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ISEE Upper Level Quantitative : How to find the solution to an equation

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

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Example Questions

Example Question #151 : How To Find The Solution To An Equation

$\sqrt{x} + \frac{1}{2} = \sqrt{y}$

Which is the greater quantity?

(a)

(b)

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

It can be deduced that both and are nonnegative, since both are radicands of square roots.

$\sqrt{y} = \sqrt{x} + \frac{1}{2}$ , so

$( \sqrt{y})^{2} = \left (\sqrt{x} + \frac{1}{2} \right )^{2}$

$y = \left (\sqrt{x}\right ) ^{2} + 2 \cdot \sqrt{x} \cdot \frac{1}{2} + \left ( \frac{1}{2} \right ) ^{2}$

$y = x + \sqrt{x} + \frac{1}{4}$

$\sqrt{x} > 0$ , so

$\sqrt{x} + \frac{1}{4} > 0$ , and

y > x .

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Example Question #151 : How To Find The Solution To An Equation

$343 x^{3} + 7 x = 49x^{2} + 1$

$y^{3} + 49 y = 7y^{2} + 343$

Which is the greater quantity?

(a)

(b)

Possible Answers:

(b) is the greater quantity

It is impossible to determine which is greater from the information given

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

$343 x^{3} + 7 x = 49x^{2} + 1$

$343 x^{3} + 7 x - 49x^{2} - 1 = 49x^{2} + 1 - 49x^{2} - 1$

$343 x^{3}- 49x^{2} + 7 x - 1 = 0$

$(343 x^{3}- 49x^{2} )+( 7 x - 1 )= 0$

$49x^{2} ( 7 x - 1 ) + 1 ( 7 x - 1 )= 0$

$(49x^{2} + 1) ( 7 x - 1 )= 0$

By the Zero Product Principle, one of the factors is equal to 0:

$49x^{2}+ 1 = 0$

$49x^{2} = - 1$

$x^{2} = -\frac{1}{49}$

which is impossible for any real value of , or

7x-1 = 0

7x = 1

$x = \frac{1}{7}$ .

$y^{3} + 49 y = 7y^{2} + 343$

$y^{3} + 49 y - 7y^{2} - 343 = 7y^{2} + 343 - 7y^{2} - 343$

$y^{3} - 7y^{2} + 49 y - 343 = 0$

$(y^{3} - 7y^{2}) + (49 y - 343 )= 0$

$y^{2} (y - 7) + 49 (y - 7) = 0$

$(y^{2} + 49) (y - 7) = 0$

By the Zero Product Principle, one of the factors is equal to 0:

$y^{2} + 49 = 0$

$y^{2} =- 49$

which is impossible for any real value of , or

y - 7 = 0

y = 7

Since $x = \frac{1}{7}$ and y = 7 , it can be determined that y > x .

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Example Question #152 : How To Find The Solution To An Equation

$x + \frac{3}{5} = \frac{7}{9}$

$y+ \frac{7}{9} = \frac{11}{13}$

Which is the greater quantity?

(a)

(b)

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

Correct answer:

(a) is the greater quantity

Explanation:

$x + \frac{3}{5} = \frac{7}{9}$

$x = \frac{7}{9} - \frac{3}{5}$

$= \frac{7\cdot 5 }{9 \cdot 5 } - \frac{3 \cdot 9}{5 \cdot 9}$

$= \frac{35 }{45 } - \frac{27}{45}$

$= \frac{8}{45 }$

$y+ \frac{7}{9} = \frac{11}{13}$

$y= \frac{11}{13} - \frac{7}{9}$

$= \frac{11 \cdot 9 }{13 \cdot 9 } - \frac{7 \cdot 13}{9 \cdot 13}$

$= \frac{99 }{117} - \frac{91}{117}$

$= \frac{8}{117}$

Between two fractions with the same numerator, the one with the lesser denominator is the greater, so

$\frac{8}{45 } > \frac{8}{117}$

and x > y .

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Example Question #152 : Equations

, , and all stand for positive quantities.

$t = \frac{1}{7} u + 63$

$t = \frac{1}{9} v+63$

Which is the greater quantity?

Possible Answers:

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

Solve the equations for and in terms of :

$\frac{1}{7} u + 63 = t$

$\frac{1}{7} u + 63 - 63 = t - 63$

$\frac{1}{7} u = t - 63$

$7 \cdot \frac{1}{7} u =7 \cdot ( t - 63)$

u = 7t - 441

$\frac{1}{9} v+63 = t$

$\frac{1}{9} v + 63 - 63 = t - 63$

$\frac{1}{9} v= t - 63$

$9 \cdot \frac{1}{9} v =9 \cdot ( t - 63)$

v = 9t - 567

Therefore, we seek to determine which of 7t - 441 and 9t - 567 is greater, bearing in mind that both of these quantities, as well as , must be positive.

We can make the following observation:

Suppose

$9t - 567 \le 7t - 441$

Then

$9t - 567 - 7t + 567 \le 7t - 441 - 7t + 567$

$2t \le 126$

$2t \div 2 \le 126 \div 2$

$t \le 63$

But if $t \le 63$ , then

$9t \le 9 \cdot 63 = 567$

and

$9t - 567 \le 567 - 567 = 0$

$v \le 0$ , a contradiction.

Therefore, it must hold that 9t - 567 > 7t - 441 , and v> u .

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Example Question #153 : Equations

, , and all stand for positive quantities.

$t = \frac{1}{5} u + 3$

$t = \frac{1}{3} v + 5$

Which is the greater quantity?

(a)

(b)

Possible Answers:

(a) and (b) are equal

(b) is the greater quantity

It is impossible to determine which is greater from the information given

(a) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

Solve the equations for and in terms of :

$\frac{1}{5} u + 3 = t$

$\frac{1}{5} u + 3 - 3 = t - 3$

$\frac{1}{5} u = t - 3$

$5 \cdot \frac{1}{5} u =5 \cdot ( t - 3)$

u = 5t - 15

$\frac{1}{3} v + 5 = t$

$\frac{1}{3} v + 5 - 5 = t- 5$

$\frac{1}{3} v = t- 5$

$3 \cdot \frac{1}{3} v =3 \cdot ( t- 5)$

v = 3t - 15

5 > 3 and is positive, so by the properties of inequality,

5 t > 3t

5 t -15 > 3t - 15

u > v

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Example Question #153 : How To Find The Solution To An Equation

Solve for :

$\frac{1}{4} t + \frac{1}{3} x= 2$

Possible Answers:

$x = 6 +\frac { 3}{4} t$

$x =8+\frac {4}{3} t$

$x =8-\frac {4}{3} t$

$x = 6 -\frac { 3}{4} t$

Correct answer:

$x = 6 -\frac { 3}{4} t$

Explanation:

$\frac{1}{4} t + \frac{1}{3} x= 2$

$\frac{1}{4} t + \frac{1}{3} x - \frac{1}{4} t = 2 - \frac{1}{4} t$

$\frac{1}{3} x = 2 - \frac{1}{4} t$

$3 \cdot \frac{1}{3} x = 3 \cdot \left ( 2 - \frac{1}{4} t \right )$

$x = 3 \cdot 2 - 3 \cdot \frac{1}{4} t$

$x = 6 -\frac { 3}{4} t$

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Example Question #154 : How To Find The Solution To An Equation

m - (-46) = -73

n - (-73) = -46

Which is the greater quantity?

(a)

(b)

Possible Answers:

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

Correct answer:

(a) and (b) are equal

Explanation:

m - (-46) = -73

m - (-46)+ (-46 ) = -73+ (-46 )

m = - (73 + 46 ) = -119

n - (-73) = -46

n - (-73) + (-73) = -46 + (-73)

n = - (46+73 ) = -119

m = n

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Example Question #151 : How To Find The Solution To An Equation

Untitled

Figure NOT drawn to scale

Above is a straight line on a graph. Which is the greater quantity?

(a)

(b) 18

Possible Answers:

(a) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(b) is the greater quantity

Correct answer:

(a) is the greater quantity

Explanation:

If we go from the point (48, 60) to (24, 42), we see that if the first coordinate decreases by 24, the second decreases by 18. Going from (24, 42) to the point on the -axis, the first coordinate again decreases by 24, so the second coordinate again decreases by 18:

y = 42 - 18 = 24> 18 .

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Example Question #156 : How To Find The Solution To An Equation

$2 < \frac{1}{Y}< 4$

The reciprocal of is between 2 and 4. Which is the greater quantity?

(a)

(b) $\frac{1}{3}$

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(a) is the greater quantity

Correct answer:

It is impossible to determine which is greater from the information given

Explanation:

$2 < \frac{1}{Y}< 4$ , so

$\frac{1}{2} > \frac{1}{ \frac{1}{Y}}> \frac{1}{4}$

$\frac{1}{2} >Y> \frac{1}{4}$

Also,

2 < 3< 4 , so

$\frac{1}{2} > \frac{1}{3}> \frac{1}{4}$

Therefore, it possible for

$\frac{1}{2} > Y >\frac{1}{3}$ ,

$Y = \frac{1}{3}$ ,

$\frac{1}{3}>Y > \frac{1}{4}$ ,

making it inconclusive whether or $\frac{1}{3}$ is the greater.

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Example Question #157 : How To Find The Solution To An Equation

Solve for :

$t = \frac{1}{4} x + 3$

Possible Answers:

x = 4t+ 12

$x = \frac{1}{4 }t+\frac{3}{4 }$

$x = \frac{1}{4 }t- \frac{3}{4 }$

x = 4t- 12

Correct answer:

x = 4t- 12

Explanation:

$\frac{1}{4} x + 3 = t$

$\frac{1}{4} x + 3- 3 = t - 3$

$\frac{1}{4} x = t - 3$

$4 \cdot \frac{1}{4} x =4 \cdot ( t - 3)$

$x = 4\cdot t - 4 \cdot 3$

x = 4t- 12

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : How to find the solution to an equation

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #151 : How To Find The Solution To An Equation

Example Question #151 : How To Find The Solution To An Equation

Example Question #152 : How To Find The Solution To An Equation

Example Question #152 : Equations

Example Question #153 : Equations

Example Question #153 : How To Find The Solution To An Equation

Example Question #154 : How To Find The Solution To An Equation

Example Question #151 : How To Find The Solution To An Equation

Example Question #156 : How To Find The Solution To An Equation

Example Question #157 : How To Find The Solution To An Equation

All ISEE Upper Level Quantitative Resources

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