Tag: maths

Questions Related to maths

What is the sum of two numbers?
I. The bigger of these two numbers is 6 more than the smaller number.
II. 40% of the smaller number is equal to 30% of the bigger number.
III. The ratio between half of the bigger number and one-third of the smaller number is 2 : 1

introduction to ratio and percentages comparing quantities maths

  1. Only II and III are sufficient

  2. Only I and II are sufficient

  3. I and either II or III is sufficient

  4. All, II and III together are sufficient

Show answer
Correct Option: C
Explanation:

From the given statements we can make the following equations.
(I) $\Rightarrow y=x+6$
(II) $\Rightarrow 0.4x=0.2y\Rightarrow \frac {x}{y}=\frac {x}{y}=\frac {3}{4}$
(III) $\Rightarrow \frac {y/2}{x/3}=\frac {2}{1}\Rightarrow \frac {y}{x}=\frac {4}{3}\Rightarrow \frac {x}{y}=\frac {3}{4}$
Obviously, question can be solved by using (I) and either (II) or (III) because equations (II) and (III) are same.

The difference between simple interest and compound interest on a certain sum of money for 3 years at 5% per annum is Rs. 122. Find the sum

introduction to ratio and percentages comparing quantities maths

  1. Rs. 12,200

  2. Rs. 15,000

  3. Rs. 16,500

  4. Rs. 16,000

Show answer
Correct Option: D
Explanation:

Suppose $sum=P$
Given $CI-SI=Rs. 122$
$P\left (1+\frac {5}{100}\right )^3-P-\left (\frac {P\times 5\times 3}{100}\right )=122$
$P\left (\frac {105^3}{100^3}-1-\frac {15}{100}\right )=122$
$P\left (\frac {7,625}{100^3}\right )=122$
$P=\frac {122\times 100^3}{7,625}=Rs. 16,000$

Two cats Billy and Kitty together catch 60 mice. If Billy catches three mice for every two caught by Kitty, the number of mice caught by Kitty is

introduction to ratio and percentages comparing quantities maths

  1. 24

  2. 30

  3. 36

  4. 40

Show answer
Correct Option: A
Explanation:

Ratio of catches by Billy and Kitty is 3 : 2. Therefore,
$3x+2x=60$
$5x=60$ or $x=\frac {60}{5}=12$
Number of mice caught by Kitty $=2x=2\times 12=24$

A person spent Rs 564 in buying geese and ducks. If each goose costs Rs 7, each duck Rs 3, and if the total number of birds bought was 108, how many of each did he buy?

introduction to ratio and percentages comparing quantities maths

  1. 60 and 48

  2. 48 and 36

  3. 48 and 24

  4. 60 and 30

Show answer
Correct Option: A
Explanation:

In questions of this kind, it is essential to have all quantities expressed in the same denomination; in the present instance, it will be convenient to express the money in rupees.
Let x be the number of geese. Then 108 -x is the number of ducks.
Since each goose costs 7 rupees, x geese cost 7x rupees.
And since each duck costs 3 rupees, 108 -x ducks cost 3(108 -x) rupees.
Therefore, the amount spent is 7x + 3(108 -x) rupees; but the question states that the amount is Rs 564. Hence,
$7x + 3(108 -x) = 564$
or $7x + 324 -3x = 564$
or $4x = 240$
Therefore, the number of geese, $x = 60$, and the number of ducks, $108 -x = 48$.

A person invested Rs 1,600 for 3 years and Rs 1,100 for 4 years at the same rate of simple interest. The total interest from these investments was Rs 506. Find the rate percent per annum

introduction to ratio and percentages comparing quantities maths

  1. 5%

  2. $5\frac {1}{2}$%

  3. 6%

  4. $6\frac {1}{2}$%

Show answer
Correct Option: B
Explanation:

$P _1=1,600, t _1=3$
$P _2=1,1000, t _2=4$
$SI _1+SI _2=506$
or $\frac {1,600\times 3\times r}{100}+\frac {1,100\times 4\times r}{100}=506$
or $r(4,800+4,400)=506\times 100$
or $r=\frac {506\times 100}{92,000}=5.5$%

A man saves 20% of his monthly salary. If on account of increase in prices, he is to increase his monthly expenses by 20%, he is only able to save Rs 800 per month. His monthly salary is

introduction to ratio and percentages comparing quantities maths

  1. Rs 40,000

  2. Rs 28,000

  3. Rs 24,000

  4. Rs 20,000

Show answer
Correct Option: D
Explanation:

Let the salary be Rs 100
Savings $=$ Rs 20 ; Expenditure $=$ Rs 80
New expenditure $=$ 120% of Rs 80 $=$ Rs 96.
New savings $=Rs:100-Rs:96=Rs:4$
$\implies4\%x=Rs:800\implies x=Rs:20,000$

Alfred buys an old scooter for Rs. $4700$ and spends Rs. $800$ on its repairs. If he sells the scooter for Rs. $5800$, his gain percent is:

introduction to ratio and percentages comparing quantities maths

  1. $4\dfrac {4}{7}$%

  2. $5\dfrac {5}{11}$%

  3. $10$%

  4. $12$%

Show answer
Correct Option: B
Explanation:

Cost Price $(C.P.) = Rs. (4700 + 800) = Rs. 5500$.
Selling Price $(S.P.) = Rs. 5800$.
$Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300$.
Gain % $= \left (\dfrac {300}{5500}\times 100\right )$% $= 5\dfrac {5}{11}$%

If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?

introduction to ratio and percentages comparing quantities maths

  1. 48 paise

  2. 54 paise

  3. 56 paise

  4. 72 paise

Show answer
Correct Option: A
Explanation:

Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
$\therefore 250 : 200 :: 60 : x \Leftrightarrow 250 \times x = (200 \times 60)$
$\Rightarrow x = \dfrac{(200 \times 60)}{250}$
$\Rightarrow x = 48$

A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is

introduction to ratio and percentages comparing quantities maths

  1. 4

  2. 9

  3. 12

  4. 49

Show answer
Correct Option: B
Explanation:

Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
$\therefore 14 : 6 :: 21 : x \Leftrightarrow 14 \times x = 6 \times 21$
$\Rightarrow x = \dfrac{6 \times 21}{14}$
$\Rightarrow x = 9$

A flagstaff 17.5 m high casts a shadow of length 40.25 m. The height of the building, which casts a shadow of length 28.75 m under similar conditions will be

introduction to ratio and percentages comparing quantities maths

  1. 10 m

  2. 12.5 m

  3. 17.5

  4. 21.25 m

Show answer
Correct Option: B
Explanation:

Let the height of the building x metres.
Less lengthy shadow, Less in the height (Direct Proportion)
$\therefore 40.25 : 28.75 :: 17.5 : x \Leftrightarrow 40.25 \times x = 28.75 \times 17.5$
$x = \dfrac{28.75 \times 17.5}{40.25}$
$\Rightarrow x = 12.5$