Class 12th

Four persons (1) Rohit, (2) Jugeet, (3) Hemraj and (4) Durgesh each had some initial money with them. They all were playing bridge in a way that the loser doubles the money of each of the other three persons from his share. They played four rounds and each person lost one round in the order 1, 2, 3 and 4 as mentioned above. At the end of fourth round each person had Rs. 3200.

Who had the lowest amount at any round of play throughout the tournament?

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	Mohit	Manohar	Prashant	Dinesh
Round – 4	3200	3200	3200	3200
Round – 3	1600	1600	1600	1600
Round – 2	800	800	7200	4000
Round – 1	400	6800	3600	2000
Initially	6600	3400	1800	1000

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5 months ago

What was the amount with Jugeet at the end of first round?

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	Mohit	Manohar	Prashant	Dinesh
Round – 4	3200	3200	3200	3200
Round – 3	1600	1600	1600	1600
Round – 2	800	800	7200	4000
Round – 1	400	6800	3600	2000
Initially	6600	3400	1800	1000

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New answer posted

5 months ago

Direction for Question: Answer the questions on the basis of the information given below.

What was the amount with Rohit to start with?

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	Mohit	Manohar	Prashant	Dinesh
Round – 4	3200	3200	3200	3200
Round – 3	1600	1600	1600	1600
Round – 2	800	800	7200	4000
Round – 1	400	6800	3600	2000
Initially	6600	3400	1800	1000

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5 months ago

If the shortest distance between the lines ${\vec{r}}_{1} = α \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}), λ \in R,$ a > 0 and ${\vec{r}}_{2} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}), μ \in R$ is 9, the a is equal to………

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$d i s t . = \frac{{\vec{A}}_{1} A_{2} . ({\vec{b}}_{1} * {\vec{b}}_{2})}{| {\vec{b}}_{1} * {\vec{b}}_{2} |}$

$= \frac{(- 4 - α, - 2, - 3) . (2, 2, 1)}{3}$

$| - 5 - \frac{2 α}{3} | = 9 \Rightarrow - 5 - \frac{2 α}{3} = \pm 9$

$\Rightarrow \frac{2 α}{3} = 4, - 14,$

but a > 0

a = 6

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New answer posted

5 months ago

Direction for questions: Answer the questions on the basis of the information given below.

In a school when a survey is conducted on liking of subjects out of three subjects physics, chemistry mathematics then outcome is as follow.

Total number of students who like physics, chemistry and mathematics is 30, 40, 50 respectively. 10 students shown interest in all three subjects, Number of students who like physics and mathematics both is equal to number of students who like chemistry and mathematics both and is equal to 3/4 times number of students who like physics and chemistry both 15 students like only chemistry.

When class teacher saw

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From the given condition g = 10,

Number of students who like both Physics and Maths = 10 + f

Number of students who like both chemistry and Maths = 10 + e

From given condition 10 + f = 10 + e or e = f

Similarly from 3^rd condition ¾ (10 + e) = 10 + d

Or 30 + 3e = 40 + 4d

d = 3e - 10/4

Since 15 students like only chemistry hence b = 15

Number of students who like only chemistry is 40

Hence e + g + d + b = 40

or e + 10 + (3e – 10)/4 + 15 = 40

or (7e – 10)/4 = 15 or 7e = 70 or e = 10

Hence a = 5, c = 20, f = e = 10, d = 5

Hence the distribution is as follows

As per the observation of class teacher actual numbers of students who lik

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5 months ago

If the value of $\underset{x \to 0}{l i m} {(2 - c o s x \sqrt[]{c o s 2 x})}^{(\frac{x + 2}{x^{2}})}$ is equal to e^a, then a is equal to………

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Contributor-Level 9

$\underset{x \to 0}{l i m} {(2 - c o s x . \sqrt[]{c o s 2 x})}^{\frac{x + 2}{x^{2}}}$ $(1^{\infty})$

= $\underset{x \to 0}{l i m} e^{2 (\frac{s i n x \sqrt[]{c o s 2 x} - c o s x . \frac{1 (- 2 s i n 2 x)}{2 \sqrt[]{c o s 2 x}}}{2 x})}$

$= \underset{x \to 0}{l i m} e^{2 (\frac{1}{2} + 1)} = e^{3} = e^{a}$

a = 3

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5 months ago

Let P be a plane passing through the points (1,0,1) (1,-2,1) and (0,1,-2). Let a vector $\vec{a} = α \hat{i} + β \hat{j} + γ \hat{k}$ be such $\vec{a}$ that is the parallel to the plane P, perpendicular to $(\hat{i} + 2 \hat{j} + 3 \hat{k}) a n d \vec{a} . (\hat{i} + \hat{j} + 2 \hat{k}) = 2,$ then ${(α - β + γ)}^{2}$ equals………

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Contributor-Level 9

For the plane P,

$\vec{n} = \frac{1}{2} (- 2 \hat{j}) * (- \hat{i} + \hat{j} - 3 \hat{k}) = \hat{j} * \hat{i} + 3 \hat{j} * \hat{k} = - \hat{k} + 3 \hat{i}$

$\vec{a} . \vec{n} = 0 \Rightarrow 3 α - γ = 0 . . . . . . . (i)$

$\vec{a} . (1, 2, 3) = 0 \Rightarrow α + 2 β + 3 γ = 0 . . . . . . . . . . (i i)$

$\vec{a} . (1, 1, 2) = 2 \Rightarrow α + β + 2 γ = 2 . . . . . . . . . . . (i i i)$

(i), (ii) & (iii) Þ a = 1, β = -5, $γ =$ 3

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5 months ago

Eight girls – ATHENA, BRISEIS, CALANDRA, DEMETER, ELLIE, FEO, GAIA, and HELEN are sitting in a row facing North. We know the following information about them.

I. The number of girls to the left of ATHENA is equal to the number of girls to the right of GAIA.

II. CALANDRA is sitting three places away to the left of ELLIE.

III. The number of girls sitting between FEO and DEMETER is two.

IV. If BRISEIS and HELEN interchange their positions, then HELEN is fifth to the right of BRISEIS.

V. CALANDRA is sitting at the extreme left end of the row.

VI. GAIA is sitting exactly between BRISEIS and DEMETER.

Who is sitting extreme right

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According to the information provided in the statement following arrangement will be obtained:

C	H	A	E	D	G	B	F
1	2	3	4	5	6	7	8

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5 months ago

The number of rational terms in the binomial expansion of ${(4^{\frac{1}{4}} + 5^{\frac{1}{6}})}^{120}$ is…….

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Contributor-Level 9

$T_{r + 1} =^{120} C_{r} 4^{\frac{120 - r}{4}} . 5^{r / 6}$

For to be rational

r should be multiple of 6

Number of such terms = 21

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5 months ago