# Equations and Inequations Questions and Answers updated daily – Aptitude

Equations and Inequations Questions: Solved 316 Equations and Inequations Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Equations and Inequations Questions

121. In the question, two equations numbered I and II are given. Solve both the equations and give answer

(a) x≥y

(b) x≤y

(c) x<y

(d) x>y

(e) Relationship between x and y cannot be established

I. 2x

II. y

(a) x≥y

(b) x≤y

(c) x<y

(d) x>y

(e) Relationship between x and y cannot be established

I. 2x

^{2}+ 5x+3 = 0II. y

^{2}+9y+14 = 0SHOW ANSWER

Correct Ans:D

Explanation:

**From I,**

2x

^{2}+ 5x+3 = 0

(2x+3) (x+1) = 0

x = -3/2, -1

**From II,**

y

^{2}+9y+14 = 0

(y+7) (y+2) = 0

**x>y**

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122. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

**Statment:**There are three positive numbers a, b and c. The average of a and b is less than the average of b and c by 1.**Quantity I:**value of c.**Quantity II:**value of a.SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Given that, The average of a and b is less than the average of b and c by 1.

---> (a + b)/2 = [(b + c)/2] - 1

---> [(b + c)/2] - (a + b)/2 = 1

---> (b + c - a - b) / 2 = 1

---> (c - a) / 2 = 1

--->

--->

and

Hence,

---> (a + b)/2 = [(b + c)/2] - 1

---> [(b + c)/2] - (a + b)/2 = 1

---> (b + c - a - b) / 2 = 1

---> (c - a) / 2 = 1

--->

**c - a = 2**--->

**Quantity I: c = a + 2**---> which implies 'c' is 2 more than 'a'and

**Quantity II: a = c - 2**---> which implies 'a' is 2 less than 'c'Hence,

**Quantity I > Quantity II**
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123. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

**Quantity I:**In how many ways the word 'SCOOTER' can be arranged such that 'S' and 'R' are always at two ends?**Quantity II:**A tricolor flag is to be formed having three adjacent strips of three different colors chosen from six different colors. How many different colored flags can be formed with different design in which all the three strips are always in horizontal positions?SHOW ANSWER

Correct Ans:Quantity I = Quantity II or relation cannot be established

Explanation:

**Quantity I:**

In the word SCOOTER ---- the letter 'O' is repeated 2 times.

And S and R are considered as one element.

So, (S, C, 2O, T, E, R)

Required number of ways = (5! * 2!) / 2!

= 5!

= 5 * 4 * 3 * 2 * 1

=

**120 ways**

**Quantity II:**

First strip can be coloured in 6 ways;

second strip can be coloured in 5 ways; and

third strip can be coloured in 4 ways.

Hence, all the three strips can be coloured in 6 * 5 * 4 ways =

**120 ways.**

So,

**Quantity I = Quantity II**

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124. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

A hemisphere of radius 4 cm is to be shipped in a shipping box of rectangular shape. The dimension of the box are consecutive odd numbers.

A hemisphere of radius 4 cm is to be shipped in a shipping box of rectangular shape. The dimension of the box are consecutive odd numbers.

**Quantity I:**Minimum volume of box required to have the shipment possible.**Quantity II:**960 cm^{3}SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

The minimum length of any side of rectangular box should be greater than the diameter of the hemisphere.

Given Radius of hemisphere = 4 cm.

Diameter of hemisphere = 2 * Radius = 8 cm

Also given that, the dimension of the rectangular box are consecutive odd numbers.

So for odd number, minimum length= 9, next side =11, next side=13

Hence,

**Minimum Volume of box = length * width * height**

= 9 * 11 * 13

=

**1287 cm**

^{3}Given

**Quantity II = 960 cm**

^{3}Hence,

**Quantity I > Quantity II**

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125. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

1 > a > 0 > b

1 > a > 0 > b

**Quantity I:**Value of [(a + b)^{2}- a^{2}- b^{2}] / [(a + b)^{2}- (a - b)^{2}]**Quantity II:**1/ [2(ab^{3}+ ab]SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

[(a + b)

^{2}- a

^{2}- b

^{2}] / [(a + b)

^{2}- (a - b)

^{2}]

---> [a

^{2}+ b

^{2}+ 2ab - a

^{2}- b

^{2}] / [a

^{2}+ b

^{2}+ 2ab - a

^{2}- b

^{2}+ 2ab]

---> 2ab / 4ab

--->

**1 / 2**

**Quantity II:**

1/ [2(ab

^{3}+ ab]

---> 1/ [2ab (b

^{2}+ 1)]

Given that 0 > b, so if we subsitute b = -1 and

From the statement 1 > a > 0, say a = 0.5, Then the above eqn becomes,

---> 1/ [2 * 0.5 * (-1) (-1

^{2}+ 1)]

---> 1/ [-1 (1 + 1)]

---> 1/ [-1 (1 + 1)]

--->

**1/ (-2)**

So,

**Quantity I > Quantity II**

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126.

Quantity I: At simple Interest, a sum becomes 3 times in 20 year. Find the time, in which the sum will be double at the same rate of interest.

Quantity II: Simple interest for the sum of Rs. 1500 is Rs. 30 in 4 year and Rs. 60 in 8 year. Find the rate of simple interest.

**Compare the value of 2 quantities given in the question and give answer.**Quantity I: At simple Interest, a sum becomes 3 times in 20 year. Find the time, in which the sum will be double at the same rate of interest.

Quantity II: Simple interest for the sum of Rs. 1500 is Rs. 30 in 4 year and Rs. 60 in 8 year. Find the rate of simple interest.

SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Formula for Simple Interest:**

SI = PNR/t

SI = Simple Interest

P = Principle Amount

N = Number of times compounded in one t

R = Rate

t =Time period

**Quantity-I**

Let sum = P, then for 20 year

SI = 3P- 2P = 2P

2P= P* R *T/100

R=10%

For the sum to be double

SI = 2P-p= P

P = P*10*N/100

N = 10 years

**Quantity-II**

SI in 8 years= 60

SI in 4 years= 30

so,

1500* R * 8/100 - 1500*R*4/100 = 60 -30

1500*R*4/100 = 30

6000R/100 = 30

R = 30/60 = 0.5%

**Hence, Quantity I > Quantity II**

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127. Compare the value of 2 quantities given in the question and give answer.

Quantity I: Two numbers are respectively 20% and 50% more than a 3rd number. What is the percentage of 2nd with respect to 1st?

Quantity II: From 2008 to 2009, the sales of a book decreased by 80%. If the sales in 2010 was the same as in 2008, by what percent did it increase from 2009 to 2010?

Quantity I: Two numbers are respectively 20% and 50% more than a 3rd number. What is the percentage of 2nd with respect to 1st?

Quantity II: From 2008 to 2009, the sales of a book decreased by 80%. If the sales in 2010 was the same as in 2008, by what percent did it increase from 2009 to 2010?

SHOW ANSWER

Correct Ans:Quantity II > Quantity I

Explanation:

Quantity-I:

Let us assume that the 3rd number is 100, so that 1st and 2nd numbers are 20% and 50% more. so,

First number =120

Second number= 150

Third number = 100

Required percentage of 2nd with respect to 1st = (150/120)*100

= 125%

Quantity-II

Sale in 2008= 100

Sale in 2009= 20

Sale in 2010= 100

Required percentage increase= (80/20)*100

= 400%

Hence Quantity II> Quantity I

Let us assume that the 3rd number is 100, so that 1st and 2nd numbers are 20% and 50% more. so,

First number =120

Second number= 150

Third number = 100

Required percentage of 2nd with respect to 1st = (150/120)*100

= 125%

Quantity-II

Sale in 2008= 100

Sale in 2009= 20

Sale in 2010= 100

Required percentage increase= (80/20)*100

= 400%

Hence Quantity II> Quantity I

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128. Compare the value of 2 quantities given in the question and give answer.

Two trains of lengths 50m and 65m are moving in the same direction at 18 m/s and 17m/s respectively.

Quantity I: Time taken by the faster train to cross the slower train.

Quantity II: Time taken by the slower train to cross a platform of length 156 m.

Two trains of lengths 50m and 65m are moving in the same direction at 18 m/s and 17m/s respectively.

Quantity I: Time taken by the faster train to cross the slower train.

Quantity II: Time taken by the slower train to cross a platform of length 156 m.

SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Quantity-I

Relative speed = 18-17= 1 m/s

Required time = (50+65)/1 = 115 seconds

Quantity-II

Time taken to cross a platform of length 156 m

= (65+156)/17

= 13 seconds

Hence Quantity I > Quantity II = 13 seconds

Relative speed = 18-17= 1 m/s

Required time = (50+65)/1 = 115 seconds

Quantity-II

Time taken to cross a platform of length 156 m

= (65+156)/17

= 13 seconds

Hence Quantity I > Quantity II = 13 seconds

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129. Compare the value of 2 quantities given in the question and give answer.(Compare only numerical values)

Quantity A: The least prime number greater than 24

Quantity B: The greatest prime number less than 28

Quantity A: The least prime number greater than 24

Quantity B: The greatest prime number less than 28

SHOW ANSWER

Correct Ans:if quantity A > quantity B

Explanation:

For the integers greater than 24, note that 25, 26, 27, and 28 are not prime numbers, but 29 is a prime number, as are 31 and many other greater integers.

Thus, 29 is the least prime number greater than 24, and Quantity A is 29.

For the integers less than 28, note that 27, 26, 25, and 24 are not prime numbers, but 23 is a prime number, as are 19 and several other lesser integers.

Thus, 23 is the greatest prime number less than 28, and Quantity B is 23.

The correct answer is Choice A, Quantity A is greater.

Thus, 29 is the least prime number greater than 24, and Quantity A is 29.

For the integers less than 28, note that 27, 26, 25, and 24 are not prime numbers, but 23 is a prime number, as are 19 and several other lesser integers.

Thus, 23 is the greatest prime number less than 28, and Quantity B is 23.

The correct answer is Choice A, Quantity A is greater.

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130. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 2x

II. 2y

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 2x

^{2}-7x +3 = 0II. 2y

^{2}- 7y + 6 = 0SHOW ANSWER

Correct Ans:E

Explanation:

I. 2x

2x

(x-3) (2x-1) = 0

x = 3, 1/2

II. 2y

2y

(y-2) (2y-3) = 0

y = 2, 3/2

x and y are not related in any way. The relationship between x and y cannot be determined.

^{2}-7x +3 = 02x

^{2}-6x - x +3 = 0(x-3) (2x-1) = 0

x = 3, 1/2

II. 2y

^{2}- 7y + 6 = 02y

^{2}- 4y - 3y + 6 = 0(y-2) (2y-3) = 0

y = 2, 3/2

x and y are not related in any way. The relationship between x and y cannot be determined.

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131. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 2x

II. 2y

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 2x

^{2}+ 11x + 14 = 0II. 2y

^{2}+ 15y + 28 = 0SHOW ANSWER

Correct Ans:C

Explanation:

I. 2x

2x

(x+2) (2x+7) = 0

x = -2,-7/2

II. 2y

2y

(y+4) (2y+7) = 0

y = -4,-7/2

x ≥ y

^{2}+ 11x + 14 = 02x

^{2}+ 4x + 7x + 14 = 0(x+2) (2x+7) = 0

x = -2,-7/2

II. 2y

^{2}+ 15y + 28 = 02y

^{2}+ 8y + 7y + 28 = 0(y+4) (2y+7) = 0

y = -4,-7/2

x ≥ y

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132. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 9x

II. 4y

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 9x

^{2}- 45x + 56 = 0II. 4y

^{2}- 17y + 18 = 0SHOW ANSWER

Correct Ans:D

Explanation:

**I. 9x**

^{2}- 45x + 56 = 09x

^{2}- 45x + 56 = 0

3x(3x-8) - 7 (3x-8) = 0

(3x-8) (3x-7) = 0

x = 8/3, 7/3

**II. 4y<sup>2</sup> — 17y + 18 = 0**

4y<sup>2</sup> — 8y - 9y + 18 = 0

(y-2) (4y-9) = 0

y =2, 9/4

**x>y**

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133. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 4x

II. 2y

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I. 4x

^{2}+ 16x + 15 = 0II. 2y

^{2}+3y +1 = 0SHOW ANSWER

Correct Ans:A

Explanation:

I. 4x

4x

2x(2x+5) + 3(2x+5) = 0

(2x+5) + (2x+3) = 0

x = -5/2, -3/2

II. 2y

2y

(y+1) (2y+1) = 0

y=-1, -1/2

y>x

^{2}+ 16x + 15 = 04x

^{2}+ 10x + 6x + 15 = 02x(2x+5) + 3(2x+5) = 0

(2x+5) + (2x+3) = 0

x = -5/2, -3/2

II. 2y

^{2}+ 3y + 1 = 02y

^{2}+ 2y + y + 1 = 0(y+1) (2y+1) = 0

y=-1, -1/2

y>x

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134. In the following question two equations are given. Solve the equations and find the answer from the 5 options given below.

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I.9x

II. 2y

(a) if x < y

(b) if x ≤ y

(c) if x ≥ y

(d) if x > y

(e) relationship between x and y cannot be determined

I.9x

^{2}- 36x + 35 = 0II. 2y

^{2}- 15y- 17 = 0SHOW ANSWER

Correct Ans:E

Explanation:

I. 9x

9x

3x (3x-7) - 5 (3x-7) = 0

(3x-7) (3x-5) = 0

x = 5/3 , 7/3

II. 2y

2y

(y+1) (2y-17) = 0

y = -1, 17/2

x and y are not related in any way. The relationship between x and y cannot be determined.

^{2}- 36x + 35 = 09x

^{2}- 21x - 15x + 35 = 03x (3x-7) - 5 (3x-7) = 0

(3x-7) (3x-5) = 0

x = 5/3 , 7/3

II. 2y

^{2}- 15y- 17 = 02y

^{2}-17y + 2y - 17 = 0(y+1) (2y-17) = 0

y = -1, 17/2

x and y are not related in any way. The relationship between x and y cannot be determined.

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135. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

**Quantity I:**How many numbers greater than 1000 but less than 5000 can be made from digits 0, 1, 3, 5, 6?**Quantity II:**There are 5 blue flags, 4 red flags and 3 green flags, in Debu’s wardrobe. He has to select 4 flags from this set. In how many ways can he select these four flags such that there is at least one blue flag and exactly one green flag in them (Do not consider that the flags are in pairs)?SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

When not written about repetition, take that repetition to be allowed.

Numbers between 1000 to 5000 means 4 digit numbers. So 4 boxes.

Now 0, 5 and 6 cannot be placed in Box 1 (Thousandth place). Because, 0 will make it 3-digit number and 5 & 6 will make number greater than 5000.

So there are 2 choices (ie., numbers 1 and 3) for Box 1 (Thousandth place), and all 5 choices (ie., numbers 0, 1, 3, 5, 6) for Box 2, 3, and 4 (Hundredths, Tenths and units places).

So total numbers = 2C1 * 5C1 * 5C1 * 5C1 ---> (since 4 digit number)

= 2*5*5*5

=

**250**

**Quantity II:**

No. of Blue flags = 5;

No. of Red flags = 4;

No. of Green flags = 3

In the selected four flags ---> there is

**at least one blue flag**and

**exactly one green flag**

So, the possible selections are:

Way 1: 1 blue flag out of 5 * 1 green flag out of 3 * 2 Red flags out of 4

----> 5C1 * 3C1 * 4C2

----> 5 * 3 * 6

----> 90

Way 2: 2 blue flag out of 5 * 1 green flag out of 3 * 1 Red flags out of 4

----> 5C2 * 3C1 * 4C1

----> 10 * 3 * 4

----> 120

Way 3: 3 blue flag out of 5 * 1 green flag out of 3

----> 5C3 * 3C1

----> 10 * 3

----> 30

So, Total No. of ways = 90 + 120 + 30 =

**240**

here,

**Quantity I > Quantity II.**

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136. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.

Ram invested Rs. P in scheme A and Rs. 2P in scheme B, for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% per annum. Respective ratio between the interest earned from scheme A and that earned from scheme B was 8 : 21.

Ram invested Rs. P in scheme A and Rs. 2P in scheme B, for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% per annum. Respective ratio between the interest earned from scheme A and that earned from scheme B was 8 : 21.

**Quantity I:**Rate of interest offered by scheme A.**Quantity II:**Rate of interest offered by scheme C (simple interest per annum), when Rs. 1600 is invested for 3 years earns an interest of Rs. 384.SHOW ANSWER

Correct Ans:Quantity I = Quantity II or Relation cannot be established

Explanation:

Let the rate of interest offered by scheme A ie., simple interest = R% per annum.

Given Time, (n) = 2 years.

Then, Simple interest,

= (P * 2 * R)/100

=

Now, Compound Interest, C.I = Amount - Principal

Amount = P x [1+ (r/100)]

So, Compound Interest received by Scheme B, C.I = 2P{[1+ (10/100)]

= 2P{[1+ (1/10)]

= 2P{[(10+ 1)/10]

= 2P{[11/10]

= 2P{121/100 - 1}

= 2P{121 - 100}/100

= 2P{21/100}

=

Given that, S.I from scheme A : C.I from scheme B = 8 : 21

----> (2PR/100) : (42P/100) = 8 : 21

----> R : 21 = 8 : 21

---->

Now,

Simple interest --> Principal, P = Rs. 1600

n = 3 years

S.I = Rs. 384

Then, Simple interest,

---> 384 = (1600 * 3 * r)/100

---> 384 = 48r

--->

Hence,

Here

Given Time, (n) = 2 years.

Then, Simple interest,

**S.I = (p x n x r)/100**= (P * 2 * R)/100

=

**2PR/100**Now, Compound Interest, C.I = Amount - Principal

Amount = P x [1+ (r/100)]

^{n}**C.I = P{[1+ (r/100)]**^{n}- 1}So, Compound Interest received by Scheme B, C.I = 2P{[1+ (10/100)]

^{2}- 1}= 2P{[1+ (1/10)]

^{2}- 1}= 2P{[(10+ 1)/10]

^{2}- 1}= 2P{[11/10]

^{2}- 1}= 2P{121/100 - 1}

= 2P{121 - 100}/100

= 2P{21/100}

=

**42P/100**Given that, S.I from scheme A : C.I from scheme B = 8 : 21

----> (2PR/100) : (42P/100) = 8 : 21

----> R : 21 = 8 : 21

---->

**R = 8****Hence, Quantity I: Rate of interest offered by scheme A, R = 8% per annum.**Now,

**Quantity II:**Simple interest --> Principal, P = Rs. 1600

n = 3 years

S.I = Rs. 384

Then, Simple interest,

**S.I = (p x n x r)/100**---> 384 = (1600 * 3 * r)/100

---> 384 = 48r

--->

**r = 8**Hence,

**Rate of interest offered by scheme C = 8% per annum.**Here

**Quantity I = Quantity II**.
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137. In the following question, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly

**I.**3x^{2}"“ 25x + 52 = 0**II.**3y^{2}"“ 17y + 20= 0SHOW ANSWER

Correct Ans:If x≥y

Explanation:

**I.**3x

^{2}â€“ 25x + 52 = 0

3x

^{2}- 12x - 13x + 52 = 0

3x(x - 4) - 13(x - 4) = 0

(x - 4)(3x - 13) = 0

x = 4, 13/3 = 4, 4.3

**II.**3y

^{2}â€“ 17y + 20= 0

3y

^{2}- 12y - 5y + 20= 0

3y(y - 4) - 5(y - 4) = 0

(y - 4)(3y - 5) = 0

y = 4, 5/3 = 4, 1.7

Hence, x≥y

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138. If x = 332, y = 333, z = 335 then the value of x

^{3}+ y^{3}+ z^{3}"“ 3xyz = ?SHOW ANSWER

Correct Ans:7000

Explanation:

(x

Also we know that,

(x

On putting the values, we get

= 1000 Ã— 1/2[(332 â€“ 333)

= 1000 Ã— 1/2[1 + 4 + 9]

= 7000

Hence, option C is correct.

^{3}+ y^{3}+ z^{3}â€“ 3xyz) = (x + y + z) (x^{2}+ y^{2}+ z^{2}â€“ xy â€“ yz â€“ zx)Also we know that,

(x

^{3}+ y^{3}+ z^{3}â€“ 3xyz) = (x + y + z) * 1/2[(x â€“ y)2 + (y â€“ z)2 + (z â€“ x)2]On putting the values, we get

= 1000 Ã— 1/2[(332 â€“ 333)

^{2}+ (333 â€“ 335)^{2}+ (335 â€“ 332)^{2}= 1000 Ã— 1/2[1 + 4 + 9]

= 7000

Hence, option C is correct.

Workspace

139.

**Quantity 1:**Perimeter of a Square having its side is equal to hypotenuse of a right-angled triangle having its sides 8mtrs and 6mtrs respectively.**Quantity 2:**Perimeter of a Rhombus having its diagonals d1 and d2 are 12mtr and 16mtr respectively.SHOW ANSWER

Correct Ans:Quantity 1 = Quantity 2 or Relationship can’t be established.

Explanation:

Quantity 1:

Hypotenuse of a triangle = √(a²+b²)

= √(8²+6²)

= √(64+36)

= √100

= 10 mtrs

And Perimeter of the square = 4a = 4 × 10 = 40mtrs.

Quantity 2:

Diagonals of rhombus intersect at 90°angle.

The diagonals and the sides of the rhombus form right triangle.

One side of the right triangle is 16/2= 8mtrs

Other side of the right triangle is 12/2= 6mtrs

Then the third side of the triangle = √(a²+b²)

= √(8²+6²)

= 10mtrs

And Perimeter of the rhombus = 4a = 4 × 10 = 40mtrs.

So Quantity 1 is equal to Quantity 2.

Hypotenuse of a triangle = √(a²+b²)

= √(8²+6²)

= √(64+36)

= √100

= 10 mtrs

And Perimeter of the square = 4a = 4 × 10 = 40mtrs.

Quantity 2:

Diagonals of rhombus intersect at 90°angle.

The diagonals and the sides of the rhombus form right triangle.

One side of the right triangle is 16/2= 8mtrs

Other side of the right triangle is 12/2= 6mtrs

Then the third side of the triangle = √(a²+b²)

= √(8²+6²)

= 10mtrs

And Perimeter of the rhombus = 4a = 4 × 10 = 40mtrs.

So Quantity 1 is equal to Quantity 2.

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140. Study the following question carefully and choose the right answer.

If x = âˆš3 + âˆš4 + âˆš5 than x

If x = âˆš3 + âˆš4 + âˆš5 than x

^{4}"“ 8x^{3}+ 8x^{2}+ 32x = ?SHOW ANSWER

Correct Ans:44

Explanation:

x = âˆš3 + âˆš4 + âˆš5

x â€“ 2 = âˆš3 + âˆš5

Squaring on both sides we get,

x

x

Squaring on both sides we get,

(a - b - c)Â² = aÂ² + bÂ² + cÂ² - 2ab + 2bc - 2ac

x

x

Hence, option C is correct.

x â€“ 2 = âˆš3 + âˆš5

Squaring on both sides we get,

x

^{2}â€“ 4x + 4 = 8 + 2âˆš15x

^{2}â€“ 4x â€“ 4 = 2âˆš15Squaring on both sides we get,

(a - b - c)Â² = aÂ² + bÂ² + cÂ² - 2ab + 2bc - 2ac

x

^{4}+ 16x^{2}+ 16 â€“ 8x^{3}+ 32x â€“ 8x^{2}= 60x

^{4}â€“ 8x^{3}+ 8x^{2}+ 32x = 44Hence, option C is correct.

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