In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
370 + x = 420
x = 50
Then the percentage of silver contained in 500 g of rod = 50/500 *100 =10%
LET ACTUAL PRICE BE 100 RS(FOR CONVINIENCE)
THEREFORE IF PRICE IS CUT BY 20%
THEREFORE NEW PRICE=100-20=80 RS
THEREFORE WE NEED TO ADD X% TO 80 RS TO MAKE IT 20 RS AND ADD IT TO 80 TO MAKE IT 100RS
THEREFORE, 80X/100 = 20
I.E. 4X/5=20
THEREFORE x=20*5/4
=5*5
=25
HENCE 25% OF PRICE SHOULD BE ADDED TO MAKE THE PRICE EQUAL TO THE ACTUAL PRICE.
d
10C3 = 120
The answer cannot be determined as there is a particular formula where the consecutive numbers start.
(60/100)*60 + (40/100)*40
=52%
4, 5, 15, 49, 201, 1011, 6073
Correct option is A)
2nd term =(1stterm×1+2)=(4×1+2)=6
3rd term =(2ndterm×2+3)=(6×2+3)=15
4th term =(3rdterm×3+4)=(15×3+4)=49
5th term =(4thterm×4+5)=(49×4+5)=201 and so on
∴5 is wrong.
500 = Total+50
Total(450) = only one paper(p) + 29+20+35 + all three (g)
285+212+127 = p + 2( 29+20+35 )+ 3g
solve above .. to get g = 45 …
( small corrctn .. i think .. questn shud be 20 read ONLY
hindu and
times of India and 29 read ONLY hindu and Indian express
and 35
read ONLY times of India and Indian express)
brother and sister
There are 18 numbers between 100 and 300 that are divisible by 11: 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297.
Distance covered by B to meet A=Total distance – Distance covered by A hrs
[using : distance = speed x time]
Putting value of from equation (1),
hrs
Therefore, time at which both A and B will meet is = 7 a.m. + 3 hrs =10 am
Ans :15
X be a present age of mudit
After 18 year age of mudit is X+18
4 year ago age was X-4
As per question,
X+18=3(X-4)
X+18=3X-12
3X-X=18+12
2X=30
X=15
RABBIT
31