10 WBCS Math Questions - 12th Solved Arithmetic Practice Set

WBCS arithmetic practice questions with answers and quick solutions

WBCS Arithmetic Practice Question Set 12 with Solutions contains WBCS arithmetic practice 10 question set, answers and quick conceptual solutions.

Solutions explain how to solve the questions easily and quickly.

It should be useful for prelims of other competitive exams also.

First take the test, verify correctness from answers and then only refer to the following detailed solutions which will show you how to solve such questions quickly in mind.

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12th WBCS Arithmetic practice question set: time to answer - 10 mins

Question 1

There is a 25% profit if an article is sold at Rs.150. At what percent should the selling price be increased so that there will be 30% profit?

1. 10%
2. 2.5%
3. 4%
4. 5%

Question 2

The ratio of principal and the final amount (principal plus interest) in 1 year is 8 : 9. Then the rate of simple interest per annum is,

1. $11\frac{1}{2}$%
2. $10\frac{1}{2}$%
3. $12\frac{1}{2}$%
4. $13\frac{1}{2}$%

Question 3

A sum of money doubles itself in 8 years at some rate of interest. In how may years would it treble itself?

1. 12 years
2. 16 years
3. 14 years
4. 15 years

Question 4

Three friends P, Q and R started a business with the capitals of Rs.15000, Rs.10000 and Rs.25000 respectively. But at the end of the year, they suffer a loss of Rs.2500. How much each have to pay for the loss?

1. Rs.500, Rs.750, and Rs.1000
2. Rs.500, Rs.750, and Rs.1150
3. Rs.1000, Rs.500, and Rs.700
4. Rs.750, Rs.500, and Rs.1250

Question 5

What least number must be subtracted from 732 to make the remainder a perfect square?

1. 6
2. 5
3. 4
4. 3

Question 6

When price of rice increases by $12\frac{1}{2}$%, a man can get 250 gm less rice for Rs.18. Find the present cost of rice per kg.

1. Rs.7
2. Rs.7.50
3. Rs.8
4. Rs.9

Question 7

The expenses of rice, fish and oil of a family are in the ratio 12 : 17 : 3. The prices of these items are increased by 20%, 30% and 50% respectively. By what percent the total expenses for these items of the family will be increased?

1. $27\frac{1}{7}$%
2. $27\frac{1}{8}$%
3. $29\frac{1}{8}$%
4. $28\frac{1}{8}$%

Question 8

A man saves 20% of his income. If his expenses be increased by 35%, by what percent his income is to be raised so that he can save 10% of his income?

1. 20%
2. 25%
3. 22%
4. 30%

Question 9

A mixture of milk and water contains $12\frac{1}{2}$% of water. How much water should be added to 200 gallons of such mixture so that the new mixture may contain $37\frac{1}{2}$% of water?

1. 100 gallons
2. 80 gallons
3. 70 gallons
4. 60 gallons

Question 10

A retailer getting a discount of 20% on marked price sells an article at the marked price. Percentage profit of the retailer is,

1. 30
2. 10
3. 20
4. 25

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Answers to the 12th WBCS math arithmetic practice question set

Q1. Answer: Option c: 4%.

Q2. Answer: Option c: $12\frac{1}{2}$%.

Q3. Answer: Option b: 16 years.

Q4. Answer: Option d: Rs.750, Rs.500, and Rs.1250.

Q5. Answer: Option d: 3.

Q6. Answer: Option d: Rs.9.

Q7. Answer: Option d: $28\frac{1}{8}$%.

Q8. Answer: Option a: 20%.

Q9. Answer: Option b: 80 gallons.

Q10. Answer: Option d: 25.

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Solution to WBCS Math Arithmetic practice question set 12: time to answer was 10 mins

Question 1

There is a 25% profit if an article is sold at Rs.150. At what percent should the selling price be increased so there will be 30% profit?

1. 10%
2. 2.5%
3. 4%
4. 5%

Solution 1: Solving in mind: Profit or loss percentage is on Cost price

As profit or loss percentage is invariably on the cost price, with 25% profit, if sale price is SP and cost price CP,

$SP=1.25CP=\text{Rs.}150$, profit of 25% is 0.25 times CP and sale price is cost price plus profit.

So $CP=\displaystyle\frac{150}{1.25}=\text{Rs.}120$.

For 30% profit, or 5% increase in profit, the sale price of Rs.150 is to be increased by 5% of cost price,

$0.05\times{120}=\text{Rs.}6$.

This means sale price is to be increased by Rs.6 for every Rs.150.

This is equivalent to a percentage increase of sale price by,

$\displaystyle\frac{6}{150}\times{100}=4$%.

Though percentage increase in profit is always on cost price, percentage increase in sale price naturally is on sale price.

Answer. Option c: 4%.

Concepts used: Percentage to decimal conversion by dividing with 100 -- Profit concept: profit is on cost price and additional to cost price -- Change analysis technique: instead of new total profit percentage, increase in profit percentage is used for calculating just the increase in sale price faster -- Percentage increase in sale price is actual increase divided by increased sale price multiplied by 100 -- Solving in mind.

Question 2

The ratio of principal and the final amount (principal plus interest) in 1 year is 8 : 9. Then the rate of simple interest per annum is,

1. $11\frac{1}{2}$%
2. $10\frac{1}{2}$%
3. $12\frac{1}{2}$%
4. $13\frac{1}{2}$%

Solution 2: Solving in mind: Ratio concept of reintroduction of cancelled out HCF and concept of percent increase in principal by the simple interest accrued

Reintroducing the cancelled out HCF as $x$ the actual values of principal at start and end of 1 year are $8x$ and $9x$ respectively.

The increase in principal is,

$9x-8x=x$.

Percentage increase is,

$\displaystyle\frac{x}{8x}\times{100}=12\frac{1}{2}$%.

This is the simple interest accrued in 1 year on the principal amount, and so is the rate of interest per annum by definition.

Answer: Option c: $12\frac{1}{2}$%.

Concepts used: Ratio concepts -- HCF reintroduction technique to get actual values of ratio terms -- Percentage concept -- Solving in mind.

Question 3

A sum of money doubles itself in 8 years at some rate of interest. In how may years would it treble itself?

1. 12 years
2. 16 years
3. 14 years
4. 15 years

Solution 3: Solving in mind: Increase proportional to number of years and unitary method

Assuming principal to be Rs.100 it becomes Rs.200 in 8 years, an increase of Rs.100.

This increase is by accumulation of annual simple interest of equal amount over 8 years.

Note: Simple interest is assumed as otherwise the problem cannot be answered with data given.

The principal to treble (3 times), it is to become Rs.300 which is an increase by another Rs.100 from Rs.200.

So by unitary method, this would happen in additional 8 years or a total of 16 years.

Answer: Option b: 16 years.

Concepts used: Simple interest concept -- Change analysis technique and unitary method -- Solving mind.

Question 4

Three friends P, Q and R started a business with the capitals of Rs.15000, Rs.10000 and Rs.25000 respectively. But at the end of the year, they suffer a loss of Rs.2500. How much each have to pay for the loss?

1. Rs.500, Rs.750, and Rs.1000
2. Rs.500, Rs.750, and Rs.1150
3. Rs.1000, Rs.500, and Rs.700
4. Rs.750, Rs.500, and Rs.1250

Solution 4: Solving in mind: Profit or loss for each investor proportional to amount invested by each

The guiding principle of multiple investors investing in a business is,

After a period of time, when profit or loss occurs, the profit or loss to be owned by each investor is proportional to the amount the investor invested in the beginning of the period.

In this case, the three investors P, Q and R suffered a loss of Rs.2500 in their business in a year. This loss is then payable by the three in the ratio of their amount of investment which is,

$15000 : 10000 : 25000=3 : 2 : 5$.

The three term ratio has total number of portions,

$3+2+5=10$.

Value of total 10 portions is 2500 loss. So each portion is worth the loss amount of,

$\displaystyle\frac{2500}{10}=\text{Rs.}250$.

That is why P, Q and R would have to pay respectively 3, 2 ad 5 portions of loss,

$3\times{250}=\text{Rs.}750$,

$2\times{250}=\text{Rs.}500$, and

$5\times{250}=\text{Rs.}1250$.

Answer: Option d: Rs.750, Rs.500, and Rs.1250.

Concepts: Multiple investors in a business -- Profit or loss apportionment to multiple investors -- Solving in mind.

Question 5

What least number must be subtracted from 732 to make the remainder a perfect square?

1. 6
2. 5
3. 4
4. 3

Solution 5: Solving in mind: Knowledge of squares of two digit numbers

From our knowledge on squares of two digit numbers, we find 729, square of 27, to be the perfect square nearest to and less than 732.

So if you subtract minimum 3, you will get a perfect square 727.

Answer: Option d: 3.

Concepts used: Knowledge of squares of two digit numbers -- Solving in mind.

Question 6

When price of rice increases by $12\frac{1}{2}$%, a man can get 250 gm less rice for Rs.18. Find the present cost of rice per kg.

1. Rs.7
2. Rs.7.50
3. Rs.8
4. Rs.9

Solution 6:

Assuming cost per kg before price rise to be $p$, with $12\frac{1}{2}$% or $\frac{1}{8}$th increase, the new price becomes $\frac{9}{8}p$.

This price rise then effectively reduces the available money of Rs.18 to $\frac{8}{9}\times{18}=16$ if purchase were made at same price $p$,

$\frac{9}{8}p(r-0.25)=18$, or, $p(r-0.25)=16$, where $r$ is the amount of rice purchased using Rs.18 before price rise.

This decrease of Rs.2 in available money caused the reduction in purchased amount by 0.25 kg with price still at $p$. So,

$0.25p=2$,

Or, $p=\text{Rs.}8$.

And present cost per kg after price rise,

$p_{new}=8+\frac{1}{8}\times{8}=\text{Rs.}9$.

Change analysis technique enabled solving the problem quickly in mind.

Otherwise, by conventional method also it is easy to form two equations (with $r$ as amount purchased before price rise),

$18=p\times{r}$, and,

$18=\frac{9}{8}p\times(r-0.25)$,

Or, $16=p(r-0.25)$

Subtract second from first,

$0.25p=18-16=2$,

Or, $p=8$.

Or, $p_{new}=8+\frac{1}{8}\times{8}=\text{Rs.}9$

Answer: Option d: Rs.9.

Concepts used: Change analysis technique -- Percentage concepts -- Solving in mind.

Question 7

The expenses of rice, fish and oil of a family are in the ratio 12 : 17 : 3. The prices of these items are increased by 20%, 30% and 50% respectively. By what percent the total expenses for these items of the family will be increased?

1. $27\frac{1}{7}$%
2. $27\frac{1}{8}$%
3. $29\frac{1}{8}$%
4. $28\frac{1}{8}$%

Solution 7: Solving in mind: Change analysis technique and portions in a ratio concept

Reintroducing the cancelled out HCF for the ratio of expenses, the actual values of expenses of rice, fish and oil for the family before the price rise were respectively,

$12x$, $17x$ and $3x$.

Assuming the consumption amounts remaining same, the increased expenses will be proportional to the price increase for a commodity as,

$\text{expense}=\text{price}\times{\text{quantity}}$.

So the new expenses for rice, fish and oil will respectively would be increments of,

$2.4x$, $5.1x$ and $1.5x$.

The increase in new total expense will be,

$2.4x+5.1x+1.5x=9x$.

The old total expenses being $12x+17x+3x=32x$, the increase as a percentage of old total expenses is,

$\displaystyle\frac{9}{32} \times{100}$%

$=\displaystyle\frac{225}{8}$%

$=28\frac{1}{8}$%.

Answer: Option d: $28\frac{1}{8}$%.

Concepts: Change analysis technique -- Ratio concept -- HCF reintroduction technique -- Expense for an item consumed -- Solving in mind.

Question 8

A man saves 20% of his income. If his expenses be increased by 35%, by what percent his income is to be raised so that he can save 10% of his income?

1. 20%
2. 25%
3. 22%
4. 30%

Solution 8: Solving in mind: Starting value of main variable technique and savings concept

As all given and desired quantities are in percentages, you can assume any convenient value as the starting value of the main variable of interest.

In the problem, income is the main variable of interest. So,

Assume existing income as Rs.100.

This technique of assuming starting value as 100 makes solution of many percentage problems dead easy.

So saving is 20% of Rs.100 or Rs.20 and expenses Rs.80.

When expenses increases by 35% it becomes,

$80\times{1.35}=108$.

As 10% savings is needed, this Rs.108 expenses will be 90% of the new income so that,

$108=0.9\times{\text{new income}}$,

Or, $\text{new income}=\displaystyle\frac{108}{0.9}=\text{Rs.}120$.

Just note that, we have avoided calculating the desired income by using 10% savings and used instead 90% expense value. We could do this because,

Savings + Expenses = Income, and,

10% + 90% = 100%, all percentages being on income.

Finally, from Rs.100 to Rs.120 increase is a 20% increase.

Answer: Option a: 20%.

Concepts: Percentage concepts -- Starting value assumption technique -- Savings concept -- Solving in mind.

Question 9

A mixture of milk and water contains $12\frac{1}{2}$% of water. How much water should be added to 200 gallons of such mixture so that the new mixture may contain $37\frac{1}{2}$% of water?

1. 100 gallons
2. 80 gallons
3. 70 gallons
4. 60 gallons

Solution 9: Solving in mind: Mixture concept and percentage concept

$12\frac{1}{2}$% is $\frac{1}{8}$th portion and $37\frac{1}{2}$% is $\frac{3}{8}$th portion.

So initially the mixture contained $\frac{1}{8}$th water and $\frac{7}{8}$th milk.

In 200 gallons of mixture there was then $\frac{7}{8}\times{200}=175$ gallons of milk.

As milk amount remains unchanged in the new mixture when water is added, this 175 gallons of milk must make up for $\frac{5}{8}$th portion of the total mixture.

By unitary method then the total new mixture volume must be,

$175\times{\displaystyle\frac{8}{5}}=35\times{8}=280$ gallons.

80 gallons of water is to be added.

Answer: Option b: 80 gallons.

Concepts: Mixture concepts -- Percentage concepts -- Solving in mind.

Question 10

A retailer getting a discount of 20% on marked price sells an article at the marked price. Percentage profit of the retailer is,

1. 30
2. 10
3. 20
4. 25

Solution 10: Solving in mind: Profit and loss concept, Discount concept, Marked price concept, Percentage concept

Buying at a 20% discount on marked price makes cost price as,

$CP=0.8MP$

So, when the retailer sales his article at marked price his profit s,

$MP-0.8MP=0.2MP$.

And his profit percentage on cost price is,

$\displaystyle\frac{0.2MP}{0.8MP}=25%$.

Answer: Option d: 25.

Concept: Profit and loss concepts -- Percentage concepts -- Discount concept -- Marked price concept -- Solving in mind.

End note

All problems could be solved in mind in a few tens of seconds, but with use of appropriate concepts, identification of patterns and application of powerful techniques.

This is systematic problem solving approach applied to math problem solving.

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Question and Solution sets on WBCS Arithmetic

For all WBCS main arithmetic question sets click here.

For all WBCS main arithmetic solution sets click here.

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