17 February 2025 · Shift 1

Train crossing a tunnel To cross a tunnel the train must cover the length of the tunnel PLUS its own length, because the last coach is clear only after it leaves the far end. Total distance =238+702=940 m Speed =36 km/h $=36\times \frac{5}{18}=10$ m/s Time $=\frac{\mathrm{distance}}{\mathrm{speed}}=\frac{940}{10}$ Time =94 seconds So the correct option is (a).

Alligation on a mixture Alligation compares COST prices, so first convert the selling price of the mixture into its cost price. $CP=\frac{SP\times 100}{100+\mathrm{gain}}=\frac{5\times 100}{125}=4$, i.e. ₹4 per kg By alligation, cheaper : dearer =(4.60-4.00):(4.00-3.80) =0.60:0.20=3:1 The dearer sugar (₹4.60) is the 1 part and it weighs 60 kg, so 1 part =60 kg. Cheaper sugar $=3\times 60=180$ kg So the correct option is (b).

Work on alternate days Q alone takes 56 days, so let the total work be 56 units and Q's one day work be 1 unit. Q begins, so Q works on days 1, 3, 5, … and R on days 2, 4, 6, … 37.5 days means 37 whole days and half of the 38th day, and day 38 is R's turn. Q's days $=1,3,\dots ,37$, that is 19 days; R's days $=2,4,\dots ,36$ plus half of day 38, that is 18+0.5=18.5 days. Work done by Q $=19\times 1=19$ units Work done by R =56-19=37 units Efficiency of R $=\frac{37}{18.5}=2$ units per day Half of the work $=\frac{56}{2}=28$ units, so R needs $\frac{28}{2}=14$ days. So the correct option is (c).

Third proportional The third proportional to a and b is the number c for which a:b=b:c. Cross-multiplying, $ac=b^2$, so $c=\frac{b^2}{a}$ $c=\frac{{\left(0.52\right)}^2}{0.13}=\frac{0.2704}{0.13}$ c=2.08 So the correct option is (d).

BODMAS with integers By BODMAS, division and multiplication are done first, and only then the addition and subtraction. $\left(-60\right)÷\left(-12\right)=5$ (a negative divided by a negative is positive) $\left(-3\right)\times 8=-24$ The expression becomes (-9)-5+(-24) =-9-5-24=-38 So the correct option is (a).

Simple interest, principal In simple interest the same amount is added every year, so 3 years at 8% per annum give 24% of the principal. $SI=\frac{P\times R\times T}{100}$ $486=\frac{P\times 8\times 3}{100}=\frac{24P}{100}$ $P=\frac{486\times 100}{24}$ $P=\frac{48600}{24}=2025$ So the correct option is (d).

LCM by prime factors Break every number into prime factors; the LCM keeps the HIGHEST power of each prime that appears. $18=2\times 3^2$ $72=2^3\times 3^2$ $184=2^3\times 23$ $144=2^4\times 3^2$ $LCM=2^4\times 3^2\times 23$ $=16\times 9\times 23=3312$ So the correct option is (c).

Percentage, literacy Find the three head counts first - females, all literates, literate males - then subtract. Females =340000-217000=123000 Total literates $=\frac{11}{100}\times 340000=37400$ Literate males $=\frac{11}{100}\times 217000=23870$ Literate females =37400-23870=13530 Required percentage $=\frac{13530}{123000}\times 100=11$ So the correct option is (d).

BODMAS order of operations By BODMAS the division and the multiplication are settled before the addition and subtraction. $10÷5=2$ $3\times 3=9$ The expression becomes 16+2-9 =18-9=9 So the correct option is (b).

Average to total weight Averages cannot be added, but totals can - so turn every average into a total first. $C+V+M=46\times 3=138$ $C+V=40\times 2=80$ $V+M=44\times 2=88$ (C+V)+(V+M)=80+88=168 This sum counts Vipul twice, so 168=(C+V+M)+V=138+V V=168-138=30 So the correct option is (b).

CI − SI difference for 3 years Only the interest earned ON the interest survives in the difference, which for three years is a fixed formula. $CI-SI=P{\left(\frac{R}{100}\right)}^2\left(3+\frac{R}{100}\right)$ $122=P\times {\left(\frac{5}{100}\right)}^2\times \left(3+\frac{5}{100}\right)$ $122=P\times \frac{1}{400}\times \frac{61}{20}=\frac{61P}{8000}$ $P=\frac{122\times 8000}{61}$ P=16000 So the correct option is (a).

Packing cubes in a box Check first that the 4 cm cube fits a whole number of times along every edge - it does, so no space is wasted. Along the depth: $36÷4=9$ Along the height: $16÷4=4$ Along the width: $12÷4=3$ Number of cubes $=9\times 4\times 3=108$ So the correct option is (c).

Simplification with brackets Open the innermost bracket first, then the curly bracket, and only then the square bracket. 9-6=3 $\frac{14}{3}\times 3=14$ $\frac{56}{4}=14$ $\left(14+14\right)=28$ $60÷5=12$ $12\times 28=336$ So the correct option is (b).

Circular track, first meeting They all start together at the SAME point, so they are together again only at a moment when each of them has completed a whole number of laps. Time for one lap by M $=\frac{150}{20}=7.5$ s Time for one lap by N $=\frac{150}{15}=10$ s Time for one lap by P $=\frac{150}{12}=12.5$ s Required time = LCM of $7.5,10,12.5=\frac{LCM\left(15,20,25\right)}{2}$ $=\frac{300}{2}=150$ s Check: $150÷7.5=20$, $150÷10=15$, $150÷12.5=12$ laps - all whole numbers. So the correct option is (c).

Profit given as selling price Let the selling price of one chair be S and its cost price be C. Gain on 153 chairs =153S-153C This gain equals the selling price of 68 chairs, so 153S-153C=68S 85S=153C $\frac{S}{C}=\frac{153}{85}=\frac{9}{5}$ Profit per cent $=\frac{S-C}{C}\times 100=\left(\frac{9}{5}-1\right)\times 100$ $=\frac{4}{5}\times 100=80$ So the correct option is (a).

CI − SI difference for 3 years The three-year gap between compound and simple interest is only the interest earned on earlier interest. $CI-SI=P{\left(\frac{R}{100}\right)}^2\left(3+\frac{R}{100}\right)$ $186=P\times {\left(\frac{10}{100}\right)}^2\times \left(3+\frac{10}{100}\right)$ $186=P\times \frac{1}{100}\times \frac{31}{10}=\frac{31P}{1000}$ $P=\frac{186\times 1000}{31}$ P=6000 So the correct option is (d).

Selling price for a given profit Profit is always reckoned on the cost price, so the selling price must be 120% of ₹540. $SP=CP\times \frac{100+20}{100}=540\times \frac{120}{100}$ $SP=540\times \frac{6}{5}=648$ So the correct option is (a).

Cost change of a mixture Take 100 units of the medicine, so that it holds 60 units of S and 40 units of T. Initial cost $=60\times 30+40\times 30=1800+1200=3000$ New cost $=60\times 40+40\times 20=2400+800=3200$ The new cost is the larger one, so the cost INCREASES - that alone rules out both 'decrease' options. Change =3200-3000=200 Percentage change $=\frac{200}{3000}\times 100=\frac{20}{3}=6.67$ So the correct option is (d).

Percentage of a number '4% of x' means $\frac{4}{100}\times x$. $\frac{4}{100}\times x=60$ $x=\frac{60\times 100}{4}$ x=1500 So the correct option is (c).

Successive discounts Successive discounts are never added: the second discount is taken on what is LEFT after the first. Let the marked price be 100. After the first discount: $100\times \frac{88}{100}=88$ After the second discount: $88\times \frac{82}{100}=72.16$ Single equivalent discount =100-72.16=27.84 (Adding them would give 30, which is why that option is a trap.) So the correct option is (b).