Arithmetic Applications (Level 1): MCQ test, Set 5

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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 school team played 36 matches and won 20 of them. The ratio of the number of matches played to the number of matches lost is
A $9 : 5$ B $9 : 4$ C $5 : 4$ D $4 : 9$

Answer and solution

Answer: (B) $9 : 4$

$\text{Matches lost} = 36 - 20 = 16$

$\displaystyle \text{Required ratio} = \frac{36}{16} = \frac{9}{4} = 9 : 4$
If ₹2100 is divided into two parts in the ratio $3 : 11$ , then the larger part is
A ₹1650 B ₹1200 C ₹450 D ₹150

Answer and solution

Answer: (A) ₹1650

$\displaystyle \text{Larger part} = \frac{11}{3 + 11} \times 2100 = \frac{11}{14} \times 2100 = 1650$
Three numbers are in the ratio $1 : 2 : 3$ and the sum of their squares is 224. The sum of the numbers is
A $12$ B $96$ C $16$ D $24$

Answer and solution

Answer: (D) $24$

Numbers: $1k, 2k, 3k$ ; $14k^2 = 224 \Rightarrow k^2 = 16 \Rightarrow k = 4$

$\text{Sum} = 6k = 24$
The fourth proportional to 5, 12 and 15 is
A $4$ B $36$ C $6.25$ D $22$

Answer and solution

Answer: (B) $36$

$5 : 12 :: 15 : x \Rightarrow 5x = 12 \times 15$

$\displaystyle x = \frac{180}{5} = 36$
If $a, b, c$ are in continued proportion, then
A $a^2 = bc$ B $b = ac$ C $b^2 = ac$ D $c^2 = ab$

Answer and solution

Answer: (C) $b^2 = ac$

$a : b :: b : c$ , so product of means $=$ product of extremes.

$b \times b = a \times c$ , i.e. $b^2 = ac$ .
If $a : b = 8 : 5$ , then $\displaystyle \frac{a+b}{a-b} =$
A $\displaystyle \frac{13}{5}$ B $\displaystyle \frac{8}{5}$ C $\displaystyle \frac{3}{13}$ D $\displaystyle \frac{13}{3}$

Answer and solution

Answer: (D) $\displaystyle \frac{13}{3}$

By componendo and dividendo:

$\displaystyle \frac{a+b}{a-b} = \frac{8 + 5}{8 - 5} = \frac{13}{3}$
A photograph of a bacterium enlarged 40,000 times has a length of 6 cm. The actual length of the bacterium is
A $1.5 \times 10^{-3}$ cm B $1.5 \times 10^{-4}$ cm C $2.4 \times 10^{5}$ cm D $1.5 \times 10^{-5}$ cm

Answer and solution

Answer: (B) $1.5 \times 10^{-4}$ cm

Enlarged 40,000 times → 6 cm, so 1 time → $\displaystyle \frac{6}{40000}$ cm.

$\displaystyle \frac{6}{40000} = 0.00015 = 1.5 \times 10^{-4}$ cm
$A$ varies directly as $B$ and inversely as $C$ . If $A = 8$ when $B = 12$ and $C = 3$ , then $A$ when $B = 15$ and $C = 5$ is
A $6$ B $1.5$ C $4.8$ D $10$

Answer and solution

Answer: (A) $6$

$\displaystyle A = \frac{kB}{C}$ : $\displaystyle 8 = \frac{12k}{3} \Rightarrow k = 2$

$\displaystyle A = \frac{2 \times 15}{5} = 6$
$\displaystyle \frac{5}{12}$ expressed as a percentage is
A $\displaystyle 4\frac{1}{6}\%$ B $2.4\%$ C $60\%$ D $\displaystyle 41\frac{2}{3}\%$

Answer and solution

Answer: (D) $\displaystyle 41\frac{2}{3}\%$

$\displaystyle \frac{5}{12} = \frac{500}{12}\% = 41\frac{2}{3}\% \approx 41.67\%$
The number which exceeds $40\%$ of itself by 36 is
A $90$ B $50.4$ C $72$ D $60$

Answer and solution

Answer: (D) $60$

$\displaystyle x - \frac{40}{100}x = 36 \Rightarrow \frac{60x}{100} = 36$

$\displaystyle x = \frac{36 \times 100}{60} = 60$
A sum of ₹2500 is decreased by $18\%$ . The new amount is
A ₹2050 B ₹2950 C ₹450 D ₹2482

Answer and solution

Answer: (A) ₹2050

$\displaystyle \frac{100 - 18}{100} \times 2500 = 2050$
If $A$ 's income is $60\%$ more than $B$ 's, then $B$ 's income is less than $A$ 's by
A $60\%$ B $62.5\%$ C $37.5\%$ D $40\%$

Answer and solution

Answer: (C) $37.5\%$

$\displaystyle \left(\frac{60}{160} \times 100\right)\% = 37.5\%$
If the present population $P$ increases at $R\%$ per annum, the population $n$ years ago was
A $\displaystyle \frac{P}{1 + \frac{nR}{100}}$ B $\displaystyle \frac{P}{\left(1 + \frac{R}{100}\right)^n}$ C $\displaystyle P\left(1 + \frac{R}{100}\right)^n$ D $\displaystyle P\left(1 - \frac{R}{100}\right)^n$

Answer and solution

Answer: (B) $\displaystyle \frac{P}{\left(1 + \frac{R}{100}\right)^n}$

If $x$ was the population $n$ years ago, $\displaystyle x\left(1 + \frac{R}{100}\right)^n = P$ , so $\displaystyle x = \frac{P}{\left(1 + \frac{R}{100}\right)^n}$ .
By selling a table for ₹1105, a man loses $15\%$ . The cost price of the table is
A ₹1270.75 B ₹1120 C ₹1300 D ₹939.25

Answer and solution

Answer: (C) ₹1300

$\displaystyle \text{C.P} = \frac{100}{100 - 15} \times 1105 = \frac{100}{85} \times 1105 = 1300$
An article marked at ₹2000 is sold with a discount of ₹300 at a profit of ₹200. Its cost price is
A ₹1900 B ₹1500 C ₹2100 D ₹1700

Answer and solution

Answer: (B) ₹1500

$\text{C.P} = \text{M.P} - \text{Discount} - \text{Profit} = 2000 - 300 - 200 = 1500$
Two fans are sold at ₹1232 each, one at $12\%$ profit and the other at $12\%$ loss. On the whole there is a
A neither profit nor loss B gain of $1.44\%$ C loss of $1.44\%$ D loss of $1.2\%$

Answer and solution

Answer: (C) loss of $1.44\%$

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

$\displaystyle \left(\frac{12}{10}\right)^2 = 1.44\%$
A shopkeeper buys pens at 20 for ₹90 and sells them at 15 for ₹81. His profit percent is
A $25\%$ B $20\%$ C $10\%$ D $\displaystyle 16\frac{2}{3}\%$

Answer and solution

Answer: (B) $20\%$

$\displaystyle \text{C.P} \text{ of one} = \frac{9}{2}, \quad \text{S.P} \text{ of one} = \frac{27}{5}$

$\displaystyle \text{Profit \%} = \frac{\frac{27}{5} - \frac{9}{2}}{\frac{9}{2}} \times 100 = 20\%$
A library has an average of 540 visitors on Sundays and 270 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is
A $300$ B $405$ C $315$ D $306$

Answer and solution

Answer: (C) $315$

A 30-day month beginning with Sunday has 5 Sundays and 25 other days.

$\displaystyle \frac{5(540) + 25(270)}{30} = \frac{2700 + 6750}{30} = 315$
A batsman scores 92 runs in his 21st innings and thus increases his average by 3. His average after the 21st innings is
A $29$ B $95$ C $32$ D $35$

Answer and solution

Answer: (C) $32$

$\displaystyle \frac{20P + 92}{21} = P + 3 \Rightarrow 20P + 92 = 21P + 63 \Rightarrow P = 29$

New average $= 29 + 3 = 32$
A train covers 315 km at 70 km/h. The time taken is
A 4 h 5 min B 4 h 30 min C 5 h 15 min D 4 h 50 min

Answer and solution

Answer: (B) 4 h 30 min

$\displaystyle T = \frac{315}{70} = 4.5$ h $= 4$ h $30$ min
A man travels 120 km in 2 hours and then 80 km in 3 hours. His average speed is
A $40$ km/h B $36$ km/h C $\displaystyle 43\frac{1}{3}$ km/h D $50$ km/h

Answer and solution

Answer: (A) $40$ km/h

Average speed $\displaystyle = \frac{120 + 80}{2 + 3} = \frac{200}{5} = 40$ km/h
Walking at 4 km/h, a boy reaches school in 45 minutes. Walking at 5 km/h, he will reach in
A $36$ minutes B $40$ minutes C $30$ minutes D $56.25$ minutes

Answer and solution

Answer: (A) $36$ minutes

$\displaystyle T_2 = \frac{4 \times 45}{5} = 36$ minutes
Two trains are moving in opposite directions at 45 km/h and 35 km/h. Their relative speed is
A $40$ km/h B $10$ km/h C $5$ km/h D $80$ km/h

Answer and solution

Answer: (D) $80$ km/h

Opposite directions: relative speed $= (x + y) = 45 + 35 = 80$ km/h
A sum gives ₹450 as simple interest in 3 years at $5\%$ per annum. The sum is
A ₹67.5 B ₹9000 C ₹2700 D ₹3000

Answer and solution

Answer: (D) ₹3000

$\displaystyle P = \frac{450 \times 100}{3 \times 5} = 3000$
A sum of ₹1000 becomes ₹1440 in 2 years at compound interest. The rate of interest per annum is
A $20\%$ B $44\%$ C $22\%$ D $12\%$

Answer and solution

Answer: (A) $20\%$

$\displaystyle \left(1 + \frac{R}{100}\right)^2 = \frac{1440}{1000} = \frac{144}{100} = \left(\frac{12}{10}\right)^2$

$\displaystyle 1 + \frac{R}{100} = \frac{6}{5} \Rightarrow R = 20\%$

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MCQ test, Set 5