Mix Test 10

The smallest number which is divided by 35, 56 and 91 is LCM (35, 56, 91).

Write 35, 56 and 91 in the prime factorization.

35 = 5 \(\times\) 7

56 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 7 = \(2 ^3\) \(\times\) 7

And 91 = 7 \(\times\) 13.

\(\therefore\) LCM (35, 56, 91) = 5 \(\times\) 7 \(\times\) \(2 ^3\) \(\times\) 13 = 35 \(\times\) 8 \(\times\) 13 = 3640.

Hence, the smallest number which is divided by 35, 56 and 91 is 3640.

\(\therefore\) The smallest number that when divided by 35, 56 and 91 leaves the same remainder 7 in each case = 3640 + 7 = 3647.

Given quadratic polynomial is \(\mathrm{4x^2-4x-3}\)

For zeros of quadratic polynomial, \(\mathrm{4x^2-4x-3} = 0\)

\(\Rightarrow \mathrm{4x^2 - 6x+2x-3=0}\)

\(\Rightarrow \mathrm{2x (2x-3)+1(2x-3)=0}\)

\(\Rightarrow \mathrm{(2x+1)(2x-3)=0}\)

\(\Rightarrow 2\mathrm x + 1=0\) or \(2\mathrm x-3=0\)

\(\Rightarrow \mathrm x=-\frac{1}{2}\) or \(\mathrm x=\frac{3}{2}\)

Hence, the zeros of the quadratic polynomial are \(-\frac{1}{2}\) and \(\frac{3}{2}\).

Relation between mean, median and mode is

Mode = 3 median – 2mean. ...........(1)

Given that median and mode of frequency distribution are 26 and 29 respectively

i.e., median = 26 and mode = 29.

Now, putting the values of median and mode in equation (1), we get \(29 = 3\times 26-2\) Mean

\(\Rightarrow \) 2 Mean = 78 - 29 = 49

\(\Rightarrow \) Mean \(= \frac{49}{2}=24.5\)

Hence, the mean of the frequency distribution is 24.5.

\(\because \) The edge of cubical ice-cream brick is a = 22 cm.

\(\therefore\) Surface area of ice-cream cube 

\(=6a^2=6\times 22^2\)

\(=6\times 484 = 2904\,cm^2\)

(\(\because \) Surface area of the cube is \(6a^2\))

\(\because \) The edge of cubical ice-cream brick is a = 22 cm.

\(\therefore\) Volume of ice-cream cube \(=a^3=22^3cm^3\) (\(\because \) The volume of the cube is \(a^3\))

Given that radius and height of the ice-cream cone are 2cm and 7cm, respectively.

\(\therefore \) Volume of each cone 

\(=\frac{1}{3}\pi r^2 h\)

\(=\frac{1}{3}\times \frac{22}{7}\times 2\times 2\times 7\)

\(=\frac{22\times 4}{3}=\frac{88}{3}cm^3\)

\(\left(\because r= 2cm, h = 7\,cm \,\&\, \pi = \frac{22}{7}\right) \)

\(\because\) The edge of cubical ice-cream brick is a =22 cm.

\(\therefore\) Volume of ice-cream cube \(= a^3 = 22^3\,cm^3\) (\(\because\) The volume of the cube is \(a^3\))

Given that radius and height of the ice-cream cone are 2cm and 7cm, respectively.

The volume of each cone \(=\frac{1}{3}\pi r^2h\) \(=\frac{1}{3}\times \frac{22}{7}\times 2\times 2 \times 7\) \(=\frac{22\times 4}{3}=\frac{88}{3}cm^3\)

\((\because r = 2cm, h = 7cm\,\&\, \pi = \frac{22}{7})\)

Let n number of children will get the ice-cream cones which are filled from cubical ice-brick.

\(\therefore\) n \(\times\) Volume of one cone = Volume of ice-cream cube

\(\Rightarrow n\times \frac{88}{3}=22^3\) (\(\because\) The volume of ice– cream cube \(=22^3\) & the volume of one ice cone \(=\frac{88}{3}\))

\(\Rightarrow n = \frac{22\times 22\times 22\times 3}{22\times 4}\) \(=11\times 11\times 3 = 363\)

\(\therefore\) Total ice-cones are 363.

\(\therefore\) The number of children who will get the ice-cream cones is 363.

Given that radius and height of the ice-cream cone are 2cm and 7cm, respectively.

\(\therefore\) The slant height of the cone is \(l=\sqrt {r^2+h^2}\) \(=\sqrt{2^2+7^2}\) \(=\sqrt{4+49}\) \(=\sqrt{53} \,cm\)

Let the radius of the wheel is r cm.

Given that the diameter of the wheel is 2r = 84 cm.

\(\therefore\) The radius of the wheel is r = 42 cm.

Total distance cover by wheel in a single revolution is the circumference of the wheel = 2πr \(= 2\times \frac{22}{7}\times 42\) \(=2\times 22\times 6\) \(=12\times 22\) \(=264\,cm\)

Let n revolutions are required to cover the distance 792 m = 79200 cm

\(\therefore\) \(n\times 264 = 79200\)

\(\Rightarrow n =\frac{79200}{264}=300\)

Hence, 300 revolutions of wheels are required to cover the distance 792m.

Given that lady has total 50 coins in her purse.

Let lady has x 25-paisa coins.

\(\therefore\) Lady has ‘‘50 – x’’ 50-paisa coin in her purse.

(\(\because\) Lady has only 25 paisa and 50 paisa coins and total coins are 50 in her purse)

Given that totaling of the value of these 50 coins are 19.50 rupees = 1950 paisa,

\(\therefore\) 25x + (50 – x)50 = 1950

\(\Rightarrow\) 25x + 2500 – 50x = 1950

\(\Rightarrow\) 25x = 2500 –1950 = 550

\(\Rightarrow\) x \(=\frac{550}{25}=22\)

Hence, the lady has total 22 coins of “25-paisa”.

Let the radius of the base of original cylinder is r.

And the height of the original cylinder is h.

\(\therefore\) The volume of the original cylinder is \(\pi r^2 h\)

Given that radius of base of reduced cylinder is half of radius of original cylinder and height will be same.

\(\therefore\) The radius of reduced cylinder is \(\frac{r}{2}\)

And the height of the reduced cylinder is h.

\(\therefore\) The volume of the reduced cylinder is \(\pi\left(\frac{r}{2}\right)^2h\)

Now, the ratio of the volume of reduced cylinder and the volume of original cylinder is

\(\frac{Volume\,of\,reduced\,cylinder}{Volume\,of\,original\,cylinder}\) \(=\frac{\pi\left(\frac{r}{2}\right)^2h}{\pi r^2 h}\) \(=\frac{\pi r^2 h}{4\pi r^2 h }=\frac{1}{4}\)

Hence, the required ratio is 1 : 4.

Let numbers a, b and c are in AP.

\(\therefore\) 2b = a + c.  ...........(1)

Given that sum of these numbers is 3.

\(\therefore\) a + b + c = 3

\(\Rightarrow\) 2b + b = 3  (From equation (1), a + c = 2b)

\(\Rightarrow\) 3b = 3 \(\Rightarrow\) b = 1.

Now putting b = 1, in equation (1), we get

a + c = 2.  ............(2)

Given that the product of these numbers is –35.

\(\therefore\) abc = –35

\(\Rightarrow\) ac = –35. (\(\because\) b = 1)

Now \((a-c)^2\) \(=(a+c)^2 - 4ac\) \(=2^2 - (4\times -35)\) \(=4+140 = 144\)

(\(\because\) a + c = 2, ac = -35)

\(\Rightarrow\) a – c = \(\sqrt{144}\) = 12

\(\Rightarrow\) a – c = 12. .............(3)

Now, adding equations (2) and (3), we get (a + c)+(a – c) = 2 + 12 = 14

\(\Rightarrow\) 2a = 14

\(\Rightarrow\) a = \(\frac{14}{2}=7\)

Now, putting a = 7 in equation (2), we get 7 + c = 2

\(\Rightarrow\) c = 2 –7 = –5.

\(\therefore\) a = 7, b = 1 and c = –5.

Hence, the numbers are 7, 1 and –5.