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20 people (men,woman,children) earn 20 coins between them.Each man earns 3 coins,each woman earns 11/2 coins and each child earns 1/2 coin. How many men, women and children are there?

Answers:

Ans. let the number of men, women and children be denoted by M, W and C respectively.
Then
M + W + C = 20 ....... (1)
3M + 1.5 W + 0.5 C = 20 ........ (2)
Multiply equation (2) by (2)
6M + 3W + C = 40 ........(3)
From (1) and (2)
W + C = 20  M ..........(4)
3 W + C = 40  6M ...... (5)
The above equations will be solved for W and C in terms of M
W = 10  {tex}{5M\over 2} {/tex} and C = 10 + {tex} {3M\over 2}{/tex}
Because W and C have to be positive integers greater than 1 we conclude
(a) M is an even number.
(b) W {tex}\ge{/tex} 1 implies {tex}{5M\over 2 } \le 9{/tex}
it is seen W = 5 and C = 13 when M = 2 . Also {tex}{5M\over 2} = 5 \lt 9{/tex}
but When M is greater than 2 the condition is false . SoThe only solution is M = 2, W =5 and C = 13
So Number of Men = 2
Number of Women = 5
Number of Children = 13
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Solve:
{m(m1/2)}= {1(m2/3)}

Answers:

{tex}\{ m(m{1\over 2})\} = \{ 1(m{2\over 3})\}{/tex}
=> m  m + {tex}1\over 2{/tex}= 1  m + {tex}2\over 3{/tex}
=> m = {tex}{5\over 3}  {1\over 2}{/tex}
=> m = {tex}7\over 6{/tex}
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Solve :
3(5x7)2(9x11)=4(8x13)17

Thank you so much.
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Answers:

Ans.
3(5x7)2(9x11)=4(8x13)17=> 15x  21  18x + 22 = 32x  52  17
=> 15x  18x  32x =  52  17 + 21  22
=> 35x =  70
=> x = {tex}{70\over 35} = 2{/tex}
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While three watchman were guarding an onchard,a thief slipped into and stole some apple. On this way out,he met the three watchman one after another and to each in turn he gave half the apples he had and two besides. In this way he managed to flee away with one apple. How many apples had he stolen originally?

Answers:

Ans. let the number of apples he stole = x
Case 1 : Apples given to the first watchmen = {tex}{x\over 2} + 2 = {x+4\over 2}{/tex}
Apples remained = {tex}x  {x+4\over 2} = {2xx4\over 2} = {x4\over 2}{/tex}
Case 1 : Apples given to the second watchmen = {tex}{{x4\over 2}\over 2} + 2 = {x4\over 4} +2 {/tex}{tex}= {x4 + 8\over 4} = {x+4\over 4}{/tex}
Apples remained = {tex}{x4\over 2}  {x+4\over 4} = {2x8x4\over 4} = {x12\over 4}{/tex}
Case 1 : Apples given to third watchmen = {tex}{{x12\over 4}\over 2} + 2 = {x12\over 8}+2 {/tex}{tex}= {x12+16\over 8} = {x+4\over 8}{/tex}
Apples Remained = {tex}{x12\over 4} {x+4\over 8} = {2x24x4\over 8} = {x28\over 8}{/tex}
According to question :
{tex}{x28\over 8} = 1 {/tex}=> x  28 = 8
=> x = 8+28
=> x = 36
So No of Apples he stole = 36
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Two planes are flying towards each other and the difference between their speeds is 60km/have. If they are 2400km apart and passes each other after 5 hours,find their speeds.

Answers:

Let the speed of First plane is x km/hr and second plane is (x + 60) km/hr.
Time taken to pass each other = 5 hours.
Than, Distance covered by first plane in 5 hours = speed × time = x × 5 = 5x.
Distance covered by second plane in 5 hours = speed × time = (x + 60) 5 = 5x + 300
Distance coverec by first plane + distance covered by second plane = 2400 km
5x + (5x + 300) = 2400
10x + 300 = 2400
10x = 2400  300 = 2100
x = {tex}{{2100} \over {10}}\, = \,210{/tex}
Hence, the speed of first plane is = x = 210 km/hr
and the speed of second plane is = x + 60 = 210 + 60 = 270 km/hr
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The speed if boat in still water is 30 km/hr. it takes the same time for the boat to travel 5 km upstream and 10 km downstream find the speed of the stream.

Answers:

Let the speed of the stream be {tex}x{/tex} km/hr.
We know that, Speed = {tex}{{{\rm{Distance}}} \over {{\rm{Time}}}}{/tex}
Speed in upstream = (x 30) Km/h and Speed in downstream =(x + 30) Km/h
Then, Time = {tex}{{{\rm{Distance}}} \over {{\rm{Speed}}}}{/tex}
According to question,
{tex}{5 \over {x  30}} = {{10} \over {x + 30}}{/tex}
=> {tex}5\left( {x + 30} \right) = 10\left( {x  30} \right){/tex}
=> {tex}5x+150=10x300{/tex}
=> {tex}10x5x=150+300{/tex}
=> {tex}5x=450{/tex}
=> {tex}x=90{/tex}
Therefore, the speed of stream is 90 km/hr
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A boat travels 20 km upstream in a river in the same period of time as it takes to travel 30 km downstream if the rate of the stream is 4 km/hr find the speed of boat in still water.

Answers:

Ans. Let the speed of boat in still water = x Km/h
rate of stream = 4 Km/h
So, Speed of Boat downstream = (x+4) Km/h
Speed of Boat upstream = (x3) Km/h
Time Taken to travel 20 Km Upstream = {tex}{20\over x4 }\space hour {/tex}
Time Taken to travel 30 Km Downstream = {tex}{30\over x+4} \space hour{/tex}
ATQ,
{tex}=> {20\over x4} = {30\over x+4}{/tex}
{tex}=> 20x+ 80 = 30x  120{/tex}
{tex}=> 30x  20x = 120 + 80 {/tex}
{tex}=> 10 x = 200 {/tex}
=> x = 20
Speed of Boat in still water = 20 Km/h
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A certain two digit number is equal to 9 times the sum of its digits.If 63 were subtracted from the number the digits would be reversed.Find the number.

Answers:

Let the ones place digit be y and tens place be x, then the number be 10x + y.
According to quesiton,
10x + y = 9(x + y)
=> 10x + y = 9x + 9y
=> 10x  9x = 9y  y
=> x = 8y ..........(i)
Again according to second condition,
10x + y  63 = 10y + x
=> 10 x  x + y  10y = 63
=> 9x  9y = 63
=> x  y = 7
Putting the value of x from eq. (i), we get
8y  y = 7
=> 7y = 7
=> y = 1
Putting the value of y in eq. (i), we get
x = 8 x 1 = 8
Therefore, the number is 10 x 8 + 1 = 81.
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Mayank visits three shops one by one.At each shop,he spent two third of the money he had at that shop.If he is left with 32 with him after visiting the shop, find the amount of money he had in the beginning.

Answers:

Solution:
Let money he had in beginning = Rs. x
Money spend on first shop = {tex}2x\over 3{/tex}
Remaining Money = {tex}x  {2x\over 3} = {x \over 3}{/tex}
Money spend on 2nd shop = {tex} {2\over 3}\times {x \over 3} = {2x\over 9}{/tex}
Remaining Money = {tex}{x\over 3}  {2x\over 9} = {x \over 9}{/tex}
Money Spend on 3rd shop = {tex}{2\over 3} \times {x \over 9} = {2x\over 27}{/tex}
Remaining Money = {tex}{x \over 9} {2x\over 27} = {x \over 27}{/tex}
ATQ,
=> {tex}{x\over 27} = 32{/tex}
=> x = 864
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Mahesh sells an article to David at aprofit of 8%, David sells it to mukesh at a loss of 5%.If Mukesh paid Rs 51700, how much did it cost to mahesh.

Answers:

{tex}\eqalign{ & Let\,the\,C.P.\,ofarticle\,for\,Mahesh\,be\,Rs.\,100 \cr & \,the\,S.P.\,of\,article\,for\,Mahesh\,at\,8\% \,profit\, = \,Rs.\,(100\, + \,8)\, = \,Rs.\,108 \cr & Mahesh\,sells\,article\,to\,David\,at\,Rs.\,108 \cr & So,\,the\,C.P.\,ofarticle\,for\,David\,is\,Rs.\,108 \cr & Now,\,David\,sells\,the\,article\,to\,Mukesh\,at\,5\% \,loss. \cr & So,\,S.P.\,of\,article\,for\,David\,at\,5\% \,loss\, = \,\left( {{{100\,  \,loss\% } \over {100}}} \right)\, \times \,C.P.\, = \,\left( {{{100\,  \,5} \over {100}}} \right)\,108\, = \,{{95} \over {100}}\, \times \,108\, = \,Rs.\,102.60 \cr & It\,means\,Mukesh\,paid\,Rs.\,102.60\,to\,David. \cr & Now,\,C.P.\,of\,article\,for\,Mahesh\,is\,Rs.\,100,\,when\,Mukesh\,pays\,Rs.\,102.60\,to\,David. \cr & If\,Mukesh\,pays\,Rs.\,51700\,to\,David,\,than\,C.P.\,of\,article\,for\,Mahesh\,will\,be\, = \,{{51700\, \times \,\,100} \over {102.60}}\, = \,Rs.\,50389.86 \cr} {/tex}
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One third of a number increased by 5 is equal to its half part decreased by 5.find the number.

Answers:

Solution:
Let Number = x
ATQ,
{tex}{x\over 3} + 5={x\over 2} 5{/tex}
{tex}=> {x\over 3}  {x\over 2} = 5+5{/tex}
{tex}=> {3x2x\over 6} = 10{/tex}
{tex}=> {x\over 6} = 10{/tex}
=> x = 60
Number = 60
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The perimeter of a rectangular field is 84 meters .If the breadth of the field is increased by 2 meters and it's length is decreased by 4 meters ,the area is decreased by 32sq m.Find the length and breadth of the rectangular field.

Answers:

Ans. Perimeter of Field = 84 meter
Let Length of Field = x meter
Perimeter = {tex}2(x+breadth\space of \space field ){/tex}
=> {tex}84 = 2(x+breadth\space of \space field ){/tex}
{tex}=> 42 = x+breadth\space of \space field {/tex}
{tex}=> breadth\space of \space field = (42  x ) meter{/tex}
Old Area = {tex}x\times (42x) = 42x  x^2{/tex}
According to Ques,
New breadth = 42  x + 2 = (44  x) meter
New length = (x  4 ) meter
New Area = {tex}(44  x) (x4) = 44x  176x^2+4x {/tex}
So, New Area = Old Area  32
=> {tex}44x  176 x^2 +4x = 42xx^2 32{/tex}
=> {tex}44x +4x  42x x^2 + x^2 = 32 + 176{/tex}
=> {tex}6x = 144{/tex}
=> x = 24
So, Length of Field = 24 meter
Breadth of Field = 42 24 = 18 meter
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Prabhat's age is one fourth of his father's age. Five years from now prabhat's age will be one third of his father's present age .Find their age.

Answers:

Ans. Let Present Age of Parbhat's Father = x
Then Present age of Parbhat = {tex}x\over 4{/tex}
Age of Prabhat after 5 years = {tex}{x\over 4 }+ 5 = {x+20\over 4}{/tex}
According to Ques,
=> {tex}{x+20\over 4} = {x\over 3}{/tex}
=> 3x + 60 = 4x
=> 4x  3x = 60
=> x = 60
So, Present Age of Father = 60 years
Present age of Prabhat = {tex}{60\over 4 } = 15 {/tex} years
Thanks (1)
Length of a rectangle is 6 cm more than its width.If it's length and breadth each is decreased by 3cm the area of new rectangle is decreased by 36 Sq cm.Find the original length and breadth of the original rectangle.

Answers:

Let the original Breadth of rectanlge be = x cm
and the original Length of rectangle be = (x + 6) cm
Therefore, Original Area = Length × Breadth = (x + 6) × x = x^{2} + 6x
Now, New Length = (x + 6)  3 = x + 6  3 = x + 3
New Breadth = x  3
Therefore, New Area = Length × Breadth = (x + 3) × (x  3) = (x)^{2}  (3)^{2} = x^{2}  9
According to Question
Original Area  New Area = 36 sq cm.
(x^{2} + 6x)  (x^{2}  9) = 36
x^{2} + 6x  x^{2} + 9 = 36
6x + 9 = 36
6x = 36  9
6x = 27
x = {tex}{27\over6} = {9\over2}{/tex}
Hence, Original Length of Rectangle = x + 6 = {tex}{9\over 2} + 6 = {9 + 12\over 2} = {21\over 2} = 10.5{/tex}cm
Original Breadht of Rectangle = x = {tex}{9\over 2} = 4.5{/tex} cm
Thanks (1)
The sum of the digits of a two digit number is 14.If the number formed by reversing the digits is less than the original number by 18.Find the original numbers .

Answers:

Let x = the 10's digit
Let y = the units
:
10x + y = the original two digit number
:
The sum of the digits of a twodigit number is 14.
x + y = 14
:
"If the digits are reversed, the number is 18 less than the original number."
Rev no. = orig no.  18
10y + x = 10x + y  18
10y  y = 10x  x  18
9y = 9x  18
simplify, divide by 9
y = x  2
:
In the sum equation replace y with (x2)
x + (x2) = 14
2x = 14 + 2
2x = 16
x = 8
obviously y = 6
:
86 is the original number
:
:
Check solution in the statement:
"If the digits are reversed, the number is 18 less than the original number."
68 = 86  18Thanks (0)
In a colony 6/8 of the total people were females.If the number of females is 36 more than that of the males ,how many males were there in the colony.

Answers:

Solution : let total number of people = x
Number of female = {tex}6x\over 8{/tex}
No of male = Total people  no of female
= {tex}x  {6x \over 8} = {2x\over 8}{/tex}
According To Ques,
{tex}=> {6x\over 8} = 36 + {2x\over 8}{/tex}
{tex}=> {6x\over 8}  {2x\over 8} = 36{/tex}
{tex}=> {4x\over 8} = 36 {/tex}
=> x = 72
So No.of male = {tex}=> {2\over 8} \times 72 = 18{/tex}
Thanks (1)
After travelling a distance of 32 4/5kmamera found tht she is still to cover 5/7 of the whole distance. What is the total distance covered by Amera?

Answers:

Let the total distance covered be x km.
Then according to question,
{tex}32{4 \over 5} + {5 \over 7}x = x{/tex}
=> {tex}x  {5 \over 7}x = 32{4 \over 5}{/tex}
=> {tex}{{7x  5x} \over 7} = {{164} \over 5}{/tex}
=> {tex}{{2x} \over 7} = {{164} \over 5}{/tex}
=> {tex}x = {{164} \over 5} \times {7 \over 2}{/tex}
=> {tex}x = {{1148} \over {10}}{/tex}= {tex}114{8 \over {10}}{/tex}km = 114.8 km
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0.99999999........ Is a rational number or irrational number??

It can be written in the form of p/q where q is not 0
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Answers:

It is a rational number.
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0.999999999...........Is a rational number or irrational number ??

Answers:

Rational no. Yes check itt
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223223223+23453332

Answers:

246676555
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1/2+(1/6)+(2/3)+2/6 evaluate the following by grouping the rational no using the associative and communicative properties of addition

What is the answer
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Answers:

{tex}{1 \over 2} + \left( {{{  1} \over 6}} \right) + \left( {{{  2} \over 3}} \right) + {2 \over 6}{/tex}
= {tex}{1 \over 2}  {1 \over 6}  {2 \over 3} + {2 \over 6}{/tex}
= {tex}{1 \over 2}  {2 \over 3} + \left( {{2 \over 6}  {1 \over 6}} \right){/tex}
= {tex}{1 \over 2}  {2 \over 3} + {1 \over 6}{/tex}
= {tex}{1 \over 2} + \left( {{1 \over 6}  {2 \over 3}} \right){/tex}
= {tex}{1 \over 2} + \left( {{{1  4} \over 6}} \right){/tex}
= {tex}{1 \over 2}  {3 \over 6}{/tex}
= {tex}{1 \over 2}  {1 \over 2}{/tex}
= 0
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I spent 5% of my total income in charity, 22% in buisness and 53% in family. Now i Had Rs. 2163 and forty paisa. Find my total income

Answers:

Total expenditure = 5% + 22% + 53% = 80%
Money left = 100%  80% = 20%
Let total income be Rs. x.
According to question,
20% of x = 2163.40
20/100 × x = 2163.40
x = 2163.40 × 100 ÷ 20
x = Rs. 10817
Therefore total income is Rs. 10817.
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A person has deposited amount at 12% rate find out years to make principle double.

Answers:

Let amount deposited be P. Then amount after required years will be 2P.
Now, S.I. = P x R x T / 100
S.I. = P x 12 x T / 100 = 2PT/25
Amount = P + S.I
2P = P + (2PT/25)
P = 2PT/25
2T = 25
T = 12.5
Therefore the time required for principal double is 12.5 years.
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At what rate , Rs 9900 amount to 13365 in 5 years

Answers:

SI = (P X R X T )/100
P = Rs. 9900
A= Rs. 13,365
SI = AP = Rs.3465
T = 5years
3465 X 100 = 9900 X R X 5
R = 7%
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A cylindrical roller has perimeter 3.5 m and dimeter 2.8 m if roller rolled 200 times , find area of playground

Answers:
The cube of metal 3,4,5cm are melted and made in to single cube. Find the side of cube.

Answers:

Volume of single cube = 3^{3}+4^{3}+5^{3}
a^{3} = 27+64+125
a^{3}= 216
a^{3}= 6^{3}
a=6cm
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A person throw bucjet if water of square shape if total water is 8 litre . Find out side of bucket.

Answers:

Ans. Volume of water = 8 L = 0.008m^{3}
Let Side of bucket = x m
Volume of bucket = Volume of water
=> x^{3} = 0.008
=> x^{3} = (0.2)^{3}
=> x = 0.2 m = 20 cm
Side of bucket is 20 cm
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There are chikens with color 50 yellow 50 black and 50 red. Find inique probability of all colors.

Answers:

Number of yellow chickens = 50
Number of black chickens = 50
Number of red chickens = 50
Total number of chickens = 50 + 50 + 50 = 150
Therefore, Probability of yellow colour chickens = {tex}{{50} \over {150}}\, = \,{1 \over 3}{/tex}
Probability of black colour chickens = {tex}{{50} \over {150}}\, = \,{1 \over 3}{/tex}
Probability of red colour chickens = {tex}{{50} \over {150}}\, = \,{1 \over 3}{/tex}
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At what rate % per annum will rs 50000 to rs 86400 in 3 years if interest compounded annually

Answers:

Ans. Amount (A)= 86400
Time(n) = 3 years
Principal (P)= 50000
Let rate (r) = r%
Then We Know,
{tex}A = P \times ({1 +{r\over 100}})^n{/tex}
{tex}=> 86400 = 50000\times ({1+{r\over 100}})^3{/tex}
{tex}=> {86400\over 50000} = ({1+{r\over 100}})^3{/tex}
{tex}=> {216\over 125} = ({1+{r\over 100}})^3{/tex}
{tex}=> ({6\over 5})^3= ({1+{r\over 100}})^3{/tex}
On comparing, We get
{tex}=> ({6\over 5})= ({1+{r\over 100}}){/tex}
{tex}=> {6\over 5}  1= {r\over 100}{/tex}
{tex}=> {1\over 5} = {r\over 100} => r = 20\%{/tex}
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Find compound interest on rs 25000 for 1 and half year at 20% rate compounded half yearly.

Answers:

Principal = Rs. 25000
Time = {tex}1{1 \over 2}\,year{/tex} = 3 Half Years
Rate = 20% p.a. = 10% Half yearly.
{tex}Amount\,\, = \,\,P{\left( {1\,\, + \,\,{R \over {100}}} \right)^n}\,\, = \,\,25000\,{\left( {1\,\, + \,\,{{10} \over {100}}} \right)^3}\,\, = \,\,25000\,{\left( {1\, + \,{1 \over {10}}} \right)^3}\, = \,25000\,{\left( {{{10\, + \,1} \over {10}}} \right)^3}{/tex}
{tex}= \,25000\, \times \,\,{{11} \over {10}}\,\, \times \,\,{{11} \over {10}}\,\, \times \,\,{{11} \over {10}}\,\, = \,\,Rs. 33275{/tex}
Compound Interest = Amount  Principal = Rs (33275  25000) = Rs. 8275
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