Hello Aspirants. Welcome to Online Quant Section in AffairsCloud.com. We are starting **IBPS Clerk course 2015 **and we are creating sample questions in **Quantitative Aptitude** section, type of which will be asked in IBPS Clerk Prelims Exam.

**Stratus – IBPS Clerk Course 2015 **

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**x**^{2}= 4, 2y^{2}– 13y + 20 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined**B) x < y**

Explanation:

x^{2}= 4

x = 2 , -2

2y^{2}– 13y + 20 = 0

2y^{2}– 8y – 5y +20 = 0

2y(y-4) – 5(y-4) = 0

y = 4, 5/2

put on number line

-2 2 5/2 4**2x + 5y = 11, x – 20y = -8**

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined**A) x > y**

Explanation:

To solve the equations, multiply 2nd equation by 2, and subtact both equations:

2x + 5y = 11

2x – 40y = -16

On subtraction – 0x + 45y = 27

y = 3/5

now put y = 3/5 in any of the equations to find x, x= 4**2x**^{2}+ x – 36 = 0, y^{2}+ y – 6 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined**E) x = y or relationship cannot be determined**

Explanation:

2x^{2}+ x – 36 = 0

2x^{2}– 8x + 9x – 36 = 0

x = 4 , -9/2

y^{2}+ y – 6 = 0

y = -3, 2

put on number line

-9/2 -3 2 4**x**^{2}– 5x = 0, 2y^{2}– y – 3 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined**E) x = y or relationship cannot be determined**

Explanation:

x^{2}– 5x = 0

x(x-5) = 0

x = 0, 5

2y^{2}– y – 3 = 0

2y^{2}+2y – 3y – 3 = 0

y = -1, 3/2

put on number line

-1 0 3/2 5**2x**^{2}– 9x + 4 = 0, 2y^{2}+ 7y – 4 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relationship cannot be determined**C) x ≥ y**

Explanation:

2x^{2}– 9x + 4 = 0

2x^{2}– 8x – x + 4 = 0

x = 4 , 1/2

2y^{2}+ 7y – 4 = 0

2y^{2}+ 8y – y – 4 = 0

y = -4, 1/2

put on number line

-4 1/2 4**A mixture contains A and B in the ratio 5 : 3. 16 litres of the mixture is taken out and 16 litres of liquid B is poured in and after this the ratio of the amount of both liquids in the mixture is 15 : 17. Find the amount of liquid A present in the mixture.**

A) 24 litres

B) 40 litres

C) 30 litres

D) 32 litres

E) None of these**B) 40 litres**

Explanation:

Method of doing these type of questions when same quantity is taken out and poured in:

Initial ratio = 5 : 3, 16 litres taken out

Let total quantity = 5x+3x + 16 = 8x+16

After pouring 16 litres of B, ratio is 15 : 17. So

5x/(3x+16) = 15/17

Solve, x = 6

So total quantity = 8*6 + 16 = 64

So A = 5/8 * 64 = 40 litres**A and B can complete a work in 10 days and 14 days respectively. Both start this work and A left after 4 days. B completes the remaining work in how many days?**

A) 9 days

B) 9 4/5 days

C) 5 3/5 days

D) 4 2/5 days

E) None of these**D) 4 2/5 days**

Explanation:

A+B)’s 1 day’s work = 1/10 + 1/14 = 12/70

So work done in 4 days = 4 * 12/70 = 24/35

Now A left.

Work left = 1 – 24/35 = 11/35

B does 1 work in 14 days so 11/35 work in 14 * 11/35 = 22/5 days**P and Q starts from point A to B 21 km away at the same time. P travels at 3 km/hr and Q at 4 km/hr. Q reaches point B and then returns towards point A and he meets P in between. Find the distance travelled by Q when he meets P.**

A) 20 km

B) 24 km

C) 22 km

D) 18 km

E) 26 km**B) 24 km**

Explanation:

Since both start at same time, when they meet they have travelled for same time

Let both meet at point C and let BC = x, then A has travelled (21-x) km, and B has travelled (21+x) km

Now times are same, so

(21-x)/3 = (21+x)/4

Solve, x = 3

So B has travelled = 21+3**A man buys a certain number of marbles at 2 for a Rupee and the same number of marbles at 4 for a Rupee. He sells all of them at 6 for Rs 2. What is his gain/loss percent in the whole transaction?**

A) 11% loss

B) 11 1/9% gain

C) 13 1/9% gain

D) 11 1/9% loss

E) None of these**D) 11 1/9% loss**

Explanation:

Buys certain number at 2/Re, let b1 = 2. Buys certain number at 4/Re, let b2 = 4

He sells 6 marbles for Rs 2, so this means he sells 3 marbles for Re 1, let s1 = 3

Next we have a direct formula for it:

Gain/loss % = [(2*b1*b2/s1(b1+b2s)) – 1] * 100

Gain/loss % = [(2*2*4/3(2+4)) – 1] * 100 = [(8/9) – 1] * 100 = -100/9 %**At a certain rate of interest Rs 1000 becomes Rs 1320 in 4 years. Now the rate of interest is increased by 4% per annum. What will Rs 1000 amount to now?**

A) Rs 1440

B) Rs 1480

C) Rs 1280

D) Rs 1400

E) None of these**B) Rs 1480**

Explanation:

When rate increase by 4% fir 4 years, Increase in interest = (1000*4*4)/100 = Rs 160

So increased amount = 1320+160

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