Calculation:
⇒ If 1000 divided by 112, the remainder is 104. ⇒ 112 – 104 = 8 ⇒ If 8 is added to 1000 it will become the smallest four-digit number and a multiple of 112. ⇒ 1000 + 8 = 1008 ∴ The required result will be 1008.
Statements :
Most teachers are boys. Some boys are students.
Conclusions :
I. Some students are boys.
II. Some teachers are students.
6 yeara old
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
20000
=12.05×5.4/0.6
=12.05×9.0
=108.45
By 1hour both trains meet, so the distance travel by fly in
1hr is 120km.
speed = distence / time
60km/hr=distance / 9 sec
km / hr to m / s==> 60*1000 / 1* 3600 = 16.667
16.667*9 = distance
150 m is the answer
108
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540 liters
8+3+4=15
15+(7)=22
22 is divisible by 11.
So,7 is right answer. Option C
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%
0.000256
400%
e.g:-
l = 5 b = 2
Area= l*b =10
New after 100% increament
l=10 b = 4
Area = 10*4
Ans 200
that he will lost 1.5 ltr in an hour
so he will lost that 10 ltr in 10/1.5=6.67hour
distance travelled in 6.67 hour is=30*6.67=200 km
One day work = 1 / 20
One mans one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 =
30
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
EALIME