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(x-2)(x+3) ×(x+2)(x-3)- 2x(x+1)=0

 

Posted by Hunar Chhabada (Jul 14, 2017 8:46 p.m.) (Question ID: 6643)

  • Please correct the question

    Posted by Poulami Dasgupta (Jul 17, 2017 8:32 p.m.)
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find the square root of 23 192/729

Posted by Shahista Anjum (Jul 14, 2017 2:15 p.m.) (Question ID: 6636)

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  • {tex}\sqrt{23192\over 729}{/tex}

    ={tex}152.28\over27{/tex}

    =5.6

    Answered by Poulami Dasgupta (Jul 16, 2017 3:46 p.m.)
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Find the four rational number -1/2and 3/4

Posted by Shahista Anjum (Jul 13, 2017 4:23 p.m.) (Question ID: 6614)

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  • {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}

    Answered by Poulami Dasgupta (Jul 16, 2017 10:52 a.m.)
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find the correct value  of {tex}{ \sqrt{0.9}}{/tex} 

Posted by Aanvi Gupta (Jul 09, 2017 12:01 p.m.) (Question ID: 6504)

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square root upto 2 decimal places:

  • {tex}{{5} \over 6}{/tex}
  • {tex}7{{3} \over 16}{/tex}
  • {tex}1{{7} \over 11}{/tex}
Posted by Aanvi Gupta (Jul 09, 2017 11:43 a.m.) (Question ID: 6502)

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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?

 

Posted by Naveen Satheesh (Jul 08, 2017 6:56 p.m.) (Question ID: 6486)

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  • 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.

    Answered by Maulik Mehta (Jul 09, 2017 7:09 a.m.)
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The side of a rectangles are 22cm and 120cm respectively Find the length of its diogonal 

Posted by Shahista Anjum (Jul 08, 2017 11:57 a.m.) (Question ID: 6474)

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  •  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.

    Answered by Sahdev Sharma (Jul 08, 2017 1 p.m.)
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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

    Answered by Payal Singh (Jul 08, 2017 12:56 p.m.)
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  • Let ABCD be a rectangle.

    Given: AB = 22 cm and BC = 120 cm

    To find: Diagonal AC.

    Using Pythagoras theorem,

    AC2 = AB2 + BC2

           = (22)2 + (120)2

           = 484 + 14400

           = 14884

    AC = 122 cm

    The length of given rectangle is 122 cm.

    Answered by Rashmi Bajpayee (Jul 08, 2017 12:56 p.m.)
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Find the square root of the following  4004001

Posted by Shahista Anjum (Jul 08, 2017 10:12 a.m.) (Question ID: 6473)

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  •   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

    Answered by Rashmi Bajpayee (Jul 08, 2017 11:41 a.m.)
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the area of a square is 59536 sq meter find the length of its side 

Posted by Shahista Anjum (Jul 08, 2017 8:10 a.m.) (Question ID: 6470)

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  • 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

     

    Answered by Arun Soni (Jul 08, 2017 9:06 a.m.)
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Find the greatest number of four digits which is a perfect square 

Posted by Shahista Anjum (Jul 08, 2017 8:07 a.m.) (Question ID: 6469)

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  • 9801

     

    Answered by Suraj Shringi (Jul 08, 2017 10:40 a.m.)
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Find the least number of four digit which is a perfect square 

Posted by Shahista Anjum (Jul 08, 2017 8:06 a.m.) (Question ID: 6468)

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  • 1024

    Answered by Suraj Shringi (Jul 08, 2017 10:40 a.m.)
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square root to 2 decimal places {tex} {{3} \over 5}{/tex}

Posted by Aanvi Gupta (Jul 05, 2017 8:02 p.m.) (Question ID: 6398)

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  • {tex}\sqrt {3\over 5}= \sqrt{0.6} = 0.77{/tex}

    Answered by Sahdev Sharma (Jul 06, 2017 5:47 a.m.)
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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

    Answered by Payal Singh (Jul 06, 2017 5:44 a.m.)
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x = 4/3 ( x + 10 )

 

Posted by Goutham Krishna (Jul 04, 2017 8:45 p.m.) (Question ID: 6380)

  • URGENT PLEASE

     

    Posted by Goutham Krishna (Jul 04, 2017 8:46 p.m.)
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  • Given : x= {tex}{4\over 3}(x+10){/tex}

    Cross multiply 

    => 3x= 4x+40

    => - 40 = 4x-3x

    => -40 = x

    => x =-40

    Answered by Payal Singh (Jul 05, 2017 5:57 a.m.)
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  • x = {tex}{4\over 3}(x+10){/tex}

    => 3x= 4x+40

    => 3x-4x= 40

    => -x= 40

    => x =-40

    Answered by Sahdev Sharma (Jul 05, 2017 5:53 a.m.)
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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

Posted by Goutham Krishna (Jul 04, 2017 8:16 p.m.) (Question ID: 6374)

  • URGENT PLEASE

     

    Posted by Goutham Krishna (Jul 04, 2017 8:17 p.m.)
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  • 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

    Answered by Sahdev Sharma (Jul 05, 2017 6:04 a.m.)
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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

    => 5x-2x= 42-21

    => 3x= 21

    => x = 7

    First number = 7

    Second number = 7×5 =35

    Answered by Payal Singh (Jul 05, 2017 6:01 a.m.)
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How to find side of cube when its volume is given?

Posted by Munazza Sahr (Jul 04, 2017 1:45 a.m.) (Question ID: 6358)

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  • 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 125cm3, then its side will be {tex}(125)^{1\over 3}=5 cm{/tex}

    Answered by Payal Singh (Jul 04, 2017 6:19 a.m.)
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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

Posted by Goutham Krishna (Jul 03, 2017 10:22 p.m.) (Question ID: 6353)

  • ITS URGENT PLEASE

     

    Posted by Goutham Krishna (Jul 03, 2017 10:24 p.m.)
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  • 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}160-4x+6x+5x=300{/tex}

    =>     {tex}160+7x=300{/tex}

    =>     {tex}7x=300-160{/tex}

    =>     {tex}7x=140{/tex}

    =>     {tex}x = 20{/tex}

    Therefore,

    Number of Rs.1 coin = {tex}160-4x{/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 

    Answered by Rashmi Bajpayee (Jul 04, 2017 12:02 p.m.)
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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 

Posted by Goutham Krishna (Jun 30, 2017 8:48 p.m.) (Question ID: 6283)

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  • 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=74-11

    9X=63

    X=7

    So the three consecutive numbers are 7, 7+1=8 and 7+2=9

    Answered by Supriyo Ghosh (Jun 30, 2017 9:36 p.m.)
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3x - 3x=0

Posted by Goutham Krishna (Jun 30, 2017 8:46 p.m.) (Question ID: 6282)

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  • 3x-3x=0

    =>3 (×-×)=0

    =>3 (0)=0

    Answered by Estherrani Ponthurai (Jul 02, 2017 1:36 a.m.)
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  • 3X-3X=0

    3 (X-X)=0

    3 (0)=0

    Answered by Estherrani Ponthurai (Jul 02, 2017 1:31 a.m.)
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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

Posted by Japneet Chawla (Jun 30, 2017 2:38 p.m.) (Question ID: 6273)

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  • <pre>2352 = 24∙3∙72 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 = 24∙32∙72 The answer is 3. Then the perfect square will be 7056, which is 24∙32∙72 and (22∙3∙7)2 = (4∙3∙7)2 = 842</pre>
    Answered by Shweta Gulati (Jun 30, 2017 11:51 p.m.)
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Show that -216/42875 is the cube of a rational number also find the rational number. 

Posted by Dinesh Kumar (Jun 29, 2017 9:59 p.m.) (Question ID: 6262)

  • Answer it fast because I need it

     

    Posted by Dinesh Kumar (Jun 30, 2017 11:42 a.m.)
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  • 216 is the cube of 6.

    42875 is the cube  of 35.

    So, the rational number is 6/35.

    Answered by Shweta Gulati (Jul 01, 2017 12:14 a.m.)
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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? 

Posted by Dinesh Kumar (Jun 29, 2017 11:28 a.m.) (Question ID: 6243)

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  • Volume of the cuboid of sides 5 cm, 2 cm, 5 cm = 5 cm ☓ 2 cm ☓ 5 cm = (5 ☓ 5 ☓ 2) cm3

    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.

    Answered by Payal Singh (Jun 29, 2017 2:28 p.m.)
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solve the following linear equations:

  • 3(5z - 7) - 2(9z -11) = 4(8z - 13) -17
Posted by Aanvi Gupta (Jun 27, 2017 4:58 p.m.) (Question ID: 6211)

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    • 3(5z - 7) - 2(9z -11) = 4(8z - 13)-17
    • 15z-21-18z+22       = 32z-52-17
    • 15z-18z-32z+1       =  -52-17
    •  15z-50z                =-52-17-1
    • - 35z                      = -70
    •       z                     =-70/-35
    •                              = 2
    Answered by Renu Gandhi (Jun 27, 2017 7:13 p.m.)
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A number divided by three ,plus nine equals twice the number minus sixteen. find the number

Posted by Aanvi Gupta (Jun 27, 2017 4:43 p.m.) (Question ID: 6209)

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  • Let the no. be  = x 

    According to ques.

     x/3+9  =  2x-16

    x/3-2x  =-16-9

    -5x/3  = -25

    -5x  =  -75

    x  =15

    Answered by Renu Gandhi (Jun 27, 2017 7:21 p.m.)
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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

 

Posted by Aryaman Vijayran (Jun 30, 2017 12:33 p.m.) (Question ID: 6116)

  • C 6

    Posted by Hans Raj (Jun 23, 2017 4:46 p.m.)
  • Please answer fast if anybody knows the answer!!

     

    Posted by Aryaman Vijayran (Jun 23, 2017 1:13 p.m.)
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  • (c) is the correct answer. 

    Answered by Dinesh Kumar (Jun 30, 2017 11:37 a.m.)
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  • b

    Answered by Dhivya N (Jun 26, 2017 12:34 p.m.)
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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}

    Answered by Payal Singh (Jun 30, 2017 12:33 p.m.)
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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. 

Posted by Shiv Shankar Sahu (Jun 22, 2017 7:56 p.m.) (Question ID: 6102)

  • Our solution totally matches Rajendra.

    Posted by Sagar Tripathi (Jun 22, 2017 8:58 p.m.)
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  • Let Amy have x notes of ₹ 50.

    Then no of ₹100 she have will have is (25-x).

    According to question equation will be

    50x+100(25-x)=2000

    2500-50x=2000

    x=500/50

    x=10

    Then she have 10 notes of ₹50 and 25-10=15 notes of ₹ 100.

     

    Answered by Sagar Tripathi (Jun 22, 2017 8:33 p.m.)
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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 = 25-10 = 15 notes

     

    Answered by Rajendra Singh (Jun 22, 2017 8:32 p.m.)
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One third of square root of which no..is 0.001

Posted by Vandana Malhotra (Jun 17, 2017 6:19 p.m.) (Question ID: 5985)

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  • 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}

     

    Answered by Payal Singh (Jun 17, 2017 7:12 p.m.)
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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.

Posted by Upakshi Dutta Baruah (Jun 15, 2017 11:09 p.m.) (Question ID: 5926)

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  • Let present age of son = x

    Then present age of Aman = 3x

    Ten year ago

    age of son = x-10

    Age of Aman = 3x-10

    According to question

    => 3x-10 = 5(x-10)

    => 3x-10 = 5x-50

    => -10+50 = 5x-3x

    => 40=2x

    => x= 20

    So present age of son = 20

    Present age of Aman = 20*3= 60

     

    Answered by Payal Singh (Jun 16, 2017 5:44 a.m.)
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Solve using suitable properties :

-2/3 * 3/5 + 5/2 - 3/5 * 1/6

Posted by Upakshi Dutta Baruah (Jun 14, 2017 2:53 p.m.) (Question ID: 5898)

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  • {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 ({ -4-1\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

    Answered by Payal Singh (Jun 14, 2017 2:53 p.m.)
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Solve :

0.5 (y - 0.4) -  0.6 (y - 2.71) - y + 6.1

       0.35                0.42

 

Posted by Upakshi Dutta Baruah (Jun 08, 2017 9:24 p.m.) (Question ID: 5771)

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  • {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

     

    Answered by Payal Singh (Jun 08, 2017 11:08 p.m.)
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What is the smallest number which must be added to 15370 to make it a perfect square? (Show workings)

Posted by Sanjeev Banerjee (May 31, 2017 11:38 p.m.) (Question ID: 5612)

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  • 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.

    Answered by Naveen Sharma (Jun 01, 2017 9:35 a.m.)
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