Let the total amount of money is x ..
Cost of 1 Pineapple = x/40
Cost of 1 Apple= x/30
15% for fares so money spent on fruits is .85x
Now (.85x-(14*x)/40)/(x/30) shall give you the number of apples which shall come out to be 15..
@saurav5517 said:Let the total amount of money is x ..Cost of 1 Pineapple = x/40Cost of 1 Apple= x/3015% for fares so money spent on fruits is .85xNow (.85x-(14*x)/40)/(x/30) shall give you the number of apples which shall come out to be 15..
arre han yar sahi hai maine calculation mistake ki baar baar..thanx
@quant89 said:arre han yar sahi hai maine calculation mistake ki baar baar..thanx
You are welcome..
You can also do such a problem by taking the LCM of the 2 entities and taking an amount equal to that..this approach is used by one of the puy in SNAP thread..
@quant89 said:guys pls help me with dis attached Q..I don't kno d answer..
x = pineapple y=apple
Total money=40x=30y ------- eq1
Amt with her=85/100(40x)=34x
cost of pineapples=14x
amt left=20x which is half of eq1 ..Hence 15 y =>15 apples
@Cat.Aspirant123 said:C-A=2C-B=1B-A=1so u had done dis trough trial n error
Yaa a kind of hit and trial....bcoz dese eqs vill lead u to C>B>A for sure......and a quick look at d options will giv u d indication dat only 123 has d potential to satisfy d above condition..
For each p>1 ,sequence {An} is defined by A0=1 and An=pn+ (-1)^n A(n-1) for n>=1
For how many integer values of p, 1000 is a term of the sequence ?
a)8
b)7
c)5
d)4
P.S A0 is Asubstcipt 0 ..... A(n-1) is A subscript n-1
the distribution of 9 scores is 0,1,2,5,3,9,9,9,3,3. the 10th score is a mystery. the 10th score is more than 5 but is not 9.
can u find the mod of this distribution. if yes, wat is the mode?
also find the median and mean of this distribution?
Right triangle PQR is to be constructed in the xy-plane
so that the right angle is at P and PR is parallel to the
x-axis. The x- and y-coordinates of P, Q, and R are to
be integers that satisfy the inequalities –4 ≤ x ≤ 5 and
6 ≤ y ≤ 16. How many different triangles with these
properties could be constructed?
so that the right angle is at P and PR is parallel to the
x-axis. The x- and y-coordinates of P, Q, and R are to
be integers that satisfy the inequalities –4 ≤ x ≤ 5 and
6 ≤ y ≤ 16. How many different triangles with these
properties could be constructed?
@SAHILVANSIL said:the distribution of 9 scores is 0,1,2,5,3,9,9,9,3,3. the 10th score is a mystery. the 10th score is more than 5 but is not 9.can u find the mod of this distribution. if yes, wat is the mode?also find the median and mean of this distribution?
edit it..there are already 10 scores.
@Gaul said:Right triangle PQR is to be constructed in the xy-planeso that the right angle is at P and PR is parallel to thex-axis. The x- and y-coordinates of P, Q, and R are tobe integers that satisfy the inequalities –4 ≤ x ≤ 5 and6 ≤ y ≤ 16. How many different triangles with theseproperties could be constructed?
45*55 = 2475 ?
@Gaul said:Right triangle PQR is to be constructed in the xy-planeso that the right angle is at P and PR is parallel to thex-axis. The x- and y-coordinates of P, Q, and R are tobe integers that satisfy the inequalities –4 ≤ x ≤ 5 and6 ≤ y ≤ 16. How many different triangles with theseproperties could be constructed?
my take 9900?
no of x coordinates=10 points
no of y coordinates= 11 points
Let us choose P,the right angle which can have any x,y coordinate mentioned above
so 10*11=110 possible coordinates --------1
Now PR is parallel to X axis =>R should have same y coordinate as P
x coordinate of R can be anything out of the 10 except P's X coordinate(in which case P,R will coiincide which we dont want )
So no of possible x coordinats for R=10-1=9
no of possible y coordinats for R=1
Total possible coordinates=1*9=9----------2
Now Q should have same x coordinate as P so as to form a right triangle
no of possible x coordinates for Q=1
no of possible y coordinates for Q=11-1=10
Total possible coordinates=1*10=10 ---------3
Now from 1,2,3
we get 110*9*10=9900
@Budokai001 said:For each p>1 ,sequence {An} is defined by A0=1 and An=pn+ (-1)^n A(n-1) for n>=1For how many integer values of p, 1000 is a term of the sequence ?a)8b)7c)5d)4P.S A0 is Asubstcipt 0 ..... A(n-1) is A subscript n-1
5, was solved yesterday i think
@Budokai001 said:For each p>1 ,sequence {An} is defined by A0=1 and An=pn+ (-1)^n A(n-1) for n>=1For how many integer values of p, 1000 is a term of the sequence ?a)8b)7c)5d)4P.S A0 is Asubstcipt 0 ..... A(n-1) is A subscript n-1
0--1
1--p-1
2--3p-1
3--1
4--4p+1 and so on with 2 as 3p-1 and 6 as 7p-1 and so on
So for p=1001 it is possible
and factors of 1001=1,7,11,13,77,91,143,1001;
factor+1 should be of 4k type
possible factors=7,11,91,143
Total 5 ways?
1--p-1
2--3p-1
3--1
4--4p+1 and so on with 2 as 3p-1 and 6 as 7p-1 and so on
So for p=1001 it is possible
and factors of 1001=1,7,11,13,77,91,143,1001;
factor+1 should be of 4k type
possible factors=7,11,91,143
Total 5 ways?
@Gaul said:Right triangle PQR is to be constructed in the xy-planeso that the right angle is at P and PR is parallel to thex-axis. The x- and y-coordinates of P, Q, and R are tobe integers that satisfy the inequalities –4 ≤ x ≤ 5 and6 ≤ y ≤ 16. How many different triangles with theseproperties could be constructed?
P(x,y) ways= 10*11 = 110
no of ways for x cordinate of R = 9
no of ways for y cordinate of Q = 10
Total ways = 110*9*10 = 9900?
no of ways for x cordinate of R = 9
no of ways for y cordinate of Q = 10
Total ways = 110*9*10 = 9900?
@Cat.Aspirant123 said:In a three digit number ABC, if the number is reversed, it increases by 198. Also, if the tenth andunit €™s place digits are reversed, the number increases by 9. If hundredth and tenth place digits arereversed, the number increases by 90. What is the number?a) 132 b) 321 c) 213 d) 312 e) 123
Let number be abc
reversed number be cba
a+10b+100c=198+c+10b+100a
c-a=2------------1
b+10c=c+10b+9
9(c-b)=9
c-b=1----------2
b-a=1-----------3
123
a>b>c
If both the roots of quadratic equation ax^2 + bx + c =0 lie in the interval (0,3) then a lies in
a) (1,3)
b) (-1,3)
c) (-1,3)
d) (-121/91, -8)
e) None of these
Share approach to find the range
a) (1,3)
b) (-1,3)
c) (-1,3)
d) (-121/91, -8)
e) None of these
Share approach to find the range
@Cat.Aspirant123 said:The average weight of 8 people decreases by 2 kgs, when two new people are added to thegroup. If the difference between the original weight and new weight is 36 kgs, find the new totalweight of the people.
(A-2)*(10)-A*8=36
2A=56
A=28
26*10=260
@Torque024 said:If both the roots of quadratic equation ax^2 + bx + c =0 lie in the interval (0,3) then a lies ina) (1,3)b) (-1,3)c) (-1,3)d) (-121/91, -8)e) None of theseShare approach to find the range
option A hai kya?