Please solve the following questions for me :-?

Question

The cost price of a shirt and a tie is Rs. 516. if the tie costs 30% more than the one third of the cost of the shirt, find the cost price of each.

Divide 32 into two parts such that if the larger is divided by the smaller, the quotient is 2 and the remainder is 5.

pl. solve in steps.

Answer 1

(1) If the shirt costs X, then one third of the shirts price is 1/3 X = X/3. The tie is 30% more then one third os the shirt (or X/3), therefore: tie = X/3 + 30 % * X/3 = X/3+ 30/100 * X/3 = X/3 + 30X/300 = X/3+ X/10 = 10X/30 + 3X/30 = 13X/30.

Shirt + tie = 516
X + 13X/30 = 516
30x/30 + 13X/30 = 516
43X/30 = 516
43X = 516*30 = 15480
X = 15480/43 = 360, that's the shirt's price
The tie's price is 13X/30 = 13*360/30 = 156.

Let us do the check: 360 + 156 = 516

(2) Let the smaller part be x, the larger 32-x. If we divide 32-x by x, then we'll get the quotient 2, and the remainder 5. That means

(32-x):x=2 (5) =>
32-x = 2x+5 =>
32=2x+5+x =>
32=3x+5 =>
3x+5=32 =>
3x = 32-5 =>
3x=27 =>
x=27:3 =>
x=9

The smaller part is 9, the larger is 32-9=23. If we divide 23:9, we'll get the quotient 2, and the remainder 5, because 23=2*9+5.

I hope this could help.

Answer 2

For the second half, let's use L to be largest part and S to be smaller part. Then we have
S + L = 32.
Next when we divide L by S we get 2 remainder 5. This means
L = 2S + 5
Substitute this into the first equation.
S + L = 32
S + (2S + 5) = 32
3S + 5 = 32 [adding S terms together]
3S = (32 -5) [take 5 from both sides]
3S = 27
S = 9 [divide 3 through both sides]

therefore L = 32 - 9 = 23.
Check 23 divide 9 = 2 with 5 remainder! Yeah, we got it!!!

Now for Part 1)

Cost of shirt and tie is 516. Use T for cost of Tie and S for cost of shirt. So we have
S + T = 516.

Next, the tie costs 30% more than one third the cost of the shirt. Okat this IS a bit tricky. Break it down ...
The tie cost 30% more than (S/3).
T = (S/3) * [130%] it's 100% of (s/3) PLUS another 30%
we need to change 130% to 1.3. This gives
T = (S/3)*1.3
Now SUBSTITUTE into first equation

S + T = 516
S + (S/3)*1.3 = 516.

USE YOUR CALCULATOR HERE. The total number of S's on the RHS are 1 + (1/3)*1.3 = 1.4333333

So we have 1.433333 S = 516. Divide by 1.4333333
S = 516/1.4333333 = 360.

So T = 516 - 360 = 156. Now, it says the tie costs 30% more than a third the cost of the shirt. Okay, a third the cost of the shirt is 360/3 = 120. Now is 156 30% more than 120. Let's see
30% of 120 = 0.3 * 120 = 36. So the Tie = 120 + 36 = 156! WE GOT IT!!!! A tricky one too.

Our answer: The tie costs 156 and the shirt costs 360.

Answer 3

Let a be the cost of the shirt and b the cost of the tie.

a + b = 516

b = 1.3 a / 3or

3b = 1.3a

a + 1.3a/3 = 516

4.3a = 516 x 3 = 1548

a = 1548/4.3 = 360

Shirt costs 360 and the tie 156

Next question:

Let a and b the two parts of 32.

a + b = 32

b/a = 2 + 5/a = (2a + 5) / a

b = 2a + 5 = 32 - a

so, 32 - a = 2a + 5 or 3a = 32 - 5 = 27 and a = 27/3 = 9

b = 23

23 and 9 are the two numbers that satisfy the condition.

Answer 4

1) Call cost of shirt S and cost of tie T
T = 1.3 * (1/3) * S
T + S = 516
1.3 * 1/3 *S + S = 516
1.43333S = 516
S = $360
T = (plug back in into first equation) $156

2) Use logic
Your first number must be at least twice as big as your second
32/3 = 10.6667
10.6667 * 2 = 21.3333
so, your first number must be bigger than 21.333 and your second number is smaller than 10.6667
23 and 9 work perfectly
23/9 = 2 Remainder 5

Answer 5

s+t=516
and t= (1/3)*s*0.30

therefore
s+(1/3)*s*0.30=516
s+0.1s=516
1.1s=516
s=469

Hence, t from equation 1 is 516-469=47

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For the second question
the numbers are 23 and 9

solution is

x+y=32
(x-5)/y=2 as quotient is 2

solving it
from second equation,
2y=x-5
and y=(x-5)/2
putting in ist equation,

x+ (x-5)/2=32

solving for x u get 23,

from i

23+y=32
therefore y=9.

Answer 6

t + s = 516
t = 1.3 * (s/3)

Substitute and solve for s, then solve for t.
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32 = a + b

b = 2a + 5

Again, substitute and solve.

Answer 7

do your own homework!