UPSC CSE Prelims
Algebra Previous Year Questions (PYQs)
Showing solved Previous Year Questions for Chapter: Algebra
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A toy $T$ jumps forward or backward. In each forward jump, it moves $5'$ forward whereas in each backward jump, it moves $2'$ backward. If in $31$ jumps, $T$ moves exactly $15'$ forward, then what is the difference of the number of forward and backward jumps?
Detailed Explanation:
Let f = forward jumps and b = backward jumps. Total jumps: f + b = 31. Net displacement: 5f - 2b = 15.
Substitute b = 31 - f into the second equation: 5f - 2(31 - f) = 15 → 5f - 62 + 2f = 15 → 7f = 77 → f = 11. Therefore, b = 20.
Therefore, the difference is |11 - 20| = 9, making Option 4 the correct answer.
The weight of $X$, in kg, is denoted by $X$. The weights of $A$, $B$, $C$, $D$, $P$, $Q$, $R$ and $S$ are measured. Given : $$A + B + C + D = 17$$ $$A + C = 6$$ $$P + Q + S + D = 15$$ $$P + Q + R + B = 17$$ $$P = R \text{ and } Q = S$$
Which one of the following statements is correct?
Detailed Explanation:
From A + B + C + D = 17 and A + C = 6, we get B + D = 11.
Using P = R and Q = S, equations (3) and (4) become: P + 2Q + D = 15 and 2P + Q + B = 17. Adding these: 3P + 3Q + (B + D) = 32. Substituting B + D = 11 gives 3(P + Q) = 21, so P + Q = 7.
Since P + Q = 7 and A + C = 6, we have 7 > 6. Options 3 and 4 cannot be determined as individual values of P and Q are unknown.
Therefore, P and Q together weigh more than A and C, making Option 2 the correct answer.
A shopkeeper employs a delivery boy and gives him a motorcycle for home delivery. For every delivery, the boy is given ₹ $5$. At the end of the day, he also gets ₹ $2$ for every kilometre of the distance covered in the day. The boy wants to earn more than ₹ $500$ a day, but does not want to travel more than $100$ km. Which of the following numbers of deliveries would definitely meet his target?
Detailed Explanation:
Earnings formula: E = 5d + 2k, where d = deliveries and k = km travelled. Target: E > 500, with constraint k ≤ 100.
To definitely meet the target using d alone, the inequality must hold even in the worst case (k = 0), requiring 5d > 500 or d > 100. For the given options (d = 80, 85, 90), worst-case earnings are ₹400, ₹425, and ₹450 respectively—all fall short of ₹500. Since actual km travelled is unknown and could be anywhere from 0 to 100, none of the delivery counts guarantees the target.
Therefore, without knowing the distance covered, the question cannot be answered due to insufficient data, making Option 4 the correct answer.
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