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Class 12th English

Some children were taking free throws at the basketball court in school during lunch break. Below are some facts about how many baskets these children shot.

I. Giana shot 8 baskets less than Amlia.

II. Drew and Raddix together shot 37 baskets.

III. Joanna shot 8 baskets more than Drew.

IV. Amlia shot 5 baskets more than Drew.

V. Amlia and Giana together shot 40 baskets.

How many shots were thrown by Drew?

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Payal Gupta

Contributor-Level 10

According to given condition, we conclude:

I. Amlia – Giana = 8

II. Drew + Raddix = 37

III. Joanna = Drew + 8

IV. Amlia = Drew + 5

V. Amlia + Giana = 40

Using (I) and (IV),

Amlia = 24, Giana = 16

Using (IV),

Drew = 19

Using (III),

Joanna = 27

Using (II),

Raddix = 18

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New answer posted

5 months ago

Class 12th English

Some children were taking free throws at the basketball court in school during lunch break. Below are some facts about how many baskets these children shot.

I. Giana shot 8 baskets less than Amlia.

II. Drew and Raddix together shot 37 baskets.

III. Joanna shot 8 baskets more than Drew.

IV. Amlia shot 5 baskets more than Drew.

V. Amlia and Giana together shot 40 baskets.

Which of the following statements is false?

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Payal Gupta

Contributor-Level 10

According to given condition, we conclude:

I. Amlia – Giana = 8

II. Drew + Raddix = 37

III. Joanna = Drew + 8

IV. Amlia = Drew + 5

V. Amlia + Giana = 40

Using (I) and (IV),

Amlia = 24, Giana = 16

Using (IV),

Drew = 19

Using (III),

Joanna = 27

Using (II),

Raddix = 18

0 0

New answer posted

5 months ago

Class 12th Ncert Solutions Maths class 12th

If the following system of linear equations

2x + y + z = 5

x – y + z = 3

x + y + az = b

has no solution, then:

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Raj Pandey

Contributor-Level 9

$Δ = | \begin{matrix} 2 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & a \end{matrix} | = 2 (- a - 1) + (1 - a) + 2 = - 3 a + 1$

$Δ_{3} = | \begin{matrix} 2 & 1 & 5 \\ 1 & - 1 & 3 \\ 1 & 1 & b \end{matrix} | = 2 (- b - 1) + (3 - b) + 5 * 2 = - 3 b + 7$

For a = $\frac{1}{3}, b \neq \frac{7}{3},$ $\frac{}{} \frac{}{}$ system has no solution

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New answer posted

5 months ago

Class 12th English

Some children were taking free throws at the basketball court in school during lunch break. Below are some facts about how many baskets these children shot.

I. Giana shot 8 baskets less than Amlia.

II. Drew and Raddix together shot 37 baskets.

III. Joanna shot 8 baskets more than Drew.

IV. Amlia shot 5 baskets more than Drew.

V. Amlia and Giana together shot 40 baskets.

Which of the following statements is true?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

According to given condition, we conclude:

I. Amlia – Giana = 8

II. Drew + Raddix = 37

III. Joanna = Drew + 8

IV. Amlia = Drew + 5

V. Amlia + Giana = 40

Using (I) and (IV),

Amlia = 24, Giana = 16

Using (IV),

Drew = 19

Using (III),

Joanna = 27

Using (II),

Raddix = 18

Drew and Joanna = 19 + 27 = 46

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New answer posted

5 months ago

Class 12th Physics

A cube is placed inside an electric field, The side of the cube is 0.5m and is placed in the field as shown in the given figure. The change inside the cube is:

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alok kumar singh

Contributor-Level 10

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m So

$E = 150 * {(0.5)}^{2} = \frac{150}{4}$

$f l u x f l o w i n g ? = E A = \frac{150}{4} * {(0.5)}^{2}$

$= \frac{150}{16}$

Gausses law $? = \frac{q}{ε_{0}}$

$\frac{150}{16} = \frac{q}{ε_{0}}$

$= 8.3 * 1 0^{- 11} C$

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New answer posted

5 months ago

Class 12th English

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 this survey report he to

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alok kumar singh

Contributor-Level 10

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

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 like Physics in

...more

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New answer posted

5 months ago

Class 12th English

Directions for questions: Read the information carefully and answer the questions given below.

Some children were taking free throws at the basketball court in school during lunch break. Below are some facts about how many baskets these children shot.

I. Giana shot 8 baskets less than Amlia.

II. Drew and Raddix together shot 37 baskets.

III. Joanna shot 8 baskets more than Drew.

IV. Amlia shot 5 baskets more than Drew.

V. Amlia and Giana together shot 40 baskets.

Which of the following statements is true?

0 Follower 1 View

Payal Gupta

Contributor-Level 10

According to given condition, we conclude:

I. Amlia – Giana = 8

II. Drew + Raddix = 37

III. Joanna = Drew + 8

IV. Amlia = Drew + 5

V. Amlia + Giana = 40

Using (I) and (IV),

Amlia = 24, Giana = 16

Using (IV),

Drew = 19

Using (III),

Joanna = 27

Using (II),

Raddix = 18

Raddix shot 18 baskets and Drew shot 19 baskets.

0 0

New answer posted

5 months ago

Class 12th English

Four married couples competed in a singing competition. Each couple had a unique team name. Points scored by the teams were 2, 4, 6 and 8. The “Sweet Couple” won 2 points. The “Bindas Singers” won two more points than Laxman's team. Mukesh's team won four points more than Lina's team, but Lina's team didn't score the least amount of points. “Just Singing” won 6 points. Waheda wasn't on the team called “New Singers”. Sanjeev's team won 4 points. Divya wasn't on the “Bindas Singers” team. Tapas and Sania were on the same team, but it wasn't the “Sweet Couple.”

Mukesh's teammate and team's name were?

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Payal Gupta

Contributor-Level 10

As per the given information we have four teams, four couples and 4 different points

I. “Sweet Couple” won 2 points

II. Mukesh's team won 4 points more than Lina's team and Lina's team has not scored the least points. Therefore, only possible options is Lina has 4 points and Mukesh has 8 points. It also means that Mukesh (with point 8) is not with Lina, and both of them does not belongs to “Sweet Couple”

III. “Just Singing” won 6 points. Hence Mukesh and Lina do not belongs to “Just Singing”

IV. Sanjeev's team scored 4 points. While from (II) we have seen that Lina has also 4 points hence Sanjeev and Lina belongs to s

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