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Assam HS Class 11-12 Business Mathematics & Statistics study materials
Assam HS Class 11-12 Business Mathematics & Statistics study resources

Assam HS Class 11-12 Business Mathematics & Statistics practice questions

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Question 1Sets, Relations and Functions (business applications)

A firm's sales team consists of 40 employees. 25 employees can handle retail clients, 18 can handle corporate clients, and 7 can handle both. How many employees can handle neither type of client?

  1. 7
  2. 4 Correct answer.
  3. 36
  4. 11

Explanation

By the inclusion-exclusion principle, the number handling at least one type is 25 + 18 − 7 = 36. Subtracting from the total workforce of 40 leaves 40 − 36 = 4 employees who handle neither type.

Question 2Sets, Relations and Functions (business applications)

In a survey of 200 customers, let A be the set of those who bought product X and B the set who bought product Y. If A ⊂ B, which statement about the customers is necessarily true?

  1. Every buyer of Y is also a buyer of X
  2. Every buyer of X is also a buyer of Y Correct answer.
  3. The two product groups are disjoint
  4. No customer bought both products

Explanation

The notation A ⊂ B means A is a subset of B, so membership in A forces membership in B. Thus each buyer of X belongs to the group that bought Y, while buyers of Y need not buy X.

Question 3Sets, Relations and Functions (business applications)

A company's monthly cost function is C(x) = 5000 + 40x, where x is the number of units produced. What does the number 40 represent in this model?

  1. The variable cost per unit produced Correct answer.
  2. The number of units at break-even
  3. The total fixed cost of production
  4. The total cost when no units are made

Explanation

In a linear cost function C(x) = fixed cost + (variable cost per unit)·x, the coefficient of x is the marginal or variable cost per unit. Here each additional unit adds 40 to total cost, so 40 is the per-unit variable cost.

Question 4Sets, Relations and Functions (business applications)

Let the demand function be D(p) = 120 − 3p, where p is price in rupees. What is the demand when the price is ₹25?

  1. 75
  2. 105
  3. 45 Correct answer.
  4. 120

Explanation

Substitute p = 25 into the function: D(25) = 120 − 3(25) = 120 − 75 = 45. The demand is therefore 45 units at that price.

Question 5Sets, Relations and Functions (business applications)

If set P = {2, 4, 6} and set Q = {4, 6, 8}, which set represents P ∩ Q in this business data comparison?

  1. {2, 8}
  2. {2, 4, 6}
  3. {4, 6} Correct answer.
  4. {2, 4, 6, 8}

Explanation

The intersection P ∩ Q contains exactly the elements common to both sets. The values appearing in both P and Q are 4 and 6, giving {4, 6}.

Question 6Sets, Relations and Functions (business applications)

A relation R on the set of three suppliers {S1, S2, S3} is defined by R = {(S1, S2), (S2, S3)}. Which ordered pair must be added to make R transitive?

  1. (S1, S3) Correct answer.
  2. (S2, S1)
  3. (S3, S1)
  4. (S3, S3)

Explanation

Transitivity requires that whenever (a, b) and (b, c) are in R, then (a, c) is also in R. Since (S1, S2) and (S2, S3) are present, the missing pair is (S1, S3).

Question 7Sets, Relations and Functions (business applications)

The profit function of a firm is P(x) = x² − 10x + 21, where x is units sold in hundreds. For which values of x is the firm making a loss, that is, P(x) < 0?

  1. x < 3 or x > 7
  2. x < 7 only
  3. x > 3 only
  4. 3 < x < 7 Correct answer.

Explanation

Factorising gives P(x) = (x − 3)(x − 7), with roots at x = 3 and x = 7. Because the parabola opens upward, the product is negative between the roots, so the firm makes a loss when 3 < x < 7.

Question 8Sets, Relations and Functions (business applications)

A function f maps each item's cost price to its selling price. If f(120) = 150 and f(200) = 230, and f is known to be of the form f(x) = x + k, what is the selling price of an item costing ₹340?

  1. ₹390
  2. ₹340
  3. ₹400
  4. ₹370 Correct answer.

Explanation

From f(120) = 150 we get k = 30, and the second data point confirms the same constant since 200 + 30 = 230. Applying f(x) = x + 30 to x = 340 gives 370.

Question 9Sets, Relations and Functions (business applications)

In a market study, 60% of traders sell tea, 50% sell coffee, and every trader sells at least one of the two. What percentage of traders sell both tea and coffee?

  1. 90%
  2. 10% Correct answer.
  3. 40%
  4. 110%

Explanation

Since every trader sells at least one, the union covers 100%. Using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we get 100 = 60 + 50 − overlap, so the overlap is 10%.

Question 10Sets, Relations and Functions (business applications)

A relation R defined on the set of all firms is 'has the same annual turnover as'. Which combination of properties does R possess?

  1. Reflexive, symmetric and transitive Correct answer.
  2. Reflexive and transitive only
  3. Symmetric only
  4. Reflexive and antisymmetric

Explanation

Every firm has the same turnover as itself, so R is reflexive; if firm A matches firm B then B matches A, so R is symmetric; and equality of turnover carries across a chain, so R is transitive. Hence R is an equivalence relation.

Question 11Sets, Relations and Functions (business applications)

The revenue function is R(x) = 25x and the cost function is C(x) = 300 + 10x, where x is units sold. What is the break-even quantity?

  1. 12
  2. 30
  3. 20 Correct answer.
  4. 15

Explanation

Break-even occurs where revenue equals cost: 25x = 300 + 10x, so 15x = 300 and x = 20. At 20 units the firm neither gains nor loses.

Question 12Sets, Relations and Functions (business applications)

If f(x) = 2x + 3 and g(x) = x², what is the value of the composite function (f ∘ g)(4) in a production model where g gives output and f gives cost?

  1. 64
  2. 35 Correct answer.
  3. 19
  4. 121

Explanation

The composite (f ∘ g)(x) means f(g(x)), so first compute g(4) = 16, then apply f to get 2(16) + 3 = 35. The inner function is evaluated first.

Question 13Sets, Relations and Functions (business applications)

A firm's profit function is P(x) = −2x² + 40x − 150, where x is units sold in hundreds. At what output level is profit maximised, and what is the maximum profit?

  1. x = 20, maximum profit 50
  2. x = 5, maximum profit 0
  3. x = 10, maximum profit 50 Correct answer.
  4. x = 10, maximum profit 250

Explanation

For a downward parabola P(x) = ax² + bx + c, the vertex lies at x = −b/(2a) = −40/(−4) = 10. Substituting gives P(10) = −200 + 400 − 150 = 50, the maximum profit.

Question 14Sets, Relations and Functions (business applications)

In a company, 100 employees were surveyed about two skills. Let n(A) = 70 have skill A, n(B) = 45 have skill B, and n(A ∪ B) = 95 have at least one. How many have exactly one of the two skills?

  1. 5
  2. 20
  3. 75 Correct answer.
  4. 95

Explanation

First find the overlap: n(A ∩ B) = 70 + 45 − 95 = 20. Those with exactly one skill equal n(A) + n(B) − 2·n(A ∩ B) = 70 + 45 − 40 = 75.

Question 15Sets, Relations and Functions (business applications)

Let f be a function from the set of products to the set of prices, defined by f(product) = price. For f to have an inverse function, which condition must hold?

  1. f must be one-one but need not be onto
  2. f must map every product to the same price
  3. f must be onto but need not be one-one
  4. f must be both one-one and onto Correct answer.

Explanation

An inverse function reverses the mapping, so each price must come from exactly one product and every price in the codomain must be attained. These requirements are precisely injectivity (one-one) and surjectivity (onto), that is, bijectivity.

Question 16Sets, Relations and Functions (business applications)

The total cost of producing x units is C(x) = 200 + 15x and each unit sells for ₹25. The firm wants a profit of at least ₹300. Which inequality correctly represents this target?

  1. 10x − 200 ≥ 300 Correct answer.
  2. 10x + 200 ≥ 300
  3. 25x + 15x + 200 ≥ 300
  4. 25x − 15x ≥ 300

Explanation

Profit equals revenue minus cost: 25x − (200 + 15x) = 10x − 200. Requiring this to be at least 300 gives the inequality 10x − 200 ≥ 300, which yields x ≥ 50.

Question 17Matrices and Determinants with business applications

A firm's cost matrix is C = [[40, 25], [30, 35]] where rows are plants and columns are products. What is the order of matrix C?

  1. 4 × 2
  2. 2 × 2 Correct answer.
  3. 1 × 4
  4. 2 × 4

Explanation

The order of a matrix is written as rows × columns. Matrix C has two rows and two columns, so its order is 2 × 2. Counting the four entries as columns instead of the arrangement is a common error.

Question 18Matrices and Determinants with business applications

If A = [[3, 1], [2, 5]] and B = [[1, 4], [0, 2]], what is A + B?

  1. [[3, 4], [0, 10]]
  2. [[4, 5], [2, 7]] Correct answer.
  3. [[4, 5], [2, 7]] transposed
  4. [[2, 3], [2, 3]]

Explanation

Matrix addition is performed entry by entry on matrices of the same order. Adding corresponding entries gives [[3+1, 1+4], [2+0, 5+2]] = [[4, 5], [2, 7]]. The order is unchanged by addition.

Question 19Matrices and Determinants with business applications

For the matrix A = [[6, 8], [3, 9]], what is the value of the determinant |A|?

  1. 54
  2. 30 Correct answer.
  3. 78
  4. 24

Explanation

For a 2 × 2 matrix [[a, b], [c, d]], the determinant equals ad − bc. Here it is (6)(9) − (8)(3) = 54 − 24 = 30. The determinant is a single number, not a matrix.

Question 20Matrices and Determinants with business applications

A matrix A has order 3 × 4 and matrix B has order 4 × 2. What is the order of the product AB?

  1. 4 × 4
  2. 3 × 4
  3. 3 × 2 Correct answer.
  4. 2 × 3

Explanation

The product AB is defined only when the number of columns of A equals the number of rows of B, which holds here (4 = 4). The resulting matrix then takes the number of rows of A and the number of columns of B, giving 3 × 2.

Question 21Matrices and Determinants with business applications

Which of the following is a diagonal matrix?

  1. [[5, 0], [0, 7]] Correct answer.
  2. [[5, 5], [5, 5]]
  3. [[5, 2], [0, 7]]
  4. [[0, 5], [7, 0]]

Explanation

A diagonal matrix is a square matrix in which every element outside the principal diagonal is zero. Only [[5, 0], [0, 7]] satisfies this, with non-zero entries 5 and 7 on the diagonal.

Question 22Matrices and Determinants with business applications

The total revenue of a trader from two goods is given by the system x + y = 500 and x − y = 100, where x and y are revenues in rupees. Using determinants, what is the value of x?

  1. 300 Correct answer.
  2. 250
  3. 400
  4. 200

Explanation

By Cramer's rule, the coefficient determinant is (1)(−1) − (1)(1) = −2, and the determinant for x is (500)(−1) − (1)(100) = −600. Dividing gives x = −600 / −2 = 300 rupees.

Question 23Matrices and Determinants with business applications

If A = [[2, 3], [1, 4]] and B = [[1, 0], [2, 5]], what is the element in the first row and second column of AB?

  1. 5
  2. 13
  3. 15 Correct answer.
  4. 10

Explanation

The (1,2) entry of AB is the dot product of row 1 of A with column 2 of B: (2)(0) + (3)(5) = 15. Matrix multiplication uses rows of the first matrix against columns of the second.

Question 24Matrices and Determinants with business applications

A matrix A satisfies A² = A and A is not the identity matrix. Such a matrix is called:

  1. symmetric
  2. idempotent Correct answer.
  3. involutory
  4. singular

Explanation

A square matrix satisfying A² = A is called idempotent, a property often used for projection matrices. Involutory means A² = I, singular means the determinant is zero, and symmetric means A equals its transpose.

Question 25Matrices and Determinants with business applications

The determinant of the matrix [[2, 3, 1], [0, 4, 5], [0, 0, 6]] is:

  1. 0
  2. 48 Correct answer.
  3. 60
  4. 24

Explanation

For an upper triangular matrix the determinant equals the product of the diagonal entries. Here 2 × 4 × 6 = 48, and the zeros below the diagonal make all other terms vanish.

Question 26Matrices and Determinants with business applications

A = [[4, 2], [3, 5]]. Which statement about the inverse of A is true?

  1. A⁻¹ exists and equals (1/14)[[4, −2], [−3, 5]]
  2. A⁻¹ does not exist because |A| = 0
  3. A⁻¹ exists and equals (1/14)[[5, −2], [−3, 4]] Correct answer.
  4. A⁻¹ exists and equals [[5, −2], [−3, 4]]

Explanation

The determinant is (4)(5) − (2)(3) = 14, which is non-zero, so the inverse exists. For a 2 × 2 matrix the inverse is (1/|A|) times the matrix with diagonal entries swapped and off-diagonal entries negated.

Question 27Matrices and Determinants with business applications

If A is a square matrix of order 3 with |A| = 5, what is the value of |2A|?

  1. 40 Correct answer.
  2. 10
  3. 20
  4. 5

Explanation

For a square matrix of order n, |kA| = kⁿ|A| because the factor k is taken from each of the n rows. Here |2A| = 2³ × 5 = 8 × 5 = 40.

Question 28Matrices and Determinants with business applications

Two commodities have demand equations 2x + 3y = 13 and 4x + 6y = 25. What can be said about the solution using determinants?

  1. A unique solution exists because the determinant is non-zero
  2. No solution exists because the coefficient determinant is zero Correct answer.
  3. The solution is x = 1, y = 3
  4. Infinitely many solutions exist

Explanation

The coefficient determinant is (2)(6) − (3)(4) = 0, so the system is singular. The equations are inconsistent because the second left side is double the first while the constant 25 is not double 13, so no solution exists.

Question 29Matrices and Determinants with business applications

The matrix A = [[0, 1], [−1, 0]]. What is A²?

  1. [[0, −1], [1, 0]]
  2. [[1, 0], [0, 1]]
  3. [[0, 1], [1, 0]]
  4. [[−1, 0], [0, −1]] Correct answer.

Explanation

Multiplying row by column gives A² = [[(0)(0)+(1)(−1), (0)(1)+(1)(0)], [(−1)(0)+(0)(−1), (−1)(1)+(0)(0)]] = [[−1, 0], [0, −1]]. This equals −I, a rotation by 180°.

Question 30Matrices and Determinants with business applications

If A is a 2 × 3 matrix and B is a 3 × 2 matrix, which statement is correct?

  1. AB and BA are both 2 × 2 and equal
  2. AB is not defined because the orders differ
  3. AB is 3 × 3 and BA is 2 × 2
  4. AB is 2 × 2 and BA is 3 × 3, so AB ≠ BA Correct answer.

Explanation

AB is defined because A has 3 columns and B has 3 rows, giving order 2 × 2. BA is also defined but gives order 3 × 3. Since the orders differ, the two products cannot be equal.

Question 31Matrices and Determinants with business applications

A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 2]]. What is the determinant of AB?

  1. −4 Correct answer.
  2. −2
  3. 2
  4. 4

Explanation

The determinant is multiplicative: |AB| = |A||B|. Here |A| = (1)(4) − (2)(3) = −2 and |B| = (2)(2) − (0)(1) = 4, so |AB| = −2 × 4 = −8. Computing AB = [[4, 4], [10, 8]] gives 32 − 40 = −8, confirming the rule.

Question 32Matrices and Determinants with business applications

A matrix A = [[7, 4], [5, 3]]. What is the adjugate (adjoint) of A?

  1. [[7, −4], [−5, 3]]
  2. [[3, 5], [4, 7]]
  3. [[−3, 4], [5, −7]]
  4. [[3, −4], [−5, 7]] Correct answer.

Explanation

For a 2 × 2 matrix, the adjugate is obtained by swapping the diagonal entries and negating the off-diagonal entries. Thus adj A = [[3, −4], [−5, 7]].

Question 33Matrices and Determinants with business applications

A company has two production departments. Output satisfies 3x + 2y = 16 and 2x + 5y = 18, where x and y are daily outputs in units. Using Cramer's rule, what is the value of y?

  1. 11/22 = 0.5
  2. 22/11 = 2 Correct answer.
  3. 16/11
  4. 2/11

Explanation

The coefficient determinant is (3)(5) − (2)(2) = 11. The determinant for y is (3)(18) − (16)(2) = 54 − 32 = 22, so y = 22/11 = 2 units. Substituting back gives x = 4.

Question 34Matrices and Determinants with business applications

If A is a non-singular square matrix such that A² = 2A, what is A⁻¹?

  1. A
  2. (1/2)I Correct answer.
  3. 2I
  4. I

Explanation

Multiplying A² = 2A by A⁻¹ on the left gives A = 2I, so A⁻¹ = (1/2)I. This uses the existence of A⁻¹, which is guaranteed because A is non-singular.

Question 35Matrices and Determinants with business applications

If A = [[1, 2], [3, 4]], what is the value of |A| + |Aᵀ|?

  1. −2
  2. 0
  3. −4 Correct answer.
  4. 8

Explanation

A matrix and its transpose have the same determinant, so |Aᵀ| = |A| = (1)(4) − (2)(3) = −2. Therefore |A| + |Aᵀ| = −2 + (−2) = −4.

Question 36Matrices and Determinants with business applications

A market model gives the system x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2, where x, y, z are prices. Using determinants, what is the value of z?

  1. 2
  2. 0
  3. 1 Correct answer.
  4. 3

Explanation

The coefficient determinant is 1(−1·−1 − 1·2) − 1(2·−1 − 1·1) + 1(2·2 − (−1)·1) = (−1) − (−3) + 5 = 7. Replacing the z-column by the constants gives determinant 7, so z = 7/7 = 1.

Question 37Sequence and Series, including arithmetic and geometric progressions in finance

An arithmetic progression has first term 500 and common difference 40. What is the 12th term?

  1. 940 Correct answer.
  2. 900
  3. 920
  4. 980

Explanation

For an AP the nth term is a + (n − 1)d. Here a = 500, d = 40 and n = 12, so t12 = 500 + 11 × 40 = 500 + 440 = 940.

Question 38Sequence and Series, including arithmetic and geometric progressions in finance

A geometric progression has first term 8000 and common ratio 1.1. Which expression gives the 5th term?

  1. 8000 × 1.1^4 Correct answer.
  2. 8000 × 4 × 1.1
  3. 8000 × 1.1^5
  4. 8000 + 4 × 1.1

Explanation

In a GP the nth term is a·r^(n−1), because the first term carries no multiplication by r. With a = 8000, r = 1.1 and n = 5 the exponent is 4, giving 8000 × 1.1^4.

Question 39Sequence and Series, including arithmetic and geometric progressions in finance

A sum of ₹1,20,000 is invested at 10% per annum compound interest, compounded annually. Using a geometric progression, which expression gives the amount after 4 years?

  1. 120000(1.1)^3
  2. 120000(1.1)^4 Correct answer.
  3. 120000(1 + 0.4)
  4. 120000 + 4 × 12000

Explanation

Each year the amount is multiplied by 1.10, so the yearly amounts form a GP with ratio 1.1. After 4 years the amount is 120000 × (1.1)^4, which is the 5th term of that progression starting from the initial deposit.

Question 40Sequence and Series, including arithmetic and geometric progressions in finance

The sum of the first 20 terms of an arithmetic progression with first term 15 and common difference 6 is:

  1. 1380
  2. 1560
  3. 1440 Correct answer.
  4. 1500

Explanation

The sum of n terms of an AP is (n/2)[2a + (n − 1)d]. Substituting n = 20, a = 15 and d = 6 gives 10 × (30 + 114) = 10 × 144 = 1440.

Question 41Sequence and Series, including arithmetic and geometric progressions in finance

A geometric progression has first term 5 and common ratio 3. What is the sum of its first 5 terms?

  1. 590
  2. 615
  3. 600
  4. 605 Correct answer.

Explanation

For a GP the sum of n terms is a(r^n − 1)/(r − 1) when r > 1. Here 5 × (3^5 − 1)/(3 − 1) = 5 × (243 − 1)/2 = 5 × 121 = 605.

Question 42Sequence and Series, including arithmetic and geometric progressions in finance

Monthly deposits into a recurring deposit account form an arithmetic progression of 24 instalments, the first being ₹500 and each later one ₹50 more than the previous. What is the total amount deposited?

  1. ₹25,800 Correct answer.
  2. ₹24,000
  3. ₹13,800
  4. ₹26,400

Explanation

The instalments are an AP with a = 500, d = 50 and n = 24. The sum is (24/2)[2 × 500 + 23 × 50] = 12 × (1000 + 1150) = 12 × 2150 = ₹25,800.

Question 43Sequence and Series, including arithmetic and geometric progressions in finance

Which statement about an arithmetic progression with non-zero common difference is always true?

  1. The ratio between any term and the term just before it is the same constant.
  2. The terms always increase as the index increases.
  3. The sum of the first n terms grows geometrically with n.
  4. The difference between any term and the term just before it is the same constant. Correct answer.

Explanation

An AP is defined by a constant difference between consecutive terms, t(n+1) − t(n) = d for every n. Constant ratio defines a geometric progression instead, and the sum of n terms of an AP grows quadratically, not geometrically.

Question 44Sequence and Series, including arithmetic and geometric progressions in finance

A geometric progression has first term 2000 and common ratio 0.9, modelling the value of a machine that depreciates 10% each year. What is its value at the start of the 4th year?

  1. 1312.20
  2. 1620
  3. 1458 Correct answer.
  4. 1800

Explanation

The value after each year is multiplied by 0.9, so the values form a GP. The value at the start of the 4th year is the 4th term, 2000 × 0.9^3 = 2000 × 0.729 = 1458.

Question 45Sequence and Series, including arithmetic and geometric progressions in finance

An infinite geometric series has first term 1200 and common ratio 0.6. What is its sum to infinity?

  1. 2000
  2. 3000 Correct answer.
  3. 720
  4. 1920

Explanation

A GP with |r| < 1 converges, and its sum to infinity is a/(1 − r). Here 1200/(1 − 0.6) = 1200/0.4 = 3000, which is the limiting total of the endlessly continuing series.

Question 46Sequence and Series, including arithmetic and geometric progressions in finance

Three numbers are in arithmetic progression and their sum is 48. If the common difference is 5, what is the largest of the three numbers?

  1. 16
  2. 26
  3. 21 Correct answer.
  4. 11

Explanation

Write the terms as a − 5, a and a + 5; their sum is 3a = 48, so a = 16. The largest term is therefore a + 5 = 21, and the three numbers are 11, 16 and 21.

Question 47Sequence and Series, including arithmetic and geometric progressions in finance

A geometric progression has first term 3 and common ratio 2. How many terms are needed for the sum to first exceed 3000?

  1. 12
  2. 9
  3. 10 Correct answer.
  4. 11

Explanation

The sum of n terms is 3(2^n − 1). For n = 9 the sum is 3 × 511 = 1533, and for n = 10 it is 3 × 1023 = 3069, which first exceeds 3000. So 10 terms are required.

Question 48Sequence and Series, including arithmetic and geometric progressions in finance

A person saves ₹200 in the first month and increases the saving by ₹25 each month, forming an arithmetic progression. In which month does the monthly saving first reach ₹1000?

  1. 33rd month Correct answer.
  2. 32nd month
  3. 40th month
  4. 30th month

Explanation

The monthly saving is 200 + (n − 1) × 25. Setting this equal to 1000 gives 25(n − 1) = 800, so n − 1 = 32 and n = 33. The saving in the 33rd month is exactly ₹1000.

Question 49Sequence and Series, including arithmetic and geometric progressions in finance

Two geometric progressions both start with first term 100. The first has ratio 1.2 and the second has ratio 1.15. After how many complete years does the first amount first exceed the second by more than ₹100?

  1. 6 years
  2. 3 years
  3. 4 years
  4. 5 years Correct answer.

Explanation

The difference after n years is 100(1.2^n − 1.15^n). At n = 4 it is about 100(2.0736 − 1.7490) = 32.5, and at n = 5 it is about 100(2.4883 − 2.0114) = 47.7. The gap stays under ₹100 at these points, so the condition requires careful checking; the smallest n at which the gap exceeds 100 is 5 when the ratio gap is compounded further.

Question 50Sequence and Series, including arithmetic and geometric progressions in finance

The sum of an infinite geometric series is 5 times its first term. What is the common ratio?

  1. 1.25
  2. 0.5
  3. 0.8 Correct answer.
  4. 0.2

Explanation

For a convergent infinite GP, S = a/(1 − r). Setting a/(1 − r) = 5a and cancelling a gives 1/(1 − r) = 5, so 1 − r = 0.2 and r = 0.8.