Ques:- The cost of 17 apples, 9 oranges , 13 bananas is rs 130 , whereas cost of 13 oranges, 7 apples and 10 bananas is rs 100. find the cost of 1 apple , 1 banana and 1 orange put together?
Recent Answer : Added by Admin On 2022-09-27 18:08:58:
Let cost for apple be a Cost for banana be b and Orange be c
So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100
Further if you solve you get value of b:
b = 10 – 2c ———-4
put value of b from 4 in 1
17a + 13 [10 – 2c] + 9c = 130
Further if you solve you find value of a
a = c ———-5
Put 5 in 3
13c + 7c + 10b = 100
further solve you get: c = 1 ———-6
from 5 and 6
a = c = 1 ———-7
Substitute value of c in expression 4
b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8
therefore a + b + c = 10
Ques:- YOU HAVE GAP OF 4 YEARS AFTER BETWEEN 10 ND 12 CLASS ..WHAT ANSWER I SHOULD GIV IF THEY ASK ME ABOUT THE GAP? WHAT U DID IN GAP?
Ques:- Four Members A,B,C,D are playing a game .A person losing a game should double the amount of others .B,C,D are losing in order after three games .The amount after 3 games are A&B having 40,D is having 16&C 80. Each questions carry one mark:
a) who started with small amount of money?
b) who started with greatest amount of money?
c) what amount did B have?
Recent Answer : Added by Admin On 2020-05-17 12:02:11:
status after 3 games
A-40
B-40
c-80
d-16
status after 2 games
A-20
B-20
C-40
D-96
status after 1 game
A-10
B-10
C-108
D-48
status after 0 games
A-5
B-93
C-54
D-24
so answers are
a)A
b)B
c)93
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A. 10 : 10 : 9
B. 19 : 19 : 18.
C. 20 : 19 : 18
D. 20 : 20 : 19
Recent Answer : Added by narendra On 2022-09-18 09:02:28:
b
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Ques:- When x is real what is the least value of (x**2-6*x+5)/(x**2+2*x+1)
Recent Answer : Added by Admin On 2020-05-17 12:01:23:
(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
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Let cost for apple be a Cost for banana be b and Orange be c
So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100
Further if you solve you get value of b:
b = 10 – 2c ———-4
put value of b from 4 in 1
17a + 13 [10 – 2c] + 9c = 130
Further if you solve you find value of a
a = c ———-5
Put 5 in 3
13c + 7c + 10b = 100
further solve you get: c = 1 ———-6
from 5 and 6
a = c = 1 ———-7
Substitute value of c in expression 4
b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8
therefore a + b + c = 10