In this article you will find 10 questions from various topics like Time & Work, Time & Distance, Ages, Average, Boats & Stream, Pipes & Cistern, Progression, Profit & Loss and Mensuration. I have tried my best to provide the detailed explanation to each and every question along with the important formulas and concepts that are used to solve the problem. At the end of this article you will find a quiz which consists of 5 questions ( 3 marks each) to test your preparation level.
 The sum of the age of Father and his daughter is 62 years. After 5 years, the age of father will be five times that of the daughter. What is the age of the daughter?
A. 9 years
B. 12 years
C. 5 years
D. 7 years
E. None of these
Ans: Let the present age of father & his daughter be X & Y.
Sum of their ages X + Y = 62 ———————(1)
After 5 years,
Age of father will be (X + 5) years
Age of daughter will be (Y + 5) years
According to question:
(X + 5) = 5 (Y + 5)
X + 5 = 5 Y + 25
X – 5 Y = 20 ——————————————–(2)
Substituting the value of X from equation (1) into equation (2) we get,
(62 – Y) – 5 Y = 20
6 Y = 42
Y = 7 years
Therefore, the age of daughter is 7 years (Option D)
2. The sum of the ages of a father and daughter is 45 years. Five years ago, the product of their ages was 4 times the father’s age at that time. What is the present age of the father & daughter:
A. 39 years, 6 years
B. 42 years, 3 years
C. 36 years, 9 years
D. 33 years, 12 years
E. None of these.
Ans: Let the present age of father be X years then, daughter’s age will be (45 – X) years.
Five years ago,
Father’s age = (X – 5) years
Daughter’s age = (45 – X – 5) = (40 – X) years
According to question:
(X – 5) × (40 – X) = 4 (X – 5)
40X – X^{2} – 200 + 5X = 4X – 20
X^{2} – 41X + 180 = 0
X^{2} – 36X – 5 X + 180 = 0
(X – 36) (X – 5) = 0
X = 36, 5
The age of father cannot be less than the age of daughter. Therefore, father’s age will be X = 36 years.
And daughter’s age = (45 – 36) = 9 years. (Option C)
3. A pilot flies an aircraft at a certain speed for a distance of 800 km. He could have saved 40 minutes by increasing the average speed of the plane by 40 km/hr. Find the average speed of the aircraft?
A. 200 km/hr
B. 240 km/hr
C. 300 km/hr
D. 360 km/hr
E. 420 km/hr
Ans: Let the average speed of the aircraft be V km/hr.
Distance = 800 km
Time taken by the aircraft T = 800/V ———————(1)
If the average speed is increased by 40 km/hr, so new average speed = (V+40) km/hr.
So, time taken will be 40 minutes i.e 2/3 hr less than the time taken by the aircraft when it is flying with the average speed V km/hr.
(T – 2/3) = 800/(V+40) ————————————(2)
Now putting the value of ‘T’ from equation (1) into equation (2), we get
{800/V} − {800/(V+40)} = 2/3
V(V+40) = 1200×40
V² + 40V – 48000 = 0
Once you have arrived at the desired equation you can easily check with the help of options.
V (average speed of the aircraft) = 200 km/hr. (Option A)
4. There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, if the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted, if there was no leakage and it was completely used for eating purpose?
A. 56 days
B. 72 days
C. 64 days
D. 120 days
E. 116 days.
Ans: Let Y kg of oil is used for eating purpose everyday.
Case 1:
If 11 kg of oil is leaked out per day, then oil will last for 50 days
So total kg of oil in the container = 11×50 + 50×Y = 550 + 50Y —————(1)
Case 2:
If the leakage was 15 kg/day, then oil will last for 45 days
Total kg of oil in the container = 15×45 + 45×Y = 675 + 45Y ——————(2)
Total Kg of oil or quantity of oil is same in both the cases, so equating (1) & (2), we get
550 + 50Y = 675 + 45Y
Y = 25 Kg ( quantity of oil used for eating purpose everyday)
Now putting the value of Y in any of the above equation:
Total Kg of oil in the container = 550 + 50×25 = 1800 kg
So, if there is no leakage then oil will last for (only used for eating purpose) = 1800/25 = 72 days (Option B)
5. 3 men and 4 women can complete a work in 16 days but 4 men and 3 women can complete the same work in 12 days accordingly in how many days will 2 men and 5 women complete this work?
A. 20 days
B. 22 days
C. 24 days
D. 32 days
E. 36 days
Ans: Let man & woman complete M% and W% of work in 1 day.
Case 1
3 men and 4 women complete 100% of work in 16 days
3×16×M + 4×16×W = 100
12M + 16W = 25 ————————(1)
Case 2
4 men & 3 women complete the same work in 12 days
4×12×M + 3×12×W = 100
12M + 9W = 25 ————————–(2)
Solving equation (1) & (2) we get
W = 0
Now putting the value of W=0 in equation (1) to get the value of M.
M = (25/12)% of work completed by a man in 1 day.
2 men & 5 women will complete = 2×25/12 + 5×0 = (25/6)% of work in 1 day
Therefore to complete 100% of work, 2 men & 5 women will require = 100/(25/6) = 24 days. (Option C)
6. Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?
A. 48
B. 56
C. 66
D. 72
E. Cannot be determined.
Ans:
7. Ajit can swim 13 km in 2 hours with the flow. He can swim with the speed of 5 km/hr in still water. How much time he needs to swim 10.5 km against the flow of river?
A. 2.5 hours
B. 3 hours
C. 3.2 hours
D. 3.5 hours
E. 4 hours
Ans: With the flow means along the direction of stream i.e downstream.
Speed of Ajit downstream = 13/2 = 6.5 km/hr
Speed of Ajit in still water = 5 km/hr
Speed of stream = 6.5 − 5 = 1.5 km/hr
Against the flow of river means upstream
Speed of Ajit upstream = 5 − 1.5 = 3.5 km/hr
To cover 10.5 km against the flow of river, Ajit needs = Distance/Upstream Speed = 10.5/3.5 = 3 hrs. (Option B)
8. Selling price of three articles M, N, P is in the ratio 3 : 4 : 5 and profit percent earned on selling these chargers is in the ratio of 4 : 12 : 5. If the cost price of article M and N is equal and cost price of article P is Rs 120. Find the overall gain on three articles?
A. Rs 90
B. Rs 120
C. Rs 150
D. Rs 180
E. Cannot be determined
Ans: Let the selling price of three articles M, N & P be 3X, 4X and 5X.
and, profit earned on these articles be 4Y% , 12Y% and 5Y%.
According to question, C.P of M = C.P of N
S.P of M/(1+4Y/100) = S.P of N/(1+12Y/100)
3X/(1+4Y/100) = 4X/(1+12Y/100)
3(100 + 12Y) = 4(100 + 4Y)
Y = 5
So, profit percent earned by M, N & P = 4Y% , 12Y% & 5Y% = 20%, 60% & 25%
Now C.P of P = Rs 120
SP of P/(1+5Y/100) = 120
5X/(1+25/100) = 120
X = 30
Selling price of M, N & P = 3X , 4X and 5X = Rs 90, Rs 120 and Rs 150
Cost Price of M, N & P = 90/(1+20/100) , 120/(1+60/100), 120 = Rs 75, Rs 75 & Rs 120
Overall profit = (90 − 75) + (120 − 75) + (150 − 120) = Rs 90 (Option A)
9. A circle and a square have equal areas, what is the ratio of square’s diagonal to the circle’s diameter?
Ans:
10. The radius of a circular field is equal to the side of a square field. If the difference between the perimeter of the circular field and that of the square field is 128 m. What is the perimeter of the square field? (in metres)
Ans:
Quantitative Aptitude Quiz on Miscellaneous Problems
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Question 1 of 5
1. Question
3 pointsA sold a TV to B for Rs 4800 by losing 20% . B sells it to C at a price which would have given A a profit of 15%. B’s gain is:
Correct
S.P of TV for A = Rs 4800
C. P of TV for A = 4800 × 100/80 = Rs 6000
S.P of TV for B = 6000 × 115/100 = Rs 6900
Therefore, B’s gain = Rs (6900 – 4800) = Rs 2100
Incorrect
S.P of TV for A = Rs 4800
C. P of TV for A = 4800 × 100/80 = Rs 6000
S.P of TV for B = 6000 × 115/100 = Rs 6900
Therefore, B’s gain = Rs (6900 – 4800) = Rs 2100

Question 2 of 5
2. Question
3 pointsRahul and Abhijit can do a piece of work in 21 and 24 days respectively. They start the work together and after some days Rahul leaves the work and Abhijit completes the remaining work in 9 days. After how many days Rahul leave?
Correct
In 1 day Rahul and Abhijit will complete 1/21 and 1/24 part of the work.
Let after ‘X’ days Rahul leave the work.
X/21 + (X + 9)/24 = 1
X/21 + X/24 = 1 − 3/8
Therefore, X = 7 days
Incorrect
In 1 day Rahul and Abhijit will complete 1/21 and 1/24 part of the work.
Let after ‘X’ days Rahul leave the work.
X/21 + (X + 9)/24 = 1
X/21 + X/24 = 1 − 3/8
Therefore, X = 7 days

Question 3 of 5
3. Question
3 pointsA steel sphere of radius 3 cm is drawn into a wire of diameter 4 mm. Find the length of wire?
Correct
Steel sphere radius R = 3cm = 30 mm
Wire diameter = 2r = 4mm
so, wire radius r = 2mm
Let the length of wire be ‘h’.
Volume of the steel sphere = Volume of wire
4ΠR³/3 = Πr²h
h = {4 × (30)³}/(3 × 2²)
h = 9000 mm
Incorrect
Steel sphere radius R = 3cm = 30 mm
Wire diameter = 2r = 4mm
so, wire radius r = 2mm
Let the length of wire be ‘h’.
Volume of the steel sphere = Volume of wire
4ΠR³/3 = Πr²h
h = {4 × (30)³}/(3 × 2²)
h = 9000 mm

Question 4 of 5
4. Question
3 pointsFor a certain article, if the discount is 25% the profit is 25%. If the discount is 10%, then the profit is
Correct
Let the Marked price be Rs 100
If discount is 25% and profit is 25% , then
S.P will be Rs 75
C.P will be Rs 60
If discount is 10% then S.P = Rs 90
Profit % = {(90 − 60)/60}× 100 = 50%
Incorrect
Let the Marked price be Rs 100
If discount is 25% and profit is 25% , then
S.P will be Rs 75
C.P will be Rs 60
If discount is 10% then S.P = Rs 90
Profit % = {(90 − 60)/60}× 100 = 50%

Question 5 of 5
5. Question
3 pointsA man can row 6 km/hr in still water. If the speed of the current is 2 km/hr, it takes 3 hours more in upstream than in the downstream for the same distance. The distance is:
Correct
Let the distance be ‘S’ km.
Upstream Speed = 62 = 4 km/hr
Downstream Speed = 6 + 2 = 8 km/hr
S/4 − S/8 = 3
S = 24 km
Incorrect
Let the distance be ‘S’ km.
Upstream Speed = 62 = 4 km/hr
Downstream Speed = 6 + 2 = 8 km/hr
S/4 − S/8 = 3
S = 24 km
Leaderboard: Quantitative Aptitude Quiz on Miscellaneous Problems
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good collection of questions and well explained.
Thank You…..:)
Thank you sir, great explanation. Please include more questions on your upcoming post🙏