Quantitative Aptitude PYQs

Given expression is (17²⁹ + 19²⁹)

It can also be written as 289¹⁴ × 17 + 19²⁹

Remainder of 289 when divided by 18 is 1.

& remainder of 19 when divided by 18 is also 1

so, remainder of 289¹⁴ × 17 is 1¹⁴ × 17 = 17 & for 19²⁹ is 1²⁹ = 1

So, remainder of 289¹⁴ × 17 + 19²⁹ = 17 + 1 = 18

And 18 is completely divisible by 18 So overall remainder is zero

Now factor of 10 is = 5 × 2

Now to break the base number into 5 or 2

So, for 2⁵ = 2⁵

3³ = 3³

4⁸ = (2 × 2)⁸ = 2¹⁶

5³ = 5³

6⁷ = (2 × 3)⁷ = 2⁷ × 3⁷

7⁶ = 7⁶

8¹² = (2 × 2 × 2)¹² = 2³⁶

9⁹ = 9⁶

10⁶ = (2 × 5)⁶ = 2⁶ × 5⁶

15¹² = (5 × 3)¹² = 5¹² × 3¹²

20¹⁴ = (5 × 2 × 2)¹⁴ = 5¹⁴ × 2²⁸

22¹¹ = (11 × 2)¹¹ = 11¹¹ × 2¹¹

25¹⁵ = (5 × 5)¹⁵ = 5³⁰

So total = 2¹⁰⁹ × 5⁶⁵

We can ignore rest of the factors

⇒ 10⁶⁵ × 2⁴⁴

So, power of 10 is 65

⇒ n = 65

Condition for number is divisible by 9 is sum of numbers must be divisible 9

⇒ 4 + 7 + 9 + 8 + 6 + 5 + A + B

= 39 + A + B

Now values of A can be between 1 to 9

But the values of B or at one’s place must be add number, 1, 3, 5, 7, 9

So maximum values of A & B can be 9 & 9 which is 39 + 9 + 9 = 57

So, numbers divisible by 9 less than 57 & greater than 39 are 45 & 54

So, number 39 + A + B must be either 45 & 54

Let’s make a table when this number can become 45 & 54

Now all above 5 cases by putting A & B in (39 + A + B) then this number must be divisible by 9.

So Ans is 5 pairs of A & B

4. 999 × abc = def132

Now, we can write it as abc (1000 – 1)

= abc × 1000 – abc

⇒ abc000 – abc

\(\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} a&b&c&0&0&0 \end{array}}\ {\begin{array}{*{20}{c}} - &{}&{}&a&b&c \end{array}}\ {\overline {a;b;\left( {c - 1} \right)\left( {9 - a} \right)\left( {9 - b} \right)\left( {10 - c} \right)} } \end{array}\)

Now, ab (c – 1) (9-a) (9 – b) (10 – c) = def 132

So, compare the digits

10 – c = 2 ⇒ c = 8

9 – b = 3 ⇒ b = 6

9 – a = 1 ⇒ a = 8

c – 1 = f ⇒ f = 7

b = e ⇒ e = 6

a = d ⇒ d = 8

So, number is 998 × 868 = 867132

Option 4 is correct

Given:

Cars A, B, and C started from a point at 5 p.m., 6 p.m., and 7 p.m

Speed of car A = 60 Km/Hr

Speed of car B = 80 Km/Hr

Speed of car C = x Km/Hr

Formula used:

\( {Time = \frac{{Distance}}{{Speed}}} \)

Calculation:

Let’s say car A, B & C meet “y” km far from the point.

Now time taken by car A, B & C  are T_A, T_B, T_C hrs respectively.

We know that All three cars meet at the same point and

Time = Distance/speed

Therefore,

T = y/60     ------(1)

T - 1 = y/80     ------(2)

T - 2 = y/x        -----(3) 

From equation (1) & (2)

y/60 - 1 = y/80

⇒ y/240 = 1

⇒  y =  240 Km/Hr

Put this value in equation (1)

⇒ T = 240/60

⇒ T = 4 Hr

Now, put the value of T & Y in equation (3), we get

4 - 2 = 240/x

⇒ 240/x = 2

⇒ x = 120 Km/hr

∴ The value of x is 120 Km/hr.

Let's assume that priya's present age was T year.

So that (T-4) should be cube of integral number as given in data.

\(\therefore\) Possible value for (T-4) are 8, 27, 64, ..... etc.

\(\therefore\) Possible value for T are 12, 31, 69, .... etc

\(\therefore\) Possible value for (T+4) are 16, 35, 73, ..... etc

But from given data (T+4) should be square of an integral number.

So T +  4 = 16 years

\(\therefore\) Present age T = 12 years

Possible square of a number greater than T = 16, 25, 36, 49......

Possible cube of a number greater than T = 27, 64, 125, ....

\(\therefore\)An age when it becomes square of a number in previous year and cube of a number in the next year = 26 years

So wait = 26 - 12 = 14 Years

Option 1:

7 - 5 = 2 which is divisible by 2

Option 2:

Let m be 4 and n be 6 and p be 3

So, (4 × 6) is divisible by 3 and 6 is also divisible by 3

Option 3:

we need to find the wrong statement.

For n = 6, 6n – 1 = 35 which is not a prime number.

Option 4:

3, 5, and 7 are three prime numbers such that there is a gap of 2 between two adjacent prime numbers

Option 3 is not true

∴ The required answer is Option 3.

Concept:

AM of a, b = (a + b)/2

GM of ab = √(ab)

AM ≥ GM

Calculation:

\(x + \frac{{x + 2}}{{2x}}\)

This can be written as

\(= x + \frac{1}{x} + \frac{1}{2}\)

We know that, A. m ≥ G.M

\(\frac{x + \frac{1}{x}}{2} ≥ \sqrt {(x\times \frac{1}{x})}\)

⇒ \(x + \frac{1}{x} \ge 2\)

Adding (1/2) both side 

\(x + \frac{1}{x} + \frac{1}{2} ≥ 2 + \frac{1}{2}\)

\(x + \frac{1}{x} + \frac{1}{2} ≥ 2\frac{1}{2}\)

∴ Minimum value is \(2\frac{1}{2}\).

\(\sqrt {\frac{{1; - ;qx}}{{1; + ;qx}}} = \frac{{1; -; px}}{{1; + ;px}}\)

Squaring both sides

\(\frac{{1; - ;qx}}{{1; + ;qx}} = {\left( {\frac{{1; - ;px}}{{1; + ;px}}} \right)^2}\)

\(\frac{{\left( {1; - ;qx} \right)}}{{\left( {1; + ;qx} \right)}} = \frac{{1; + ;{p^2}{x^2} - 2px}}{{1; + ;{p^2}{x^2}; + ;2px}}\)

By applying componendo and dividendo

\(\frac{{\left( {1 - ;qx; + ;1; + ;qx} \right)}}{{\left( {1; -; qx; - ;1; - ;qx} \right)}} = \frac{{1; + ;{p^2}{x^2} - 2px; + ;1; + ;2px; + ;{p^2}{x^2}}}{{1; + ;{p^2}{x^2} - 2px - 1 - 2px - {p^2}{x^2}}}\)

\(\frac{2}{{ - 2qx}} = \frac{{2\left( {1; + ;{p^2}{x^2}} \right)}}{{2\left( { - 2px} \right)}}\)

\(\frac{{1; + ;{p^2}{x^2}}}{{2px}} = \frac{1}{{qx}}\)

qx*(1 + p²x²) = 2px

x*(p²qx² – 2p + q) = 0

⇒ either x = 0

Or p²qx² = 2p – q

\(x^2 = \frac{{2p - q}}{{{p^2}q}}\)

\(x = \pm \frac{1}{p}\sqrt {\frac{{2p - q}}{q}} or;x = 0\)

Option ‘1’ is correct

Suppose, number of rooms in hostel is x & rent per room is y

So total cash collection is x * y

Now there is an increase of room & rent by 20%

So now Total rooms are x + 0.2x = 1.2x and Rent per person = y + 0.2y = 1.2y

So, percentage change in total collection

\( = \frac{{1.2x; \times ;1.2y - xy}}{{xy}} \times 100\)

\( = \frac{{1.44xy - xy}}{{xy}} \times 100\)

= 44 %

Let’s say school is “x” km away from their house

Time taken by Radha \(= \frac{x}{8}\) and for Hema \(= \frac{x}{{10}}\)

Now Hema takes 9 minutes less than Radha

\(\frac{x}{8} - \frac{x}{{10}} = \frac{9}{{60}}\)

⇒ x = 6 km

3^{x + 2} + 3^-x = 10

\({3^x}*{3^2} + \frac{1}{{{3^x}}} = 10\)

\(9*{3^x} + \frac{1}{{{3^x}}} = 10\)

Now put 3^x = y

\(9y + \frac{1}{y} = 10\)

9y² + 1 = 10y

9y² – 10y + 1 = 0

9y² – 9y – y + 1 = 0

9y*(y - 1) – (y - 1) = 0

(y - 1) *(9y - 1) = 0

\(\Rightarrow ;y = 1,;\frac{1}{9}\)

So, for y = 1 ⇒ 3^x = 1

⇒ x = 0

\(& ;for;y = \frac{1}{9} \Rightarrow ;{3^x} = \frac{1}{9}\)

⇒ x = -2

So, values of x = 0, -2

Shortcut

We can directly put x = 0, 1, 2, -2, -1 from the options into the equation

So, from that x = 0 & -2 satisfy the above equation

So, factors of 108 or 2² × 3³

So, let’s say y = (108)¹⁰

Take log both sides

Log y = 10 log (108)

Log y = 10[log 2² × 3³]

= 10[2 log 2 + 3 log 3]

= 10[0.602 + 1.431]

⇒ Log y = 20.33

So, from the property if log y = abc.efg Then number of digits in y = abc + 1

⇒ y has 20 + 1 = 21 digits

Sum of three prime number is 100

But we know that except 2 all prime numbers are odd

& sum of 3 odd number is always an odd number

But 100 is an even number

So, one of the numbers is 2

Let’s say other number are x and x + 36

⇒ 2 + x + x + 36 = 100

⇒ 2x = 62

⇒ x = 31 & x + 36 = 67

So, all three numbers are 2, 31, 67

\( Given: R.H.S =16/23 \)

\(⇒ 1/(23/16)= 1/(1+7/16) \)

\(⇒1/(1+1/(16/7))\)

\(⇒1/(1+1/(2+2/7))\)

\(⇒ 1/(1+1/(2+1/(7/2)))\)

\(1/(1+1/(2+1/(3+1/2)))\)

On comparing this with the given expression of the question, we get

a = 1, b = 2 and c = 3

Now, we have to find the mean of a, b and c 

⇒ Mean = (1 + 2 + 3)/3

∴ Mean = 2

Total population = 1, 00, 000

So, the vein diagram of above data is

So, number of people who read only one newspaper are 1 + 19 = 20%

20% of 1, 00, 000

⇒ 0.2 × 1, 00, 000

= 20, 000

Total population = 1,00,000

So, the vein diagram of the above data is

Number of people who read at least two Newspaper are (1 + 7 + 3 + 1) % = 12% of 1, 00, 000

= 12000

Total population = 1,00,000

So, the vein diagram of the above data is

So according to the diagram, the percentage of people that read only one newspaper = 1 + 19 + 0 = 20%

And the percentage of people that read any 2 newspaper = 7 + 1 + 3 = 11% 

And number of people that read all three newspaper = 1% 

∴ Number of people who do not read any of these three newspapers are [100 – (20 + 11 + 1)]%

⇒ (100 - 32) % of 1, 00, 000

= 68, 000

Calculation:

7¹  =  7 ⇒ unit digit = 7

7²  =  49 ⇒ unit digit = 9

7³  =  343 ⇒ unit digit = 3

7⁴  =  2401 ⇒ unit digit = 1

The unit digit pattern is 7, 9, 3, 1, 7, 9, 3, 1.........

Now

7⁷³ = 7⁷² × 7

7⁷³ = (7⁴)¹⁸ × 7

7⁷³ = (2401)¹⁸ × 7

∴The digit in the unit place is 7.

For n = 1 ⇒ 1² + 48 = 49 is a perfect square

n = 2 ⇒ 2² + 48 = 52 is not a perfect square

n = 3 ⇒ 3² + 48 = 57 is not a perfect square

n = 4 ⇒ 4² + 48 = 64 is a perfect square

So, for n = 1 & 4

So, number of such n is 2