Linear Inequalities in One Unknown: Paper 1 Challenge

The claim is not correct. $(\ast )$ means $x>5$ and $x<3$, which no real number satisfies, so $(\ast )$ has no solution.

The numbers 3 and 5 are right; the directions are not. Test x = 4: the chain reads 9 < 8 < 6, which is false.

(a) $-4<x\le -3$    (b) No. $-3$ satisfies $(\ast )$.    (c) All real numbers

The range is only one unit wide, but its right-hand end, −3, is included. With “or”, x ≤ −3 and x > −4 together cover everything.

(a) $-8\le y<13$    (b) $21$    (c) No. $y=13$ needs $x=-2$, which is not allowed, so $y$ is always less than $13$ and has no greatest value.

Multiplying by −3 reverses both signs, so the included end (x = 5) becomes the bottom of the range and the excluded end (x = −2) becomes the top.

(a) $x>3$    (b) No. For example, when $a=1$ the inequality becomes $x+6>2x+3$, whose solution is $x<3$.

The inequality rearranges to (a − 2)x > 3(a − 2). Dividing by a − 2 keeps the sign only when a > 2; when a < 2 it reverses, and when a = 2 there is no solution at all.

(a) Must be true: multiplying $a>b$ by $-3$ gives $-3a<-3b$, and adding $1$ keeps the sign.    (b) Not necessarily: $a=1$, $b=-2$ gives $1>4$, which is false.    (c) Not necessarily: $a=1$, $b=-1$ gives $1<-1$, which is false.

Statements (b) and (c) are true when a and b are both positive. They fail when b is negative, and the question never says it is not.

$-5<k\le -3$

The seven integers must be −4, −3, …, 2. So 2 must be below the upper boundary (k + 9)/2 and 3 must not be. When the boundary equals 3 exactly, 3 is still excluded, so that end is allowed.

$\frac{11}{3}<k\le \frac{13}{3}$

The inequality gives x ≤ (5 − 3k)/2. For −4 to be the greatest integer allowed, that boundary must be at least −4 but less than −3.

(a) $0,1,2,3,4,5$    (b) $k\ge 2$

There is no gap exactly when the left-pointing ray reaches at least as far as the start of the right-pointing one: 2k − 1 ≥ 5 − k. At k = 2 the rays meet at 3 with nothing left out.

(a) $17$    (b) $12$    (c) No. The lighter worker can still take only $17$ boxes per trip ($n\le 17.5$), so $12$ trips are still needed.

Boxes per trip round down (17.19 gives 17); trips round up (11.76 gives 12). Saving 15 kg is less than one box, so nothing changes.

(a) $90$    (b) No. She would need at least $110$ in the fifth test, which is more than the full score.    (c) $2$

The four scores add up to 300. In (c), n + 4 is positive, so multiplying both sides by it keeps the sign: 300 + 100n ≥ 82(n + 4) gives n ≥ 1.56, so n = 2.