Quantitative Aptitude Test - 7

First let us find out sum of all the natural numbers between 30 and 260

Total number of terms= 260-30-1= 229 (subtracting 1 because exclusive of the end)

Using the formula for finding out sum of first n terms of an Arithmetic Progression

Now finding out the sum of all the numbers between 30 and 260 which are divisible by 6

i.e. 36 + 42 +…….+ 246 + 252 + 258

Using the formula for the n­­th term of an A.P.

t_n^th =a + (n-1) x d {where a = first term and d = common difference}

258 = 36 +(n-1) x 6

∴ n=38

Now finding out the sum of all the numbers between 30 and 260 which are divisible by 9

i.e. 36 + 45 +…….+ 243 + 252

Using the formula for the n­­th term of an A.P.

t_n^th =a + (n-1) x d {where a = first term and d = common difference}

252 = 36 +(n-1) x 9

∴ n=25

Now finding out the sum of all the numbers between 30 and 260 which are divisible by both 6 and 9

i.e. 36 + 54 +……. + 252

Using the formula for the n­­th term of an A.P.

t_n^th =a + (n-1) x d {where a = first term and d = common difference}

252 = 36 +(n-1) x 18

∴ n=13

Now, Subtracting Eq. 2 , 3 and 4 from 1, to get the answer

∴Sum of all the terms between 30 and 260,neither divisible by 6 nor by 9 is=

= 33205 – (5586+ 3600 -1872)

= 25891

Hence mark option b

When the time is constant, Distance is directly proportional to the speed

When A finishes the race, distance covered will be 400m

Since A gives B a start of 50m and beats him by 30m, distance covered will be 400m - 50m - 30 m= 320m

Ratio of speed is same as ratio of distance

Hence mark option a

Any point on the circumference always makes an angle of 90° with the diameter

∴ Using Pythagoras theorem, Hypotenuse² = Base² + Perpendicular²

∴ YZ² = 15² - 12² = 9²

Hence mark option c

There are 4 ways to give each chocolate (as Priya has 4 friends)

Hence there are 4 x 4 x 4 x 4 x 4 x…………………….(15 times) =4¹⁵ ways

If a 31 day month starts either on Friday or Saturday or Sunday then there are 5 Sundays in that month, otherwise there are 4 Sundays in that month.

The graph x² +(y-1)²=4 is a circle with the center (0, 1) and radius 2

As a curve (x-p)² +(y-q)²=r²

Where (p, q) are the center coordinates and r is the radius of the circle

Hence mark option a

Minimum value of any entity inside modulus function can be zero (as it can never be negative)

∴ Minimum value of x at which the expression 10 + |3x-5|, the value of 3x-5=0

Hence mark option b