4 February 2025 · Shift 1

LCM by prime factorisation To build an LCM, break every number into primes and then keep the highest power of each prime that appears anywhere. $28=2^2\times 7$ $60=2^2\times 3\times 5$ $120=2^3\times 3\times 5$ $135=3^3\times 5$ The highest powers available are $2^3$, $3^3$, $5^1$ and $7^1$. LCM $=2^3\times 3^3\times 5\times 7=8\times 27\times 35$ $=216\times 35=7560$ So the correct option is (b).

Weighted percentage of marks Percentages of different subjects cannot be averaged directly, because the subjects carry different totals - turn every percentage into marks first. Maximum marks $=3\times 100+2\times 100+1\times 100=600$ Mathematics $=\frac{60}{100}\times 300=180$ English $=\frac{70}{100}\times 200=140$ Science $=\frac{50}{100}\times 100=50$ Total marks scored =180+140+50=370 Overall percentage $=\frac{370}{600}\times 100=\frac{185}{3}=61.67\%$ So the correct option is (c).

Successive percentage change Successive percentage changes multiply, they are never added. Let the original number be x. After a fall of $15\%$ it becomes $x\times \frac{85}{100}$. After a rise of $20\%$ it becomes $x\times \frac{85}{100}\times \frac{120}{100}=\frac{102x}{100}$. Net increase $=\frac{102x}{100}-x=\frac{2x}{100}$ Given that this increase is 78: $\frac{2x}{100}=78$ $x=\frac{78\times 100}{2}=3900$ So the correct option is (a).

Difference of CI and SI In simple interest every year earns the same amount, so one year's interest $=\frac{42}{2}=21$, i.e. ₹21. Compound interest exceeds simple interest only by the second-year interest earned on the first year's interest. Difference =51-42=9, i.e. ₹9. So ₹9 is one year's interest on ₹21: $\frac{21\times R\times 1}{100}=9$ $R=\frac{900}{21}=\frac{300}{7}\%$ Now use $SI=\frac{P\times R\times T}{100}$ with SI=42, T=2: $42=\frac{P\times \frac{300}{7}\times 2}{100}=\frac{6P}{7}$ $P=\frac{42\times 7}{6}=49$ So the principal is ₹49 and the correct option is (b).

Compound interest sharing Ketan's money is invested for 22-11=11 years and Tushar's for 22-12=10 years, and the two maturity amounts are equal. Amount $=P{\left(1+\frac{R}{100}\right)}^n$, and here $1+\frac{8}{100}=\frac{27}{25}$. $K{\left(\frac{27}{25}\right)}^{11}=T{\left(\frac{27}{25}\right)}^{10}$ Cancelling ${\left(\frac{27}{25}\right)}^{10}$ from both sides: $K\times \frac{27}{25}=T$ $\frac{K}{T}=\frac{25}{27}$ - the boy with the extra year of growth simply needs one factor less. Total parts =25+27=52, and these 52 parts stand for ₹24,700. 1 part $=\frac{24700}{52}=475$ Tushar's share $=27\times 475=12825$ So Tushar was given ₹12,825 and the correct option is (c).

Cost price from profit percent Profit percent is always reckoned on the cost price, so the selling price is $\frac{108}{100}$ of the cost price. $SP=CP\times \frac{100+8}{100}$ $1566=CP\times \frac{108}{100}$ $CP=\frac{1566\times 100}{108}=\frac{156600}{108}$ CP=1450 So the cost price is ₹1,450 and the correct option is (c).

Mean proportional The mean proportional between a and b is $\sqrt{ab}$, because it is the middle term x of a:x=x:b. Let the numbers be 4x and 9x. $\sqrt{4x\times 9x}=24$ $\sqrt{36x^2}=24$ $6x=24\Rightarrow x=4$ The numbers are $4\times 4=16$ and $9\times 4=36$. Difference =36-16=20 So the correct option is (d).

Trains crossing in same direction When two bodies move in the same direction, the gap between them closes at the difference of their speeds. Relative speed =150-80=70 km/h Convert to metres per second: $70\times \frac{5}{18}=\frac{175}{9}$ m/s To cross completely, the faster train must cover both lengths: 210+140=350 m. Time $=\frac{\mathrm{distance}}{\mathrm{relative}\mathrm{speed}}=350÷\frac{175}{9}=350\times \frac{9}{175}=18$ s The options are in minutes: $\frac{18}{60}=0.3$ So the correct option is (b).

Price, consumption and expenditure Expenditure = Price $\times$ Consumption, so fix both at 100 and follow what each becomes. Original expenditure $=100\times 100=10000$ New price =100+85=185 New expenditure $=10000\times \frac{148}{100}=14800$ New consumption $=\frac{14800}{185}=80$ Fall in consumption =100-80=20 on a base of 100. Percentage reduction $=\frac{20}{100}\times 100=20\%$ So the correct option is (c).

Time and work Take the total work as the LCM of the two times, so that every day's work is a whole number. Total work =96 units, the LCM of 32 and 48. A's rate $=\frac{96}{32}=3$ units/day and B's rate $=\frac{96}{48}=2$ units/day. Work done together in 8 days $=\left(3+2\right)\times 8=40$ units Remaining work =96-40=56 units $60\%$ of the remaining work $=\frac{60}{100}\times 56=33.6$ units Time taken by B $=\frac{33.6}{2}=16.8=16\frac{4}{5}$ days So the correct option is (c).

BODMAS simplification Simplify by BODMAS: innermost bracket first, then division and multiplication (left to right), then addition. Innermost bracket: 7-3=4 Multiplication inside the curly bracket: $\frac{40}{4}\times 4=10\times 4=40$ Curly bracket: $\frac{49}{7}+40=7+40=47$ Round bracket: $48÷8=6$ Whole expression $=6\times 47=282$, so the correct option is (a).

LCM and divisibility A number divisible by each of 15, 25, 40 and 75 must be a multiple of their LCM, so find the LCM first. $15=3\times 5,25=5^2,40=2^3\times 5,75=3\times 5^2$ LCM $=2^3\times 3\times 5^2=600$ The largest four-digit number is 9999; divide it by 600: $9999=600\times 16+399$ Subtract the remainder: 9999-399=9600 9600 is the largest four-digit multiple of 600, so the correct option is (a).

Volume ratio of cones Volume of a cone: $V=\frac{1}{3}\pi r^{2}h$, so in a ratio the constant $\frac{1}{3}\pi$ cancels and $V\propto r^{2}h$. Take $h_1:h_2=4:3$ and $r_1:r_2=1:2$. $\frac{V_1}{V_2}=\frac{r12h_1}{r22h_2}=\frac{1^2\times 4}{2^2\times 3}$ $=\frac{4}{12}=\frac{1}{3}$ The required ratio is 1:3, so the correct option is (b).

False weight profit He sells at cost price, so his whole gain comes from the goods he holds back - compare what the customer pays for with what he actually hands over. A weight marked 1000 g really weighs $46\%$ less, i.e. 1000-460=540 g; in ratio form the customer pays for 100 units of goods but receives only 54 units. Gain (in goods) =100-54=46 units, and the cost of what he actually gave is 54 units. Profit% = gain ÷ cost of goods actually given $\times 100=\frac{46}{54}\times 100$ $=\frac{4600}{54}=\frac{2300}{27}$ $=85\frac{5}{27}\%$, so the correct option is (b).

Averages and totals Turn every average into a total, because totals can be added and subtracted: total = average × number of persons. $G+A+T=46\times 3=138$ $G+A=40\times 2=80$ $A+T=45\times 2=90$ Add the last two: (G+A)+(A+T)=80+90=170, i.e. (G+A+T)+A=170 138+A=170 A=170-138=32, so Akshay weighs 32 kg and the correct option is (c).

Successive discounts For two successive discounts do not add them - multiply the fractions of the price that still have to be paid: price left $=\left(1-d_1\right)\left(1-d_2\right)$. The smallest 'price left' is the biggest discount. I: $0.80\times 0.80=0.64$, so the net discount is $36\%$ II: $0.75\times 0.85=0.6375$, so the net discount is $36.25\%$ III: $0.70\times 0.90=0.63$, so the net discount is $37\%$ IV: $0.95\times 0.65=0.6175$, so the net discount is $38.25\%$ Useful check: for a fixed sum of the two discounts the offer is better when the two parts are more unequal, which is why $5\%+35\%$ beats $20\%+20\%$. The greatest net discount, $38.25\%$, is offered by scheme IV, so the correct option is (c).

Average speed both ways Average speed is always total distance ÷ total time, never the plain average of the two speeds. The two legs cover the same distance, so the formula reduces to the harmonic mean: average speed $=\frac{2xy}{x+y}$ Substitute x=9 and y=18: $\frac{2\times 9\times 18}{9+18}=\frac{324}{27}$ =12 Check with a convenient distance d=18 km each way: time $=\frac{18}{9}+\frac{18}{18}=2+1=3$ hours for 36 km, giving $\frac{36}{3}=12$ km/hr. So the average speed is 12 km/hr and the correct option is (b).

Simple interest rate Simple interest: $SI=\frac{P\times R\times T}{100}$. Since R and T are the same for both men, the extra interest comes only from the extra principal. Extra principal =9400-5400=4000, i.e. ₹4,000 $\frac{4000\times R\times 5}{100}=840$ 200R=840 $R=\frac{840}{200}=4.2$ So the rate is $4.2\%$ per annum and the correct option is (d).

HCF and co-prime pairs If the HCF of the two numbers is 11, write them as 11a and 11b, where a and b must be co-prime (otherwise the HCF would be more than 11). 11a+11b=132 $a+b=\frac{132}{11}=12$ Co-prime pairs adding to 12 are (1,11) and (5,7); (2,10),(3,9),(4,8),(6,6) all share a factor, so they are rejected. (1,11) gives the numbers 11 and 121, but 11 is not greater than 42 - rejected. (5,7) gives $11\times 5=55$ and $11\times 7=77$, and both are greater than 42. Difference =77-55=22, so the correct option is (b).

Third proportional The third proportional to a and b is the number c for which a:b::b:c, i.e. the middle term is repeated. $\frac{a}{b}=\frac{b}{c}\Rightarrow c=\frac{b^2}{a}$ $c=\frac{{48}^2}{16}=\frac{2304}{16}$ c=144, so the correct option is (a).