Class 7 Maths Ganita Prakash Part 1 Chapter 4 MCQs – Expressions using Letter-Numbers

Shabnam’s age is found by adding the age difference to Aftab’s age.
Shabnam’s age = a + 3
For example, if a = 23, her age is 23 + 3 = 26 years. The letter a allows the same relationship to work for different ages.

Each L uses two sticks, so n letters need n groups of two sticks.
Number of matchsticks = 2 × n = 2n
For instance, five letters need 2 × 5 = 10 sticks. The multiplication sign may be omitted between a number and a letter.

The cost formula is 35c + 60j, where c is the number of coconuts and j is the jaggery mass in kilograms.
35 × 7 + 60 × 4
= 245 + 240
= ₹485

A regular hexagon has six equal sides. Its perimeter is the sum of all six side lengths.
Perimeter = s + s + s + s + s + s
= 6s cm
The letter s represents the length of one side, not the number of sides.

The mill starts once, so the starting time contributes a fixed 10 seconds. Grinding y kilograms takes 8y seconds.
Total time = 10 + 8y
Multiplying the starting time by y would incorrectly count a new start for every kilogram.

The expression 3d means 3 × d; it does not mean placing the digits 3 and d side by side.
When d = 6,
3d = 3 × 6 = 18
The omitted multiplication sign must be restored when substituting the number.

Replacing a with −4 gives subtraction of a negative number.
10 − a = 10 − (−4)
= 10 + 4
= 14
The brackets help keep the sign of the substituted value clear, so −4 is not mistaken for 4.

The expression inside the brackets is evaluated first.
3(m + 1) = 3(−6 + 1)
= 3 × (−5)
= −15
The factor 3 multiplies the whole bracket, so it multiplies the result −5 rather than only one part of the sum.

The first term is 4 × 1, the second is 4 × 2, and the nth term is 4n.
29th term = 4 × 29 = 116
The position number gives the number of groups of 4 needed.

A rectangle has two sides of length l and two sides of length b.
Perimeter = l + b + l + b
= l + l + b + b
= 2l + 2b
Grouping the equal side lengths gives a shorter form of the same total.

The expression s = a + 3 means Shabnam’s age is 3 more than Aftab’s. ‘s’ represents Shabnam’s age and ‘a’ represents Aftab’s age. It models the situation where Shabnam is always 3 years older than Aftab.

Replacing a with 23 in the expression s = a + 3 gives s = 23 + 3 = 26. This use of substitution helps evaluate expressions with letter-numbers by replacing the variable with a known value.

Aftab is 3 years younger than Shabnam. To find Aftab’s age from Shabnam’s age, we subtract 3. The expression becomes a = s – 3, reversing the earlier relationship between a and s.

Each L-shaped figure is made with 2 matchsticks. So if n represents the number of L’s, then 2 × n gives the total number of matchsticks needed. This is a multiplication-based expression.

Like terms contain the same letter-number part. In 12n and −4n, both terms contain n, so their numerical multipliers can be combined.
12n − 4n = 8n
Terms involving different letters, or a letter and a number alone, cannot be combined this way.

Total pencils = 5 + 3 + 10 = 18
Total erasers = 4 + 6 + 1 = 11
Revenue = 18c + 11d rupees
The pencil and eraser terms remain separate because the two prices are represented by different letter-numbers.

Initial payment = 40x + 75y
Refund = 6x + 10y
Final cost = (40x + 75y) − (6x + 10y)
= (40 − 6)x + (75 − 10)y
= 34x + 65y
The refund is deducted for every rented item.

The whole expression for Charu’s score is subtracted.
(23p − 7q) − (21p − 9q)
= 23p − 7q − 21p + 9q
= 2p + 2q
The −9q becomes +9q when the brackets after the minus sign are removed.

Original area = 12n square units
Removed area = 4n square units
Remaining area = 12n − 4n = 8n square units
This also follows from the remaining length, 12 − 4 = 8, multiplied by the unchanged width n.

Using the expression 35c + 60j and substituting c = 10 and j = 5 gives 35×10 + 60×5 = 350 + 300 = ₹650. This method applies algebraic expressions to real-life cost calculations.

A square has four equal sides. So the perimeter is 4 times the side length. Using q to represent side length, the perimeter is given by the expression 4 × q.

Perimeter of a rectangle is the sum of all sides: l + b + l + b = 2l + 2b. This algebraic expression is a simplified form using the rule of combining like terms.

All terms have the same letter-number c. So we can combine like terms: 5c + 3c + 10c = (5 + 3 + 10)c = 18c. This shows use of the distributive property in algebra.

The factor 4 multiplies both terms inside the brackets.
4(x + y) − y = 4x + 4y − y
= 4x + (4 − 1)y
= 4x + 3y
Only the y terms combine; the x term remains separate.

The minus sign before the brackets changes both signs inside when the brackets are removed.
5 − (2 − 6z) = 5 − 2 + 6z
= 3 + 6z
The numerical terms combine, while the term involving z stays as it is.

Subtracting the first expression from the second gives:
(6a + 9b − 18) − (9a − 6b + 14)
= 6a + 9b − 18 − 9a + 6b − 14
= −3a + 15b − 32
Each term in the subtracted bracket changes sign.

The expressions are being added, so their term signs remain unchanged.
4d − 7c + 9 + 8c − 11 + 9d
= (4 + 9)d + (−7 + 8)c + (9 − 11)
= 13d + c − 2

Expanding the bracket gives 10(y − 3) = 10y − 30.
Difference = (10y − 3) − (10y − 30)
= −3 + 30 = 27
The letter terms cancel, so the difference is always 27, whatever value y takes.

Area = length × width = (4 + 3) × v = 7v. This shows how expressions can describe areas by combining numeric addition and multiplication using variables.

12n and 4n are like terms. Subtracting them gives 12n – 4n = 8n. Like terms can be added or subtracted by combining their coefficients while keeping the letter-number unchanged.

Distributing the negative sign: (40x + 75y) – (6x + 10y) = 40x – 6x + 75y – 10y = (40 – 6)x + (75 – 10)y = 34x + 65y. This is simplification using removal of brackets and combining like terms.

Adding Charu’s scores from three rounds: (7p – 3q) + (8p – 4q) + (6p – 2q) = 7p + 8p + 6p – (3q + 4q + 2q) = 21p – 9q. This uses grouping and combining like terms.

If the inputs are a and b, the rule is 2a − b.
For a = 6 and b = 4,
2a − b = 2 × 6 − 4 = 12 − 4 = 8
Only the first input is doubled before the subtraction.

The B designs occur at positions 2, 5, 8, 11, … . The nth C is at position 3n, and the nth B is one place before it.
Position of the nth B = 3n − 1
For n = 1, this gives position 2.

The four entries are a, a + 1, a + 7 and a + 8.
One diagonal sum = a + (a + 8) = 2a + 8
Other diagonal sum = (a + 1) + (a + 7) = 2a + 8
Thus, the two sums agree.

Adding all five entries gives:
a + (a − 1) + (a + 1) + (a − 7) + (a + 7)
= 5a + (−1 + 1 − 7 + 7)
= 5a
The opposite changes cancel, leaving five copies of the centre number.

The first triangle needs 3 sticks. The other 32 triangles each add 2 sticks.
Total = 3 + 2(33 − 1)
= 3 + 64 = 67
The equivalent formula 2n + 1 also gives 2 × 33 + 1 = 67.

Distribute 4 to both terms: 4(x + y) = 4x + 4y. Then subtract y: 4x + 4y – y = 4x + 3y. This uses distributive property followed by combining like terms.

In algebra, 5u means 5 × u. It represents five times the number u. This form is a product of a number and a variable and omits the multiplication sign.

10y – 3 means subtracting 3 after multiplying y by 10. 10(y – 3) means multiplying the difference (y – 3) by 10. The operations differ so the values will differ except at specific values.

p + p + p = 3p, and since q is a different term, it remains unchanged. Combining them gives 3p + q. Only like terms can be combined using addition.

There is one first square and w − 1 later squares.
Number of sticks = 4 + 3(w − 1)
= 4 + 3w − 3
= 3w + 1
Shared sides are already present, so each additional square needs only three new sticks.

Each week contains seven days.
Total distance = 7 × 5 + 7(5 + z) + 7(5 + 2z)
= 35 + 35 + 7z + 35 + 14z
= 105 + 21z km
The weekly increase applies to every day of that week.

Three intermediate stations divide the journey into four travel sections.
Travelling time = 4t minutes
Stopping time = 3 × 2 = 6 minutes
Total journey time = 4t + 6 minutes
Only the three intermediate stops are included in the stated waiting time.

The three groups are separate: customers buying only champak, only marigold, or both. Each person receives one flag.
Total customers = p + q + r
Therefore, the number of flags is p + q + r. Buying both flowers does not give a customer two flags.

Each completed row contains four numbers.
147 = 4 × 36 + 3
The first 36 rows therefore contain 144 numbers. Row 37 contains 145, 146, 147 and 148, so 147 is its third entry, in column 3.

Combine like terms: 5c + 10c = 15c, then 15c – 2c = 13c. All terms have the same letter-number so they can be combined using coefficients.

Combine like terms: 2y + 3y = 5y. The constant –6 remains unchanged. The final expression is 5y – 6. This is a basic example of addition of like terms.

First simplify inside the bracket: 2 × (y – 1) = 2y – 2. Then add 3: 3 + 2y – 2 = 2y + 1. This expression gives the matchsticks needed in step y.

Group and subtract like terms: (40x – 6x) + (75y – 10y) = 34x + 65y. This is done using distributive and associative properties to remove brackets and combine similar terms.