share= 5:7
i.e.,
=9600*5/12
= 4000
( A ) 124
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74
ANS = 74
11 14
12 15
13 16
=======
14 17
the answers is 14
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.
23. 15s squre is 225and 8 squre is 64 addition is 289 product also 120 addition 15+8=23
Take one fruit from box with label mixture. If we see
orange, because the basket lies (it cant have a mixture),
then it has only oranges. The other 2 are labeled apples and
oranges. The one labeled apples,
cannot have oranges inside, cos they are allready been
identified, and because it lies, it cannot have apples either.
So it has a mixture. And we are left with the one labeled
oranges that lies and has apples.
Day 1: 30-3ft = 27+2ft = 29
Day 8: 23 jumps 3ft = 20 slips 2ft = 22
Day 16: 13 jumps 3ft = 10 slips 2ft = 12
Day 24: 3 jumps 3ft = Out?
x=12;y=7
ans : 101 bcoz after every two matches one team is eliminated…so to eliminate 50 team there will be require 100 matches…+ 1 for deciding winner…
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
4400+(97 leap year )=4497
TOTAL NUMBER OF BOOKS 40+40=80
ISSUED BOOKS 30
REMAINING BOOKS 80-30=50
RATE=50/80=5/8
How breadth is 19.84..?
Apply Pythagoras therom,
a^2+b^2=c^2,
Lets Assume
c is hypotenuse or diagonal,
a is length or opposite side,
b is breadth or adjacent side,
Then,
256+b^2=400
b^2=144
b=12,
So the breadth is 12 cm
1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0
Palwal