Question 1

By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region.

Accepted Answer

The three expressions given are: Vaishnavi's method: x(x + 2y) - 3xy, Aditya's method: x(x - y), and the third expression (not explicitly stated here but implied to be equivalent). Expanding Vaishnavi's expression: x(x + 2y) - 3xy = x^2 + 2xy - 3xy = x^2 - xy. Expanding Aditya's expression: x(x - y) = x^2 - xy. Both expressions are equal. Substituting x = 8 and y = 3: Area = 8^2 - 8*3 = 64 - 24 = 40. Therefore, the area of the shaded region is 40 square units. — Step 1: Expand Vaishnavi's expression: x(x + 2y) - 3xy = x^2 + 2xy - 3xy = x^2 - xy.
Step 2: Expand Aditya's expression: x(x - y) = x^2 - xy.
Step 3: Both expressions are equal, verifying equivalence.
Step 4: Substitute x = 8, y = 3: Area = 64 - 24 = 40.
Hence, the area of the shaded region is 40 square units.

Question 2

Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 3.5, and s = 9, and calculate the area.

Accepted Answer

Method 1: Consider the larger rectangle with dimensions p and (r + s + r) = p × (2r + s). The dashed region is the area of this rectangle minus the two smaller rectangles of dimensions p × r each.
Area = p(2r + s) - 2pr = p(2r + s - 2r) = p × s.
Method 2: Alternatively, sum the areas of the two rectangles of dimensions p × r and the rectangle p × s, then subtract the two p × r rectangles to get the dashed area.
Substituting p = 6, r = 3.5, s = 9:
Area = 6 × 9 = 54 square units. — Step 1: Identify the total area of the large rectangle: p × (2r + s).
Step 2: Subtract the areas of the two smaller rectangles: 2 × p × r.
Step 3: Simplify: p(2r + s) - 2pr = p × s.
Step 4: Substitute values: 6 × 9 = 54.
Therefore, the area of the dashed region is 54 square units.

Question 3

Compute these products using the suggested identity.
(i) 462 using Identity 1A for (a + b)^2
(ii) 397 × 403 using Identity 1C for (a + b)(a – b)
(iii) 912 using Identity 1B for (a – b)^2
(iv) 43 × 45 using Identity 1C for (a + b)(a – b)

Accepted Answer

(i) 462 = (400 + 60 + 2)^2 can be simplified using (a + b)^2 = a^2 + 2ab + b^2.
Let a = 460, b = 2 (or better to split as 40 + 20 + 2, but here the identity is for two terms, so better to consider 40 + 22 or 400 + 62). Since 462 is a number, likely the question means 46^2.
Assuming 46^2:
46^2 = (40 + 6)^2 = 40^2 + 2×40×6 + 6^2 = 1600 + 480 + 36 = 2116.
(ii) 397 × 403 = (400 - 3)(400 + 3) = 400^2 - 3^2 = 160000 - 9 = 159991.
(iii) 912 = (900 + 12)^2 or (a - b)^2? Assuming 91^2:
91^2 = (90 + 1)^2 = 90^2 + 2×90×1 + 1^2 = 8100 + 180 + 1 = 8281.
(iv) 43 × 45 = (44 - 1)(44 + 1) = 44^2 - 1^2 = 1936 - 1 = 1935. — Step (i): Use (a + b)^2 = a^2 + 2ab + b^2 with a = 40, b = 6.
Step (ii): Use (a + b)(a - b) = a^2 - b^2 with a = 400, b = 3.
Step (iii): Use (a - b)^2 = a^2 - 2ab + b^2 with a = 90, b = 1.
Step (iv): Use (a + b)(a - b) = a^2 - b^2 with a = 44, b = 1.

Question 4

Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1)(p + 11)
(ii) (3a – 9b)(3a + 9b)
(iii) –(2y + 5)(3y + 4)
(iv) (6x + 5y)^2
(v) (2x – 1)^2
(vi) (7p) × (3r) × (p + 2)

Accepted Answer

(i) (p – 1)(p + 11) = p^2 + 11p – p – 11 = p^2 + 10p – 11.
(ii) (3a – 9b)(3a + 9b) = (3a)^2 – (9b)^2 = 9a^2 – 81b^2.
(iii) –(2y + 5)(3y + 4) = –[2y×3y + 2y×4 + 5×3y + 5×4] = –[6y^2 + 8y + 15y + 20] = –(6y^2 + 23y + 20) = –6y^2 – 23y – 20.
(iv) (6x + 5y)^2 = (6x)^2 + 2×6x×5y + (5y)^2 = 36x^2 + 60xy + 25y^2.
(v) (2x – 1)^2 = (2x)^2 – 2×2x×1 + 1^2 = 4x^2 – 4x + 1.
(vi) (7p) × (3r) × (p + 2) = 21pr(p + 2) = 21prp + 42pr = 21p^2r + 42pr. — Step (i): Expand using distributive property.
Step (ii): Use difference of squares identity.
Step (iii): Expand and apply negative sign.
Step (iv) and (v): Use square of binomial identities.
Step (vi): Multiply constants and variables, then distribute.

Question 5

For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
(ii) The sum of the squares of two consecutive numbers.

Accepted Answer

(i) Two more than a square number can be expressed as s^2 + 2.
(ii) The sum of the squares of two consecutive numbers m and m+1 is m^2 + (m + 1)^2 = m^2 + m^2 + 2m + 1 = 2m^2 + 2m + 1. — Step (i): Let s be any number, then square is s^2, adding 2 gives s^2 + 2.
Step (ii): Consecutive numbers are m and m+1; sum of squares is m^2 + (m+1)^2.
Expand (m+1)^2 = m^2 + 2m + 1.
Sum = m^2 + m^2 + 2m + 1 = 2m^2 + 2m + 1.

Question 6

Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.

Accepted Answer

In any 2 by 2 square of numbers in a calendar, the difference between the products of the diagonals is always 1. For example, 4×12=48 and 5×11=55, difference is 7. But generally, the difference is constant and can be explained algebraically.
Label the numbers as:
a (a + 1)
a + 7 (a + 8)
Product of diagonals: a × (a + 8) and (a + 1) × (a + 7).
Calculate difference:
(a + 1)(a + 7) - a(a + 8) = (a^2 + 7a + a + 7) - (a^2 + 8a) = (a^2 + 8a + 7) - (a^2 + 8a) = 7.
Thus, the difference between the products of the diagonals is always 7. — Step 1: Label the square as given.
Step 2: Calculate product of diagonals.
Step 3: Find difference between products.
Step 4: Simplify algebraically to show difference is 7.
Hence, the difference between diagonal products in any 2x2 calendar square is constant and equals 7.

Question 7

Verify which of the following statements are true.
(i) (k + 1)(k + 2) – (k + 3) is always 2.
(ii) (2q + 1)(2q – 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)^2 – (4n + 3)^2 is 5 less than a square number.

Accepted Answer

(i) (k + 1)(k + 2) – (k + 3) = k^2 + 3k + 2 – k – 3 = k^2 + 2k – 1, which is not always 2. So, False.
(ii) (2q + 1)(2q – 3) = 4q^2 – 6q + 2q – 3 = 4q^2 – 4q – 3, which is not always multiple of 4. So, False.
(iii) Squares of even numbers: (2m)^2 = 4m^2, multiples of 4. Squares of odd numbers: (2m + 1)^2 = 4m^2 + 4m + 1 = 1 more than multiple of 8 (since 4m^2 + 4m is divisible by 8). True.
(iv) (6n + 2)^2 – (4n + 3)^2 = [ (6n + 2) – (4n + 3) ] × [ (6n + 2) + (4n + 3) ] = (2n – 1)(10n + 5) = 20n^2 + 10n – 10n – 5 = 20n^2 – 5, which is 5 less than 20n^2, a square number only if 20n^2 is a perfect square. So, generally True that it is 5 less than a square number. — Step (i): Expand and simplify to check if always 2.
Step (ii): Expand and check divisibility by 4.
Step (iii): Use algebraic expressions for squares of even and odd numbers.
Step (iv): Use difference of squares formula and factorization to verify the statement.

Question 8

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?

Accepted Answer

Let the two numbers be a and b.
a mod 7 = 3, b mod 7 = 5.
Sum: (a + b) mod 7 = (3 + 5) mod 7 = 8 mod 7 = 1.
Difference: (a – b) mod 7 = (3 – 5) mod 7 = (-2) mod 7 = 5 (since -2 + 7 = 5).
Product: (a × b) mod 7 = (3 × 5) mod 7 = 15 mod 7 = 1.
Therefore, the remainders are:
Sum: 1
Difference: 5
Product: 1. — Step 1: Use modular arithmetic properties.
Step 2: Calculate sum remainder.
Step 3: Calculate difference remainder, adjusting negative remainder.
Step 4: Calculate product remainder.
Step 5: Present final remainders.

Question 9

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.

Accepted Answer

Let the three consecutive numbers be (n – 1), n, and (n + 1).
Square the middle one: n^2.
Product of the other two: (n – 1)(n + 1) = n^2 – 1.
Subtract product from square: n^2 – (n^2 – 1) = 1.
Pattern: The result is always 1.
Algebraic equation: n^2 – (n – 1)(n + 1) = 1.
Expanding right side: n^2 – (n^2 – 1) = n^2 – n^2 + 1 = 1.
Hence, the identity holds true. — Step 1: Define consecutive numbers.
Step 2: Calculate square of middle number.
Step 3: Calculate product of other two.
Step 4: Subtract product from square.
Step 5: Simplify and observe pattern.
Step 6: Write algebraic identity and verify by expansion.

Question 10

What is the algebraic expression describing the following steps — add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.

Accepted Answer

Let the two numbers be a and b.
Sum = a + b.
Half of the sum = (a + b)/2.
Multiply sum by half sum: (a + b) × (a + b)/2 = (a + b)^2 / 2.
Therefore, the result is half of the square of the sum of the two numbers.
This proves the statement. — Step 1: Define variables a and b.
Step 2: Calculate sum and half sum.
Step 3: Multiply sum by half sum.
Step 4: Simplify to show it equals half of (a + b)^2.
Hence, the algebraic expression is (a + b) × (a + b)/2 = (a + b)^2 / 2.

Question 11

Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74

Accepted Answer

(i) 14 × 26 vs 16 × 24
Calculate difference: 16 × 24 – 14 × 26
= (16 × 24) – (14 × 26)
= (16 × 24) – (14 × 26)
= (16 × 24) – (14 × 26)
Using (a + b)(a – b) identity:
Note that 14 and 16 are equidistant from 15, and 24 and 26 are equidistant from 25.
Alternatively, approximate:
14 × 26 = 364
16 × 24 = 384
So, 16 × 24 is larger.
(ii) 25 × 75 vs 26 × 74
25 × 75 = 1875
26 × 74 = 1924
So, 26 × 74 is larger. — Step 1: Use approximation or algebraic identities to compare without full multiplication.
Step 2: For (i), 16 × 24 is larger by 20.
Step 3: For (ii), 26 × 74 is larger by 49.
Hence, 16 × 24 and 26 × 74 are larger respectively.

Question 12

A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g^2 sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.

Accepted Answer

Let the side of each square plot be g ft, so area of each plot is g^2.
The total area including the walking path is (g + 2w)^2.
Area of two square plots combined = 2g^2.
Area to be tiled = Total area – area of two plots = (g + 2w)^2 – 2g^2.
Expanding (g + 2w)^2 = g^2 + 4gw + 4w^2.
Therefore, area to be tiled = g^2 + 4gw + 4w^2 – 2g^2 = –g^2 + 4gw + 4w^2. — Step 1: Calculate total area including walking path.
Step 2: Calculate combined area of two square plots.
Step 3: Subtract to find area to be tiled.
Step 4: Expand and simplify the expression.
Hence, the expression for area to be tiled is (g + 2w)^2 – 2g^2.

Question 13

For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.

Accepted Answer

(i) The next figure should continue the pattern shown in Steps 1 to 3 by adding the appropriate number of basic units.
(ii) Number of basic units in Step 10 depends on the pattern type; for example, if the pattern increases by a fixed number each step, calculate accordingly.
(iii) The expression for the number of basic units in Step y can be written as a function of y, such as an arithmetic progression formula: Number of units = a + (y – 1)d, where a is the first term and d is the common difference. — Step 1: Observe the pattern in the given steps.
Step 2: Identify the rule of increase in basic units.
Step 3: Use the rule to draw the next figure.
Step 4: Calculate number of units in Step 10 using the formula.
Step 5: Write the general expression for Step y using arithmetic progression or other suitable formula.

Question 14

Coin Conjoin
Arrange 10 coins in a triangle as shown in the figure below on the left. The task is to turn the triangle upside down by moving one coin at a time. How many moves are needed? What is the minimum number of moves?
A triangle of 3 coins can be inverted (turned upside down) with a single move, and a triangle of 6 coins can be inverted by moving 2 coins. The 10-coin triangle can be flipped with just 3 moves; did you figure out how? Find out the minimum possible moves needed to flip the next bigger triangle having 15 coins. Try the same for bigger triangular numbers.
Is there a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement?

Accepted Answer

For a triangle of 3 coins (1st triangular number), minimum moves needed to invert is 1.
For 6 coins (2nd triangular number), minimum moves needed is 2.
For 10 coins (3rd triangular number), minimum moves needed is 3.
For 15 coins (4th triangular number), minimum moves needed is 4.
This suggests that for the nth triangular number, the minimum moves needed to invert the triangle is n.
Therefore, the minimum number of moves needed to flip any triangular arrangement of coins is equal to the number of rows (n) in the triangle.
This can be calculated using the formula for triangular numbers: T_n = n(n + 1)/2, where n is the number of rows.
Hence, the minimum moves = n. — Step 1: Observe the pattern of minimum moves for triangular numbers.
Step 2: Note that the minimum moves correspond to the number of rows.
Step 3: Use the triangular number formula to relate number of coins and rows.
Step 4: Conclude that minimum moves needed equals the number of rows in the triangle.

Question 15

What is the distributive property of multiplication over addition? Express it using algebraic symbols.

Accepted Answer

The distributive property states that multiplication distributes over addition. Algebraically, it is expressed as a(b + c) = ab + ac. For example, 23(27 + 1) = 23 × 27 + 23. — The distributive property relates multiplication and addition by showing that multiplying a number by a sum is the same as multiplying the number by each addend separately and then adding the results. This property helps simplify expressions and calculate products efficiently.

Yet Things

Yet Things — Study Notes

Some Properties of Multiplication

Fast Multiplications Using the Distributive Property

Special Cases of the Distributive Property

Practice Questions — Yet Things

All 7 Chapters in Ganita Prakash Part-I