A. 12
B. 16
C. 24
D. 48
Explanation:
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48.
A. Rs 100000
B. Rs 11000
C. Rs 120000
D. Rs 170000
Explanation:
Let the sum be Rs x.
Then S. I = Rs (x × 8/100 × 2) = Rs 4x/25
C. I = Rs [x × (1 + 8/100)2 – x]
= Rs (x × 27/25 × 27/25 – x)
=Rs 104x/625 (C.I) – (S.I)
= Rs(104x/625 – 4x /25)
= Rs 4x /625
Therefore 4x/625 = 768
=> x = ((768 ×625)/4) = 120000
Therefore sum = Rs 120000
A. 12.25%
B. 12%
C. 6%
D. 13.75%
Simple Interest, SI = (3875 – 2500) = Rs.1375
Principal, P = Rs. 2500
Time, T = 4 years
R = ?
R=100×SI/PT=100×13752/500×4
=100×1375/10000=13.75%
A. 21
B. 22
C. 23
D. 24
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
A. 2 : 5
B. 3 : 5
C. 4 : 5
D. 6 : 7
Let the third number be x.
Then, first number = 120% of x = 120x/100 = 6x/5 second numbet = 150% of x = 150x/100 = 3x/2 Ratio of first two numbers = 6x/5 : 3x/2 =12x : 15x = 4 : 5.
A. Rs. 31
B. Rs. 32.10
C. Rs. 40.40
D. Rs. 64.10
Explanation:
S.I. = (1000 * 10 * 4)/100 = Rs. 400
C.I. = [1000 * (1 + 10/100)4 – 1000] = Rs. 464.10 Difference = (464.10 – 400) = Rs. 64.10
A. 5days
B. 6days
C. 7days
D. 8days
Explanation:
Let us assume that Arun uses X units of petrol everyday.
So the amount of petrol in the tank when it is fuel will be 10X.
If he started using 25% more petrol every day, then the amount of petrol he how uses everyday will be
X (1 +25/100) =1.25x
Therefore, number of days his petrol will how last = Amount of petrol in tank / amount of petrol used everyday = 10x/1.25x = 10/1.25 = 8 Days