sectors of the circle is 27.72 cm².

Given:

Average weight of A, B and C = 65 kg

Average weight of C and B = 61.5 kg

Average weight of A and C = 68.5 kg

Formula used:

Average = sum of weights/No of persons

Calculation:

Weight of A, B and C

⇒ (A + B + C)/3 = 65

⇒ A + B + C = 195

Weight of C and B

⇒ (C + B)/2 = 61.5

⇒ C + B = 123

Weight of A and C

⇒ (A + C)/2 = 68.5

⇒ A + C = 137

Weight of A

⇒ 195 - 123

⇒ 72

Weight of C

A + C = 137

⇒ C = 137 - 72

⇒ 65 Kg

∴ The weight of C is 65 Kg.