Quantitative Aptitude PYQs

Given:

Rate of interest changes from 11% to \(17\frac{1}{2}\)%

Due to change of rate of interest yearly income increases by Rs. 1071.20

Formula used:

Simple Interest = (P × R × T)/100

Where,

P → Principal 

R → Rate 

T → Time

Calculation:

Difference between two interest rates = \(17\frac{1}{2}\)% – 11% = (13/2)%

⇒ Increased income = (P × R × T)/100

⇒ 1071.20 = (P × (13/2) × 1)/100

⇒ (107120 × 2)/13 = P

⇒ P = 16480 

Simple Interest = (P × R × T)/100

⇒ (16480 × 10 × 5)/100

⇒ 1648 × 5

⇒ 8240 Rs.

∴ The simple interest for five years at 10% is 8240 Rs.

Alternate Method Difference between two interest rates = \(17\frac{1}{2}\)% – 11% = (13/2)%

Increased income = (P × R × T)/100

⇒ 1071.20 = (P × (13/2) × 1)/100

⇒ P = 16480 

Simple Interest = (P × R × T)/100

⇒ (16480 × 10 × 5)/100

⇒ 8240

Ans: 3

Given:

5x2y6z is divisible by 7, 11 and 13

Concept used:

7 × 11 × 13 = 1001, when any 3 digit number multiplied by 1001 then it repeats itself and always completely divisible by 7,11 and 13.

Calculation:

Let 3 digit number be pqr.

⇒ pqr × 1001 = 5x2y6z

⇒ pqrpqr = 5x2y6z

Compare both sides

⇒ p = 5, q = 6 and r = 2

Hence, that number is 562562.

⇒ 5x2y6z = 562562

by comparing both numbers, x = 6, y = 5, and z = 2 

Required value = (x – y + 3z)

⇒ 6 – 5 + 3 × 2

⇒ 7

∴ The required value is 7.

Alternate Method When any 3 digit number multiplied by 1001 then it repeats itself and always completely divisible by 7,11 and 13.

Let 3 digit number be PQR.

⇒ PQRPQR = 5x2y6z

⇒ p = 5, q = 6 and r = 2

Hence, that number is 562562.

⇒ 5x2y6z = 562562

⇒ x = 6, y = 5, and z = 2 

Required value = (x – y + 3z)

⇒ 6 – 5 + 3 × 2

⇒ 7

Given:

(72 × 73 × 78 × 76) divided by 35

Concept Used:

We divide every number individually by 35 and remainder left will be operated as signs given in the original equation, and if the operated value is more than 35 then we will divide it further till we get values less than 35 and that value will be remainder for the given equation.

Calculation:

(72 × 73 × 78 × 76)/35

⇒ (2 × 3 × 8 × 6)/35

⇒ 288/35

⇒ 8/35

Since 8 cannot be divided further by 35

∴ The remainder of the given equation is 8

Given: 

AB is a diameter.

ABCD is a cyclic quadrilateral in which ∠ADC = 118°

Concept used:

The angle formed by the diameter of the circle at the circumference of the circle is always 90°   

ABCD is a cyclic quadrilateral 

So, ∠ADC + ∠ABC = 180° 

Calculation:

In a cyclic quadrilateral, the sum of each pair of opposite angles is 180 degrees.

⇒ ∠ADC + ∠ABC = 180°

⇒ 118° + ∠ABC = 180°

⇒ ∠ABC = 62°

∠ACB = 90°      ----(The angle formed by the diameter of the circle at the circumference of the circle always 90°)

In ΔABC,

∠BAC + ∠ACB + ∠ABC = 180°

⇒ ∠BAC + 90° + 62° = 180°

⇒ ∠BAC = 28°

∴ The value of ∠BAC is 28°

Given :

Asha fort height  = 9 cm

Suman fort height = 16 cm

Distance between the top of the forts = 25 cm 

Formula Used:

Pythagoras theorem: (Base)² + (Perpendicular)²  = (Hypotenuous)²  

Calculation:

Let the distance between the fort  CD be x

From the above diagram

Height of the Suman fort AC = 16 cm 

Height of the Asha fort ED = 9 cm 

Here the BCDE forms a rectangle 

The opposite side is equal BE = CD = x and BC = DE = 9 cm                                            

By Pythagoras theorem 

AB2 + BE2  = AE2 

⇒ 7² + x² = 25²

⇒ x² = 625 – 49 

⇒ x = √576

⇒ x = 24 

∴ The Distance between forts = 24 cm

Given:

Class donated a sum of Rs. 2,809

Each student donated as many rupees as the number of students in the class.

Calculation:

Let the total number of the students be 'x'

Each student will donate 'x' amount of rupees.

So, total class fund = Donation of each student × Total number of students

⇒ 2,809 = x × x

⇒ 2,809 = x² 

⇒ x = 53

∴ Total number of students in class is 53

Given:

\(\frac{{{{\sin }^2}30^\circ + {{\cos }^2}60^\circ + sec45^\circ .sin45^\circ }}{{sec60^\circ + cosec30^\circ }}\)

Important Trigonometric Identities:

​

Calculation:

\(\frac{{\left[ {{{\left( {\frac{1}{2}} \right)}^2} + ;{{\left( {\frac{1}{2}} \right)}^2} + \left( {\sqrt 2 \ \times\ \frac{1}{{\sqrt 2 }}} \right)} \right]}}{{2\ + \ 2}} \)

= (3/2)/4

= 3/8

∴ Simplified value of given equation is 3/8

Given:

sin3x = cos(3x – 45°)

Formula Used :

sin(90° – θ) = cosθ 

Calculations:

sin3x = cos(3x – 45°)

⇒ cos(90° – 3x) = cos(3x – 45°)

⇒ 90° – 3x = 3x – 45°

⇒ 6x = 135°

⇒ x = 22.5

∴  The value of x is 22.5°

Given:

Δ ABC ∼ Δ RPQ

\(\frac{{ar\left( {ABC} \right)}}{{ar\left( {RPQ} \right)}}\) =  \(\frac{4}{9}\)

AB = 3 cm

BC = 4 cm

AC = 5 cm

Concept used:

Δ ABC ∼ Δ RPQ

 \(\frac{{AB}}{{RP}} = ;\frac{{BC}}{{PQ}} = ;\frac{{AC}}{{QR}} = √\frac{{ar\left( {ABC} \right)}}{{ar\left( {RPQ} \right)}}\)

Calculation: 

AB/RP = √(4/9)

⇒ AB/RP = 2/3 

⇒ RP = (3/2) × AB

⇒ (3/2) × 3 = 4.5

∴ The value of RP is 4.5 cm.

Shortcut TrickΔ ABC ∼ Δ RPQ

AB/RP = √(4/9) 

RP = 3/2 × 3  (∵ AB = 3)

RP = 4.5

Given:

L₁ = 17 cm

L₂ = 15 cm

Height (H) = 6 cm

Concept used:

Area of trapezium = (1/2) × (Sum of parallel side) × Height 

1 cm = (1/100) m 

Calculation:

Let two parallel side be L1 and L2

Area of trapezium = (1/2) × (17 + 15) × 6

⇒ 96 cm²

⇒ 96/10000 m² = 0.0096 m²

∴ The area of the trapezium is 0.0096 m².

Shortcut Trick Area of trapezium = (1/2) × (17 + 15) × 6

⇒ 96 cm2

= 0.0096 m2

Given: 

Breadth of rectangle (B) = 12 m

Length of rectangle (L) = 2 × Breadth = 24 m

Formula used:

Area of rectangle = L × B

Area of semicircle = πr²/2

Where, 

L → Length of the rectangle 

B → Breadth of the rectangle 

r → Radius of circle 

Calculation:

Four semi-circular portions are constructed on the sides of the rectangle

Let radius of small semi-circle be r and big semi-circle be R

⇒ AD = 2r and AB = 2R

⇒ r = 6 and R = 12

Total area of the newly constructed lawn = Area of rectangle + Area of the four semi-circles

⇒ L × B + 2(small semi-circle area) + 2(Big semi-circle area)

⇒ 12 × 24 + 2(3.14 × 36)/2 + 2(3.14 × 144)/2

⇒ 288 + (3.14 × 36) + (3.14 × 144)

⇒ 288 + 3.14 × 180

⇒ 288 + 565.2

⇒ 853.2 m²

∴ Total area of lawn is 853.2 m²

Alternate Method Breadth of rectangle (B) = 12 m

Length of rectangle (L) = 2 × Breadth = 24 m

⇒ AD = 2r and AB = 2R

⇒ r = 6 and R = 12

Total area of the newly constructed lawn = Area of rectangle + Area of the four semi-circles

⇒ L × B + 2(small semi-circle area) + 2(Big semi-circle area)

⇒ 12 × 24 + 2(3.14 × 36)/2 + 2(3.14 × 144)/2

⇒ 288 + (3.14 × 36) + (3.14 × 144)

⇒ 288 + 565.2

⇒ 853.2 m2

Given:

Area of the hexagon field = 1944\(√ {3}\) m²  

Rate of fencing the area = 11.50 per metre 

Formula Used :

Area of the hexagon = 3√3 × (a²/2)

The perimeter of the hexagon = 6a

Cost of the fencing = 6a × Rate per metre

Where,

a → side of the hexagon

Calculation:

Substituting the values in the Formula

3√3 × (a2/2) = 1944√3

⇒ a = 36 m

The perimeter of the hexagon = 6a = 6 × 36 = 216 m 

Cost of fencing = 216 × 11.50 = Rs. 2,484

∴ Cost of fencing field in the shape of a hexagon is Rs. 2,484

Given:

\(2 \frac{1}{3} ÷ 2 \frac{1}{2} \ of \ 1 \frac{3}{5} + (\frac{3}{8} + \frac{1}{7} \times 1 \frac{3}{4})\)

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Calculations:

\(2 \frac{1}{3} ÷ 2 \frac{1}{2} \ of \ 1 \frac{3}{5} + (\frac{3}{8} + \frac{1}{7} \times 1 \frac{3}{4})\)

⇒ (7/3) ÷ (5/2) of (8/5) + {(3/8) + (1/7) × (7/4)}                

⇒ (7/3)/(4) + {(3/8) + (1/4)}

⇒ (7/12) + (5/8)

⇒ (14 + 15)/24 = 29/24

∴ The correct value is (29/24).

Given:

Percentage reduction on the amount of bill if paid before due date = 12%

Reduction amount = Rs.2100

Formula Used: 

Total Tax Amount = (100/reduction%) × reduction amount 

Tax amount paid = Total Tax amount - reduced amount

Calculation:

Total tax amount = (100/12) × 2100 = Rs .17500

Total amount paid = 17500 – 2100 = Rs.15400

∴ The amount of house tax paid was is Rs. 15400

Given:

Numbers 4, 5, 8, 10, 12

Concept Used:

Least Number divisible by a set of Number = Least common multiple (LCM) of that set of numbers

Calculation:

4 = 2²

5 = 5¹

8 = 2³

10 = 2¹ × 5¹

12 = 2² × 3¹

Least Common Multiple (LCM) of 4, 5, 8, 10, 12 = 2³ × 3¹ × 5¹ 

⇒ 8 × 3 × 5 = 120

∴ 120 is the least number which is exactly divisible by 4, 5, 8, 10, 12.

Additional Information

Least common multiple: The least common multiple of numbers is the smallest non-zero common numbers which are the multiple of all the numbers.

Calculations:

Percentage of boys in school A = (600/1100) × 100 = 54.54%

Percentage of boys in school B = (450/1000) × 100 = 45%

Percentage of boys in school C = (750/1400) × 100 = 53.57%

Percentage of boys in school D = (700/1300) × 100 = 53.84%

Percentage of boys in school E = (800/1450) × 100 = 55.68%

∴ In school E percentage of boys is more than 55%

Given:

x² + 8y² + 12y – 4xy + 9 = 0

Formula used:

(A - B)² = A² – 2AB + B²

(A + B)² = A² + 2AB + B²

Calculation:

x² + 8y² + 12y – 4xy + 9 = 0

⇒ x² – 4xy + 4y² + 4y² + 12y + 9 = 0

⇒ (x – 2y)² + (2y + 3)² = 0

⇒ x – 2y = 0, 2y + 3 = 0  

⇒ x = 2y, 2y = -3  

⇒ x = -3, y = -3/2

Now,

7x + 8y = 7 × (-3) + 8 × (-3/2)

⇒ -21 – 12 = -33

∴ The correct answer is -33

Given :

Total sum of money = 8200

A had Rs. 500 more than B and C had Rs. 300 more than A.

Concept used:

Using linear equation. 

Calculation :

Let the money of B be 'x'

So, money of A = x + 500

Now, money of C = x + 500 + 300 = x + 800

According to the Question ,

x + x + 500 + x + 800 = 8200

⇒ 3x + 1300 = 8200

⇒ 3x = 8200 – 1300

⇒ 3x = 6900

⇒ x = 2300

So, C's share = 2300 + 800 = Rs. 3100

∴ C's share is Rs. 3100.

Given:

Successive discount = 6%, 15% and 14%

Concept used:

S.P = M.P × (100 – D₁)/100 × (100 – D₂)/100 × (100 – D₃)/100

Discount = M.P – S.P 

Discount% = ((Discount)/(M.P)) × 100

Calcualtions:

Let MP be 100x

⇒ S.P = 100x × (100 – 6)/100 × (100 – 15)/100 × (100 – 14)/100

⇒ (100x × 94 × 85 × 86)/100 × 100 × 100)

⇒ 68.714x

Discount = 100x – 68.714x = 31.286x

Discount% = (31.286x/100x) × 100 = 31.286%

∴ The equivalent discount% is 31.286%

Alternate Method Let MP be 100x

⇒ S.P = 100x × (100 – 6)/100 × (100 – 15)/100 × (100 – 14)/100

⇒ (100x × 94 × 85 × 86)/100 × 100 × 100)

⇒ 68.714x

Discount = 100x – 68.714x = 31.286x

Discount% = (31.286x/100x) × 100 = 31.286%

Given:

Days taken by  A and  B together to do total work = 25 days

Days taken by B to do \(66\frac{2}{3}\)% of total work = 30 days

Formula used:

No of days taken = (Total work)/(Efficiency)

Calculation:

Days taken by B to do (\(66\frac{2}{3}\)% = 2/3) ​of total work = 30

⇒ Days taken by B to do total work = 30 × 3/2 = 45 days

Total work = LCM of (25, 45) = 225 

The efficiency of A and B together = 225/25 = 9

The efficiency of  B  = 225/45 = 5

The efficiency of  A = 9 – 5 = 4

Days taken by A to do (4/15) part of total work = (225/4) × (4/15) = 15 days

∴ Time taken by A to do (4/15) part of the total work is 15 days

Alternate Method 

 Days taken by  A and  B together to do total work = 25 days

 Days taken by B to do total work = 30 × 3/2 = 45 days

Total work = LCM of (25, 45) = 225

The efficiency of A and B together = 225/25 = 9

The efficiency of  B  = 225/45 = 5

The efficiency of  A = 9 – 5 = 4

Days taken by A to do (4/15) part of total work = (225/4) × (4/15) = 15 days