A. 0.0256
B. 0.00256
C. 0.000256
D. 0.0000256
Explanation:
4 * 0.4 * 0.04 * 0.004 => 4 * (4/10) * (4/100) * (4/1000) = (256/1000000) = 0.000256
A. If a > b
B. If a ≥ b
C. If a < b
D. If a ≤ b
Explanation:
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ b
A. 50km
B. 60km
C. 70km
D. 80 km
Explanation:
Let the distance between the two parts be x km.
Then speed downstream = x/4 km/hr.
Speed Upstream = x/5 km/hr
Speed of the stream = ½ (x/4 –x/5)
Therefore, ½(x/4 –x/5) = 2. => x/4 –x/5 = 4 => x = 80.
Hence, the distance between the two ports is 80km.
A. 1 year
B. 2 years
C. 3 years
D. 4 years
28 * 2 = 56
20 * 3 = 60
4 years
A. 40 %
B. 32 %
C. 22 %
D. 8 %
Explanation:
Let A = set of students who play football and
B = set of students play cricket.
Then n(A) = 40, n (B) = 50 and
n(A U B) = (100 – 18) = 82
n(A U B) = n(A) + n(B) – n(A ∩ B)
n(A∩B) = n(A) + n(B) – n(AUB) = (40 + 50 -82) = 8
Percentage of the students who play both = 8%
A. 216
B. 217.50
C. 236.50
D. 245
Explanation:
Given expression = (45/100 * 750) – (25/100 * 480) = (337.50 – 120) = 217.50
A. 6 %
B. 7 %
C. 8 %
D. 9 %
Explanation:
AC = 3cm, CB = (5 – 3)cm = 2 cm
New length AC = 106 % of 3 cm
= (106/100 ×3) cm = 3.18 cm.
New length CB = (5 – 3.18) cm = 1.82 cm
CB Decreased on 2cm = (2 – 1.82)cm = 0.18cm
CB Decrease % = (0.18/2 ×100) % = 9 %