if initially the price was Re 1.the new price nw bcoms to b
0.7..
so ,if the price z 0.7 ,the increment required is 0.3
so % in crement= 0.3/0.7*100=42.87…
30%
suppose the area is 100 and it was increased by 69%, then
the area is 169 which it indicates the side of the square
is 13.. which means 30% increase in its side.
Initially, potatoes have 99% water by weight ; which means they have 1% solid non-water content.
1% of 100 kg = 1 kg
Now even when they dehydrate,this 1kg solid mass remains constant.
It is given that finally, 98% is water by weight , which implies that 2% is non-water solid.
This means 2% of total weight = 1 kg
Total weight *2/100 = 1 kg
Therefore, total weight finally = 50 kg.
sow
8 days
Let n be the number of days it takes A and B, working together, to finish. And we know B=A+10 and B=3A, so:
3A=A+10
2A=10
A=5
Then B=15
So:
1/A + 1/B=1/n where n is the total amount of days. So:
1/5 + 1/15=1/n
3n+n=15
n=15/4 days
$57.30
Reduction of 40% or 4/10th in price of bananas will lead to an increase of 4/(10 – 4) = 2/3rd part in quantity if expenditure is constant.
If original quantity is q, then :
2q/3 = 64
=> q = 64 x 3/2 = 96 = 8 dozens.
=> Original price per dozen = 40/8 = Rs 5
ans=13.33
solution: (710/12)*8 – 460
3, 4, 9, 22.5, 67.5, 202.5, 810
A. 4
Let 10’splace digit is x and unit’s place digit y
First milestone : 10x+y
Second milestone : 10y+x
Third milestone: 100x+y
Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
After solving, we get —-> y=6x but since x and y are digits so only possible combination is x=1 and y=6,
So average speed = 45 KM/HR
( d ) 8404
(a) 29.12 (actual)
first year rate of paper (inflation rate) : 25 + 25 *
6.5/100 = 26.625
paper rate in first year is 1.5 % more . ..thr4
inflated cost in 1 yr = 26.625 + 1.5 / 100 * 26.625 =
27.024.
second year : 27.024 + 27.024 * 6.5 / 100 = 28.78
paper cost : 1.5 /100 * 28.78 + 28.78 = 29.12 ( approx..)
It’s choice A because you take the last to letters and move them to the front then the previous two letters go after them and so on.
The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.
Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?
=12.05×5.4/0.6
=12.05×9.0
=108.45