4 February 2025 · Shift 2

BODMAS with signed numbers BODMAS fixes the order of work: Brackets, Of, Division, Multiplication, Addition, Subtraction - every division and multiplication is cleared before any addition or subtraction. Expression: $\left(-9\right)-\left(-60\right)÷\left(-12\right)+\left(-3\right)\times 9$ Division first: $\left(-60\right)÷\left(-12\right)=+5$, because a negative divided by a negative gives a positive. Multiplication next: $\left(-3\right)\times 9=-27$. The expression now reads -9-5+(-27) =-9-5-27=-41 Hence the value is -41, so the correct option is (c).

Simple and compound interest For 2 years at $R\%$ per annum, compound interest is the same as a single rate of $\left(2R+\frac{R^2}{100}\right)\%$, while simple interest is a flat $2R\%$. Here R=4, so the two-year CI rate $=2\left(4\right)+\frac{4^2}{100}=8+0.16=8.16\%$ and the two-year SI rate $=2\left(4\right)=8\%$. Find the principal from the compound interest: $8.16\%$ of P=153 $P=\frac{153\times 100}{8.16}=1875$, so the sum invested is ₹1,875. Simple interest $=\frac{P\times R\times T}{100}=\frac{1875\times 4\times 2}{100}=150$ Shortcut worth remembering: the principal is the same for both, so $SI=CI\times \frac{8}{8.16}=153\times \frac{8}{8.16}=150$. Hence the corresponding simple interest is ₹150, so the correct option is (c).

Simplification of brackets Peel the brackets from the inside outwards - round $\left(\right)$ first, then curly $\left\{\right\}$, then square $\left(\right)$ - applying BODMAS inside each one. Innermost round bracket: 8-6=2 Inside the curly bracket, do the division and multiplication first: $\frac{91}{7}=13$ and $\frac{16}{2}\times 2=8\times 2=16$ So the curly bracket =13+16=29 The other round bracket: $84÷6=14$ Finally the square bracket $=14\times 29=406$ Hence the value is 406, so the correct option is (d).

Simple interest rate Simple interest is $SI=\frac{P\times R\times T}{100}$, so when R and T are the same for both people the interest is proportional to the principal. The whole of the extra interest therefore comes from the extra principal: 9400-5400=₹4000. $\frac{4000\times R\times 4}{100}=480$ 160R=480 $R=\frac{480}{160}=3$ Hence the rate of interest is $3\%$ per annum, so the correct option is (b).

Time and work, LCM method Take the total work as the LCM of the three times, so that every man's daily work is a whole number: LCM of 12, 6 and 15 is 60 units. One day's work: Rajeshwar $=\frac{60}{12}=5$ units, Vikram $=\frac{60}{6}=10$ units, Tiger $=\frac{60}{15}=4$ units. Let the work be finished in D days. Rajeshwar works only the first 2 days, Vikram works (D-3) days and Tiger works right through, all D days. Sum of the work done = total work: $5\times 2+10\times \left(D-3\right)+4\times D=60$ 10+10D-30+4D=60 14D-20=60 14D=80 $D=\frac{80}{14}=\frac{40}{7}=5\frac{5}{7}$ Hence the work is completed in $5\frac{5}{7}$ days, so the correct option is (c).

Percentage, pass marks Let the maximum marks be M and the pass mark be p; write the same p twice, once from each candidate. The first candidate is 68 short of passing: $p=\frac{25M}{100}+68$ The second candidate is 42 above the pass mark: $p=\frac{50M}{100}-42$ Equate the two expressions: $\frac{25M}{100}+68=\frac{50M}{100}-42$ $68+42=\frac{50M-25M}{100}$ $110=\frac{25M}{100}$ $M=\frac{110\times 100}{25}=440$ Hence the maximum marks are 440, so the correct option is (b).

Selling price from profit per cent Profit per cent is always reckoned on the cost price, so $SP=\frac{\left(100+P\right)\times CP}{100}$, where P is the profit per cent. Here CP=520 and P=10. $SP=\frac{\left(100+10\right)\times 520}{100}$ $SP=\frac{110\times 520}{100}=\frac{57200}{100}=572$ Check: $10\%$ of 520=52, and 520+52=572. Hence the article must be sold for ₹572, so the correct option is (b).

Volume of spheres The same material is reshaped, so the volume of the big laddoo equals the total volume of the small ones. Volume of a sphere $=\frac{4}{3}\pi r^3$. Let n small laddoos be made: $\frac{4}{3}\pi {\left(7\right)}^3=n\times \frac{4}{3}\pi {\left(3.5\right)}^3$ $\frac{4}{3}\pi$ cancels from both sides: $n=\frac{{\left(7\right)}^3}{{\left(3.5\right)}^3}={\left(\frac{7}{3.5}\right)}^3$ $n={\left(2\right)}^3=8$ Rule to reuse: volume varies as the cube of the radius, so halving the radius makes the volume $\frac{1}{8}$ and exactly 8 small laddoos are obtained. Hence 8 laddoos can be made, so the correct option is (c).

Successive discounts Two successive discounts of $x\%$ and $y\%$ together equal one single discount of $\left(x+y-\frac{xy}{100}\right)\%$; the last term is the discount that is lost because the second cut is taken on an already reduced price. Here x=25 and y=15: single equivalent discount $=25+15-\frac{25\times 15}{100}$ $=40-\frac{375}{100}=40-3.75=36.25\%$ The rival offer is a flat $40\%$ discount. Difference $=40\%-36.25\%=3.75\%$ Hence the difference between the two discounts is $3.75\%$, so the correct option is (d).

Average speed Average speed is total distance divided by total time; it is never the plain average of the two speeds. First part: distance = speed $\times$ time $=60\times 2=120$ km, taking 2 hours. Second part: time $=\frac{80}{40}=2$ hours for the 80 km. Total distance =120+80=200 km Total time =2+2=4 hours Average speed $=\frac{200}{4}=50$ Hence the average speed is 50 km/hr, so the correct option is (d).

Profit from false weight He charges the cost price, so his whole gain comes from the goods he holds back while weighing. Take the marked weight as 1000 g; the real weight is 26% less, so the goods handed over =1000-260=740 g. The customer pays the cost of 1000 g but takes home only the cost of 740 g. Gain =1000-740=260 g of goods, while the shopkeeper's own cost is that of the 740 g he parted with. Rule for a short weight: profit percentage $=\frac{\mathrm{error}}{\mathrm{true}\mathrm{weight}\mathrm{given}}\times 100$ $=\frac{260}{740}\times 100=\frac{1300}{37}=35\frac{5}{37}$ So his profit is $35\frac{5}{37}$, and the correct option is (c).

LCM by prime factors Break each number into prime factors, then take the highest power of every prime that appears. $672=2^5\times 3\times 7$ $7056=2^4\times 3^2\times 7^2$ LCM $=2^5\times 3^2\times 7^2=32\times 9\times 49$ $=32\times 441=14112$ Check: $14112÷672=21$ and $14112÷7056=2$, both whole numbers. So the LCM is 14112, and the correct option is (a).

Compound interest ratio Both sons end with the same amount, but Rishi's money grows for one year longer, so Pravin must have started with more. Compound amount: $A=P{\left(1+\frac{r}{100}\right)}^n$, and here r=10, so the yearly factor is $\frac{11}{10}$. Equal amounts give $P_1{\left(\frac{11}{10}\right)}^{17}=P_2{\left(\frac{11}{10}\right)}^{18}$ Divide both sides by ${\left(\frac{11}{10}\right)}^{17}$: $P_1=P_2\times \frac{11}{10}$ So Pravin : Rishi =11:10 (the shorter time always carries the larger principal). Total parts =11+10=21, and these 21 parts stand for ₹1617, so one part $=\frac{1617}{21}=77$. Pravin's share $=11\times 77=847$ So Nitin gave ₹847 to Pravin, and the correct option is (d).

Trains crossing each other Two trains running in opposite directions approach each other, so their speeds are added. Relative speed =60+110=170 km/h $=170\times \frac{5}{18}=\frac{425}{9}$ m/s Crossing each other means covering the sum of the two lengths. Sum of the lengths = relative speed $\times$ time $=\frac{425}{9}\times 21=\frac{8925}{9}=991.66$ m Length of the other train =991.66-360=631.66 m So the other train is 631.66 m long, and the correct option is (b).

Price, consumption, expenditure Expenditure = price $\times$ consumption, so fix easy starting numbers: price =100 and consumption =100, giving expenditure =10000. New price $=100\times \frac{160}{100}=160$ The new expenditure is to be 12% more $=10000\times \frac{112}{100}=11200$ New consumption $=\frac{\mathrm{expenditure}}{\mathrm{price}}=\frac{11200}{160}=70$ Fall in consumption =100-70=30 Percentage reduction $=\frac{30}{100}\times 100=30$ So the consumption must be cut by 30%, and the correct option is (c).

Average with a new member Total = average $\times$ number of people, so work with totals rather than averages. Total age of the students $=48\times 22=1056$ years With the teacher there are 49 people, so their total age $=49\times 23=1127$ years Teacher's age =1127-1056=71 years Shortcut: teacher's age = new average + (rise in average) $\times$ (number of students) $=23+1\times 48=71$ So the teacher is 71 years old, and the correct option is (b).

Provisions and man-days Measure the food in man-days: food = number of men $\times$ number of days. Food still left after 15 days $=150\times \left(85-15\right)=150\times 70=10500$ man-days Men remaining =150-25=125 Days the food now lasts $=\frac{10500}{125}=84$ So the remaining food lasts 84 more days, and the correct option is (b).

BODMAS simplification Follow BODMAS: division and multiplication are settled first (left to right), and only then addition and subtraction. Division: $10÷5=2$ Multiplication: $2\times 3=6$ The expression becomes 16+2-6 16+2-6=18-6=12 So the value is 12, and the correct option is (c).

Time to cross each other The trains move towards each other, so the speeds are added to get the relative speed. Relative speed =50+70=120 km/h $=120\times \frac{5}{18}=\frac{100}{3}$ m/s To cross completely they must cover both lengths =180+180=360 m Time $=\frac{\mathrm{distance}}{\mathrm{relative}\mathrm{speed}}=\frac{360}{100/3}=360\times \frac{3}{100}$ $=\frac{1080}{100}=10.8$ seconds So they cross each other in 10.8 seconds, and the correct option is (b).

Ratio from equal products When several products are equal, put the common value equal to the LCM of the coefficients; the divisions then stay whole. LCM of $7,6$ and 12 is 84, so take 7A=6B=12C=84. $A=\frac{84}{7}=12$, $B=\frac{84}{6}=14$, $C=\frac{84}{12}=7$ Hence A:B:C=12:14:7 Check: $7\times 12=6\times 14=12\times 7=84$, so the correct option is (b).