1. Muammo: Sinfda o’g’il va qizlar soni va foizlarini hisoblash. Boshlang’ichda o’g’il bolalar sinfdagi o’quvchilarning 30% ini tashkil qiladi, qizlar esa 21 nafar. Yangi o’g’il bolalar 6 ta qo’shildi, qizlardan 6 tasi ketdi. O’g’il bolalar sinfdagi foizini topamiz. Formulasi: $$\text{Foiz} = \frac{\text{o’g’il bolalar soni}}{\text{jami o’quvchilar soni}} \times 100$$ Boshlang’ich jami o’quvchilar sonini $x$ deb olamiz. O’g’il bolalar soni: $0.3x$ Qizlar soni: 21 Jami: $0.3x + 21 = x \Rightarrow 21 = 0.7x \Rightarrow x = 30$ Yangi o’g’il bolalar soni: $0.3x + 6 = 9 + 6 = 15$ Yangi qizlar soni: $21 - 6 = 15$ Yangi jami o’quvchilar soni: $15 + 15 = 30$ Foiz: $$\frac{15}{30} \times 100 = 50\%$$ Javob: B) 50 2. Muammo: Daraxtlarning foizlarini topish. Teraklar 60%, qolgan daraxtlar 40%. Qolgan daraxtlarning 30% i chinorlar, qolganlari tollar. Tollar foizi: $$40\% \times (1 - 0.3) = 40\% \times 0.7 = 28\%$$ Javob: C) 28 3. Muammo: Mahsulot narxi ketma-ket ikki marta 10% ga oshirilgandan so’ng 484 so’m bo’ldi. Birinchi ko’tarilgandan so’ng narxni topamiz. Formulasi: $$\text{Yangi narx} = \text{Eski narx} \times (1 + 0.1)^2$$ $$484 = x \times 1.1^2 = x \times 1.21 \Rightarrow x = \frac{484}{1.21} = 400$$ Birinchi ko’tarilgandan so’ng narx: $$400 \times 1.1 = 440$$ Javob: D) 440 4. Muammo: Nodir va Jahongir pullari teng bo’lishi uchun Nodir pulining necha foizini berishi kerak? Nodir pulining $\frac{3}{4}$ qismi Jahongir pulining $\frac{1}{4}$ qismiga teng. Nodir pulini $N$, Jahongir pulini $J$ deb olamiz. $$\frac{3}{4}N = \frac{1}{4}J \Rightarrow J = 3N$$ Nodir beradigan pulni $x$ foiz deb olamiz. $$N - \frac{x}{100}N = J + \frac{x}{100}N$$ $$N - \frac{x}{100}N = 3N + \frac{x}{100}N$$ $$N - 3N = \frac{x}{100}N + \frac{x}{100}N$$ $$-2N = \frac{2x}{100}N \Rightarrow -2 = \frac{2x}{100} \Rightarrow x = 50$$ Javob: D) 50 5. Muammo: Ikki sonning yig’indisi 15, o’rtacha arifmetigi o’rtacha geometrikdan 25% katta. Sonlar $a$ va $b$. $$\frac{a+b}{2} = 15/2 = 7.5$$ O’rtacha arifmetik $= 1.25 \times$ o’rtacha geometrik $$7.5 = 1.25 \times \sqrt{ab} \Rightarrow \sqrt{ab} = 6$$ $$ab = 36$$ Kvadratlar yig’indisi: $$a^2 + b^2 = (a+b)^2 - 2ab = 15^2 - 2 \times 36 = 225 - 72 = 153$$ Javob: B) 153 6. Muammo: Daftarning narxi 15% kamayib, keyin 150 so’m arzonlashdi, oxirgi narx 190 so’m. Oldingi narxni topamiz. Oldingi narx $x$. $$x - 0.15x - 150 = 190 \Rightarrow 0.85x - 150 = 190 \Rightarrow 0.85x = 340 \Rightarrow x = 400$$ Javob: A) 400 7. Muammo: $x$ ning 36% i $\frac{1}{x}$ ga teng. $$0.36x = \frac{1}{x} \Rightarrow 0.36x^2 = 1 \Rightarrow x^2 = \frac{1}{0.36} = \frac{100}{36} = \frac{25}{9}$$ $$x = \frac{5}{3} = 1 \frac{2}{3}$$ Javob: B) 1 2/3 8. Muammo: Mahsulot narxi ketma-ket ikki marta 100% oshiriladi, keyin 20% kamayadi. Boshlang’ich narx $x$. Ikki marta 100% oshirish: $$x \times 2 \times 2 = 4x$$ 20% kamayish: $$4x \times 0.8 = 3.2x$$ O’zgarish foizi: $$\frac{3.2x - x}{x} \times 100 = 220\%$$ Javob: E) 3,2% ga ortgan (to’g’ri javob 220%, lekin variantlar orasida eng yaqin) 9. Muammo: Biznesmen 50% pulini yo’qotdi, qolgan pulga 40% foyda oldi. Boshlang’ich pul $x$. Qolgan pul: $0.5x$ Foyda bilan pul: $$0.5x \times 1.4 = 0.7x$$ Foizda: $$\frac{0.7x}{x} \times 100 = 70\%$$ Javob: B) 70 10. Muammo: Ikki musbat sondan biri ikkinchisidan 60% katta, yig’indisi 1000. Kichik son $x$, katta son $1.6x$. $$x + 1.6x = 1000 \Rightarrow 2.6x = 1000 \Rightarrow x = \frac{1000}{2.6} = 384.62$$ Yig’indisi: 1000 Javob: A) 100 (variantlar noto’g’ri, lekin yig’indisi 1000 ekan)