Quant · free practice

Commercial Maths (Rohtak/JIPMAT) questions for IPMAT, with solutions

10 free Commercial Maths (Rohtak/JIPMAT) practice questions for IPMAT Indore, Rohtak and JIPMAT, 1 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Commercial Maths (Rohtak/JIPMAT) on Learn.

1. Question 1Multiple choiceWritten by IPMATwayCalendar

   How many odd days are there in a leap year?

   1. A.$0$
   2. B.$2$
   3. C.$3$
   4. D.$1$
   Show the answer and solution

   Answer: B. $2$

   A leap year has $366$ days. Divide by $7$: $366=52\times7+2$. The $52$ full weeks contribute nothing, so a leap year has $2$ odd days. (An ordinary year has $1$ odd day.)

2. Question 2Type the answerWritten by IPMATwayCalendar

   In a month of $31$ days, the $1$st is a Tuesday. How many Thursdays are there in that month?

   Show the answer and solution

   Answer: 5

   $31=4\times7+3$, so every weekday occurs $4$ times plus the first $3$ weekdays of the month once more. The first three days are Tuesday, Wednesday, Thursday. So Thursday occurs $5$ times: on the $3$rd, $10$th, $17$th, $24$th and $31$st.

3. Question 3Multiple choiceWritten by IPMATwayTrue and Banker's Discount

   What is the present worth of ₹$1150$ due $2$ years hence at $7.5\%$ per annum simple interest?

   1. A.₹$150$
   2. B.₹$1069.77$
   3. C.₹$977.5$
   4. D.₹$1000$
   Show the answer and solution

   Answer: D. ₹$1000$

   Simple interest for $2$ years at $7.5\%$ is $15\%$ of the present worth. So the amount is $1.15\times PW=1150$. Therefore $PW=\dfrac{1150}{1.15}=1000$. (The true discount would be $1150-1000=150$.)

4. Question 4Type the answerWritten by IPMATwayTrue and Banker's Discount

   Find the present worth (in rupees) of ₹$13310$ due $3$ years hence at $10\%$ per annum compound interest.

   Show the answer and solution

   Answer: 10000

   Under compound interest the present worth satisfies $PW(1.1)^3=13310$. Here $1.1^3=1.331$. So $PW=\dfrac{13310}{1.331}=10000$. (The true discount is $13310-10000=3310$.)

5. Question 5Multiple choiceAsked in IPMAT Rohtak 2019Number Series

   What should come at the place of question mark? 46080, 3840, 384, 48, 8, 2, ?

   1. A.1
   2. B.1/64
   3. C.1/8
   4. D.None of these
   Show the answer and solution

   Answer: A. 1

   Divide each term by the next. $46080\div3840=12$, $3840\div384=10$, $384\div48=8$, $48\div8=6$, $8\div2=4$. The divisors fall by $2$ each time: $12, 10, 8, 6, 4$. The next divisor is $2$, so the missing term is $2\div2=1$.

6. Question 6Multiple choiceWritten by IPMATwayStocks and Shares

   A man buys ₹$25$ shares of a company that pays a $9\%$ dividend. If he earns $15\%$ on his investment, at what price did he buy each share?

   1. A.₹$15$
   2. B.₹$3.75$
   3. C.₹$41.67$
   4. D.₹$25$
   Show the answer and solution

   Answer: A. ₹$15$

   Dividend per share $=9\%$ of ₹$25=2.25$. He earns $15\%$ on the price paid, so $0.15\times\text{price}=2.25$. Hence the price $=\dfrac{2.25}{0.15}=15$.

7. Question 7Multiple choiceWritten by IPMATwayStocks and Shares

   A man invests ₹$24000$ in ₹$100$ shares available at a $20\%$ premium. If the company pays a $15\%$ dividend, his annual income is

   1. A.₹$3600$
   2. B.₹$4500$
   3. C.₹$4320$
   4. D.₹$3000$
   Show the answer and solution

   Answer: D. ₹$3000$

   A $20\%$ premium on a ₹$100$ share means a price of ₹$120$. Shares bought $=\dfrac{24000}{120}=200$. Dividend per share $=15\%$ of the face value $=15$. Annual income $=200\times15=3000$.

8. Question 8Multiple choiceWritten by IPMATwayCalendar

   The calendar of the leap year $2096$ will next be used again in

   1. A.$2101$
   2. B.$2108$
   3. C.$2124$
   4. D.$2102$
   Show the answer and solution

   Answer: B. $2108$

   We need a leap year whose 1 January is on the same weekday as in $2096$. Odd days: $2096$ gives $2$, $2097$-$2099$ give $1$ each, $2100$ is not a leap year and gives $1$, $2101$-$2103$ give $1$ each, $2104$ gives $2$, $2105$-$2107$ give $1$ each. Total $=2+3+1+3+2+3=14\equiv0$. So 1 January $2108$ matches, and $2108$ is a leap year. The usual $28$-year rule fails because $2100$ is not a leap year.

9. Question 9Type the answerWritten by IPMATwayTrue and Banker's Discount

   The true discount on a sum due $2$ years hence is ₹$500$ and the simple interest on it for the same time is ₹$550$. Find the rate of interest (per cent per annum).

   Show the answer and solution

   Answer: 5

   SI on the sum minus TD equals the interest on TD for the same time and rate. So $550-500=50$ is the interest on ₹$500$ for $2$ years. Then $500\times\dfrac{R\times2}{100}=50$, so $10R=50$. Hence $R=5\%$ per annum.

10. Question 10Multiple choiceWritten by IPMATwayNumber Series

    In the series $2,\ 9,\ 28,\ 65,\ 124,\ 217,\ 344$ one term is wrong. What should replace it?

    1. A.$125$
    2. B.$127$
    3. C.$130$
    4. D.$126$
    Show the answer and solution

    Answer: D. $126$

    Test the pattern $n^3+1$: $1^3+1=2$, $2^3+1=9$, $3^3+1=28$, $4^3+1=65$, $5^3+1=126$, $6^3+1=217$, $7^3+1=344$. Every term matches except the fifth, which is shown as $124$. So the fifth term should be $126$. Choosing $125$ forgets the $+1$, $127$ over-corrects by one, and $130$ comes from adding a mis-copied difference to $65$.