Solved Questions on Algebraic Expressions
1. What is the area of a square field if the length of each side is (ax + by − c) units?
a) [a2x2 + b2y2 + 2abxy + 2acx − 2bcy + c2] sq. units
b) [a2x2 + b2y2 + 2abxy − 2acx − 2bcy + c2] sq. units
c) [a2x2 + b2y2 + 2abxy − 2acx + 2bcy + c2] sq. units
d) [a2x2 + b2y2 − 2abxy + 2acx − 2bcy + c2] sq. units
Answer: b) [a2x2 + b2y2 + 2abxy − 2acx − 2bcy + c2] sq. units
Explanation: Length of each side = (ax + by − c) units
Area of a square field = (Length of each side)2
= (ax + by − c)2
= (ax + by − c) × (ax + by − c)
= ax (ax + by − c) + by (ax + by − c) − c (ax + by − c)
= a2x2 + abxy − acx + abxy + b2y2 − bcy − acx − bcy + c2
= a2x2 + b2y2 + abxy + abxy − acx − acx − bcy − bcy + c2
= a2x2 + b2y2 + (1 + 1)abxy − (1 + 1)acx − (1 + 1)bcy + c2
= [a2x2 + b2y2 + 2abxy − 2acx − 2bcy + c2] sq. units
2. If the length of a rectangle is (4x – 13y) units and the perimeter is (8x + 128y) units, what is the breadth of the rectangle?
a) 51y units
b) (x + 51y) units
c) 77y units
d) (x + 77y) units
Answer: c) 77y units
Explanation: Perimeter of the rectangle = (8x + 128y) units
⇒ 2(Length + Breadth) = 8x + 128y
⇒ 2(4x – 13y + Breadth) = 8x + 128y
⇒ 4x – 13y + Breadth =
⇒ 4x – 13y + Breadth = 4x + 64y
⇒ Breadth = 4x + 64y – 4x + 13y
⇒ Breadth = 77y units
3. In an auditorium, (p + t) seats are arranged in (p − t) rows. How many seats are there in the auditorium?
a) p2 − 2pt − t2
b) p2 + 2pt − t2
c) p2 − t2
d) t2 − p2
Answer: c) p2 − t2
Explanation: Number of seats = p + t
Number of rows = p − t
Total number of seats = Number of seats × Number of rows
= (p + t)(p − t)
= p(p − t) + t (p − t)
= p2 − pt + pt − t2
= p2 − t2
4. What is the cost of (q − 3) cups if the cost of q cups is $(11q3 − 9pq)?
a) $(11q3 + 33q2 − 9pq + 27p)
b) $(11q3 − 33q2 + 9pq + 27p)
c) $(11q3 + 33q2 − 9pq + 27p)
d) $(11q3 − 33q2 − 9pq + 27p)
Answer: d) $(11q3 − 33q2 − 9pq + 27p)
Explanation: Cost of q cups = $(11q3 − 9pq)
Cost of 1 cup = $[(11q3 − 9pq)/q] = $(11q2 − 9p)
Cost of (q − 3) cups = (q − 3) × (11q2 − 9p)
= q × (11q2 − 9p) − 3 × (11q2 − 9p)
= 11q3 − 9pq − 33q2 + 27p
= $(11q3 − 33q2 − 9pq + 27p)
5. What must be added to 3y3 – 5y2 + 4y + 7 to get 7y3 + 9y – 13?
a) y(4y2 – 5y – 5) + 20
b) y(4y2 + 5y + 5) – 20
c) y(4y2 + 5y – 5) – 20
d) y(4y2 + 5y – 5) + 20
Answer: b) y(4y2 + 5y + 5) – 20
Explanation: Required algebraic expressions = (7y3 + 9y – 13) − (3y3 – 5y2 + 4y + 7)
= 7y3 + 9y – 13 − 3y3 + 5y2 − 4y − 7
= 7y3 − 3y3 + 5y2 + 9y − 4y – 13 − 7
= 4y3 + 5y2 + 5y – 20
= y(4y2 + 5y + 5) – 20
