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Quantitative Aptitude Miscellaneous Problem Quiz

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. 

 

  1. 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 – X2 – 200 + 5X = 4X – 20

X2 – 41X + 180 = 0

X2 – 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:

Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?

Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?

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:

A circle and a square have equal areas, what is the ratio of square’s diagonal to the circle’s diameter?

A circle and a square have equal areas, what is the ratio of square’s diagonal to the circle’s diameter?

 

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:

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 32 m. What is the perimeter of the square field? (in metres)

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 32 m. What is the perimeter of the square field? (in metres)

 

Quantitative Aptitude Quiz on Miscellaneous Problems

Read the instructions carefully:

Total number of questions = 5
For each correct answer 3 marks will be awarded.
There is no negative marking.
Time allotted – 6 minutes.

Leaderboard: Quantitative Aptitude Quiz on Miscellaneous Problems

maximum of 15 points
Pos. Name Entered on Points Result
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Knowledge Sharing

Knowledge Sharing is an educational site where you will find quality contents for the aspirants preparing for competitive examinations like Banking , GATE , IIT JEE , MBA entrance examination. Apart from that you will also find motivational & inspirational stories/quotes which will encourage & empower you to think big in life.

  • abhishek says:

    good collection of questions and well explained.

  • Struggler says:

    Thank you sir, great explanation. Please include more questions on your upcoming post🙏


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