2 solved Identities previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.
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Q1:jipmat 2025QA › IdentitiesHardQA · MCQ
(3.07−2.15)(2.15−4.53)(4.53−3.07)2+(2.15−4.53)(4.53−3.07)(3.07−2.15)2+(4.53−3.07)(3.07−2.15)(2.15−4.53)2 is simplified to
A0
B1
C2
D3
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The Setup: Doing the decimals here would be a colossal waste of time. This is a classic algebraic identity in disguise. Let's sub in variables to reveal the true form.
Step 1: Let x=4.53−3.07, let y=3.07−2.15, and let z=2.15−4.53.
Notice that if you add them all up, everything cancels out perfectly:
x+y+z=(4.53−3.07)+(3.07−2.15)+(2.15−4.53)=0.
Step 2: Rewrite the absolute monstrosity of an expression using x,y,z.
E=yzx2+zxy2+xyz2Step 3: Find a common denominator (xyz) and add the fractions.
E=xyzx3+xyzy3+xyzz3=xyzx3+y3+z3Step 4: Deploy the master identity. If x+y+z=0, then x3+y3+z3=3xyz.
Substitute 3xyz into the numerator:
E=xyz3xyz=3Final Answer: 3
Q2:jipmat 2025QA › IdentitiesMediumQA · MCQ
If x3+y3=468 and x+y=12, then value of x4+y4 will be
A3620
B2036
C3025
D3026
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The Setup: We are stepping up the algebraic ladder here. We need to find xy, use it to find x2+y2, and then square that to finally unlock x4+y4. It's a three-phase boss fight.
Step 1: Find xy using the sum of cubes identity.
x3+y3=(x+y)((x+y)2−3xy)468=12(122−3xy)468=12(144−3xy)
Divide by 12:
39=144−3xy⟹3xy=105⟹xy=35Step 2: Find x2+y2.
x2+y2=(x+y)2−2xyx2+y2=(12)2−2(35)x2+y2=144−70=74Step 3: Find x4+y4 by squaring the squares.
(x2+y2)2=x4+y4+2x2y2
We can rewrite this as x4+y4=(x2+y2)2−2(xy)2.
Substitute the values we've farmed:
x4+y4=(74)2−2(35)2742=(70+4)2=4900+560+16=5476.
2(35)2=2(1225)=2450.
x4+y4=5476−2450=3026Final Answer: 3026