FTCE General Knowledge Math (828) practice test

The most accurate part is the one closest to its target length, so compare the absolute values of the deviations, their distances from 0 on a number line: 0.05, 0.12, 0.1 and 0.04 mm. The least of these is 0.04 mm, so Part Z, which is 0.04 mm too short, was made most accurately. The sign only tells whether a part is too long or too short.

A

Considers only the positive deviations (or treats a short part as farther off), but +0.05 mm is 0.05 mm from the target, more than Part Z's 0.04 mm.

B

Confuses the least number with the least distance from 0: −0.12 mm is the least deviation but the farthest from the target, 0.12 mm.

C

Place-value slip: compares the digits after the decimal point as whole numbers (1 < 4), but +0.1 mm = +0.10 mm is farther from 0 than 0.04 mm.

The segment is 0.9 − (−0.6) = 1.5 units long, so each of the 5 equal parts is 1.5 ÷ 5 = 0.3 unit. The second tick mark to the right of −0.6 is 2 parts away: −0.6 + 2 × 0.3 = 0.

A

Sign slip: finds the length as 0.9 − 0.6 = 0.3, so each part is 0.06 and the mark is placed at −0.6 + 0.12 = −0.48.

C

Unit-size error: divides the length 1.5 by the 4 tick marks instead of the 5 parts, getting 0.375 per part and −0.6 + 0.75 = 0.15.

D

Intermediate value: 0.6 is the distance from −0.6 to the mark (2 × 0.3), not the number located there.

Simple interest is I = prt, so the rate is r = I ÷ (pt). The interest is the growth in the balance, $2360 − $2000 = $360, earned over t = 4 years. Then r = 360 ÷ (2000 × 4) = 360 ÷ 8000 = 0.045, which is 4.5% per year.

A

Decimal not converted to a percent: r = 0.045 is the rate as a decimal, and 0.045 × 100 = 4.5%, so 0.045% is 100 times too small.

C

Intermediate value: 360 ÷ 2000 = 0.18 is the interest for all 4 years as a share of the deposit; dividing it by the 4 years gives the rate per year.

D

Balance used as interest: 2360 ÷ (2000 × 4) = 0.295 treats the whole $2360 balance as interest, but only the $360 growth is interest.

A quadrilateral with exactly one pair of parallel sides is a trapezoid, and a trapezoid whose two nonparallel sides (its legs) are equal in length is an isosceles trapezoid. Because the second property is given, isosceles trapezoid names the shape more precisely than trapezoid alone.

A

Correct but less precise: it uses only the first property and ignores that the two nonparallel sides are equal in length.

B

A right trapezoid has one leg perpendicular to the parallel sides, and that leg is shorter than the slanted one, so its two legs cannot be equal.

C

A parallelogram has both pairs of opposite sides parallel, but the tabletop has exactly one parallel pair; its equal sides are not a second parallel pair.

A scale written as a ratio compares two lengths in the same unit. The wall is 36 × 12 = 432 inches long, so the ratio of drawing length to actual length is 9 : 432. Dividing both terms by 9 gives 1 : 48, so each inch on the drawing stands for 48 inches, or 4 feet, of wall.

A

Units not converted: 1 : 4 compares 9 inches with 36 feet as if both were in the same unit; it restates 1 in. = 4 ft rather than a ratio of like lengths.

C

Ignored the drawing length: 1 : 432 pairs 1 inch with the whole 432-inch wall, but the wall is drawn 9 inches long, so both terms must be divided by 9.

D

Reversed order: 48 : 1 compares actual length to drawing length, the opposite of the ratio asked for.

Each tile is 6 inches = 0.5 foot on a side, so 9 ÷ 0.5 = 18 tiles fit along the 9-foot side and 12 ÷ 0.5 = 24 tiles fit along the 12-foot side. The floor takes 18 × 24 = 432 tiles. Check by area: the floor is 9 × 12 = 108 ft², each tile covers 0.5 × 0.5 = 0.25 ft², and 108 ÷ 0.25 = 432.

A

Intermediate value: 18 tiles make one row along the 9-foot side; the 12-foot side holds 24 such rows.

B

Squared-unit slip: counted each tile as 0.5 ft² because its side is 0.5 ft; a 0.5-by-0.5-foot tile covers only 0.25 ft².

D

Neighboring measure: converted the floor to 15,552 in² but divided by the tile's 6-inch side instead of its area, 6 × 6 = 36 in².

Both terms have the factor (x + 2), so they are like terms, and the distributive property combines them: 5(x + 2) + 3(x + 2) = (5 + 3)(x + 2) = 8(x + 2). Expanding confirms it: 5x + 10 + 3x + 6 = 8x + 16, and 8(x + 2) = 8x + 16.

B

Procedural slip: combined the coefficients 5 + 3 for x but did not multiply the 2; 8(x + 2) = 8x + 16, not 8x + 2.

C

Procedural slip: added the coefficients and also added the two copies of (x + 2); 8(2x + 4) = 16x + 32, twice the value of the expression.

D

Neighboring operation: multiplied the coefficients, 5 × 3, instead of adding them; 5 groups of (x + 2) plus 3 more groups make 8 groups.

Each term of 6x − 9y is 3 times the matching term of 2x − 3y, so by the distributive property 6x − 9y = 3(2x − 3y) for all values of x and y. Replacing 2x − 3y with 7 gives 3 × 7 = 21. The value is found without knowing x or y separately.

A

Intermediate value: 7 is the value of 2x − 3y; multiplying every term by 3 multiplies the value of the expression by 3.

C

Procedural slip: multiplied 7 by 6, the coefficient of x; the factor that turns 2x − 3y into 6x − 9y is 3, since 6 = 3 × 2 and 9 = 3 × 3.

D

Procedural slip: multiplied 7 by 9, the coefficient of y; the common factor that relates the two expressions is 3.

Each number of guests must be multiplied by the number of rooms that held it: 1 × 12 + 2 × 25 + 3 × 6 + 4 × 7 = 12 + 50 + 18 + 28 = 108 guests. The second line counts rooms, so its total, 50, is the number of occupied rooms, not the number of guests.

A

Procedural slip: added the guest counts 1 + 2 + 3 + 4 = 10 without using how many rooms had each count.

B

Another row: 12 + 25 + 6 + 7 = 50 is the number of occupied rooms, not the number of guests staying in them.

D

Unweighted average: multiplied the 50 rooms by 2.5, the mean of 1, 2, 3 and 4, which ignores that most rooms held 1 or 2 guests.

When both the height and the width of a picture are doubled, its area is multiplied by 2 × 2 = 4. Viewers tend to compare pictures by their areas, so the larger stack looks like 4 times as much paper although recycling only doubled. A fair picture would double the area, or show the amounts as bars of equal width.

A

Scale-factor slip: doubling both dimensions does not double the picture; it makes the area 2 × 2 = 4 times as large.

B

Height-only reading: viewers see the whole picture, not just its height, and the doubled width makes the larger stack 4 times the area of the smaller.

C

Wrong factor: 2 × 2 × 2 = 8 applies to a solid enlarged in three dimensions; a flat picture doubled in height and width has 4 times the area.