CDS Elementary Mathematics 3 Feb 2019 Official Paper (Previous Year Paper)

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

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

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

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

Given:

$2 + \sqrt {2 + \sqrt {2 + \sqrt { \ldots \ldots \ldots } } }$

Calculation:

$⇒ 2 + \sqrt {2 + \sqrt {2 + } } = x$

⇒  2 + √x = x

⇒  √x = x – 2

Squaring both sides

⇒  x = x² – 4x + 4

⇒ x² – 5x + 4 = 0

⇒ x² – 4x – x + 4 = 0

⇒ x(n - 4) – 1(n - 4) = 0

⇒ (x - 4) (n - 1) = 0

⇒ x = 1, 4

but 1 cannot be the answer because sum is greater than 2

∴  x = 4

Given:

52% of candidates failed in English and 42% failed in only Mathematics.

17% failed in both the subjects

Concept used:

We will be using the concept of the Venn diagram.

Solution:

From Venn diagram

So, percentage failed in exam = 35 + 17 + 42 = 94%

Hence, percentage of passed = 100 - 94 = 6%

Mistake Points

Here in question, failed percentage in only mathematics given not the total failed in mathematics. For both cases, the answer will be different.

Wife’s share = x

each son’s share = y

Each daughter’s share = z

Each son receives 3 times as much as each daughter

⇒ y = 3z       ----(1)

Each daughter receives 2 twice as much as their mother

⇒ z = 2x (2)

total sum is Rs. 3,90,000

⇒ x + 5y + 4z = 3,90,000       ----(3)

3 equation, 3 unknowns

⇒ x = 10,000

so, their mother receives Rs. 10,000

Let’s say the sum of money be Rs. X

Then amount $= P{\left( {1 + \frac{R}{n}} \right)^{nt}}$

\(P = x,;R = \frac{{40}}{{100}}& ;n = 1\)

$\Rightarrow Amount = x\left( {1 + \frac{{40}}{{100}}} \right)$

Amount = x × (1.4) ^t

Now put, t = 2

⇒ Amount = x × (1.4)²

= 1.96 x

Now put, t = 3

Amount = x (1.4)³

= 2.74 x

so, it’s more than double

• Now put t = 4

Amount = x (1.4)⁴

= 3 .84 x

So, amount is more the tripled

So complete four years for amount to be more than  trebled

Let’s say person divides it in x, y & z parts

Then x + y + z = 17, 2000      ----(1)

after two year they get same interest from all three

⇒ simple interest $= \frac{{P; \times ;R; \times ;T}}{{100}}$

⇒ x × 0.05 × 2 = y × 0.06 × 2

⇒ 5x = 6y      ----(2) 

⇒ y × 0.06 × 2 = z × 0.09 × 2

⇒ 6y = 9z      ----(3)

3 equation 3 unknown

= > x = 7200, y = 6000 & z = 4000

so, money invested at 9% is 4000.

As we know that,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab - bc - ca)

then let a = x-y

b = y – z

c = z – x

& we know a + b + c = x – y + y – z + z – x = 0

Then RHS term is zero

⇒ (x - y)³ + (y - z)³ + (z - x)³ – 3(x - y) (y - z) (z - x) = 0

$\Rightarrow \frac{{{{\left( {x - y} \right)}^3}; + ;{{\left( {y - z} \right)}^3}; + {{\left( {z - x} \right)}^3}}}{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}} = 1$

a^x = b^y = c^z = r

then a = r ^1/x, b = r^1/y, c = r^1/z

b² = ac

⇒ r^2/y = r^1/x. r^1/z

${r^{\frac{2}{y}}} = {r^{\left( {\frac{1}{x}; + ;\frac{1}{z}} \right)}}$

$\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$

Concept:

If α and β are the roots of quadratic equation ax2 + bx + c = 0

Sum of root α + β = -b/a

Product of root αβ = c/a

Calculation:

P & q are the roots of the equation x² – 15x + r = 0

the sum of roots = p + q = 15      ----(1)

p – q = 1 (given)

⇒ p = 8 & q = 7

Then product of roots = r = p × q

r = 8 × 7 = 56

∴ The value of r is 56.

Calculation:

x² – 7x + 12 > 0

x² – 4x – 3x + 12 > 0

x (x - 4) – 3 (x - 4) > 0

We have,

(x - 4) (x - 3) > 0

Which represents the product of two terms is greater than zero. 

This will be possible in two cases:

Case:1  (x - 4) > 0 and  (x - 3) > 0

⇒ x > 4 & x > 3 

⇒ x > 4 or x ∈ (4, ∞)

Case:2  (x - 4) < 0 and  (x - 3) < 0

(x - 4) < 0 and  (x - 3) < 0

⇒ x < 4 & x < 3 

⇒ x < 3 or x ∈ (-∞, 3)

Hence, corrct naswer is 

x ∈ (-∞, 3) ∪ (4, ∞)

Given:

The given expression is 5²ⁿ – 2³ⁿ

Shortcut Trick

52n – 23n

put n = 1

⇒ 5² – 2³ 

⇒ (25 – 8) = 17

∴ The required factor is 17.

Alternate MethodConcept:

aⁿ - bⁿ always have a common factor (a - b)

Calculation:

We have,

(5²)ⁿ  - (2³)ⁿ

Using the above concept, 

(5² - 2³) will always be a factor 

⇒ 25 - 8 = 17

∴ The required factor is 17.

Given:

tan x = 1 for 0 < x < 90°

Formula used:

sin 45° = $\frac{1}{{\sqrt2}}$

cos 45° = $\frac{1}{{\sqrt2}}$

Calculation:

tan x = 1 for 0 < x < 90°

then, x = 45°

⇒ 2 sin x cos x = 2 × sin 45° cos 45°

$= 2 \times \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = 1$

∴ The required value is 1.

As we know that

sin (A + B) = sin A cos B + cos A sin B

& given expression is sin 46° cos 44° + cos 46° sin 44°

by equating we get

A = 46°, B = 44°

so, sin (46 + 44) = sin 90°

= 1

Given trigonometric expression is 4 sin²θ + 1

We know that,     0 ≤  sinθ ≤  1, [∵ 0 ≤  θ ≤  90°]

⇒ 0 ≤ sin²θ ≤ 1

⇒ 0 ≤ 4sin2θ ≤ 4

⇒ 1≤ 4sin2θ + 1 ≤ 5

Now, we can write the value of (4 sin2θ + 1) belongs to 1 to 5.

Now, check by option 

Option 1^st is not possible because the vale of (4 sin2θ + 1) can lie between 1 and 2 also.

Option 3^rd is also not possible because at θ = 0° 

The value of 4cosθ = 4 and the value of (4 sin2θ + 1) is 1

⇒ 4cosθ > (4 sin2θ + 1)

Option 4^th is also not possible because at θ  90° 

4 tan θ = ∞ and the value of (4 sin2θ + 1) = 5

⇒ 4 tan θ > (4 sin2θ + 1)

∴ 2nd option is correct.

Regular hexagon has side “x” let’s say tower at B has height H_B & height at C has height H_C

$\Rightarrow \tan 30^\circ = \frac{{{H_B}}}{x}$        ----(1)

angle of elevation for A to the top of tower C is 45°

$\tan 45^\circ = \frac{{{H_c}}}{y}$       ----(2)

two sides & included angle are given

then y² = x² + x² - 2 * x * x * cos 120°

y²= 2x² - 2x² cos 120°

y² = 2x² (1 - cos 120°)

y² = 2x² (1 + 0.5)

y² = 3x²

$y = \sqrt 3 x$       ----(3)

So, form equation 1, 2 & 3

$\tan 30^\circ = \frac{{{H_B}}}{x} \Rightarrow x = \frac{{{H_B}}}{{\tan 30^\circ }}$

$\tan 45^\circ = \frac{{{H_c}}}{y} \Rightarrow \sqrt 3 x = \frac{{{H_c}}}{{\tan 45^\circ }}$

equate above 2 equation

$\frac{1}{{\sqrt 3 }} = \frac{{{H_B}}}{{{H_C}}} \times \frac{{\tan 45^\circ }}{{\tan 30^\circ }}$

$\Rightarrow \frac{{{H_B}}}{{{H_C}}} = \frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }} \times 1 = \frac{1}{3}$

so, H_B : H_C = 1 : 3

As we know $\tan \theta = \frac{1}{{\cot \theta }}$

& tan (90 - θ) = cot θ

Then

⇒ tan 1° tan 2° tan 3°-------- tan 45° -------- tan 84° tan 88° tan 99°

⇒ tan 1° tan 2° tan 3°-------- tan 45° -------- tan (90 - 3)° tan (90 - 2)° tan (90 - 1)°

⇒ tan 1° tan 2° tan 3° -------- tan 45° ------- cot 3° cot 2° cot 1°

Replace cot θ by $\frac{1}{{\tan \theta }}$

Then by cancelling all terms we get tan 45° & tan 45° = 1

So persons is standing on first street at point P facing towards north. Now he takes 150° turn & travels for 15 minutes at the steel of 20 km/hr.

So distance travelled = speed × time

$= 20\frac{{km}}{{hr}} \times \frac{1}{4}hr$

= 5 km

So, he travels 5 km & reaches point ‘q’.

Now, he takes 60° to the left & travels for 20 minutes at the speed of 30 km/hr & reaches point ‘r’.

So, distance travelled

$= 30\frac{{km}}{{hr}} \times \frac{1}{3}hr$