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.
B will use 8days to work
let the third no be x ,
then the first no is 3x ,
second no is 2*3x ie 6x
average is (3x+6x+x )/3 = 20
10x/3 = 20
x= 60/10 = 6
third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no
d
HCF= 2
LCM = 2*5*7 = 70
Father’s sister
Let the distance between each pole be x m. Then, distance up to 12th pole = 11x m
∴ Speed = 11x22m/s
∴ Time to cover total distance up to 20th pole
= 19x×2411x
= 41.45 s
Since the car has met the person 20 minutes beforehand, it has saved 10 mins of a journey
A man has started 1.30 hrs before and the car has met him 10 mins before the actual time, he takes to reach daily is 1hr and 20 mins
red
answer is 306.04
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
451 times.
Explanation: There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will
flashes 451 times in ¾ of an hour.
16 chapatis* 6 Rs= 96
5 plates of rice * 45 Rs = 225
7 plates of mixed vegetable * 70 Rs = 490
6 ice-cream * ??
The amount that Alok paid the cashier was Rs.961
96+225+490 = 811
961-811= 150
150/6 = 25
The cost of each ice-cream cup is 25 .Rs
1/2 Hr -> 80 Cheques
Per Hour -> 80 * 2 = 160
For 7 Hours -> 160 * 7 = 1120
For 7 1/2 Hour -> 1120 + 80 = 1200
clerk can process 1200 cheques in Sever & one half an hour day.