MATHS 5 Marks Questions

(a) Since the cost of each food stall = ₹ x
So, the cost of 4 food stall = 4x
Since, the cost of each game stall = ₹ y
So, the cost of 3 game stall = 3y
And, the fixed decoration cost = ₹ 500
So, expression for the total cost to organise the school fair = 4x + 3y + 500.
(b) The term 4x represents cost of 4 food stalls.
(c) Since the expression for the total cost to organise the school fair is 4x + 3y + 500
So, the total cost when the cost of each food stall is ₹ 200 and each game stall is ₹ 150 is
4 × 200 + 3 × 150 + 500
= 800 + 450 + 500 = ₹ 1750.

(i) (a) Explanation: Every algebraic expression has at least one letter-numbers which is used to represent an unknown value in an expression. Hence, both assertion and reason are true.
(ii) (c) Explanation: Since, Aditi present age is of x. 5 years ago her age was 10 years, then her present age is 15 years. We can express this situation by the expression, x – 5 = 10, so the value of x is 15 years.
Hence, the assertion is true, but the reason is false.

(a) (5a – b + 3c) – {la +8b — c)
= 5a-b + 3c-la-8b + c
= 5a-la-b-8b + 3c + c
(Rearranging like terms) = (5 – 7)a + (-1 -8)b + (3 + 1)c
= -2a -9b + 4c.

(b) (z – x) – (2x – 3y) = z – x – 2x + 3y
= z – 3x + 3y

(c) (4pqr + 7p – 2p) – (3pq – 2p + q)
= 4pq + 7p – 2p – 3pq + 2p – q
= 4pq – 3pq + 7p – 2p + 2p – q
= pq + 7p – q

(d) (15xy + y-x) – (11xy — 5y + 2x)
= 15 xy + y – x- 11 xy + 5y – 2x
= 15 xy – 11xy – x – 2x + y + 5y
= 4xy – 3x + 6y

(a) Ruhi’s age after 5 years from now = Her present age + 5 years
= x years + 5 years
= (x + 5) years

(b) Ruhi’s age 3 years back = Her present age – 3 years
= x years – 3 years
= (x – 3) years

(c) Grandfather’s age = 6 × Ruhi’s age
= 6 × x years
= 6x years

(d) Grandmother’s age = Grandfather’s age – 2 years = 6x years – 2 years
= (6x – 2) years

(e) Ruhi’s father’s age = 3 × Ruhi’s age + 5 years
= 3 × x years + 5 years
= 3x years + 5 years
= (3x + 5) years

(a)The length of the knife = x cm + 4 cm = (x + 4) cm

(b)The length of the handle of the screwdriver
= x cm – 3 cm
= (x – 3) cm

(c)Perimeter of the notebook = 2(length + breadth)
= 2(4x + 2y) units
Area of the notebook = (length × breadth) sq units
= 4x × 2y sq units
= 8xy sq units

(d)Total cost of the fruits = total cost of apples + total cost of bananas
= ₹ 5 × x + ₹ 2 × y
= ₹ 5x + ₹ 2y
= ₹ (5x + 2y)

(a)In the given figure there are 4 matchsticks. To get the two similar figures we need 8 matchsticks, for three similar figures 12 matchsticks, and so on.
So, the number of matchsticks in «th figure is An.

(b)Similar as part (a).
General term for n number of figures is 5n.

(c)To get the two similar figures, we just need to add 4 more matchsticks and so on.
The general rule is 4n + 2.

(d)To get the two similar figures, we just need to add 5 more matchsticks in the given figure, and so on.
The general rule is 5n + 2.

(e)To get the two similar figures, we just need to add 4 more matchsticks in the given figure, and so on.
The general rule is 4n + 1.

(a)It can be observed from the given figures that 2 matchsticks are fixed and 3 matchsticks are added at each step. The general term for the number of matchsticks in a figure will be 3n + 2.
The complete table is

Number of figures formed	1	2	3	4	25	n
Number of matchsticks	5	8	11	14	77	3n+2

(b)By observing the matchsticks pattern, the general term for the number of matchsticks will be 7n + 1.

Number of figures formed	1	2	3	4	25	50	n
Number of matchsticks	8	15	22	29	176	351	7n+1

(c)By observing the matchsticks pattern, the genera term for the number of matchsticks will be 3n + 2.

Number of figures formed	1	2	3	4	25	100	n
Number of matchsticks	5	8	11	14	77	302	3n+2