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.
125
10m
ans is C.
let H be present age of husaband.W be present age of wife.
H+W=91.
now let diff between their ages be x.that is H-W=x.
now when husband is W yaers old, his wife must be W-x years
old.and it is given that H=2(W-x). so 2W-H=2x. and H-
W=x.eliminating x we get 4W=3H. but H+W=91, so solving thse
two H=52 W=39.
temp at 3.oo pm is = 22.5
temp at 6.oo pm is = 30
thus, percentage rise= 7.5/30 *100
=25%
23
30
(1/2)*x*y
126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126
As fast as possible. For example, I will suitable with the working environment within one month.
The first 10 odd prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31.
Sum of the odd prime numbers = (3+5+7+11+13+17+19+23+29+31)
= 158
Number of odd prime numbers = 10
We know, Average = (sum of the 10 odd prime numbers ÷ Number of odd
prime numbers)
Average =
= 15.8
∴ The Average of first 10 prime numbers which are odd is 15.8
12%
cube root of 5 is 125 so 128 is nearer to 125 and hence 5+5+5=15 and the nearest answer amongst options is 17 so 17 is correct answer:) yipppy……..
127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)
2/3 * 15 miles = 10 miles then
time for 10 miles travel is t1 = 10/40 = 1/4
then remaning distance is 5 miles
time for 5 miles travel is t2 = 5/60 = 1/12
t1+ t2= 1/3 in seconds * 60 = 20 min
8