Quantitative Aptitude PYQs

Shortcut Trick

For an equation in the form a² + b² + c² + Constant = 2(ka + lb + mc), the values are directly given by a = k, b = l, c = m.

Rewrite the given equation: a² + b² + c² + 84 = 2(2a − 4b + 8c).

Comparing terms: a = 2, b = −4, and c = 8.

Substitute in the expression: √(ab − bc + ca) = √((2 × −4) − (−4 × 8) + (8 × 2)) = √(−8 + 32 + 16) = √40.

∴ The correct answer is 2√10.

Alternate Method

Given: a² + b² + c² + 84 = 4(a − 2b + 4c)

Formula used: (x ± y)² = x² ± 2xy + y²

⇒ a² + b² + c² + 84 = 4a − 8b + 16c

⇒ (a² − 4a + 4) + (b² + 8b + 16) + (c² − 16c + 64) = 0

⇒ (a − 2)² + (b + 4)² + (c − 8)² = 0

Since the sum of squares is zero, each individual term must be zero:

⇒ a − 2 = 0 ⇒ a = 2

⇒ b + 4 = 0 ⇒ b = −4

⇒ c − 8 = 0 ⇒ c = 8

Required value = √(ab − bc + ca)

⇒ √((2 × −4) − (−4 × 8) + (8 × 2))

⇒ √(−8 + 32 + 16) = √40

⇒ 2√10

∴ The correct answer is 2√10.

Additional Information

Sum of Squares Property

If the sum of squares of real numbers is zero (x² + y² + z² = 0), then each individual term must be zero (x = 0, y = 0, z = 0).

Completing the Square

To convert x² + kx into a perfect square, add and subtract (k/2)² to get (x + k/2)².

Square Root Simplification

To simplify √n, factor out the largest perfect square: √40 = √(4 × 10) = 2√10.

Calculation:

Total number of students = 30 + 45 + 60 + 35 + 40 +35

⇒ Total number of students = 245

And students whose marks are more than 250 but less than 300 = 60

But 245 students represents 360^°

⇒ 1 student = 360^° /245

⇒ 60 students = (360^° /245) × 60

⇒ 60 students = 88.16^°

∴ The nearest value for students corresponding to 250 or more but lesser than 300 is 88^° .

Given:

Pump can fill the tank in 3 hours.

It took \(3\frac{1}{3}\) hours for the leakage.

Concept used:

Total time = [Total units/Total efficiency of pipe and leakage]

Calculation:

Let the leakage can empty the tank in y hours

Then, leakage will empty 1/y units in 1 hour

⇒ Pipe can fill 1/3 units in 1 hour

Because of the leak, it took \(3\frac{1}{3} = \frac{{10}}{3}\) hours to fill the tank.

⇒ The tank filled in 1 hour when the leak is there = 3/10 units

⇒ \(\frac{1}{3} - \frac{1}{y} = \frac{3}{{10}}\)

 ⇒ \(\frac{1}{y} = ;\frac{{10 - ;9}}{{30}}\)

⇒ y = 30

∴ The tank will be empty in 30 hours

Alternate Method

L.C.M of 3 and \(3\frac{1}{3}\) = 30

Efficiency of the pipe = 10

Efficiency of (pipe + leakage) = 9

Efficiency of leakage = (10 - 9) = 1

Time takes to empty the tank is 30/1 = 30 hours

∴ Required time to empty the tank is 30 hours

Given:

Length of circular path = 64cm

Diameter of Circular garden = 10m = 1000 cm

Radius of Circular garden = 10/2 m = 500 cm

Quantity I:

Inner radius = 500 cm 

Outer radius =  564 cm

Area of path = Outer area - Inner area

⇒ 22/7 × (5642 - 5002)

⇒ 22/7 × (564 + 500)(564 - 500)

⇒ 22/7 × (1064)(64)

⇒ 21.4016 m2

⇒ 21 m2(approx.)

Concept:

Follow the BODMAS rule according to the table given below:

Calculation:

The given expression is

25% of 400 + 35% of 1260 + 27% of 1800 = 1020 + x

⇒ (25/100) × 400 + (35/100) × 1260 + (27/100) × 1800= 1020 + x

⇒ 100 + 441 + 486 = 1020 + x

⇒ 1027 = 1020 + x

∴ x = 7

Given:

Length of the room = 12 m

Breadth of the room = 8 m

Height of the room = 10 m

Cost of whitewashing = 25 per m²

Formula used:

Lateral surface area = 2 × h × (l + b)

Area = l × b

l = Length

b = Breadth

h = Height

Calculation:

Area of walls = 2 × 10 × (12 + 8)

⇒ 20 × 20 = 400

Area of ceiling = 12 × 8 = 96

Total cost of whitewashing = 25 ×  (400 + 96) = 12400

∴ The total cost of whitewashing all four walls of the room and also the ceiling of the room is Rs. 12400

Given:

Present age ratio of A and B = 3 : 4

Ratio of age of A and B before 12 years = 2 : 3

Calculation:

Let the present age of A and B be 3x and 4x

⇒ (3x - 12)/(4x - 12) = 2/3

⇒ 3× (3x - 12) = 2× (4x -12)

⇒ 9x - 36 = 8x - 24

⇒ 9x - 8x = 36 - 24

⇒ x = 12

The total present age is 4x + 3x

⇒ 7x =7 × 12 = 84

∴ The total present age is 84years.

Given: 

Angle of elevation of Ladder = 60º 

Base of trianlge is 3.6 cm

Calculation:

In ∆ BAC

Cos 60º = 1/2

cosθ = B/H

1/2= 3.6/ AC

AC= 7.2 cm.

 

 

H = 7.2 m Here hypotenous is equal to length of ladder 

Note: The mentioned answer is accepted in the previous year's answer key of the test conducting agency (SSC CPO). The interpretation of language may be ambiguous but the Answer key has the same answer marked as correct.

Given:

PC = 12 cm

PB = 10 cm

Formula used:

PC² = AP × PB

Calculation:

Let AB be x cm

AP = AB + PB ⇒ x + 10

PC2 = AP × PB

⇒ 12² = 10 × (x + 10)

⇒ 144/10 = x + 10

⇒ x + 10 = 14.4

⇒ x = 14.4 -10 

⇒ x = 4.4

∴ The length of chord AB is 4.4 cm

The number of students who obtained 150 to 250 marks is 30

The number of students who obtained 200 to 250 marks is 45

The number of students who obtained 250 to 300 marks is 60

The number of students who obtained 300 to 350 marks is 35

The number of students who obtained 350 to 400 marks is 40

The number of students who obtained 400 to 450 marks is 35.

∴ Number of students who obtained less than 250 marks is 30 + 45 = 75 students.

Calculation:

L.C.M of 5,6 and 7 = 210

The first number divisible by 210 between 400 and 700 is 420.

The last number divisible by 210 between 400 and 700 is 630.

Tn = a + (n - 1) d

⇒ 630 = 420 + (n - 1) 210

⇒ 210 = (n - 1) 210

⇒ (n - 1) = 1

⇒ n = 2

Alternate Method:

L.C.M of 5,6 and 7 = 210

Quotient on dividing 400 by 210 = 1

Quotient on dividing 700 by 210 = 3

Hence, 3 - 1 = 2

∴ 2 is the answer.

Given:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\)

Identity used:

Sin θ = cos(90 - θ)

Sin² θ + cos² θ = 1

Calculation:

\(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\)

using Sin θ = cos(90 - θ)

⇒ \(\frac{{si{n^2}52^\circ ; + ;2; + ;si{n^2}\left( {90 - 52} \right)^\circ }}{{4co{s^2}43^\circ - 5; + ;co{s^2}\left( {90; - 43} \right)^\circ }}\)

⇒ \(\frac{{si{n^2}52^\circ ; + ;2; + ;co{s^2}52^\circ }}{{4co{s^2}43^\circ - 5; + ;si{n^2}43^\circ }}\)

Applying Sin2 θ + cos2 θ = 1

⇒ (1 + 2)/(4 - 5) ⇒ -3

∴ The value of \(\frac{{{{\sin }^2}{}52^\circ + 2 + {{\sin }^2}{{ }}38^\circ }}{{4{}{{\cos }^2}43^\circ - 5 + 4{{\cos }^2}{{}}47^\circ }}\) is -3.

Given:

Subsidy = 21%

Formula used:

Required angle = (360/100) × given percentage 

Calculation:

Required angle = (360/100) × given percentage 

⇒ subsidy angle = (360/100) × 21

⇒ subsidy angle = 75.6° 

∴ The central angle for subsidy is 75.6°

Given:

Sum of selling price = Rs. 1995

Discount on 1^st case = 35 percent

Discount on 2^nd case (Successive) = 20 percent & 15 percent

Formula used:

Selling price = [(discount)/marked price] × 100

Or Selling price = [(100 – discount percent)/100] × marked price

Calculation:

Let marked price be 100x

[{(100 - 35)/100} × 100x] + [{(100-20)/100} × {(100-15)/100} × 100x] = 1995

⇒ 65x + [(80 × 85)/100]x = 1995

⇒ 65x + 68x = 1995

⇒ 133x = 1995

⇒ x = 15

⇒ 100x = 1500

∴ The marked price of the article is Rs. 1500

In this type of problem, we find the LCM of five numbers first and then we look for what makes the LCM a perfect square.

1^st term = 10 = 5 × 2

2^nd term = 12 = 2 × 2 × 3

3^rd term = 15 = 3 × 5

4^th term = 18 = 2 × 3 × 3

5^th term = 20 = 2 × 2 × 5

Hence, LCM ( 12,10,15,18,20)

⇒  180

Since the LCM= 180 is not a perfect square , we make it a perfect square by multiplying 5 ( which is the least value to be multiplied)

Then 180 × 5 = 900 = 30²

Therefore, 900 solider can be drawn up in the form of a square.

The HCF of a group of numbers will be always a factor of their LCM.

HCF is the product of all common prime factors using the least power of each common prime factor.

LCM is the product of the highest powers of all prime factors.

HCF of the two numbers = 29

Since HCF will be always a factor of LCM, 23 is a factor of the LCM.

Given that the other two factors in the LCM are 15 and 13

Calculation : 

Hence factors of the LCM are 29,13,15

So, numbers can be taken as (29 × 13) and (29 × 15)

Hence, largest number = 435

Given:

4θ is an acute angle,

And cot 4θ = tan (θ - 5°).

Formula used:

tan (90° - θ) = cot θ 

Calculation:

cot 4θ = tan (θ -5°)

⇒ tan (90° - 4θ) = tan (θ -5°)

⇒ 90° - 4θ = θ -5° 

⇒ 90° + 5° = θ + 4θ 

⇒ 95° = 5θ 

⇒ θ = 19° 

∴ the value of θ is 19°.

Given:

No. of students in the five sections are 25, 30, 40, 45, 60

Percentage of pass students in each section are 20% , 30%, 35%, 40%, 100% respectively

Formula Used:

% of pass student = no. of student pass/ total no. of student × 100

Calculation:

Let the following section be A, B, C, D, E

No. of students pass in section A = 25 × 20 /100 = 5

No. of students pass in section B = 30 ×  30/100 = 9

No. of students pass in section C = 40 × 35/100 = 14

No. of students pass in section D = 45 × 40/100 =18

No. of students pass in section E = 60 × 100/100 = 60

Total no. of students pass in all section = 5 + 9 + 14 + 18 + 60 = 106

Total no. of student = 25 + 30 + 40 + 45 + 60 = 200

Pass % of entire class = 106/200 × 100 = 53%

The pass percentage of the entire class is 53%

Given:

Marked price: Rs. 1,250

Discount offered 5%, 15% and 20%

Formula used:

S.P= M.P × {(100 - d1)/100} × {(100 - d2)/100} × ......× {(100 - dn)/100}

M.P = Marked price

S.P = Selling price

d1,d2....dn = discount%

Calculation:

We need to find S.P

S.P = M.P × {(100 - d1)/100} × {(100 - d2)/100} × ......× {(100 - dn)/100}

⇒ S.P = 1250 × {(100 - 5)/100} × {(100 - 15)/100} × {(100 - 20)/100}

⇒ S.P = 1250 × (95/100) × (85/100) × (80/100) 

⇒ S.P = 807.50

∴ The selling price is Rs. 807.50

Given:

Time taken by P and Q to do the work = 12 days

P alone can do the work in 18 days

Calculation:

P and Q one day work = 1/12

P's one day work = 1/18

Q's one day work = (P + Q) - P

⇒ 1/12 -1/18

⇒ 1/36

Q can do whole work in 36 days

⇒ 2/3 work can be done by Q in 36 × (2/3) = 24days.