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.

**An article is available for Rs 12,000 cash payment or Rs 7,000 as cash payment along with installments of Rs 630 for 8 months. Find the rate percent per annum.**

A. 1.4%

B. 2.4%

C. 1.2%

D. 2.6%

**C. 1.2%**

Explanation:

With installments, payment after 7,000 is 630*8 = 5,040

This implies that instead of 5,000, the person gave 5040, this gives interest of 5040-5000 = 40

So by simple interest formula,

40 = 5000*r*(8/12) /100

Solve, r = 1.2%**If a shopkeeper offers a discount of 20% on the list price of an article, then he makes a profit of 12%. What percent profit or loss will he make if he sells it at a discount of 25% on the list price?**

A. 0.6% loss

B. 5% profit

C. 4.25% loss

D. 5% loss

**B. 5% profit**

Explanation:

We have a direct formula for this:

Profit/Loss % = { (100+12) [(100-25)/(100-20)] } – 100

= +5%

So profit 5%

**A dishonest seller uses a weight of 800 gm in place of 1 kg of sugar which contains 20% impurities. What would be his profit% if he claims to sale at cost price?**

A. 50%

B. 45%

C. 45.4%

D. 56.25%

**D. 56.25%**

Explanation:

Since the shopkeeper uses 20% impurity, hence in the 800 gm which he really sells in place of 1000 gm, the original amount of pure material is

80% of 800 gm = 80/100 * 800 =640 gm

Hence gain% = (1000-640)/640 * 100= 56.25%**In a rectangle if length increases by 20% and breadth increases by 20%, then what is the percentage change in the perimeter of the rectangle?**

A. 20%

B. 45%

C. -5%

D. Cannot be determined

**A. 20%**

Explanation:

Perimeter = 2(l+b)

New peri = 2(1.2l + 1.2b)

Change in perimeter = 2 [(1.2l – l) + (1.2b – b)]

= 2[0.2l + 0.2b] = 0.2 * 2(l+b)

Percentage change = 0.2 * 100 = 20%

* When you are given same change in length and breadth, then there is that same change in perimeter also, but when you are given different changes in length and breadth, then without knowing the actual values you cannot find the % change in perimeter.**In a circular race of 700 m, P and Q start from same point, same time and in opposite directions with speeds of 18 km/hr and 27 km/hr. Find the time after which they will meet agin for the first time on the track.**

A. 1 min 56 sec

B. 56 sec

C. 1 min 10 sec

D. 50 sec

**B. 56 sec**

Explanation:

18 km/hr = 18 * (5/18) = 5 m/s

27 km/hr = 27 * (5/18) = 15/2 m/s

In opposite direction, relative speed is 5 + 15/2 = 25/2 m/s

Time = 700/ (25/2) = 56 sec**The area of a square is 450 cm**^{2}. Find the length of diagonal.

A. 30 cm

B. 40 cm

C. 25 cm

D. 35 cm

**A. 30 cm**

Explanation:

1/2 d^{2}= 450

d^{2}= 900, d = 30

**There are 3 red balls, 4 blue balls, 6 white balls. Three balls are drawn at random. Find the probability that there is at least 1 red ball.**

A. 63/143

B. 82/143

C. 83/143

D. 53/143

**B. 83/143**

Required prob = 1 – prob. none is red

Prob. that none is red =^{10}C_{3}/^{13}C_{3}= 10*9*8 / 13*12*11 = 60/143.

So prob that at least 1 red = 1 – 60/143 = 83/143

**In an election there are three candidates. Out of total 1200 cast votes, Ram received 30%, Balu received 720 votes and Kapil received the rest of the votes. Find out percent of votes which the winner got in comparison to his closest rival?**

A. 100%

B. 200%

C. 180%

D. 90%

**B. 200%**

Explanation:

Total votes = 1200

Ram received = 30% of 1200 = 360

Balu received = 720

Kapil = 1200 – (360+720) = 120

Winner balu = 720, 2nd ram = 360

Required % = 720/360 * 100 = 200%

**How many kg of sugar costing Rs 11 per kg must be mixed with 21 kg of sugar costing Rs 9 per kg, so that there is a gain of 10% by selling the mixture at Rs 10.45 per kg**

A. 6 kg

B. 7 kg

C. 9 kg

D. 10 kg

**B. 7 kg**

Explanation:

Gain of 10%, SP = 10.45

So, CP = 100/110 * 10.45 = Rs 9.50 per kg

By the method of allegation:

11 9

. 9.5

0.5 1.5

0.5 : 1.5 = 1 : 3

For 3 parts, 1 part is added, so for 21 kg, 21/3 = 7 kg of first type of sugar will be added.**A, B and C together can do a work in 15 days. After working with B and C for 5 days, A leaves and then B and C take 20 days more to finish that work. In how many days can A do the work alone?**

A. 20 days

B. 25 days

C. 30 days

D. 35 days

**C. 30 days**

Explanation:

Given 1/x + 1/B + 1/C = 1/15.

So, 1/B + 1/C = 1/15 – 1/x

Let A alone completes the work in x days.

A worked for 5 days. B and C worked for (5+20) = 25 days. And now the work is completed. So

1/x * 5 + (1/B + 1/C) * 25 = 1

1/x * 5 + (1/15 – 1/x) * 25 = 1

5/x – 25/x = 1 – 25/15

x = 30

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