Quantitative Aptitude
BASIC MATHS MCQs
Total Questions : 233
| Page 2 of 24 pages
Answer: Option C. -> 21
LET HER AGE BE X
AFTER 1 YEAR HER AGE WILL BE X+1=2(X-10)
X+1=2X-20
1+20=2X-X
X=21 YEARS
LET HER AGE BE X
AFTER 1 YEAR HER AGE WILL BE X+1=2(X-10)
X+1=2X-20
1+20=2X-X
X=21 YEARS
Answer: Option B. -> 22 ½ days
IF A COMPLETES A WORK IN 1 DAY, B COMPLETES THE SAME WORK IN 3 DAYS
HENCE, IF THE DIFFERENCE IS 2 DAYS, B CAN COMPLETE THE WORK IN 3 DAYS
=> IF THE DIFFERENCE IS 60 DAYS, B CAN COMPLETE THE WORK IN 90 DAYS
=> AMOUNT OF WORK B CAN DO IN 1 DAY= 1/90
AMOUNT OF WORK A CAN DO IN 1 DAY = 3 × (1/90) = 1/30
AMOUNT OF WORK A AND B CAN TOGETHER DO IN 1 DAY = 1/90 + 1/30 = 4/90 = 2/45
=> A AND B TOGETHER CAN DO THE WORK IN 45/2 DAYS = 22 ½ DAYS
IF A COMPLETES A WORK IN 1 DAY, B COMPLETES THE SAME WORK IN 3 DAYS
HENCE, IF THE DIFFERENCE IS 2 DAYS, B CAN COMPLETE THE WORK IN 3 DAYS
=> IF THE DIFFERENCE IS 60 DAYS, B CAN COMPLETE THE WORK IN 90 DAYS
=> AMOUNT OF WORK B CAN DO IN 1 DAY= 1/90
AMOUNT OF WORK A CAN DO IN 1 DAY = 3 × (1/90) = 1/30
AMOUNT OF WORK A AND B CAN TOGETHER DO IN 1 DAY = 1/90 + 1/30 = 4/90 = 2/45
=> A AND B TOGETHER CAN DO THE WORK IN 45/2 DAYS = 22 ½ DAYS
Answer: Option A. -> 12
WORK DONE BY P IN 1 DAY = 1/15
WORK DONE BY Q IN 1 DAY = 1/10
WORK DONE BY P AND Q IN 1 DAY = 1/15 + 1/10 = 1/6
WORK DONE BY P AND Q IN 2 DAYS = 2 × (1/6) = 1/3
REMAINING WORK = 1 – 1/3 = 2/3
TIME TAKEN BY P TO COMPLETE THE REMAINING WORK 2/3 = (2/3) / (1/15) = 10 DAYS
TOTAL TIME TAKEN = 2 + 10 = 12 DAYS
WORK DONE BY P IN 1 DAY = 1/15
WORK DONE BY Q IN 1 DAY = 1/10
WORK DONE BY P AND Q IN 1 DAY = 1/15 + 1/10 = 1/6
WORK DONE BY P AND Q IN 2 DAYS = 2 × (1/6) = 1/3
REMAINING WORK = 1 – 1/3 = 2/3
TIME TAKEN BY P TO COMPLETE THE REMAINING WORK 2/3 = (2/3) / (1/15) = 10 DAYS
TOTAL TIME TAKEN = 2 + 10 = 12 DAYS
Answer: Option D. -> 6 hrs.
X’S 1 HOUR WORK = 1/10
Y’S 1 HOUR WORK = 1/15
(X+Y)’S 1 HOUR WORK = 1/10+1/15=5/30=1/6
BOTH THE TAPS CAN FILL THE TANK IN 6 HRS.
X’S 1 HOUR WORK = 1/10
Y’S 1 HOUR WORK = 1/15
(X+Y)’S 1 HOUR WORK = 1/10+1/15=5/30=1/6
BOTH THE TAPS CAN FILL THE TANK IN 6 HRS.
Answer: Option D. -> 60
AMOUNT OF WORK DONE BY A AND B IN 1 DAY = 1/30
AMOUNT OF WORK DONE BY A AND B IN 20 DAYS = 20 × (1/30) = 20/30 = 2/3
REMAINING WORK – 1 – 2/3 = 1/3
A COMPLETES 1/3 WORK IN 20 DAYS
AMOUNT OF WORK A CAN DO IN 1 DAY = (1/3)/20 = 1/60
=> A CAN COMPLETE THE WORK IN 60 DAYS
AMOUNT OF WORK DONE BY A AND B IN 1 DAY = 1/30
AMOUNT OF WORK DONE BY A AND B IN 20 DAYS = 20 × (1/30) = 20/30 = 2/3
REMAINING WORK – 1 – 2/3 = 1/3
A COMPLETES 1/3 WORK IN 20 DAYS
AMOUNT OF WORK A CAN DO IN 1 DAY = (1/3)/20 = 1/60
=> A CAN COMPLETE THE WORK IN 60 DAYS
Answer: Option C. -> 1.2 days
TIME IS
DIRECTLY PROPORTIONAL TO NUMBER OF SOFAS
INVERSELY PROPORTIONAL TO NUMBER OF PEOPLE NO. OF DAYS
= (6/8)*(8/10)*2
= 1.2 DAYS
TIME IS
DIRECTLY PROPORTIONAL TO NUMBER OF SOFAS
INVERSELY PROPORTIONAL TO NUMBER OF PEOPLE NO. OF DAYS
= (6/8)*(8/10)*2
= 1.2 DAYS
Answer: Option C. -> 10
10 WORKERS MAKE 10 TABLES IN 10 DAYS. 1 WORKER MAKES 1 TABLE IN 10 DAYS. SIMILARLY, 5 WORKERS CAN MAKE 5 TABLES IN 10 DAYS.
10 WORKERS MAKE 10 TABLES IN 10 DAYS. 1 WORKER MAKES 1 TABLE IN 10 DAYS. SIMILARLY, 5 WORKERS CAN MAKE 5 TABLES IN 10 DAYS.
Answer: Option C. -> 20 Rs.
B’S DAILY EARNING = RS.(150-94)= RS.56
A’S DAILY EARNING = RS.(150-76)= RS.74
C’S DAILY EARNING = RS.[(150-(56+74)]=RS.20
B’S DAILY EARNING = RS.(150-94)= RS.56
A’S DAILY EARNING = RS.(150-76)= RS.74
C’S DAILY EARNING = RS.[(150-(56+74)]=RS.20
Answer: Option D. -> 18 men
FOR THE WORK TO BE COMPLETED IN 9 DAYS,
6 MEN ARE REQUIRED. TIME AND MEN ARE INVERSELY PROPORTIONAL. FOR THE WORK TO BE COMPLETED IN 3 DAYS,
6*9/3 = 18 MEN ARE REQUIRED
FOR THE WORK TO BE COMPLETED IN 9 DAYS,
6 MEN ARE REQUIRED. TIME AND MEN ARE INVERSELY PROPORTIONAL. FOR THE WORK TO BE COMPLETED IN 3 DAYS,
6*9/3 = 18 MEN ARE REQUIRED
Answer: Option C. -> 16 days
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
NO EXPLANATION IS AVAILABLE FOR THIS QUESTION!
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email