RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2

Other Exercises

Question 1.
Add the following algebraic expressions
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 1
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 2
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 3
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 4
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 5
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 6

Question 2.
Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 7
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 8
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 9
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 10
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 11

Question 3.
Take away :
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 12
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 13
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 14
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 15
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 16
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 17
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 18
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 19

Question 4.
Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and – 4X + 9y- 11z.
Solution:
Sum of x – 3y + 2z and – 4x + 9y – 11z
= x – 3y + 2z + (- 4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y – 11z
= x – 4x – 3y + 9y + 2z – 11z
= – 3x + 6y – 9z
Now (-3x + 6y – 9z) – (3x – 4y – 7z)
= -3x + 6y – 9z – 3x + 4y + 7z
= -3x – 3x + 6y + 4y -9z +7z
= -6x + 10y – 2z

Question 5.
Subtract the sum of 3l- 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3nand -3l + m + 4n2.
Solution:
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
= 9l + 2m – 3 n2 + (-3l) + m + 4n2
= 9l + 2m – 3n2 – 3l + m + 4n2
= 9l- 3l+ 2m + m – 3 n2 + 4n2
= 6l + 3m + n2
and sum of 3l – 4m – 7n2 and 2l +3m- 4n2
= 3l- 4m – 7n2 + 2l+ 3m- 4n2
= 3l + 2l – 4m + 3m- 7n2 – 4n2
= 5l -m- 11n2
Now (6l + 3m + n2) – (5l – m – 11n2)
= 6l + 3m + n2 – 5l + m + 11n2
= 6l – 5l + 3m + m + n2 + 11n2
= l + 4m+ 12n2

Question 6.
Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
5 – (2x-x2 + 5-4x-3 + 7x2)
= 5 – (2x – 4x- x2 + 7x2 + 5-3)
= 5 – (-2x + 6x2 + 2)
= 5 + 2x – 6x2 – 2
= – 6x2+2x+3
= 3 + 2x – 6x2

Question 7.
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 20
Solution:
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 21
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 22
RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 23

Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.2 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1

Other Exercises

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 1
Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 2
Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 3
Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 4
Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 5
Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 6
Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 7
Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 1\(\frac { 1 }{ 2 }\) years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 1\(\frac { 1 }{ 2 }\) years = 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 8
Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 9
Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 10
Compound interest for the third half-years or 1\(\frac { 1 }{ 2 }\) years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 11
Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 12
Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 13
Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 14
Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 12\(\frac { 1 }{ 2 }\) % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 15
Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 16
Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 17
Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 1\(\frac { 1 }{ 2 }\) years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 1\(\frac { 1 }{ 2 }\) years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 1\(\frac { 1 }{ 2 }\) years or 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 18
Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 19
Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 20
Total compound interest for 3 half-years
or 1\(\frac { 1 }{ 2 }\) years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 21
In second case
Interest for the first year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 22
Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 23
Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 24
Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 25
Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 26
Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 27
Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 28
Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 29
Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or \(\frac { 5 }{ 2 }\) % quarterly
Period (T) = 1 year = 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 30
Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 31
Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 32
= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 33
Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 34
Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 35
Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 36
Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 5\(\frac { 1 }{ 2 }\) % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 5\(\frac { 1 }{ 2 }\) % = \(\frac { 11 }{ 2 }\) % p.a.
Period (T) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 37
Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 38
Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 39
Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 40
Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 41
Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 42
Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 43
Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 44
Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3

Other Exercises

Solve each of the following cryptarithms.
Question 1.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 1
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 2
Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9
∴ A = 6 or less.
∴ 7 + B = A = 6 or less
∴ 7 + 9 or 8 = 16 or 15
∴ But it is two digit number
B = 8
Then A = 5
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 3

Question 2.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 4
Solution:
Values of A and B can be between 0 and 9
In tens digit, A + 3 = 9
∴ A = 9 – 3 = 6 or less than 6
In ones unit B + 7 = A = 6or less
∴ 7 + 9 or 8 = 16 or 15
But it is two digit number
∴ B = 8 and
∴ A = 5
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 5

Question 3.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 6
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 7
Value of A and B can be between 0 and 9 In units place.
1+B = 0 ⇒1+B = 10
∴ B = 10 – 1 = 9
and in tens place
1 + A + 1 = B ⇒ A + 2 = 9
⇒ A = 9 – 2 = 7
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 8

Question 4.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 9
Solution:
Values of A and.B can be between 0 and 9
In units place, B+1 = 8 ⇒ B = 8-1=7
In tens place A + B= 1 or A + B = 11
⇒ A + 7 = 11 ⇒ A =11-7 = 4

Question 5.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 10
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 11
Values of A and B can be between 0 and 9
In tens place, 2 + A = 0 or 2 + A=10
A = 10-2 = 8
In units place, A + B = 9
⇒ 8 + B = 9 ⇒ B = 9- 8 = 1
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 12

Question 6.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 13
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 14
Values of A and B can be between 0 and 9
In hundreds place,
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 15

Question 7.
Show that cryptarithm 4 x \(\overline { AB } =\overline { CAB }\) does not have any solution.
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 16
It means that 4 x B is a numebr whose units digit is B
Clearly, there is no such digit
Hence the given cryptarithm has no solution.

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

Other Exercises

Question 1.
Given that the number  \(\overline{35 a 64}\) is divisible by 3, where a is a digit, what are the possible volues of a ?
Solution:
The number \(\overline{35 a 64}\) is divisible by 3
∵The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be, 0, 3, 6, 9

Question 2.
If x is a digit such that the number \(\overline { 18×71 }\) is divisible by 3,’ find possible values of x.
Solution:
∵ The number \(\overline { 18×71 }\)
is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ l + 8+ x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27…………
∴ x can be 1, 4, 7 which are divisible by 3.

Question 3.
If is a digit of the number \(\overline { 66784x }\) such that it is divisible by 9, find the possible values of x.
Solution:
∵ The number 66784 x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6+6+7+8+4+x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5

Question 4.
Given that the number \(\overline { 67 y 19 }\) is divisible by 9, where y is a digit, what are the possible values of y ?
Solution:
∵ The number \(\overline { 67 y 19 }\) is divisible by 9
∴The sum of its digits will also be divisible by 9
⇒ 6 + 7+ y+ 1+ 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴y = A

Question 5.
If \(\overline { 3 x 2 }\) is a multiple of 11, where .v is a digit, what is the value of * ?
Solution:
∵ The number \(\overline { 3 x 2 }\) is multiple of 11
∴ It is divisible by 11
∴ Difference of the sum of its alternate digits is zero or multiple of 11
∴ Difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0
Then x = 5

Question 6.
If \(\overline { 98125 x 2 }\) is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.
Solution:
∵ The number \(\overline { 98125 x 2 }\) is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ \(\overline { x2 }\) is divisible by 4
∴ Possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1,3, 5, 7, 9

Question 7.
If x denotes the digit at hundreds place of the number \(\overline { 67 x 19 }\) such that the
number is divisible by 11. Find all possible values of x.
Solution:
∵ The number \(\overline { 67 x 19 }\) is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15+x-8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x=11-7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4

Question 8.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2

Question 9.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must by divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1

Question 10.
Find the remainder, without performing actual division when 798 is divided by 11.
Solution:
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6

Question 11.
Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
Let n = 928174653
= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder =10

Question 12.
Given an example of a number which is divisible by :
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not 32.
Solution:
(i) 2 but not by 4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342 etc.
(ii) 3 but not by 6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201 etc.
(iii) 4 but not by 8
A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.
∴ The number can be 244, 1356 etc.
(iv) Both 4 and 8 but not by 32
A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

Question 13.
Which of the following statements are true ?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-priirie, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) False, it is not necessarily that it must divide by 9.
(ii) Trae.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii)True.
(ix) False. It is not necessarily.
(x) True.

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2

RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1

Other Exercises

Question 1.
Find the S.P. If
(i) M.P. = Rs. 1300 and Discount = 10%
(ii) M.P. = Rs. 500 and Discount = 15%
Solution:
(i) M.P. = Rs. 1300, Discount = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 1
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 2
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 3

Question 2.
Find the M.P. If
(i) S.P. = Rs. 1222 and Discount = 6%
(ii) S.P. = Rs. 495 and Discount = 1%
Solution:
(i) S.P. = Rs. 1222, discount = 6%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 4

Question 3.
Find discount in percent when
(i) M.P. = Rs. 900 and S.P. = Rs. 873
(ii) M.P. = Rs. 500 and S.P. = Rs. 425
Solution:
(i) M.P. = Rs. 900
S.P. = Rs. 873
Discount = M.P. – S.P. = Rs. 900 – Rs. 873 = Rs. 27
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 5
(ii) M.P. = Rs. 500
S.P. = Rs. 425
Discount M.P. – S.P. = Rs. 500 – Rs. 425 = = Rs. 75
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 6

Question 4.
A shop selling sewing machines offers 3% discount on ail cash purchases. What cash amount does a customer pay for a sewing machine, the price of which is marked as Rs. 650.
Solution:
Marked price (M.P.) of one sewing machine = Rs. 650
Rate of discount = 3%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 7

Question 5.
The marked price of a ceiling fan is Rs. 720. During off season, it is sold for Rs. 684. Determine the discount percent.
Solution:
Marked price (M.P.) of fan = Rs. 720
Sale price (S.P.) = Rs. 684
Amount of discount = M.P. – S.P. = Rs. 720 – Rs. 684 = Rs. 36
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 8

Question 6.
On the eve of Gandhi Jayanti, a saree is sold for Rs. 720 after allowing 20% discount. What is the marked price ?
Solution:
S.P. of saree = Rs. 720
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 9

Question 7.
After allowing a discount of 7\(\frac { 1 }{ 2 }\) % on the marked price, an article is sold for Rs. 555. Find its marked price.
Solution:
Selling price (S.P.) = Rs. 555
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 10

Question 8.
A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 250?
Solution:
Marked price = Rs. 250
Discount allowed = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 11
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 12

Question 9.
A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 500 ?
Solution:
Marked price (M.P.) of an article = Rs. 500
Discount allowed = 20%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 13

Question 10.
A tradesman marks his goods at such a price that after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170 ?
Solution:
Rate of discount = 15% gain = 20%
Cost price (C.P.) of an article = Rs 170
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 14

Question 11.
A shopkeeper marks his goods in such a way that after allowing a discount of 25% on the marked price, he still makes a profit of 50%. Find the ratio of the C.P. to the M.P.
Solution:
Let cost price (C.P.) = Rs 100
Profit = 50%
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 15

Question 12.
A cycle dealer offers a discount of 10% and still makes a profit of 26%. What is the actual cost to him of a cycle whose marked price is Rs 840 ?
Solution:
Rate of discount = 10%
Gain = 26%
Marked Price (M.P.) = Rs 840
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 16

Question 13.
A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, find his advertised price.
Solution:
Rate of commission = 23%
Profit = 10%
Total gain = Rs 56
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 17

Question 14.
A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount ?
Solution:
Let cost price (C.P.) = Rs 100
Marked price = Rs 100 + 40 = Rs 140
Rate of discount = 5%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 18

Question 15.
By selling a pair of ear rings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price ? What is the marked price and the price at which the pair was eventually bought ?
Solution:
Total profit = Rs 48
Profit percent = 16%
Cost price = \(\frac { 48 x 100 }{ 16 }\) = Rs 300
Selling Price = C.P. + profit = Rs 300 + Rs 48 = Rs 348
Rate of discount = 25%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 19

Question 16.
A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275 ?
Solution:
Printed price of a book = Rs 275
Rate of discount = 32%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 20

Question 17.
After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price ?
Solution:
Rate of discount = 20%
Loss = 10%
Let the cost price of the lamp = Rs 100
Loss = 10%
Selling price = Rs 100 – 10 = Rs 90
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 21

Question 18.
The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15% ?
Solution:
List price of table fan (M.P.) = Rs 480
Rate of discount = 25%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 22

Question 19.
Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item ?
Solution:
Rate of discount = 25%
Selling price (S.P.) for Rohit = Rs 660
Profit = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 23
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 24

Question 20.
A cycle merchant allows 20% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle, find the marked price of the cycle.
Solution:
Rate of discount = 20%
Profit = 20%
Total gain = Rs 360
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 25

Question 21.
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs 1470.
Solution:
Rate of discount = 12.5%
Profit = 10%
Cost price = Rs 1470
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 26
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 27

Question 22.
What price should Aslam mark on a pair of shoes ? Which costs him Rs 1200 so as to gain 12% after allowing a discount of 16% ?
Solution:
Cost price of shoes = Rs 1200
Gain = 12%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 28

Question 23.
Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs 850 ?
Solution:
Marked price of a shirt = Rs 850
Discount = 4%
Selling price
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 29
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 30

Question 24.
A shopkeeper offers 10% off-season discount to the customers and still makes a profit of 26%. What is the cost price for the shopkeeper on a pair of shoes marked at Rs 1120 ?
Solution:
Marked price = Rs 1120
Rate of discount = 10%
S.P. of shoes
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 31

Question 25.
A lady shopkeeper allows her customers 10% discount on the marked price of the goods and still gets a profit of 25%. What is the cost price of a fan for her marked at Rs 1250 ?
Solution:
Marked price (M.P.) of fan = Rs 1250
Discount = 10%
S.P. of the fan
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 32
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 33
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 34

Hope given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax Ex 13.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1

RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1

Other Exercises

Question 1.
A student buys a pen for Rs. 90 and sells it for Rs. 100. Find his gain and gain percent ?
Solution:
C.P. of a pen = Rs. 90
and S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 90 (S.P > C.P.)
= Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 1

Question 2.
Rekha bought a saree for Rs. 1240 and sold it for Rs. 1147. Find her loss and loss percent.
Solution:
C.P. of saree.= Rs. 1240 and
S.P. = Rs. 1147
Loss = C.P – S.P. = Rs. 1240 – Rs. 1147 (C.P. > S.P.)
= Rs. 93
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 2

Question 3.
A boy buys 9 apples for Rs. 9.60 and sells them at 11 for Rs. 12. Find his gain or loss percent.
Solution:
L.C.M. of 9 and 11 = 99
Let 99 apples were purchased.
C.P. of 99 apples at the rate of 9 apples for Rs. 9.60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 3

Question 4.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.
Solution:
C.P. of 10 articles = S.P. of 9 articles = 90 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 4

Question 5.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300, determine his profit percent.
Solution:
Cost of radio = Rs. 225
Over head expenses = Rs. 15
Total C.P. of the ratio = Rs. 225 + 15 = Rs. 240
S.P. of radio = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 5

Question 6.
A retailer buys a cooler for Rs. 1200 and overhead expenses on it are Rs. 40. If he sells the cooler for Rs. 1550, determine his profit percent.
Solution:
Cost of cooler = Rs. 1200
Overhead expenses = Rs. 40
Total cost price of cooler = Rs. 1200 + Rs. 40 = Rs. 1240
Selling price = Rs. 1550
Gain = S.P. – C.P. = Rs. 1550 – Rs. 1240 = Rs. 310
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 6

Question 7.
A dealer buys a wristwatch for Rs. 225 and spends Rs. 15 on its repairs. If he sells the same for Rs. 300, find his profit percent.
Solution:
Cost of wristwatch = Rs. 225
Cost on repairs = Rs. 15
Total cost price = Rs. 225 + Rs. 15 = Rs. 240
Selling price = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 7

Question 8.
Ramesh bought two boxes for Rs. 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
Solution:
Total cost price of two boxes = Rs. 1300
S.P. of each box is same.
Let S.P of each box = Rs. 100
S.P. of first box = Rs. 100
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 8
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 9

Question 9.
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
Solution:
S.P. of 10 pens = C.P. of 14 pens = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 10

Question 10.
If the cost price of 18 chairs be equal to selling price of 16 chairs, And the gain or loss percent.
Solution:
C.P. of 18 chairs = S.P. of 16 chairs = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 11
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 12

Question 11.
If the selling price of 18 oranges is equal to the cost price of 16 oranges, And the loss percent.
Solution:
S.P. of 18 oranges = C.P. of 16 oranges = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 13

Question 12.
Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs. 1680 on its repairs and sold the motor cycle to Rahul for Rs. 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.
Solution:
Cost price of cycle for Rahul or Selling price for Vineet = Rs. 35910
Gain = 12.5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 14
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 15

Question 13.
By selling a book for Rs. 258, a book-seller gains 20%. For how much should he sell it to gain 30% ?
Solution:
S.P. of a book = Rs. 258
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 16

Question 14.
A defective briefcase costing Rs. 800 is being sold at a loss of 8%. If its price is further reduced by 5%, find its selling price.
Solution:
C.P. of a briefcase = Rs. 800
Loss = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 17
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 18

Question 15.
By selling 90 ball pens for Rs. 160, a person losses 20%. How many ball pens should be sold for Rs. 96 so as to have a profit of 20%.
Solution:
S.P of 90 ball pens = Rs. 160
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 19

Question 16.
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 36.75 less, he would have gained 30%. Find the cost price of the article.
Solution:
Let C.P. of the article = Rs. 100
In first case gain = 25%
S.P. = 100 + 25 = Rs. 125
In second case,
C.P. = 20% less of Rs. 100 = 100 – 20 = Rs. 80
Gain = 30%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 20

Question 17.
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 for each kilogram. Find his gain percent.
Solution:
Let C.P. of 1 kg of pulses = Rs. 100
S.P. of 950 gm = Rs. 100
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 21
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 22

Question 18.
A dealer bought two tables for Rs. 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then he found that each table was sold for the same price, find the cost price of each table.
Solution:
Cost price of two tables = Rs. 3120
Let S.P of each table = Rs. 100
Now S.P of one table = Rs. 100
Loss = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 23
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 24

Question 19.
Mariam bought two fans for Rs. 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.
Solution:
Total cost price of two fans = Rs. 3605
Let selling price of each fan = Rs. 100
Now S.P. of first fan = Rs. 100
Profit = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 25

Question 20.
Some toffees are bought at the rate of 11 for Rs. 10 and the same number at the rate of 9 for Rs. 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.
Solution:
L.C.M. of 11 and 9 = 99
Let each time 99 toffees are bought.
In first case, the C.P. of 99 toffees at the rate of 11 for Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 26

Question 21.
A tricycle is sold at a gain of 16%. Had it been sold for Rs. 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
Solution:
Let the C.P. of tricycle = Rs. 100
In first case, gain = 16%
S.P. = Rs. 100 + 16 = Rs. 116
In second case, gain = 20%
S.P. = Rs. 100 + 20 = Rs. 120
Difference in S.P.’s = Rs. 120 – Rs. 116 = Rs. 4
If difference is Rs. 4,
then C.P. of the tricycle = Rs. 100
and if difference is Re. 1, then C.P.
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 27

Question 22.
Shabana bought 16 dozen ball pens and sold then at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs. 576.
Solution:
C.P. of 16 dozen ball pens = S.P. of 16 dozen pens – loss
C.P. of 16 x 12 pens = S.P. of 16 dozen pens x S.P. of 8 pens
C.P. of 192 pens = S.P. of 16 x 12 pens x S.P. of 8 pens
S.P. of 192 pens + S.P. of 8 pens = S.P. of 200 pens
(i) Let C.P. of 1 pen = Re. 1
Then C.P. of 192 pens = Rs. 192
and S.P. of 200 pens = Rs. 192
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 28

Question 23.
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs. 6. Find
(i) C.P. of the shirt
(ii) The two selling prices of the shirt
Solution:
The difference of two selling prices of shirt = Rs. 6
Difference in profits of 4% and 5% = 5 – 4 = 1%
(i) C.P. = 1 x 6 x 100 = Rs. 600
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 29

Question 24.
Toshiba bought 100 hens for Rs. 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole ?
Solution:
Total number of hens bought = 100
C.P. of 100 hens = Rs. 8000
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 30

Hope given RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax Ex 13.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1

Other Exercises

Question 1.
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by
(i) 11
(ii) 15
Solution:
Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9
(i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
(ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11

Question 2.
Without performing actual computations, find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Two given numbers are 94 and 49. Whose digits are reversed.
(i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5
(ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9

Question 3.
If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.
Solution:
The given number is 985
The other two numbers by arranging its digits
in cyclic order, will be 859, 598 of the form
\(\overline{ abc } ,\overline{ bca } ,\overline{ cba }\)
Therefore,
If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then the quotient = 111
and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66

Question 4.
Find the quotient when difference of 985 and 958 is divided by 9.
Solution:
The numbers of three digits are
985 and 958 in which tens and ones digits are reversed, then
\(\overline{ abc } -\overline{ acb }\) = 9 (b – c)
985 – 958 = 9 (8 – 5) = 9 x 3
i. e., it is divisible by 9, then quotient = b-c =8-5=3

 

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2

Other Exercises

Question 1.
Find :
(i) 22% of 120
(ii) 25% of Rs. 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metres
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 1

Question 2.
Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) \(\frac { 1 }{ 2 }\) % of a is 50
(iv) 100% of a is 100
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 2
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 3

Question 3.
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
x = 5% of y, y = 24% of z.
x = 480
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 4

Question 4.
A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.
Solution:
Let his monthly income = Rs. x
15% of x = Rs. 150
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 5

Question 5.
Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.
Solution:
Let total marks of the examination = x
86.875% of x = 695
=> 86.875 x \(\frac { 1 }{ 100 }\) x x = 695
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 6

Question 6.
Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ?
Solution:
Let the school opened for = x days = 90% of x = 216
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 7

Question 7.
A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Solution:
Number of total trees = 2000
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 8
Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400
Number of orange trees = 1400

Question 8.
Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take.
Solution:
Balance diet contains
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Number of total calories = 2600
Number of calories of proteins = 12% of 2600 = \(\frac { 12 }{ 100 }\) x 2600 = 312
Number of calories of fats = 25% of 2600 = \(\frac { 25 }{ 100 }\) x 2600 = 650
Number of calories of carbohydrates = 63% of 2600 = \(\frac { 63 }{ 100 }\) x 2600 = 1638

Question 9.
A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in :
(i) Sixes
(ii) Fours
(iii) Twos
(iv) Singles
Solution:
Total score of a cricketer = 62 runs
(z) Number of sixes = 3
Run from 3 sixes = 3 x 6 = 18
Percentage = \(\frac { 18 }{ 62 }\) x 100 = 29.03%
(ii) Number of fours = 8
Total run from 8 fours = 4 x 8 = 32
Percentage = \(\frac { 32 }{ 62 }\) x 100 = 51.61%
(iii) Number of twos = 2
Total score from 2 twos = 2 x 2 = 4
Percentage = \(\frac { 4 }{ 62 }\) x 100 = \(\frac { 400 }{ 62 }\) = 6.45%
(iv) Number of single run = 8
Percentage = \(\frac { 8 }{ 62 }\) x 100 = \(\frac { 800 }{ 62 }\) = 12.9%

Question 10.
A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in :
(i) 6’s
(ii) 4’s
(iii) 2’s
(iv) singles
What % of his shots were scoring ones ?
Solution:
Total runs scored by a cricketer =120
(i) Number of runs from sixes (6’s) = 20% of 120
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 9

Question 11.
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ?
Solution:
Total earning from investment = Rs. 187
Percent earning = 22%
Let his investment = x
Then 22% of x = Rs. 187
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 10

Question 12.
Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ?
Solution:
Deposit in the bank = Rs. 1440
Percentage = 12% of his total income
Let his total income = Rs. x
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 11

Question 13.
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ?
Solution:
(i) In gunpowder,
Nitre = 75%
Sulphur = 10%
Let total amount of gunpowder = x kg
Nitre = 9 kg
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 12

Question 14.
An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ?
Solution:
In an alloy,
Number of parts of tin = 15
and number of parts of copper = 105
Total parts = 15 + 105 = 120
Percentage of copper in the alloy = \(\frac { 105 }{ 120 }\) x 100 = 87.5%

Question 15.
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
In an alloy,
Copper = 32%
Nickel = 40%
Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28%
Mass of zinc in 1 kg = 28% of 1 kg = \(\frac { 28 }{ 100 }\) x 100 gm = 280 gm.

Question 16.
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ?
Solution:
Distance travelled before first stop = 122 km
Let total journey = x km
10% of x = 122
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 13

Question 17.
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ?
Solution:
Total numbers of teachers = 30
Number of male teachers = 12
Number of female teacher = 30 – 12 = 18
Percentage of female teachers = \(\frac { 18 x 100 }{ 30 }\) = 60%

Question 18.
Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ?
Solution:
Let Anil’s income = Rs. 100
Then Aman’s income = Rs, 100 – 20 = Rs. 80
Now, difference of both’s incomes = 100 – 80 = Rs. 20
Anil income is Rs. 20 more than that of Aman’s
Percentage = \(\frac { 20 x 100 }{ 80 }\) = 25%

Question 19.
The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ?
Solution:
Present value of machine = Rs. 100000
Rate of depreciation per year = 5%
Period = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 14

Question 20.
The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ?
Solution:
Present population of the town = 60000
Increase annually = 10%
Period = 2 years
Population after 2 years will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 15

Question 21.
The population of a town increases 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Let the population of the town a year ago was = x
Increase in population = 10%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 16

Question 22.
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ?
Solution:
Let the salary of Ankit before increment = x
Increment given = 10% of the salary
Salary after increment will be
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 17

Question 23.
In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ?
Solution:
Let price of petrol before budged = Rs. 100
Increase = 10%
Price after budget = Rs. 100 + 10 = Rs. 110
Let the consumption of petrol before budget = 100 l
Price pf 100 l = Rs. 110
Now of new price is Rs. 110, consumption = 100 l
are of new price will be 100, then
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 18

Question 24.
Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ?
Solution:
Mohan’s income = Rs. 15500
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 19
We see that the savings is same
There is no change in savings.

Question 25.
Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ?
Solution:
Let Shalu’s income = Rs. 100
Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160
Now difference in their incomes = Rs. 160 – 100 = Rs. 60
Shalu’s income is less than Shikha’s income by Rs. 60
Percentage less = \(\frac { 60 x 100 }{ 160 }\) = \(\frac { 75 }{ 2 }\) % = 37.5%

Question 26.
Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ?
Solution:
Let the third person gets = Rs. x
Then second person will get
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 20
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 21

Question 27.
After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ?
Solution:
Let the original price of the vase = Rs. x
Hike in price = 20%
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.2 22
Original price of the vase = Rs. 1666.66

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RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1

Other Exercises

Question 1.
Write each of the following as percent: Solution—
(i) \(\frac { 7 }{ 25 }\)
(ii) \(\frac { 16 }{ 625 }\)
(iii) \(\frac { 5 }{ 8 }\)
(iv) 0.8
(v) 0.005
(vi) 3 : 25
(vii) 11 : 80
(viii) 111 : 125
(ix) 13 : 75
(x) 15 : 16
(xi) 0.18
(xii) \(\frac { 7 }{ 125 }\)
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 1
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 2

Question 2.
Convert the following percentages to fractions and ratios :
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 3
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 4

Question 3.
Express the following as decimal fractions :
(i) 27%
(ii) 6.3%
(iii) 32%
(iv) 0.25%
(v) 7.5%
(vi) \(\frac { 1 }{ 8 }\) %
Solution:
RD Sharma Class 8 Solutions Chapter 12 Percentage Ex 12.1 5

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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Other Exercises

Making use of the cube root table, find the cubes root of the following (correct to three decimal places) 
Question 1.
7
Solution:
\(\sqrt [ 3 ]{ 7 }\) =1.913 (From the table)

Question 2.
70
Solution:
\(\sqrt [ 3 ]{ 70 }\) =4.121 (From the table)

Question 3.
700
Solution:
\(\sqrt [ 3 ]{ 700 } =\sqrt [ 3 ]{ 7\times 100 } \)= 8.879 (from \(\sqrt [ 3 ]{ 10x }\))

Question 4.
7000
Solution:
\(\sqrt [ 3 ]{ 7000 } =\sqrt [ 3 ]{ 70\times 100 }\) = 19.13 (from \(\sqrt [ 3 ]{ 100x }\))

Question 5.
1100
Solution:
\(\sqrt [ 3 ]{ 1100 } =\sqrt [ 3 ]{ 11\times 100 }\) = 10.32 (from \(\sqrt [ 3 ]{ 100x }\))

Question 6.
780
Solution:
\(\sqrt [ 3 ]{ 780 } =\sqrt [ 3 ]{ 78\times 100 }\) = 9.205 (from \(\sqrt [ 3 ]{ 10x }\))

Question 7.
7800
Solution:
\(\sqrt [ 3 ]{ 7800 } =\sqrt [ 3 ]{ 78\times 100 }\) = 19.83 (from \(\sqrt [ 3 ]{ 100x }\))

Question 8.
1346
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 1

Question 9.
250
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 2
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 3

Question 10.
5112
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 4

Question 11.
9800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 5

Question 12.
732
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 6

Question 13.
7342
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 7
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 8

Question 14.
133100
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 9

Question 15.
37800
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 10

Question 16.
0.27
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 11

Question 17.
8.6
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 12
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 13

Question 18.
0.86
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 14

Question 19.
8.65
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 15

Question 20.
7532
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 16

Question 21.
833
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 17
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 18

Question 22.
34.2
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 19

Question 23.
What is the length of the side of a cube whose volume is 275 cm3. Make use of the table for the cube root.
Solution:
RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5 20

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RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1

Question 1.
Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ?
Solution:
Rakesh can do it in 20 days = 1
his 1 day’s work = \(\frac { 1 }{ 20 }\)
and his 4 days work = \(\frac { 1 }{ 20 }\) x 4 = \(\frac { 1 }{ 5 }\) th work

Question 2.
Rohan can paint \(\frac { 1 }{ 3 }\) of a painting in 6 days. How many days will he take to complete the painting ?
Solution:
Rohan can paint \(\frac { 1 }{ 3 }\) of painting in = 6 days
he will complete the painting in = \(\frac { 6 x 3 }{ 1 }\) = 18 days

Question 3.
Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ?
Solution:
Anil’s 1 day’s work = \(\frac { 1 }{ 5 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 1

Question 4.
Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 2

Question 5.
Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
Sita can do a work in 1 hour = \(\frac { 1 }{ 9 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 3

Question 6.
A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 4

Question 7.
A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 5

Question 8.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 6
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 7

Question 9.
A, B and C can reap a field in 15\(\frac { 3 }{ 4 }\) days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 8
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 9

Question 10.
A and B can polish the floors of a building in 10 days A alone can do \(\frac { 1 }{ 4 }\) th of it in 12 days. In how many days can B alone polish the floor ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 10

Question 11.
A and B can finish a work in 20 days. A alone can do \(\frac { 1 }{ 5 }\) th of the work in 12 days. In how many days can B alone do it ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 11
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 12

Question 12.
A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 13

Question 13.
A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 14
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 15

Question 14.
Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 16
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 17

Question 15.
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 18
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 19

Question 16.
6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ?
Solution:
6 men can complete the work in = 7 days
1 man will complete the same work in = 7 x 6 days (Less men, more days)
21 men will finish the work in = \(\frac { 7 x 6 }{ 21 }\) days (More men, less days) = 2 days

Question 17.
8 men can do a piece of work in 9 days. In how many days will 6 men do it ?
Solution:
8 men can do a work in = 9 days
1 men will do the work in = 9 x 8 days (Less men, more days)
6 men will do the work in = \(\frac { 9 x 8 }{ 6 }\) days (More men, less days)
= \(\frac { 72 }{ 6 }\) = 12 days

Question 18.
Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?
Solution:
Reema can weave 35 baskets in = 25 days
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 20

Question 19.
Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ?
Solution:
Neha types pages in 14 hours = 75 pages
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 21

Question 20.
If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ?
Solution:
12 boys in 7 days earn an amount of = Rs. 840
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 22

Question 21.
If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ?
Solution:
25 men can earn in 10 days = Rs. 1000
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 23

Question 22.
Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ?
Solution:
Ashu can copy a book in 18 days working in a day = 8 hours
He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day)
He will copy the book in 12 days working in a day = \(\frac { 8 x 18 }{ 12 }\) hours
(More days, less hours a day)
= \(\frac { 144 }{ 12 }\) = 12 hours a day

Question 23.
If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour.
Solution:
135 garlands in 3 hours are prepared by = 9 girls
1 garland in 3 hours will be prepared by
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 24

Question 24.
A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ?
Solution:
First tap’s 1 hour work to fill the cistern = \(\frac { 1 }{ 8 }\)
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 25

Question 25.
Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 26

Question 26.
A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 27

Question 27.
A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 28
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 29

Question 28.
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ?
Solution:
RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 30

Hope given RD Sharma Class 8 Solutions Chapter 11 Time and Work Ex 11.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.