Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS exams. We have included Some questions that are repeatedly asked in exams.

**A train starts from station A at 6:00 AM and reaches station B at 9:00 AM. Another train starts from station B at 7:00 AM and reaches station A at 11:00 AM. At what time the two trains will meet?**

A. 8:00 AM

B. 8:08 AM

C. 8:48 AM

D. 9:08 AM**A. 8:00 AM**

Explanation:

Let the distance from A to B is ‘d’ km.

First train reaches destination in 3 hrs, so speed of first train = d/3 km/hr.

Speed of second train = d/4 km/hr

Speed of first train = d/3 km/hr, so after 1 hr, i.e. at 7:AM, the first train has covered d/3 km.

The distance left = (d – d/3) km = 2d/3 km. relative speed = d/3 + d/4.

So time they will meet at = 7:00 AM + (2d/3) / (d/3 + d/4)

= 7:00 AM + 1 (1/7)*60 = 8:08 AM**A man starts at 8:00 AM with speed 25 km/hr. Another man 300 km away from first man starts at 9:00 AM at 30 km/hr in opposite direction of the first one. At what time will they meet?**

A. 11: 00 AM

B. 2:00 PM

C. 1:00 PM

D. 3:30 PM**B. 2:00 PM**

Explanation:

Distance is 300 km

Speed of first = 25 km/hr, so after 1 hr, i.e. at 9:00 AM, the first has covered 25 km.

The distance left = (300 – 25) km = 275 km. relative speed = 25 + 30 = 55

So time they will meet at = 9:00 AM + 275 / 55

= 9:00 AM + 5 hrs = 2:00 PM**A vessel is full of milk. 6 litres are drawn from vessel and filled with water. This operation is performed two more times. The ratio of the quantities of milk and water in vessel is 8 : 117. How much milk did the vessel contain originally?**

A. 15 litres

B. 10 litres

C. 20 litres

D. 30 litres**B. 10 litres**

Explanation:

When the original quantity is x litres, and y litres of mixture is poured out and again filled with water with operation being performed n times.

The quantity left is = x [1 – y/x]^{n}

Here the final ratio is 8 : 117

So,

8/8+117 = x [1 – 6/x]^{3}/ x

8/125 = [1 – 6/x]^{3}

2/5 = [1 – 6/x]

X = 10**A motorboat whose speed is 15 km/hr in still water goes 30 km upstream and back again to the starting point in 4 hours 03 minutes. Find the speed (km/hr) of the stream.**

A) 3

B) 5

C) 5/3

D) 10**C) 5/3**

Explanation:

4 hrs 03 min = 4 3/60 = 4 1/20 = 81/20Distance = time * [B^2 – R^2] / 2*B

30 = 81/20 * [15^2 – R^2] / 2*15**A man travel a distance of 80 km in 7 hrs. Some part he travels by foot and some by car. If speed with foot is 8 km/hr and with car is 14 km/hr, how much distance is travelled by car?**

A. 12 km

B. 24 km

C. 56 km

D. 60 km**C. 56 km**

Explanation:

By allegation method

8 14

. 80/7

14 – 80/7 = 18/7 80/7 – 8 = 24/7

Ratio of times = 18/7 : 24/7 = 3 : 4

So out of 7 hrs, 3 hrs by foot, and 4 hrs by car

Distance by car = 14*4 = 56 km**A and B jointly invest Rs. 2100 and Rs. 3100 respectively in a firm. A is an active partner and he gets 25% of the profit separately. If their business yields them total Rs. 1040 as profit, what will be the gain of each of them?**

A) Rs. 415, Rs. 625

B) Rs. 575, Rs. 465

C) Rs. 515, Rs. 425

D) Rs. 465, Rs. 575**B) Rs. 575, Rs. 465****A, B, and C invested some money in ratio 3 : 4 : 5. After some time, they receive a profit which is divided among them in ratio 4 : 5 : 6. Find the ratio of the time of their respective investments.**

A. 80 : 72 : 75

B. 80 : 75 : 72

C. 75 : 72 : 80

D. 72 : 80 : 75**B. 80 : 75 : 72**

Explanation:

Let their times are a months, b months, c months respectively.

Investment ratio is 3 : 4 : 5. Then the ratio of profits should be 3*a : 4*b : 5*c

And this ratio is given to be 4 : 5 : 6. So 3a : 4b : 5c = 4 : 5 : 6

So 3a = 4, 4b = 5, 5c = 6

a = 4/3, b = 5/4 , c = 6/5

a : b : c = 4/3 : 5/4 : 6/5 = 80 : 75 : 72**A shopkeeper sells an item at 20% profit but uses a weight of 1200 gm instead of 1 kg. Find hid net profit/loss %.**

A. 15%

B. 12.5%

C. 10%

D. 0%**D. 0%**

Explanation:

Uses 1200 gm in place of 1 kg.

So loss in this = (1200-1000)/1200 * 100 = 50/3%

Now profit of 20% and successive loss of 50/3%.

So overall profit/loss = 20 – 50/3 + (20)(-50/3)/100 = 20 – 50/3 – 10/3 = 0%**x**^{2}= 1369; y = √1369

A. x ≤ y

B. x ≥ y

C. x > y

D. No relationship**A. x ≤ y**

Explanation:

x^{2}= 1369

x = 37, -37

y = √1369

y = 37

put on number line

-37 37

When y = 37, y ≥ x

So y ≥ x**A man went to market and purchased an item of Rs 15,000 and sold it at a loss of 10%, from that amount he purchased another amount and sold it at a profit of 16%. What is the amount of overall profit made by him in all dealing?**

A. Rs 720

B. Rs 660

C. Rs 650

D. Rs 610**B. Rs 660**

Explanation:

By successive formula, overall profit % = -10 + 16 + (-10)(16)/100 [ – for loss]

= 6 – 1.6 = 4.4%

So, overall profit = 4.4 *15000/100 = 660

**AffairsCloud Recommends Oliveboard Mock Test**

**AffairsCloud Ebook - Support Us to Grow**

**Govt Jobs by Category**

**Bank Jobs Notification**