Quantitative Aptitude PYQs

(X + Y) = 100 – 20 – 15 – 10 = 55     ---- (1)

(2X + Y) = 100 – 4 – 8 – 4 = 84     ---- (2)

From (1) and (2):

X = 29 and Y = 26

Mean = ∑x_if_i/∑f_i = 3760/100 = 37.6

Mode \(= l; + ;\frac{{{f_1} - {f_0}}}{{\left( {{f_1} - {f_0}} \right); + ;\left( {{f_1} - {f_2}} \right)}} \times h\)

h = Class difference = 10

l = Lower limit of model class = 40

Now,

Mode \(= 40; + ;\frac{{58 - 4}}{{\left( {58 - 4} \right); + ;\left( {58 - 26} \right)}} \times 10\)

\(= 40; + ;\frac{{54}}{{54; + ;32}} \times 10; = ;40; + ;\frac{{54}}{{86}} \times 10\)

= 40 + 6.28 = 46.28 = 46 (Approx)

Number of scooters of company Y sold by showroom E = (12% of 6400) – (8% of 3000) = 768 – 240 = 528

Number of scooters of both companies sold by showroom C = 15% of 6400 = 960

Required percent = \(\frac{{528}}{{960}} \times 100; = ;55%\)

Number of scooters of both the companies sold by showroom B = 21% of 6400 = 1344

Number of scooters of company X sold by showroom A = 24% of 3000 = 720

Difference = 1344 – 720 = 624

Required percent = \(\frac{{624}}{{720}} \times 100; = ;86\frac{2}{3}\)

Number of scooters of company Y sold by the showrooms A, C, and E = (scooters of company X and Y sold by A, C, and E) - (scooters of company x sold by A, C, and E)

Number of scooters of company Y sold by the showrooms A, C and E = [(19 + 15 + 12)% of 6400] – [(24 + 20 + 8)% of 3000] = 2944 – 1560 = 1384

Required average = \(\frac{{1384}}{3} = 461\frac{1}{3}\)

Number of scooters of company X and Y sold by showroom A = 19% of 6400 = 1216

Number of scooters of company X sold by showrooms B and E together = (18 + 8)% of 3000 = 780

Required difference = 1216 – 780 = 436

Total days in a year = 365

Average number of pedestrians killed pedestrians killed per day in the year 2017 = 20475/365 = 56.095 ≈ 56

Pedestrians’ fatalities in 2014 = 12330

Pedestrians’ fatalities in 2017 = 20475

Difference = 20475 – 12330 = 8145

Required percent = (8145/12330) × 100 = 66.05% ≈ 66%

Total days in a year = 365

Average number of bikers killed daily in road accidents in the year 2017 = 48746/365 = 133.55 ≈ 134

Total days in a year = 365

Average number of cyclists killed daily in road accidents in 2017 = 3559/365 = 9.75 ≈ 10

Formula used:

1. a^m aⁿ = a^m+n
2. (a + b)³ = a³ + b³ + 3ab(a + b)
3. a³ + b³ = (a + b)(a² - ab + b²)

 

Calculation:

Suppose a = x² and b = y²

We have , a√a + b√b = 32

⇒ x3 + y3 = 32         ---- (1)

Also, a√b + b√a = 31.

⇒ √a√b(√a + √b) = 31

⇒ xy(x + y) = 31      ---- (2)

Using the identity (2)

⇒ (x + y)³ = x³ + y³ + 3xy(x + y)

Using equation (1) & (2)

⇒ (x + y)³ = 32 + 3 × 31 = 125 = 5³

⇒ (x + y) = 5

Again using identity (3)

⇒ x³ + y³ = (x + y)(x² + y² – xy) = 32

⇒  5(x² + y²) – 31 = 32

⇒ (x² + y²) = 63/5

⇒ (a + b) = 63/5

⇒ 5(a + b)/7 = (5/7) × (63/5) = 9

∴  5(a + b)/7 is equal to 9.

Formula used:

(a + b)² = a² +2ab + b²

Calculation:

Given that,

\(x = \frac{{1; + ;\sqrt 3 }}{2}\)

\({x^2} = {\left( {\frac{{1; + ;\sqrt 3 }}{2}} \right)^2}\)

\({x^2} = \frac{1}{4}{\left( {1; + ;\sqrt 3 } \right)^2} = \frac{1}{4}\left( {1; + ;3; + ;2\sqrt 3 } \right) = \frac{{2; + ;\sqrt 3 }}{2}\)

\(y = {x^3} = {x^2} \times x = \frac{{2; + ;\sqrt 3 }}{2} \times \frac{{1; + ;\sqrt 3 }}{2} = \frac{{5; + ;3\sqrt 3 }}{4}\)

\({y^2} = {\left( {\frac{{5; + ;3\sqrt 3 }}{4}} \right)^2} = \frac{{25; + ;27; + ;30\sqrt 3 }}{{16}} = \frac{{52; + ;30\sqrt 3 }}{{16}}\)

\(8{y^2} = \frac{{8\left( {52; + ;30\sqrt 3 } \right)}}{{16}} = \left( {26; + ;15\sqrt 3 } \right)\)

\(20y = \frac{{20\left( {5; + ;3\sqrt 3 } \right)}}{4} = 25; + ;15\sqrt 3\)

⇒ 8y² – 20y – 1 = (26 + 15√3) – (25 + 15√3) – 1

⇒ 8y2 – 20y – 1 = 0

Given:

HCF of two numbers = 12

Concept used:

HCF = the greatest number which divides each of the two or more numbers

Calculation:

Let the two numbers be ‘12a’ and ‘12b’ respectively where ‘a’ and ‘b’ are co-prime integers.

LCM of both the numbers = 12ab which must be divisible by 12.

Only 80 is not divisible by 12.

So, 80 can never be the HCF of both the numbers.

∴ The required answer is 80.

1. Unit digit = 17¹⁴⁷ = 7¹⁴⁷ = 7² = 49 = 9 [Since, cyclicity of 7 is ‘4’.]
2. Let the two odd numbers be ‘x’ and ‘x + 2’ respectively.

Now,

(x + 2)² – x² = x² + 4 + 4x – x² = 4(1 + x).

Hence, it will be divisible by 8 as ‘x’ is an integer. 3. Let the two consecutive odd numbers be ‘x’ and ‘x + 2’ respectively.

Now,

x(x + 2) + 1 = x² + 2x + 1 = (x + 1)².

Hence, it will be a perfect square.

Given :

Rate of interest = 20% per annum

First interest compounded half - yearly

Second interest compounded yearly

Time = 2 years

Formula used :

Amount = \(P(1+\frac{R}{100})^t\)        ----- (1)

Where, P = principal

R = rate and t = time

Calculations :

Let the amount invested in each scheme be 100x.

When interest in compounded half - yearly then

t = 2t and R = R/2, using equation (1)

Total amount from the first scheme

⇒ \(100{\rm{x}} \times {\left( {1; + ;\frac{{20}}{{2 \times 100}}} \right)^{2 \times 2}}; = ;146.41{\rm{x}}\)

Total amount from the second scheme 

⇒ \(100{\rm{x}} \times {\left( {1; + ;\frac{{20}}{{100}}} \right)^2}; = ;144{\rm{x}}\)

The required difference between two schemes

⇒ 146.41x – 144x = 482

⇒ 2.41x = 482

⇒ x = 200

⇒ 100x = Rs 20000

∴ Principle amount in each scheme is Rs. 20000.

Put k = 0 in the equation:

x³ – 7x + 6 = (x – 2)(x – 1)(x + 3)

Put k = 1 in the equation:

x³ + x² – 7x + 6 = x(x² + x – 7) + 6

Put k = 2 in the equation:

x³ + 2x² – 7x + 6 = x(x² + 2x – 7) + 6

Put k = 3 in the equation:

x³ + 3x² – 7x + 6 = x(x² + 3x – 7) + 6

Hence, k = 0

Time at which X completes a round = 252 seconds

Time at which Y completes a round = 308 seconds

Time at which Z completes a round = 198 seconds

Time after which they meet again at the starting point after the start = LCM of 252, 308, 198 = 2772 seconds = 46 minutes 12 seconds

Given:

Numbers are 1/3, 5/6, 2/9, 4/27

Concept used:

LCM of fractions = LCM of numerator/HCF of denominator 

Calculation:

LCM of 1/3, 5/6, 2/9, 4/27

⇒ LCM of (1, 5, 2, 4)/HCF of (3, 6, 9, 27)

⇒ 20/3

∴ Required answer is 20/3

x² + 5x + 6 = 0

⇒ x² + 3x + 2x + 6 = 0

⇒ x(x + 3) + 2(x + 3) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = -2 and -3

Put x = -2 in x² + kx + 1 = 0

⇒ (-2)² + k × (-2) + 1 = 0

⇒ 2k = 5

⇒ k = 5/2

Put x = -3 in x² + kx + 1 = 0

⇒ (-3)² + k × (-3) + 1 = 0

⇒ 3k = 10

⇒ k = 10/3

Given :

A lent Rs. 25000 to B

Rate = 7%

Time = 4 years

Formula used :

Simple interest = \(\frac{PRT}{100}\)        ----- (1)

P = principal, R = rate and T = time

Calculations :

Let the amount given to C be ‘x’.

According to the question, we have

Interest = Interest when  A lent Rs 25000 to B + 'x' lent to C

Using equation (1), we get

⇒ \( \frac{(25000 × 7 × 4)}{100} + \frac{(x × 7 × 4)}{100} = 11200\)

⇒ 0.28x = 11200 – 7000

⇒ x = Rs.15000

∴ Amount of money A lend to C is Rs 15000.