EPFO EO/AO 2023 General Ability Test · Question 103 of 120

Correct — A, (a) ₹ 16. The quickest route does not solve the pair of equations at all; it reads what changes between the two baskets. The first basket is ten pens and eight pencils for ₹ 200. The second basket is five pens and twenty-four pencils for the same ₹ 200. Going from the first to the second, the buyer gives up five pens and takes on sixteen extra pencils, and the bill does not move. So the five pens surrendered are worth exactly the sixteen pencils gained: 5 x (price of a pen) = 16 x (price of a pencil). Now substitute into the first basket. Ten pens are two lots of five pens, so ten pens are worth 32 pencils. Ten pens and eight pencils therefore cost the same as 32 + 8 = 40 pencils, and 40 pencils cost ₹ 200, so a pencil costs ₹ 5. Five pens are worth sixteen pencils, that is 16 x 5 = ₹ 80, so one pen costs ₹ 16. The formal route reaches the same place. Writing x for the price of a pen and y for the price of a pencil, the two purchases give 10x + 8y = 200 and 5x + 24y = 200. Halving the first gives 5x + 4y = 100. Subtracting that from the second removes x in one step: 20y = 100, so y = 5, and then 5x = 100 - 20 = 80, so x = 16. Check both baskets before moving on. Ten pens and eight pencils: 10 x 16 + 8 x 5 = 160 + 40 = ₹ 200. Five pens and twenty-four pencils: 5 x 16 + 24 x 5 = 80 + 120 = ₹ 200. Both come to the stated amount, which no other option manages.

• (b)₹ 15 — If a pen cost ₹ 15, the first basket fixes the pencil: 10 x 15 = 150, leaving ₹ 50 for eight pencils, so a pencil would be ₹ 6.25. Test that pair on the second basket: 5 x 15 + 24 x 6.25 = 75 + 150 = ₹ 225, which is more than the ₹ 200 the stem says the second basket also cost. The pair fails, so ₹ 15 is out. Notice the direction of the failure: the lower the pen price, the higher the pencil price forced by the first equation, and because the second basket is pencil-heavy its total rises. That monotonic behaviour means exactly one of the four options can work, which is worth knowing when you decide to test options rather than solve.
• (c)₹ 14 — A pen at ₹ 14 puts 10 x 14 = 140 into the first basket, so eight pencils must account for ₹ 60 and a pencil costs ₹ 7.50. The second basket then comes to 5 x 14 + 24 x 7.50 = 70 + 180 = ₹ 250, half as much again as it should be. The option is attractive to a candidate who has set up the equations correctly but subtracted them in the wrong order, or who has divided 200 by a wrong coefficient at the last step. The safeguard is the substitution check: any candidate answer must satisfy both purchases, not just the one it was derived from.
• (d)₹ 13 — At ₹ 13 a pen, the first basket leaves 200 - 130 = ₹ 70 for eight pencils, so a pencil is ₹ 8.75, and the second basket becomes 5 x 13 + 24 x 8.75 = 65 + 210 = ₹ 275, further still from ₹ 200. This is the cheapest option offered and it fails by the widest margin, which fits the structure of the item: the second basket trades pens for pencils, so any answer that makes pencils dearer than the true ₹ 5 must overshoot. A second sanity check is available without algebra at all. The exchange 5 pens for 16 pencils means a pen is worth 16/5 = 3.2 pencils, so the pen price must be a multiple-of-3.2 relationship with a pencil price that divides ₹ 200 sensibly, and only ₹ 16 with ₹ 5 does that in whole rupees.

Two purchases of the same two items at the same unit prices give two linear equations in two unknowns, and any of three standard methods will solve them: elimination, in which one variable is removed by scaling and subtracting; substitution, in which one equation is rearranged and fed into the other; and cross-multiplication. What makes an examination version quicker than the textbook version is noticing that a pair of equations with the same right-hand side carries an extra piece of information for free. If two different bundles cost the same, then whatever was given up in going from one bundle to the other is exactly equal in value to whatever was taken on. Here the two totals are both ₹ 200, so 5 pens = 16 pencils in value, and the problem collapses into a single-variable one. The same reading works whenever a question says 'using the same amount', 'for the same money' or 'the cost is unchanged'. Where the totals differ, elimination is the reliable route, and the scaling to choose is the one that makes the coefficients of one variable match with the least arithmetic. In every version, the last step is substitution back into both original equations, because that is what distinguishes a solved system from a plausible-looking answer.

Simultaneous linear equations dressed as a shopping problem are a staple of the quantitative blocks in the EO/AO General Ability Test, partly because they can be solved in three lines and partly because they punish a candidate who reaches for algebra without first reading the structure of the data. The examiner has arranged the numbers so that both totals are ₹ 200, and that repetition is not decoration: it is the shortcut. The habit rewarded is to look for what the two statements share before writing down a single symbol, and then to verify the answer against both statements rather than the one it came from.

• Two bundles that cost the same amount give a direct exchange rate: what is given up equals in value what is taken on.
• Here 5 pens = 16 pencils in value, because that is the only change between the two ₹ 200 baskets.
• The equations are 10x + 8y = 200 and 5x + 24y = 200, with x the pen price and y the pencil price.
• Halving the first gives 5x + 4y = 100; subtracting it from the second gives 20y = 100, so a pencil is ₹ 5.
• A pen is therefore ₹ 16, and 10 x 16 + 8 x 5 = 200 while 5 x 16 + 24 x 5 = 200.
• A pair of linear equations in two unknowns has a unique solution unless the two lines are parallel or identical.
• Substituting a candidate answer back into both original conditions is the check that eliminates every wrong option here.

• Solving one equation and never testing the answer against the other.
• Scaling only one equation and then subtracting as though both had been scaled.
• Assuming the prices must be whole rupees; here they are, but the check must not rest on that assumption.
• Missing the shortcut hidden in the fact that both baskets cost the same amount.
• Answering with the pencil price after solving for it first; the question asks for the price of a pen.

EO/AO papers set this family either as two purchases of the same two goods, as here, or as one purchase plus a stated relation between the two prices, such as 'a pen costs three times a pencil'. Occasionally the second condition is hidden in a phrase like 'using the same amount' or 'for the same money', which is a same-total condition and therefore an exchange rate. Read for the second condition first, decide whether it is an equation or a ratio, and then eliminate. Testing the four options against the stem is a legitimate second method here, because the totals move monotonically with the pen price and only one option can survive.

No directly related past PYQ was found.

• practice — not a real PYQ

  A shopper buys eight apples and six oranges for ₹ 96. With the same ₹ 96 she could instead have bought four apples and fifteen oranges of the same kinds. What is the price of one apple ?

  • (a)₹ 6
  • (b)₹ 8
  • (c)₹ 9
  • (d)₹ 12

  Answer(c) ₹ 9

• practice — not a real PYQ

  Three chairs and two tables cost ₹ 4,500, while two chairs and three tables cost ₹ 5,000. What is the cost of one table ?

  • (a)₹ 900
  • (b)₹ 1,100
  • (c)₹ 1,200
  • (d)₹ 1,400

  Answer(c) ₹ 1,200