Direction: Study the following information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.

In how many ways this can be done if 1 trainees and 4 engineers be included in a committee

If 1 trainees and 4 engineers include in committee

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Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 8x2 + 30x + 28 = 0

II. 9y2 + 11y+2 =0

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option C**

**Explaination / Solution: **

I. 8x2 + 30x + 28 = 0

II. 9y2 + 11y+2 =0

I. 8x^{2} + 30x + 28 = 0

x = (-7/4, -2)

II. 9y^{2} + 11y+2 =0

y = (-1, -2/9)

So x<y

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Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 5x2- 24x +16 = 0

II. 5y2 + 29y + 20 = 0

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option A**

**Explaination / Solution: **

I. 5x2- 24x +16 = 0

II. 5y2 + 29y + 20 = 0

I. 5x^{2}- 24x +16 = 0

x = 4/5, 4

II. 5y^{2} + 29y + 20 = 0

y = (-4/5, -5)

So X>Y

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Direction: In the following question, there are two equations. Solve the equations and answer accordingly:

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option E**

**Explaination / Solution: **

I. 6x2 + 46x +60 = 0

II. 4y2+ 29y + 45= 0

II. 4y2+ 29y + 45= 0

I. 6x^{2} + 46x +60 = 0

3x^{2} + 23x + 30 = 0

3x^{2} + 18x + 5x + 30 = 0

3x(x+6)+5(x+6)=0

(x+6)(3x+5)=0

x = -5/3, -6

II. 4y^{2} + 29y + 45= 0

4y^{2} + 20y + 9y + 45= 0

4y(y+5)+9(y+5)=0

(4y+9)(y+5)=0

y = -5, -9/4

So Relationship cannot be established

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Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 8x – y = 40

II. 4x + y = 32

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option C**

**Explaination / Solution: **

After solving the both equation got values of x = 6 and y = 8 So X

I. 8x – y = 40

II. 4x + y = 32

After solving the both equation got values of x = 6 and y = 8 So X

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Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 4x2 - 24x + 32 = 0

II. y2 = (12)2 – 140

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option B**

**Explaination / Solution: **

I. 4x2 - 24x + 32 = 0

II. y2 = (12)2 – 140

I. 4x^{2} - 24x + 32 = 0

x = (4, 2)

II. y = (+2,-2)

So X ≥Y

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Directions: Study the given information carefully and answer the questions that follow—

A store contains 5 red, 4 blue, 5 green shirts.

If two shirts are picked at random, what is the probability that both are blue?

Probabilities if both are blue

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Directions: Study the given information carefully and answer the questions that follow—

A store contains 5 red, 4 blue, 5 green shirts.

If three shirts are picked at random, what is the probability that two are blue and one is red?

Probability if two are Blue and one are Red

[(^{4}C_{2}*^{5}C_{1})/^{14}C_{3}] = (6*5)/364 -->15/182

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Directions: Study the given information carefully and answer the questions that follow—

A store contains 5 red, 4 blue, 5 green shirts.

If two shirts are picked at random, what is the probability that at least one is green?

Probability if at least one is Green

[1-(^{9}C_{2}/^{14}C_{2})] = 55/91

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Directions: Study the given information carefully and answer the questions that follow—

A store contains 5 red, 4 blue, 5 green shirts.

If two shirts are picked at random, what is the probability that either both are red or both are green?

Probabilities if both either are Red or either are green

(^{5}C_{2} + ^{5}C_{2})/ ^{14}C_{2} = (10+10)/91 --> 20/91

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