Algebra Questions and Answers

Q: If the sum of two numbers is 9 and the sum of their reciprocals is \(\dfrac{1}{2}\), then those numbers will be

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.

Q: The cost of 10 pears, 8 guavas and 6 mangoes is ₹ 44. If the cost of each of the fruits, in rupees, is a natural number and cost of no two fruits is the same, find the cost of 4 mangoes and 3 guavas.

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.

Q: If \(a^{3} - b^{3}\) = 91 and a - b = 1, what is the value of (ab) ?

\(ab = \) 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.

Q: On simplifying expression 13y + 2z + 3y, answer will be

Only the \( y \) terms are alike.\[ 13y + 3y = 16y \]\[ \therefore\ 16y + 2z \]The \( z \) term is unlike and stays as it is; it cannot be merged with the \( y \) terms.

Q: Product of 4 and -3b is

\[ 4 \times (-3b) = -12b \]A positive times a negative is negative, and the numbers multiply as 4 and 3 to give 12.

Q: By factorizing 54x - 45y, answer must be

Both terms share a factor of 9.\[ 54x - 45y = 9 \times 6x - 9 \times 5y = 9(6x - 5y) \]9 is the highest common factor of 54 and 45; taking out 3 alone would leave \( 3(18x - 15y) \), which is not fully factorised.

Q: By factorizing following expression 5ab + 10ac + 15ad, answer must be

Every term shares the factor \( 5a \).\[ 5ab + 10ac + 15ad = 5a(b + 2c + 3d) \]Checking: \( 5a \times b = 5ab \), \( 5a \times 2c = 10ac \), \( 5a \times 3d = 15ad \). Taking out 5 alone would leave the \( a \) behind in each term.

Q: If algebraic expression is [4(a + 2b - 3c) - 3(2b - 3c)] + 2a then answer will be

Expand each bracket, then collect.\[ 4(a + 2b - 3c) = 4a + 8b - 12c \]\[ -3(2b - 3c) = -6b + 9c \]\[ (4a + 8b - 12c) + (-6b + 9c) + 2a = 6a + 2b - 3c \]The \( -3 \) turns \( -3c \) into \( +9c \), which is what leaves only \( -3c \) at the end rather than \( -21c \).

Q: Simplify (1 ⁄ (x² - 1)) ÷ (1 ⁄ (x + 1))

Dividing by a fraction means multiplying by its reciprocal.\[ \frac{1}{x^{2}-1} \div \frac{1}{x+1} = \frac{1}{x^{2}-1} \times (x+1) \]Factorising the denominator:\[ = \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1} \]The \( (x+1) \) cancels, which is why the difference of squares has to be split before the simplification becomes visible.

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81.

If the sum of two numbers is 9 and the sum of their reciprocals is \(\dfrac{1}{2}\), then those numbers will be

A.1 and 8
B.2 and 7
C.6 and 3
D.4 and 5

Exam Year(s)

MPPSC State Service - 2023, MPPSC Forest Service - 2023,

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.

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82.

The cost of 10 pears, 8 guavas and 6 mangoes is ₹ 44. If the cost of each of the fruits, in rupees, is a natural number and cost of no two fruits is the same, find the cost of 4 mangoes and 3 guavas.

A.₹ 17
B.₹ 18
C.₹ 14
D.Cannot be determined

Exam Year(s)

Punjab Civil Service - 2021,

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.

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83.

If \(a^{3} - b^{3}\) = 91 and a - b = 1, what is the value of (ab) ?

A.27
B.6
C.9
D.30

Exam Year(s)

Punjab Civil Service - 2025,

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.

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84.

Simplify (3x ⁄ 2y)(x ⁄ 3y)

A.x ⁄ 2y
B.(3x²) ⁄ (2y²)
C.x² ⁄ (6y²)
D.x² ⁄ (2y²)

Answer: D

Multiply numerators and denominators separately, then cancel.

\[ \frac{3x}{2y} \cdot \frac{x}{3y} = \frac{3x^{2}}{6y^{2}} = \frac{x^{2}}{2y^{2}} \]

The 3 in the numerator cancels one factor of the 6 - leaving it in gives \( \tfrac{x^{2}}{6y^{2}} \).

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85.

On simplifying expression 13y + 2z + 3y, answer will be

A.39xy + 2z
B.16y + 2z
C.18xyz
D.16y + 1

Answer: B

Only the \( y \) terms are alike.

\[ 13y + 3y = 16y \]

\[ \therefore\ 16y + 2z \]

The \( z \) term is unlike and stays as it is; it cannot be merged with the \( y \) terms.

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86.

Product of 4 and -3b is

A.12b
B.−12b
C.3b
D.−3b

Answer: B

\[ 4 \times (-3b) = -12b \]

A positive times a negative is negative, and the numbers multiply as 4 and 3 to give 12.

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87.

By factorizing 54x - 45y, answer must be

A.9(6x - 5y)
B.5(7x + 6y)
C.8(6x + 3y)
D.9(6x + 5y)

Answer: A

Both terms share a factor of 9.

\[ 54x - 45y = 9 \times 6x - 9 \times 5y = 9(6x - 5y) \]

9 is the highest common factor of 54 and 45; taking out 3 alone would leave \( 3(18x - 15y) \), which is not fully factorised.

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88.

By factorizing following expression 5ab + 10ac + 15ad, answer must be

A.8a(3b + 2c + 3d)
B.5a(b + 2c + 3d)
C.−5a(b + 2c + 3d)
D.5b(a + c + d)

Answer: B

Every term shares the factor \( 5a \).

\[ 5ab + 10ac + 15ad = 5a(b + 2c + 3d) \]

Checking: \( 5a \times b = 5ab \), \( 5a \times 2c = 10ac \), \( 5a \times 3d = 15ad \). Taking out 5 alone would leave the \( a \) behind in each term.

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89.

If algebraic expression is [4(a + 2b - 3c) - 3(2b - 3c)] + 2a then answer will be

A.7a - 3b +4c
B.6a + 2b - 3c
C.8a + 3b - 4c
D.9a + 2b + 5c

Answer: B

Expand each bracket, then collect.

\[ 4(a + 2b - 3c) = 4a + 8b - 12c \]

\[ -3(2b - 3c) = -6b + 9c \]

\[ (4a + 8b - 12c) + (-6b + 9c) + 2a = 6a + 2b - 3c \]

The \( -3 \) turns \( -3c \) into \( +9c \), which is what leaves only \( -3c \) at the end rather than \( -21c \).

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90.

Simplify (1 ⁄ (x² - 1)) ÷ (1 ⁄ (x + 1))

A.1 ⁄ (x - 1)
B.1 ⁄ (x + 1)
C.1 ⁄ (x² - 1)
D.None of the above

Answer: A

Dividing by a fraction means multiplying by its reciprocal.

\[ \frac{1}{x^{2}-1} \div \frac{1}{x+1} = \frac{1}{x^{2}-1} \times (x+1) \]

Factorising the denominator:

\[ = \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1} \]

The \( (x+1) \) cancels, which is why the difference of squares has to be split before the simplification becomes visible.

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