Answer: 40
Explanation:
800*(25/100) = 200 —- 5
? —- 1 => Rs.40
1, 1, 2, 6, 24, 96, 720
C. 96
Please let me know answer.
8+3+4=15
15+(7)=22
22 is divisible by 11.
So,7 is right answer. Option C
let no of total crockery =x
2x/3 crockery broken
x/2 handle broken
x/4 both brocken
2x/3+x/2-x/4 = no of total broken(crockery or handle)
=11x/12
unbroken =x-11x/12=x/12
x/12=2(given in question)
x=24
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
6.40 am
avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27
c
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
0237
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answer :
Let the distance which I walk to the station be x km Then Time needed to reach the station at 3 km/hr = (3x+602) hrs and
Time needed to reach the station at 4 km/hr = (4x+602) hrs But 3x+301=4x−301⇒3x−4x=−302=151
⇒124x−3x=151⇒12x=151⇒x=1512=54 km
8/15 left.
1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15
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When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
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