21 February 2025 · Shift 1

Average speed Average speed is total distance divided by total time - never the plain average of the two speeds, because the two stretches take different times. Time for the first stretch $=\frac{120}{48}=2.5$ hours Time for the second stretch $=\frac{180}{45}=4$ hours Total distance =120+180=300 km and total time =2.5+4=6.5 hours Average speed $=\frac{300}{6.5}=\frac{3000}{65}=46.15$ km/hr So the correct option is (a).

Compound interest ratio Equal amounts with the same rate mean the money invested for the shorter time must be the larger principal. Let Vijay get V (for 19 years) and Ajay get A (for 20 years). Amount $=P{\left(1+\frac{R}{100}\right)}^T$, so $V{\left(\frac{11}{10}\right)}^{19}=A{\left(\frac{11}{10}\right)}^{20}$ $\frac{V}{A}={\left(\frac{11}{10}\right)}^{20-19}=\frac{11}{10}$ So Vijay : Ajay =11:10, a total of 11+10=21 parts. 21 parts $=1617\Rightarrow 1$ part $=\frac{1617}{21}=77$ Vijay's share $=11\times 77=847$, i.e. ₹847 So the correct option is (c).

BODMAS rule By BODMAS, division and multiplication are settled before addition and subtraction. $16+12÷4-2\times 3$ =16+3-6 =19-6=13 So the correct option is (c).

Third proportional The third proportional x to a and b satisfies a:b=b:x, so $x=\frac{b^2}{a}$. Here a=16 and b=28. $x=\frac{{28}^2}{16}=\frac{784}{16}=49$ Number to be added =49-40=9 So the correct option is (b).

Averages of groups Turn every average into a total, because totals can be added and subtracted but averages cannot. $T+S+M=3\times 45=135$ $T+S=2\times 43=86$ $S+M=2\times 44=88$ Subtracting the second from the first: M=135-86=49 From the third: S=88-M=88-49=39 So the correct option is (c).

Work done in a cycle Take the total work as the LCM of 5, 8 and 12, that is 120 units. Ajit does $\frac{120}{5}=24$ units a day, Bimal $\frac{120}{8}=15$ units a day, and Chandan removes $\frac{120}{12}=10$ units a day. One cycle of 3 days =24+15-10=29 units. In 9 days (3 cycles) $=3\times 29=87$ units are done, so 120-87=33 units remain. The work cannot finish earlier: the highest running total before day 10 is 97 units, on day 8. Day 10 belongs to Ajit: 87+24=111 units, leaving 120-111=9 units. Day 11 belongs to Bimal, who needs $\frac{9}{15}=\frac{3}{5}$ of that day. Total time $=10+\frac{3}{5}=10\frac{3}{5}$ days So the correct option is (b).

BODMAS with brackets Open the brackets from the inside out: first $\left(\right)$, then $\left\{\right\}$, then $\left(\right)$. $79-2\times \left(10+8\times \{16-8\left(6-5\right)\times 2\}÷37\right)$ $=79-2\times \left(10+8\times \{16-8\times 1\times 2\}÷37\right)$ $=79-2\times \left(10+8\times \left\{16-16\right\}÷37\right)$ $=79-2\times \left(10+8\times 0÷37\right)=79-2\times \left(10+0\right)$ =79-20=59 So the correct option is (b).

BODMAS with fractions Clear the innermost bracket first, then the curly bracket, and multiply last. $\left(\left(16÷8\right)\times \left(\frac{40}{2}+\frac{11}{6}\times \left(8-2\right)\right)\right)$ $=\left(2\times \left(20+\frac{11}{6}\times 6\right)\right)$ $=\left(2\times \{20+11\}\right)$ $=2\times 31=62$ So the correct option is (b).

Simple interest Formula: $SI=\frac{P\times R\times T}{100}$ $540=\frac{P\times 2\times 4}{100}=\frac{8P}{100}$ $P=\frac{540\times 100}{8}$ $P=\frac{54000}{8}=6750$, i.e. ₹6,750 So the correct option is (c).

Third proportional ratio The third proportional to a and b is $\frac{b^2}{a}$. $T_1=\frac{{36}^2}{16}=\frac{1296}{16}=81$ $T_2=\frac{{60}^2}{24}=\frac{3600}{24}=150$ $T_1:T_2=81:150$ Dividing both terms by 3: =27:50 So the correct option is (c).

Compound interest ratio Baban's money grows from age 11 to 24, that is for 24-11=13 years. Rakesh's money grows from age 12 to 24, that is for 24-12=12 years. Equal amounts: $B{\left(1+\frac{5}{100}\right)}^{13}=R{\left(1+\frac{5}{100}\right)}^{12}$ $\frac{B}{R}={\left(\frac{21}{20}\right)}^{12-13}=\frac{20}{21}$ So Baban : Rakesh =20:21, a total of 20+21=41 parts. 41 parts $=24600\Rightarrow 1$ part $=\frac{24600}{41}=600$ Rakesh's share $=21\times 600=12600$, i.e. ₹12,600 So the correct option is (a).

HCF of three numbers The greatest number that divides all three exactly is their HCF, so factorise each. $210=2\times 3\times 5\times 7$ $1245=3\times 5\times 83$ $2160=2^4\times 3^3\times 5$ Only 3 and 5 appear in all three, each to the first power. HCF $=3\times 5=15$ 20, 25 and 10 are all even or multiples of 25, so they cannot divide 1245. So the correct option is (b).

Profit and loss A loss of 20% means the selling price is only 80% of the cost price, so recover the cost price first. CP of 50 bananas $=\frac{160\times 100}{80}=200$, i.e. ₹200 CP of 1 banana $=\frac{200}{50}=4$, i.e. ₹4 For a 10% profit, SP of 1 banana $=4\times \frac{110}{100}=4.40$, i.e. ₹4.40 Number of bananas sold for ₹1,540 $=\frac{1540}{4.4}=350$ So the correct option is (d).

False weight profit He charges the price of the written weight but parts with less goods, so his profit sits on the shortfall. For every 1000 g written on the weight, the customer actually receives $1000-\frac{48}{100}\times 1000=520$ g. The shopkeeper therefore takes the cost of 1000 g while giving away goods that cost him 520 g. Profit is always measured on what he gave up, i.e. on 520 g. Profit $\%=\frac{1000-520}{520}\times 100=\frac{480}{520}\times 100$ $=\frac{1200}{13}=92\frac{4}{13}\%$ So the correct option is (d).

Mark-up and discount Fix the cost price at ₹100 - the answer is a percentage, so any convenient cost price works. For a 25% gain, SP $=100\times \frac{125}{100}=125$ A 20% discount means the SP is 80% of the marked price: $125=\frac{80}{100}\times MP$ $MP=\frac{125\times 100}{80}=156.25$ Mark-up above cost =156.25-100=56.25, i.e. $56.25\%$ So the correct option is (d).

Average of a remaining group Convert each average into a total weight, subtract, and then average again. Total weight of the class $=50\times 34=1700$ kg Total weight of 39 students $=39\times 32=1248$ kg Total weight of the other 11 =1700-1248=452 kg Required average $=\frac{452}{11}=41.09$ kg So the correct option is (b).

Speed and time inverse The distance is fixed, so time is inversely proportional to speed. Speed becomes $\frac{1}{4}$ of normal, hence time becomes 4 times normal. Let the normal time be t minutes; the new time is 4t. Delay =4t-t=3t $3t=51\Rightarrow t=\frac{51}{3}=17$ minutes So the correct option is (b).

Price, consumption, expenditure Expenditure = price $\times$ consumption, so compare the multipliers. Take price =100 and consumption =100, so expenditure $=100\times 100=10000$. New price =150 and new expenditure $=10000\times \frac{132}{100}=13200$ New consumption $=\frac{13200}{150}=88$ Reduction in consumption =100-88=12 Required reduction $\%=\frac{12}{100}\times 100=12\%$ So the correct option is (b).

Area of a triangle Formula: area of a triangle $=\frac{1}{2}\times b\times h$, where b is the base and h the height. The ratio b:h=5:4, so put b=5k and h=4k. $\frac{1}{2}\times 5k\times 4k=250$ $10k^2=250\Rightarrow k^2=25\Rightarrow k=5$ Base $=5k=5\times 5=25$ cm So the correct option is (d).

Constant area percentage Area = length $\times$ breadth is to stay the same, so breadth is inversely proportional to length. Length, old : new =100:115=20:23 Hence breadth, old : new =23:20 Decrease in breadth =23-20=3 parts out of the original 23. Required decrease $\%=\frac{3}{23}\times 100=\frac{300}{23}$ $=13.04\%$ So the correct option is (c).