Algebra Questions and Answers

Answer: C

The numbers are 6 and 3.

Let the numbers be a and b, with a + b = 9. The sum of reciprocals is

\[ \begin{aligned} \frac{1}{a} + \frac{1}{b} &= \frac{a + b}{ab} = \frac{1}{2} \\ ab &= 2 \times 9 = 18 \end{aligned} \]

Two numbers with sum 9 and product 18 are 6 and 3, and indeed 1/6 + 1/3 = 1/2.

Each other pair also adds up to 9 but has the wrong product: 1 × 8 = 8, 2 × 7 = 14 and 4 × 5 = 20, so their reciprocals sum to 9/8, 9/14 and 9/20.

Answer: B

4 mangoes and 3 guavas cost ₹ 18.

Let a pear, a guava and a mango cost p, g and m rupees, all different natural numbers. Halving the given total:

\[ 10p + 8g + 6m = 44 \;\Rightarrow\; 5p + 4g + 3m = 22 \]

Try p = 1: 4g + 3m = 17 gives g = 2, m = 3. Try p = 2: 4g + 3m = 12 has no solution in natural numbers. Try p = 3: 4g + 3m = 7 gives g = m = 1, but no two prices may be equal. Larger p leaves too little. So p = 1, g = 2, m = 3 is the only solution.

\[ 4m + 3g = 12 + 6 = 18 \]

Because the prices are unique, 'Cannot be determined' is wrong; 17 and 14 would need different prices.

Answer: D

\(ab = \) 30.

Use the identity for the difference of cubes:

\[ \begin{aligned} a^{3} - b^{3} &= (a - b)^{3} + 3ab(a - b) \\ 91 &= 1 + 3ab \\ ab &= 30 \end{aligned} \]

Check with \(a = 6\), \(b = 5\): \(216 - 125 = 91\) and \(6 - 5 = 1\), with \(ab = 30\). 27 and 9 come from slips in the identity.

Another route: from \(a - b = 1\), \((a - b)^{2} = 1\), so \(a^{2} + b^{2} = 1 + 2ab\); then \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) = 1 + 3ab = 91\), giving the same result.

Answer: D

The expression simplifies to \(\frac{1}{3x-3}\).

Turn the division into multiplication by the reciprocal and factorise, using \(x^{3} - 1 = (x-1)(x^{2}+x+1)\) and \(3x + 9 = 3(x+3)\):

\[ \begin{aligned} &\frac{x+3}{(x-1)(x^{2}+x+1)} \times \frac{x^{2}+x+1}{3(x+3)} \\ &= \frac{1}{3(x-1)} = \frac{1}{3x-3} \end{aligned} \]

\(3x - 3\) is the reciprocal of the answer, and the forms with \(3x + 1\) come from factorising \(x^{3} - 1\) wrongly.

Answer: B

The fraction is \(\frac{5}{7}\).

Let the fraction be \(\frac{x}{y}\). The two conditions give

\[ \begin{aligned} \frac{2x}{y - 4} = \frac{10}{3} &\Rightarrow 6x = 10y - 40 \\ \frac{x + 6}{2y} = \frac{11}{14} &\Rightarrow 14x + 84 = 22y \end{aligned} \]

From the first, \(x = \frac{5y - 20}{3}\). Substituting in the second and multiplying through by 3 gives \(70y - 280 + 252 = 66y\), so \(4y = 28\), \(y = 7\) and \(x = 5\).

Check: \(\frac{10}{3}\) and \(\frac{11}{14}\) both come out. \(\frac{7}{5}\) is the reciprocal, and the other options fail the first condition.

Answer: A

The values are \(x\) = 2, \(y\) = 1.

Put \(u = \frac{1}{2x+3y}\) and \(v = \frac{1}{3x-2y}\). The equations become

\[ \begin{aligned} \frac{u}{2} + \frac{12v}{7} &= \frac{1}{2} \\ 7u + 4v &= 2 \end{aligned} \]

Solving gives \(u = \frac{1}{7}\) and \(v = \frac{1}{4}\), so \(2x + 3y = 7\) and \(3x - 2y = 4\), whose solution is \(x = 2\), \(y = 1\).

Check directly: \(\frac{1}{14} + \frac{12}{28} = \frac{1}{2}\) and \(1 + 1 = 2\). The pair 1, 2 swaps the values, and the negative pairs make \(2x + 3y\) negative.

Answer: B

The total area is 4515 cm\(^{2}\).

The total is \(11^{2} + 12^{2} + \cdots + 24^{2}\). Use \(\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}\):

\[ \begin{aligned} \sum_{1}^{24} k^{2} &= \frac{24 \cdot 25 \cdot 49}{6} = 4900 \\ \sum_{1}^{10} k^{2} &= \frac{10 \cdot 11 \cdot 21}{6} = 385 \\ \text{total} &= 4900 - 385 = 4515 \end{aligned} \]

Subtracting the squares up to 11 instead of up to 10 would leave out the 11 cm square and give 4394, which is not offered; the other options do not come from the formula.

Answer: C

The value is 2.

Pick the right piece of the function for each argument:

\[ \begin{aligned} f(-3) &= -3 + 5 = 2 \\ f(4) &= 4 - 1 = 3 \\ f(-6) &= 36 - 12 + 1 = 25 \\ f(1) &= 1 + 5 = 6 \end{aligned} \]

\[ \frac{4(2) - 2(3)}{25 - 4(6)} = \frac{8 - 6}{25 - 24} = 2 \]

\(\frac{1}{2}\) is the answer turned upside down, and the other values come from using the wrong piece of \(f\) for one of the arguments, for example \(x + 5\) for \(f(-6)\).

Answer: B

Together they have 80 bangles.

Let Mala have \(m\) and Latha \(l\) bangles.

\[ \begin{aligned} m + 4 &= 3(l - 4) \Rightarrow m = 3l - 16 \\ l + 36 &= 3(m - 36) \\ l + 36 &= 9l - 48 - 108 \Rightarrow l = 24 \end{aligned} \]

Then \(m = 56\), and the total is 80. Check: if Latha gives 4, Mala has 60 and Latha 20; if Mala gives 36, Latha has 60 and Mala 20. Both times one has three times the other.

Answer: B

From the first ratio, \(4(6x+2y)=3(3x+4y)\) gives \(24x+8y=9x+12y\), so \(15x=4y\) and x : y = 4 : 15. From the second, \(7(x+z)=3(2x+3z)\) gives \(x=2z\), so x : z = 2 : 1 = 4 : 2.

Hence x : y : z = 4 : 15 : 2, option B. The other options do not satisfy both equations.

The two equations are linear and have only the ratios as unknowns. The first gives 15x = 4y and the second gives x = 2z, so with x = 4 we get y = 15 and z = 2. Checking, (24 + 30)/(12 + 60) = 54/72 = 3/4 and (4 + 2)/(8 + 6) = 6/14 = 3/7, as required.