let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
P1/P2 = C1*T1/ C2*T2
where P1 = Partner 1’s Profit.
C1 = Partner 1’s Capital.
T1 = Time period for which Partner 1 contributed his capital.
P2 = Partner 2’s Profit.
C2 = Partner 2’s Capital.
T2 = Time period for which Partner 2 contributed his capital
6000/3000= 20,000*6/x*12
x=5000
31
Speed of train = 54km/hr
= 54 x (5/18) m/s
= 15 m/s
Length of train = 165m
Time required to cross a bridge of 660m in length = (660+165) / 15
= 55 seconds
8+1=9
9+2=11
11+3=14
14+4=18
18+5=23
and finally
23+6=29
C
let no. of boys not participating be x
then the no. of girls not participating = x+5
no. of boys : girls participating = 3:2
given no.of boys participating = 15
therefore, the ratio is now 15:y(say)
then 3:2 = 15 : x
on solving 3/2 =15/y ie.., 3y =30 we get y =10
hence no. of girls participating =10
therefore total no of students paricipating = 15+10=25
total no of students in class =60 given
hence no. of students not participating = 60-25=35
therefore x+(x+5)=35
2x=30
x=15
therfore no of girls not participating =15+5=20
therefore total no of girls in class = no of girls
participating + no of girls not participating
=10+20
=30 is the answer
X×x-x=272
X^2-x-272=0
(X-16) or (x-17)
X=16 or x=17
17^2-17=272
Answer Is 17
The answer is W. Since, the direction left to him, would be west.
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.
414 Rs
1.2*37.8=45.36=46 sq.feet
46*9=414 RS
10, 25, 45, 54, 60, 75, 80
54
According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.
20&130