## MP Board Class 6th Maths Solutions Chapter 11 Algebra Ex 11.4

Question 1.

Answer the following:

(a) Take Sarita’s present age to be y years

(i) What will be her age 5 years from now?

(ii) What was her age 3 years back?

(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?

(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?

(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?

(b) The length of a rectangular hall is 4 metres less than 3 times the breadth of the hall. What is the length, if the breadth is b metres?

(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.

(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.

Solution:

(a) Sarita’s present age = y years

(i) After 5 years, her age will be (y + 5) years.

(ii) 3 years ago, her age was (y – 3) years.

(iii) Age of her grandfather = 6 × (Sarita’s age) = (6 × y) years = 6y years

(iv) Age of her grandmother = age of her grandfather – 2 years = (6y – 2) years

(v) Age of her father = 3 (Sarita’s age) + 5 years = (3y + 5) years

(b) Breadth of the hall = b m

Length of the hall = 3 (breadth) – 4m = 3b m – 4 m = (3b – 4) m

(c) Height of the box = h cm

Length of the box = 5 × height = 5 h cm

Breadth of the box = Length – 10 cm = (5h – 10) cm

(d) Meena’s position = s

Beena’s position = 8 steps ahead = s + 8

Leena’s position = 7 steps behind = s – 7

∴ Total number of steps = 4s – 10

(e) Speed of the bus = v km/h

Distance travelled in 5 hours = 5v km

Remaining distance = 20 km

Therefore, total distance = (5v + 20) km

Question 2.

Change the following statements using expressions into statements in ordinary language.

(For example. Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim.)

(a) A notebook costs ₹ p. A book costs ₹ 3 p.

(b) Tony puts q marbles on the table. He has 8 q marbles in his box.

(c) Our class has n students. The school has 20n students.

(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.

(e) In an arrangement of dots there are r rows. Each row contains 5 dots.

Solution:

(a) A book costs 3 times the cost of a notebook.

(b) The number of marbles in Tony’s box is 8 times the marbles on the table.

(c) Total number of students in the school is 20 times the number of students in our class.

(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.

(e) The total number of dots is 5 times the number of rows.

Question 3.

(a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?

{Hint: Think of Munnu’s younger brother.)

Can you guess what (x + 4) may show? What (3x+ 7) may show?

(b) Given Sara’s age today to be y years. Think of her age in the future or in the past.

What will the following expression indicate? y + 7, y – 3, y + \(4 \frac{1}{2}\), y – \(2 \frac{1}{2}\).

(c) Given n students in the class like football, what may 2n show? What may \(\frac{n}{2}\) show?

(Hint: Think of games other than football).

Solution:

(a) Munnu’s age = x years Age of his. younger brother who is 2 years younger than him = (x – 2) years

Age of his elder brother who is 4 years elder than him = (x + 4) years

Age of his father whose age is 7 years more than thrice of his age = (3x + 7) years

(b) Sara’s present age = y years

Her age in past = (y – 3), (y – \(2 \frac{1}{2}\))

Her age in future = (y + 7), (y + \(4 \frac{1}{2}\))

(c) Number of students like football = n

Number of students like hockey is twice the number of students liking football,

i. e., 2n

Number of students like tennis is half the number of students liking football,

i.e., \(\frac{n}{2}\)