GMAT Math

How many 3-digit even number can be formed if one of the digit is 5,the following digit must be 7 ?

1. 5
2. 405
3. 365
4. 495

correct answer:365

i think answer should be 570,572,574,576,758 i.e 5

In Sivaski,each boy's quota of matchsticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to the boy?

a) 200
b) 150
c) 125
d) 175

correct answer:150

please tell the logic,i can't get it

Alord got an order from a garment manufacturer for 480 Denim Shirts. He bought 12 sewing machines and appointed some expert tailors to do the job.However many didn't report so each of those who did, had to stich 32 more shirts than originally,with equal distribution of work.How many Tailors had been appointed earlier and how many had nt reported to work ?

1. 12,4
2. 10,3
3. 10,4
4. None of these

answer:3

Please explain the answer

A man sells chocolates in boxes. Either full boxes or half a box of chocolates can be bought from him. A customer comes and buys half the number of boxes the seller has plus half a box. A second customer comes and buys half a box. After this, the seller is left with no chocolate box. How many chocolate boxes did the seller have before the first customer came initially?

1. A. 2
2. B. 3
3. C. 1
4. D. Cannot be determined

correct answer: 3

Please explain the answer

maxpain5 wrote:How many 3-digit even number can be formed if one of the digit is 5,the following digit must be 7 ?

1. 5
2. 405
3. 365
4. 495

3 digit even no. will be formed if the last digit is any of the following 0,2,4,6,8;

case 1) when 0 is at the unit position,- - 0; now 5 can occur only at the hundredth place because if it occurs at the tens position or unit position than the initial condition of 7 following 5 won't be fulfilled; so if 5 occur at the hundreds position then ten's position will contain 7; hence one possible no. would be 570;
now consider all those cases when 5 won't occur at the hundred's position than we have these (1,2,3,4,6,7,8,9) numbers to fill this place - - 0; which can be done in 8 ways and the ten's place can be filled by (0,1,2,3,4,6,7,8,9) any of the these numbers in 9 ways hence we have total 8*9*1+1=73 ways;

case 2) when 2 is at the unit position, - - 2; by giving same reasoning number possible when 5 occur at the unit place is 572; and when 5 won't occur at the hundred's position we have (1,2,3,4,6,7,8,9) numbers to fill this place--2; when can be done in 8 ways; and now similarly for ten's place we have
(0,1,2,3,4,6,7,8,9) numbers to fill this place which can be done in 9 ways hence we have total 8*9*1+1=73;

we can see that we will have 73 no. of ways for each of the number ending with 0,2,4,6,8 hence total no. of ways= 73*5=365; hence C

maxpain5 wrote:In Sivaski,each boy's quota of matchsticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to the boy?

a) 200
b) 150
c) 125
d) 175

let no. of matchstick assigned to boy=n; no. of matchsticks he put in each box=x; no. of boxes=y;

initially we have;
n/x=y;----1)
after he reduces no. of matchsticks in each box by 25 we have;
n/x-25=y+3;------2)

solving 1 and 2 we have;
n= 3x(x-25)/25; i.e. no. of stick must be a multiple of 25; let x=50 we have;
=3*50*(50-25)/25;
=3*50=150;
hence n=150 ; hence B