Quantitative Aptitude PYQs

\(x = \frac{{4\sqrt 6 }}{{\sqrt 2 ; + ;\sqrt 3 }}\)

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

By applying componendo and dividendo

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

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

Now for second term

\(x = \frac{{4\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}\)

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

By applying componendo and dividendo

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

\(= ;\frac{{3\sqrt 2 ; + ;\sqrt 3 }}{{\sqrt 2 - \sqrt 3 }}or\frac{{ - \left( {3\sqrt 2 ; + ;\sqrt 3 } \right)}}{{\sqrt 3 - \sqrt 2 }} \ldots \left( 2 \right)\)

Add equation 1^st & 2^nd

We get

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

Given :

x = 30% of z

y = 40% of z

x = p% of y

Calculations :

x, y & z are three numbers such that,

⇒ x = \(\frac{30}{100} × z\) = 0.3z       ---- (1)

⇒ y = \(\frac{40}{100} × z\) =  0.4z        ----- (2)

⇒ \(x = \frac{p}{{100}}y\)          ----- (3)

Using equation (1), (2) and (3)

⇒ \( ;0.3 × z = \frac{p}{{100}} × 0.4 × z\)

⇒ \(\frac{3}{10} × z = \frac{p}{100} × \frac{4}{10} × z\)

⇒ p = 3 × 25

⇒ p = 75

∴ The value of p is 75.

Let’s say plane is at point “A” vertically above point P Let’s say at point B the plane is at t seconds & from point P angle of elevation to the plane at point B & C are same it means point opposite to A which is point O is exactly mid of point C & B means time taken by plane from point B to O is equal to time taken by plane from point O to C then plane reaches point O at (t + 15) seconds

Now it takes 3 minutes to complete one round.

Then from point O to A it will take \(\frac{3}{2} = 1.5\) minutes or 90 seconds

From B to O it takes 15 seconds.

So, 90 + 15 = 105 seconds

⇒ in 105 seconds, plane goes from point B which is at time t to point A

So that entire x second are 105 seconds

p & q are relatively prime

Relative prime = Two integers are relatively prime if there is no integer greater than one that divides them both

1. so, two prime numbers don’t have common factors so pair of any two of any two prime numbers are always relatively prime

2. Let’s p = 4 & q = 9 which are composite numbers

Factor of p = 4 = 2 × 2 × 1

Factors of q = 9 = 3 × 3 × 1

They don’t have any common factors

So, they are relatively prime

3. Let’s say p = 3 & q = 8

Factors of P = 3 × 1

q = 8 = 2 × 2 × 2 × 1

So, they don’t have any common factors, so they are relatively prime

Answer option 1, 2 & 3 all are correct

Let’s say number of girls is the class is “x”

So, number of boys in the class is “x - 100”

Average weight of class is 30

⇒ total weight of class is 30 × 100 = 3000

Total weight of girls is 24 × x^-

Total weight of boys is (100 - x) × 32

⇒ 24x + (100 - x) 32 = 3000

⇒ x = 25

So, number of girls in the class is 25

\(\sqrt {{{\left( {a - b} \right)}^2}} + \sqrt {{{\left( {b - a} \right)}^2}} \)

Example 1

So, for a = 2 & b = 1

Value of expression is 2

⇒ option “1” is eliminated

Example 2

For a = b = 0

Value of expression is zero

⇒ option 2 is eliminated

Example 3

For b = 2 & a = 1

Value of expression is positive

⇒ option d is eliminated

& from example 1 & 3

⇒ option c is correct

Given:

\(\frac{a}{b} = \frac{c}{d} = \frac{1}{6}\)     

 Shortcut Trick)

Let a = c = 1 & b = d = 6

⇒ \(\frac{{{a^2}; + ;{c^2}}}{{{b^2}; + ;{d^2}}} = \frac{1^2+1^2}{6^2+6^2}\)

⇒ \(=\frac{2}{{72}}=\frac{1}{36}\)

∴ 1/36 is correct

Alternate MethodWe have,

\(\frac{a}{b} = \frac{c}{d} = \frac{1}{6}\)

⇒ \(\frac{a}{b} = \frac{1}{6}\)

⇒ a = \( \frac{b }{6}\)  Similarly c = \( \frac{d }{6}\)

⇒ \(\frac{{{a^2}; + ;{c^2}}}{{{b^2}; + ;{d^2}}} = \frac{\frac{b^{2}}{6^{2}}+\frac{d^{2}}{6^{2}}}{{b^{2}+d^{2}}}\)

⇒ \(\frac{{{a^2}; + ;{c^2}}}{{{b^2}; + ;{d^2}}} = \frac{\frac{1}{36}(b^{2}+d^{2})}{b^{2}+d^{2}}\) = \( \frac{1 }{36}\)

∴ 1/36 is correct

\(\begin{array}{*{20}{c}} {0.\overline {53} ; = ;0.53535353 \ldots }\ {; + ;0.5\bar 3; = ;0.53333333 \ldots }\ {\overline {1.0686868} \ldots } \end{array}\)

Or \(1.0\overline {68}\)

From options its closer that value of inequality holds when at least N is greater than 3

Now check for values less than equal to 3

For N = 3 = 3³ = 27

3³ = 27

So 3^N = N³ inequality doesn’t hold

For N = 2 3² = 9

2³ = 8

⇒ 3^N > N³ inequality holds.

For N = 1 3¹ = 3

1³ = 1

⇒ 3^N > N³ equality holds

So N is natural number except 3

Concept:

• Rational numbers are those numbers that show the ratio of numbers or the number which we get after dividing it with any two integers.
• Irrational numbers are those numbers that we can not represent in the form of simple fractions a/b, and b is not equal to zero.

 

Explanation:

a) \(= \sqrt {59049} ; = ;243\) which is an rational number.

b) \(\frac{{231}}{{593}} = \) any number expressed into ratio’s are rational number

c) 0.4545….

repeating decimals can be expressed into ratio’s & are rational number

d) 0.121122111…

nonrepeating, non-terminating decimals are irrational numbers

\(Average;speed = \frac{{total;distance}}{{total;time}}\)

Covert time into hours

So, 50 minutes \( = \frac{{50}}{{60}}\) hr & so on.

Average speed \( = \frac{{\frac{5}{6}; \times ;9; + ;\frac{8}{6}; \times ;8; + ;\frac{{10}}{6}; \times ;7.5}}{{\frac{5}{6}; + ;\frac{8}{6}; + ;\frac{{10}}{6}}}\)

\(= \frac{{45; + ;64; + ;75}}{{5; + ;8; + ;10}} = 8.00kmph\)

Calculation:

\(\frac{a}{{b; + ;c}} = \frac{b}{{c; + ;a}} = \frac{c}{{a; + ;b}}\) = k (let)

⇒ a = k(b + c)            ------(1)

⇒ b = k(c + a)            ------(2)

⇒ c = k(a + b)            ------(c)

Adding equations (1), (2) & (3), we get

⇒ a + b + c = 2k [(a + b) + (b + c) + (c + a)]

⇒ a + b + c = 2k (a + b + c)

⇒  2k(a + b + c) - (a + b + c) = 0

⇒ (a + b + c)(2k - 1) = 0

⇒ a + b + c = 0 & 2k - 1 = 0

⇒ k = 1/2

∴ \(\frac{a}{{b; + ;c}} = \frac{b}{{c; + ;a}} = \frac{c}{{a; + ;b}}\) = 1/2.

Number 3⁵²¹ can be written as 9²⁶⁰ × 3

Remainder when 9 is divided by 8 is 1

Remainder when 3 is divided by 8 is 3

So, remainder when 9²⁶⁰ × 3 is divisible by 8 is 1²⁶⁰ × 3 = 3

Concept used:

Prime number:

Prime numbers are numbers that have only 2 factors: 1 and themselves.

Calculation:

11 is a prime number one’s place is 1

2 is a prime number one’s place is 2

3 is a prime number one’s place is 3

5 is a prime number one’s place is 5

7 is a prime number one’s place is 7

19 is a prime number one’s place is 9

So, Answer is 6

Cost price of article is “x”

Then sold at 6% gain ⇒ x + 0.06x = 1.06x

Sold at 6% loss ⇒ x – 0.06x = 0.94x

& their difference is Rs. 6

= > 1.06x – 0.94x = 6

= > x = Rs. 50

12 men reaps field in 14 days

Then 1 man will reap it in 12 × 14 days

Work done by 1 man in 1 day is \(\frac{1}{{12; \times ;14}}\)

Same for

18 women reaps field in 14 days

Then 1 woman will reap it in 18 × 14 days

⇒ work done by 1 woman in 1 day is \(\frac{1}{{18; \times ;14}}\)

Now 8 men & 16 women reaps the field in “n” days

\( = > ;n \times \left[ {\frac{8}{{12; \times ;14}} + \frac{{16}}{{18; \times ;14}}} \right] = 1\)

⇒ n = 9 days

Let’s say 3^x = 4^y = 12^z = r

3 = r^1/x, 4 = r^1/y & 12 = r^1/z

3 × 4 = 12

r^1/x × r^1/y = r^1/z

\({r^{\frac{1}{x}; + ;\frac{1}{y}}} = {r^{1/z}}\)

\( \Rightarrow ;\frac{1}{z} = \frac{1}{x} + \frac{1}{y}\)

\(\Rightarrow ;z = \frac{{xy}}{{x; + ;y}}\)

\(\frac{{4a; + ;7b}}{{4c; +; 7d}} = \frac{{4a; - ;7b}}{{4c; -; 7d}}\)

\(\Rightarrow ;\frac{{4a; + ;7b}}{{4a - 7b}} = \frac{{4c; + ;7d}}{{4c - 7d}}\)

By applying componendo and dividendo

\(\frac{{8a}}{{14b}} = \frac{{8c}}{{14d}}\)

\(\Rightarrow \frac{a}{b} = \frac{c}{d}\)

When x² + ax + b is divided by x + 1 then remainder is

b – a + 1

When x² + an + b is divided by x – 1 then remainder is a + b + 1

Equating both terms

= > b – a + 1 = a + b + 1

⇒ a = 0

So, 1 term is zero & other any integer

We don’t have enough data to calculate ‘b’

Given:

Tushar takes 6 hours to complete a piece of work,

Amar completes the same work in 10 hours.

Calculation:

Work done by Tushar in 1 day is \(\frac{1}{6}\)

Work done by Amar in 1 day is \(\frac{1}{{10}}\)

If they work together, they finish it in “n” days

\(\Rightarrow n \times \left[ {\frac{1}{6} + \frac{1}{{10}}} \right] = 1\)

\( \Rightarrow n = \frac{{15}}{4} = 3.75\)  hr 

∴  3 hours & 45 minutes