Let Mango be M, Apple be = A, Orange be = O
3M + 2A = 5(8) = 40, 3A + 3O = 27, 3M + 3O = 42
(3M + 5A + 3O) = 67 â€“ (3M + 3O) = 42, => 5A = 25 => A = Rs.5, Mango = Rs.10, Orange = Rs.4
As all are walking together/simultaneously then 6 personâ€™s walk 5 km in 1 hour.
Let present age be x then
2(x - 8) => (x + 12) => 2x â€“ 16 = x + 12 => x = 16 + 12 = 28.
Let Hens be H, Goats be = G
H + G = 80, legs = 80 x 2.5 = 200, Hen has 2 legs, Goat has 4 legs.
2(H + G = 80)
2H + 4G = 200
2G = 40 => G = 20 or Goats = 20
Hens = 80 â€“ 20 = 60
A + B = 52, A + C = 34, B + C = 30.
(A + B) + (A + C) = 52 + 34 => 2A + B + C = 86 - (B + C) = 30 => 2A = 56.
A = $\displaystyle \frac{56}{2}$ = 28
A + C = 34
28 + C = 34
C = 34 â€“ 28 = 6
Let father be F, son = S.
Sum of ages of 6 years ago => F + S = 2 x 18 = 36.
Present ages of F + S = 36 + 12 = 48.
Father = 48 $\displaystyle \times \frac{3}{4}$ = 36 years, Son = 48 $\displaystyle \times \frac{1}{4}$ = 12 years.
Sum of the numbers = 2x, one of the numbers = n
Other number = 2x â€“ n
Sum of ages = 2 x 25 = 50, Father to son = 4 : 1.
Father = 50 $\displaystyle \times \frac{4}{5}$ = 40, son = 40 $\displaystyle \times \frac{1}{5}$ = 10.
128 - 2 = 126.
Total age of remaining 7 persons = 126 => $\displaystyle \frac{126}{7}$ = 18 years.
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Average is the mean value of a set of numbers.
1. Arithmetic Mean (A.M):
Average = $\displaystyle \frac{\text{Sum of observations}}{\text{Number of observations}}$
2. Harmonic Mean:
Average = $\displaystyle \frac{2[\text{v}_1 \times \text{v}_2]}{\text{v}_1 \ + \ \text{v}_2}$
3. Average of first â€˜nâ€™ Natural numbers = (n + 1 ) Ã· 2.
4. Average of first â€˜nâ€™ Even numbers = n + 1.
5. Average of first â€˜nâ€™ Odd Numbers = n.
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