Numerical Inequalities for CUET Applied Maths

AnswerB · $-3a > -3b$

1. Multiplying both sides by a negative number reverses the inequality: $a < b$ gives $-3a > -3b$.

2. The others fail: $-3 < 1$ but $9 > 1$; $a = -1$, $b = 1$ give reciprocals $-1$ and 1, still in the same order; and subtracting 5 keeps $<$.

AnswerB · $\frac{1}{a} < \frac{1}{b}$

1. For positive numbers, taking reciprocals reverses the order: the larger number has the smaller reciprocal.

2. So $\frac{1}{a} < \frac{1}{b}$. The others reverse true facts: $a^{2} > b^{2}$, $-a < -b$, $a - b > 0$.

AnswerC · $2 < xy < 15$

1. All four ends are positive, so the inequalities may be multiplied.

2. $2 \times 1 < xy < 5 \times 3$, that is $2 < xy < 15$.

AnswerC · $199 + 2x < 299$

1. Plan A costs $199 + 2x$ rupees for $x$ GB.

2. Cheaper than Plan B: $199 + 2x < 299$.

AnswerB · $x < 50$

1. $199 + 2x < 299 \Rightarrow 2x < 100$.

2. $x < 50$ GB.

AnswerD · I, II and IV only

1. Adding any number, or multiplying by a positive one, keeps the direction (I and II). For positive numbers, the larger has the smaller reciprocal (IV).

2. III fails when $c < 0$: $1 < 2$ but $\frac{1}{-1} > \frac{2}{-1}$.

AnswerA · $x^{2} > y^{2}$

1. Both are negative and $x$ is further from 0, so $x^{2} > y^{2}$: for example $-3 < -2$ and $9 > 4$.

2. Also $\frac{1}{-3} > \frac{1}{-2}$, and the product of two negatives is positive.

AnswerA · Only I

1. I: inequalities in the same direction can be added. True.

2. II: $5 > 1$ and $4 > 0$, but $5 - 4 = 1$ is not more than $1 - 0 = 1$. They cannot be subtracted. False.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. Dividing by $-2$ turns $>$ into $<$: $x < \frac{6}{-2} = -3$. A is true.

2. R is the rule used, so R explains A.

AnswerD · $ab < 0$

1. A negative number times a positive number is negative: $ab < 0$.

2. The others can fail: $a = -5$, $b = 1$ gives $a^{2} > b^{2}$ and $a + b < 0$; and $\frac{1}{a} < 0 < \frac{1}{b}$ always.

AnswerB · Only II

1. I is false: $5 > 4$ and $10 > 1$, but $5 - 10 = -5$ is less than $4 - 1 = 3$. Inequalities may be added, not subtracted.

2. II is true: $ac > bc > bd$, since all the numbers are positive.

AnswerD · A is false, but R is true.

1. R is true: $x^{2} - 9 = (x - 3)(x + 3) > 0$ outside the roots.

2. A is false: $x = -4$ gives $x^{2} = 16 > 9$ but $x < 3$.

AnswerA · I, II and IV only

1. Multiplying by $a$, a number between 0 and 1, makes things smaller: $a^{2} < a$ and $a^{3} < a^{2}$. And $\sqrt{a} > a$, as with $\sqrt{0.25} = 0.5$.

2. III is false: $\frac{1}{a} > 1$, as with $\frac{1}{0.5} = 2$.

AnswerA · $-1 < x - y < 4$

1. $-3 < -y < -1$. Add this to $2 < x < 5$.

2. $2 - 3 < x - y < 5 - 1$, that is $-1 < x - y < 4$.

AnswerC · $xy < -2$

1. $x > 2 > 0$ and $-y > 1 > 0$, so multiplying, $-xy > 2$.

2. That is $xy < -2$. The others fail for $x = 3$, $y = -5$.

AnswerD · Plan A, by ₹10

1. She uses $1.5 \times 30 = 45$ GB, so Plan A costs $199 + 90 = 289$ rupees.

2. $45 < 50$: Plan A is cheaper, by $299 - 289 = 10$ rupees.

AnswerC · 40 GB

1. $199 + 2.5x < 299 \Rightarrow 2.5x < 100$.

2. $x < 40$ GB.

AnswerB · $\frac{7}{9}$

1. Compare how far each is from 1: $\frac{2}{9}$, $\frac{1}{4}$, $\frac{2}{7}$ and $\frac{1}{3}$.

2. $\frac{2}{9} \approx 0.22$ is the smallest gap, so $\frac{7}{9}$ is the greatest.

AnswerB · $0.3$

1. $(0.3)^{2} = 0.09$.

2. For a number between 0 and 1, squaring makes it smaller, so $0.3 > 0.09$.

AnswerA · They are equal

1. 25% of 80 is $\frac{25 \times 80}{100} = 20$.

2. 80% of 25 is $\frac{80 \times 25}{100} = 20$: the same product over 100.

AnswerC · $\frac{4}{9}$

1. $\frac{3}{7} \approx 0.4286$ and $\frac{4}{9} = 0.4444\ldots$

2. So $\frac{4}{9}$ is the largest, ahead of 0.43, $\frac{3}{7}$ and 0.42.

AnswerB · $\frac{3}{5}$

1. As decimals: $\frac{3}{5} = 0.6$, $\frac{5}{8} = 0.625$, $\frac{7}{11} \approx 0.636$, $\frac{9}{14} \approx 0.643$.

2. The smallest is $\frac{3}{5}$.

AnswerB · $\frac{12}{13}$

1. Cross-multiply: $12 \times 12 = 144$ and $11 \times 13 = 143$.

2. $144 > 143$, so $\frac{12}{13} > \frac{11}{12}$: each falls short of 1 by $\frac{1}{13}$ and $\frac{1}{12}$.

AnswerA · At least 240

1. An average of at least 60 over four tests means a total of at least $4 \times 60$.

2. At least 240.

AnswerB · $165 + x \ge 240$

1. Her three scores total $62 + 48 + 55 = 165$.

2. $\frac{165 + x}{4} \ge 60 \Rightarrow 165 + x \ge 240$.

AnswerC · $3^{20}$

1. Write each as a tenth power: $2^{30} = 8^{10}$, $3^{20} = 9^{10}$, and $5^{10}$, $6^{10}$.

2. With the same power, the largest base wins: $9^{10} = 3^{20}$.

AnswerD · $\sqrt{10} - \sqrt{9}$

1. $\sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}}$, on multiplying top and bottom by $\sqrt{n + 1} + \sqrt{n}$.

2. The denominator $\sqrt{10} + \sqrt{9}$ is smaller than $\sqrt{11} + \sqrt{10}$, so $\sqrt{10} - \sqrt{9}$ is larger.

AnswerC · $(0.7)^{2} < 0.7 < \sqrt{0.7}$

1. For a number between 0 and 1, squaring makes it smaller and taking the square root makes it larger.

2. $(0.7)^{2} = 0.49$, and $\sqrt{0.7} \approx 0.84$.

AnswerD · They are equal

1. Multiply top and bottom by $\sqrt{3} + \sqrt{2}$: the bottom becomes $3 - 2 = 1$.

2. So $\frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2}$.

AnswerB · $0.41$

1. $\frac{2}{5} = 0.4$ and $\frac{3}{7} \approx 0.4286$.

2. Only 0.41 lies between them; 0.43 is just above $\frac{3}{7}$.

AnswerD · $3^{20}$

1. Write both as tenth powers: $2^{30} = (2^{3})^{10} = 8^{10}$ and $3^{20} = (3^{2})^{10} = 9^{10}$.

2. $9 > 8$, so $3^{20} > 2^{30}$.

AnswerD · $75 \le x \le 100$

1. $165 + x \ge 240$ gives $x \ge 75$; she also needs $x \ge 40$, which 75 already beats.

2. A test is out of 100, so $75 \le x \le 100$.

AnswerC · $60$

1. The first three total $70 + 72 + 38 = 180$, so $180 + x \ge 240$ gives $x \ge 60$.

2. The least such score is 60, though the classmate has already failed the course on the 38.

AnswerC · $\sqrt[6]{6} < \sqrt{2} < \sqrt[3]{3}$

1. Raise each to the sixth power, which keeps the order of positive numbers: $(\sqrt{2})^{6} = 8$, $(\sqrt[3]{3})^{6} = 9$, $(\sqrt[6]{6})^{6} = 6$.

2. $6 < 8 < 9$, so $\sqrt[6]{6} < \sqrt{2} < \sqrt[3]{3}$.

AnswerA · $\sqrt{5} + \sqrt{3}$

1. Both are positive, so compare their squares: $8 + 2\sqrt{15}$ and $8 + 2\sqrt{12}$.

2. $\sqrt{15} > \sqrt{12}$, so $\sqrt{5} + \sqrt{3}$ is greater.

AnswerD · $x > 4$

1. Add 7 to both sides: $3x > 12$.

2. Divide by the positive number 3, which keeps the direction: $x > 4$.

AnswerD · $x < 3$

1. Subtract 7: $-2x > -6$.

2. Divide by $-2$ and reverse the sign: $x < 3$.