#### Quantitative Aptitude Problems With Answers - Free Online Practice Test

- Latest model Quant Questions for SBI PO - Free online Practice Test 1
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- Quantitative Aptitude Problems With Answers All Chapters with multiple Examples

- 1. If a boat goes 7 km upstream in 42 minuters and the speed of the stream is 3 kmph, then the speed of the boat in still water is

**Answer: Option 'C'**

**Speed Upstream = 7/42 km/min = 7/42 × 60 km/hr
i.e, b = 10 km/hr
speed of the stream = 3 km/hr
Let speed in sttil water is x km/hr
Then, speed upstreeam = (x - 3) km/hr
x - 3 = 10 or x = 13 km/hr**

- 2. Charan is an much younger than Raju as he is older than Tharun. If the sum of the ages of Raju and Tharun is 50 years, what is definitely the difference between Raju and Charan's age?

**Answer: Option 'D'**

**Given that
1. The difference of age between Raju and Charan = The difference of age between Charan and Tharun.
2. Sum of age of Raju and Tharun is 50 i.e (Raju + Tharun) = 50
Question : Raju - Charan = ?
Solution :
Raju - Charan = Charan - Tharun
(Raju + Tharun) = 2Charan
Now given that, (Raju + Tharun) = 50
So, 50 = 2Charan and therefore Charan = 25.
The question is (Raju - Charan) = ?
Here we know the value (age) of Charan(25), but we don't know the age of Raju.
Therefore, (Raju - Charan) cannot be determined**

- 3. Vamsi started a company investing Rs. 200000. After 8 months, Krishna joined him with a capital of Rs. 120000. After 2 years, they earned a profit of Rs. 42000, what is Krishna's share of the profit?

**Answer: Option 'C'**

**Profits of Vamsi and Krishna are in the ratio 200000 × 24 : 120000 × 16 = 5 : 2
Krishna's share = (2/7) × 42000 = Rs. 12000/-**

- 4. How many seconds will a 500 meter long train moving with a speed of 63 km/hr, take to cross a man walking with a speed of 3 of 3 km/hr is the direction of the train?

**Answer: Option 'B'**

**Distance = 500m
Speed = 63 - 3 km/hr = 60 km/hr
= 600/36 m/s = 50/3 m/s
Time taken = distance/speed = 500/(50/3) = 30 sec**

- 5. A shopkeeper sells 1 kg of apples for Rs. 160 at a loss of 20%. At what price should he sell a kg of apples to gain 20%?

**Answer: Option 'D'**

**Let the cost price be Rs. (160 + x) Loss = Rs. x Loss% = (x/160+x)) × 100 = 20%
x = Rs. 40
Cost Price = Rs. 200 Gain% = (Gain/ Cost Price) × 100 = 20% Gain = Rs. 40
Selling Price = Gain + Cost Price = Rs. 240/-**

- 6. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

**Answer: Option 'C'**

**Let the original rate be R%. Then, new rate = (2R)%
Note : Here, Original rate is for 1 year(s), the new rate is for only 4 months
that is 1/3 years
[(725 × r × 1)/100] + [(362.50 × 2r × 1)/(100 × 3)] = 33.50
R = 3.46%**

- 7. Excluding stoppages, the average speed of a buss is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?

**Answer: Option 'C'**

**Due to stoppages, the bus travels only 45 kms in an hour (9 kms less). To cover a distance of 9 km at a speed of 54 kmph, time taken = 9/54 = 1/6 hrs = 10 mins.**

- 8. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to from a third alloy C, the ratio of gold and copper in C will be?

**Answer: Option 'B'**

**Let 1 kg of each one of A and B be taken to from 2 kg of C.
Gold in 1 kg of A = 7/9, copper in 1 kg of A = 2/9
Gold in 1 kg of B = 7/18, copper in 1 kg og B = 11/18
Gold in 2 kg of C = (7/9 + 7/18) = 21/18 = 7/6
Copper in 2 kg of C = (2/9 + 11/18) = 15/18 = 5/6
Ratio of gold and copper in C = 7/6 : 5/6 = 7 : 5
**

- 9. Two trains travelling in opposite directions at speeds of 100 kmph and 90 kmph cross each other for 9s. If the length of the first train is 225m, find the length of the second train.

**Answer: Option 'C'**

**Relative speed of the trains = 110 + 90 = 200 kmph = 200 × 5/18 m/s 200×5/18 = (225+I)/9 I = 275 m**

- 10. Pipe A and Pipe B can fill a tank in 20 minutes and 30 minutes respectively. Pipe C can empty the tank in 60 minutes. The tank is initially empty. Both the pipes A and B are opened. Pipe C is opened after 6 minutes. How much time does it take to fill the tank?

**Answer: Option 'D'**

**1/20th of the tank is filled by Pipe A in 1 minute.
1/30th of the tank is filled by Pipe B in 1 minute.
1/60th of the tank is emptied by Pipe C in 1 minute.
Part of tank filled in the first 6 minuters = 6 × [(1/20) + (1/30)] = 1/2 When all pipes are opened,
part of tank filled in 1 minute = (1/12) - (1/60) = 1/15
Time required to fill the remaining 1/2 of the tank = (1/2)/(1/15) = 7.5
Total time = 6 + 7.5 = 13.5 minutes**

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