**Rational Numbers** Class- 7th RS Aggarwal Exe-4 E 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 E to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-7 Mathematics. in this post we would learn about Reciprocal or Multiplicative Inverse of a Rational number such as What is the multiplicative inverse of rational numbers? Is the multiplicative inverse of a rational number.

## Rational Numbers Class- 7th RS Aggarwal Exe-4 E Goyal Brothers ICSE Maths 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 E |

Academic Session | 2023 – 2024 |

**Exercise – 4 E**

Rational Numbers Class- 7th RS Aggarwal Goyal Brothers ICSE Maths Solution

**1. Multiply :**

**(i) (2/3) by (4/5)**

**Solution:** (2/3) by (4/5)

(2/3) × (4/5)

= (8/15)

**(ii) (7/6) by (9/2)**

**Solution:** (7/6) by (9/2)

(7/6) × (9/2)

= (63/12)

**(iii) (5/6) by 30**

**Solution:** (5/6) by 30

(5/6) × 30

= 25

**(iv) (-3/4) by (8/7)**

**Solution:** (-3/4) by (8/7)

(-3/4) × (8/7)

= (-6/7)

**(v) (-16/9) by (12/-5)**

**Solution:** (-16/9) by (12/-5)

(-16/9) × (12/-5)

= (-64/-15)

= (64/15)

**(vi) (35/-8) by (12/-5)**

**Solution:** (35/-8) by (12/-5)

(35/-8) × (12/-5)

= (21/2)

**(vii) (-3/10) by (-40/9)**

**Solution:** (-3/10) by (-40/9)

(-3/10) × (-40/9)

= (12/9)

= (4/3)

**(viii) (-32/5) by (15/-16)**

**Solution:** (-32/5) by (15/-16)

(-32/5) × (15/-16)

= (-6/-1)

= 6

**(ix) (-8/15) by (-25/32)**

**Solution:** (-8/15) by (-25/32)

(-8/15) × (-25/32)

= (5/12)

**2. Simplify :**

**(i) (7/15) × (5/6)**

**Solution:** (7/15) × (5/6)

= (7/18)

**(ii) (-5/24) × (6/25)**

**Solution:** (-5/24) × (6/25)

= (-1/20)

**(iii) (7/-18) × (-9/14)**

**Solution:** (7/-18) × (-9/14)

= (-1/-4)

= (1/4)

**(iv) (-9/5) × (-10/3)**

**Solution:** (-9/5) × (-10/3)

= (-3 × (-2))

= 6

**(v) (-28) × (-8/7)**

**Solution:** (-28) × (-8/7)

= (-4 × (-8)

= 32

**(vi) (8/-21) × (-14/3)**

**Solution:** (8/-21) × (-14/3)

= (-16/-9)

= (16/9)

**3. Simplify :**

**(i) (5/12) × (-36)**

**Solution:** (5/12) × (-36)

= -15

**(ii) (-17/18) × 12**

**Solution:** (-17/18) × 12

= (-34/3)

**(iii) (-5/6) × (6/5)**

**Solution:** (-5/6) × (6/5)

= -1

**(iv) (-14) × (9/28)**

**Solution:** (-14) × (9/28)

= (-9/2)

**(v) -4(4/5) × (-7(1/2))**

**Solution:** -4(4/5) × (-7(1/2))

= (-24/5) × (-15/2)

= (-12 × (-3))

= 36

**(vi) (-8/15) × (-25/32)**

**Solution:** (-8/15) × (-25/32)

= (5/12)

**4. Simplify : **

**(i) [(2/5) × (5/8)] + [(-3/7) × (14/-15)]**

**Solution:** [(2/5) × (5/8)] + [(-3/7) × (14/-15)]

= (1/4) + (-2/-5)

= (5 + 8)/20

= (13/20)

**(ii) [(-14/3) × (-12/7)] + [(-6/25) × (15/8)]**

**Solution:** [(-14/3) × (-12/7)] + [(-6/25) × (15/8)]

= (8/1) + (-9/20)

= (8/1) – (9/20)

= (160 – 9)/20

= (151/20)

**(iii) [(6/25) × (-15/8)] – [(13/100) × (-25/26)]**

**Solution:** [(6/25) × (-15/8)] – [(13/100) × (-25/26)]

= (-9/20) – (-1/8)

= (-9/20) + (1/8)

= (-18 + 5)/40

= (-13/40)

**(iv) ((-14/5) × (-10/7)) – ((-8/9) × (3/16))**

**Solution:** [(-14/5) × (-10/7)] – [(-8/9) × (3/16)]

= (4/1) – (-1/6)

= (4/1) + (1/6)

= (24 + 1)/6

= (25/6)

**5. Find the cost of 3(1/3) kg of rice at ₹40(1/2) per kg.**

**Solution:** Cost of 1 kg of rice = ₹40(1/2) = (81/2)

Cost of 3(1/3) kg of rice = (10/3)

So, (81/2) × (10/3)

= ₹135

Cost of 3(1/3) kg of rice is ₹135.

**6. Find the distance covered by a car in 2(2/5) hours at a speed of 46(2/3) km per hour.**

[**Hint.** Required distance = 46(2/3) × 2(2/5) km]

**Solution:** Speed of 1 hour = 46(2/3) = (140/3)

Distance covered by the car in 2(2/5) hour = (12/5)

So, (12/5) × (140/3)

= 112 km

Distance covered by the car in (12/5) hour =112km.

**7. Write the multiplicative inverse of :**

**The reciprocal or multiplicative inverse of a non zero rational number (a/b) is (b/a).**

**(i) (5/6)**

**Solution:** (5/6)

= (6/5)

**(ii) (-3/7)**

**Solution:** (-3/7)

= (-7/3)

**(iii) -8**

**Solution:** -8

= (-1/8)

**(iv) (-11/3)**

**Solution:** (-11/3)

= (-3/11)

**(v) (-1/8)**

**Solution:** (-1/8)

= -8

**— : end of Rational Numbers Class- 7th RS Aggarwal Exe-4 E Goyal Brothers ICSE Maths Solution:–**

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

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