Question
A sum of Rs. 21500 was lent partly at 4% p.a. and
remaining at 6% p.a. simple interest. The total interest received after 3(1/2) years is Rs. 3521. The sum (in Rs.) lent at 6% p.a. is:Solution
Let amount Rs. x lent at 6% p. a Amount lent at 4% p. a = (21500 – x) As we know, SI = P × r × t/100 Simple interest for 7/2 year = Rs.3521 Simple interest for 1 year = 3521 × 2/7 = 1006 According to the question (x × 6 × 1)/100 + [(21500 – x) × 4 × 1]/100 = 1006 ⇒ 6x + 86000 – 4x = 100600 ⇒ 2x = 100600 – 86000 ⇒ x = 14600/2 ⇒ x = 7300 ∴ The sum lent at 6% per annum is Rs. 7300
Statement: M < N ≤ O = P, Q ≥ O ≤ R ≤ Z
Conclusion: I. Q > MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â II. Z > M
...Statements: J > K > M ≤ N < O; M > P > L
Conclusions:
I. J > L
II. P < O
III. J > O
Statements: V > P < L = O, R > N; Q > V > R
Conclusions:
I. R > L
II. Q > N
III. L > N
Statements: R ≤ K ≤ H = O ≥ D > Q; K > P
Conclusions:I. O ≥ Q II. Q > P
In the question, assume the given statements to be true. Find which of the following conclusion(s) among the three conclusions is/ are definitely true ...
Which one the following symbols should replace the question mark in the given expression, in order to make the expressions M ≤ J as well as O > L ...
Statement- 1 - 6# ≥ 9# ≥ 10# ≤ 5#
2 - 7* ≤ 10# ˃ 5* ≥ 6*
Conclusions:
1) 5# > 5*
2) 9# ˃ 6*
3) ...
Statements: Â T # L # C $ X & Y % E % F
Conclusions:
I. Y @ F
II. Y & F
III. Y & C
...Statements:
M < K ≤ G ≤ Z; P = J > Z; I ≥ R > P;
Conclusions:
I. K ≤ P
II. M < R
Statements: M ≤ N > O; P > M; Q < R < P
Conclusions:
I. M < P
II. O < P
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