NCERT Solutions: Class 11 Maths Chapter 5 - Linear Inequalities

$24x<100$ gives $x<\dfrac{25}{6}$. Natural-number solutions are $1,2,3,4$. Integer solutions are all integers less than $\dfrac{25}{6}$, i.e. $x\le4$.

Dividing $-12x>30$ by $-12$ reverses the inequality: $x<-\dfrac52$. No natural number satisfies this. Integer solutions are $\ldots,-5,-4,-3$, i.e. $x\le-3$.

$5x-3<7$ gives $5x<10$, so $x<2$. Thus the integer solutions are all integers at most 1.

$3x+8>2$ gives $3x>-6$, so $x>-2$. The integer solutions are $-1,0,1,\ldots$.

$4x+3<5x+7$ gives $-4<x$, hence $x>-4$.

$3x-7>5x-1$ gives $-2x>6$, so $x<-3$.

$3x-3\le2x-6$, so $x\le-3$.

$6-3x\ge2-2x$ gives $4\ge x$, so $x\le4$.

$x+\dfrac{x}{2}+\dfrac{x}{3}=\dfrac{11x}{6}$. Thus $\dfrac{11x}{6}<11$, giving $x<6$.

Multiplying by 6 gives $2x>3x+6$, so $x<-6$.

Multiplying by 15 gives $9x-18\le50-25x$. Thus $34x\le68$, so $x\le2$.

Multiplying by 30 gives $15\left(\dfrac{3x}{5}+4\right)\ge10(x-6)$, i.e. $9x+60\ge10x-60$. Hence $x\le120$.

$4x+6-10<6x-12$ gives $4x-4<6x-12$, so $8<2x$ and $x>4$.

$37-(3x+5)=32-3x$ and $9x-8(x-3)=x+24$. Thus $32-3x\ge x+24$, so $8\ge4x$ and $x\le2$.

Multiplying by 60 gives $15x<20(5x-2)-12(7x-3)=16x-4$. Therefore $x>4$.

Multiplying by 60 gives $20(2x-1)\ge15(3x-2)-12(2-x)$, so $40x-20\ge57x-54$. Hence $34\ge17x$ and $x\le2$.

$3x-2<2x+1$ gives $x<3$. On the number line, use an open circle at 3 and shade to the left.

$5x-3\ge3x-5$ gives $2x\ge-2$, so $x\ge-1$. On the number line, use a closed circle at $-1$ and shade to the right.

$3-3x<2x+8$ gives $-5<5x$, so $x>-1$. On the number line, use an open circle at $-1$ and shade to the right.

Multiplying by 30 gives $15x\le10(5x-2)-6(7x-3)=8x-2$. Thus $7x\le-2$, so $x\le-\dfrac27$. On the number line, use a closed circle at $-\dfrac27$ and shade left.

Let the third-test mark be $x$. Then $\dfrac{70+75+x}{3}\ge60$, so $145+x\ge180$ and $x\ge35$.

Let the fifth mark be $x$. Required total for average 90 in five exams is $450$. Since $87+92+94+95=368$, we need $368+x\ge450$, so $x\ge82$.

The possible pairs of consecutive odd positive integers smaller than 10 are $(1,3)$, $(3,5)$, $(5,7)$ and $(7,9)$. Their sums are 4, 8, 12 and 16. The pairs with sum more than 11 are $(5,7)$ and $(7,9)$.

Consecutive even positive pairs larger than 5 begin $(6,8)$, $(8,10)$, $(10,12)$, etc. Their sums are 14, 18, 22, ... . The sums less than 23 are from $(6,8)$ and $(8,10)$; $(10,12)$ has sum 22, which is also less than 23, so it is included. Thus the pairs are $(6,8)$, $(8,10)$ and $(10,12)$.

Let the shortest side be $x$ cm. Then the longest side is $3x$ and the third side is $3x-2$. The perimeter condition gives $x+3x+(3x-2)\ge61$, so $7x\ge63$ and $x\ge9$. Therefore the minimum shortest side is 9 cm.

Using the hint, total length gives $x+(x+3)+2x\le91$, so $4x+3\le91$ and $x\le22$. The third piece condition gives $2x\ge(x+3)+5$, so $x\ge8$. Hence $8\le x\le22$.

Add 4 throughout: $6\le3x\le9$. Divide by 3 to get $2\le x\le3$.

$6\le-6x+12<12$. Subtract 12: $-6\le-6x<0$. Divide by $-6$ and reverse inequalities: $1\ge x>0$, i.e. $0<x\le1$.

Subtract 4: $-7\le-\dfrac{7x}{2}\le14$. Multiply by 2: $-14\le-7x\le28$. Divide by $-7$ and reverse inequalities to get $2\ge x\ge-4$, i.e. $-4\le x\le2$.

Multiply by 5: $-75<3x-6\le0$. Add 6: $-69<3x\le6$. Divide by 3 to get $-23<x\le2$.

Since $-\dfrac{3x}{-5}=\dfrac{3x}{5}$, the inequality is $-12<4+\dfrac{3x}{5}\le2$. Subtract 4: $-16<\dfrac{3x}{5}\le-2$. Multiply by $\dfrac53$ to get $-\dfrac{80}{3}<x\le-\dfrac{10}{3}$.

Multiply by 2: $14\le3x+11\le22$. Subtract 11: $3\le3x\le11$. Divide by 3 to get $1\le x\le\dfrac{11}{3}$.

$5x+1>-24$ gives $x>-5$, and $5x-1<24$ gives $x<5$. The intersection is $-5<x<5$. On the number line, use open circles at $-5$ and $5$ and shade between them.

$2(x-1)<x+5$ gives $x<7$. Also $3(x+2)>2-x$ gives $4x>-4$, so $x>-1$. The intersection is $-1<x<7$.

$3x-7>2x-12$ gives $x>-5$. Also $6-x>11-2x$ gives $x>5$. The common solution is $x>5$.

$5(2x-7)-3(2x+3)\le0$ gives $4x-44\le0$, so $x\le11$. Also $2x+19\le6x+47$ gives $x\ge-7$. Hence $-7\le x\le11$.

Use $68<\dfrac95C+32<77$. Subtract 32: $36<\dfrac95C<45$. Multiply by $\dfrac59$: $20<C<25$.

Let $x$ litres of 2% solution be added. Acid amount is $0.08(640)+0.02x=51.2+0.02x$, and total volume is $640+x$. The condition is $0.04(640+x)<51.2+0.02x<0.06(640+x)$. Solving gives $x<1280$ and $x>320$. Thus $320<x<1280$.

Let $x$ litres of water be added. The acid amount remains $0.45(1125)=506.25$ litres and the total volume is $1125+x$. We need $0.25(1125+x)<506.25<0.30(1125+x)$. Solving gives $x<900$ and $x>562.5$.

Here $CA=12$, so $80\le\dfrac{MA}{12}\times100\le140$. Multiplying by $\dfrac{12}{100}$ gives $9.6\le MA\le16.8$.