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JIPMAT Linear Equations — PYPs

3 solved Linear Equations previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

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Q1:jipmat 2025QA › Linear EquationsMediumQA · MCQ

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:

A5997
B5979
C5994
D2997

Pick an option to attempt

The Setup: We need to evaluate the sum of six mixed fractions. Doing this the traditional way by converting to improper fractions is major NPC energy. Let's break it down using the property of mixed fractions:

A\frac{B}{C}=A+\frac{B}{C}

. Step 1: Separate the whole numbers from the fractions. Since

999

appears

6

times, we can group them up.

S=(999 \times 6)+(\frac{1}{7}+\frac{2}{7}+\frac{3}{7}+\frac{4}{7}+\frac{5}{7}+\frac{6}{7})

Step 2: Calculate the whole number part. Let's use girl math:

999

is practically

1000

999 \times 6=(1000-1) \times 6=6000-6=5994

Step 3: Sum up the fractions. The denominators are all matching vibes (they are all

7

\frac{1+2+3+4+5+6}{7}=\frac{21}{7}=3

Step 4: Add the two parts together for the ultimate W.

5994+3=5997

Final Answer: 5997

Q2:jipmat 2025QA › Linear EquationsMediumQA · MCQ

Which of the following statements is/are correct? A. If

2^x=3^y=6^{-z}

, then

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0

(243)^{0.16} \times (9)^{0.1}=0.3

C. If

3.105 \times 10^P=0.00239+0.000715

, then

P=-3

AB only
BB and C only
CA and B only
DA and C only

Pick an option to attempt

The Setup: We've got a classic 'two truths and a lie' situation. Let's audit each statement mathematically to see which ones are actually valid and which one is straight capping. Step 1: Fact-check Statement A. Let

2^x=3^y=6^{-z}=k

. This implies

2=k^{\frac{1}{x}}

3=k^{\frac{1}{y}}

, and

6=k^{-\frac{1}{z}}

. We know

2 \times 3=6

. Substituting the

k

terms:

k^{\frac{1}{x}} \times k^{\frac{1}{y}}=k^{-\frac{1}{z}}

k^{\frac{1}{x}+\frac{1}{y}}=k^{-\frac{1}{z}}

Equating the powers:

\frac{1}{x}+\frac{1}{y}=-\frac{1}{z} \implies \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0

. Statement A is a W (True). Step 2: Fact-check Statement B. Convert bases to powers of

3

243=3^5

and

9=3^2

(3^5)^{0.16} \times (3^2)^{0.1}=3^{0.8} \times 3^{0.2}

3^{0.8+0.2}=3^1=3

The statement claims it equals

0.3

. Total cap. Statement B is False. Step 3: Fact-check Statement C. Add the decimals on the right side:

0.00239+0.000715=0.003105

. Equation:

3.105 \times 10^P=0.003105

. To get from

3.105

0.003105

, we move the decimal left by

3

places, meaning multiplying by

10^{-3}

. So

P=-3

. Statement C is True. Final Answer: A and C only

Q3:jipmat 2025QA › Linear EquationsMediumQA · MCQ

a+b=2c

, then the value of

\frac{a}{a-c}+\frac{b}{b-c}

A1/2
B1
C2
D3

Pick an option to attempt

The Setup: You could do full algebra to solve this, but picking smart numbers is a massive brain play because if the expression holds true for variables, it holds true for any real numbers that satisfy the condition. Step 1: Choose values for

a

b

, and

c

that satisfy

a+b=2c

. Let's make sure the denominators don't become zero (

a \neq c

b \neq c

). Let

a=3

and

b=1

3+1=4

, so

2c=4 \implies c=2

. Step 2: Plug these values straight into the expression.

E=\frac{a}{a-c}+\frac{b}{b-c}

E=\frac{3}{3-2}+\frac{1}{1-2}

Step 3: Evaluate.

E=\frac{3}{1}+\frac{1}{-1}

E=3-1=2

*Bonus Algebra Flex:* If you rearrange

a+b=2c

, you get

b-c=c-a

. Substitute

b-c

with

-(a-c)

. The expression becomes

\frac{a}{a-c}-\frac{b}{a-c}=\frac{a-b}{a-c}

. Since

b=2c-a

, then

a-b=a-(2c-a)=2a-2c=2(a-c)

. So the fraction simplifies perfectly to

\frac{2(a-c)}{a-c}=2

. Final Answer: 2

JIPMAT Linear Equations — PYPs

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