Samacheer Kalvi 7th Maths (Term 2) Information Processing Exercise Solutions
How to Use This Page
This page provides complete, step-by-step solutions for every exercise in Chapter 5 — Information Processing of the Samacheer Kalvi 7th Maths Term 2 textbook. Work through each problem on your own first, then check your working against the solutions below. Focus on understanding why each step is taken, not just the final answer.
Key Concepts (Quick Recap)
- Linear relation: A relationship of the form y = mx + c, where m is the rate of change (slope) and c is the starting value (y-intercept)
- Rate of change (m): The constant change in y for every unit increase in x; found by finding the difference in y-values divided by the difference in x-values
- Arithmetic sequence: A sequence where each term increases by a constant difference (d); the nth term is a + (n − 1)d
- Pascal's triangle: A triangular arrangement of numbers where each number is the sum of the two numbers directly above it; starts with 1 at the top
- Row sum property: The sum of numbers in row n of Pascal's triangle equals 2<sup>n−1</sup>
- Binomial expansion: (a + b)<sup>n</sup> can be expanded using the coefficients from row (n + 1) of Pascal's triangle
- Combination notation: Each element of Pascal's triangle at row n, position r is C(n, r) = n! / (r! × (n − r)!)
Exercise 5.1 — Linear Relations and Patterns
Q1. Find the relationship between x and y:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | −5 | −10 | −15 | −20 | −25 |
Step 1: Check the rate of change. As x increases by 1, y changes by −5 each time:
- From x = 1 to x = 2: y changes from −5 to −10 (change = −5)
- From x = 2 to x = 3: y changes from −10 to −15 (change = −5)
- From x = 3 to x = 4: y changes from −15 to −20 (change = −5)
The rate of change is constant at m = −5.
Step 2: Find the y-intercept (value of y when x = 0).
From the pattern, if we go back from x = 1, y = −5, adding 5 gives y = 0 when x = 0.
So c = 0.
Step 3: Write the relation: y = −5x
Step 4: Verify: x = 1 → y = −5(1) = −5; x = 2 → y = −5(2) = −10; x = 3 → y = −5(3) = −15.
All values match.
Q2. Find the relationship between x and y:
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y | 2 | 5 | 8 | 11 | 14 |
Step 1: Check the rate of change. As x increases by 1, y increases by 3 each time:
- From x = 0 to x = 1: y changes from 2 to 5 (change = +3)
- From x = 1 to x = 2: y changes from 5 to 8 (change = +3)
- From x = 2 to x = 3: y changes from 8 to 11 (change = +3)
The rate of change is constant at m = 3.
Step 2: Find the y-intercept (value of y when x = 0).
From the table, when x = 0, y = 2. So c = 2.
Step 3: Write the relation: y = 3x + 2
Step 4: Verify: x = 0 → y = 3(0) + 2 = 2; x = 1 → y = 3(1) + 2 = 5; x = 4 → y = 3(4) + 2 = 14.
All values match.
Q3. Find the next three terms: 2, 5, 8, 11, 14, ...
Step 1: Find the common difference.
5 − 2 = 3
8 − 5 = 3
11 − 8 = 3
14 − 11 = 3
Common difference d = 3.
Step 2: Add the common difference to the last given term to find each subsequent term.
Next term after 14: 14 + 3 = 17
Next term: 17 + 3 = 20
Next term: 20 + 3 = 23
Q4. A taxi charges Rs 20 for the first km and Rs 12 for each additional km. Write the relation and find the charge for 10 km.
Step 1: Express the charge as a function of distance.
Total charge = Charge for first km + Charge for remaining km
The first km costs Rs 20. For the remaining (x − 1) km, the charge is Rs 12 per km.
Charge = 20 + 12(x − 1)
Step 2: Simplify.
Charge = 20 + 12x − 12 = 12x + 8
So the relation is: Charge = 12x + 8, where x is the total distance in km.
Step 3: Find the charge for 10 km.
Charge = 12(10) + 8 = 120 + 8 = Rs 128
Verification: First km = Rs 20. Remaining 9 km at Rs 12 each = 9 × 12 = Rs 108. Total = 20 + 108 = Rs 128. Correct.
Exercise 5.2 — Pascal's Triangle
Q1. Find the sum of elements of each row of Pascal's triangle and identify the relationship.
Let us write out Pascal's triangle and find the row sums:
| Row | Elements | Sum |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 1, 1 | 2 |
| 3 | 1, 2, 1 | 4 |
| 4 | 1, 3, 3, 1 | 8 |
| 5 | 1, 4, 6, 4, 1 | 16 |
| 6 | 1, 5, 10, 10, 5, 1 | 32 |
| 7 | 1, 6, 15, 20, 15, 6, 1 | 64 |
The row sums are: 1, 2, 4, 8, 16, 32, 64, ...
These are powers of 2:
- Row 1: 2<sup>0</sup> = 1
- Row 2: 2<sup>1</sup> = 2
- Row 3: 2<sup>2</sup> = 4
- Row 4: 2<sup>3</sup> = 8
- Row 5: 2<sup>4</sup> = 16
- Row 6: 2<sup>5</sup> = 32
- Row 7: 2<sup>6</sup> = 64
Relationship: The sum of elements in row n of Pascal's triangle = 2<sup>n−1</sup>
Q2. Construct Pascal's triangle up to Row 7.
Pascal's triangle is built by starting with 1 at the top. Each subsequent row starts and ends with 1, and each interior number is the sum of the two numbers directly above it.
| Row | Pascal's Triangle |
|---|---|
| 1 | 1 |
| 2 | 1   1 |
| 3 | 1   2   1 |
| 4 | 1   3   3   1 |
| 5 | 1   4   6   4   1 |
| 6 | 1   5   10   10   5   1 |
| 7 | 1   6   15   20   15   6   1 |
Verification of Row 7: Each number is the sum of two numbers above it:
- 1 (boundary)
- 6 = 1 + 5
- 15 = 5 + 10
- 20 = 10 + 10
- 15 = 10 + 5
- 6 = 5 + 1
- 1 (boundary)
Q3. In Pascal's triangle, what is the sum of numbers in the 8th row?
Using the relationship that the sum of row n equals 2<sup>n−1</sup>:
Sum of Row 8 = 2<sup>8−1</sup> = 2<sup>7</sup> = 128
Verification: Row 8 of Pascal's triangle is: 1, 7, 21, 35, 35, 21, 7, 1
Sum = 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128. Correct.
Exercise 5.3 — Pascal's Triangle Applications
Q1. Find the missing numbers in Pascal's triangle:
(i) The partial row given is: 9, 1, ___
This appears to be referencing Pascal's triangle elements. Let us identify the row context.
Looking at Row 10 of Pascal's triangle: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1
The element after 9 (second element) in Row 10 is 36.
But let us also consider: Row 11 is 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1
The context given mentions "45, 1, 11" suggesting Row 11 elements.
Looking at Row 11: the third element is 45 (= 10 + 35 from row above).
Actually, let me reconsider. The question provides partial context: "9, 1, ___ = 45, 1, 11"
Row 11 is: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1
The missing number in the context of Row 10/11 is 45.
(ii) The partial row given references: 1, 6, ___
Row 6 of Pascal's triangle is: 1, 5, 10, 10, 5, 1
Row 7 is: 1, 6, 15, 20, 15, 6, 1
The element after 6 in Row 7 is 15. But the context mentions "1, 4, 15" suggesting a different reference.
Looking more carefully, Row 5 is: 1, 4, 6, 4, 1
The element between 1 and 4 (or in the context of the question between 6 and what follows) in Row 6 is 10.
The missing number is 10.
(iii) The partial row references: 21, 8, 1 = ___, 6, 28
Row 8 of Pascal's triangle is: 1, 7, 21, 35, 35, 21, 7, 1
Row 9 is: 1, 8, 28, 56, 70, 56, 28, 8, 1
Looking at Row 9: the elements are 1, 8, 28, 56, ...
The context "21, 8, 1" refers to the right portion of Row 8, and "___, 6, 28" appears to reference a different row.
The missing number at the beginning of Row 9 (which starts with 1) or in the appropriate context is 1.
Q2. Using Pascal's triangle, expand (a + b)<sup>3</sup>
Step 1: Identify the correct row of Pascal's triangle.
For (a + b)<sup>n</sup>, we use Row (n + 1) of Pascal's triangle.
For (a + b)<sup>3</sup>, n = 3, so we use Row 4: 1, 3, 3, 1
Step 2: Write the expansion using the coefficients.
The coefficients are: 1, 3, 3, 1
(a + b)<sup>3</sup> = 1·a<sup>3</sup>b<sup>0</sup> + 3·a<sup>2</sup>b<sup>1</sup> + 3·a<sup>1</sup>b<sup>2</sup> + 1·a<sup>0</sup>b<sup>3</sup>
= a<sup>3</sup> + 3a<sup>2</sup>b + 3ab<sup>2</sup> + b<sup>3</sup>
Q3. Using Pascal's triangle, expand (x + 1)<sup>4</sup>
Step 1: Identify the correct row of Pascal's triangle.
For (x + 1)<sup>4</sup>, n = 4, so we use Row 5: 1, 4, 6, 4, 1
Step 2: Write the expansion using the coefficients.
The coefficients are: 1, 4, 6, 4, 1
(x + 1)<sup>4</sup> = 1·x<sup>4</sup>(1)<sup>0</sup> + 4·x<sup>3</sup>(1)<sup>1</sup> + 6·x<sup>2</sup>(1)<sup>2</sup> + 4·x<sup>1</sup>(1)<sup>3</sup> + 1·x<sup>0</sup>(1)<sup>4</sup>
= x<sup>4</sup> + 4x<sup>3</sup> + 6x<sup>2</sup> + 4x + 1
Multiple Choice Questions (MCQs)
Q1. The sum of elements in the 6th row of Pascal's triangle is:
(a) 16   (b) 64   (c) 32   (d) 8
Sum of row n = 2<sup>n−1</sup>
Sum of Row 6 = 2<sup>6−1</sup> = 2<sup>5</sup> = 32
Q2. In the pattern 3, 7, 11, 15, 19, ..., the next term is:
(a) 21   (b) 23   (c) 25   (d) 27
Common difference = 7 − 3 = 4
Next term = 19 + 4 = 23
Q3. If y = 4x + 1, then when x = 3, y equals:
(a) 11   (b) 12   (c) 14   (d) 13
y = 4x + 1
When x = 3: y = 4(3) + 1 = 12 + 1 = 13
Q4. Pascal's triangle Row 4 is:
(a) 1, 3, 3, 1   (b) 1, 2, 1   (c) 1, 4, 6, 4, 1   (d) 1, 1
Row 1: 1 Row 2: 1, 1 Row 3: 1, 2, 1 Row 4: 1, 3, 3, 1
Q5. The missing number in Pascal's triangle row "1, 4, ___, 4, 1" is:
(a) 4   (b) 6   (c) 8   (d) 10
This is Row 5 of Pascal's triangle: 1, 4, 6, 4, 1
The middle number is 6 (it equals 3 + 3 from the row above: Row 4 is 1, 3, 3, 1).
True or False
1. Each number in Pascal's triangle is the sum of the two numbers directly above it.
2. The sum of each row in Pascal's triangle is a power of 2.
3. In a linear relation y = mx + c, m represents the y-intercept.
4. The first and last number in each row of Pascal's triangle is always 1.
5. Pascal's triangle has 10 rows total.
Fill in the Blanks
1. The sum of the 5th row of Pascal's triangle is ___
2. In the relation y = 3x + 2, when x = 4, y = ___
3. The next term in 1, 4, 9, 16, 25, ... is ___
This is a sequence of perfect squares: 1<sup>2</sup>, 2<sup>2</sup>, 3<sup>2</sup>, 4<sup>2</sup>, 5<sup>2</sup>, ...
The next term is 6<sup>2</sup> = 36.
4. Row 3 of Pascal's triangle is ___
5. Each element of Pascal's triangle can be expressed as ___
Common Mistakes
1. Confusing the row numbering in Pascal's triangle. Row 1 is just "1", not "1, 1". Row 2 is "1, 1". Row 3 is "1, 2, 1". Always start counting from the single 1 at the top as Row 1.
2. Adding incorrectly when constructing Pascal's triangle. Each interior number must equal the sum of the two numbers directly above it from the previous row. Common errors include adding the wrong pair or skipping a number.
3. Forgetting to check linear relations by substituting at least two pairs of values. After finding y = mx + c, always verify by substituting at least two (x, y) pairs from the table to confirm the formula is correct.
4. Mixing up the slope and y-intercept in y = mx + c. The coefficient m is the slope (rate of change, or how much y increases per unit of x). The constant c is the y-intercept (the value of y when x = 0).
5. Not recognizing that the sum of each row equals 2<sup>n−1</sup>. This is a key property for quick verification and for answering exam questions about row sums without writing out the entire row.
6. Confusing arithmetic sequences with geometric sequences. In an arithmetic sequence, each term increases by a constant difference (e.g., 2, 5, 8, 11). In a geometric sequence, each term is multiplied by a constant ratio (e.g., 3, 6, 12, 24). Make sure you identify which type of sequence you are working with.
7. Making calculation errors in pattern continuation. Always double-check the common difference or ratio before computing subsequent terms. A small error early in the pattern propagates through all later terms.
Exam Checklist
Before your exam, make sure you can confidently:
- [ ] Find linear relations from tables using y = mx + c (identify m as rate of change and c as y-intercept)
- [ ] Construct Pascal's triangle up to any given row
- [ ] Calculate the sum of any row of Pascal's triangle using 2<sup>n−1</sup>
- [ ] Find missing numbers in Pascal's triangle by adding adjacent numbers from the row above
- [ ] Use Pascal's triangle coefficients for binomial expansions like (a + b)<sup>n</sup>
- [ ] Identify and continue number patterns (arithmetic and geometric sequences)
- [ ] Distinguish between arithmetic sequences (constant difference) and other patterns
- [ ] Verify linear relations by substituting known values
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