Quantitative Aptitude
AVERAGES MCQs
Averages
Total Questions : 3752
| Page 370 of 376 pages
Answer: Option D. -> 19
AVERAGE OF 20 NUMBERS = 0
SUM OF 20 NUMBERS = (0 * 20) = 0
IT IS QUITE POSSIBLE THAT 19 OF THESE NUMBERS MAY BE POSITIVE AND IF THEIR SUM IS A, THEN 20TH NUMBER IS (-A).
AVERAGE OF 20 NUMBERS = 0
SUM OF 20 NUMBERS = (0 * 20) = 0
IT IS QUITE POSSIBLE THAT 19 OF THESE NUMBERS MAY BE POSITIVE AND IF THEIR SUM IS A, THEN 20TH NUMBER IS (-A).
Answer: Option B. -> 20
AVERAGE = (10 + 15 + 20 + 25 + 30)/5 = 100/5 = 20
AVERAGE = (10 + 15 + 20 + 25 + 30)/5 = 100/5 = 20
Answer: Option D. -> None of these
AVERAGE = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75
AVERAGE = (76 + 65 + 82 + 67 + 85)/5 = 375/5 = 75
Answer: Option B. -> 60, 60
LET THE NUMBER OF BOYS BE X.
NUMBER OF GIRLS IS 120 – X.
TOTAL NUMBER OF CHOCOLATES RECEIVED BY BOYS AND GIRLS = 2X + 3(120 – X) = 300
=> 360 – X = 300 => X = 60.
SO, THE NUMBER OF BOYS OR GIRLS IS 60
LET THE NUMBER OF BOYS BE X.
NUMBER OF GIRLS IS 120 – X.
TOTAL NUMBER OF CHOCOLATES RECEIVED BY BOYS AND GIRLS = 2X + 3(120 – X) = 300
=> 360 – X = 300 => X = 60.
SO, THE NUMBER OF BOYS OR GIRLS IS 60
Answer: Option A. -> Rs. 21000
LET THE COST PRICES OF THE COLOUR TELEVISION SOLD AT 30% PROFIT AND 40% PROFIT BE RS. X AND RS. (35000 – X) RESPECTIVELY.
TOTAL SELLING PRICE OF TELEVISIONS = X + 30/100 X + (35000 – X) + 40/100 (35000 – X)
=> 130/100 X + 140/100 (35000 – X) = 35000 + 32/100 (35000)
X = 28000
35000 – X = 7000
DIFFERENCE IN THE COST PRICES OF TELEVISIONS = RS. 21000
LET THE COST PRICES OF THE COLOUR TELEVISION SOLD AT 30% PROFIT AND 40% PROFIT BE RS. X AND RS. (35000 – X) RESPECTIVELY.
TOTAL SELLING PRICE OF TELEVISIONS = X + 30/100 X + (35000 – X) + 40/100 (35000 – X)
=> 130/100 X + 140/100 (35000 – X) = 35000 + 32/100 (35000)
X = 28000
35000 – X = 7000
DIFFERENCE IN THE COST PRICES OF TELEVISIONS = RS. 21000
Answer: Option C. -> 23
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.
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.
Answer: Option A. -> 5 : 1
LET THE AMOUNT LENT AT 7% BE RS. X
AMOUNT LENT AT 10% IS RS. (6000 – X)
TOTAL INTEREST FOR ONE YEAR ON THE TWO SUMS LENT
= 7/100 X + 10/100 (6000 – X) = 600 – 3X/100
=> 600 – 3/100 X = 450 => X = 5000
AMOUNT LENT AT 10% = 1000
REQUIRED RATIO = 5000 : 1000 = 5 : 1
LET THE AMOUNT LENT AT 7% BE RS. X
AMOUNT LENT AT 10% IS RS. (6000 – X)
TOTAL INTEREST FOR ONE YEAR ON THE TWO SUMS LENT
= 7/100 X + 10/100 (6000 – X) = 600 – 3X/100
=> 600 – 3/100 X = 450 => X = 5000
AMOUNT LENT AT 10% = 1000
REQUIRED RATIO = 5000 : 1000 = 5 : 1
Answer: Option A. -> 40
LET THE MARKS OBTAINED BY THE STUDENT IN MATHEMATICS, PHYSICS AND CHEMISTRY BE M, P AND C RESPECTIVELY.
GIVEN , M + C = 60 AND C – P = 20 M + C / 2 = [(M + P) + (C – P)] / 2 = (60 + 20) / 2 = 40
LET THE MARKS OBTAINED BY THE STUDENT IN MATHEMATICS, PHYSICS AND CHEMISTRY BE M, P AND C RESPECTIVELY.
GIVEN , M + C = 60 AND C – P = 20 M + C / 2 = [(M + P) + (C – P)] / 2 = (60 + 20) / 2 = 40
Answer: Option B. -> 151
TOTAL HEIGHT = 150 * 50 = 7500 CM.
NEW AVERAGE = [7500 – 5 * 146 + 5 * 156 ] / 50 = 151 CM
TOTAL HEIGHT = 150 * 50 = 7500 CM.
NEW AVERAGE = [7500 – 5 * 146 + 5 * 156 ] / 50 = 151 CM
Answer: Option D. -> Data inadequate
LET THE PRESENT AGES OF A, B AND C BE A, B AND C RESPECTIVELY.
GIVEN, [(A – 5) + (B – 5)] / 2 = 40 => A + B = 90 — (1)
(B + C)/2 = 48 => B + C = 96 — (2)
FROM (1) AND (2), WE CANNOT FIND B
LET THE PRESENT AGES OF A, B AND C BE A, B AND C RESPECTIVELY.
GIVEN, [(A – 5) + (B – 5)] / 2 = 40 => A + B = 90 — (1)
(B + C)/2 = 48 => B + C = 96 — (2)
FROM (1) AND (2), WE CANNOT FIND B
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