WeCP Number System, Decimal, and Power Quiz 1

Numbers that are divisible by 11 between 124 and 298,

\Rightarrow 132, 143.........297

Using AP:

a_{n} = a+(n-1)d

297= 132+ (n - 1) \times 11

n = 16

\therefore There are 16 numbers

Since\ P\ and\ Q\ are\ roots\ of\ the\ equation\ ax^2 + bx + c = 0,

\Rightarrow Sum\ of\ roots,\ P + Q = -b/a

\Rightarrow Product\ of\ roots,\ PQ = c/a

If the roots are (PQ + P + Q) and (PQ - P - Q),

\Rightarrow Sum\ of\ roots = 2PQ = 2c/a

\Rightarrow Multiplication\ of\ roots = P^2Q^2 - P^2Q - PQ^2 +P^2Q - P^2 - PQ + PQ^2 - PQ -Q^2

P^2Q^2 - P^2 - Q^2 - 2PQ

P^2Q^2 - (P +Q)^2

Putting the value of PQ and (P + Q),

\Rightarrow \frac{c^2}{a^2} - \frac{b^2}{a^2}

\Rightarrow \frac{c^2-b^2}{a^2}

Now the equation according to the roots,

\Rightarrow  x^2 + (-Sum\ of\ roots)x + (Multiplication\ of\ roots) = 0

\Rightarrow  x^2 - (\frac{2c}{a})x + [\frac{c^2-b^2}{a^2}] = 0

\therefore  a^2x^2 - 2acx + c^2-b^2 = 0

LCM or least common multiple, is the least number which is a multiple of both the original numbers. Hence, the LCM would factor out the HCF of the numbers.

LCM \times HCF =  m \times n

i.e.\ n = \frac{LCM \times HCF}{m}

\therefore The\ value\ of\ n = 3

Let the divisor be X.

Using the standard identity,

Dividend = (Divisor × Quotient) + Reminder

We have,

159 = (X \times Q_1) + 6

\Rightarrow (X \times Q_1) = 153\ [Where\ Q_1\ is\ a\ quotient]

Similarly,

124 = (X \times Q_2) + 5

\Rightarrow (X \times Q_2) = 119\ [Where\ Q_2\ is\ a\ quotient]

208 = (X \times Q_3) + 4

\Rightarrow (X \times Q_3) = 204\ [Where\ Q_3\ is\ a\ quotient]

Now, since X is the largest number satisfying the given condition, X will be the HCF of 153, 119 and 204.

HCF [153, 119, 204] = 17

Hence, the largest number which satisfies the given requirements is 17.

 

From given data

When the number is divided by 18, it leaves a remainder of 7.

When the number is divided by 21, it leaves a remainder of 10.

When the number is divided by 24, it leaves a remainder of 13.

Observing the constant difference of 11 between divisor and remainder:

We can conclude that if 11 is subtracted from the LCM of 18, 21 and 24, then it will leave a remainder of 7 when divided by 18, a remainder of 10 when divided by 21 and a remainder of 13 when divided by 24.

When 11 is subtracted from any multiple of LCM, it will give the same result.

Now,

18=2\times 3\times 3

21=3\times 7

24=2\times 2\times 2\times 3

\therefore LCM\ of\ 18,\ 21\ and\ 24 = 504

So, any number of the form (504n – 11) will give the required remainders, when n is an integer.

But, this number also has to be divisible by 23.

Using trial and error method, for n = 6, we find that-

504n-11=504\times 6-11=3024-11=3013

Let us try to make sense of the data given,

A_1= 3 =3\times 1

A_2= 33 =3\times 11

A_3= 333 =3\times 111

.

.

.

A_n= 3333...(n\ times) =3\times 1111...(n\ times)

Looking at the options given, the first and the fourth are straightaway ruled out, because they do not give us the required answer.

Looking at the third option, let us try n = 2,

(\frac{3}{9})^{n}(10^n-1)=\frac{9}{81}(100-1)=11

This is also not the right answer.

Now, look at option 2, and trying for n = 3,

\frac{3}{9}(1000-1)=\frac{1}{3}(999)=333

This is giving us the right result.

Given, 1^3 + 2^3 + 3^3 +.... + 10^3 = 3025

m=2^3+ 4^3+ 6^3.... + 20^3

\Rightarrow m=2^3(1^3 + 2^3 + 3^3 +.... + 10^3)

\Rightarrow m=8\times 3025

\Rightarrow m=24200

\therefore 2m=2\times 24200=48400

As the number leaves same remained in each case,

required numbers = 5 + (common multiples of 4, 7 and 9)

Let's Consider the LCM of 4, 7 and 9.

LCM(4,7,9) = 252

\Rightarrow One\ of\ the\ required\ numbers = 252 + 5 = 257

Next common multiple of 4, 7 and 9 in the given range = 252 x 2 = 504

\Rightarrow Second\ required\ number = 504+5 = 509

Next common multiple of 4, 7 and 9 in the given range = 252 × 3 = 756

\Rightarrow Third\ required\ number = 756+5 = 761

Therefore, Sum of required numbers = 1527

Hence, Average of required numbers = 1527/3 = 509

Let denominator of fraction = x

Then, numerator = x – 5

\therefore Fraction=\frac{(x-5)}{x}

Now, according to the question,

\frac{(x-5)+3}{x-6}=\frac{3}{1}

\Rightarrow x-2=3(x-6)

\Rightarrow x-2=3x-18

\Rightarrow 2x=16

\therefore x=8

Hence,\ fraction=\frac{8-5}{8}=\frac{3}{8}

A=2^{30}=4^{15}

B=2^{29}+2^{28} + 2^{27} + ... + 2^0\ [Series\ are\ in\ GP]

\Rightarrow B=a\times \frac{1-r^n}{1-r}

\Rightarrow B=2^{29}\times \frac{1-(1/2)^{30}}{1-(1/2)}

\Rightarrow B=2^{29}\times \frac{2^{30}-1}{2^{30}\times 2^{-1}}

\Rightarrow B=2^{29}\times \frac{2^{30}-1}{2^{29}}

\Rightarrow B=2^{30}-1=4^{15}-1

\Rightarrow A>B

 

C=3^{14} + 3^{13} + 3^{12} +... +3^0\ [Series\ are\ in\ GP]

\Rightarrow C=a\times \frac{1-r^n}{1-r}

\Rightarrow C=3^{14}\times \frac{1-(1/3)^{15}}{1-(1/3)}

\Rightarrow C=3^{14}\times \frac{3^{15}-1}{3^{15}\times (2/3)}

\Rightarrow C=3^{14}\times \frac{3^{15}-1}{3^{14}\times 2}

\Rightarrow C=\frac{3^{15}-1}{2}