A person bought an article at 15% discount and sold it for Rs 1600. What is the % profit? (I) The marked price is 1500 (II) By selling the article at Rs 153 more, he would get 12% more
IMO D.
my expn:
take 1)clearly we can calc the profit.
1500*0.85=1275 % profit = 1600-1275/1275 * 100
take 2) here 153 is 12% so 100% is 1275 so we know the cost price and we can calc the % profit.
What is the proportion of women in Austria? (I) Number of men in Austria is 5% more than the number of women (II) Austria's population is 5% more than that of one of its neighboring countries, where women are 52% of the population.
IMO A
take1) W/M+W = W/2.05W so 1) Alone is suff.
take 2) onlly percentages present no information on the actual demographics.
Puys, a long list of problems.. But the pain is I donthave the OA. I found these questions unanswered in earlier posts.
2. Is the five digited number ABCDE divisible by 13? 1. The number CD is divisible by 13. 2. 10A+B+4C+3D-E is divisible by 13.
Very interesting ...I am not sure if divisibility rules of 13 are expected on GMAT..
Nevertheless, couple of things to learn from this sum : Div rules of 13 ;
a) Delete the last digit from the given number. Then subtract nine times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.
b) To find out, if a number is divisible by 13, take the last digit, Multiply it with 4, and add it to the rest of the number.
If you get an answer divisible by 13 (including zero), then the original number is divisible by 13.
If you don't know the new number's divisibility, you can apply the rule again.
Was aware of the first one ...and to generalize, modualr technique is used ...just learnt abt it ...can google to get the details ..
St 1 : It does not depend on any 2 specific digits ...not suff..
St 2 : generalised form of modular algebra ...Suff
Puys, a long list of problems.. But the pain is I donthave the OA. I found these questions unanswered in earlier posts.
4. In triangle ABC angle A is the greatest angle. D is the foot of the perpendicular dropped on to BC from A. Is triangle ABC right-angled? 1. AD^2= BD x DC. 2. AD/DC
If triangle ABC is rt angled at A, and AD is perpendicular to hypotenuse, then triangle ACD and BAD is similar So AD^2 = CD *BD
Reverse is true as well ...given this ratio , triangle is a rt angled triangle at A..
St 1 : ratio given, hence, rt angled at A ...suff
St 2: unequal ratios, hence not a rt angled at A ..suff
6.If x and y are positive integers such that x= 8y +12 ,what is the GCD of x and y? a) x=12u , where u is an integer b) y=12z , where z is an integer
statement1.8y=12u-12=12(u-1) y=12(u-1)/8=3(u-1)/2 to make y an integer ,value of u will be 3,5,7,9.....giving different values of x,y.not suffecient. statement2.x=12(z+1) =>x and y both are multiples of 12.for every value of Z will give even and odd values of x/y. so B
If N is a positive integer, what is the last digit of the result for
1! + 2! + ....+ N! ?
1) N is divisible by 4
2) (N2 + 1 ) / 5 is an odd integer
Puys attack this one.. OA later
If N =1 then last digit would be 1 If N =2 then last digit would be (1+2 )= 3 If N =3 then last digit would be (1+2+6) =9 If N =4 then last digit would be (1+2+6+24) =3 If N >4 then last digit would be (1+2+6++4+0) =3
Option 1) N is divisible by 4 so N = 4,8,12,16 etc...last digit of all the numbers is 3 ..so option 1alone is suff Option 2) (N^2 +1)/5 is an odd integer ... N =1 , (N^2 +1)/5 = fraction N=2 , (N^2 +1)/5 =1 N=3 , (N^2 +1)/5 =2 N=4 , (N^2 +1)/5 = fraction
So the number N can be 2 or any number greater than 4...but if v see above for N=2 , N >4 the last digit is 3...
So v can say they Option 2 alone is suff...!!
Please let me know if i am on the right track of solving 😉
