Let's say the required number is $n$. When $n$ is divided by $12$, $21$ and $35$, remainder is $6$ in each case. This means: $n = 12k_1 + 6$ $n = 21k_2 + 6$ $n = 35k_3 + 6$ where $k_1$, $k_2$, $k_3$ are some integers Therefore, $n - 6$ is divisible by $12$, $21$ and $35$. In other words, $n - 6$ is divisible by $LCM(12, 21, 35)$ Let's find $LCM(12, 21, 35)$: $12 = 2^2 × 3$ $21 = 3 × 7$ $35 = 5 × 7$ Therefore, $LCM(12, 21, 35) = 420$ So, $n - 6 = 420k$, where $k$ is a non-negative integer The least value of $n$ will be when $k = 1$ Therefore, $n = 420 + 6 = 426$.

<ol node="[object Object]"> <li node="[object Object]">Let's say maximum marks is $M$</li> </ol> <ul node="[object Object]"> <li node="[object Object]">Then $x\%$ of $M = \frac{xM}{100}$</li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">Let pass marks be $P$</li> </ol> <ol start="3" node="[object Object]"> <li node="[object Object]">For first candidate:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">Score = $\frac{xM}{100}$ <ul node="[object Object]"> <li node="[object Object]">Fails by $a$ marks</li> <li node="[object Object]">Therefore: $\frac{xM}{100} = P - a$</li> </ul> </li> </ul> <ol start="4" node="[object Object]"> <li node="[object Object]">For second candidate:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">Score = $\frac{yM}{100}$ <ul node="[object Object]"> <li node="[object Object]">Exceeds by $b$ marks</li> <li node="[object Object]">Therefore: $\frac{yM}{100} = P + b$</li> </ul> </li> </ul> <ol start="5" node="[object Object]"> <li node="[object Object]">From equations:</li> </ol> $\frac{xM}{100} = P - a$ ...(1) $\frac{yM}{100} = P + b$ ...(2) <ol start="6" node="[object Object]"> <li node="[object Object]">Subtracting (1) from (2):</li> </ol> $\frac{(y-x)M}{100} = a+b$ $M = \frac{100(a+b)}{y-x}$ Therefore, maximum marks = $\frac{100(a+b)}{y-x}$

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/YJOXn55yKQWRfhL.webp">When an equilateral triangle in inscribed in a circle, the centroid is the same as the circle's center.Hence, the radius of the circle $=$ vertex of triangle to centroid $= 20 \sqrt{3}$ cm.

Let the initial amount be $P$ and rate of interest be $r\%$ Given that amount doubles in 5 years: $P(1 + \frac{r}{100})^5 = 2P$ $(1 + \frac{r}{100})^5 = 2$ Now, let's say it takes $x$ years to make the amount 16 times: $P(1 + \frac{r}{100})^x = 16P$ $(1 + \frac{r}{100})^x = 16$ We know that $16 = 2^4$ $[(1 + \frac{r}{100})^5]^{\frac{x}{5}} = 2^4$ $2^{\frac{x}{5}} = 2^4$ Therefore: $\frac{x}{5} = 4$ $x = 20$

<ol node="[object Object]"> <li node="[object Object]">For successive discounts, we need to calculate final price after both discounts:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">If original price is $100$, then after two successive discounts of $x\%$ and $y\%$: <ul node="[object Object]"> <li node="[object Object]">Final price = $100 × (1-\frac{x}{100}) × (1-\frac{y}{100})$</li> </ul> </li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">Let's calculate for each option:</li> </ol> A) $25\%$ and $15\%$: = $100 × (1-0.25) × (1-0.15)$ = $100 × 0.75 × 0.85$ = $63.75$ B) $30\%$ and $10\%$: = $100 × (1-0.30) × (1-0.10)$ = $100 × 0.70 × 0.90$ = $63.00$ C) $35\%$ and $5\%$: = $100 × (1-0.35) × (1-0.05)$ = $100 × 0.65 × 0.95$ = $61.75$ D) $15\%$ and $20\%$: = $100 × (1-0.15) × (1-0.20)$ = $100 × 0.85 × 0.80$ = $68.00$ <ol start="3" node="[object Object]"> <li node="[object Object]">Arranging in ascending order (lowest to highest final price):</li> </ol> C ($61.75$) &lt; B ($63.00$) &lt; A ($63.75$) &lt; D ($68.00$) Therefore, the correct order is (C), (B), (A), (D)

<ol node="[object Object]"> <li node="[object Object]">Given:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$\frac{3}{5}$ of articles sold at 20% gain <ul node="[object Object]"> <li node="[object Object]">Remaining $\frac{2}{5}$ sold at cost price (0% gain)</li> </ul> </li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">Let's say total cost price is $100$ for calculation ease</li> </ol> <ol start="3" node="[object Object]"> <li node="[object Object]">At cost price $100$:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$\frac{3}{5}$ part = $60$ worth of goods sold at 20% gain <ul node="[object Object]"> <li node="[object Object]">$\frac{2}{5}$ part = $40$ worth of goods sold at no gain</li> </ul> </li> </ul> <ol start="4" node="[object Object]"> <li node="[object Object]">Calculating gains:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">On $60$ worth of goods: 20% gain = $12$ <ul node="[object Object]"> <li node="[object Object]">On $40$ worth of goods: 0% gain = $0$</li> <li node="[object Object]">Total gain = $12$</li> </ul> </li> </ul> <ol start="5" node="[object Object]"> <li node="[object Object]">Therefore on $100$ cost price, gain = $12$</li> </ol> So gain percentage = $12\%$

Let me solve this step by step. <ol node="[object Object]"> <li node="[object Object]">Given ratios:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$a:b = 5:3$ <ul node="[object Object]"> <li node="[object Object]">$b:c = 2:5$</li> </ul> </li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">From first ratio $\frac{a}{b} = \frac{5}{3}$</li> </ol> <ul node="[object Object]"> <li node="[object Object]">If we take $b = 6$ (to make calculations easier) <ul node="[object Object]"> <li node="[object Object]">Then $a = 10$</li> <li node="[object Object]">These values maintain the ratio $5:3$</li> </ul> </li> </ul> <ol start="3" node="[object Object]"> <li node="[object Object]">From second ratio $\frac{b}{c} = \frac{2}{5}$</li> </ol> <ul node="[object Object]"> <li node="[object Object]">With $b = 6$ <ul node="[object Object]"> <li node="[object Object]">Then $c = 15$</li> <li node="[object Object]">These values maintain the ratio $2:5$</li> </ul> </li> </ul> <ol start="4" node="[object Object]"> <li node="[object Object]">Now we can find $(c-a)$:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$c-a = 15-10 = 5$</li> </ul> <ol start="5" node="[object Object]"> <li node="[object Object]">Let's check Statement I:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$10(a+c)$ <ul node="[object Object]"> <li node="[object Object]">$10(10+15)$</li> <li node="[object Object]">$10(25)$</li> <li node="[object Object]">$250$</li> </ul> </li> </ul> This is not equal to $(c-a) = 5$ <ol start="6" node="[object Object]"> <li node="[object Object]">Let's check Statement II:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$10a+25b$ <ul node="[object Object]"> <li node="[object Object]">$10(10)+25(6)$</li> <li node="[object Object]">$100+150$</li> <li node="[object Object]">$250$</li> </ul> </li> </ul> This is not equal to $(c-a) = 5$ Therefore, both statements are false.

<ol node="[object Object]"> <li node="[object Object]">Given:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">HCF($x,y$) = $6$ <ul node="[object Object]"> <li node="[object Object]">LCM($x,y$) = $210$</li> <li node="[object Object]">$x + y = 72$</li> </ul> </li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">We know that for any two numbers:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$x × y = \text{HCF} × \text{LCM}$ <ul node="[object Object]"> <li node="[object Object]">Therefore, $x × y = 6 × 210 = 1260$</li> </ul> </li> </ul> <ol start="3" node="[object Object]"> <li node="[object Object]">Now we have two equations:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$x + y = 72$ <ul node="[object Object]"> <li node="[object Object]">$x × y = 1260$</li> </ul> </li> </ul> <ol start="4" node="[object Object]"> <li node="[object Object]">To find $\frac{1}{x} + \frac{1}{y}$:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">$\frac{1}{x} + \frac{1}{y} = \frac{y + x}{xy}$ <ul node="[object Object]"> <li node="[object Object]">= $\frac{72}{1260}$</li> <li node="[object Object]">= $\frac{72}{1260}$</li> <li node="[object Object]">= $\frac{36}{630}$</li> <li node="[object Object]">= $\frac{2}{35}$</li> </ul> </li> </ul> Therefore, $\frac{1}{x} + \frac{1}{y} = \frac{2}{35}$

<ol node="[object Object]"> <li node="[object Object]">Let's say first number is n. Then:</li> </ol> <ul node="[object Object]"> <li node="[object Object]">First 5 numbers: n, n+1, n+2, n+3, n+4 <ul node="[object Object]"> <li node="[object Object]">Next 2 numbers: n+5, n+6</li> </ul> </li> </ul> <ol start="2" node="[object Object]"> <li node="[object Object]">First 5 numbers average = a</li> </ol> = $\frac{n + (n+1) + (n+2) + (n+3) + (n+4)}{5} = a$ = $\frac{5n + 10}{5} = a$ = $n + 2 = a$ <ol start="3" node="[object Object]"> <li node="[object Object]">All 7 numbers average:</li> </ol> = $\frac{n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6)}{7}$ = $\frac{7n + 21}{7}$ = $n + 3$ <ol start="4" node="[object Object]"> <li node="[object Object]">Difference between averages:</li> </ol> = New average - Original average = $(n + 3) - (n + 2)$ = $1$ Therefore, the average will increase by 1.

Time is of the essence. For these questions, just solve one or two equations and match them with the options, and mark the answer then and there! Solving the first equation: $\sqrt{15\times163\div5-89}$ Use DMAS - Division, then Multiplication, then Addition, then Subtraction $=\sqrt{15\times32.6-89}=\sqrt{489-89}=\sqrt{400}=20$ So we know that A-(II), which is either Option 2 or Option 3. Onto the second equation: $\sqrt{15^2+11\times3^2}=\sqrt{225+11\times9}=\sqrt{225+99}=\sqrt{324}=18$ Hence, B-(IV), which is Option 3.

jipmat 2024 Complete Paper Solution | QA

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Instructions

Question 10.

jipmat 2024 Complete Paper Solution | QA

Question 1.

<div><div class="latex-wrapper">What will be the least number which when divided by 12, 21 and 35 leaves 6 as a remainder in each case?<br /></div></div>

Question 2.

<div><div class="latex-wrapper">A candidate scoring $x \%$ marks in an examination fails by ' $a$ ' marks while another candidate who scores $y \%$ marks, gets ' $b$ ' marks more than the minimum required pass marks. What is the maximum marks for the examination ?<br /></div></div>

Question 3.

<div><div class="latex-wrapper">An equilateral triangle $A B C$ is inscribed in a circle of radius $20 \sqrt{3} \mathrm{~cm}$. The centroid of the triangle $A B C$ is at a distance $d$ from the vertex $A$. then $d$ is equal to ?</div></div>

Question 4.

<div><div class="latex-wrapper">An amount doubles itself at compound interest in 5 years. How many years will it take to make the amount 16 times itself ?<br /></div></div>

Question 5.

Question 6.

<div><div class="latex-wrapper">A man sold $\frac{3}{5}^{\text {th }}$ of his articles at a gain of $20 \%$ and the remaining at the cost price. The gain earned in the transaction is<br /></div></div>

Question 7.

Question 8.

<div><div class="latex-wrapper">The HCF and LCM of two numbers $\mathrm{x}$ and $\mathrm{y}$ are respectively 6 and 210 .<br />If $x+y=72$, then $\frac{1}{x}+\frac{1}{y}$ is equal to :<br /></div></div>

Question 9.

<div><div class="latex-wrapper">The average of five consecutive numbers is a. If the next two numbers are also included, how will the average vary?<br /></div></div>

Instructions

Question 10.

<div><div class="latex-wrapper">Choose the correct answer from the options given below :<br /></div></div>