Ratio and Proportion — ICSE Class 10 Maths Important Questions

Answer: (a) $9:16$9:16.

The duplicate ratio of $a:b$a:b is $a^2:b^2=3^2:4^2=9:16$a^2:b^2=3^2:4^2=9:16.

Answer: (a) $12$12.

Mean proportional $=\sqrt{9\times16}=\sqrt{144}=12$=√9×16=√144=12.

Answer: (a) $5:6$5:6.

The sub-duplicate ratio of $a:b$a:b is $\sqrt a:\sqrt b=\sqrt{25}:\sqrt{36}=5:6$a: b=√25:√36=5:6.

Answer: (a) $24$24.

If $6,12,x$6,12,x are in continued proportion then $12^2=6x\Rightarrow x=\dfrac{144}{6}=24$12^2=6x x=144/6=24.

Answer: (d) A is false but R is true.

By R, the compound ratio is $(2\times3):(3\times4)=6:12=1:2$(2×3):(3×4)=6:12=1:2. The value $5:7$5:7 comes from wrongly adding the terms.

If $3:12=5:x$3:12=5:x, then $3x=12\times5=60$3x=12×5=60.

$x=\dfrac{60}{3}=\mathbf{20}$x=60/3=20.

Compound ratio $=\dfrac{2\times6\times11}{3\times11\times2}=\dfrac{132}{66}=\dfrac{2}{1}$=2×6×11/3×11×2=132/66=2/1.

So the compound ratio is $\mathbf{2:1}$2:1.

Since $a,b,c$a,b,c are in continued proportion, $b^2=ac$b^2=ac.

Take the right-hand side and substitute $b^2=ac$b^2=ac:

$\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a(a+c)}{c(a+c)}=\frac{a}{c}.$a^2+b^2/b^2+c^2=a^2+ac/ac+c^2=a(a+c)/c(a+c)=a/c.

Hence $\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}$a/c=a^2+b^2/b^2+c^2, as required.

Since $x$x is the mean proportional between $(x-2)$(x-2) and $(x+6)$(x+6):

$x^2=(x-2)(x+6).$x^2=(x-2)(x+6).

$x^2=x^2+4x-12\Rightarrow 0=4x-12\Rightarrow x=\mathbf{3}.$x^2=x^2+4x-12 0=4x-12 x=3.

(Check: mean proportional between $1$1 and $9$9 is $\sqrt{9}=3$√9=3.)

Make $B$B common: $A:B=8:12$A:B=8:12 and $B:C=12:15$B:C=12:15, so $A:B:C=8:12:15$A:B:C=8:12:15.

Total parts $=8+12+15=35$=8+12+15=35; value of one part $=\dfrac{1400}{35}=40$=1400/35=40.

A $=8\times40=$=8×40= ₹$\mathbf{320}$320, B $=12\times40=$=12×40= ₹$\mathbf{480}$480, C $=15\times40=$=15×40= ₹$\mathbf{600}$600.

By componendo and dividendo:

$\frac{(3x+5y)+(3x-5y)}{(3x+5y)-(3x-5y)}=\frac{7+3}{7-3}\Rightarrow\frac{6x}{10y}=\frac{10}{4}.$(3x+5y)+(3x-5y)/(3x+5y)-(3x-5y)=7+3/7-36x/10y=10/4.

So $\dfrac{x}{y}=\dfrac{10}{4}\times\dfrac{10}{6}=\dfrac{100}{24}=\dfrac{25}{6}$x/y=10/4×10/6=100/24=25/6, i.e. $x:y=\mathbf{25:6}$x:y=25:6.

(Check: with $x=25$x=25, $y=6$y=6: $\dfrac{75+30}{75-30}=\dfrac{105}{45}=\dfrac{7}{3}$75+30/75-30=105/45=7/3 ✓.)

The ratio equals $\dfrac{9}{1}$9/1. Apply componendo and dividendo:

$\frac{(\sqrt{3x+4}+\sqrt{3x-5})+(\sqrt{3x+4}-\sqrt{3x-5})}{(\sqrt{3x+4}+\sqrt{3x-5})-(\sqrt{3x+4}-\sqrt{3x-5})}=\frac{9+1}{9-1}.$(√3x+4+√3x-5)+(√3x+4-√3x-5)/(√3x+4+√3x-5)-(√3x+4-√3x-5)=9+1/9-1.

$\frac{2\sqrt{3x+4}}{2\sqrt{3x-5}}=\frac{10}{8}\Rightarrow\frac{\sqrt{3x+4}}{\sqrt{3x-5}}=\frac{5}{4}.$2√3x+4/2√3x-5=10/8√3x+4/√3x-5=5/4.

Squaring: $\dfrac{3x+4}{3x-5}=\dfrac{25}{16}$3x+4/3x-5=25/16.

$16(3x+4)=25(3x-5)\Rightarrow 48x+64=75x-125\Rightarrow 189=27x$16(3x+4)=25(3x-5) 48x+64=75x-125 189=27x.

So $x=\mathbf{7}$x=7.

Let $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$x/a=y/b=z/c=k. Then $x=ak$x=ak, $y=bk$y=bk, $z=ck$z=ck.

Left-hand side:
$\frac{x^3}{a^3}+\frac{y^3}{b^3}+\frac{z^3}{c^3}=\frac{a^3k^3}{a^3}+\frac{b^3k^3}{b^3}+\frac{c^3k^3}{c^3}=k^3+k^3+k^3=3k^3.$x^3/a^3+y^3/b^3+z^3/c^3=a^3k^3/a^3+b^3k^3/b^3+c^3k^3/c^3=k^3+k^3+k^3=3k^3.

Right-hand side:
$\frac{3xyz}{abc}=\frac{3(ak)(bk)(ck)}{abc}=\frac{3abc\,k^3}{abc}=3k^3.$3xyz/abc=3(ak)(bk)(ck)/abc=3abc\,k^3/abc=3k^3.

Since LHS $=$= RHS $=3k^3$=3k^3, the identity is proved.

(i) $\dfrac{\text{boys}}{\text{girls}}=\dfrac{5}{4}\Rightarrow\dfrac{500}{\text{girls}}=\dfrac{5}{4}\Rightarrow\text{girls}=\dfrac{4\times500}{5}=\mathbf{400}$boys/girls=5/4500/girls=5/4=4×500/5=400.

(ii) Total $=500+400=\mathbf{900}$=500+400=900 students.

(iii) New girls $=400+50=450$=400+50=450. New ratio $=500:450=\mathbf{10:9}$=500:450=10:9.