Read RD Sharma Solutions Class 7 Chapter 5 Operation on Decimal Numbers below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations
Exercise 5.1
Question 1: Add the following rational numbers:
- ((-5)/7) and (3/7)
- ((-15)/4) and (7/4)
- (((-8)/4) and (-(-4)/11)
- ((6/13) and ((-9)/13)
Solution:
(i) ((-5)/7) and (3/7)
Denominators of the values are equal so add the numerator
= ((-5+3))/7
= (-2)/7
(ii) ((-15)/4) and (7/4)
Denominators of the values are equal so add the numerator
= ((-15+7))/4
= (-8)/4 = -2
(iii) ((-8)/4) and ((-4)/11)
Denominators of the values are equal so add the numerator
= (-8+(-4))/11
= (-12)/11
(iv) (6/13) and ((-9)/13)
Denominators of the values are equal so add the numerator
= ((6+(-9))/13)
= ((-3)/13)
Question 2: Add the following rational numbers:
- (3/4) and ((-3)/5)
- -3 and (3/5)
- ((-7)/27) and (11/18)
- (31/(-4)) and ((-5)/8)
Solution:
(i) (3/4) and ((-3)/5)
If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then
p/q + r/s = (p ×s + r × q)/(q × s )
(3/4) + ((-3)/5)
= ((3×5+(-3)×4)/20)
= (15-12)/20
= 3/20
(ii) -3 and (3/5)
If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then
p/q + r/s = (p ×s + r × q)/(q × s )
((-3)/1) + (3/5) = ((-3×5+3×1))/((1×5))
= ((-15+3)/5)
= ((-12)/5)
Question 3: Simplify:
- (8/9) + ((-11)/6)
- ((-5)/16) + (7/24)
- (1/(-12)) + (2/(-15))
- ((-8)/19) + ((-4)/57)
Solution:
Question :4. Add and express the sum as mixed fraction:
(i) ((-12)/5) + (43/10)
(ii) (24/7) + ((-11)/4)
(iii) ((-31)/6) + ((-27)/8)
Solution:
((-12)/5) + (43/10)
Take LCM of 5 and 10
Exercise 5.2
Question 1: Subtract the first rational number from the second in each of the following:
(i) (3/8), (5/8)
(ii) ((-7)/9), (4/9)
(iii) ((-2)/11), ((-9)/11)
(iv) (11/13), ((-4)/13)
Solution 1:
(i) (3/8), (5/8)
= (5/8) – (3/8) =((5-3))/8
= (2/8) = (1/4)
(ii) ((-7)/9), (4/9)
= (4/9) – ((-7)/9) = (4/9) + (7/9)
= ((4+7))/9 = (11/9)
(iii) ((-2)/11), ((-9)/11)
= ((-9)/11)– ((-2)/11) = ((-9)/11) + (2/11)
= ((-9+2))/11 = ((-7)/11)
(iv) (11/13), ((-4)/13)
= ((-4)/13) – (11/13) = ((-4-11))/13
= ((-15)/13)
Question 2: Evaluate each of the following:
(i) (2/3) – (3/5)
(ii) ((-4)/7) – (2/(-3))
(iii) (4/7) – ((-5)/(-7))
(iv) -2 – (5/9)
Solution 2:
Question 3: The sum of the two numbers is (5/9). If one of the numbers is (1/3), find the other.
Solution 3:
We found that the sum of two numbers are = (5/9)
One of them is = (1/3 )
Let be x the unknown number
Question 4: The sum of two numbers is ((-1)/3). If one of the numbers is ((-12)/3), find the other.
Solution 4:
We found that the sum of two numbers = ((-1)/3)
One of them is ((-12)/3)
Let be x the required number
x + ((-12)/3) = ((-1)/3)
x = ((-1)/3 – (-12)/3)
x = ((-1)/3 + 12/3)
x = ((-1+12))/3
x = (11/3)
Question 5: The sum of two numbers is ((-4)/3). If one of the numbers is -5, find the other.
Solution 5:
We found that the sum of two numbers = ((-4)/3)
One of them is -5
Let be x the required number
x + (-5) = ((-4)/3)
((-5)/1) = ((-5 ×3)/(1×3)) = ((-15)/3)
On substituting
x + ((-15)/3) = ((-4)/3)
x = ((-4)/3 – (-15)/3)
x = ((-4)/3 + 15/3)
x = (((-4+15))/3)
x = (11/3)
Question :6. The sum of two rational numbers is – 8. If one of the numbers is ((-15)/7), find the other.
Solution 6:
We found that the sum of two numbers is -8
One of them is ((-15)/7)
Let the required number be x
x + ((-15)/7) = -8
Consider ((-8)/1) = ((-8 ×7)/(1 ×7)) = (-56)/7
On substituting
x + ((-15)/7) = ((-56)/7)
x = ((-56)/7 – (-15)/7)
x = ((-56)/7 + 15/7)
x = ((-56+15))/7
x = ((-41)/7)
Question :7. What should be added to ((-7)/8) so as to get (5/9)?
Solution 7:
We found that the = ((-7)/8)
Let be x the required number
Question :8. What number should be added to ((-5)/11) so as to get (26/33)?
Solution 8:
As we found the number = ((-5)/11)
Let be the x required number
x + ((-5)/11) = (26/33)
x = (26/33) – ((-5)/11)
x = (26/33) + (5/11)
=(5/11 × (3.)/3) = (15/33)
On substituting
x = (26/33) + (15/33)
x = (41/33)
Question :9. What number should be added to ((-5)/7) to get ((-2)/3)?
Solution 9:
We found that the number =(-5)/7
Let be the x required number
Question 10: What number should be subtracted from ((-5)/3) to get (5/6)?
Solution 10:
We found that the number is = ((-5)/3)
Let be x the required number
((-5)/3) – x = (5/6)
– x = (5/6) – ((-5)/3)
– x = (5/6) + (5/3)
= (5/3) × (2/2) = (10/6)
On substituting
– x = (5/6) + (10/6)
– x = (15/6)
x = ((-15)/6)
Question :11. What number should be subtracted from (3/7) to get (5/4)?
Solution 11:
According to the question we found that= (3/7)
Let be x the required number
(3/7) – x = (5/4)
– x = (5/4) – (3/7)
The LCM of 4 and 7 is 28
= (( 5 ×7)/(4 ×7)) -((3×4)/(7×4))
= (35/28)- (12/28)
On substituting
- x = (35/28) – (12/28)
– x = ((35-12))/28
– x = (23/28)
x = ((-23)/28)
Question :12. What should be added to ((2/3) + (3/5)) to get ((-2)/15)?
Solution 12:
According to the question we found that: ((2/3) + (3/5))
Let be x be the required number
Question :13. What should be added to ((1/2) + (1/3) + (1/5)) to get 3?
Solution 13:
We found that ((1/2) + (1/3) + (1/5))
Let be x the required number
Question :14. What should be subtracted from ((3/4) - (2/3)) to get ((\-1)/6)?
Solution 14:
We found that ((3/4) - (2/3))
Let be x the required number
((3/4) - (2/3)) – x = ((-1)/6)
– x = ((-1)/6) – ((3/4) - (2/3))
Take LCM of 3,4 and 6
Question: 15. Simplify:
(i) ((-3)/2) + (5/4) – (7/4)
(ii) (5/3) – (7/6) + ((-2)/3)
(iii) (5/4) – (7/6) – ((-2)/3)
(iv) ((-2)/5) – ((-3)/10) – ((-4)/7)
Solution:
(i) ((-3)/2) + (5/4) – (7/4)
((-3)/2) = ((-3× 2)/(2 ×2)) ((-6)/4)
On substituting
((-3)/2) + (5/4)– (7/4) = ((-6)/4) + (5/4) – (7/4)
= ((-6+5-7))/4
= ((-13+5))/4
= ((-8)/4)
= -2
(ii) (5/3) – (7/6) + ((-2)/3)
(5/3) = ((5×2)/(3×2)) = (10/6)
((-2)/3) = ((-2 ×2)/(3 ×2)) = ((-4)/6)
(5/3) – (7/6) +((-2)/3) = (10/6) – (7/6) – (4/6)
= ((10 -7 - 4))/6
= ((10-11))/6
= ((-1)/6)
(iii) (5/4) – (7/6) – ((-2)/3)
Take LCM of 4,6 and 3
5 × 5 × 7 = 70
= ((-2 ×14)/(5×14)) - ((-3 ×7)/(10 ×7)) - ((-4 × 10)/(7×10))
= ((-28)/70)- ((-21)/70)- ((-40)/70)
On substituting
((-2)/5) – ((-3)/10) – ((-4)/7) = ((-28)/70) + (21/70) + (40/70)
= ((-28 + 21 + 40))/70
= (33/70)
Question :16. Fill in the blanks:
(i) ((-4)/13) – ((-3)/26) = …..
(ii) ((-9)/14) + ….. = -1
(iii) ((-7)/9) + ….. = 3
(iv) ….. + (15/23) = 4
Solution:
(i) ((-5)/26)
Description:
= ((-4)/13) – ((-3)/26)
=((-4)/13) = ((-4×2)/(13 ×2) ) = ((-8)/26)
=((-4)/13) – ((-3)/26)
= ((-8)/26) – ((-3)/26)
= ((-5)/26)
(ii) ((-5)/14)
Description:
((-9)/14) + ….. = -1
((-9)/14) + 1 = ….
((-9)/14) + (14/14) = (5/14)
((-9)/14) + ((-5)/14) = -1
(iii) (34/9)
Description:
Given ((-7)/9) + ….. = 3
((-7)/9) + x = 3
x = 3 + (7/9)
(3/1) = ((3×9)/(1 ×9 )) = (27/9)
x = (27/9) + (7/9) = (34/9)
(iv) (77/23)
Description:
Given ….. + (15/23) = 4
x + (15/23) = 4
x = 4 – (15/23)
(4/1) = ((4×23)/(1 ×23)) = (92/23)
x = (92/23) – (15/23)
= (77/23)
Exercise 5.3
Question 1: Multiply:
(i) (7/11) by (5/4)
(ii) (5/7) by ((-3)/4)
(iii) ((-2)/9) by (5/11)
(iv) ((-3)/13) by ((-5)/(-4))
Solution 1:
(i) (7/11) by (5/4)
= (7/11) × (5/4) = (35/44)
(ii) (5/7) by ((-3)/4)
= (5/7) × ((-3)/4) = ((-15)/28)
(iii) ((-2)/9) by (5/11)
= ((-2)/9) × (5/11) = ((-10)/99)
(iv) ((-3)/13) by ((-5)/(-4))
= ((-3)/13) × ((-5)/(-4)) = ((-15)/68)
Question 2: Multiply:
(i) ((-5)/17) by (51/(-60))
(ii) ((-6)/11) by ((-55)/36)
(iii) ((-8)/25) by ((-5)/16)
(iv) (6/7) by ((-49)/36)
Solution 2:
(i) ((-5)/17) by (51/(-60))
=((-5)/17) × (51/(-60)) = ((-225)/(-1020))
= (225/1020) = (1/4)
(ii) ((-6)/11) by ((-55)/36)
=((-6)/11) × ((-55)/36) = (330/396)
= (5/6)
(iii) ((-8)/25) by ((-5)/16)
=((-8)/25) × ((-5)/16) = (40/400)
= (1/10)
(iv) (6/7) by ((-49)/36)
=(6/7) × ((-49)/36) = ((-294)/252)
= ((-7)/6)
Question 3: Simplify each of the following and express the result as a rational number in standard form:
(i) ((-16)/21) × (14/5)
(ii) (7/6) × ((-3)/28)
(iii) ((-19)/36) × 16
(iv) ((-13)/9) × (27/(-26))
Solution 3:
(i) ((-16)/21) × (14/5)
= ((-16)/21) × (14/5)
= ((-224)/105)
= ((-32)/15)
(ii) (7/6) × ((-3)/28)
= (7/6) × ((-3)/28)
= ((-21)/168)
= ((-1)/8)
(iii) ((-19)/36) × 16
= ((-19)/36) × 16 = ((-304)/36)
= ((-76)/9)
(iv) ((-13)/9) × (27/(-26))
= ((-13)/9) × (27/(-26)) = ((-351)/234)
= (3/2)
Question 4: Simplify:
(i) (-5 × (2/15)) – (-6 × (2/9))
(ii) (((-9)/4) × (5/3)) + ((13/2) × (5/6))
Solution 4:
(i) (-5 × (2/15)) – (-6 × (2/9))
= (-5 × 2/15) – (-6 × 2/9) = (10/15) – ((-12)/9)
= ((-2)/3) + (12/9)
= ((-6)/9) + (12/9)
= (6/9) = (2/3)
(ii) (((-9)/4) × (5/3)) + ((13/2) × (5/6))
(((-9)/4 × (5/3)) + ((13/2)× 5/6)) = ((-3)/4× 5) + (13/2) × (5/6)
= ((-15)/4) + (65/12)
= ((-15)/4) × (3/3) + (65/12)
= ((-45)/12) + (65/12)
= (((65-45))/12)
= (20/12) = (5/3)
Question 5: Simplify:
(i) ((13/9) × ((-15)/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2))
(ii) ((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15))
Solution 5:
(i) ((13/9) × ((-15)/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2))
((13/9) × ((-15)/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2)) = ((-195)/18) + (56/15) + (3/10)
= ((-65)/6) + (56/15) + (3/10)
= ((-65)/6) × (5/5) + (56/15) × (2/2) + (3/10) × (3/3).
= ((-325)/30) + (112/30) + (9/30)
= ((-325+112+9))/30
= ((-204)/30) = ((-34)/5)
(ii) ((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15))
((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15)) = (66/55) – (36/36) + (30/195)
= (5/22) – (12/12) + (1/11)
= (5/22) – 1 + (2/13)
= (5/22) × (13/13) + (1/1) × (286/286) + (2/13) × (22/22)
= (65/286) – (286/286) + (44/286) = ((-177)/286)
Exercise 5.4
Question 1: Divide:
(i) 1 by (1/2)
(ii) 5 by ((-5)/7)
(iii) ((-3)/4) by (9/(-16))
(iv) ((-7)/8) by ((-21)/16)
(v) (7/(-4)) by (63/64)
(vi) 0 by ((-7)/5)
(vii) ((-3)/4) by -6
(viii) (2/3) by ((-7)/12)
Solution 1:
(i) 1 by (1/2)
= 1 ÷ (1/2) = 1 × 2 = 2
(ii) 5 by ((-5)/7)
= 5 ÷ ((-5)/7) = 5 × ((-7)/5)
= -7
(iii) ((-3)/4) by (9/(-16))
= ((-3)/4) ÷ (9/(-16)) = ((-3)/4) × ((-16)/9)
= ((-4)/(-3)) =(4/3)
(iv) ((-7)/8) by ((-21)/16)
= ((-7)/8) ÷ ((-21)/16) = ((-7)/8) × (16/(-21))
= ((-2)/(-3)) = (2/3)
(v) (7/(-4)) by (63/64)
= (7/(-4)) ÷ (63/64) = (7/(-4)) × (64/63) = ((-16)/9)
(vi)((-7)/5)
= 0 ÷ ((-7)/5) = 0 × (5/7)
= 0
(vii) ((-3)/4) by -6
= ((-3)/4) ÷ -6 = ((-3)/4) × (1/(-6))
= ((-1)/(-8)) = (1/8)
(viii) (2/3) by ((-7)/12)
= (2/3) ÷ ((-7)/12) = (2/3) × (12/(-7))
= (8/(-7))
Question 2: Find the value and express as a rational number in standard form:
(i) (2/5) ÷ (26/15)
(ii) (10/3) ÷ ((-35)/12)
(iii) -6 ÷ ((-8)/17)
(iv) (40/98) ÷ (-20)
Solution 2:
(i) (2/5) ÷ (26/15)
= (2/5) ÷ (26/15) = (2/5) × (15/26)
= (3/13)
(ii) (10/3) ÷ ((-35)/12)
= (10/3) ÷ ((-35)/12) = (10/3) × (12/(-35))
= ((-40)/35) = ((-8)/7)
(iii) -6 ÷ ((-8)/17)
= -6 ÷ ((-8)/17) = -6 × (17/(-8))
= (102/8) = (51/4)
(iv) (40/98) ÷ (-20)
= (40/98) ÷ (-20) = (40/98) × (1/(-20))
= ((-2)/98) = ((-1)/49)
Question 3: The product of two rational numbers is 15. If one of the numbers is -10, find the other.
Solution 3:
Let be x the required number
= x – 10 = 15
x = (15/(-10))
x = (3/(-2))
x x = ((-3)/2)
Hence the number is ((-3)/2)
Question 4: The product of two rational numbers is ((-8)/9). If one of the numbers is ((-4)/15), find the other.
Solution 4:
We found the product of two numbers = ((-8)/9)
One of them is ((-4)/15)
Let the required number be x
x × ((-4)/15) = ((-8)/9)
x = ((-8)/9) ÷ ((-4)/15)
x = ((-8)/9) × (15/(-4))
x = ((-120)/(-36))
x = (10/3)
Question: 5. By what number should we multiply ((-1)/6) so that the product may be ((-23)/9)?
Solution 5:
We found that the product = ((-23)/9)
One number is ((-1)/6)
Let be x the required number
x × ((-1)/6) = ((-23)/9)
x = ((-23)/9) ÷ ((-1)/6)
x = ((-23)/9) × ((-6)/1)
x = (138/9)
x = (46/3)
Question 6: By what number should we multiply ((-15)/28) so that the product may be ((-5)/7)?
Solution 6:
We found the product = ((-5)/7)
One number is ((-15)/28)
Let the required number be x
x × ((-15)/28) = ((-5)/7)
x = ((-5)/7) ÷ ((-15)/28)
x = ((-5)/7) × (28/(-15))
x = ((-4)/(-3))
x = (4/3)
Question 7: By what number should we multiply ((-8)/13) so that the product may be 24?
Solution 7:
We found that the product = 24
One of the number is = ((-8)/13)
Let be x the required number
x × ((-8)/13) = 24
x = 24 ÷ ((-8)/13)
x = 24 × (13/(-8))
x = -39
Question 8: By what number should ((-3)/4) be multiplied in order to produce ((-2)/3)?
Solution 8:
We found that the product = ((-2)/3)
One of the number is = ((-3)/4)
Let be x the required number
x × ((-3)/4) = ((-2)/3)
x = ((-2)/3) ÷ ((-3)/4)
x = ((-2)/3) × (4/(-3))
x = ((-8)/(-9))
x = (8/9)
Question 9: Find (x + y) ÷ (x – y), if
(i) x = (2/3), y = (3/2)
(ii) x = (2/5), y = (1/2)
(iii) x = (5/4), y = ((-1)/3)
Solution 9:
(x + y) ÷ (x – y)
(i) x = (2/3), y = (3/2)
= ((2/3) + (3/2)) ÷ ((2/3) – (3/2))
= ((4 + 9))/6 ÷ ((4 - 9)/6)
= ((4 + 9))/6 × 6/((4-9))
= ((4 + 9))/((4-9)) = (13/(-5))
(ii) x = (2/5), y = (1/2)
= ((2/5) + (1/2)) ÷ ((2/5) – (1/2))
= ((4 + 5))/10 ÷ ((4-5))/10
= ((4 + 5))/10 × 10/((4 - 5))
= ((4 + 5))/((4 - 5)) = (9/(-1))
(iii) x = (5/4), y = ((-1)/3)
= ((5/4) + ((-1)/3)) ÷ ((5/4) – ((-1)/3))
= ((15- 4))/12 ÷ ((15 + 4))/12
= ((15- 4))/12 × 12/((15+4))
= ((15- 4))/((15+4)) = (11/19)
Question 10: The cost of 723 meters of rope is Rs. 1234. Find its cost per meter.
Solution 10:
Given cost of 723 = (23/3) meters of rope is Rs. 1234 = (51/4)
Cost per meter = (51/4) ÷ (23/3)
= (51/4) × (3/23)
= (153/92)
= Rs 1 61/92
Question :11. The cost of 213 meters of cloth is Rs.7514. Find the cost of cloth per meter.
Solution 11:
The cost of 213 meters of rope = Rs. 7514
Cost of cloth per meter = 7514 ÷ 213
= (301/4) ÷ (7/3)
= (301/4) × (3/7)
= (129/4)
= Rs 32 1/4
Question 12: By what number should ((-33)/16) be divided to get ((-11)/4)?
Solution:
Let the required number be x
((-33)/16) ÷ x = ((-11)/4)
x = ((-33)/16) ÷ ((-11)/4)
x = ((-33)/16) × (4/(-11))
x = (3/4)
Question :15. If 24 trousers of equal size can be prepared in 54 meters of cloth, what length of cloth is required for each trouser?
Solution:
According to the question for 24 trousers cloth required = 54m
for 1 trouser cloth required = (54/24) = (9/4) meters
Exercise 5.5
Question 1: Find six rational numbers between ((-4)/8) and (3/8)
Solution 1:
Rational numbers between -4 and -8 are: -3, -2, -1, 0, 1, 2.
According to definition of rational numbers are in the form of (p/q) where q not equal to zero.
Thus, six rational numbers between ((-4)/8) and (3/8) are
(-3)/8, (-2)/8, (-1)/8, 0/8, 1/8, 2/8, 3/8
Question 2: Find 10 rational numbers between (7/13) and ((-4)/13)
Solution 2:
Rational numbers Between 7 to -4 are: -3, -2, -1, 0, 1, 2, 3, 4, 5, 6.
According to definition of rational numbers are in the form of (p/q) where q not equal to zero.
Thus, six rational numbers between (7/13) and ((-4)/13) are
(-3)/13, (-2)/13, (-1)/13, 0/13, 1/13, 2/13, 3/13, 4/13, 5/13, 6/13
Question 3: State true or false:
(i) Between any two distinct integers there is always an integer.
(ii) Between any two distinct rational numbers there is always a rational number.
(iii) Between any two distinct rational numbers there are infinitely many rational numbers.
Solution 3:
(i) False
Description:
One integer does not need to be between any two distinct integers.
(ii) True
Description:
According to the properties of rational numbers between any two distinct rational numbers there is always a rational number.
(iii) True
Description:
There are infinitely many rational numbers between any two distinct rational numbers, according to the characteristics of rational numbers.