Linear Inequalities in One Unknown: Paper 1 Long Questions

(a) $x<\frac{13}{2}$    (b) $6$

The last step divides by −2, so the sign turns round. Since 13/2 = 6.5, the greatest integer below it is 6.

(a) $-3\le x<5$    (b) $8$

“And” keeps only the overlap. The integers are −3, −2, …, 4: −3 is included, 5 is not.

(a) $x<-9$ or $x\ge -5$    (b) $-6$

The two parts do not overlap, so the answer stays as two separate ranges. The numbers left out are −9 ≤ x < −5, and the greatest integer there is −6.

(a) $x<-2$    (b) $-3$

Every number with x ≤ −5 already has x < −2, so “or” gives the wider range. −2 itself is not allowed, so the greatest integer is −3.

(a) $x<\frac{13}{3}$    (b) $1,2,3,4$

Both inequalities together give −2 ≤ x < 13/3. The integers there are −2 to 4, but only 1, 2, 3 and 4 are positive. Zero is not positive.

(a) All real numbers    (b) Yes. $-3$ does not satisfy $x>-3$, but it satisfies $x\le -1$, and one part is enough for “or”.

The first part gives x > −3 and the second gives x ≤ −1. Together they cover the whole number line.

(a) $-3<x\le 5$    (b) $12$

A chain a < b ≤ c means a < b and b ≤ c. The integers are −2 to 5; the pairs −2 and 2, −1 and 1 cancel, leaving 3 + 4 + 5.

(a) $21<x<26$    (b) $23$

Multiply through by 10, and remember the minus sign applies to the whole of 2(2x + 3). The integers in the range are 22, 23, 24, 25; only 23 is prime.

(a) $x>80$    (b)(i) $108$    (ii) Yes, since $108>80$. (Shop A charges $\$498$; Shop B would charge $\$540$.)

In (b)(i) the inequality gives x ≤ 108.57…, and a number of posters must be rounded down, not off: 109 posters would cost $501.5.

(a) $23<n<\frac{94}{3}$    (b) $4$    (c) $35$

The three numbers are n, n + 2 and n + 4, with sum 3n + 6. Only the odd values n = 25, 27, 29, 31 count; the largest number is then 31 + 4.