101
11
110
111
10
Correct answer is B
\(\frac{11}{x}\) = \(\frac{1000}{x + 101}\) = 11(x + 101)
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = \(\frac{1111}{101}\)
x = 11
N180.00
N165.00
N150.00
N250.00
N200.00
Correct answer is D
S.I = \(\frac{PTR}{100}\)
T = 4yrs, R = 8%, a = N330.00
330 - P = \(\frac{PTR}{100}\), A = P + I
i.e. A = P + \(\frac{PTR}{100}\)
330 = P + \(\frac{P(4) (8)}{100}\)
33000 = 32P + 100p
132P = 33000
P = N250.00
\(\frac{6}{7}\) < 0.865 < \(\frac{13}{15}\)
\(\frac{13}{15}\) < \(\frac{6}{7}\) < 0.865
\(\frac{6}{7}\) < \(\frac{13}{15}\) < 0.865
0.865 < \(\frac{6}{7}\) < \(\frac{13}{15}\)
Correct answer is A
\(\frac{6}{7}\), \(\frac{13}{15}\), 0.8650
In ascending order, we have 0.8571, 0.8650, 0.8666
i.e. \(\frac{6}{7}\) < 0.8650 < \(\frac{13}{15}\)
If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of \(\frac{3p - q}{4q + r}\)
\(\frac{3}{2}\)
\(\frac{2}{3}\)
2
3
Correct answer is B
P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = \(\frac{6}{15}\), q = \(\frac{4}{15}\), r = \(\frac{5}{15}\) = \(\frac{1}{3}\)
To find \(\frac{3p - q}{4q + r}\)
3p - q = 3 x \(\frac{6}{15}\) - \(\frac{4}{15}\)
\(\frac{18}{15}\) - \(\frac{4}{15}\) = \(\frac{14}{15}\)
∴ 4q + r = 4 x \(\frac{4}{15}\) + \(\frac{5}{15}\)
\(\frac{16}{15}\) = \(\frac{16}{15}\) + \(\frac{5}{15}\)
= \(\frac{21}{15}\)
\(\frac{14}{15}\) x \(\frac{15}{21}\) = \(\frac{14}{21}\)
= \(\frac{2}{3}\)
(27, 4)
(14,4)
(13,4)
(4,4)
Correct answer is A
M = mode = the number having the highest frequency = 4
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)