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find the square root of 23 192/729

Answers:

{tex}\sqrt{23192\over 729}{/tex}
={tex}152.28\over27{/tex}
=5.6
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Find the four rational number 1/2and 3/4

Answers:

{tex}{1\over2 }and{3\over4}{/tex}
LCM of 2 and 4 = 4
{tex}{1\over2 }={1\times2\over2\times2}{/tex}={tex}2\over4{/tex}
{tex}{3\over4}={3\times1\over4\times1}{/tex}={tex}3\over4{/tex}
Four rational numbers between {tex}{2\over4} and {3\over4}{/tex}
are {tex}{1\over4},{0},{1\over4},{2\over4}{/tex}
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find the correct value of {tex}{ \sqrt{0.9}}{/tex}

Answers:
square root upto 2 decimal places:
 {tex}{{5} \over 6}{/tex}
 {tex}7{{3} \over 16}{/tex}
 {tex}1{{7} \over 11}{/tex}

Answers:
I have a total of 300 Rs in coins of denominations 1Rs 2 Rs and 5 Rs. The number of 2 coins is 3 Times the number of $5 coins. The total number of coins is 160 . How many Coins of each denomination are with me?

Answers:

Given
Total no. of coins =160
Let the no of 5₹ coins be x
Therefore no of ₹2 coins = 3x
Now let no of ₹1 coins be y.
So,
3x + x + y = 160
So 4x +y = 160 ....(1)
Since total money in ₹5 coins = 5 × x = 5x
Similarly total money in ₹2 coins = 2 × 3x = 6x
And total money in ₹1 coins = 1×y = y
So we get
5x + 6x + y = 300
11x + y = 300
or, y = 300  11x ....(2)
Putting value of y in ....(1)
4x + 300  11x = 160
7x = 140
x = 20
Now you can find the no of ₹1, ₹2 and ₹5 coins using above equations and value of x.
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The side of a rectangles are 22cm and 120cm respectively Find the length of its diogonal

Answers:

Given: sides of rectangle = 22 cm 120 cm To find:
Diagonal of rectangle =
Using Pythagoras theorem,
Diagonal = {tex}\sqrt {(22)^2+(120)^2}{/tex}
= {tex}\sqrt {14884}{/tex}
= 122 cm.
So length of diagonal is 122 cm.
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Side of rectangle = 22 cm, 120 cm
Diagonal of triangle = {tex}\sqrt {(22)^2+(120)^2}{/tex}
= {tex}\sqrt {484+14400}{/tex}
= {tex}\sqrt {14884}{/tex}= 122 cm
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Let ABCD be a rectangle.
Given: AB = 22 cm and BC = 120 cm
To find: Diagonal AC.
Using Pythagoras theorem,
AC^{2} = AB^{2} + BC^{2}
= (22)^{2} + (120)^{2}
= 484 + 14400
= 14884
AC = 122 cm
The length of given rectangle is 122 cm.
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Find the square root of the following 4004001

Answers:

3  4004001
3  1334667
23  444889
23  19343
29  841
29  29
 1
{tex}\sqrt {4004001} = {/tex} {tex}\sqrt {3 \times 3 \times 23 \times 23 \times 29 \times 29} {/tex}
= 3 x 23 x 29
= 2001
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the area of a square is 59536 sq meter find the length of its side

Answers:

As We Know,
Area Of Square=(side)^{2}
Area Of Square=59536(Given)
So, It Means
59536=(side)^{2}
Side=√59536
So, Side=244 meter
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Find the greatest number of four digits which is a perfect square

Answers:

9801
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Find the least number of four digit which is a perfect square

Answers:

1024
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square root to 2 decimal places {tex} {{3} \over 5}{/tex}

Answers:

{tex}\sqrt {3\over 5}= \sqrt{0.6} = 0.77{/tex}
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{tex}\sqrt {3\over 5}{/tex}
= {tex}1.73\over 2.23{/tex}
= {tex}173\times 100\over 223\times 100{/tex}
= 0.77
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x = 4/3 ( x + 10 )

URGENT PLEASE
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Answers:

Given : x= {tex}{4\over 3}(x+10){/tex}
Cross multiply
=> 3x= 4x+40
=>  40 = 4x3x
=> 40 = x
=> x =40
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x = {tex}{4\over 3}(x+10){/tex}
=> 3x= 4x+40
=> 3x4x= 40
=> x= 40
=> x =40
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A POSITIVE NUMBER IS 5 TIMES ANOTHER NUMBER IF 21 IS ADDED TO BOTH THE NUMBERS THEN ONE OF THE NEW NUMBER BECOMES BECOMES TWICE THE OTER NEW NUMBER BEOMES TWICE THE OTHER NEW NUMBER WHAT ARE NUMBERS

URGENT PLEASE
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Answers:

suppose first number = y
then as per question Second number = 5y
according to conditions given in question
5y+21 = 2(y+21)
=> 5y+21= 2y+42
=> 3y= 21
=> y = 7
First number = 7
Second number = 7*5 =35
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Let first number = x
Second number = 5x
Adding 21 in both number,
We get
First number = x+21
Second number = 5x+21
According to question
5x+21 = 2(x+21)
=> 5x+21= 2x+42
=> 5x2x= 4221
=> 3x= 21
=> x = 7
First number = 7
Second number = 7×5 =35
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How to find side of cube when its volume is given?

Answers:

Volume of cube = {tex}(side)^3{/tex}
So side = {tex}(volume \ of \ cube )^{1\over 3}{/tex}
Means cube root of volume.
For example: if volume of cube is 125cm^{3}, then its side will be {tex}(125)^{1\over 3}=5 cm{/tex}
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I HAVE A TOTAL OF RS.1 ,RS.2 AND RS.5THE NUMBER OF RS.2 IS 3 TIMES THE NUMBER OF RS.5 THE TOTAL NUMBER OF COINS IS 160.HOW MANY NOTES OF EACH DENOMINATION WITH ME

ITS URGENT PLEASE
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Answers:

The question is incomplete.
This question is "I have a total of Rs.300. Rs.1, Rs.2........
Then,
Let the number of Rs.5 coin be x, then number of Rs.2 coin is 3x.
=> Number of Rs.1 coin is 160  (Total number of Rs.2 and Rs.5 coins) = 160  (x + 3x) = 160  4x
Now,
Amount of Rs.1 coin = Rs.1 x (160  4x) = Rs. (160  4x)
Amount of Rs.2 coin = Rs.2 x {tex}3x{/tex} = Rs. {tex}6x{/tex}
Amount of Rs.5 coin = Rs.5 x {tex}x {/tex} = Rs. {tex}5x{/tex}
According to question,
{tex}1604x+6x+5x=300{/tex}
=> {tex}160+7x=300{/tex}
=> {tex}7x=300160{/tex}
=> {tex}7x=140{/tex}
=> {tex}x = 20{/tex}
Therefore,
Number of Rs.1 coin = {tex}1604x{/tex} = 160  4 x 20 = 160  80 = 80
Numebr of Rs.2 coin = {tex}3x{/tex} = 3 x 20 = 60
Numebr of Rs.5 coin = {tex}x {/tex} = 20
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three consicutive integers are such that when they are taken in incresing order and multiplied by 2 ,3 and 4 respectively, they add up to get 74 find these numbers

Answers:

Let the three consecutive integers be X, X+1, X+2.
According to condition,
2(X)+3(X+1)+4(X+2)=74
2X+3X+3+4X+8=74
9X+11=74
9X=7411
9X=63
X=7
So the three consecutive numbers are 7, 7+1=8 and 7+2=9
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3x  3x=0

Answers:

3x3x=0
=>3 (××)=0
=>3 (0)=0
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3X3X=0
3 (XX)=0
3 (0)=0
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is 2352 a perfect square? if not, find the smallest multiple of 2352 which is a perfect square. find the square root of the new number

Answers:

<pre>2352 = 2^{4}∙3∙7^{2} The primes 2 and 7 already have even powers, so we must multiply by 3 so that the 3 will also have an even power. So we must multiply by 3: 2352∙3 = 2^{4}∙3^{2}∙7^{2} The answer is 3. Then the perfect square will be 7056, which is 2^{4}∙3^{2}∙7^{2} and (2^{2}∙3∙7)^{2} = (4∙3∙7)^{2} = 84^{2}</pre>Thanks (0)
Show that 216/42875 is the cube of a rational number also find the rational number.

Answer it fast because I need it
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Answers:

216 is the cube of 6.
42875 is the cube of 35.
So, the rational number is 6/35.
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Parikshit makes a cuboid of plasticine of 5cm, 2cm,5cm. How many such cuboids will he need to form a cube?

Answers:

Volume of the cuboid of sides 5 cm, 2 cm, 5 cm = 5 cm ☓ 2 cm ☓ 5 cm = (5 ☓ 5 ☓ 2) cm^{3}
Here, two 5s and one 2 are left which do not form a triplet.
To make it a pefect cube, multiply this expression by 2 ☓ 2 ☓ 5 = 20,
then it will become a perfect cube. i.e. (5 ☓ 5 ☓ 2 ☓ 2 ☓ 2 ☓ 5) = (5 ☓ 5 ☓ 5 ☓ 2 ☓ 2 ☓ 2) = 1000 is a perfect cube.
Therefore 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.
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solve the following linear equations:
 3(5z  7)  2(9z 11) = 4(8z  13) 17

Answers:

 3(5z  7)  2(9z 11) = 4(8z  13)17
 15z2118z+22 = 32z5217
 15z18z32z+1 = 5217
 15z50z =52171
  35z = 70
 z =70/35
 = 2
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A number divided by three ,plus nine equals twice the number minus sixteen. find the number

Answers:

Let the no. be = x
According to ques.
x/3+9 = 2x16
x/32x =169
5x/3 = 25
5x = 75
x =15
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To find the square root of a number, correct upto 3 places of decimal, the number of digits after decimal point should be??
(a). 3 (b). 4 (c). 6 (d). None

C 6

Please answer fast if anybody knows the answer!!
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Answers:

(c) is the correct answer.
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b
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(c) 6
Number of decimal points in square root of number = {tex}Number\ of\ decimal\ points\ in\ number \over 2{/tex}
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amy went to a bank to withdrewal Rs2000 she asked to casher to give Rs 50 and Rs100 note only she got 25 notes in all. Find how many 50 and 100 rupees note she received.

Our solution totally matches Rajendra.
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Answers:

Let Amy have x notes of ₹ 50.
Then no of ₹100 she have will have is (25x).
According to question equation will be
50x+100(25x)=2000
250050x=2000
x=500/50
x=10
Then she have 10 notes of ₹50 and 2510=15 notes of ₹ 100.
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Let no of Rs. 50 notes = x
No of Rs. 100 notes = 25  x.
50 x + 100(25 x) = 2000
50x + 2500  100x = 2000
50x = 500
x = 10 notes
y = 2510 = 15 notes
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One third of square root of which no..is 0.001

Answers:

Let number is = x
Then according to question
{tex}{1\over 3}\sqrt x= 0.001{/tex}
Squaring both sides
{tex}=> {1\over 9}x= 0.000001{/tex}
{tex}=> x= 0.000009{/tex}
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Aman's age is three times his son's age. Ten years ago he age was five times his son's age. Find their present ages.

Answers:

Let present age of son = x
Then present age of Aman = 3x
Ten year ago
age of son = x10
Age of Aman = 3x10
According to question
=> 3x10 = 5(x10)
=> 3x10 = 5x50
=> 10+50 = 5x3x
=> 40=2x
=> x= 20
So present age of son = 20
Present age of Aman = 20*3= 60
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Solve using suitable properties :
2/3 * 3/5 + 5/2  3/5 * 1/6

Answers:

{tex}{2\over 3} \times {3\over 5} +{5\over 2}{3\over 5} \times {1\over 6}{/tex}
= {tex}{2\over 3} \times {3\over 5}{3\over 5} \times {1\over 6} +{5\over 2} {/tex}
Using distributive property of multiplication,
{tex}= {3\over 5}\times \left ( {2\over 3}{1\over 6}\right ) +{5\over 2} {/tex}
{tex}= {3\over 5}\times \left ({ 41\over 6}\right ) +{5\over 2} {/tex}
{tex}= {3\over 5}\times { 5\over 6} +{5\over 2} {/tex}
= {tex}{1\over 2} + {5\over 2}{/tex}
{tex}= {1+5\over 2} = {4\over 2}{/tex}
= 2
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Solve :
0.5 (y  0.4)  0.6 (y  2.71)  y + 6.1
0.35 0.42

Answers:

{tex}{0.5 (y  0.4)\over0.35} { 0.6 (y  2.71) \over 0.42} y + 6.1{/tex}
= {tex}{ (y  0.4){10\over 7}} { (y  2.71){10 \over 7}} y + 6.1{/tex}
= {tex}{ 10(y  0.4)} { (y  2.71)10} 7y + 42.7\over 7{/tex}
= {tex} 10y  4  10y +27.1 7y + 42.7\over 7{/tex}
= {tex}4 7y + 69.8\over 7{/tex}
= {tex}65.8 7y\over 7{/tex}
= 9.4  y
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What is the smallest number which must be added to 15370 to make it a perfect square? (Show workings)

Answers:

Ans. First let us find the square of 15370 using long division method.
The remainder is 241. Hence, (123)^{2} < 15370
The next perfect square number is (124)^{2} = 15376 > 15370
Hence, the number to be added is 15376  15370 = 6
Therefore 6 should be added to 15370 to get a perfect square.Thanks (1)