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Binomial Theorem questions for IPMAT, with solutions

10 free Binomial Theorem practice questions for IPMAT Indore, Rohtak and JIPMAT, 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 Binomial Theorem on Learn.

1. Question 1Type the answerWritten by IPMATwayBinomial Coefficients and Terms

   The sum of the coefficients in the expansion of $(3x-1)^5$ is

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   Answer: 32

   The sum of all coefficients of a polynomial $p(x)$ is $p(1)$. Here $p(x)=(3x-1)^5$, so the sum is $(3\cdot1-1)^5=2^5$. That is $32$.

2. Question 2Multiple choiceWritten by IPMATwayBinomial Coefficients and Terms

   The coefficient of the middle term in the expansion of $(x+2y)^6$ is

   1. A.$120$
   2. B.$240$
   3. C.$20$
   4. D.$160$
   Show the answer and solution

   Answer: D. $160$

   $(x+2y)^6$ has $7$ terms, so the single middle term is $T_4$ (here $r=3$). $T_4=\binom{6}{3}x^{3}(2y)^{3}=20\cdot8\,x^3y^3$. So the coefficient of the middle term is $160$. Forgetting the $2^3$ gives the trap $20$.

3. Question 3Type the answerWritten by IPMATwayBinomial Applications

   The value of $\binom{12}{2}+\binom{12}{3}+\cdots+\binom{12}{10}$ is

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   Answer: 4070

   The sum of all coefficients is $\binom{12}{0}+\cdots+\binom{12}{12}=2^{12}=4096$. The required sum leaves out $\binom{12}{0}=1$, $\binom{12}{1}=12$, $\binom{12}{11}=12$ and $\binom{12}{12}=1$, which total $26$. So the sum is $4096-26=4070$.

4. Question 4Type the answerWritten by IPMATwayBinomial Coefficients and Terms

   The number of terms in the expansion of $(x+y+z)^8$ is

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   Answer: 45

   Each term of $(x+y+z)^8$ is $x^ay^bz^c$ with $a+b+c=8$ and $a,b,c\ge0$. The number of such triples is $\binom{8+3-1}{3-1}=\binom{10}{2}$ (stars and bars). $\binom{10}{2}=45$. So there are $45$ terms.

5. Question 5Multiple choiceWritten by IPMATwayBinomial Applications

   The value of $\binom{40}{0}+\binom{41}{1}+\binom{42}{2}+\cdots+\binom{60}{20}$ is

   1. A.$\binom{60}{21}$
   2. B.$\binom{60}{20}$
   3. C.$\binom{61}{21}$
   4. D.$\binom{61}{20}$
   Show the answer and solution

   Answer: D. $\binom{61}{20}$

   Replace $\binom{40}{0}$ by $\binom{41}{0}$ (both are $1$). Then $\binom{41}{0}+\binom{41}{1}=\binom{42}{1}$ by Pascal's rule. Next $\binom{42}{1}+\binom{42}{2}=\binom{43}{2}$, and so on. Each step raises both indices by one. After adding $\binom{60}{20}$ the total becomes $\binom{61}{20}$.

6. Question 6Type the answerWritten by IPMATwayBinomial Applications

   The remainder when $5^{99}$ is divided by $13$ is

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   Answer: 8

   Note $5^2=25=26-1\equiv-1\pmod{13}$. Write $5^{99}=5\cdot5^{98}=5\cdot(25)^{49}=5\cdot(26-1)^{49}$. Every term of $(26-1)^{49}$ except $(-1)^{49}$ has a factor $13$. So $5^{99}\equiv5\cdot(-1)=-5\pmod{13}$. A negative remainder is shifted up by $13$: $-5+13=8$.

7. Question 7Multiple choiceWritten by IPMATwayBinomial Coefficients and Terms

   In the expansion of $(1+ax)^n$ the first three terms are $1$, $12x$ and $64x^2$. Then $n$ is

   1. A.$8$
   2. B.$6$
   3. C.$12$
   4. D.$9$
   Show the answer and solution

   Answer: D. $9$

   $(1+ax)^n=1+nax+\binom{n}{2}a^2x^2+\cdots$. So $na=12$ and $\frac{n(n-1)}{2}a^2=64$. Divide the square of the first by the second: $\frac{n^2a^2}{\frac{n(n-1)}{2}a^2}=\frac{2n}{n-1}=\frac{144}{64}=\frac{9}{4}$. Then $8n=9n-9$, so $n=9$ and $a=\frac{12}{9}=\frac{4}{3}$. Check: $\binom{9}{2}\cdot\frac{16}{9}=36\cdot\frac{16}{9}=64$.

8. Question 8Multiple choiceWritten by IPMATwayBinomial Coefficients and Terms

   In the expansion of $(a-b)^n$, $n\ge5$, the sum of the $5$th and $6$th terms is zero. Then $\frac{a}{b}$ equals

   1. A.$\frac{5}{n-4}$
   2. B.$\frac{n-5}{6}$
   3. C.$\frac{n-4}{5}$
   4. D.$\frac{6}{n-5}$
   Show the answer and solution

   Answer: C. $\frac{n-4}{5}$

   In $(a-b)^n$, $T_5=\binom{n}{4}a^{n-4}(-b)^4=\binom{n}{4}a^{n-4}b^4$ and $T_6=\binom{n}{5}a^{n-5}(-b)^5=-\binom{n}{5}a^{n-5}b^5$. Setting $T_5+T_6=0$ and dividing by $a^{n-5}b^4$ gives $\binom{n}{4}a=\binom{n}{5}b$. So $\frac{a}{b}=\frac{\binom{n}{5}}{\binom{n}{4}}=\frac{n-4}{5}$.

9. Question 9Multiple choiceWritten by IPMATwayBinomial Coefficients and Terms

   The number of terms in the expansion of $\left[(x+2y)^2(x-2y)^2\right]^5$ is

   1. A.$21$
   2. B.$25$
   3. C.$6$
   4. D.$11$
   Show the answer and solution

   Answer: D. $11$

   Combine first: $(x+2y)^2(x-2y)^2=[(x+2y)(x-2y)]^2=(x^2-4y^2)^2$. So the expression is $\left[(x^2-4y^2)^2\right]^5=(x^2-4y^2)^{10}$. A binomial to the power $10$ has $10+1=11$ terms.

10. Question 10Multiple choiceWritten by IPMATwayBinomial Applications

    Which is greater: $101^{50}$ or $100^{50}+99^{50}$?

    1. A.they are equal
    2. B.$101^{50}$
    3. C.$100^{50}+99^{50}$
    4. D.it cannot be decided without a calculator
    Show the answer and solution

    Answer: B. $101^{50}$

    Compare $101^{50}-99^{50}$ with $100^{50}$. Write $101=100+1$ and $99=100-1$. In $(100+1)^{50}-(100-1)^{50}$ the even-index terms cancel and the odd ones double: it equals $2\left[\binom{50}{1}100^{49}+\binom{50}{3}100^{47}+\cdots\right]$. The first term alone is $2\cdot50\cdot100^{49}=100^{50}$, and the rest are positive. So $101^{50}-99^{50}>100^{50}$, i.e. $101^{50}>100^{50}+99^{50}$.