Arithmetic Applications (Level 1): MCQ test, Set 8

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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.

Find the ratio of the price of cashew nuts to that of almonds, when cashew nuts costs ₹96 per 100 g and almonds costs ₹1200 per kg.
A $8 : 1$ B $5 : 4$ C $2 : 25$ D $4 : 5$

Answer and solution

Answer: (D) $4 : 5$

100 g of cashew nuts costs ₹96, so 1000 g (1 kg) costs ₹960.

$\displaystyle \text{Required ratio} = \frac{960}{1200} = \frac{4}{5} = 4 : 5$
₹5400 is divided among three persons in the ratio $4 : 5 : 9$ . The second share is
A ₹2700 B ₹1200 C ₹300 D ₹1500

Answer and solution

Answer: (D) ₹1500

Sum of the terms $= 4 + 5 + 9 = 18$

$\displaystyle \text{Second share} = \frac{5}{18} \times 5400 = 1500$
The ratio of boys to girls in a school with 640 students is $9 : 7$ . If 30 more boys are admitted, the new ratio of boys to girls is
A $9 : 7$ B $36 : 31$ C $39 : 28$ D $10 : 7$

Answer and solution

Answer: (C) $39 : 28$

$\displaystyle \text{Boys} = \frac{9}{16} \times 640 = 360, \quad \text{Girls} = 640 - 360 = 280$

$\displaystyle \text{New ratio} = \frac{390}{280} = 39 : 28$
The fourth proportional to 9, 6 and 15 is
A $3.6$ B $12$ C $22.5$ D $10$

Answer and solution

Answer: (D) $10$

$9 : 6 :: 15 : x \Rightarrow 9x = 6 \times 15$

$\displaystyle x = \frac{90}{9} = 10$
If 49, $x$ , $x$ , 81 are in proportion, then $x =$
A $63$ B $32$ C $3969$ D $65$

Answer and solution

Answer: (A) $63$

$\displaystyle \frac{49}{x} = \frac{x}{81} \Rightarrow x^2 = 3969 \Rightarrow x = 63$
If $x : y = 11 : 4$ , then $\displaystyle \frac{x-y}{x+y} =$
A $\displaystyle \frac{15}{7}$ B $\displaystyle \frac{7}{15}$ C $\displaystyle \frac{11}{4}$ D $\displaystyle \frac{7}{4}$

Answer and solution

Answer: (B) $\displaystyle \frac{7}{15}$

By componendo and dividendo:

$\displaystyle \frac{x-y}{x+y} = \frac{11 - 4}{11 + 4} = \frac{7}{15}$
A car covers 156 km on 12 litres of petrol. On 20 litres it will cover
A $93.6$ km B $260$ km C $164$ km D $13$ km

Answer and solution

Answer: (B) $260$ km

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle y = \frac{20 \times 156}{12} = 260$
If $x$ is inversely proportional to $y$ , then
A $\displaystyle \frac{x}{y} = k$ (a constant) B $x = ky$ C $xy = k$ (a constant) D $x + y = k$ (a constant)

Answer and solution

Answer: (C) $xy = k$ (a constant)

$\displaystyle x \propto \frac{1}{y} \Rightarrow x = \frac{k}{y} \Rightarrow xy = k$
$\displaystyle 2\frac{1}{2}\%$ as a fraction is
A $\displaystyle \frac{5}{2}$ B $\displaystyle \frac{1}{40}$ C $\displaystyle \frac{1}{400}$ D $\displaystyle \frac{1}{4}$

Answer and solution

Answer: (B) $\displaystyle \frac{1}{40}$

$\displaystyle 2\frac{1}{2}\% = \frac{5}{2}\% = \frac{5}{2 \times 100} = \frac{1}{40}$
What percent of 2 hours is 18 minutes?
A $9\%$ B $30\%$ C $900\%$ D $15\%$

Answer and solution

Answer: (D) $15\%$

$2$ hours $= 120$ minutes

$\displaystyle \left(\frac{18}{120} \times 100\right)\% = 15\%$
A number increased by $20\%$ becomes 168. The number is
A $148$ B $134.4$ C $140$ D $201.6$

Answer and solution

Answer: (C) $140$

$\displaystyle \frac{120}{100} \times x = 168$

$\displaystyle x = \frac{168 \times 100}{120} = 140$
The price of rice increases by $25\%$ . The reduction in consumption so as not to increase the expenditure is
A $\displaystyle 33\frac{1}{3}\%$ B $25\%$ C $80\%$ D $20\%$

Answer and solution

Answer: (D) $20\%$

$\displaystyle \left(\frac{25}{125} \times 100\right)\% = 20\%$
A population of 12500 grows at $20\%$ per annum. After 3 years it will be
A $21{,}600$ B $20{,}000$ C $18{,}000$ D $25{,}920$

Answer and solution

Answer: (A) $21{,}600$

$\displaystyle 12500\left(\frac{6}{5}\right)^3 = 12500 \times \frac{216}{125} = 21600$
A shopkeeper buys 50 eggs at ₹5 each. 2 eggs break. He sells the rest at ₹66 a dozen. His profit is
A ₹25 B ₹264 C ₹14 D ₹250

Answer and solution

Answer: (C) ₹14

C.P $= 50 \times 5 = 250$ ; 1 egg sells for $\displaystyle \frac{66}{12} = 5.5$

S.P of 48 eggs $= 48 \times 5.5 = 264$ ; Profit $= 264 - 250 = 14$
A shopkeeper marks his goods $25\%$ above the cost price and allows a discount of $10\%$ . His profit percent is
A $35\%$ B $12.5\%$ C $10\%$ D $15\%$

Answer and solution

Answer: (B) $12.5\%$

Let C.P $= 100$ : M.P $= 125$ , S.P $\displaystyle = \frac{90}{100} \times 125 = 112.5$

Profit $= 12.5\%$
Two articles have the same selling price; one is sold at $a\%$ profit and the other at $a\%$ loss. Taking a gain as positive, the overall gain percent is
A $\displaystyle -\left(\frac{a}{10}\right)^2\%$ B $-a\%$ C $\displaystyle \left(\frac{a}{10}\right)^2\%$ D $0\%$

Answer and solution

Answer: (A) $\displaystyle -\left(\frac{a}{10}\right)^2\%$

There is always a loss, of $\displaystyle \left(\frac{a}{10}\right)^2\% = \frac{a^2}{100}\%$ , so the gain is $\displaystyle -\left(\frac{a}{10}\right)^2\%$ .
If the cost price of 9 articles equals the selling price of 10 articles, then there is a
A $1\%$ loss B $\displaystyle 11\frac{1}{9}\%$ loss C $10\%$ loss D $10\%$ gain

Answer and solution

Answer: (C) $10\%$ loss

$\displaystyle 9x = 10y \Rightarrow \frac{y}{x} = \frac{9}{10}$

$\displaystyle \left(\frac{y}{x} - 1\right) \times 100 = -10\%$ (negative, so a loss)
The average of 11 observations is 20. The average of the first 6 is 18 and that of the last 6 is 23. The 6th observation is
A $23$ B $18$ C $26$ D $20$

Answer and solution

Answer: (C) $26$

First 6 $= 108$ , last 6 $= 138$ , all 11 $= 220$

$x_6$ is counted twice: $108 + 138 - 220 = 26$
The average age of a family of 6 members is 26 years. The youngest member is 6 years old. The average age of the family at the birth of the youngest member was (in years)
A $24$ B $25$ C $30$ D $20$

Answer and solution

Answer: (A) $24$

Other 5 members: $6(26) - 6 = 150$

6 years ago: $\displaystyle \frac{150 - 5(6)}{5} = \frac{120}{5} = 24$
A man sees the flash of a gun fired 1.36 km away and hears the sound 4 seconds later. The speed of sound is
A $0.34$ m/s B $340$ m/s C $34$ m/s D $3400$ m/s

Answer and solution

Answer: (B) $340$ m/s

$\displaystyle S = \frac{1.36 \times 1000}{4} = \frac{1360}{4} = 340$ m/s
A man goes to a place at 15 km/h and returns by the same route at 10 km/h. His average speed for the whole journey is
A $12.5$ km/h B $6$ km/h C $12$ km/h D $25$ km/h

Answer and solution

Answer: (C) $12$ km/h

$\displaystyle \frac{2 \times 15 \times 10}{15 + 10} = \frac{300}{25} = 12 \text{ km/h}$

(Not the simple average of the two speeds.)
A truck takes 9 hours for a journey at 40 km/h. A car at 60 km/h will take
A $5$ hours B $6$ hours C $7$ hours D $13.5$ hours

Answer and solution

Answer: (B) $6$ hours

$\displaystyle T_2 = \frac{40 \times 9}{60} = 6$ hours
A man crosses a 720 m long bridge in 4 minutes. His speed is
A $10.8$ km/h B $3$ km/h C $18$ km/h D $180$ km/h

Answer and solution

Answer: (A) $10.8$ km/h

$\displaystyle S = \frac{720}{4 \times 60} = 3$ m/s

$\displaystyle = 3 \times \frac{18}{5} = 10.8$ km/h
The amount is ₹1380 and the simple interest is ₹180 at $5\%$ per annum. The time (in years) is
A $3$ B $2.6$ C $4$ D $2$

Answer and solution

Answer: (A) $3$

$P = 1380 - 180 = 1200$

$\displaystyle T = \frac{180 \times 100}{1200 \times 5} = 3$ years
When interest is compounded half-yearly, the amount on $P$ at $R\%$ per annum for $n$ years is
A $\displaystyle P\left(1 + \frac{R}{200}\right)^{n}$ B $\displaystyle P\left(1 + \frac{R}{100}\right)^{2n}$ C $\displaystyle P\left(1 + \frac{R}{100}\right)^{n/2}$ D $\displaystyle P\left(1 + \frac{R}{200}\right)^{2n}$

Answer and solution

Answer: (D) $\displaystyle P\left(1 + \frac{R}{200}\right)^{2n}$

$\displaystyle A = P\left(1 + \frac{R}{100k}\right)^{nk}$ with $k = 2$ gives $\displaystyle P\left(1 + \frac{R}{200}\right)^{2n}$

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MCQ test, Set 8