6 February 2025 · Shift 3

LCM of numbers The smallest number divisible by several numbers is their LCM. $24=2^3\times 3$, $58=2\times 29$, $48=2^4\times 3$, $12=2^2\times 3$ LCM $=2^4\times 3\times 29$ $=16\times 3\times 29$ $=48\times 29$ =1392 Hence the required smallest natural number is 1392.

Compound interest ratio Let Rishi get ₹x, so Shan gets ₹(1612-x). Both amounts are equal, so $x{\left(1+\frac{8}{100}\right)}^{17}=\left(1612-x\right){\left(1+\frac{8}{100}\right)}^{18}$ $x=\left(1612-x\right)\times \frac{27}{25}$ $25x=27\times 1612-27x$ 52x=43524 $x=\frac{43524}{52}=837$ Hence Pravin gave ₹837 to Rishi.

Scheme discount The buyer pays for 20 pencils but takes home 25 pencils. So 5 pencils out of the 25 received are free. Discount $\%=\frac{\mathrm{free}}{\mathrm{total}\mathrm{received}}\times 100$ $=\frac{5}{25}\times 100$ $=20\%$

Percentage of a number $2\%$ of x=288 $\frac{2}{100}\times x=288$ $x=288\times \frac{100}{2}$ $x=288\times 50$ x=14400

Surface area of cube Total surface area of a cube $=6a^2$ Here a=19 cm $=6\times {\left(19\right)}^2$ $=6\times 361$ =2166 ${\mathrm{cm}}^2$

Loss percentage When there is a loss, $SP=CP\times \frac{100-\mathrm{loss}\%}{100}$ $SP=78200\times \frac{100-44}{100}$ $=78200\times \frac{56}{100}$ $=782\times 56$ =43792 Hence the selling price is ₹43,792.

BODMAS simplification By BODMAS, do division and multiplication before addition and subtraction. $16+12÷4-3\times 3$ =16+3-9 =19-9 =10

Average speed Total distance =150+250=400 km Total time =4+4=8 hours Average speed $=\frac{\mathrm{total}\mathrm{distance}}{\mathrm{total}\mathrm{time}}$ $=\frac{400}{8}$ =50 km/hr

Mean proportional The mean proportional between a and b is $\sqrt{ab}$. $=\sqrt{36\times 9}$ $=\sqrt{324}$ =18 Check: 36:18=18:9=2:1

Simple interest rate Simple interest $=\frac{P\times R\times T}{100}$ The rate and time are the same, so the extra interest comes only from the extra principal. Extra principal =9400-5400=₹4000 $960=\frac{4000\times R\times 4}{100}$ 960=160R $R=\frac{960}{160}=6$ Hence the rate of interest is $6\%$ per annum.

Profit and selling price Cost price of 100 pears =₹1230 Cost price of 400 pears $=4\times 1230=₹4920$ Selling price of 400 pears =4920+980=₹5900 Selling price of 1 pear $=\frac{5900}{400}=₹14.75$ Selling price per dozen $=14.75\times 12$ =₹177

Difference of CI and SI For 2 years, CI - SI =45-44=₹1 Simple interest for 1 year $=\frac{44}{2}=₹22$ The extra ₹1 is one year's interest on that first year's interest of ₹22. $1=22\times \frac{R}{100}$ $R=\frac{100}{22}=\frac{50}{11}\%$ $P=\frac{SI\times 100}{R\times T}=\frac{44\times 100}{\frac{50}{11}\times 2}$ $=\frac{4400\times 11}{100}$ =484 Hence the principal is ₹484.

Trains crossing each other Moving in opposite directions, the speeds add up. Relative speed =60+50=110 km/h $=110\times \frac{5}{18}=\frac{275}{9}$ m/s Sum of the lengths = relative speed $\times$ time $=\frac{275}{9}\times 44$ $=\frac{12100}{9}=1344.44$ m Length of the other train =1344.44-430 =914.44 m

Combined average Total spent by group A $=49\times 106=₹5194$ Total spent by group B $=35\times 141=₹4935$ Total spent by group C $=31\times 226=₹7006$ Total members =49+35+31=115 Total amount =5194+4935+7006=₹17135 Overall average $=\frac{17135}{115}$ =149

Average and new member 700 grams =0.7 kg, so the new average =53+0.7=53.7 kg Total weight of 27 students $=27\times 53=1431$ kg Total weight of 28 persons $=28\times 53.7=1503.6$ kg Weight of the teacher =1503.6-1431 =72.6 kg

HCF with remainders The number leaves remainder 6 in 1657 and 5 in 2037, so it divides (1657-6) and (2037-5) exactly. Required number = HCF of 1651 and 2032 $2032=1\times 1651+381$ $1651=4\times 381+127$ $381=3\times 127+0$ HCF =127 Hence the greatest such number is 127.

BODMAS simplification By BODMAS, do division and multiplication before addition and subtraction. $16+9÷3-3\times 4$ =16+3-12 =19-12 =7

Alternate day work Let the total work = LCM of 15 and 20 =60 units. X does $\frac{60}{15}=4$ units per day and Y does $\frac{60}{20}=3$ units per day. In one pair of days (X then Y) =4+3=7 units In 8 such pairs, that is 16 days $=8\times 7=56$ units Remaining work =60-56=4 units The 17th day is X's turn and X does exactly 4 units that day. Hence the work is completed in 17 days.

Comparing ratios Write each ratio as a fraction and compare its decimal value. $\frac{12}{19}=0.63$, $\frac{15}{19}=0.79$, $\frac{21}{26}=0.81$, $\frac{19}{13}=1.46$ In only one ratio is the first term greater than the second, so only that value exceeds 1. Hence 19:13 is the highest ratio.

Successive percentage change Let the original price be ₹100. Multiply by the factor of each change, one after the other. Final price $=100\times \frac{85}{100}\times \frac{60}{100}\times \frac{40}{100}\times \frac{145}{100}$ $=100\times 0.85\times 0.60\times 0.40\times 1.45$ =29.58 Change =29.58-100=-70.42 Required change $=\frac{70.42}{100}\times 100=70.42\%$ decrease