Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution

Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution . We provide step by step Solutions of lesson-4 Rational Numbers for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan . Our Solutions contain all type Questions of Exe-4 F to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics. How do you divide rational numbers? What is the definition of division in rational numbers of class-7th / 6th?

we mentioned that multiplying by a number’s reciprocal is the same as dividing by the number. That’s how we can divide rational numbers; to divide by a rational number, just multiply by that number’s reciprocal.

Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution

Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution

Board ICSE
Publications Goyal brothers Prakashan
Subject Maths
Class 7th
Chapter-4 Rational Numbers
Writer RS Aggrawal
Book Name Foundation
Topics Solution of Exe-4 F
Academic Session 2023 – 2024

Exercise – 4 F

Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution

1. Find the multiplicative inverse (or reciprocal) of each of the following rational numbers :

(i) (6/25)

Solution: (6/25)

= (25/6)

(ii) -2(3/11)

Solution: -2(3/11)

= (-25/11)

= (-11/25)

(iii) (-8/9)

Solution: (-8/9)

= (-9/8)

(iv) (-23/16)

Solution: (-23/16)

= (-16/23)

(v) 12

Solution: 12

= (1/12)

(vi) (1/10)

Solution: (1/10)

= 10

(vii) -6

Solution: -6

= (-1/6)

(viii) -1

Solution: -1

= (-1/1)

= 1

(ix) (-1/5)

Solution: (-1/5)

= -5

(x) (-7/-9)

Solution: (-7/-9)

= (9/7)

2. Evaluate :

(i) (7/12) ÷ (-4/3)

Solution: (7/12) ÷ (-4/3)

= (7/12) × (3/-4)

= (7/-16)

(ii) (-12/25) ÷ (-5/6)

Solution: (-12/25) ÷ (-5/6)

= (-12/25) × (6/-5)

= (-72/-125)

= (72/125)

(iii) (-27/32) ÷ (-9/16)

Solution: (-27/32) ÷ (-9/16)

= (-27/32) × (16/-9)

= (-3/-2)

= (3/2)

(iv) -2(4/7) ÷ (6/35)

Solution: -2(4/7) ÷ (6/35)

= -2(4/7) × (35/6)

= (-18/7) × (35/6)

= -3 × 5

= -15

(v) 26 ÷ (-1/13)

Solution: 26 ÷ (-1/13)

= (26/1) ÷ (13/-1)

= -338

(vi) (1/25) ÷ (-5)

Solution: (1/25) ÷ (-5)

= (1/25) × (1/-5)

= (-1/125)

3. The product of two rational numbers is (2/5). If one of them is (-8/25), find the other.

Solution: Product of two rational no. (2/5)

One no. = (-8/25)

Let the other no. be x.

According to questions –

(-8/25) × x = (2/5)

x = (2/5) × (-25/8)

x = (-5/4)

Other no. is (-5/4).

4. The product of two rational numbers is (-2/3). If one of them is (16/39), find the other.

Solution: Product of two rational no. (-2/3)

One no. = (16/39)

Let the other no. be x.

According to questions –

(16/39) × x = (-2/3)

x = (-2/3) × (39/16)

x = (-13/8)

Other no. is (-13/8).

5. By what rational number should (-9/35) be multiplied to get (3/5)?

[Hint. (-9/35) × Other number = (3/5) ⇒ Other number = (3/5) ÷ (-9/35)]

Solution: Product of two rational no. (3/5)

One no. = (-9/35)

Let the other no. be x.

According to questions –

(-9/35) × x = (3/5)

x = (3/5) × (-35/9)

x = (-7/3)

Other no. is (-7/3).

6. By what rational number should (25/8) be multiplied to get (-20/7)?

Solution: Product of two rational no. (-20/7)

One no. = (25/8)

Let the other no. be x.

According to questions –

(25/8) × x = (-20/7)

x = (-20/7) × (8/25)

x = (-32/35)

Other no. is (-32/35).

7. The  cost of 17 pencils is ₹59(1/2). Find the cost of each pencil.

Solution: Cost of 17 pencil = ₹59(1/2)  = (119/2)

So, cost of 1 pencil = (119/2) ÷ 17

= (119/2) × (1/17)

= (7/2)

= 3(1/2)

Hance, cost of 1 pencil = ₹3(1/2).

8. The cost of 20 meters of ribbon is ₹335. Find the cost of each meter of it.

Solution: Cost of 20 meters of ribbon = ₹335

So, cost of 1 meter = ₹335 ÷ 20

= 335 × (1/20)

= (67/4)

Cost of 1 meter ribbon = ₹(67/4)

9. How many pieces, each of length 2(3/4) m, can be cut from a rope of length 66 m?

[Hint. Number of pieces = 66 ÷ (11/4)]

Solution: Total length  = 66

Length of each pieces = 2(3/4) = (11/4) m

So, 66 ÷ (11/4)

= 66 × (4/11)

= 24 pieces

10. Fill in the blanks :

(i) (….25….) ÷ (-5/6) = -30

(ii) (….6….) ÷ (-8) = (-3/4)

(iii) (-15/14) ÷ (….(-3/7)….) = (5/2)

(iv) (-16) ÷ (….(-8/3)….) = 6

— : end of Rational Numbers Class- 7th RS Aggarwal Exe-4 F Goyal Brothers ICSE Math Solution:–

Return to- ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions

Thanks

Please share with yours friends if you find it helpful. 

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.