Arithmetic Applications (Level 1): MCQ test, Set 4

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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 300 g to 1.5 kg is
A $5 : 1$ B $1 : 50$ C $1 : 5$ D $2 : 1$

Answer and solution

Answer: (C) $1 : 5$

$1.5$ kg $= 1.5 \times 1000$ g $= 1500$ g

$\displaystyle \text{Ratio} = \frac{300}{1500} = \frac{1}{5} = 1 : 5$
₹4800 is divided among three persons in the ratio $1 : 2 : 5$ . The third share is
A ₹1200 B ₹3000 C ₹960 D ₹600

Answer and solution

Answer: (B) ₹3000

Sum of the terms $= 1 + 2 + 5 = 8$

$\displaystyle \text{Third share} = \frac{5}{8} \times 4800 = 3000$
Metal P is 15 times as heavy as water and metal Q is 5 times as heavy as water. To get an alloy 9 times as heavy as water, P and Q must be mixed in the ratio
A $2 : 5$ B $3 : 5$ C $3 : 2$ D $2 : 3$

Answer and solution

Answer: (D) $2 : 3$

Take 1 g of P and $x$ g of Q: $15 + 5x = 9(1 + x)$

$\displaystyle \Rightarrow 6 = 4x \Rightarrow x = \frac{3}{2}$

$\displaystyle \text{Ratio} = 1 : \frac{3}{2} = 2 : 3$
The ratio of the length to the width of a rectangular sheet is $9 : 5$ . If its width is 35 cm, its length is
A $63$ cm B $49$ cm C $\displaystyle 19\frac{4}{9}$ cm D $315$ cm

Answer and solution

Answer: (A) $63$ cm

$\displaystyle \frac{l}{35} = \frac{9}{5} \Rightarrow l = \frac{9 \times 35}{5} = 63$
The mean proportional between 27 and 48 is
A $1296$ B $37.5$ C $36$ D $21$

Answer and solution

Answer: (C) $36$

$\text{Mean proportional} = \sqrt{27 \times 48} = \sqrt{1296} = 36$
If $\displaystyle \frac{a+b}{a-b} = \frac{5}{3}$ , then $a : b =$
A $5 : 3$ B $4 : 1$ C $8 : 3$ D $1 : 4$

Answer and solution

Answer: (B) $4 : 1$

By componendo and dividendo:

$\displaystyle \frac{2a}{2b} = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$

So $a : b = 4 : 1$
The cost of 5 dozen bananas is ₹150. The number of bananas that can be bought for ₹45 is
A $15$ B $18$ C $20$ D $200$

Answer and solution

Answer: (B) $18$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle y = \frac{45 \times 60}{150} = 18$
A man has money to buy 40 cycles at ₹600 each. If each cycle costs ₹200 more, the number of cycles he can buy is
A $32$ B $120$ C $36$ D $30$

Answer and solution

Answer: (D) $30$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$600 \times 40 = 800 \times y \Rightarrow y = 30$
The ratio $9 : 20$ expressed as a percentage is
A $90\%$ B $4.5\%$ C $45\%$ D $\displaystyle 222\frac{2}{9}\%$

Answer and solution

Answer: (C) $45\%$

$\displaystyle 9 : 20 = \left(\frac{9}{20} \times 100\right)\% = 45\%$
If $30\%$ of a number is 24 less than $50\%$ of that number, then the number is
A $120$ B $80$ C $30$ D $48$

Answer and solution

Answer: (A) $120$

$\displaystyle \frac{50x}{100} - \frac{30x}{100} = 24 \Rightarrow \frac{20x}{100} = 24$

$\displaystyle x = \frac{24 \times 100}{20} = 120$
If a quantity $P$ is increased by $R\%$ , the final quantity is
A $P + R$ B $\displaystyle \frac{100 - R}{100} \times P$ C $\displaystyle \frac{R}{100} \times P$ D $\displaystyle \frac{100 + R}{100} \times P$

Answer and solution

Answer: (D) $\displaystyle \frac{100 + R}{100} \times P$

Increase $\displaystyle = \frac{R}{100}P$ , so final $\displaystyle = P + \frac{R}{100}P = \frac{100 + R}{100}P$
The price of milk falls by $10\%$ . Without changing the expenditure, consumption can be increased by
A $10\%$ B $20\%$ C $\displaystyle 11\frac{1}{9}\%$ D $\displaystyle 9\frac{1}{11}\%$

Answer and solution

Answer: (C) $\displaystyle 11\frac{1}{9}\%$

$\displaystyle \left(\frac{10}{100 - 10} \times 100\right)\% = \frac{100}{9}\% = 11\frac{1}{9}\%$
The population of a village 2 years ago was 80000. It decreases at $5\%$ per annum. Its present population is
A $76{,}000$ B $72{,}200$ C $72{,}000$ D $88{,}200$

Answer and solution

Answer: (B) $72{,}200$

$\displaystyle 80000\left(1 - \frac{5}{100}\right)^2 = 80000 \times \frac{19}{20} \times \frac{19}{20} = 72200$
An article bought for ₹1500 is sold for ₹1800. The profit percent is
A $20\%$ B $120\%$ C $\displaystyle 83\frac{1}{3}\%$ D $\displaystyle 16\frac{2}{3}\%$

Answer and solution

Answer: (A) $20\%$

Profit $= 300$

Profit $\displaystyle \% = \frac{300}{1500} \times 100 = 20\%$ (always on C.P)
An article is marked at ₹1035 with a markup of $15\%$ . Its cost price is
A ₹879.75 B ₹1020 C ₹1190.25 D ₹900

Answer and solution

Answer: (D) ₹900

$\displaystyle x + \frac{15}{100}x = 1035 \Rightarrow \frac{115}{100}x = 1035 \Rightarrow x = 900$
A trader who professes to sell at cost price but uses false weights gains
A $\displaystyle \frac{\text{True value} - \text{Error}}{\text{Error}} \times 100\%$ B $\displaystyle \frac{\text{Error}}{\text{True value}} \times 100\%$ C $\displaystyle \frac{\text{True value}}{\text{Error}} \times 100\%$ D $\displaystyle \frac{\text{Error}}{\text{True value} - \text{Error}} \times 100\%$

Answer and solution

Answer: (D) $\displaystyle \frac{\text{Error}}{\text{True value} - \text{Error}} \times 100\%$

He pays for (true value $-$ error) but is paid for the true value, so the gain $f$ is on cost $(p - f)$ : $\displaystyle \frac{f}{p - f} \times 100$ .
If the cost price of 22 articles equals the selling price of 20 articles, then there is a
A $10\%$ gain B $2\%$ gain C $\displaystyle 9\frac{1}{11}\%$ gain D $10\%$ loss

Answer and solution

Answer: (A) $10\%$ gain

$\displaystyle 22x = 20y \Rightarrow \frac{y}{x} = \frac{11}{10}$

$\displaystyle \left(\frac{y}{x} - 1\right) \times 100 = 10\%$ (a gain)
The average age of 40 students is 15 years. The average age of the first 24 is 14 years. The average age of the remaining 16 is
A $16.5$ B $15$ C $16$ D $17$

Answer and solution

Answer: (A) $16.5$

Remaining sum $= 40(15) - 24(14) = 600 - 336 = 264$

Average $\displaystyle = \frac{264}{16} = 16.5$ years
The average weight of 35 students is 45 kg. When the teacher's weight is included, the average rises by 0.5 kg. The teacher's weight (in kg) is
A $45.5$ B $81$ C $62.5$ D $63$

Answer and solution

Answer: (D) $63$

$T = 36(45.5) - 35(45) = 1638 - 1575 = 63$ kg
A car moving at 48 km/h travels for 2 hours 15 minutes. The distance covered (in km) is
A $108$ B $96$ C $120$ D $105.6$

Answer and solution

Answer: (A) $108$

$\displaystyle T = 2\frac{15}{60} = \frac{9}{4}$ h

$\displaystyle D = 48 \times \frac{9}{4} = 108$ km
A traveller covers half of his journey by train at 100 km/h and the other half by car at 150 km/h. His average speed is
A $250$ km/h B $120$ km/h C $130$ km/h D $125$ km/h

Answer and solution

Answer: (B) $120$ km/h

$\displaystyle \frac{2 \times 100 \times 150}{100 + 150} = \frac{30{,}000}{250} = 120 \text{ km/h}$

(Not the simple average of the two speeds.)
For the same distance, if speeds $s_1$ and $s_2$ take times $T_1$ and $T_2$ , then $s_1 : s_2 =$
A $T_1 T_2 : 1$ B $T_1 : T_2$ C $T_2 : T_1$ D $(T_1 + T_2) : T_1$

Answer and solution

Answer: (C) $T_2 : T_1$

$\displaystyle s_1 = \frac{d}{T_1}$ , $\displaystyle s_2 = \frac{d}{T_2}$ , so $\displaystyle \frac{s_1}{s_2} = \frac{T_2}{T_1}$
When a train crosses a pole, the distance it covers is equal to
A zero B the length of the pole C twice its own length D its own length

Answer and solution

Answer: (D) its own length

A pole has no length, so the train only has to move its own length past it.
The simple interest on ₹800 at $8\%$ per annum is ₹192. The time (in years) is
A $1.92$ B $6$ C $3$ D $4$

Answer and solution

Answer: (C) $3$

$\displaystyle T = \frac{192 \times 100}{800 \times 8} = 3$ years
The amount on ₹20000 at $12\%$ per annum for 6 months, compounded quarterly, is
A ₹25,088 B ₹21,218 C ₹20,600 D ₹21,200

Answer and solution

Answer: (B) ₹21,218

$k = 4$ , $\displaystyle nk = \frac{1}{2} \times 4 = 2$ : $\displaystyle A = 20000\left(1 + \frac{12}{400}\right)^2 = 20000 \times (1.03)^2 = 21218$

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MCQ test, Set 4