300+ Arithmetic Questions and Answers for Competitive Exams [100% Free ]

300+ Most Asked Arithmetic Questions and Answers

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1. Monty’s and Bunty’s present ages are in 9:4 ratio. After ‘Y’ years, age of Bunty becomes half of age of Monty. The ratio of ages of Monty after 6 years and Bunty 4 years ago, is 6: 1. Find the value of Y.

Ans:3
Let, present ages of Monty = 9m and Bunty = 4m
(9m + 6)/(4m – 4) = 6/1
9m + 6 = 24m – 24
15m = 30
m =2
So, Monty = 18 years
Bunty = 8 years
Now,
(18 + Y) = 2*(8 + Y)
18 – 16 = Y = 2

2. 4 men and 6 women working together can finish the complete work in 10 days. 3 men and 3 women working together will complete the same work in 16 days. In how many days will 5 women complete the work?

Ans:2
Let’s suppose 1 man and 1 woman can complete the entire work in m and n days respectively while working alone.

 

Hence according to 1st condition; 4/m+6/n=1/10…. (1)

Similarly, 3/m+3/n=1/16…. (2)

Solving (1) and (2), we will get n = 120 days

Hence, time taken by 5 women to complete the work = 24 days

3. A vessel of volume 100 lit had milk and water in the ratio 4:1. 25% of it was drawn out and replaced with water. The same process was repeated again. What was the final ratio of milk to water?

Ans:3
Original ratio of milk: water = 4:1 => Original volume: Milk = 80l and Water = 20l

25l was replaced in the first time. The ratio of milk: water in the replaced amount would also be 4:1 => Volume of milk lost = (4/5)*25 = 20l and volume of water lost = 25-20=5l

=> After replacing 25l of liquid with 25l of water:

Volume of milk = 80-20 = 60l and Volume of water = 20-5+25 = 40l => Ratio = 3:2

So when the next 25l is replaced, the volume replaced is also in the same ratio

=> Volume of milk lost = (3/5)*25 = 15l and volume of water lost = 25-15= 10l

=> After the 2nd replacement:

Volume of milk left = 60-15 = 45l and volume of water left = 40-10+25 = 55l

=> Ratio = 9:11

Alternate:

Volume of milk left = Original volume of milk *(1 – Volume of liquid replaced/Total volume of liquid)n

=> Milk left = 80*(1- 25/100)2 = 45l

=> Ratio = 45/(100-45) = 9/11

4. If the length of a rectangle is decreased by 6 cm and its width is increased by 5 cm, then we will get a square. If the area of the square which we got now is equal to the area of the previous rectangle then what was the perimeter of the original rectangle?

Ans:4
Let the original length and breadth of the rectangle be m and n respectively.
According to the question; (m – 6) = (n + 5)
=> m – n = 11 .. (1)
Also; mn = (m – 6) (n + 5)
=> mn = mn + 5m – 6n – 30
=> 5m – 6n = 30 .. (2)
Solving (1) & (2) we will get; m = 36 and n = 25
Therefore, area of the original perimeter = 2(m + n) cm = 2(36 + 25) cm = 122 cm

5. From the income of an employee, 20% is deducted as PF, 20% of the rest he spends as house rent and 25% of the balance, he spends on TA. After all this, he is left with Rs.13000. approximately, what is his monthly income?

Ans:4
Let his monthly salary be S.
After PF deduction, salary left = (1 – 20/100) * S = 80/100 * S
After house rent, salary left = (1 – 20/100) * 80/100 * S = 80/100 * 80/100 * S
After expenditure on TA, Salary left = (1 – 25/100) * 80/100 * 80/100 * S = 75/100 * 80/100 * 80/100 * S
Now,
75/100 * 80/100 * 80/100 * S = 13000
S = 27083.33 = Rs.27084 (approx)

6. Three persons A, B and C entered into a partnership with initial investments Rs.10000, Rs.24000 and Rs.12000 respectively and hired three employees for the business and paid them Rs.500, Rs.1200 and Rs.200 per month which was paid by A, B and C respectively from their respective share from total profit of Rs.46000. If time period of their investment is in ratio 4:5:2, then find the ratio of sum of profit left with A and C together to the profit left with B after 1 year.

Ans:2
Let the time of investment of A, B and C = 4t, 5t and 2t
Ratio of share of profit is
10000*4t: 24000*5t: 12000*2t = 5: 15: 3
Profit share of A = 5*46000/23 = 10000
Profit share of B = 15*46000/23 = 30000
Profit share of C = 3*46000/23 = 6000
A paid salary = 500*12 = 6000
B paid salary = 1200*12 = 14400
C paid salary = 200*12 = 2400
Required ratio = [(10000 – 6000) + (6000 – 2400)]/ (30000 – 14400) = 7600/15600 = 19: 39

7. Keki Fanwalla is offering a discount of 16% on each fan. Fanwalla went to meet Panwalla and in his absence, his assistant sold a fan for ₹756 at an additional 10% discount over the discounted price set by Fanwalla. What was the marked price of the fan?

Ans:3
Let marked price of fan be M.
Selling price of fan when 10% additional discount is given over the 16% discount = (1 – 16/100) * M * (1 – 10/100) = 84/100 * 90/100 * M
Now,
84/100 * 90/100 * M = 756
M = Rs.1000

8. An election was contested by only two candidates A and B. A got 41% of total votes and B got 35% of total votes. The remaining votes were declared invalid. The difference between the votes obtained by B and invalid votes is 968. Calculate the total numbers of votes.

Ans:4
Let total votes be x
Invalid votes percentage = 100 – (41 +35) = 24%
(35% – 24%) of x = 968
11% of x = 968
=> x = 8800

9. Abhay invested Rs 12500 in a chit fund for 3 years and got a total amount of Rs 20000. How much money he will get if he invested the same amount for 2 years but now instead of simple interest; interest is compounded annually?

Ans:3
Principal amount = Rs 12500
Hence total interest that Abhay obtained = Rs (20000 – 12500) = Rs 7500
Hence rate of interest in this case = (7500 X 100)/(12500 X 3) = 20%
Total amount that Abhay will get after 2 years when interest is compounded annually = Rs (12500 X 1.2 X 1.2) = Rs 18000

10. A man sailing a boat having upstream speed 20 kmph has to cover a route of certain distance but after covering 18% of distance (actual distance which he has to travel) downstream, he realised that he went wrong way so he returned back to start point and then covered the right path having 80% upstream and rest downstream journey. Find the distance of the route to be covered if total time taken is 38.9 hours. (speed of stream = 2kmph)

Ans:2
Let the distance be ‘d’ km
Upstream speed = 20 km/h
Speed of stream = 2 km/h
Speed of boat in still water = 20 + 2 = 22 km/h
Downstream speed = 22 + 2 = 24 km/h
Time taken till he returned to starting point = 0.18d/24 + 0.18d/20= 0.18d*(1/24+1/20)
Time taken for the right path = 0.8d/20 + 0.2d/24 = 0.2d*(1/5 + 1/24)
Now,
0.18d*(1/24 + 1/20) + 0.2d*(1/5 + 1/24) = 38.9
d = 600 km