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Ans 1:
Class : Class 8
Tom and jasmine have Rs 41 altogether .
let,
Jasmine has Rs x..
therefore tom has Rs( 41-x) {the rest of the money}
according to the question,
1/4 of Tom's money is 2 more than 1/7 of jasmine's money.
therefore,
1/4 x (41-X) = (1/7 x X)+2
or, ( 41-X)/4=X/7+2
or, (41-X)/4=(X+14)/7
or; (7x41)-7X=4X+(4x14)
or; 287-7X=4X+56
or; 287-56=4X+7X
or, 231=11X: 11X=231
X=231/11 =21 THE ANSWER IS 21 OR TOM HAS RS 21................
Ans 2:
Class : Class 8
Let tom has Rs. X
Let Jasmine has Rs y
A.T.Q
x+y = 41
x = 41-y
1/4 * x = 2 + 1/7*y
1/4*(41-y) = 2 + 1/7*y
41/4 - y/4 = 2 + y/7
41/4 - 2 = y/7 + y/4
33/4 = 11y/28
33/4*28/11 = y
21 = y
x+21=41
X = 41-21 = 20
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Ans 1:
Class : Class 10
There exist two such numbers : (1-sqrt(5))/2 and (1+sqrt(5))/2
This can be taken out by making a quadratic equation and using the discriminant formula,
which is:
x=(b*sqrt(b^2-4ac))/2a