Arithmetic Applications (Level 1): MCQ test, Set 9

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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 40 matches and won 25 of them. The ratio of the number of matches played to the number of matches lost is
A $8 : 3$ B $3 : 8$ C $5 : 3$ D $8 : 5$

Answer and solution

Answer: (A) $8 : 3$

$\text{Matches lost} = 40 - 25 = 15$

$\displaystyle \text{Required ratio} = \frac{40}{15} = \frac{8}{3} = 8 : 3$
If ₹2925 is divided into two parts in the ratio $4 : 9$ , then the larger part is
A ₹1125 B ₹2025 C ₹900 D ₹225

Answer and solution

Answer: (B) ₹2025

$\displaystyle \text{Larger part} = \frac{9}{4 + 9} \times 2925 = \frac{9}{13} \times 2925 = 2025$
Metal P is 18 times as heavy as water and metal Q is 6 times as heavy as water. To get an alloy 14 times as heavy as water, P and Q must be mixed in the ratio
A $1 : 2$ B $1 : 3$ C $2 : 3$ D $2 : 1$

Answer and solution

Answer: (D) $2 : 1$

Take 1 g of P and $x$ g of Q: $18 + 6x = 14(1 + x)$

$\displaystyle \Rightarrow 4 = 8x \Rightarrow x = \frac{1}{2}$

$\displaystyle \text{Ratio} = 1 : \frac{1}{2} = 2 : 1$
If $15 : 25 :: x : 40$ , then $x =$
A $9.375$ B $\displaystyle 66\frac{2}{3}$ C $24$ D $30$

Answer and solution

Answer: (C) $24$

Product of means $=$ product of extremes.

$25x = 15 \times 40 \Rightarrow x = 24$
The third proportional to 25 and 15 is
A $5\sqrt{15}$ B $\displaystyle 41\frac{2}{3}$ C $375$ D $9$

Answer and solution

Answer: (D) $9$

$25 : 15 :: 15 : c \Rightarrow 25c = 15 \times 15$

$\displaystyle c = \frac{225}{25} = 9$
If $\displaystyle \frac{x+y}{x-y} = \frac{11}{7}$ , then $x : y =$
A $9 : 2$ B $18 : 7$ C $11 : 7$ D $2 : 9$

Answer and solution

Answer: (A) $9 : 2$

By componendo and dividendo:

$\displaystyle \frac{2x}{2y} = \frac{11 + 7}{11 - 7} = \frac{18}{4} = \frac{9}{2}$

So $x : y = 9 : 2$
If $x$ and $y$ are in direct proportion, then
A $x + y$ is a constant B $xy$ is a constant C $x - y$ is a constant D $\displaystyle \frac{x}{y}$ is a constant

Answer and solution

Answer: (D) $\displaystyle \frac{x}{y}$ is a constant

$\displaystyle x \propto y \Rightarrow x = ky \Rightarrow \frac{x}{y} = k$ , a constant.
There is food for 500 men for 24 days. After 4 days, the number of extra men who must join so that the remaining food lasts 16 more days is
A $100$ B $625$ C $250$ D $125$

Answer and solution

Answer: (D) $125$

Remaining food $= 500 \times 20 = 10000$ man-days

$\displaystyle \frac{10000}{16} = 625$ men, so $625 - 500 = 125$ more
$\displaystyle \frac{6}{7}$ expressed as a percentage is
A $\displaystyle 116\frac{2}{3}\%$ B $\displaystyle \frac{6}{7}\%$ C $\displaystyle 85\frac{5}{7}\%$ D $\displaystyle 71\frac{3}{7}\%$

Answer and solution

Answer: (C) $\displaystyle 85\frac{5}{7}\%$

$\displaystyle \frac{6}{7} = \frac{600}{7}\% = 85\frac{5}{7}\% \approx 85.71\%$
What percentage of 2 kg is 250 g?
A $8\%$ B $125\%$ C $1.25\%$ D $12.5\%$

Answer and solution

Answer: (D) $12.5\%$

$2$ kg $= 2000$ g

$\displaystyle \left(\frac{250}{2000} \times 100\right)\% = 12.5\%$
The population of a village is 15000. If it decreases by $6\%$ , the new population is
A $14{,}100$ B $900$ C $15{,}900$ D $14{,}400$

Answer and solution

Answer: (A) $14{,}100$

$\displaystyle \frac{100 - 6}{100} \times 15000 = \frac{94}{100} \times 15000 = 14100$
The price of oil decreases by $25\%$ . The increase in consumption so as not to decrease the expenditure is
A $75\%$ B $\displaystyle 33\frac{1}{3}\%$ C $25\%$ D $20\%$

Answer and solution

Answer: (B) $\displaystyle 33\frac{1}{3}\%$

$\displaystyle \left(\frac{25}{75} \times 100\right)\% = 33\frac{1}{3}\%$
The population of a town is 90000 and it decreases at $10\%$ per annum. After 2 years it will be
A $81{,}000$ B $72{,}900$ C $1{,}08{,}900$ D $72{,}000$

Answer and solution

Answer: (B) $72{,}900$

$\displaystyle 90000\left(1 - \frac{10}{100}\right)^2 = 90000 \times \frac{81}{100} = 72900$
An article costing ₹840 is sold at a loss of $5\%$ . Its selling price is
A ₹798 B ₹882 C ₹42 D ₹835

Answer and solution

Answer: (A) ₹798

$\displaystyle \text{S.P} = \frac{100 - 5}{100} \times 840 = 798$
An article marked at ₹1200 is sold at a $20\%$ discount, still giving a $20\%$ profit. Its cost price is
A ₹800 B ₹1152 C ₹960 D ₹768

Answer and solution

Answer: (A) ₹800

S.P $\displaystyle = \frac{80}{100} \times 1200 = 960$

C.P $\displaystyle = \frac{100}{120} \times 960 = 800$
A trader claims to sell tea at cost price but gives only 980 g for every kg. His gain percent is
A $98\%$ B $\displaystyle 2\frac{2}{49}\%$ C $\displaystyle 102\frac{2}{49}\%$ D $2\%$

Answer and solution

Answer: (B) $\displaystyle 2\frac{2}{49}\%$

Gain $\displaystyle \% = \frac{\text{Error}}{\text{True value} - \text{Error}} \times 100$

$\displaystyle = \frac{20}{980} \times 100 = 2\frac{2}{49}\%$
If the cost price of 14 articles equals the selling price of 12 articles, then there is a
A $\displaystyle 14\frac{2}{7}\%$ gain B $\displaystyle 16\frac{2}{3}\%$ loss C $\displaystyle 16\frac{2}{3}\%$ gain D $2\%$ gain

Answer and solution

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

$\displaystyle 14x = 12y \Rightarrow \frac{y}{x} = \frac{7}{6}$

$\displaystyle \left(\frac{y}{x} - 1\right) \times 100 = 16\frac{2}{3}\%$ (a gain)
The average of four consecutive even numbers is 25. The largest of them is
A $26$ B $31$ C $28$ D $22$

Answer and solution

Answer: (C) $28$

$\displaystyle \frac{x + (x+2) + (x+4) + (x+6)}{4} = 25 \Rightarrow x + 3 = 25 \Rightarrow x = 22$

Largest $= 22 + 6 = 28$
The average age of 30 students is 15 years. When the teacher's age is included, the average increases by 1. The teacher's age (in years) is
A $46$ B $31$ C $16$ D $45$

Answer and solution

Answer: (A) $46$

$31(16) - 30(15) = 496 - 450 = 46$
If a man travels at 36 km/h, he reaches his destination 10 minutes late, and at 45 km/h he reaches 5 minutes early. The distance travelled is
A $36$ km B $45$ km C $54$ km D $40$ km

Answer and solution

Answer: (B) $45$ km

Let $T$ be the correct time (hours): $\displaystyle D = 36\left(T + \frac{10}{60}\right) = 45\left(T - \frac{5}{60}\right)$

$\displaystyle \Rightarrow T = \frac{13}{12} \text{ h}, \quad D = 36 \times \frac{5}{4} = 45 \text{ km}$
A man goes to a place at 70 km/h and returns by the same route at 30 km/h. His average speed for the whole journey is
A $50$ km/h B $42$ km/h C $21$ km/h D $100$ km/h

Answer and solution

Answer: (B) $42$ km/h

$\displaystyle \frac{2 \times 70 \times 30}{70 + 30} = \frac{4200}{100} = 42 \text{ km/h}$

(Not the simple average of the two speeds.)
The speeds of two trains are 72 km/h and 48 km/h. The ratio of the times taken by them for the same distance is
A $5 : 3$ B $2 : 5$ C $2 : 3$ D $3 : 2$

Answer and solution

Answer: (C) $2 : 3$

$T_1 : T_2 = s_2 : s_1 = 48 : 72 = 2 : 3$
A train 120 m long running at 54 km/h crosses a bridge in 30 seconds. The length of the bridge is
A $300$ m B $450$ m C $330$ m D $570$ m

Answer and solution

Answer: (C) $330$ m

Distance $= 15 \times 30 = 450$ m (speed $= 15$ m/s)

Bridge $= 450 - 120 = 330$ m
The formula for simple interest is
A $\displaystyle \text{S.I} = \frac{P \times T \times R}{100}$ B $\displaystyle \text{S.I} = \frac{P \times T}{100 \times R}$ C $\text{S.I} = 100 \times P \times T \times R$ D $\displaystyle \text{S.I} = P\left(1 + \frac{R}{100}\right)^T$

Answer and solution

Answer: (A) $\displaystyle \text{S.I} = \frac{P \times T \times R}{100}$

$\displaystyle \text{S.I} = \frac{PTR}{100}$ , where $P$ is the principal, $R$ the rate and $T$ the time.
A trader invests ₹40000. He loses $10\%$ in the first year and gains $20\%$ in the second year. His net profit is
A ₹4000 B ₹2800 C ₹3600 D ₹3200

Answer and solution

Answer: (D) ₹3200

$\displaystyle 40000 \times \frac{90}{100} \times \frac{120}{100} = 43200$

Net profit $= 43200 - 40000 = 3200$

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MCQ test, Set 9