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

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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 30 matches and won 18 of them. The ratio of the number of matches played to the number of matches lost is
A $2 : 5$ B $5 : 2$ C $3 : 2$ D $5 : 3$

Answer and solution

Answer: (B) $5 : 2$

$\text{Matches lost} = 30 - 18 = 12$

$\displaystyle \text{Required ratio} = \frac{30}{12} = \frac{5}{2} = 5 : 2$
If ₹1560 is divided into two parts in the ratio $5 : 7$ , then the larger part is
A ₹260 B ₹650 C ₹130 D ₹910

Answer and solution

Answer: (D) ₹910

$\displaystyle \text{Larger part} = \frac{7}{5 + 7} \times 1560 = \frac{7}{12} \times 1560 = 910$
If $\displaystyle 1\frac{1}{2} : 2\frac{1}{3} : 1\frac{3}{4}$ is $A : B : C$ , then
A $C > B > A$ B $B > C > A$ C $A > B > C$ D $B > A > C$

Answer and solution

Answer: (B) $B > C > A$

$\displaystyle A : B : C = \frac{3}{2} : \frac{7}{3} : \frac{7}{4}$

L.C.M of the denominators $= 12$

$\displaystyle = \frac{18}{12} : \frac{28}{12} : \frac{21}{12} = 18 : 28 : 21$

$\therefore B > C > A$
The fourth proportional to 3, 7 and 12 is
A $1.75$ B $16$ C $28$ D $\displaystyle 5\frac{1}{7}$

Answer and solution

Answer: (C) $28$

$3 : 7 :: 12 : x \Rightarrow 3x = 7 \times 12$

$\displaystyle x = \frac{84}{3} = 28$
The mean proportional between 18 and 32 is
A $576$ B $14$ C $25$ D $24$

Answer and solution

Answer: (D) $24$

$\text{Mean proportional} = \sqrt{18 \times 32} = \sqrt{576} = 24$
If $a : b = 5 : 2$ , then $\displaystyle \frac{a+b}{a-b} =$
A $\displaystyle \frac{7}{3}$ B $\displaystyle \frac{7}{2}$ C $\displaystyle \frac{5}{2}$ D $\displaystyle \frac{3}{7}$

Answer and solution

Answer: (A) $\displaystyle \frac{7}{3}$

By componendo and dividendo:

$\displaystyle \frac{a+b}{a-b} = \frac{5 + 2}{5 - 2} = \frac{7}{3}$
If 5 kg of rice can be served to 40 people, the rice required for 72 such people is
A $\displaystyle 2\frac{7}{9}$ kg B $0.125$ kg C $576$ kg D $9$ kg

Answer and solution

Answer: (D) $9$ kg

Direct proportion: $x_1 y_2 = x_2 y_1$

$\displaystyle y = \frac{72 \times 5}{40} = 9$
42 men can do a work in 30 days. 35 men will do the same work in
A $36$ days B $25$ days C $37$ days D $49$ days

Answer and solution

Answer: (A) $36$ days

Inverse proportion: $x_1 y_1 = x_2 y_2$

$42 \times 30 = 35 \times y \Rightarrow y = 36$
$\displaystyle \frac{3}{16}$ expressed as a percentage is
A $16\%$ B $1.875\%$ C $18.75\%$ D $\displaystyle \frac{3}{16}\%$

Answer and solution

Answer: (C) $18.75\%$

$\displaystyle \frac{3}{16} = \left(\frac{3}{16} \times 100\right)\% = \frac{300}{16}\% = 18.75\%$
In an election between two candidates, one got $60\%$ of the valid votes and $15\%$ of the votes were invalid. If the total number of votes was 9000, the number of valid votes the other candidate got is
A $3060$ B $3600$ C $1350$ D $4590$

Answer and solution

Answer: (A) $3060$

$\text{Valid votes} = 85\% \text{ of } 9000 = 7650$

$\text{Other candidate} = 40\% \text{ of } 7650 = 3060$
The price of a cooler is ₹1600. If it is increased by $15\%$ , the new price is
A ₹1360 B ₹1615 C ₹1840 D ₹240

Answer and solution

Answer: (C) ₹1840

$\displaystyle \frac{100 + 15}{100} \times 1600 = 1840$
If $A$ is $50\%$ more than $B$ , then $B$ is less than $A$ by
A $25\%$ B $50\%$ C $\displaystyle 66\frac{2}{3}\%$ D $\displaystyle 33\frac{1}{3}\%$

Answer and solution

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

$\displaystyle \left(\frac{50}{150} \times 100\right)\% = 33\frac{1}{3}\%$
The population of a town is 40000 and it increases at $5\%$ per annum. Its population after 2 years will be
A $44{,}000$ B $44{,}100$ C $42{,}000$ D $46{,}305$

Answer and solution

Answer: (B) $44{,}100$

$\displaystyle 40000\left(1 + \frac{5}{100}\right)^2 = 40000 \times \frac{21}{20} \times \frac{21}{20} = 44100$
By selling 24 pens, a vendor loses the selling price of 4 pens. His loss percent is
A $4\%$ B $20\%$ C $\displaystyle 16\frac{2}{3}\%$ D $\displaystyle 14\frac{2}{7}\%$

Answer and solution

Answer: (D) $\displaystyle 14\frac{2}{7}\%$

Let S.P of 1 pen $=$ ₹1. S.P of 24 pens $= 24$ , Loss $= 4$

$\text{C.P} = 24 + 4 = 28$ , Loss $\displaystyle \% = \frac{4}{28} \times 100 = 14\frac{2}{7}\%$
The marked price of a saree is ₹1800 and the discount is $15\%$ . The selling price is
A ₹1785 B ₹2070 C ₹1530 D ₹270

Answer and solution

Answer: (C) ₹1530

$\displaystyle \text{S.P} = \frac{100 - 15}{100} \times 1800 = 1530$
Two radios are sold at ₹960 each, one at $20\%$ profit and the other at $20\%$ loss. On the whole there is a
A loss of $4\%$ B neither profit nor loss C loss of $2\%$ D gain of $4\%$

Answer and solution

Answer: (A) loss of $4\%$

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

$\displaystyle \left(\frac{20}{10}\right)^2 = 4\%$
If the cost price of 25 articles equals the selling price of 20 articles, then there is a
A $25\%$ loss B $25\%$ gain C $5\%$ gain D $20\%$ gain

Answer and solution

Answer: (B) $25\%$ gain

$\displaystyle 25x = 20y \Rightarrow \frac{y}{x} = \frac{5}{4}$

$\displaystyle \left(\frac{y}{x} - 1\right) \times 100 = 25\%$ (a gain)
The average of five consecutive odd numbers is 47. The largest of them is
A $49$ B $51$ C $55$ D $43$

Answer and solution

Answer: (B) $51$

Let them be $x, x + 2, x + 4, x + 6, x + 8$ : $\displaystyle \frac{5x + 20}{5} = 47 \Rightarrow x + 4 = 47 \Rightarrow x = 43$

Largest $= 43 + 8 = 51$
The average of 15 numbers is 32. If each number is increased by 6, the new average is
A $38$ B $122$ C $32$ D $32.4$

Answer and solution

Answer: (A) $38$

If each $x_i$ becomes $x_i + b$ , the average becomes $\bar{x} + b$ .

$32 + 6 = 38$
$90$ km/h in m/s is
A $15$ B $50$ C $25$ D $324$

Answer and solution

Answer: (C) $25$

$\displaystyle 90 \times \frac{5}{18} = 25$ m/s
A man goes to a place at 40 km/h and returns by the same route at 60 km/h. His average speed for the whole journey is
A $48$ km/h B $50$ km/h C $52$ km/h D $100$ km/h

Answer and solution

Answer: (A) $48$ km/h

$\displaystyle \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h}$

(Not the simple average of the two speeds.)
The speeds of two cyclists are 30 km/h and 50 km/h. The ratio of the times they take to cover the same distance is
A $3 : 5$ B $8 : 5$ C $5 : 3$ D $3 : 8$

Answer and solution

Answer: (C) $5 : 3$

$T_1 : T_2 = s_2 : s_1 = 50 : 30 = 5 : 3$
A 250 m long train crosses a signal post in 10 seconds. Its speed is
A $\displaystyle 6\frac{17}{18}$ km/h B $90$ km/h C $25$ km/h D $45$ km/h

Answer and solution

Answer: (B) $90$ km/h

$\displaystyle S = \frac{250}{10} = 25$ m/s $\displaystyle = 25 \times \frac{18}{5} = 90$ km/h
The simple interest on ₹3500 at $6\%$ per annum for 4 years is
A ₹4340 B ₹210 C ₹1050 D ₹840

Answer and solution

Answer: (D) ₹840

$\displaystyle \text{S.I} = \frac{3500 \times 4 \times 6}{100} = 840$
The amount on ₹10000 at $10\%$ per annum for 2 years, compounded annually, is
A ₹11,000 B ₹12,000 C ₹12,100 D ₹2100

Answer and solution

Answer: (C) ₹12,100

$\displaystyle A = 10000\left(\frac{11}{10}\right)^2 = 12100$

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MCQ test, Set 1