# Tag Info

## New answers tagged quantitative-aptitude

0

Remaining payment= $14000\$$. Let each monthly instalment be x. Monthly rate of interest=\frac{30}{100*12}=\frac1{40} So, 14000+\frac{14000*5}{40}=(x+\frac{x*4}{40})+(x+\frac{x*3}{40})+(x+\frac{x*2}{40})+(x+\frac{x*1}{40})+x\implies 14000+1750=5x+\frac{10x}{40}\implies 15750=\frac{21x}{4}\implies x=3000 0 Cash price = 34000\$$, Cash down payment =$20000\$$, Balance to be paid in 5 equal instalments = 14000\$$, Let each instalment be x. So, interest charged under instalment plan = $(5x – 14000)$. The buyer owes to the seller for $1st$ month=$14000$, $2nd$ month=$(14000 –x)$, $3rd$ month=$(14000 –2x)$, $4th$ month=$(14000 –3x)$, $5th$ month= $(14000 –4x)$ ...

3

This kind of problem can be easier to handle if you rotate the clock backwards just fast enough to stop the hour hand. Then the minute hand will be seen to rotate at $330^{\circ}$ per hour, and the second hand at $60 \times 360 - 30 = 21570^{\circ}$ per hour. The minute hand is at the $120^{\circ}$ position at $\frac{360m+120}{330}$ hours, and the second ...

1

A reasonable reading of the question is whether there are any times where the hands are spaced at angles of $120^\circ$. All angles will be in degrees. The hour hand moves $\frac 1{120}$ per second, the minute hand moves $\frac 1{10}$ per second and the second hand moves 6 per second. Starting at noon, the minute hand gains $\frac {11}{120}$ per second ...

0

Here’s a strategy, not a solution, always assuming that the hands are of equal length and the “triangle” in question is formed by the tips. The desired condition happens when the angles between different hands are $120^\circ$. Now it’s a standard high-school algebra problem to determine when there’s a $120^\circ$ angle from the hour to the minute hand, it ...

1

You’ve set it up wrong, I’m afraid, because you’re not being careful enough with your indices. Let $n$ be the number of innings that he batted before the last match, and for $k=1,\ldots,n+2$ let $x_k$ be his score in the $k$-th innings. Then $\sum_{k=1}^nx_k=750$, more or less as you had it, and his total for the $n+2$ innings including those of the last ...

1

Sum of the entries in first column$=2+7+8+3+5=25$. And that of second column$=3+6+7+0+4=20$ So, Sum of entries of third column$=15=4+5+1+Z+3\implies z=2$

0

B + C + D + E = 4A ------1 C + F = 3A ----------------2 C + D + E = 2F -----------3 F = 2D --------------------4 E + F = 2C + 1-----------5 Also A can 13,17 or 19 Eliminating D from all the equations we get B+2F = 4A ------6 C+F = 3A -------7 C+E = 3F/2 -----8 E+F = 2C+1 ----9 Adding 7 and 8 2C+E+F = 3(A+F/2) 4C+1 = 3(A+F/2) Substituting value ...

3

Notice that for each number listed, the sum of the first and last digit is the sum of the middle two digits. The only number in the options satisfying this property is $3456$. Notice the question is not concerned with the sequence of numbers, but the set of numbers.

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