Basic learning: $1 + 0.5 = 1.5$ This can be written as a fraction: $1.5 = 1 \frac{1}{2}$ This can also be written as: $1 \frac{1}{2} = 1 + \frac{1}{2}$ <hr class="my-[18px]" node="[object Object]"> Hence, question states: $999\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + 999\frac{4}{7} + 999\frac{5}{7} + 999\frac{6}{7}$ This can be written as: $999 + \frac{1}{7} + 999 + \frac{2}{7} + 999 + \frac{3}{7} + 999 + \frac{4}{7} + 999 + \frac{5}{7} + 999 + \frac{6}{7}$ We can add the fractions separately. Also, $999$ appears six times: $\Rightarrow 6 \times 999 + \left( \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} \right)$ $\Rightarrow 5994 + \left( \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} \right)$ $\Rightarrow 5994 + \frac{21}{7}$ $\Rightarrow 5994 + 3 = \boxed{5997}$

There will be two cases: <ol node="[object Object]"> <li node="[object Object]"> A speaks the truth and B lies </li> <li node="[object Object]"> A lies and B speaks the truth. </li> </ol> Hence, $\Rightarrow [75\% \times (100-80)\%]+[(100-75)\%\times80\%]$ $\Rightarrow \dfrac{75\times20+25\times80}{10000}=\dfrac{3500}{1000}=\dfrac{35}{100}=35\%$

For equation: $ax^2+bx+c=0$ Sum of roots $=-\frac{b}{a}$ Product of roots $= \frac{c}{a}$ Let's assume the equation here to be $x^2+bx+c=0$, we need not worry ourself with the value of $a$ as it doesn't matter here! Amit got the constant term wrong, but he has gotten the coefficient of $x$ right. So from his equation, we can derive $-b=4+3=7 \Rightarrow \boxed{-b=7}$ <hr class="my-[18px]" node="[object Object]"> Similarly, Alok has gotten the coefficient of $x$ wrong, but he has gotten the constant term right. So from his equation, we can derive $c=3\times2=6 \Rightarrow \boxed{c=6}$ <hr class="my-[18px]" node="[object Object]"> Now we know sum of roots is $7$ and product of roots is $6$. We can go through the options to see what matches these (add both the numbers for sum of roots; multiply both the numbers for product). Roots as per Option 2: $(6,1)$ Sum $= 6+1 = 7$ Product $= 6 \times 1 = 6$

Note: Repetition is not allowed For a three-digit even, the unit place has two possibilities. Case 1: The unit place is 2 i.e. _ _ 2 Now, Hundreds place can be filled in 4 ways (1,3,4,5) and Tens place can be filled in 3 ways. Case 1 $=4 \times 3 = 12$ ways Case 2: The unit place is 4 i.e. _ _ 4 Now, Hundreds place can be filled in 4 ways (1,2,3,5) and Tens place can be filled in 3 ways. Case 2 $=4 \times 3 = 12$ ways Total cases $= 12 + 12 = 24$ Hence, the correct option is Option 3.

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/J7gM8tVBm84wuco.webp">The above diagram shows the rectangle of dimensions $25 \times 20$ cm.When squares of side $2$ cm are cut out from each side.The new length $= 25 - 2 -2 = 21$ cmThe new breadth $= 20 - 2 -2 = 16$ cmWhen folded, the height will be $2$ cmHence, the volume will be $21 \times 16 \times 2 = 672 \text{ cm}^3$Hence, the correct option is Option 2.

Let's simplify : $2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}$ $= 2^{122} + 2^{2 \times{62}} + 2^{3 \times{42}} + 2^{2 \times {64}} + 2^{130}$ $= 2^{122} + 2^{124} + 2^{126} + 2^{128} + 2^{130}$ $= 2^{122} + 2^{122} \times 2^2 + 2^{122} \times 2^4 + 2^{122} \times 2^6 +2^{122} \times 2^8$ $= 2^{122} (1 + 4 + 16 + 64 + 256)$ $= 2^{122} (341)$ $341$ is divisible by $11$ , hence, entire expression will be divisible by 11. Hence, the correct answer is Option 4.

The cost of fencing an equilateral triangular park and a square park is the same, implying the perimeter of both parts is equal. Let $a$ be the side of the equilateral triangular park and $s$ be the side of the square park. Thus, $3a = 4s \rightarrow s = \dfrac{3}{4}a$ Now, area of triangular park : $\dfrac{\sqrt3}{4} a^2 = 16 \sqrt3$ Hence, $a^2 = 64$ $a = 8$ units (since side of an equilateral triangle can't be negative) Thus, $s = \dfrac{3}{4} \times 8 = 6$ units. Now, length of diagonal $= \sqrt 2 \times s = \sqrt2 \times 6 = 6\sqrt2$ units Hence, the correct option is Option 4.

Let the sum of money be $x$. The rate is $10 \%$ per annum, compounded semi-annually, so the rate per half-year is $5\%$, The number of periods in $1.5$ years is $3$. Using the compound interest formula: $CI =$ $x\left(1+\dfrac{5}{100}\right)^{3} -x$ $CI =x\left(\dfrac{21}{20}\right)^{3}-x$ $CI=x\left(\dfrac{441 \times 21}{8000}\right)-x$ $CI=x\left(\dfrac{9261}{8000}\right)-x$ $CI=x\left(\dfrac{1261}{8000}\right)$ Now, simple interest: $SI = \dfrac{x \cdot 3 \cdot 5}{100}=\dfrac{15 x}{100}=\dfrac{3 x}{20}$. The difference between Cl and SI is given as ₹183, so: $\Rightarrow \dfrac{1261 x}{8000}-\dfrac{3 x}{20}=183$ $\Rightarrow \dfrac{1261 x-1200 x}{8000}=183$ $\Rightarrow \dfrac{61 x}{8000}=183$ $\Rightarrow x=3 \times 8000=24,000$ Hence, the correct option is Option 2.

It is best to first check the (B) statement from the given options. B. $(243)^{0.16} \times (9)^{0.1} = 0.3$ LHS = $(3)^{5 \times{0.16}} \times (3)^{2 \times {0.1}} = 3^{(0.80)} \times 3^{(0.20)} = 3^1 = 3$ Thus, B is an incorrect statement. Hence, the only possible option is Option 4. <hr class="my-[18px]" node="[object Object]"> A. If $2^x = 3^y = 6^{-z}$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ Given: $2^{x}=3^{y}=6^{-z}$ Let: $2^{x}=3^{y}=6^{-z}=k \quad \text { (say) }$ Now take logs: <ul node="[object Object]"> <li node="[object Object]">From $2^{x}=k$, we get $x=\log _{2} k$ <ul node="[object Object]"> <li node="[object Object]">From $3^{y}=k$, we get $y=\log _{3} k$</li> <li node="[object Object]">From $6^{-z}=k$, take log: $-z=\log _{6} k \Rightarrow z=-\log _{6} k$</li> </ul> </li> </ul> Now evaluate: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{\log _{2} k}+\frac{1}{\log _{3} k}+\frac{1}{-\log _{6} k}$ Use change of base: $\frac{1}{\log _{b} k}=\log _{k} b$ So: $=\log _{k} 2+\log _{k} 3-\log _{k} 6=\log _{k}\left(\frac{2 \times 3}{6}\right)=\log _{k}(1)=0$ <hr class="my-[18px]" node="[object Object]"> C. $3.105 \times 10^{P}=0.00239+0.000715=0.003105$ Thus, $3.105 \times 10^{P}=0.003105 \Rightarrow 10^{P}=\frac{0.003105}{3.105}=10^{-3} \Rightarrow P=-3$

Let $x, y$ and $z$ be the no. of litres purchased in three consecutive years, respectively. <table class="question-table"><tbody><tr><th>Years</th><th>1</th><th>2</th><th>3</th></tr><tr><td node="[object Object]">No. of litres</td><td node="[object Object]">x</td><td node="[object Object]">y</td><td node="[object Object]">z</td></tr><tr><td node="[object Object]">Rate</td><td node="[object Object]">17</td><td node="[object Object]">19</td><td node="[object Object]">20</td></tr><tr><td node="[object Object]">Total Expense</td><td node="[object Object]">6460</td><td node="[object Object]">6460</td><td node="[object Object]">6460</td></tr></tbody></table> From the above table, $x = \dfrac{6460}{17} = 380$ $y = \dfrac{6460}{19} = 340$ $z= \dfrac{6460}{20} = 323$ Thus, the average cost per litre $= \dfrac{\text{Total expense}}{\text{Total No. of litres}} = \dfrac{6460 + 6460 + 6460}{380 + 340 +323}$ $= \dfrac{3 \times 6460}{1043} = \dfrac{19380}{1043} = \dfrac{19.38}{1.043}$ From here, we understand that the average has to be lower than $₹ 19.38$. However, we still have two options, $₹18.49$ or $₹18.58$ To get a unique answer, let's assume the average is $₹18.5$ Thus, total expense $= 18.5 \times 1043 = ₹ 19295.5$ which is lower than actual expense of $₹ 19380$ Hence, the average is greater than $18.5$, which gives us $18.58$ as the unique solution. Hence, the correct option is Option 2.

jipmat 2025 Complete Paper Solution | QA

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

jipmat 2025 Complete Paper Solution | QA

Question 1.

<div><div class="latex-wrapper">The value of $999\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + 999\frac{4}{7} + 999\frac{5}{7} + 999\frac{6}{7}$ is equal to :<br /></div></div>

Question 2.

<div><div class="latex-wrapper">A speaks truth in 75% cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident?<br /></div></div>

Question 3.

Question 4.

<div><div class="latex-wrapper">How many three-digit even numbers can be formed using the digits 1, 2, 3, 4 and 5, when repetition of digits is not allowed?<br /></div></div>

Question 5.

<div><div class="latex-wrapper">From the four corners of a rectangular sheet of dimensions 25 cm $\times$ 20 cm, square of side 2 cm is cut off from four corners and a box is made. The volume of the box is</div></div>

Question 6.

<div><div class="latex-wrapper">$2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}$ is divisible by which one of the following integers?<br /></div></div>

Question 7.

<div><div class="latex-wrapper">The cost of fencing of an equilateral triangular park and a square park is the same. If the area of the triangular park is $16\sqrt{3}$ m$^2$, then the length of the diagonal of the square park is :<br /></div></div>

Question 8.

<div><div class="latex-wrapper">The difference between compound and simple interests on a certain sum of money at the interest rate of $10\%$ per annum for $1\frac{1}{2}$ years is $₹183$, when the interest is compounded semi-annually, then the sum of money is :<br /></div></div>

Question 9.

<div><div class="latex-wrapper">Which of the following statements is/are correct?<br />A. If $2^x = 3^y = 6^{-z}$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$<br />B. $(243)^{0.16} \times (9)^{0.1} = 0.3$<br />C. If $3.105 \times 10^P = 0.00239 + 0.000715$, then P = -3<br /></div></div>

Question 10.

<div><div class="latex-wrapper">A car owner buys petrol at the rate ₹ 17, ₹ 19 and ₹ 20 per litre, respectively for three consecutive years. Compute the average cost per litre, if he spends ₹ 6,460 per year for the three consecutive years.<br /></div></div>