CBSE Class 7 Mathematics Expressions using Letter Numbers Assignment Set 01

Short Answer Questions

Question 1. What is meant by a letter-number? What is an algebraic expression? Write one algebraic expression of your own and name the letter-number used in it.
Answer:
A letter that stands for a number is called a letter-number. Any letter may be chosen, and it can take different values.
An expression that contains letter-numbers is called an algebraic expression.
Example: a + 3. Here a is the letter-number; when a = 10, the expression has the value 13.
In simple words: A letter-number stands for a number, and an expression with such letters is algebraic.

Teacher's Note:
1. Give your own example, not one from the book.
2. Show one value of your expression too.

Question 2. Ravi is 5 years younger than his sister Nita. Taking t for Nita's age in years and r for Ravi's age in years:
(i) Write an algebraic expression for Ravi's age in terms of t.
(ii) Write an algebraic expression for Nita's age in terms of r.
Also find Ravi's age when Nita is 19 years old.
Answer:
Ravi is 5 years younger, so Nita is 5 years older than Ravi.
(i) Ravi's age: r = t − 5.
(ii) Nita's age: t = r + 5.
When Nita is 19: r = 19 − 5 = 14. So Ravi is 14 years old.
In simple words: Subtract 5 for the younger one, add 5 for the older one.

Teacher's Note:
1. The same relation can be read both ways.
2. Check: 14 + 5 = 19.

Question 3. Using a letter of your own choice, write an algebraic expression for each of the following and state what your letter stands for.
(i) 7 more than a number
(ii) 9 less than 4 times a number
(iii) 4 less than 9 times a number
Answer:
Let the letter n stand for the number.
(i) 7 more than the number: n + 7.
(ii) 9 less than 4 times the number: 4n − 9.
(iii) 4 less than 9 times the number: 9n − 4.
(ii) and (iii) are different expressions, because '4 times' and '9 times' cannot be swapped.
In simple words: Choose a letter, then turn the words into + , − and × step by step.

Teacher's Note:
1. Always say what your letter stands for.
2. 'Less than' means subtract from the second quantity.

Question 4. Write each arithmetic expression below as a sum of terms and then find its value.
(i) 23 − 5 × 4
(ii) 60 − (15 + 12)
Answer:
Each expression is read as a sum of terms, and the sign in front of a term goes with it.
(i) 23 − 5 × 4 = 23 + (−5 × 4) = 23 + (−20) = 3.
(ii) 60 − (15 + 12) = 60 + (−15) + (−12) = 60 − 27 = 33.
In simple words: Turn each subtraction into adding a negative term, then add.

Teacher's Note:
1. 5 × 4 is one term, so multiply first.
2. The minus before the bracket goes to both 15 and 12.

Question 5. Look at the algebraic expression 7m − 4n + 9. Write down (i) all its terms, (ii) the coefficient of m and the coefficient of n, (iii) the constant term, and (iv) the letter-numbers used in it.
Answer:
(i) Terms: 7m, −4n and 9.
(ii) Coefficient of m is 7; coefficient of n is −4.
(iii) Constant term: 9.
(iv) Letter-numbers used: m and n.
In simple words: Split the expression at the signs, keeping each sign with its term.

Teacher's Note:
1. The coefficient of n is −4, not 4.
2. A constant term has no letter.

Question 6. Why do we write 4 × n simply as 4n? State the usual way of writing such products, and rewrite these without the multiplication symbol: 4 × n, y × 15, 3 × (p + 2).
Answer:
In algebra the multiplication sign between a number and a letter-number is left out, because a letter can never be mistaken for a digit. It also saves writing.
The usual practice is to write the number first and then the letter.
4 × n = 4n; y × 15 = 15y; 3 × (p + 2) = 3(p + 2).
In simple words: Drop the × sign and write the number before the letter.

Teacher's Note:
1. Write 15y, not y15.
2. We cannot drop × between two numbers: 4 × 5 is not 45.

Question 7. Find the value of each expression for the number given.
(i) 6t − 5 when t = 4
(ii) 3(u + 2) when u = 5
Answer:
Replace the letter-number by the given number and work it out.
(i) 6t − 5 = 6 × 4 − 5 = 24 − 5 = 19.
(ii) 3(u + 2) = 3 × (5 + 2) = 3 × 7 = 21.
In simple words: Put the number in place of the letter and calculate.

Teacher's Note:
1. Work out the bracket first in (ii).
2. Write the × sign back when you substitute.

Question 8. From the terms 6a, 4b, −2a, 9, a, pick out the like terms and say why they are called like terms. Then simplify 8c + 3c − 4c and check your answer by taking c = 2.
Answer:
The like terms are 6a, −2a and a, because they contain the same letter-number a. 4b has a different letter and 9 has no letter, so they are unlike.
8c + 3c − 4c = (8 + 3 − 4)c = 7c.
Check with c = 2: 16 + 6 − 8 = 14 and 7 × 2 = 14. Both are equal.
In simple words: Like terms share the same letter, so their numbers can be combined.

Teacher's Note:
1. The term a has coefficient 1.
2. Always check by putting a small value.

Question 9. The perimeter of a rectangle of length l and breadth b can be written as l + b + l + b. Explain in two or three steps why this simplifies to 2l + 2b, and check that both expressions give the same value when l = 9 and b = 5.
Answer:
Terms can be added in any order, so group the two l terms and the two b terms:
l + b + l + b = l + l + b + b.
l + l = 2l and b + b = 2b, so the perimeter is 2l + 2b.
Check with l = 9, b = 5: 9 + 5 + 9 + 5 = 28 and 2 × 9 + 2 × 5 = 18 + 10 = 28. Equal.
In simple words: Put the same letters together and count them.

Teacher's Note:
1. Swapping terms does not change a sum.
2. 2l + 2b can also be written 2(l + b).

Question 10. Simplify (9x + 5y) − (4x + 2y), showing how the bracket is removed. Verify your simplified expression by putting x = 3 and y = 2 in both forms.
Answer:
A minus sign before a bracket changes the sign of every term inside it:
(9x + 5y) − (4x + 2y) = 9x + 5y − 4x − 2y.
Grouping like terms: (9 − 4)x + (5 − 2)y = 5x + 3y.
Check with x = 3, y = 2: original = (27 + 10) − (12 + 4) = 37 − 16 = 21; simplified = 15 + 6 = 21. Equal.
In simple words: Flip the signs inside the second bracket, then combine like terms.

Teacher's Note:
1. Both 4x and 2y become negative.
2. Matching values confirm the work.

Question 11. Simplify 4(m + 3) − m and state the property you used to remove the bracket. Check your answer by taking m = 5.
Answer:
The bracket is opened using the distributive property (a multiple of a sum is the sum of the multiples).
4(m + 3) = 4m + 12.
So 4(m + 3) − m = 4m + 12 − m = (4 − 1)m + 12 = 3m + 12.
Check with m = 5: 4 × 8 − 5 = 32 − 5 = 27 and 3 × 5 + 12 = 27. Equal.
In simple words: Multiply 4 into the bracket, then take away one m.

Teacher's Note:
1. 4 multiplies both m and 3.
2. m means 1m.

Question 12. Mohit says that 7u and 7 + u are the same expression. Do you agree? Compare the values of the two expressions for u = 2 and u = 5, and write your conclusion.
Answer:
No, Mohit is not correct. 7u means 7 times u, while 7 + u means 7 added to u.
For u = 2: 7u = 14 but 7 + u = 9.
For u = 5: 7u = 35 but 7 + u = 12.
The two expressions give different values, so they are not equal.
In simple words: 7u is multiplication and 7 + u is addition, so they give different answers.

Teacher's Note:
1. Equal expressions must agree for every value.
2. One mismatch is enough to disprove.

Question 13. A tailor's machine takes 15 minutes to be set up before stitching begins, and after that each shirt takes 20 minutes to stitch.
(i) Write an expression for the total time in minutes taken to stitch x shirts.
(ii) Use it to find the time taken to stitch 6 shirts.
Answer:
The set-up time is fixed; the stitching time depends on the number of shirts.
(i) Time for x shirts = 15 + 20x minutes.
(ii) For x = 6: 15 + 20 × 6 = 15 + 120 = 135 minutes, that is 2 hours 15 minutes.
In simple words: Add the one-time set-up to 20 minutes for each shirt.

Teacher's Note:
1. The fixed part is added only once.
2. Convert to hours only if asked.

Question 14. Look at the sequence 7, 11, 15, 19, 23, …
(i) Write an algebraic expression for its nth term.
(ii) Use the expression to find the 20th term, and check that your expression gives 19 for n = 4.
Answer:
Each term is 4 more than the one before, and the first term 7 is 3 more than 4.
(i) The nth term is 4n + 3.
(ii) 20th term = 4 × 20 + 3 = 83.
Check: for n = 4, 16 + 3 = 19, which is the fourth term.
In simple words: The terms go up by 4, so start from 4n and adjust by 3.

Teacher's Note:
1. Compare the sequence with the multiples of 4.
2. Always check with a known term.

Question 15. Rhea simplified 5 − (2 − 3z) and wrote 3 − 3z. Identify her mistake, explain what went wrong, and write the correct simplest form. Check your answer by taking z = 1.
Answer:
Her mistake is in the sign of the second term inside the bracket. A minus sign before a bracket changes the sign of every term inside, so −(−3z) is +3z, not −3z.
Correct working: 5 − (2 − 3z) = 5 − 2 + 3z = 3 + 3z.
Check with z = 1: 5 − (2 − 3) = 5 − (−1) = 6 and 3 + 3 = 6. Correct.
In simple words: She changed only the first sign inside the bracket; both must change.

Teacher's Note:
1. Minus times minus gives plus.
2. Her answer gives 0 for z = 1, not 6.

Question 16. In a calendar month, a 2 × 3 block of dates is chosen (2 rows and 3 columns). The date in the bottom middle cell is w. Write expressions in terms of w for the dates in the other five cells, and explain how you got the numbers in the top row.
Answer:
In a calendar, moving one cell to the right adds 1, and moving one cell up subtracts 7.
Bottom row: w − 1, w, w + 1.
Top row: w − 8, w − 7, w − 6.
Each date in the top row is 7 less than the date just below it, because a week has 7 days. So we subtract 7 from w − 1, w and w + 1.

In simple words: Left and right change by 1; up changes by 7.

Teacher's Note:
1. Draw the block and fill it in.
2. Check with a real calendar.

Question 17. A lizard climbs the inside wall of a well. Each day it climbs up u cm and each night it slips down d cm. This goes on for 12 days and 12 nights.
(i) Write an expression for how far the lizard is above its starting point at the end of the 12th night.
(ii) What can you say about the lizard's movement if d > u?
(iii) Find the distance climbed when u = 9 and d = 4.
Answer:
In one full day and night the lizard gains u − d cm.
(i) After 12 days and nights it is 12(u − d) cm above its starting point, which can also be written as 12u − 12d.
(ii) If d > u, then u − d is negative, so the lizard slips down more each night than it climbs each day. It keeps going below its starting point and can never climb out of the well.
(iii) For u = 9 and d = 4: 12(9 − 4) = 12 × 5 = 60 cm above the start.
In simple words: Find the net gain for one day and night, then multiply by 12.

Teacher's Note:
1. Work out one day-night cycle first.
2. A negative gain means the lizard goes down.

Question 18. Simplify 3(2r + 4s + 5) − 6s completely. Then write how many terms and how many different letter-numbers your simplest form has, and state the relation you notice between these two counts. Verify the simplification for r = 1 and s = 1.
Answer:
Open the bracket first, then collect like terms.
3(2r + 4s + 5) = 6r + 12s + 15.
6r + 12s + 15 − 6s = 6r + (12 − 6)s + 15 = 6r + 6s + 15.
The simplest form has 3 terms and 2 different letter-numbers. The number of letter-numbers can never be more than the number of terms, because in the simplest form each letter-number has a term of its own (here the extra term is the constant 15).
Check with r = 1, s = 1: 3(2 + 4 + 5) − 6 = 33 − 6 = 27 and 6 + 6 + 15 = 27. Equal.
In simple words: Open the bracket, combine the s-terms, then count.

Teacher's Note:
1. The constant term is also a term.
2. Like terms are joined, so each letter appears once.

Question 19. A number machine takes two numbers a and b and gives one output. For the inputs (4, 5) it gives 7, for (6, 2) it gives 16 and for (7, 10) it gives 11. Describe the rule of the machine in words, write it as an algebraic expression, and find the output for the inputs (9, 4).
Answer:
Compare the outputs with the inputs: 3 × 4 − 5 = 7, 3 × 6 − 2 = 16 and 3 × 7 − 10 = 11. All three agree.
Rule in words: three times the first number, minus the second number.
As an expression: 3a − b.
For (9, 4): 3 × 9 − 4 = 27 − 4 = 23.
In simple words: The machine triples the first number and takes away the second.

Teacher's Note:
1. A rule must fit every pair given.
2. Try simple operations on the first number first.

Question 20. Shreya claims that the sum of any three consecutive whole numbers is always 3 times the middle number. Take the middle number to be a, write the other two numbers in terms of a, and use algebra to show that her claim is always true. Check it once with the numbers 14, 15 and 16.
Answer:
Let the middle number be a. The number before it is a − 1 and the number after it is a + 1.
Sum = (a − 1) + a + (a + 1) = a − 1 + a + a + 1.
= (a + a + a) + (−1 + 1) = 3a + 0 = 3a.
Since a stands for any number, the sum is always 3 times the middle number. Shreya's claim is true.
Check: 14 + 15 + 16 = 45 and 3 × 15 = 45.
In simple words: The −1 and +1 cancel, leaving three middle numbers.

Teacher's Note:
1. Algebra proves the claim for all numbers at once.
2. One example is only a check.