A candidate with **quantitative aptitude** knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an **important** measure for a prospective business executive’s **abilities**.

The article **IBPS Clerk Quantitative Aptitude Quiz 1** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. The article **IBPS Clerk Quantitative Aptitude Quiz 1** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. 15

B. 18

C. 16

D. 20

**Answer**: Option C

**Explanation**:

According to the question let n be the number of articles.

=> Cost price of n articles = n

=> Selling price of n articles = 20

=> Profit % = 25%

=> Profit % = \(\frac{(SP – CP)}{CP }\) X 100

Putting the values in the formula

\(\frac{(20-n)}{n }\) x 100 = 25

=> \(\frac{5n}{4 }\) = 20

=> n = 16

**2. Akshay buys a motorcycle for Rs.25000. If he decides to sell the motorcycle for a profit of 7%, find the selling price. **

- A. 27000

B. 26750

C. 26000

D. 25750

**Answer**: Option B

**Explanation**:

Profit = \(\frac{%profit}{100 }\) * cost price

P =\(\frac{7}{100 }\) * 25000 = 1750.

Selling price = cost price + profit

= 25000 + 1750

= Rs.26750

**3. If the money was put in for the same duration of time by A,B and C, three business partners and four times Aâ€™s capital is equal to 6 times Bâ€™s capital is equal to 10 times Câ€™s capital. Determine that out of a total profit of Rs 4650 what is Câ€™s share? **

- A. Rs 2250

B. Rs 1550

C. Rs 450

D. Rs 900

**Answer**: Option D

**Explanation**:

It is given that business partnership is independent of time in this case as all the investors have invested their money for same amount of time.

Let Aâ€™s capital be x, Bâ€™s capital is y and Câ€™s capital be z.

=>x+y+z = 4650

=> 4x = 6y = 10z

=>Câ€™s share is z so let us solve the question w.r.t. z

=>x = \(\frac{5z}{2 }\)

=>y = \(\frac{5z}{3 }\)

=>(\(\frac{5z}{2 }\))+(\(\frac{5z}{3 }\))+(z) = 4650

=>31z = 4650*6

=>z = 900

**4. Shweta buys a product at a 25% discount rate. At what % over the cost price should she sell the product to make an overall 25% profit over the listed price? **

- A. 22

B. 33.333

C. 66.667

D. 40

**Answer**: Option C

**Explanation**:

Hence at 25% discount rate the Cost Price will become: 3x/4

Now to make an overall profit of 25% over the listed price:

SP = CP+(25/100)CP

=> SP = X + \(\frac{X}{4 }\)

=>SP = 5 \(\frac{X}{4 }\)

=>Selling Price over Cost price: (5 \(\frac{X}{4 }\))-(3 \(\frac{X}{4 }\))

=>x/2

=>%Selling price over Cost price: ( \(\frac{X}{2 }\))/(3 \(\frac{X}{4 }\))*100

=>66.67%

**5. A man sells 45 lemons for Rs 40 and loses 20%. At how much price should he sell 24 lemons to the next customer to make a 20% profit?**

- A. 32

B. 20

C. 24

D. 16

**Answer**: Option A

**Explanation**:

Let cost price of lemons be x.

The selling price of lemons becomes CP – loss.

=>SP = x- (\(\frac{20}{100 }\))x

=>40 = x(\(\frac{80}{100 }\))

=>50

So 45 lemons cost Rs 50

Cost of 1 lemon = \(\frac{50}{45 }\) Rs

Cost of 24 lemons = (\(\frac{50}{45 }\))*24 = \(\frac{80}{3 }\) Rs

Selling Price of 24 lemons =(\(\frac{80}{3 }\))+(\(\frac{20}{100 }\))(\(\frac{80}{3 }\))=Rs 32

- A. 99.6

B. 96.2

C. 97.5

D. 98.25

**Answer**: Option B

**Explanation**:

Use the formula,

CP = 100 â€“ discount + brokerage%

CP = 100 – 4 + \(\frac{1}{5 }\)

96.2

Thus the CP is Rs 96.2.

**2. An employer pays Rs. 30 for each day a worker works and forfeits Rs. 5 for each day he is idle. At the end of 60 days, a worker gets Rs. 500. For how many days did the worker remain idle? **

- A. 35

B. 48

C. 52

D. 58

**Answer**: Option C

**Explanation**:

Suppose the worker remained idle for m days. Then, he worked for (60 – m) days.

30 (60 – m) â€“ 5m = 500

1800 â€“ 25m = 500

25m = 1300

m = 52

So, the worker remained idle for 52 days.

**3. P and Q together can complete a piece of work in 4 days. If P alone can complete the same work in 20 days, in how many days can Q alone complete that work? **

- A. 8

B. 7

C. 4

D. 5

**Answer**: Option D

**Explanation**:

(P + Q)’s 1 day’s work = \(\frac{1}{4 }\), P’s 1 day’s work = \(\frac{1}{20 }\)

Q’s 1 day’s work = (\(\frac{1}{4 }\) â€“ \(\frac{1}{20 }\)) = (\(\frac{4}{20 }\)) = (\(\frac{1}{5 }\))

Hence, Q alone can complete the work in 5 days.

**4. If 10 bulls can plough 20 identical fields in 3 days working 10 hours a day, then in how many days can 30 bulls plough 32 same identical fields working 8 hours a day? **

- A. 2

B. 4

C. 8

D. 10

**Answer**: Option B

**Explanation**:

M1*D1*W2 = M2*D2*W1

10*3*10*32 = 30*d*8*20

d = 2 days

**5. A pack of people can complete a certain amount of work in 12 days. Two times the same number of persons will complete half of the work in? **

- A. 12 days

B. 3 days

C. 4 days

D. 6 days

**Answer**: Option D

**Explanation**:

More no of people: Less days (Inverse Relationship)

More work: More days (Direct Relationship)

The ratio is given:

Persons: 1:: 2

Work 1:: \(\frac{1}{2 }\)

Time 12:: x

The solution of this can be explained by solving this ratio by the sort of relationships they possess with time.

1*\(\frac{1}2 }\)*12 = 2*1*x

=> x = 3 days

- A. 42

B. 46

C. 48

D. 38

** Answer**: Option A

**Explanation**:

1 man is equivalent to 2 boys in work capacity. 12 men + 18 boys = 12 x 2 + 18 = 42 boys.

Let the required number of boys be x. So a Total number of people doing work = 21 men x 2 (according to work capacity) + x boys.

Taking respective ratios as required:

Given:

=> Hours 50:60

=> Minutes per hour 9: \(\frac{15}2 }\)

=> Work 1:2

=> People: 42: 42 + x

=> 50 x 9 x 1 x (42+x) = 60 x \(\frac{15}{2 }\) x 2 x 42

=> 42 + x = 84

x = 42

**2. 5/8th of a job is completed in 10 days. If a person works at the same pace, how many days will he take to complete the job? **

- A. 4

B. 5

C. 6

D. 7

** Answer**: Option C

**Explanation**:

Solution: It is given that 5/8th of the work is completed in 10 days.

=> Remaining work = \(\frac{3}{8 }\)th of total

Applying the unitary method:

Total work will be completed in 10 * \(\frac{8}{5 }\) days

=> It takes 16 days to complete total work

=> Hence, remaining work days = 16 – 10 = 6 days

**3. If the ratio of present ages of Jeet and Jay is 5:7 and after 6 years the ratio will be 3:4, what is the present age of Jay? **

- A. 42

B. 30

C. 36

D. None of these

** Answer**: Option A

**Explanation**:

As the present age of Jeet and Jay are in the ratio 5:7, let their ages be 5x and 7x respectively.

Therefore, their ages after 6 years will be (5x+6) and (7x+6) respectively.

Now, it is given that \(\frac{(5x+6)}{(7x+6) }\) = \(\frac{3}{4 }\)

4*(5x+6) = 3*(7x+6)

x = 6

Hence, the present age of Jay is 7x = 7*6 = 42 years

**4. Find the value of x when y = 5, if x varies directly as 4y-1 and x = 14 when y = 2. **

- A. 38

B. 35

C. 10

D. 28

** Answer**: Option A

**Explanation**:

Let z = 4y-1

When x = 14, y = 2, z = (4*2) â€“ 1 = 7

Now, x varies directly as z = 4y-1

When y = 5, z = (4*5) â€“ 1 = 19

x 14 7

y – 19

Therefore, x = \(\frac{(14*19)}{7 }\)= 38

**5. In what ratio must one add water to milk so as to gain 16.666% on selling this mixture at the cost price? **

- 1:6

B. 1:3

C. 6:1

D. 3:1

** Answer**: Option A

**Explanation**:

To start off this question let us assume that cost price of 1 litre milk is Rs 1

No need to make a mixture and sell this mixture at 1 Rs per liter such that the total gain on the mixture is 16.667%.

Therefore, CP of 1 liter of the mixture becomes (quantity of milk)/ (quantity of mixture containing 1 L milk)*(the price of 1-liter milk).

=>(100/(100+50/3))*1

=>CP of 1-litre milk of mixture: Rs 5/6

As the price of any amount of water is zero, and as 1-liter milk costs Rs 1.5/6litre of the mixture will comprise entirely of cost of milk which means,1 liter of the mixture will

contain a 5/6th amount of milk.

=>Water is added in the ratio of (1-5/6)=1/6