Quantitative Aptitude
BASIC MATHS MCQs
Total Questions : 233
| Page 3 of 24 pages
Answer: Option A. -> Rs.1104
ASSUME THAT COST OF KEEPING A COW FOR 1 DAY = C ,
COST OF KEEPING A GOAT FOR 1 DAY = G
COST OF KEEPING 20 COWS AND 40 GOATS FOR 10 DAYS = 460
COST OF KEEPING 20 COWS AND 40 GOATS FOR 1 DAY = 460/10 = 46
=> 20C + 40G = 46
=> 10C + 20G = 23 —(1)
GIVEN THAT 5G = C
HENCE EQUATION (1) CAN BE WRITTEN AS 10C + 4C = 23 => 14C =23
=> C=23/14
COST OF KEEPING 50 COWS AND 30 GOATS FOR 1 DAY
= 50C + 30G
= 50C + 6C (SUBSTITUTED 5G = C)
= 56 C = 56×23/14
= 92
COST OF KEEPING 50 COWS AND 30 GOATS FOR 12 DAYS = 12×92 = 1104
ASSUME THAT COST OF KEEPING A COW FOR 1 DAY = C ,
COST OF KEEPING A GOAT FOR 1 DAY = G
COST OF KEEPING 20 COWS AND 40 GOATS FOR 10 DAYS = 460
COST OF KEEPING 20 COWS AND 40 GOATS FOR 1 DAY = 460/10 = 46
=> 20C + 40G = 46
=> 10C + 20G = 23 —(1)
GIVEN THAT 5G = C
HENCE EQUATION (1) CAN BE WRITTEN AS 10C + 4C = 23 => 14C =23
=> C=23/14
COST OF KEEPING 50 COWS AND 30 GOATS FOR 1 DAY
= 50C + 30G
= 50C + 6C (SUBSTITUTED 5G = C)
= 56 C = 56×23/14
= 92
COST OF KEEPING 50 COWS AND 30 GOATS FOR 12 DAYS = 12×92 = 1104
Answer: Option D. -> 12 days
2 CHILDREN = 1 MAN
(8 CHILDREN + 12 MEN ) = 16 MEN
NOW, LESS MEN, MORE DAYS
12 : 16 : 9 : X
X = 16×9/12=12 DAYS
2 CHILDREN = 1 MAN
(8 CHILDREN + 12 MEN ) = 16 MEN
NOW, LESS MEN, MORE DAYS
12 : 16 : 9 : X
X = 16×9/12=12 DAYS
Answer: Option A. -> 10 days
PART OF WORK COMPLETED BY C IN 1 DAY = 1/12 PART OF WORK COMPLETED BY D IN 1 DAY = 1/15 PART OF WORK COMPLETED BY C IN 4 DAYS
= 1/3 WORK REMAINING = 2/3 TIME TAKEN BY D TO COMPLETE THE WORK
= (2/3)÷(1/15) = 10 DAYS
PART OF WORK COMPLETED BY C IN 1 DAY = 1/12 PART OF WORK COMPLETED BY D IN 1 DAY = 1/15 PART OF WORK COMPLETED BY C IN 4 DAYS
= 1/3 WORK REMAINING = 2/3 TIME TAKEN BY D TO COMPLETE THE WORK
= (2/3)÷(1/15) = 10 DAYS
Answer: Option A. -> 15 days
RAIO OF TIMES TAKEN BY A AND B = 1 : 3
IF THE DIFFERENCE OF TIME IS 2 DAYS, B TAKES 3 DAYS. IF THE DIFFERENCE OF TIME IS 10 DAYS.
B TAKES [3/2×10]=15 DAYS
RAIO OF TIMES TAKEN BY A AND B = 1 : 3
IF THE DIFFERENCE OF TIME IS 2 DAYS, B TAKES 3 DAYS. IF THE DIFFERENCE OF TIME IS 10 DAYS.
B TAKES [3/2×10]=15 DAYS
Answer: Option D. -> 20 days
TIME AND MEN ARE INVERSELY PROPORTIONAL. TIME TAKEN BY 28 MEN TO FINISH THE WORK = 15 DAYS TIME TAKEN BY A MAN TO FINISH THE WORK
= 15*28 DAYS TIME TAKEN BY 21 MEN = (15*28)/21 = 20 DAYS
TIME AND MEN ARE INVERSELY PROPORTIONAL. TIME TAKEN BY 28 MEN TO FINISH THE WORK = 15 DAYS TIME TAKEN BY A MAN TO FINISH THE WORK
= 15*28 DAYS TIME TAKEN BY 21 MEN = (15*28)/21 = 20 DAYS
Answer: Option C. -> 15 days
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option C. -> 40
WORK DONE BY 4 MEN AND 6 WOMEN IN 1 DAY = 1/8
WORK DONE BY 3 MEN AND 7 WOMEN IN 1 DAY = 1/10
LET 1 MAN DOES M WORK IN 1 DAY AND 1 WOMAN DOES W WORK IN 1 DAY. THE ABOVE EQUATIONS CAN BE WRITTEN AS
4M + 6W = 1/8 —(1)
3M + 7W = 1/10 —(2)
SOLVING EQUATION (1) AND (2) , WE GET M=11/400 AND W=1/400
AMOUNT OF WORK 10 WOMEN CAN DO IN A DAY = 10 × (1/400) = 1/40
IE, 10 WOMEN CAN COMPLETE THE WORK IN 40 DAYS
WORK DONE BY 4 MEN AND 6 WOMEN IN 1 DAY = 1/8
WORK DONE BY 3 MEN AND 7 WOMEN IN 1 DAY = 1/10
LET 1 MAN DOES M WORK IN 1 DAY AND 1 WOMAN DOES W WORK IN 1 DAY. THE ABOVE EQUATIONS CAN BE WRITTEN AS
4M + 6W = 1/8 —(1)
3M + 7W = 1/10 —(2)
SOLVING EQUATION (1) AND (2) , WE GET M=11/400 AND W=1/400
AMOUNT OF WORK 10 WOMEN CAN DO IN A DAY = 10 × (1/400) = 1/40
IE, 10 WOMEN CAN COMPLETE THE WORK IN 40 DAYS
Answer: Option B. -> 20 min
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Answer: Option D. -> 6
LET P TAKES X DAYS TO COMPLETE THE WORK
THEN Q TAKES X/2 DAYS AND R TAKES X/3 DAYS TO FINISH THE WORK
AMOUNT OF WORK P DOES IN 1 DAY = 1/X
AMOUNT OF WORK Q DOES IN 1 DAY = 2/X
AMOUNT OF WORK R DOES IN 1 DAY = 3/X
AMOUNT OF WORK P,Q AND R DO IN 1 DAY = 1/X + 2/X + 3/X = 1/X (1 + 2 + 3) = 6/X
6/X = 2
=> X = 12
=> Q TAKES 12/2 DAYS = 6 DAYS TO COMPLETE THE WORK
LET P TAKES X DAYS TO COMPLETE THE WORK
THEN Q TAKES X/2 DAYS AND R TAKES X/3 DAYS TO FINISH THE WORK
AMOUNT OF WORK P DOES IN 1 DAY = 1/X
AMOUNT OF WORK Q DOES IN 1 DAY = 2/X
AMOUNT OF WORK R DOES IN 1 DAY = 3/X
AMOUNT OF WORK P,Q AND R DO IN 1 DAY = 1/X + 2/X + 3/X = 1/X (1 + 2 + 3) = 6/X
6/X = 2
=> X = 12
=> Q TAKES 12/2 DAYS = 6 DAYS TO COMPLETE THE WORK
Answer: Option D. -> 36 days
NO. OF DAYS IS DIRECTLY PROPORTIONAL TO THE LENGTH OF THE WALL AND INVERSELY PROPORTIONAL TO THE NUMBER OF MEN NO. OF DAYS = (1200/500)*(15/30)*30 = 36 DAYS
NO. OF DAYS IS DIRECTLY PROPORTIONAL TO THE LENGTH OF THE WALL AND INVERSELY PROPORTIONAL TO THE NUMBER OF MEN NO. OF DAYS = (1200/500)*(15/30)*30 = 36 DAYS