# IBPS PO 2018 Prelims: Quants Test Day 14

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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**The sum of two numbers is 520. If the bigger number is decreased by 4% and the smaller number is increased by 12% then the numbers obtained are equal. Find the smallest number?**CorrectAnswer –

**1) 240**

Explanation:

Let the bigger number = A

The smaller number= 520 – A

According to question,

A * ((100-4)/100) = (520 – A) * ((100 + 12)/100)

96A/100 = (520 – A) * (112/100)

96A = (520 – A) * 112

13A = 3640

A = 280

Therefore, Bigger Number = 280

Smaller Number = 240IncorrectAnswer –

**1) 240**

Explanation:

Let the bigger number = A

The smaller number= 520 – A

According to question,

A * ((100-4)/100) = (520 – A) * ((100 + 12)/100)

96A/100 = (520 – A) * (112/100)

96A = (520 – A) * 112

13A = 3640

A = 280

Therefore, Bigger Number = 280

Smaller Number = 240 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Equal sum of money are lent to X and Y at 7.5% per annum for a period of 4 years and 5 years respectively. If the difference in interest, paid by them was Rs 150. What was the sum lent to each?**CorrectAnswer

**4) 2000**

Explanation:

Total Interest Rate for X= 7.5 * 4= 30%

Total Interest Rate for Y= 7.5 * 5= 37.5%

Difference in Rates= (37.5 – 30)% = 7.5%

According to question,

7.5% of sum= 150

1% of sum= 150/7.5

Individual Sum= (150/7.5) * 100

= Rs 2000

Therefore, Sum lent by X and Y each = Rs 2000IncorrectAnswer

**4) 2000**

Explanation:

Total Interest Rate for X= 7.5 * 4= 30%

Total Interest Rate for Y= 7.5 * 5= 37.5%

Difference in Rates= (37.5 – 30)% = 7.5%

According to question,

7.5% of sum= 150

1% of sum= 150/7.5

Individual Sum= (150/7.5) * 100

= Rs 2000

Therefore, Sum lent by X and Y each = Rs 2000 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Two trains 125 meters and 115 meters in length are running towards each other on parallel lines, one at the rate of 33 Km/hr and the other at 39 Km/hr. How much time (in seconds) will they take to pass each other from the moment they meet?**CorrectAnswer

**– 2) 12 seconds**

Explanation:

Time taken by trains to cross each other in opposite direction

= (Total Distance/ Relative Speed in opposite direction)

= (125 + 115)/ ((33 + 39) * 5/18)

= (240 * 18)/ (72 * 5)

Time = 12 secondsIncorrectAnswer

**– 2) 12 seconds**

Explanation:

Time taken by trains to cross each other in opposite direction

= (Total Distance/ Relative Speed in opposite direction)

= (125 + 115)/ ((33 + 39) * 5/18)

= (240 * 18)/ (72 * 5)

Time = 12 seconds - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B fill it in 7 hours more. How much time taken by C alone to fill the cistern?**CorrectAnswer –

**3) 14 hours**

Explanation:

Let Total capacity= 42 units

Therefore, (A + B + C) per hour work = 42/6 = 7 units

(A + B + C) fills the tank in 7 units/hr

They all worked for 2 Hours.

Total water filled= 7 * 2= 14 units

Capacity Left= 42 – 14= 28 units

A + B= 28/7= 4 units/hr

(A + B) efficiency 4 units

C’s efficiency= ((A + B + C) – (A + B))= 7 – 4= 3 units/hr

C can alone fill the cistern= Total Capacity/ Efficiency

= 42/3

= 14 hrsIncorrectAnswer –

**3) 14 hours**

Explanation:

Let Total capacity= 42 units

Therefore, (A + B + C) per hour work = 42/6 = 7 units

(A + B + C) fills the tank in 7 units/hr

They all worked for 2 Hours.

Total water filled= 7 * 2= 14 units

Capacity Left= 42 – 14= 28 units

A + B= 28/7= 4 units/hr

(A + B) efficiency 4 units

C’s efficiency= ((A + B + C) – (A + B))= 7 – 4= 3 units/hr

C can alone fill the cistern= Total Capacity/ Efficiency

= 42/3

= 14 hrs - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**The tea costing Rs 192 per kg is to be mixed with tea costing Rs 150 per kg. In what ratio tea is mixed, when sold for Rs 194.40 per kg gives a profit of 20%?**CorrectAnswer –

**5) 2:5**

Explanation:

IncorrectAnswer –

**5) 2:5**

Explanation:

- Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Find the wrong number in the series**

1, 9, 32, 114, 478, 2400CorrectAnswer –

**5) 478**

Explanation

1 x1 +8 x 1 =9

9 x2 +7 x 2=32

32×3 +6 x3=114

114×4 +5×4 =476

476×5 +4×5=2400IncorrectAnswer –

**5) 478**

Explanation

1 x1 +8 x 1 =9

9 x2 +7 x 2=32

32×3 +6 x3=114

114×4 +5×4 =476

476×5 +4×5=2400 - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**What will come in the place of question mark(?) in the following question.**

34 x 41 +65 x 29 -45 x 39 = ? x26 +16CorrectAnswer –

**3) 58**

Explanation

? x 26 = 34 x 41 + 65 x 29 – 45 x 39 -16

? x 26 = 1394 +1885 -1775-16

= 1508/26

= 58IncorrectAnswer –

**3) 58**

Explanation

? x 26 = 34 x 41 + 65 x 29 – 45 x 39 -16

? x 26 = 1394 +1885 -1775-16

= 1508/26

= 58 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Shyam went to the market to buy 1.5kg of dried peas having 20% water content. He went home and soaked them for some time and the water content in the peas becomes 60%. Find the final weight of soaked peas.**CorrectAnswer –

**4) 3 kg**

Explanation

Weight of dried peas = 1.5 kg

Water content = 20%

Hence, non-water content = 80% of 1.5

=0.8 x 1.5 = 1.2 kg (1)

Let the weight of the soaked pees = x kg

Water content = 60%

Non-water content = 40% of x kg = 0.4x (2)

From (1) and (2) an equal

Hence, 1.2x = 0.4x

X = 3IncorrectAnswer –

**4) 3 kg**

Explanation

Weight of dried peas = 1.5 kg

Water content = 20%

Hence, non-water content = 80% of 1.5

=0.8 x 1.5 = 1.2 kg (1)

Let the weight of the soaked pees = x kg

Water content = 60%

Non-water content = 40% of x kg = 0.4x (2)

From (1) and (2) an equal

Hence, 1.2x = 0.4x

X = 3 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**Each question is followed by two statements, I and II . Indicate your response based on the following instruction.****Which scheme between P and Q doubles the investment faster?**

Statement I Scheme P – 16% per annum with interest compounded quarterly.

Statement II Scheme Q -18 % per annum with interest compounded half yearly.CorrectAnswer –

**3) If the question can be answered using I and II together but not using I or II alone**

Explanation

Statement I

In scheme P money grows at (1 + 4/100)2 = (1.04)2 = 1.0816

i.e the investment becomes 1.0816 times ever half year. But nothing is mentioned about Scheme Q. Hence I alone is not sufficient.

Statement II

In scheme Q money grows at (1 +9/100)2 .i.e. the investment becomes 1.09 times every half year. But nothing is mentioned about scheme P, hence II alone is sufficient.

From I and II together, we can answer the question as the rates in both schemes is known.IncorrectAnswer –

**3) If the question can be answered using I and II together but not using I or II alone**

Explanation

Statement I

In scheme P money grows at (1 + 4/100)2 = (1.04)2 = 1.0816

i.e the investment becomes 1.0816 times ever half year. But nothing is mentioned about Scheme Q. Hence I alone is not sufficient.

Statement II

In scheme Q money grows at (1 +9/100)2 .i.e. the investment becomes 1.09 times every half year. But nothing is mentioned about scheme P, hence II alone is sufficient.

From I and II together, we can answer the question as the rates in both schemes is known. - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**Solve the two statements to get quantities and then compare the two quantities.****Quantity I**: Rohan sold two shirts, each for Rs 500. One is sold at a profit of 25% and another at a loss of 20%. What is overall percentage of profit/loss made by Rohan.

**Quantity II**: Ramesh sold a watch at Rs 1000 and made a profit of 25% whereas he sold table for Rs 1500 at a loss of 20%. What is his overall percentage of profit/loss.CorrectAnswer –

**1) Quantity I < Quantity II**

Explanation

Quantity I

Total S.P of the two shirts =Rs 1000

C.P of one shirt with 25% profit = Rs 500/1.25 = Rs400

C.P of second shirt = Rs 500/.8 = Rs 625

Total C.P of both the shirts = Rs (400 + 625) = Rs 1025

Therefore loss % = (1025-1000)/1025 x 100 = 2.44%

Quantity II

C.P of the watch = Rs 1000/1.25 = Rs 800

C.P of the table = Rs 1500/.8 = Rs 1875

Total C.P of both the items = Rs (800 + 1875) = Rs 2675

S.P of both the items = Rs (1000 +1500) = Rs 2500

Hence Loss % = (2675-2500)/2675 x 100 = 6.54%IncorrectAnswer –

**1) Quantity I < Quantity II**

Explanation

Quantity I

Total S.P of the two shirts =Rs 1000

C.P of one shirt with 25% profit = Rs 500/1.25 = Rs400

C.P of second shirt = Rs 500/.8 = Rs 625

Total C.P of both the shirts = Rs (400 + 625) = Rs 1025

Therefore loss % = (1025-1000)/1025 x 100 = 2.44%

Quantity II

C.P of the watch = Rs 1000/1.25 = Rs 800

C.P of the table = Rs 1500/.8 = Rs 1875

Total C.P of both the items = Rs (800 + 1875) = Rs 2675

S.P of both the items = Rs (1000 +1500) = Rs 2500

Hence Loss % = (2675-2500)/2675 x 100 = 6.54%

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