A. 18 3/4 %
B. 20 %
C. 16 3/4 %
D. 19 %
Explanation:
Required % = (9/48 × 100)% = 75/4 % = 18 ¾ %
A. 8 m
B. 10 m
C. 6 m
D. 12 m
A. 93
B. 39
C. 75
D. 48
E. Either (a) or (b)
Explanation:
Let the two-digit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93.
A. Rs.1014
B. Rs.1140
C. Rs.999
D. Rs.1085
Explanation:
a2 = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.5 = 999
A. 2.5
B. 2.0
C. 1.75
D. 1.5
E. 1.25
One moves at 60km/h, so in an hour, he’ll travel 60km
The other moves at 40km/h, so in an hour, he’ll travel 40km
This is towards each other, so, in an hour, we can say that they’ll get 60+40=100km closer.
This means that they get closer at 100km/h
Given that we have the distance of 150km, and the speed of closure at 100kmh, we can now use the equation
S=D/T
Rearrange to find T (times both sides by T)
S x T = D (now divide both sides by S)
T=D/S
Time = 150km divided by 100km/h
Time = 1.5 hours
Or one hour and 30 minutes
A. 2:5
B. 4:5
C. 6:5
D. 8:5
A:C=(A:B)*(B:C) =(2/3)*(3/4)= 1/2 C:D=(C:A)*(A:D) =(2)*(4/5) = 8:5
A. Rs.4000
B. Rs.9000
C. Rs.5000
D. Rs.6000
Let sum = P and original rate = R. Then
[(P * (R+2) * 3)/100] – [ (P * R * 3)/100] = 360
3P*(R+2) – 3PR = 36000
3PR + 6P – 3PR = 36000
6P = 36000
P = 6000