Arithmetic Applications (Level 1): MCQ test, Set 7

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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.

A box contains 60 bulbs, out of which 24 are defective. If the ratio of defective bulbs to good bulbs is $x : y$ in simplest form, then $x + y =$
A $8$ B $5$ C $60$ D $7$

Answer and solution

Answer: (B) $5$

$\text{Good bulbs} = 60 - 24 = 36$

$\displaystyle \frac{\text{Defective}}{\text{Good}} = \frac{24}{36} = \frac{2}{3} \Rightarrow x : y = 2 : 3$

$x + y = 2 + 3 = 5$
If ₹1680 is divided into two parts in the ratio $5 : 9$ , then the smaller part is
A ₹600 B ₹1080 C ₹480 D ₹120

Answer and solution

Answer: (A) ₹600

$\displaystyle \text{Smaller part} = \frac{5}{5 + 9} \times 1680 = \frac{5}{14} \times 1680 = 600$
In a school library, the ratio of comic books to story books is the same as the ratio of story books to Hindi books. If there are 360 story books and 240 Hindi books, then the number of comic books is
A $540$ B $480$ C $240$ D $600$

Answer and solution

Answer: (A) $540$

$\displaystyle \frac{\text{Comic}}{\text{Story}} = \frac{\text{Story}}{\text{Hindi}} = \frac{360}{240} = \frac{3}{2}$

$\displaystyle \text{Comic books} = 360 \times \frac{3}{2} = 540$
Which of the following are in proportion?
A $3, 5, 9, 25$ B $2, 5, 4, 8$ C $3, 5, 9, 15$ D $4, 6, 9, 12$

Answer and solution

Answer: (C) $3, 5, 9, 15$

$a, b, c, d$ are in proportion when $ad = bc$ .

$3 \times 15 = 45$ and $5 \times 9 = 45$ , so $3, 5, 9, 15$ are in proportion.
The mean proportional between 20 and 45 is
A $32.5$ B $30$ C $900$ D $25$

Answer and solution

Answer: (B) $30$

$\text{Mean proportional} = \sqrt{20 \times 45} = \sqrt{900} = 30$
If $\displaystyle \frac{p}{q} = \frac{13}{8}$ , then $\displaystyle \frac{p-q}{q} =$
A $\displaystyle \frac{13}{8}$ B $\displaystyle \frac{5}{13}$ C $\displaystyle \frac{21}{8}$ D $\displaystyle \frac{5}{8}$

Answer and solution

Answer: (D) $\displaystyle \frac{5}{8}$

By dividendo, $\displaystyle \frac{p-q}{q} = \frac{13 - 8}{8} = \frac{5}{8}$
$x$ and $y$ are in direct proportion. If $y = 15$ when $x = 6$ , then $y$ when $x = 10$ is
A $9$ B $25$ C $19$ D $4$

Answer and solution

Answer: (B) $25$

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle y = \frac{10 \times 15}{6} = 25$
If $x$ varies inversely as $y$ and $x = 18$ when $y = 5$ , then $x$ when $y = 15$ is
A $8$ B $54$ C $90$ D $6$

Answer and solution

Answer: (D) $6$

Inverse proportion: $x_1 y_1 = x_2 y_2$

$5 \times 18 = 15 \times y \Rightarrow y = 6$
$85\%$ expressed as a decimal is
A $8.5$ B $0.085$ C $0.85$ D $85$

Answer and solution

Answer: (C) $0.85$

$\displaystyle 85\% = \frac{85}{100} = 0.85$
In an election between two candidates, one got $65\%$ of the valid votes and $25\%$ of the votes were invalid. If the total number of votes was 6400, the number of valid votes the other candidate got is
A $1600$ B $3120$ C $2240$ D $1680$

Answer and solution

Answer: (D) $1680$

$\text{Valid votes} = 75\% \text{ of } 6400 = 4800$

$\text{Other candidate} = 35\% \text{ of } 4800 = 1680$
By what percent is 45 less than 54?
A $\displaystyle 16\frac{2}{3}\%$ B $\displaystyle 83\frac{1}{3}\%$ C $9\%$ D $20\%$

Answer and solution

Answer: (A) $\displaystyle 16\frac{2}{3}\%$

$\displaystyle \left[\frac{\text{Greater} - \text{smaller}}{\text{Greater}} \times 100\right]\% = \left(\frac{9}{54} \times 100\right)\% = 16\frac{2}{3}\%$
$X$ 's weight is $30\%$ less than $Y$ 's. Then $Y$ 's weight is more than $X$ 's by
A $\displaystyle 23\frac{1}{13}\%$ B $\displaystyle 42\frac{6}{7}\%$ C $70\%$ D $30\%$

Answer and solution

Answer: (B) $\displaystyle 42\frac{6}{7}\%$

$\displaystyle \left(\frac{30}{70} \times 100\right)\% = 42\frac{6}{7}\%$
The present population of a city is 36300. It has been increasing at $10\%$ per annum. Its population 2 years ago was
A $29{,}403$ B $29{,}040$ C $33{,}000$ D $30{,}000$

Answer and solution

Answer: (D) $30{,}000$

$\displaystyle \frac{36300}{\left(\frac{11}{10}\right)^2} = 36300 \times \frac{100}{121} = 30000$
An article costs ₹2000. To gain $12\%$ , it must be sold for
A ₹1760 B ₹2240 C ₹2012 D ₹240

Answer and solution

Answer: (B) ₹2240

$\displaystyle \text{S.P} = \frac{100 + 12}{100} \times 2000 = 2240$
If a discount of $D\%$ is given on the marked price, the selling price is
A $\displaystyle \frac{100 + D}{100} \times \text{M.P}$ B $\displaystyle \frac{100 - D}{100} \times \text{M.P}$ C $\displaystyle \frac{100}{100 - D} \times \text{M.P}$ D $\displaystyle \frac{D}{100} \times \text{M.P}$

Answer and solution

Answer: (B) $\displaystyle \frac{100 - D}{100} \times \text{M.P}$

$\displaystyle \text{S.P} = \text{M.P} - \frac{D}{100}\text{M.P} = \frac{100 - D}{100} \times \text{M.P}$
Two phones are sold at ₹1820 each, one at $30\%$ profit and the other at $30\%$ loss. On the whole there is a
A loss of $9\%$ B loss of $3\%$ C gain of $9\%$ D neither profit nor loss

Answer and solution

Answer: (A) loss of $9\%$

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

$\displaystyle \left(\frac{30}{10}\right)^2 = 9\%$
If the cost price of 30 articles equals the selling price of 25 articles, then there is a
A $20\%$ loss B $5\%$ gain C $20\%$ gain D $\displaystyle 16\frac{2}{3}\%$ gain

Answer and solution

Answer: (C) $20\%$ gain

$\displaystyle 30x = 25y \Rightarrow \frac{y}{x} = \frac{6}{5}$

$\displaystyle \left(\frac{y}{x} - 1\right) \times 100 = 20\%$ (a gain)
The sum of five numbers is 640. The average of the first two is 110 and the third number is 150. The average of the last two numbers is
A $128$ B $270$ C $160$ D $135$

Answer and solution

Answer: (D) $135$

$x_1 + x_2 = 220$ ; $x_4 + x_5 = 640 - 220 - 150 = 270$

Average $\displaystyle = \frac{270}{2} = 135$
The average of a batsman in 15 innings is 36. To raise his average by 2, the runs he must score in the 16th innings are
A $70$ B $66$ C $68$ D $38$

Answer and solution

Answer: (C) $68$

$16(38) - 15(36) = 608 - 540 = 68$
A train leaves at 8.50 am at 60 km/h. It covers 210 km at
A 12.20 pm B 11.55 am C 11.50 am D 12.40 pm

Answer and solution

Answer: (A) 12.20 pm

$\displaystyle T = \frac{210}{60} = 3\frac{1}{2}$ h $= 3$ h $30$ min

8.50 am $+ 3$ h $30$ min $=$ 12.20 pm
A man goes to a place at 24 km/h and returns by the same route at 40 km/h. His average speed for the whole journey is
A $28$ km/h B $32$ km/h C $30$ km/h D $64$ km/h

Answer and solution

Answer: (C) $30$ km/h

$\displaystyle \frac{2 \times 24 \times 40}{24 + 40} = \frac{1920}{64} = 30 \text{ km/h}$

(Not the simple average of the two speeds.)
The speeds of two runners are 25 km/h and 35 km/h. The ratio of the times they take for the same distance is
A $5 : 7$ B $7 : 5$ C $7 : 12$ D $12 : 7$

Answer and solution

Answer: (B) $7 : 5$

$T_1 : T_2 = s_2 : s_1 = 35 : 25 = 7 : 5$
When a train crosses a bridge, the distance it covers is
A length of the bridge − length of the train B length of the train only C length of the train + length of the bridge D length of the bridge only

Answer and solution

Answer: (C) length of the train + length of the bridge

The engine must travel the whole bridge and then the whole train must clear it: train $+$ bridge.
A sum amounts to ₹960 in 3 years and to ₹1040 in 5 years at simple interest. The sum is
A ₹760 B ₹880 C ₹800 D ₹840

Answer and solution

Answer: (D) ₹840

S.I for 2 years $= 1040 - 960 = 80$ , so 40 per year

$P = 960 - 3 \times 40 = 840$
The population of a town is 50000. It increases by $2\%$ in the first year and by $4\%$ in the second. The population after 2 years is
A $53{,}040$ B $54{,}080$ C $52{,}000$ D $53{,}000$

Answer and solution

Answer: (A) $53{,}040$

$\displaystyle 50000 \times \frac{102}{100} \times \frac{104}{100} = 53040$

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MCQ test, Set 7