Arithmetic Applications (Level 1): MCQ test, Set 10

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

The ratio of 25 minutes to 1.5 hours is
A $1 : 6$ B $50 : 3$ C $18 : 5$ D $5 : 18$

Answer and solution

Answer: (D) $5 : 18$

$1.5$ hours $= 1.5 \times 60$ minutes $= 90$ minutes

$\displaystyle \text{Ratio} = \frac{25}{90} = \frac{5}{18} = 5 : 18$
₹8400 is divided among three persons in the ratio $5 : 6 : 9$ . The third share is
A ₹2520 B ₹420 C ₹3780 D ₹2100

Answer and solution

Answer: (C) ₹3780

Sum of the terms $= 5 + 6 + 9 = 20$

$\displaystyle \text{Third share} = \frac{9}{20} \times 8400 = 3780$
Three numbers are in the ratio $2 : 3 : 6$ and the sum of their squares is 441. The sum of the numbers is
A $18$ B $33$ C $99$ D $9$

Answer and solution

Answer: (B) $33$

Numbers: $2k, 3k, 6k$ ; $49k^2 = 441 \Rightarrow k^2 = 9 \Rightarrow k = 3$

$\text{Sum} = 11k = 33$
The ratio of the length to the width of a rectangular sheet is $8 : 3$ . If its width is 27 cm, its length is
A $32$ cm B $216$ cm C $72$ cm D $10.125$ cm

Answer and solution

Answer: (C) $72$ cm

$\displaystyle \frac{l}{27} = \frac{8}{3} \Rightarrow l = \frac{8 \times 27}{3} = 72$
The mean proportional between 28 and 63 is
A $35$ B $42$ C $45.5$ D $1764$

Answer and solution

Answer: (B) $42$

$\text{Mean proportional} = \sqrt{28 \times 63} = \sqrt{1764} = 42$
If $a : b = c : d$ , then $\displaystyle \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d}$ . This property is called
A Subtrahendo B Addendo C Dividendo D Componendo

Answer and solution

Answer: (B) Addendo

Adding the numerators and the denominators, $\displaystyle \frac{a+c}{b+d}$ , is addendo.
If $x \propto y$ and $x = 8$ when $y = 20$ , then the value of $x$ when $y = 35$ is
A $\displaystyle 4\frac{4}{7}$ B $87.5$ C $14$ D $23$

Answer and solution

Answer: (C) $14$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle y = \frac{35 \times 8}{20} = 14$
A town of 1500 people has food for 36 days. After 6 days, 300 more people arrive. The remaining food will last for
A $30$ days B $25$ days C $19$ days D $31$ days

Answer and solution

Answer: (B) $25$ days

Food left $= 1500 \times 30$ person-days

$\displaystyle y = \frac{1500 \times 30}{1800} = 25$ days
$0.4$ expressed as a percentage is
A $40\%$ B $0.4\%$ C $400\%$ D $4\%$

Answer and solution

Answer: (A) $40\%$

$0.4 = (0.4 \times 100)\% = 40\%$
If $25\%$ of 80 is greater than $40\%$ of a number by 4, then the number is
A $10$ B $16$ C $40$ D $60$

Answer and solution

Answer: (C) $40$

$\displaystyle \frac{25}{100} \times 80 = \frac{40}{100}x + 4 \Rightarrow 20 - 4 = \frac{40x}{100}$

$\displaystyle x = \frac{16 \times 100}{40} = 40$
Percentage of decrease is
A $\displaystyle \left(\frac{\text{decrease}}{\text{new value}} \times 100\right)\%$ B $\displaystyle \left(\frac{\text{new value}}{\text{original value}} \times 100\right)\%$ C $\displaystyle \left(\frac{\text{decrease}}{\text{original value}} \times 100\right)\%$ D $\displaystyle \left(\frac{\text{original value}}{\text{decrease}} \times 100\right)\%$

Answer and solution

Answer: (C) $\displaystyle \left(\frac{\text{decrease}}{\text{original value}} \times 100\right)\%$

The change is always compared with the original value: $\displaystyle \% \text{ of decrease} = \left(\frac{\text{decrease}}{\text{original value}} \times 100\right)\%$
If $A$ is $R\%$ more than $B$ , then $B$ is less than $A$ by
A $R\%$ B $\displaystyle \left[\frac{R}{100 + R} \times 100\right]\%$ C $\displaystyle \left[\frac{R}{100 - R} \times 100\right]\%$ D $\displaystyle \left[\frac{100 + R}{R} \times 100\right]\%$

Answer and solution

Answer: (B) $\displaystyle \left[\frac{R}{100 + R} \times 100\right]\%$

Take $B = x$ , $\displaystyle A = \frac{100 + R}{100}x$ ; the difference $\displaystyle \frac{R}{100}x$ is compared with $A$ .

Required $\displaystyle = \left[\frac{R}{100 + R} \times 100\right]\%$
The present population of a town is 34560. It has grown at $20\%$ per annum. Its population 3 years ago was
A $24{,}000$ B $28{,}800$ C $20{,}000$ D $21{,}600$

Answer and solution

Answer: (C) $20{,}000$

$\displaystyle \frac{34560}{\left(\frac{6}{5}\right)^3} = 34560 \times \frac{125}{216} = 20000$
A dealer sells 50 quintals of wheat at $8\%$ profit. If a quintal cost him ₹2000, his total profit is
A ₹1,08,000 B ₹800 C ₹160 D ₹8000

Answer and solution

Answer: (D) ₹8000

C.P $= 50 \times 2000 = 100000$

Profit $\displaystyle = \frac{8}{100} \times 100000 = 8000$
A lamp marked at ₹500 is sold at a $28\%$ discount and at a loss of $10\%$ . Its cost price is
A ₹400 B ₹360 C ₹396 D ₹324

Answer and solution

Answer: (A) ₹400

S.P $\displaystyle = \frac{72}{100} \times 500 = 360$

C.P $\displaystyle = \frac{100}{90} \times 360 = 400$
Two tables are sold at ₹1242 each, one at $8\%$ profit and the other at $8\%$ loss. On the whole there is a
A loss of $0.64\%$ B neither profit nor loss C gain of $0.64\%$ D loss of $0.8\%$

Answer and solution

Answer: (A) loss of $0.64\%$

Same S.P, $a\%$ profit and $a\%$ loss: always a loss of $\displaystyle \left(\frac{a}{10}\right)^2\%$ .

$\displaystyle \left(\frac{8}{10}\right)^2 = 0.64\%$
A shopkeeper buys oranges at 6 for ₹5 and sells them at 5 for ₹6. His profit percent is
A $\displaystyle 30\frac{5}{9}\%$ B $36\%$ C $20\%$ D $44\%$

Answer and solution

Answer: (D) $44\%$

$\displaystyle \text{C.P} \text{ of one} = \frac{5}{6}, \quad \text{S.P} \text{ of one} = \frac{6}{5}$

$\displaystyle \text{Profit \%} = \frac{\frac{6}{5} - \frac{5}{6}}{\frac{5}{6}} \times 100 = 44\%$
The average of seven consecutive natural numbers is 33. The smallest of them is
A $36$ B $33$ C $30$ D $27$

Answer and solution

Answer: (C) $30$

$\displaystyle \frac{7x + 21}{7} = 33 \Rightarrow x + 3 = 33 \Rightarrow x = 30$
The average of 20 numbers is 50. If 5 is subtracted from each number, the new average is
A $45$ B $50$ C $55$ D $49.75$

Answer and solution

Answer: (A) $45$

Average of $x_i - 5$ is $\bar{x} - 5 = 45$
$1$ km/h is equal to
A $\displaystyle \frac{1000}{60}$ m/s B $\displaystyle \frac{18}{5}$ m/s C $\displaystyle \frac{1}{36}$ m/s D $\displaystyle \frac{5}{18}$ m/s

Answer and solution

Answer: (D) $\displaystyle \frac{5}{18}$ m/s

$1$ km/h $\displaystyle = \frac{1000 \text{ m}}{60 \times 60 \text{ s}} = \frac{5}{18}$ m/s
A cyclist covers three equal distances at 10 km/h, 15 km/h and 30 km/h. His average speed is
A $\displaystyle 18\frac{1}{3}$ km/h B $18$ km/h C $12$ km/h D $15$ km/h

Answer and solution

Answer: (D) $15$ km/h

$\displaystyle \frac{3 \times 10 \times 15 \times 30}{150 + 450 + 300} = \frac{13{,}500}{900} = 15$
The speeds of A and B are in the ratio $2 : 3$ . If B takes 4 hours for a journey, A takes
A $6$ hours B $\displaystyle 2\frac{2}{3}$ hours C $5$ hours D $7$ hours

Answer and solution

Answer: (A) $6$ hours

$s_A : s_B = T_B : T_A \Rightarrow 2 : 3 = 4 : T_A$

$\displaystyle T_A = \frac{3 \times 4}{2} = 6$ hours
A train running at 108 km/h crosses a signal post in 8 seconds. The length of the train is
A $240$ m B $300$ m C $225$ m D $864$ m

Answer and solution

Answer: (A) $240$ m

Speed $\displaystyle = 108 \times \frac{5}{18} = 30$ m/s

Length $= 30 \times 8 = 240$ m
The difference between the compound interest (year by year) and the simple interest on ₹5000 for 2 years at $10\%$ per annum is
A ₹100 B ₹0 C ₹1050 D ₹50

Answer and solution

Answer: (D) ₹50

S.I $= 500 + 500 = 1000$

C.I $= 500 + 550 = 1050$ (2nd year on ₹5500)

Difference $= 50$
The population of a village rose from 6400 to 8100 in 2 years. The rate of growth per annum (compound) is
A $25\%$ B $12.5\%$ C $\displaystyle 13\frac{9}{32}\%$ D $15\%$

Answer and solution

Answer: (B) $12.5\%$

$\displaystyle \left(1 + \frac{R}{100}\right)^2 = \frac{8100}{6400} = \left(\frac{90}{80}\right)^2$

$\displaystyle 1 + \frac{R}{100} = \frac{9}{8} \Rightarrow R = 12.5\%$

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MCQ test, Set 10