Circles - ISEE Upper Level Quantitative Reasoning

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Question

The clock in the classroom reads 5:00pm. What is the angle that the hands are forming?

Answer

Since the clock is a circle, you can determine that the total number of degrees inside the circle is 360. Since a clock has 12 numbers, we can divide 360 by 12 to see what the angle is between two numbers that are right next to each other. Thus, we can see that the angle between two numbers right next to each other is $30^{\circ}$ . However, the clock is reading 5:00, so there are five numbers we have to take in to account. Therefore, we multiply 30 by 5, which gives us $150^{\circ}$ as our answer.

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Question

The time on a clock reads 5:00. What is the measure of the central angle formed by the hands of the clock?

Answer

First, remember that the number of degrees in a circle is 360. Then, figure out how many degrees are in between each number on the face of the clock. Since there are 12 numbers, there are $30^{\circ}$ between each number. Since the time reads 5:00, multiply $30\cdot 5$ , which yields $150^{\circ}$ .

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Question

Refer to the above figure. Which is the greater quantity?

(a) $\frac{a}{d}$

(b) $\frac{c}{b}$

Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

Divide both sides of this equation by , then cancelling:

$\frac{ab }{bd}= \frac{cd}{bd}$

$\frac{a }{d}= \frac{c}{b}$

The two quantities are equal.

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Question

Refer to the above figure. Which is the greater quantity?

(a)

(b) 3

Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

$a \cdot b = 1.5 \cdot 1.5$

or

Therefore, .

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Question

Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b) 7

Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

$t \cdot 4t = 8 \cdot 24$

Solving for :

$4t ^{2} = 192$

$4t ^{2} \div 4 = 192 \div 4$

$t ^{2} = 48$

Since $t ^{2} < 49$ , it follows that $t < \sqrt{49 }$ , or .

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Question

In the above figure, $\overline{NT}$ is a tangent to the circle.

Which is the greater quantity?

(a)

(b) 32

Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

$NA \cdot NB = (NT )^{2}$

$NA \cdot (NA + AB)= (NT )^{2}$

$t \cdot (t+t) = 16 ^{2}$

Simplifying, then solving for :

$t \cdot (2t) = 256$

$2t ^{2} = 256$

$2t ^{2} \div 2 = 256 \div 2$

$t ^{2} = 128$

To compare to 32, it suffices to compare their squares:

, so, applying the Power of a Product Principle, then substituting,

$(NB)^{2} = (2t) ^{2} = 2^{2} \cdot t^{2} = 4 t ^{2} = 4 \cdot 128 = 512$

$32^{2} = 1,024$

, so

$(NB)^{2} < 32 ^{2}$ ;

it follows that

.

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Question

Figure NOT drawn to scale

In the above figure, $\overline{NT}$ is a tangent to the circle.

Which is the greater quantity?

(a)

(b) 8

Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle intersected by the secant; in other words,

$(NT)^{2}= NA \cdot NB$

$(NT)^{2}= NA \cdot (NA+AB)$

$(2t) ^{2} = t (t + 24)$

Simplifying and solving for :

$2^{2} \cdot t^{2} = t \cdot t + t \cdot 24$

$4 t^{2} = t ^{2} + 24t$

$4 t^{2} - t ^{2} - 24t = t ^{2} + 24t - t ^{2} - 24t$

$3 t^{2}- 24t = 0$

Factoring out :

Either - which is impossible, since must be positive, or

, in which case .

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Question

Figure NOT drawn to scale

In the above figure, is the center of the circle, and $\overline{NT}$ is a tangent to the circle. Also, the circumference of the circle is $14 \pi$ .

Which is the greater quantity?

(a)

(b) 25

Answer

$\overline{OT}$ is a radius of the circle from the center to the point of tangency of $\overline{NT}$ , so

$\overline{OT} \perp \overline{NT}$ ,

and $\bigtriangleup NTO$ is a right triangle. The length of leg $\overline{NT}$ is known to be 24. The other leg $\overline{OT}$ is a radius radius; we can find its length by dividing the circumference by $2 \pi$ :

$OT = r = \frac{C}{2\pi} = \frac{14 \pi}{2\pi} = 7$

The length hypotenuse, $\overline{NO}$ , can be found by applying the Pythagorean Theorem:

$NO = \sqrt{(NT)^{2} + (OT)^{2}}= \sqrt{24^{2} + 7^{2}}=\sqrt{576 + 49} = \sqrt{625} = 25$ .

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Question

A giant clock has a minute hand five yards in length. Since noon, the tip of the minute hand has traveled $200 \pi$ feet. Which is the greater quantity?

(A) The amount of time that has passed since noon

(B) The amount of time until midnight

Answer

Five yards is equal to fifteen feet, which is the length of the minute hand. Subsequently, fifteen feet is the radius of the circle traveled by its tip in one hour; the circumference of this circle is $2 \pi$ times this, or

$C = 2 \pi \cdot 15 = 30 \pi$ feet.

In one six-hour period, the minute hand revolves six times, so its tip travles six times the circumference, or

$30 \pi \times 6 = 180 \pi$

The clock has traveled farther than this, so the time is later than 6:00 PM, and more time has elapsed since noon than is left until midnight. This makes (A) greater.

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Question

Compare the two quantities:

Quantity A: The area of a circle with radius

Quantity B: The circumference of a circle with radius

Answer

Recall for this question that the formulae for the area and circumference of a circle are, respectively:

$A=\pi * r^2$

$C=2\pir$

For our two quantities, we have:

Quantity A: $\pi * 5^2 = 25\pi$

Quantity B: $2 * \pi * 8.1 = 16.2\pi$

Therefore, quantity A is greater.

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Question

The circumferences of eight circles form an arithmetic sequence. The smallest circle has radius two inches; the second smallest circle has radius five inches. Give the radius of the largest circle.

Answer

The circumference of a circle can be determined by multiplying its radius by $2 \pi$ , so the circumferences of the two smallest circles are

$C_{1} = 2 \cdot 2 \pi = 4 \pi$

and

$C_{2} = 2 \cdot 5 \pi = 10 \pi$

The circumferences form an arithmetic sequence with common difference

$C_{2}- C_{1}= 10 \pi - 4 \pi = 6 \pi$

The circumference of a circle can therefore be found using the formula

$C_{n} = C_{1} + (n-1)d$

where $C_{1} = 4 \pi$ and $d = 6 \pi$ ; we are looking for that of the th smallest circle, so

$C_{8} = 4 \pi + (8-1)6 \pi$

$= 4 \pi + 7 \cdot 6 \pi$

$= 4 \pi + 42 \pi$

$= 46 \pi$

Since the radius of a circle is the circumference of the circle divided by $2 \pi$ , the radius of this eighth circle is

$r = \frac{46 \pi}{2 \pi} = 23$ inches, or 1 foot 11 inches.

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Question

The track at Truman High School is shown above; it is comprised of a square and a semicircle.

Veronica begins at Point A, runs three times around the track counterclockwise, and continues until she reaches Point B. Which of the following comes closest to the distance Veronica runs?

Answer

First, it is necessary to know the length of the semicircle connecting Points B and D, which has diameter 500 feet; this length is about

$\frac{1}{2} C = \frac{1}{2} \cdot \pi d \approx \frac{1}{2} \cdot 3.14 \cdot 500 = 785$ feet.

The distance around the track is about

feet.

Veronica runs around the track three complete times, for a distance of about

$3 \times 2,285 = 6,855$ feet.

She then runs from Point A to Point E, which is another 500 feet; Point E to Point D, which is yet another 500 feet, and, finally Point D to Point B, for a final 785 feet. The total distance Veronica runs is about

feet.

Divide by 5,280 to convert to miles:

$8,640 \div5,280 \approx 1.6$

The closest answer is $1 \frac{1}{2}$ miles.

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Question

The track at Simon Bolivar High School is a perfect circle of radius 500 feet, and is shown in the above figure. Manuel starts at point C, runs around the track counterclockwise three times, and continues to run clockwise until he makes it to point D. Which of the following comes closest to the number of miles Manuel has run?

Answer

The circumference of a circle with radius 500 feet is

$C = 2 \pi r \approx 2 \cdot 3.14 \cdot 500 = 3,140$ feet.

Manuel runs this distance three times, then he runs from Point C to D, which is about four-fifths of this distance. Therefore, Manuel's run will be about

$3,140 \times 3 \frac{4}{5} = 11,932$ feet.

Divide by 5,280 to convert to miles:

$11,932 \div 5,280 \approx 2.26$ ,

making $2\frac{1}{4}$ miles the response closest to the actual running distance.

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Question

The track at James Buchanan High School is shown above; it is comprised of a square and a semicircle.

Diane wants to run two miles. If she begins at Point A and begins running counterclockwise, when she is finished, which of the five points will she be closest to?

Answer

First, it is necessary to know the length of the semicircle connecting Points B and D, which has diameter 400 feet; this length is

$\frac{1}{2} C = \frac{1}{2} \cdot \pi d \approx \frac{1}{2} \cdot 3.14 \cdot 400 = 628$ feet.

The distance around the track is about

feet.

Diane wants to run two miles, or

$5,280 \times 2 = 10, 560$ feet.

She will make about

$10,560 \div 1,828 \approx 5.78$

circuits around the track.

Equivalently, she will run the track 5 complete times for a total of about

$5 \times 1,828 = 9,140$ feet,

so she will have

feet to go.

She is running counterclockwise, so she will proceed from Point A to Point D, running another 800 feet, leaving

feet.

She will almost, but not quite, finish the 628 feet from Point D to Point B.

The correct response is Point B.

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Question

The track at Monroe High School is a perfect circle of radius 600 feet, and is shown in the above figure. Quinnella wants to run around the track for one and a half miles. If Quinnella starts at point C and runs counterclockwise, which of the following is closest to the point at which she will stop running?

(Assume the five points are evenly spaced)

Answer

A circle of radius 600 feet will have a circumference of

$C = 2 \pi r \approx 2 \cdot 3.14 \cdot 600 = 3,768$ feet.

Quinnella will run one and a half miles, or

$5,280 \times 1\frac{1}{2} = 7,920$ feet,

which is about times the circumference of the circle.

Quinnella will run around the track twice, returning to Point C; she will not quite make it to Point B a third time, since that is one-fifth of the track, or 0.2. The correct response is that she will be between Points B and C.

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Question

The track at Monroe Elementary School is a perfect circle of radius 400 feet, and is shown in the above figure.

Evan and his younger brother Mike both start running from Point A. Evan runs counterclockwise, running once around the track and then on to Point E; Mike runs clockwise, meeting Evan at Point E and stopping.

Which of the following is the greater quantity?

(a) Twice Mike's average speed.

(b) Evan's average speed.

(Assume the five points are evenly spaced)

Answer

It is not actually necessary to know the radius or length of the track if we know the points are equally spaced. Evan runs once around the track counterclockwise and then on to Point E, which is the next point after A; this means he runs around the track $1\frac{1}{5} = \frac{6}{5}$ times. Mike runs around the track clockwise from Point A to Point E, in the same time, meaning he runs around the track $\frac{4}{5}$ times.

Therefore, Evan's speed is $\frac{ \frac{6}{5}}{\frac{4}{5}} = \frac{3}{2}$ times Mike's speed. As a result, Twice Mike's speed would be greater than Evan's speed, making (b) the greater.

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Question

The track at Monroe High School is a perfect circle of radius 600 feet, and is shown in the above figure.

Jerry begins his one-mile run at Point A, then runs counterclockwise around the track. At the end of his one-mile run, which is the greater quantity?

(a) The additional distance he would have to run if he were to continue to run counterclockwise to Point A.

(b) The additional distance he would have to run if he were to turn back and run clockwise to Point A.

Answer

The circumference of a circle with radius 600 feet is

$C = 2 \pi r \approx 2 \cdot 3.14 \cdot 600 = 3,768$ feet.

A one mile run would be

$\frac{5,280 }{3,768} \approx 1.4$

times the length of the track.

Therefore, Jerry's run takes him around the track once, and about 0.4 times the length of the track. Since he is running counterclockwise, but has not made it halfway around the track yet, the longer of the two paths is to proceed counterclockwise and run the remaining 0.6 of the track. This makes (a) greater.

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Question

The radii of six circles form an arithmetic sequence. The radius of the second-smallest circle is twice that of the smallest circle. Which of the following, if either, is the greater quantity?

(a) The circumference of the largest circle

(b) Twice the circumference of the third-smallest circle

Answer

Call the radius of the smallest circle . The radius of the second-smallest circle is then , and the common difference of the radii is .

The radii of the six circles are, from least to greatest:

The largest circle has circumference

$C_{6} = 2 \pi \cdot 6r = 12 \pi r$

The third-smallest circle has circumference:

$C_{3} = 2 \pi \cdot 3r = 6 \pi r$

Twice this is

$2 C_{3} = 2 \cdot 6 \pi r = 12 \pi r$

The circumference of the sixth circle is equal to twice that of the third-smallest circle, so the correct choice is that that (a) and (b) are equal.

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Question

In the above figure, .

Which is the greater quantity?

(a) The sum of the circumferences of the inner and outer circles

(b) The sum of the circumferences of the second-largest and third-largest circles

Answer

For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length.

The four concentric circles have radii 1, 2, 3, and 4, respectively, and their circumferences can be found by multiplying these radii by $2 \pi$ .

The inner and outer circles have circumferences $2 \pi \cdot 1 = 2 \pi$ and $2 \pi \cdot 4 =8 \pi$ , respectively; the sum of these circumferences is $2 \pi + 8 \pi = 10 \pi$ . The other two circles have circumferences $2 \pi \cdot 2 = 4 \pi$ and $2 \pi \cdot 3 =6 \pi$ ; the sum of these circumferences is $4 \pi + 6 \pi = 10 \pi$ .

The two sums are therefore equal.

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Question

Circle B has a radius $\frac{2}{3}$ as long as that of Circle A.

Which is the greater quantity?

(a) The area of Circle A

(b) Twice the area of Circle B

Answer

If we call the radius of Circle A , then the radius of Circle B is $\frac{2}{3} r$ .

The areas of the circles are:

(a) $\pi r^{2}$

(b) $\pi \left( \frac{2}{3} r \right ) ^{2} = \frac{4}{9} \pi r^{2}$

Twice the area of Circle B is

$2 \cdot \pi \left( \frac{2}{3} r \right ) ^{2} =2 \cdot \frac{4}{9} \pi r^{2} = \frac{8}{9} \pi r^{2} < \pi r^{2}$ ,

making (a) the greater number.

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